{ "date": "2026-10-20", "seed": 202610, "perType": 5, "thresholds": { "exact": 0.001, "correct": 0.01, "confident": 80 }, "costUsd": 3.9152, "byModel": [ { "name": "chat_gpt:gpt-6.1-sol", "n": 105, "errors": 0, "exact": 103, "correct": 104, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.01 }, { "name": "chat_gpt:gpt-5.4-mini", "n": 105, "errors": 0, "exact": 84, "correct": 103, "wrong": 2, "none": 0, "confidentlyWrong": 2, "brier": 0.018 }, { "name": "claude:claude-opus-5-5", "n": 105, "errors": 0, "exact": 103, "correct": 105, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.012 }, { "name": "claude:claude-haiku-5-5", "n": 105, "errors": 0, "exact": 87, "correct": 98, "wrong": 7, "none": 0, "confidentlyWrong": 2, "brier": 0.094 }, { "name": "gemini:gemini-4-argon", "n": 105, "errors": 0, "exact": 96, "correct": 99, "wrong": 6, "none": 0, "confidentlyWrong": 6, "brier": 0.054 }, { "name": "gemini:gemini-3.8-flash", "n": 105, "errors": 0, "exact": 101, "correct": 104, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.009 } ], "byTier": [ { "name": "spec", "n": 420, "errors": 0, "exact": 392, "correct": 416, "wrong": 4, "none": 0, "confidentlyWrong": 3, "brier": 0.017 }, { "name": "realistic", "n": 210, "errors": 0, "exact": 182, "correct": 197, "wrong": 13, "none": 0, "confidentlyWrong": 9, "brier": 0.064 } ], "byModelTier": [ { "name": "chat_gpt:gpt-6.1-sol / spec", "n": 70, "errors": 0, "exact": 70, "correct": 70, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0 }, { "name": "chat_gpt:gpt-6.1-sol / realistic", "n": 35, "errors": 0, "exact": 33, "correct": 34, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.029 }, { "name": "chat_gpt:gpt-5.4-mini / spec", "n": 70, "errors": 0, "exact": 55, "correct": 69, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.015 }, { "name": "chat_gpt:gpt-5.4-mini / realistic", "n": 35, "errors": 0, "exact": 29, "correct": 34, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.026 }, { "name": "claude:claude-opus-5-5 / spec", "n": 70, "errors": 0, "exact": 69, "correct": 70, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.01 }, { "name": "claude:claude-opus-5-5 / realistic", "n": 35, "errors": 0, "exact": 34, "correct": 35, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.015 }, { "name": "claude:claude-haiku-5-5 / spec", "n": 70, "errors": 0, "exact": 60, "correct": 67, "wrong": 3, "none": 0, "confidentlyWrong": 2, "brier": 0.076 }, { "name": "claude:claude-haiku-5-5 / realistic", "n": 35, "errors": 0, "exact": 27, "correct": 31, "wrong": 4, "none": 0, "confidentlyWrong": 0, "brier": 0.13 }, { "name": "gemini:gemini-4-argon / spec", "n": 70, "errors": 0, "exact": 69, "correct": 70, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0 }, { "name": "gemini:gemini-4-argon / realistic", "n": 35, "errors": 0, "exact": 27, "correct": 29, "wrong": 6, "none": 0, "confidentlyWrong": 6, "brier": 0.161 }, { "name": "gemini:gemini-3.8-flash / spec", "n": 70, "errors": 0, "exact": 69, "correct": 70, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0 }, { "name": "gemini:gemini-3.8-flash / realistic", "n": 35, "errors": 0, "exact": 32, "correct": 34, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.026 } ], "byDomain": [ { "name": "options", "n": 210, "errors": 0, "exact": 185, "correct": 208, "wrong": 2, "none": 0, "confidentlyWrong": 1, "brier": 0.024 }, { "name": "forex", "n": 60, "errors": 0, "exact": 58, "correct": 60, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.006 }, { "name": "risk", "n": 150, "errors": 0, "exact": 131, "correct": 140, "wrong": 10, "none": 0, "confidentlyWrong": 9, "brier": 0.067 }, { "name": "prediction", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.004 }, { "name": "perps", "n": 180, "errors": 0, "exact": 170, "correct": 175, "wrong": 5, "none": 0, "confidentlyWrong": 2, "brier": 0.028 } ], "byType": [ { "name": "bs_call_price", "n": 30, "errors": 0, "exact": 21, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.029 }, { "name": "bs_put_delta", "n": 30, "errors": 0, "exact": 29, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.007 }, { "name": "bs_gamma", "n": 30, "errors": 0, "exact": 26, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.007 }, { "name": "bs_vega", "n": 30, "errors": 0, "exact": 29, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.01 }, { "name": "implied_vol", "n": 30, "errors": 0, "exact": 26, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.028 }, { "name": "fx_position_size", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.005 }, { "name": "var_parametric", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.006 }, { "name": "cvar_parametric", "n": 30, "errors": 0, "exact": 29, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.008 }, { "name": "sharpe_annualized", "n": 30, "errors": 0, "exact": 28, "correct": 29, "wrong": 1, "none": 0, "confidentlyWrong": 1, "brier": 0.03 }, { "name": "kelly_binary", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.004 }, { "name": "inverse_liquidation", "n": 30, "errors": 0, "exact": 28, "correct": 28, "wrong": 2, "none": 0, "confidentlyWrong": 2, "brier": 0.063 }, { "name": "inverse_avg_entry", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.006 }, { "name": "inverse_pnl_fees", "n": 30, "errors": 0, "exact": 27, "correct": 29, "wrong": 1, "none": 0, "confidentlyWrong": 0, "brier": 0.027 }, { "name": "linear_pnl_fees", "n": 30, "errors": 0, "exact": 29, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.003 }, { "name": "r_fx_lots_usd_account", "n": 30, "errors": 0, "exact": 28, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.007 }, { "name": "r_inverse_pnl_plain", "n": 30, "errors": 0, "exact": 26, "correct": 28, "wrong": 2, "none": 0, "confidentlyWrong": 0, "brier": 0.057 }, { "name": "r_inverse_avg_plain", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.012 }, { "name": "r_var_from_returns", "n": 30, "errors": 0, "exact": 19, "correct": 24, "wrong": 6, "none": 0, "confidentlyWrong": 5, "brier": 0.179 }, { "name": "r_sharpe_daily", "n": 30, "errors": 0, "exact": 25, "correct": 27, "wrong": 3, "none": 0, "confidentlyWrong": 3, "brier": 0.109 }, { "name": "r_iv_then_delta", "n": 30, "errors": 0, "exact": 24, "correct": 28, "wrong": 2, "none": 0, "confidentlyWrong": 1, "brier": 0.07 }, { "name": "r_deribit_straddle_breakeven", "n": 30, "errors": 0, "exact": 30, "correct": 30, "wrong": 0, "none": 0, "confidentlyWrong": 0, "brier": 0.017 } ], "tasks": [ { "id": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "unit": "USD", "expected": 40.790551141452084, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 382.5, strike 350, 18 days to expiry, volatility 60.7% per year, risk-free rate 5.97% per year. What is the call option price?" }, { "id": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "unit": "USD", "expected": 3.2317412561362406, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 22.89, strike 23, 171 days to expiry, volatility 51.8% per year, risk-free rate 1.26% per year. What is the call option price?" }, { "id": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "unit": "USD", "expected": 16.773153687059427, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 89.61, strike 95, 152 days to expiry, volatility 78.8% per year, risk-free rate 5.77% per year. What is the call option price?" }, { "id": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "unit": "USD", "expected": 0.6945968823131317, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 384.5, strike 427, 9 days to expiry, volatility 42.6% per year, risk-free rate 2.52% per year. What is the call option price?" }, { "id": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "unit": "USD", "expected": 23.323917391138792, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 314.3, strike 294, 20 days to expiry, volatility 35.5% per year, risk-free rate 0.06% per year. What is the call option price?" }, { "id": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "unit": "delta", "expected": -0.280639388201964, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 114.52, strike 106, 50 days to expiry, volatility 46.6% per year, risk-free rate 5.85% per year. What is the delta of the put option (a number between -1 and 0)?" }, { "id": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "unit": "delta", "expected": -0.31093085044630486, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 46.9, strike 43, 41 days to expiry, volatility 72.7% per year, risk-free rate 3.27% per year. What is the delta of the put option (a number between -1 and 0)?" }, { "id": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "unit": "delta", "expected": -0.5641608283248785, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 215.92, strike 231, 65 days to expiry, volatility 52.8% per year, risk-free rate 3.76% per year. What is the delta of the put option (a number between -1 and 0)?" }, { "id": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "unit": "delta", "expected": -0.3853788948381438, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 291.06, strike 286, 113 days to expiry, volatility 72.9% per year, risk-free rate 5.94% per year. What is the delta of the put option (a number between -1 and 0)?" }, { "id": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "unit": "delta", "expected": -0.4080759766740296, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 146.31, strike 152, 162 days to expiry, volatility 79.9% per year, risk-free rate 4.56% per year. What is the delta of the put option (a number between -1 and 0)?" }, { "id": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "unit": "gamma", "expected": 0.007721610843754648, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 168.77, strike 150, 34 days to expiry, volatility 84.4% per year, risk-free rate 0.3% per year. What is the option gamma (change in delta per 1 unit change in the spot price)?" }, { "id": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "unit": "gamma", "expected": 0.011719035620003572, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 350.55, strike 375, 71 days to expiry, volatility 16.1% per year, risk-free rate 4.48% per year. What is the option gamma (change in delta per 1 unit change in the spot price)?" }, { "id": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "unit": "gamma", "expected": 0.00332788170697725, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 471.11, strike 474, 107 days to expiry, volatility 46.4% per year, risk-free rate 5.03% per year. What is the option gamma (change in delta per 1 unit change in the spot price)?" }, { "id": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "unit": "gamma", "expected": 0.015589164355628841, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 87.11, strike 100, 147 days to expiry, volatility 44.6% per year, risk-free rate 5.14% per year. What is the option gamma (change in delta per 1 unit change in the spot price)?" }, { "id": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "unit": "gamma", "expected": 0.00596435584251129, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 335.65, strike 300, 103 days to expiry, volatility 24% per year, risk-free rate 0.03% per year. What is the option gamma (change in delta per 1 unit change in the spot price)?" }, { "id": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "unit": "USD per vol point", "expected": 0.646446181061521, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 424.81, strike 380, 74 days to expiry, volatility 55% per year, risk-free rate 0.25% per year. What is the vega, expressed as the change in the option price for a 1 percentage point increase in volatility?" }, { "id": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "unit": "USD per vol point", "expected": 0.26781898170848095, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 146.99, strike 155, 104 days to expiry, volatility 16.3% per year, risk-free rate 0.24% per year. What is the vega, expressed as the change in the option price for a 1 percentage point increase in volatility?" }, { "id": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "unit": "USD per vol point", "expected": 0.45587809695630876, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 469.67, strike 512, 47 days to expiry, volatility 25.4% per year, risk-free rate 1.39% per year. What is the vega, expressed as the change in the option price for a 1 percentage point increase in volatility?" }, { "id": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "unit": "USD per vol point", "expected": 0.6097213395050691, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 383.12, strike 429, 61 days to expiry, volatility 72.6% per year, risk-free rate 2.03% per year. What is the vega, expressed as the change in the option price for a 1 percentage point increase in volatility?" }, { "id": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "unit": "USD per vol point", "expected": 0.2403754082675673, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. Spot price 158.91, strike 148, 65 days to expiry, volatility 56.4% per year, risk-free rate 5.99% per year. What is the vega, expressed as the change in the option price for a 1 percentage point increase in volatility?" }, { "id": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "unit": "% per year", "expected": 61.799447576373446, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. A European call with spot price 27.09, strike 29, 136 days to expiry and risk-free rate 5.4% per year trades at 3.5195. What is its implied volatility, in percent per year?" }, { "id": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "unit": "% per year", "expected": 20.900010973832025, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. A European put with spot price 490.18, strike 490, 53 days to expiry and risk-free rate 5.08% per year trades at 13.6926. What is its implied volatility, in percent per year?" }, { "id": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "unit": "% per year", "expected": 81.50015130720656, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. A European put with spot price 79.84, strike 86, 161 days to expiry and risk-free rate 0.04% per year trades at 20.922. What is its implied volatility, in percent per year?" }, { "id": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "unit": "% per year", "expected": 87.70003216471197, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. A European put with spot price 340.53, strike 362, 105 days to expiry and risk-free rate 1.06% per year trades at 75.8825. What is its implied volatility, in percent per year?" }, { "id": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "unit": "% per year", "expected": 33.39992865638508, "question": "Use the Black-Scholes model for a European option on a non-dividend-paying asset, a continuously compounded risk-free rate, and time to expiry T = days / 365. A European call with spot price 140.66, strike 143, 161 days to expiry and risk-free rate 1.9% per year trades at 11.908. What is its implied volatility, in percent per year?" }, { "id": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "unit": "standard lots", "expected": 6.438461538461515, "question": "A long AUD/USD trade enters at 0.94593 with a stop-loss at 0.94203. The trader wants to risk exactly 2511 USD if the stop is hit (ignore spread and commission). One standard lot is 100,000 units of the base currency. What position size is that, in standard lots?" }, { "id": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "unit": "standard lots", "expected": 12.3258666666662, "question": "A long EUR/JPY trade enters at 154.513 with a stop-loss at 154.363. The trader wants to risk exactly 184888 JPY if the stop is hit (ignore spread and commission). One standard lot is 100,000 units of the base currency. What position size is that, in standard lots?" }, { "id": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "unit": "standard lots", "expected": 0.4756567164179193, "question": "A long EUR/JPY trade enters at 191.077 with a stop-loss at 190.407. The trader wants to risk exactly 31869 JPY if the stop is hit (ignore spread and commission). One standard lot is 100,000 units of the base currency. What position size is that, in standard lots?" }, { "id": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "unit": "standard lots", "expected": 5.01704545454548, "question": "A long USD/JPY trade enters at 142.806 with a stop-loss at 142.366. The trader wants to risk exactly 220750 JPY if the stop is hit (ignore spread and commission). One standard lot is 100,000 units of the base currency. What position size is that, in standard lots?" }, { "id": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "unit": "standard lots", "expected": 1.6141304347826113, "question": "A long EUR/USD trade enters at 0.76163 with a stop-loss at 0.75243. The trader wants to risk exactly 1485 USD if the stop is hit (ignore spread and commission). One standard lot is 100,000 units of the base currency. What position size is that, in standard lots?" }, { "id": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.023819567833442343, "question": "Daily portfolio returns are normally distributed with mean -0.0003 and standard deviation 0.012 (as decimal fractions). What is the 1-day parametric Value at Risk at 97.5% confidence, expressed as a positive loss fraction of portfolio value (VaR = -mean + z * stdev)?" }, { "id": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.08208473980663146, "question": "Daily portfolio returns are normally distributed with mean -0.0005 and standard deviation 0.0496 (as decimal fractions). What is the 1-day parametric Value at Risk at 95% confidence, expressed as a positive loss fraction of portfolio value (VaR = -mean + z * stdev)?" }, { "id": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.10032876732331801, "question": "Daily portfolio returns are normally distributed with mean 0.0011 and standard deviation 0.0436 (as decimal fractions). What is the 1-day parametric Value at Risk at 99% confidence, expressed as a positive loss fraction of portfolio value (VaR = -mean + z * stdev)?" }, { "id": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.07275151453115127, "question": "Daily portfolio returns are normally distributed with mean -0.0011 and standard deviation 0.0308 (as decimal fractions). What is the 1-day parametric Value at Risk at 99% confidence, expressed as a positive loss fraction of portfolio value (VaR = -mean + z * stdev)?" }, { "id": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.11150152403343133, "question": "Daily portfolio returns are normally distributed with mean -0.001 and standard deviation 0.0475 (as decimal fractions). What is the 1-day parametric Value at Risk at 99% confidence, expressed as a positive loss fraction of portfolio value (VaR = -mean + z * stdev)?" }, { "id": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.04404168455292251, "question": "Daily portfolio returns are normally distributed with mean 0.0018 and standard deviation 0.0172 (as decimal fractions). What is the 1-day parametric Conditional Value at Risk (Expected Shortfall) at 99% confidence, expressed as a positive loss fraction of portfolio value?" }, { "id": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.02233967069948375, "question": "Daily portfolio returns are normally distributed with mean -0.0013 and standard deviation 0.0102 (as decimal fractions). What is the 1-day parametric Conditional Value at Risk (Expected Shortfall) at 95% confidence, expressed as a positive loss fraction of portfolio value?" }, { "id": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.09524684038587, "question": "Daily portfolio returns are normally distributed with mean 0.0023 and standard deviation 0.0366 (as decimal fractions). What is the 1-day parametric Conditional Value at Risk (Expected Shortfall) at 99% confidence, expressed as a positive loss fraction of portfolio value?" }, { "id": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.12233732788143936, "question": "Daily portfolio returns are normally distributed with mean 0.0018 and standard deviation 0.0531 (as decimal fractions). What is the 1-day parametric Conditional Value at Risk (Expected Shortfall) at 97.5% confidence, expressed as a positive loss fraction of portfolio value?" }, { "id": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "unit": "fraction of portfolio", "expected": 0.1205670856534704, "question": "Daily portfolio returns are normally distributed with mean 0.0017 and standard deviation 0.0523 (as decimal fractions). What is the 1-day parametric Conditional Value at Risk (Expected Shortfall) at 97.5% confidence, expressed as a positive loss fraction of portfolio value?" }, { "id": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "unit": "Sharpe ratio", "expected": 1.6781883884002948, "question": "A strategy had these 12 monthly returns (decimal fractions): -0.0353, 0.0402, 0.0555, 0.0622, 0.0058, 0.0795, 0.0049, 0.0435, 0.0415, 0.0469, -0.0363, -0.0547. Using a risk-free rate of 0, the sample standard deviation (divide by n - 1), and annualizing by multiplying the monthly Sharpe ratio by sqrt(12), what is the annualized Sharpe ratio?" }, { "id": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "unit": "Sharpe ratio", "expected": -0.48328609065675004, "question": "A strategy had these 12 monthly returns (decimal fractions): -0.0474, 0.0376, 0.0074, -0.0481, -0.0542, 0.0097, -0.0123, 0.0621, -0.0496, 0.0357, 0.0095, -0.0158. Using a risk-free rate of 0, the sample standard deviation (divide by n - 1), and annualizing by multiplying the monthly Sharpe ratio by sqrt(12), what is the annualized Sharpe ratio?" }, { "id": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "unit": "Sharpe ratio", "expected": 2.241475231769422, "question": "A strategy had these 12 monthly returns (decimal fractions): 0.0236, 0.0642, 0.02, -0.029, 0.0274, 0.0497, -0.0065, 0.0325, 0.0352, 0.0197, 0.0796, -0.0422. Using a risk-free rate of 0, the sample standard deviation (divide by n - 1), and annualizing by multiplying the monthly Sharpe ratio by sqrt(12), what is the annualized Sharpe ratio?" }, { "id": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "unit": "Sharpe ratio", "expected": 1.9830100176589147, "question": "A strategy had these 12 monthly returns (decimal fractions): 0.05, 0.044, 0.0617, 0.0753, -0.0527, -0.0366, 0.0167, 0.046, -0.0225, -0.0093, 0.0695, 0.0687. Using a risk-free rate of 0, the sample standard deviation (divide by n - 1), and annualizing by multiplying the monthly Sharpe ratio by sqrt(12), what is the annualized Sharpe ratio?" }, { "id": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "unit": "Sharpe ratio", "expected": 1.3915400475486268, "question": "A strategy had these 12 monthly returns (decimal fractions): 0.0603, -0.0347, 0.0392, 0.0484, -0.0114, -0.0247, 0.0144, 0.0301, 0.0712, 0.0025, -0.0155, -0.0103. Using a risk-free rate of 0, the sample standard deviation (divide by n - 1), and annualizing by multiplying the monthly Sharpe ratio by sqrt(12), what is the annualized Sharpe ratio?" }, { "id": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "unit": "fraction of bankroll", "expected": 0.4848484848484847, "question": "A binary prediction-market contract costs $0.67 and pays $1 if the event happens, $0 otherwise (no fees). You estimate the probability of the event at 0.83. What fraction of your bankroll does the full Kelly criterion say to stake on buying this contract?" }, { "id": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "unit": "fraction of bankroll", "expected": 0.28985507246376807, "question": "A binary prediction-market contract costs $0.31 and pays $1 if the event happens, $0 otherwise (no fees). You estimate the probability of the event at 0.51. What fraction of your bankroll does the full Kelly criterion say to stake on buying this contract?" }, { "id": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "unit": "fraction of bankroll", "expected": 0.09859154929577463, "question": "A binary prediction-market contract costs $0.29 and pays $1 if the event happens, $0 otherwise (no fees). You estimate the probability of the event at 0.36. What fraction of your bankroll does the full Kelly criterion say to stake on buying this contract?" }, { "id": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "unit": "fraction of bankroll", "expected": 0.8333333333333334, "question": "A binary prediction-market contract costs $0.16 and pays $1 if the event happens, $0 otherwise (no fees). You estimate the probability of the event at 0.86. What fraction of your bankroll does the full Kelly criterion say to stake on buying this contract?" }, { "id": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "unit": "fraction of bankroll", "expected": 0.576271186440678, "question": "A binary prediction-market contract costs $0.41 and pays $1 if the event happens, $0 otherwise (no fees). You estimate the probability of the event at 0.75. What fraction of your bankroll does the full Kelly criterion say to stake on buying this contract?" }, { "id": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 24920.824742268043, "question": "An isolated-margin short position in an inverse (coin-margined) BTCUSD perpetual is opened at 24173.2 with 25x leverage. Position size is in USD contracts, margin and PnL are in BTC. Initial margin = position value in BTC at entry / leverage. The position is liquidated when margin plus unrealized PnL (in BTC) falls to the maintenance margin, which is 1% of the position value in BTC at entry. Ignore fees and funding. What is the liquidation price?" }, { "id": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 53342.9702970297, "question": "An isolated-margin long position in an inverse (coin-margined) BTCUSD perpetual is opened at 53876.4 with 50x leverage. Position size is in USD contracts, margin and PnL are in BTC. Initial margin = position value in BTC at entry / leverage. The position is liquidated when margin plus unrealized PnL (in BTC) falls to the maintenance margin, which is 1% of the position value in BTC at entry. Ignore fees and funding. What is the liquidation price?" }, { "id": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 53194.15525114154, "question": "An isolated-margin long position in an inverse (coin-margined) BTCUSD perpetual is opened at 58247.6 with 10x leverage. Position size is in USD contracts, margin and PnL are in BTC. Initial margin = position value in BTC at entry / leverage. The position is liquidated when margin plus unrealized PnL (in BTC) falls to the maintenance margin, which is 0.5% of the position value in BTC at entry. Ignore fees and funding. What is the liquidation price?" }, { "id": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 170355.2216748768, "question": "An isolated-margin short position in an inverse (coin-margined) BTCUSD perpetual is opened at 115273.7 with 3x leverage. Position size is in USD contracts, margin and PnL are in BTC. Initial margin = position value in BTC at entry / leverage. The position is liquidated when margin plus unrealized PnL (in BTC) falls to the maintenance margin, which is 1% of the position value in BTC at entry. Ignore fees and funding. What is the liquidation price?" }, { "id": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 115003.24607329845, "question": "An isolated-margin short position in an inverse (coin-margined) BTCUSD perpetual is opened at 109828.1 with 20x leverage. Position size is in USD contracts, margin and PnL are in BTC. Initial margin = position value in BTC at entry / leverage. The position is liquidated when margin plus unrealized PnL (in BTC) falls to the maintenance margin, which is 0.5% of the position value in BTC at entry. Ignore fees and funding. What is the liquidation price?" }, { "id": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 2398.0256188582507, "question": "On an inverse (coin-margined) ETHUSD perpetual, where position size is in USD contracts and PnL is settled in ETH, a trader bought these fills: 7000 USD of contracts at 2309.36; 34000 USD of contracts at 3563.54; 36000 USD of contracts at 1854.34; 9000 USD of contracts at 2321.56. What is the average entry price of the combined position, i.e. the single entry price that gives the same ETH PnL at any exit price?" }, { "id": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 3712.5303509582454, "question": "On an inverse (coin-margined) ETHUSD perpetual, where position size is in USD contracts and PnL is settled in ETH, a trader bought these fills: 34000 USD of contracts at 3528.19; 16000 USD of contracts at 4176.2. What is the average entry price of the combined position, i.e. the single entry price that gives the same ETH PnL at any exit price?" }, { "id": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 2495.29829080704, "question": "On an inverse (coin-margined) ETHUSD perpetual, where position size is in USD contracts and PnL is settled in ETH, a trader bought these fills: 41000 USD of contracts at 3517.25; 10000 USD of contracts at 2976.86; 27000 USD of contracts at 4229.06; 40000 USD of contracts at 1545.09. What is the average entry price of the combined position, i.e. the single entry price that gives the same ETH PnL at any exit price?" }, { "id": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 2610.72093572179, "question": "On an inverse (coin-margined) ETHUSD perpetual, where position size is in USD contracts and PnL is settled in ETH, a trader bought these fills: 10000 USD of contracts at 3282.44; 14000 USD of contracts at 2147.97; 12000 USD of contracts at 1744.23; 23000 USD of contracts at 3736.8. What is the average entry price of the combined position, i.e. the single entry price that gives the same ETH PnL at any exit price?" }, { "id": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "unit": "USD price", "expected": 3424.594098647905, "question": "On an inverse (coin-margined) ETHUSD perpetual, where position size is in USD contracts and PnL is settled in ETH, a trader bought these fills: 31000 USD of contracts at 3315.48; 36000 USD of contracts at 3410.8; 40000 USD of contracts at 3481.71; 5000 USD of contracts at 3813.27. What is the average entry price of the combined position, i.e. the single entry price that gives the same ETH PnL at any exit price?" }, { "id": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "unit": "BTC", "expected": 0.02840777213808475, "question": "A short position of 44000 USD in an inverse (coin-margined) BTCUSD perpetual is opened at 42887.5 and closed at 41686.7. PnL is settled in BTC. A fee of 0.055% is charged at opening and at closing, each on that fill's position value in BTC (size / fill price). What is the net PnL in BTC after fees (negative for a loss)?" }, { "id": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "unit": "BTC", "expected": -0.06525146382766053, "question": "A long position of 45000 USD in an inverse (coin-margined) BTCUSD perpetual is opened at 49446 and closed at 46182.6. PnL is settled in BTC. A fee of 0.05% is charged at opening and at closing, each on that fill's position value in BTC (size / fill price). What is the net PnL in BTC after fees (negative for a loss)?" }, { "id": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "unit": "BTC", "expected": 0.22922018359225318, "question": "A long position of 189000 USD in an inverse (coin-margined) BTCUSD perpetual is opened at 91422.1 and closed at 102932.1. PnL is settled in BTC. A fee of 0.05% is charged at opening and at closing, each on that fill's position value in BTC (size / fill price). What is the net PnL in BTC after fees (negative for a loss)?" }, { "id": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "unit": "BTC", "expected": -0.026151408867684533, "question": "A long position of 14000 USD in an inverse (coin-margined) BTCUSD perpetual is opened at 89747.2 and closed at 76940.3. PnL is settled in BTC. A fee of 0.055% is charged at opening and at closing, each on that fill's position value in BTC (size / fill price). What is the net PnL in BTC after fees (negative for a loss)?" }, { "id": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "unit": "BTC", "expected": -0.011586282591352507, "question": "A short position of 44000 USD in an inverse (coin-margined) BTCUSD perpetual is opened at 102777.6 and closed at 105593.7. PnL is settled in BTC. A fee of 0.02% is charged at opening and at closing, each on that fill's position value in BTC (size / fill price). What is the net PnL in BTC after fees (negative for a loss)?" }, { "id": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "unit": "USDT", "expected": 6057.334255752001, "question": "A short position of 1.703 BTC in a linear USDT-margined BTCUSDT perpetual is opened at 26380.89 and closed at 22814.19. A fee of 0.02% is charged at opening and at closing, each on that fill's notional value (size x fill price). What is the net PnL in USDT after fees (negative for a loss)?" }, { "id": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "unit": "USDT", "expected": 2844.8890875599996, "question": "A long position of 1.164 BTC in a linear USDT-margined BTCUSDT perpetual is opened at 36718.19 and closed at 39204.01. A fee of 0.055% is charged at opening and at closing, each on that fill's notional value (size x fill price). What is the net PnL in USDT after fees (negative for a loss)?" }, { "id": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "unit": "USDT", "expected": -3783.556934676003, "question": "A long position of 0.612 BTC in a linear USDT-margined BTCUSDT perpetual is opened at 57972.37 and closed at 51850.49. A fee of 0.055% is charged at opening and at closing, each on that fill's notional value (size x fill price). What is the net PnL in USDT after fees (negative for a loss)?" }, { "id": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "unit": "USDT", "expected": -1108.5578196799993, "question": "A long position of 2.648 BTC in a linear USDT-margined BTCUSDT perpetual is opened at 4857.54 and closed at 4440.76. A fee of 0.02% is charged at opening and at closing, each on that fill's notional value (size x fill price). What is the net PnL in USDT after fees (negative for a loss)?" }, { "id": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "unit": "USDT", "expected": 3351.620043262996, "question": "A short position of 1.949 BTC in a linear USDT-margined BTCUSDT perpetual is opened at 28983.12 and closed at 27232.54. A fee of 0.055% is charged at opening and at closing, each on that fill's notional value (size x fill price). What is the net PnL in USDT after fees (negative for a loss)?" }, { "id": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "unit": "standard lots", "expected": 0.11472881355932601, "question": "My account is 5,000 USD and I risk 1% per trade. I'm buying USD/CAD at 1.3538 with a stop at 1.3479. How many standard lots should I trade? Convert the pip value to USD at the entry price." }, { "id": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "unit": "standard lots", "expected": 0.5939090909090843, "question": "My account is 25,000 USD and I risk 1% per trade. I'm buying USD/CAD at 1.3066 with a stop at 1.3011. How many standard lots should I trade? Convert the pip value to USD at the entry price." }, { "id": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "unit": "standard lots", "expected": 1.1982352941177008, "question": "My account is 25,000 USD and I risk 1% per trade. I'm buying USD/CHF at 0.8148 with a stop at 0.8131. How many standard lots should I trade? Convert the pip value to USD at the entry price." }, { "id": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "unit": "standard lots", "expected": 1.0178846153846508, "question": "My account is 25,000 USD and I risk 1% per trade. I'm buying USD/JPY at 158.79 with a stop at 158.4. How many standard lots should I trade? Convert the pip value to USD at the entry price." }, { "id": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "unit": "standard lots", "expected": 7.182727272727309, "question": "My account is 100,000 USD and I risk 1% per trade. I'm buying USD/JPY at 158.02 with a stop at 157.8. How many standard lots should I trade? Convert the pip value to USD at the entry price." }, { "id": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "unit": "BTC", "expected": -0.23573231643213824, "question": "I went short 103,000 contracts of the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) at 49553.7 and closed at 55827.2, paying the 0.055% taker fee on both fills. What's my realized PnL in BTC after fees?" }, { "id": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "unit": "BTC", "expected": -0.0669895883004742, "question": "I went long 78,000 contracts of the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) at 95195.8 and closed at 88094.2, paying the 0.055% taker fee on both fills. What's my realized PnL in BTC after fees?" }, { "id": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "unit": "BTC", "expected": 0.0024428835303150374, "question": "I went short 42,000 contracts of the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) at 117914.1 and closed at 116982.6, paying the 0.055% taker fee on both fills. What's my realized PnL in BTC after fees?" }, { "id": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "unit": "BTC", "expected": 0.001886499688812526, "question": "I went long 192,000 contracts of the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) at 20434 and closed at 20460.6, paying the 0.055% taker fee on both fills. What's my realized PnL in BTC after fees?" }, { "id": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "unit": "BTC", "expected": -0.0962309957129481, "question": "I went short 139,000 contracts of the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) at 59513.2 and closed at 62000.9, paying the 0.055% taker fee on both fills. What's my realized PnL in BTC after fees?" }, { "id": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "unit": "USD price", "expected": 70867.19713259465, "question": "I built a long on the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) in pieces: 76,000 contracts at 62771.7, 55,000 contracts at 86235.1. What average entry price will the exchange show for my position?" }, { "id": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "unit": "USD price", "expected": 34209.14079495059, "question": "I built a long on the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) in pieces: 76,000 contracts at 33536.1, 73,000 contracts at 24929.9, 42,000 contracts at 108003.1. What average entry price will the exchange show for my position?" }, { "id": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "unit": "USD price", "expected": 88678.94192491287, "question": "I built a long on the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) in pieces: 59,000 contracts at 82124.2, 34,000 contracts at 81534, 62,000 contracts at 101232.7. What average entry price will the exchange show for my position?" }, { "id": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "unit": "USD price", "expected": 46924.5904555392, "question": "I built a long on the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) in pieces: 42,000 contracts at 30128.4, 67,000 contracts at 72132.8. What average entry price will the exchange show for my position?" }, { "id": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "unit": "USD price", "expected": 37529.651947786224, "question": "I built a long on the Bybit BTCUSD inverse perpetual (1 contract = 1 USD) in pieces: 36,000 contracts at 23930.3, 61,000 contracts at 56468.2. What average entry price will the exchange show for my position?" }, { "id": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "unit": "% of portfolio", "expected": 4.917273424034056, "question": "My last 40 daily returns in %: -0.91, 3.56, 2.19, 1.35, 4.41, -3.67, -0.89, 0.39, 3.69, -0.92, -1.11, 3.17, -1.75, -3.56, -2.60, -3.03, 3.30, 0.28, -4.20, -3.23, -3.99, 4.97, -2.38, 1.30, -4.46, 3.09, 0.45, -3.26, 3.89, -2.96, 3.33, -0.70, -4.41, 1.07, -2.16, -1.49, 3.17, 0.65, -1.47, -1.30. What is my 1-day 95% parametric (normal) VaR, in %, using the sample mean and sample standard deviation?" }, { "id": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "unit": "% of portfolio", "expected": 4.2934299989968965, "question": "My last 40 daily returns in %: 4.59, -3.68, 2.15, -0.27, 3.28, 4.83, 2.46, -1.45, 3.40, 4.44, -0.87, -0.24, -1.92, 3.13, -0.12, 4.82, 0.68, -3.67, 0.20, 2.82, -0.87, -4.38, 0.18, 3.37, -3.79, 2.97, -1.94, 3.40, -1.52, -4.07, 4.19, 1.52, -3.76, -4.57, 0.37, -1.28, 0.73, -1.77, 2.77, 2.31. What is my 1-day 95% parametric (normal) VaR, in %, using the sample mean and sample standard deviation?" }, { "id": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "unit": "% of portfolio", "expected": 5.2422646771168795, "question": "My last 40 daily returns in %: 0.81, -3.34, -1.80, -2.90, -3.97, 2.94, 2.88, 2.34, -2.55, -2.93, 4.20, 4.36, 1.59, 3.83, 4.51, 4.60, 1.02, -3.34, 3.82, 3.43, -4.88, -4.91, 0.19, -2.96, -1.54, -3.61, -2.21, -3.84, -2.12, 2.49, 4.47, 1.13, 4.34, 1.00, -3.76, -1.06, 3.00, -3.34, 3.52, -2.41. What is my 1-day 95% parametric (normal) VaR, in %, using the sample mean and sample standard deviation?" }, { "id": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "unit": "% of portfolio", "expected": 7.037483251314367, "question": "My last 40 daily returns in %: -1.04, -4.23, 3.83, 1.02, -1.35, -2.72, -1.23, 4.44, -3.35, 3.60, -2.10, -3.18, 2.25, -3.34, 1.56, -3.57, 3.91, -4.06, -2.55, 1.55, -1.96, -2.03, -3.66, 1.01, -0.72, 2.20, -3.08, -3.86, 2.12, -3.83, 2.02, 1.40, 4.86, 2.98, 1.89, -4.40, -1.20, -2.94, 0.93, 0.00. What is my 1-day 99% parametric (normal) VaR, in %, using the sample mean and sample standard deviation?" }, { "id": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "unit": "% of portfolio", "expected": 6.7877378350475945, "question": "My last 40 daily returns in %: -2.50, 4.47, -1.71, 0.32, -3.74, 0.12, 2.74, 2.80, -0.95, -3.34, -0.21, 3.76, -0.86, -0.92, -2.76, 4.54, -4.23, 1.16, 0.50, 4.48, 4.64, 0.58, -4.75, -3.14, 4.66, 3.24, 2.86, -2.04, -1.49, -4.19, 1.99, -0.03, -3.00, -3.51, 2.36, 1.91, 2.20, 1.18, -4.75, -0.60. What is my 1-day 99% parametric (normal) VaR, in %, using the sample mean and sample standard deviation?" }, { "id": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "unit": "Sharpe ratio", "expected": 0.40527021190035356, "question": "My crypto bot's last 30 daily returns in %: -0.03, -0.14, -2.24, -0.42, -2.93, 2.82, 3.50, -3.33, -2.35, -0.77, 1.20, 0.38, -0.96, 1.19, -3.90, -2.28, 2.86, -0.20, 4.21, 2.17, -0.68, 2.42, 3.83, 0.96, -0.03, 2.59, -1.03, -1.20, -2.51, -1.70. What's the annualized Sharpe ratio? Crypto trades 365 days a year, risk-free rate 0, sample standard deviation." }, { "id": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "unit": "Sharpe ratio", "expected": 0.37258355219979333, "question": "My crypto bot's last 30 daily returns in %: 0.49, 3.50, 1.12, 2.59, -3.20, -0.05, -0.34, -2.32, -0.09, 1.99, -3.04, 2.82, 3.56, -2.59, 1.21, -3.72, -1.96, 0.68, -0.86, 2.46, -1.70, -2.02, 2.95, 0.85, -0.52, -1.58, -3.05, -1.92, 2.48, 3.61. What's the annualized Sharpe ratio? Crypto trades 365 days a year, risk-free rate 0, sample standard deviation." }, { "id": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "unit": "Sharpe ratio", "expected": -0.6855095501276544, "question": "My crypto bot's last 30 daily returns in %: 1.48, -2.22, 0.55, 4.03, -0.27, 3.13, -1.18, 0.65, 0.44, -1.96, 1.71, -3.20, 2.07, 3.86, 3.92, -1.40, -1.88, -3.08, -0.74, -1.95, -2.32, 2.23, -0.71, 2.02, -3.28, 0.61, 1.45, -2.96, 0.14, -3.64. What's the annualized Sharpe ratio? Crypto trades 365 days a year, risk-free rate 0, sample standard deviation." }, { "id": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "unit": "Sharpe ratio", "expected": 0.5941064255855656, "question": "My crypto bot's last 30 daily returns in %: -2.72, -1.97, -3.54, 0.68, -3.46, 4.40, 0.18, 0.01, 3.68, 0.26, 3.31, 2.79, 0.26, -1.31, -3.29, -0.19, -1.70, -1.24, 0.43, -2.23, -0.57, 0.39, 0.29, 2.92, -3.54, 2.00, 0.24, -0.21, 2.00, 4.31. What's the annualized Sharpe ratio? Crypto trades 365 days a year, risk-free rate 0, sample standard deviation." }, { "id": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "unit": "Sharpe ratio", "expected": 6.721949210975506, "question": "My crypto bot's last 30 daily returns in %: 2.83, 1.74, -2.40, 3.16, -2.74, -1.88, 0.21, 3.34, -0.56, 4.01, -0.04, -0.90, 4.15, 0.88, 3.74, 2.93, 1.64, -1.87, 2.92, -0.05, 1.95, -3.52, -0.10, 0.47, 1.98, -2.33, 3.66, -3.52, 1.75, 4.29. What's the annualized Sharpe ratio? Crypto trades 365 days a year, risk-free rate 0, sample standard deviation." }, { "id": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "unit": "delta", "expected": 0.44697422776459944, "question": "A call with strike 463 expiring in 23 days trades at 21.6661 while the underlying is at 447.74. Rates are zero, no dividends. What is the call's Black-Scholes delta at its implied volatility?" }, { "id": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "unit": "delta", "expected": 0.25544432164825037, "question": "A call with strike 383 expiring in 78 days trades at 5.0781 while the underlying is at 357.63. Rates are zero, no dividends. What is the call's Black-Scholes delta at its implied volatility?" }, { "id": "r_iv_then_delta-3", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "unit": "delta", "expected": 0.4262987878297406, "question": "A call with strike 162 expiring in 32 days trades at 8.6404 while the underlying is at 152.85. Rates are zero, no dividends. What is the call's Black-Scholes delta at its implied volatility?" }, { "id": "r_iv_then_delta-4", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "unit": "delta", "expected": 0.6063214899790841, "question": "A call with strike 192 expiring in 108 days trades at 28.2965 while the underlying is at 198.88. Rates are zero, no dividends. What is the call's Black-Scholes delta at its implied volatility?" }, { "id": "r_iv_then_delta-5", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "unit": "delta", "expected": 0.6017546889052645, "question": "A call with strike 107 expiring in 14 days trades at 4.3243 while the underlying is at 108.83. Rates are zero, no dividends. What is the call's Black-Scholes delta at its implied volatility?" }, { "id": "r_deribit_straddle_breakeven-1", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "unit": "USD price", "expected": 113713.12767394731, "question": "On Deribit I bought a BTC straddle at the 101000 strike (BTC options settled in BTC): the call cost 0.054 BTC and the put 0.0578 BTC. Counting my profit in BTC, above what BTC/USD price at expiry does the straddle break even?" }, { "id": "r_deribit_straddle_breakeven-2", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "unit": "USD price", "expected": 93608.77985797288, "question": "On Deribit I bought a BTC straddle at the 87000 strike (BTC options settled in BTC): the call cost 0.015 BTC and the put 0.0556 BTC. Counting my profit in BTC, above what BTC/USD price at expiry does the straddle break even?" }, { "id": "r_deribit_straddle_breakeven-3", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "unit": "USD price", "expected": 64193.23251006419, "question": "On Deribit I bought a BTC straddle at the 59000 strike (BTC options settled in BTC): the call cost 0.0196 BTC and the put 0.0613 BTC. Counting my profit in BTC, above what BTC/USD price at expiry does the straddle break even?" }, { "id": "r_deribit_straddle_breakeven-4", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "unit": "USD price", "expected": 102706.75082914304, "question": "On Deribit I bought a BTC straddle at the 96000 strike (BTC options settled in BTC): the call cost 0.022 BTC and the put 0.0433 BTC. Counting my profit in BTC, above what BTC/USD price at expiry does the straddle break even?" }, { "id": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "unit": "USD price", "expected": 73994.63806970509, "question": "On Deribit I bought a BTC straddle at the 69000 strike (BTC options settled in BTC): the call cost 0.0412 BTC and the put 0.0263 BTC. Counting my profit in BTC, above what BTC/USD price at expiry does the straddle break even?" } ], "rows": [ { "taskId": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 40.790551141452084, "answer": 40.7899, "confidence": 100, "grade": "exact", "relError": 0.00001596304619232906, "cost": 0.007372, "reply": "The Black–Scholes call price is\n\\[\nC=S_0N(d_1)-Ke^{-rT}N(d_2),\n\\]\nwhere \\(T=18/365\\), giving\n\\[\nd_1\\approx0.747977,\\qquad d_2\\approx0.613181.\n\\]\nThus,\n\\[\nC\\approx382.5N(0.747977)-350e^{-0.0597(18/365)}N(0.613181)\n\\approx40.7899.\n\\]\n\n{\"answer\":40.7899,\"confidence\":100}" }, { "taskId": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 3.2317412561362406, "answer": 3.2317, "confidence": 100, "grade": "exact", "relError": 0.00001276591563829087, "cost": 0.007212, "reply": "Using Black–Scholes,\n\\[\nC=S_0N(d_1)-Ke^{-rT}N(d_2),\n\\quad\nd_1=\\frac{\\ln(S_0/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\quad d_2=d_1-\\sigma\\sqrt T.\n\\]\n\nWith \\(T=171/365\\), we get \\(d_1\\approx0.180404\\) and \\(d_2\\approx-0.174149\\), giving a call price of approximately **3.2317**.\n\n{\"answer\": 3.2317, \"confidence\": 100}" }, { "taskId": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 16.773153687059427, "answer": 16.77315, "confidence": 100, "grade": "exact", "relError": 2.198190927336856e-7, "cost": 0.010192, "reply": "Using Black–Scholes,\n\\[\nC=S_0N(d_1)-Ke^{-rT}N(d_2),\n\\quad\nd_1=\\frac{\\ln(S_0/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\quad d_2=d_1-\\sigma\\sqrt T.\n\\]\n\nWith \\(T=152/365\\), we get \\(d_1\\approx0.186644\\) and \\(d_2\\approx-0.321868\\), giving a call price of approximately **16.77315**.\n\n{\"answer\":16.77315,\"confidence\":100}" }, { "taskId": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.6945968823131317, "answer": 0.6946, "confidence": 99, "grade": "exact", "relError": 0.000004488483820938638, "cost": 0.012632, "reply": "Using Black–Scholes,\n\\[\nC=S_0N(d_1)-Ke^{-rT}N(d_2),\n\\qquad\nd_1=\\frac{\\ln(S_0/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T}.\n\\]\n\nWith \\(T=9/365\\), we get \\(d_1\\approx-1.52453\\) and \\(d_2\\approx-1.59143\\), giving a call price of approximately **0.6946**.\n\n{\"answer\": 0.6946, \"confidence\": 99}" }, { "taskId": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 23.323917391138792, "answer": 23.324, "confidence": 100, "grade": "exact", "relError": 0.0000035418090290763216, "cost": 0.005982, "reply": "Using \\(r=0.0006\\), \\(\\sigma=0.355\\), and \\(T=20/365\\),\n\n\\[\nd_1=\\frac{\\ln(314.3/294)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T}\n\\approx0.84542,\\qquad d_2\\approx0.76232.\n\\]\n\nThe Black–Scholes call price is\n\\[\nC=314.3\\,N(d_1)-294e^{-rT}N(d_2)\\approx23.3240.\n\\]\n\n{\"answer\":23.3240,\"confidence\":100}" }, { "taskId": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.280639388201964, "answer": -0.2806, "confidence": 100, "grade": "exact", "relError": 0.00014035165276097544, "cost": 0.003274, "reply": "**Calculating expressions**\n\nI need to compute normal d1, which involves quite a few steps. First, I have to work with the values I've got: T, sigma, and the results of some logarithmic calculations. I need to take the series of values and find their rates, ultimately leading to a final sum. I also want to make sure I'm approximating correctly. Let's see how this all fits together to yield the answers I need!\nFor a non-dividend-paying asset, the put delta is\n\\[\n\\Delta_{\\text{put}}=\\Phi(d_1)-1,\\qquad\nd_1=\\frac{\\ln(114.52/106)+(0.0585+0.466^2/2)(50/365)}\n{0.466\\sqrt{50/365}}\\approx0.5809.\n\\]\nThus, \\(\\Delta_{\\text{put}}\\approx -0.2806\\).\n\n{\"answer\": -0.2806, \"confidence\": 100}" }, { "taskId": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.31093085044630486, "answer": -0.31093, "confidence": 100, "grade": "exact", "relError": 0.0000027351621868911074, "cost": 0.005104, "reply": "For a non-dividend-paying asset, put delta is\n\\[\n\\Delta_{\\text{put}}=N(d_1)-1,\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T}.\n\\]\nUsing \\(T=41/365\\) gives \\(d_1\\approx0.49321\\), so \\(\\Delta_{\\text{put}}\\approx-0.31093\\).\n\n{\"answer\": -0.31093, \"confidence\": 100}" }, { "taskId": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.5641608283248785, "answer": -0.56416, "confidence": 100, "grade": "exact", "relError": 0.0000014682424529991218, "cost": 0.003814, "reply": "The Black–Scholes put delta is\n\\[\n\\Delta_{\\text{put}}=N(d_1)-1,\\qquad\nd_1=\\frac{\\ln(215.92/231)+(0.0376+0.528^2/2)(65/365)}\n{0.528\\sqrt{65/365}}\n\\approx-0.16153.\n\\]\nThus, \\(\\Delta_{\\text{put}}\\approx-0.56416\\).\n\n{\"answer\": -0.56416, \"confidence\": 100}" }, { "taskId": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.3853788948381438, "answer": -0.38538, "confidence": 100, "grade": "exact", "relError": 0.000002867728022990249, "cost": 0.006104, "reply": "For a non-dividend-paying asset, the put delta is\n\\[\n\\Delta_{\\text{put}}=\\Phi(d_1)-1,\\qquad\nd_1=\\frac{\\ln(291.06/286)+(0.0594+0.729^2/2)(113/365)}\n{0.729\\sqrt{113/365}}\n\\approx 0.291384.\n\\]\nThus, \\(\\Delta_{\\text{put}}\\approx -0.38538\\).\n\n{\"answer\": -0.38538, \"confidence\": 100}" }, { "taskId": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.4080759766740296, "answer": -0.408076, "confidence": 100, "grade": "exact", "relError": 5.7160851681539386e-8, "cost": 0.005104, "reply": "The Black–Scholes put delta is\n\\[\n\\Delta_{\\text{put}}=N(d_1)-1,\\qquad\nd_1=\\frac{\\ln(146.31/152)+(0.0456+0.799^2/2)(162/365)}\n{0.799\\sqrt{162/365}}\n\\approx0.23250.\n\\]\nThus, \\(\\Delta_{\\text{put}}\\approx-0.408076\\).\n\n{\"answer\": -0.408076, \"confidence\": 100}" }, { "taskId": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.007721610843754648, "answer": 0.007722, "confidence": 100, "grade": "exact", "relError": 0.000050398324031885966, "cost": 0.004926, "reply": "**Calculating options parameters**\n\nI need to accurately compute parameters using the formula for gamma. I’m working with various values, like volatility represented by sigma, time T, and the stock price S. I have some calculations involving logarithms and drift, and I'm checking if I get consistent results. My goal is to derive a concise numerical answer based on these parameters, but I’m feeling a bit overwhelmed by all these calculations. I’ll strive for clarity to make the results easier to understand.\nFor a non-dividend-paying asset, Black–Scholes gamma is\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt{T}},\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nUsing \\(T=34/365\\), \\(\\sigma=0.844\\), and \\(r=0.003\\), we get \\(d_1\\approx0.58758\\), so \\(\\Gamma\\approx0.007722\\). Gamma is the same for a European call or put.\n\n{\"answer\": 0.007722, \"confidence\": 100}" }, { "taskId": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.011719035620003572, "answer": 0.011719, "confidence": 99, "grade": "exact", "relError": 0.000003039499556679529, "cost": 0.006446, "reply": "The Black–Scholes gamma is\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt{T}},\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nWith \\(T=71/365\\), \\(d_1\\approx-0.791276\\), giving \\(\\Gamma\\approx0.0117190\\).\n\n{\"answer\": 0.0117190, \"confidence\": 99}" }, { "taskId": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.00332788170697725, "answer": 0.0033279, "confidence": 100, "grade": "exact", "relError": 0.0000054968969333978925, "cost": 0.003476, "reply": "For a non-dividend-paying asset,\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt T},\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T}.\n\\]\nUsing \\(T=107/365\\), we get \\(d_1\\approx0.15996\\), giving \\(\\Gamma\\approx0.0033279\\).\n\n{\"answer\": 0.0033279, \"confidence\": 100}" }, { "taskId": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.015589164355628841, "answer": 0.01558819, "confidence": 100, "grade": "exact", "relError": 0.00006250210765720775, "cost": 0.007146, "reply": "For a non-dividend-paying asset, Black–Scholes gamma is\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt{T}},\n\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nUsing \\(T=147/365\\), we get \\(d_1\\approx-0.272902\\), giving \\(\\Gamma\\approx0.01558819\\).\n\n{\"answer\": 0.01558819, \"confidence\": 100}" }, { "taskId": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.00596435584251129, "answer": 0.0059644, "confidence": 100, "grade": "exact", "relError": 0.000007403563750354649, "cost": 0.005392, "reply": "The Black–Scholes gamma is\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt T},\n\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T}.\n\\]\nUsing \\(T=103/365\\) and \\(r=0.0003\\), we get \\(d_1\\approx0.94514\\), giving \\(\\Gamma\\approx0.0059644\\).\n\n{\"answer\": 0.0059644, \"confidence\": 100}" }, { "taskId": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.646446181061521, "answer": 0.6465, "confidence": 99, "grade": "exact", "relError": 0.00008325354848662669, "cost": 0.005002, "reply": "Black–Scholes vega per 1 percentage point increase in volatility is\n\\[\n\\frac{S\\phi(d_1)\\sqrt{T}}{100},\n\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nUsing \\(T=74/365\\) gives \\(d_1\\approx0.57598\\), so the vega is approximately **0.6465**.\n\n{\"answer\": 0.6465, \"confidence\": 99}" }, { "taskId": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.26781898170848095, "answer": 0.2678, "confidence": 100, "grade": "exact", "relError": 0.00007087514245585115, "cost": 0.005946, "reply": "Black–Scholes vega for a 1 percentage point volatility increase is\n\\[\n\\text{Vega}_{1\\%}=0.01S\\phi(d_1)\\sqrt T,\n\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T}.\n\\]\nUsing \\(T=104/365\\) gives \\(d_1\\approx-0.5585\\), so\n\\[\n\\text{Vega}_{1\\%}\\approx0.2678.\n\\]\n\n{\"answer\":0.2678,\"confidence\":100}" }, { "taskId": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.45587809695630876, "answer": 0.4559, "confidence": 100, "grade": "exact", "relError": 0.000048045834703398345, "cost": 0.004806, "reply": "Black–Scholes vega for a **1 percentage point** increase in volatility is\n\\[\n\\text{Vega}_{1\\%}=0.01S\\phi(d_1)\\sqrt{T},\n\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nWith \\(T=47/365\\), we get \\(d_1\\approx-0.8816\\), giving\n\\[\n\\text{Vega}_{1\\%}\\approx 0.4559.\n\\]\n\n{\"answer\": 0.4559, \"confidence\": 100}" }, { "taskId": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.6097213395050691, "answer": 0.60972, "confidence": 100, "grade": "exact", "relError": 0.0000021969135444301086, "cost": 0.005526, "reply": "Black–Scholes vega per 1 percentage point increase in volatility is\n\\[\n\\frac{S\\phi(d_1)\\sqrt{T}}{100},\n\\quad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nUsing \\(T=61/365\\), we get \\(d_1\\approx-0.22127\\), giving vega \\(\\approx 0.60972\\).\n\n{\"answer\": 0.60972, \"confidence\": 100}" }, { "taskId": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.2403754082675673, "answer": 0.24045, "confidence": 100, "grade": "exact", "relError": 0.00031031349242548477, "cost": 0.005336, "reply": "For a 1 percentage point increase in volatility,\n\\[\n\\text{Vega}_{1\\%}=\\frac{S\\phi(d_1)\\sqrt{T}}{100},\n\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt{T}}.\n\\]\nWith \\(T=65/365\\), \\(d_1\\approx0.46253\\), giving \\(\\text{Vega}_{1\\%}\\approx0.24045\\).\n\n{\"answer\": 0.24045, \"confidence\": 100}" }, { "taskId": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 61.799447576373446, "answer": 61.7995, "confidence": 99, "grade": "exact", "relError": 8.482863296056637e-7, "cost": 0.012244, "reply": "Using Black–Scholes,\n\\[\n3.5195=27.09\\,N(d_1)-29e^{-0.054T}N(d_2),\n\\qquad T=\\frac{136}{365},\n\\]\nwhere\n\\[\nd_1=\\frac{\\ln(27.09/29)+(0.054+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\qquad d_2=d_1-\\sigma\\sqrt T.\n\\]\nSolving for \\(\\sigma\\) gives approximately \\(0.617995\\), or **61.7995% per year**.\n\n{\"answer\":61.7995,\"confidence\":99}" }, { "taskId": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 20.900010973832025, "answer": 20.9, "confidence": 100, "grade": "exact", "relError": 5.250634576190053e-7, "cost": 0.009734, "reply": "Using Black–Scholes,\n\\[\nP=Ke^{-rT}\\Phi(-d_2)-S\\Phi(-d_1),\\qquad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\quad d_2=d_1-\\sigma\\sqrt T.\n\\]\nWith \\(T=53/365\\), solving for the volatility that gives \\(P=13.6926\\) yields \\(\\sigma\\approx0.2090\\), or **20.90% per year**.\n\n{\"answer\":20.9000,\"confidence\":100}" }, { "taskId": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 81.50015130720656, "answer": 81.5, "confidence": 99, "grade": "exact", "relError": 0.0000018565266952024242, "cost": 0.013892, "reply": "Using \\(T=161/365\\) and \\(r=0.0004\\), solve\n\\[\n20.922=86e^{-rT}\\Phi(-d_2)-79.84\\Phi(-d_1),\n\\]\nwhere\n\\[\nd_1=\\frac{\\ln(79.84/86)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\qquad d_2=d_1-\\sigma\\sqrt T.\n\\]\nThe implied volatility is approximately **81.50% per year**.\n\n{\"answer\":81.50,\"confidence\":99}" }, { "taskId": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 87.70003216471197, "answer": 87.7, "confidence": 99, "grade": "exact", "relError": 3.6675826876828525e-7, "cost": 0.008824, "reply": "**Evaluating calculations**\n\nI'm working through some calculations where I added a tiny value and considered a square root. I noticed an error in my previous log, which is tiny but notable. I'm computing the log ratio now using K/S, and the results are lining up. I found some series adjustments, and ultimately, the differences are minimal, which is good. Everything seems to be settling into place nicely, and I'm happy with the outcome.\nFor a European put,\n\\[\nP=Ke^{-rT}\\Phi(-d_2)-S\\Phi(-d_1),\n\\quad\nd_1=\\frac{\\ln(S/K)+(r+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\quad d_2=d_1-\\sigma\\sqrt T.\n\\]\nUsing \\(T=105/365\\) and \\(r=0.0106\\), solving for the volatility that gives \\(P=75.8825\\) yields \\(\\sigma\\approx0.8770\\), or **87.70% per year**.\n\n{\"answer\":87.70,\"confidence\":99}" }, { "taskId": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 33.39992865638508, "answer": 33.401, "confidence": 99, "grade": "exact", "relError": 0.000032076224651448434, "cost": 0.008782, "reply": "Using Black–Scholes,\n\\[\n11.908=140.66\\,N(d_1)-143e^{-0.019T}N(d_2),\n\\quad T=\\frac{161}{365},\n\\]\nwhere\n\\[\nd_1=\\frac{\\ln(140.66/143)+(0.019+\\sigma^2/2)T}{\\sigma\\sqrt T},\n\\qquad d_2=d_1-\\sigma\\sqrt T.\n\\]\nSolving for \\(\\sigma\\) gives approximately \\(0.33401\\), or **33.401% per year**.\n\n{\"answer\":33.401,\"confidence\":99}" }, { "taskId": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 6.438461538461515, "answer": 6.438461538, "confidence": 100, "grade": "exact", "relError": 7.168084790274904e-11, "cost": 0.002078, "reply": "**Calculating division**\n\nI need to compute 2511 divided by 390, which equals approximately 6.438461538. The explanation might involve the context, maybe related to a distance or a specific lot. The decimal portion 0.00390 could indicate some relevant detail or measurement that I might need to keep in mind. It's a simple calculation, but I want to ensure I’m capturing all key points related to this computation!\nThe stop distance is \\(0.94593-0.94203=0.00390\\), so each standard lot risks \\(100{,}000 \\times 0.00390 = 390\\) USD.\n\nPosition size: \\(2511/390 = 6.438461538\\) standard lots.\n\n{\"answer\": 6.438461538, \"confidence\": 100}" }, { "taskId": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 12.3258666666662, "answer": 12.3258666667, "confidence": 100, "grade": "exact", "relError": 2.7422427042406143e-12, "cost": 0.002178, "reply": "**Calculating the difference**\n\nI need to compute 184888 divided by 15000, which equals approximately 12.33. Next, I'm considering the difference of 0.150 yen per euro, and how that might factor into a lot loss of 15000. Finally, I'll summarize this information in a concise JSON format. It's important to ensure that I present it clearly and correctly to avoid any confusion. Let's piece it all together!\nThe stop distance is \\(154.513-154.363=0.150\\) JPY per EUR.\n\nEach standard lot risks \\(100000\\times0.150=15000\\) JPY, so the position size is \\(184888/15000=12.3258666667\\) standard lots.\n\n{\"answer\":12.3258666667,\"confidence\":100}" }, { "taskId": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.4756567164179193, "answer": 0.4756567164, "confidence": 100, "grade": "exact", "relError": 3.767270923482983e-11, "cost": 0.001878, "reply": "**Calculating division**\n\nI'm working on calculating the division of 31869 by 67000. It seems to yield approximately 0.4757 when rounded. This isn't too complicated, but it’s nice to have that figure clearly stated. Now, I guess I should think about how this result might be useful or relevant. It's always interesting to see how numbers play out! So, 31869 divided by 67000 is about 0.4757—short and sweet!\nThe stop distance is 191.077 − 190.407 = 0.670 JPY per EUR.\n\nPosition size = 31869 ÷ (0.670 × 100000) = 0.4756567164 standard lots.\n\n{\"answer\": 0.4756567164, \"confidence\": 100}" }, { "taskId": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 5.01704545454548, "answer": 5.01704545, "confidence": 100, "grade": "exact", "relError": 9.060072485877473e-10, "cost": 0.002018, "reply": "**Calculating division result**\n\nI need to compute the division of 220750 by 44000, which should give me a result of approximately 5.01704545. I'll make sure to display the result clearly for the user. It seems straightforward enough, and I want to ensure the accuracy is spot-on. Let me confirm that calculation one more time to make sure it comes out just right! I’ll show the final answer neatly.\nThe stop distance is 142.806 − 142.366 = 0.440 JPY per USD.\n\nRisk per standard lot = 0.440 × 100000 = 44000 JPY.\n\nPosition size = 220750 ÷ 44000 = **5.01704545 standard lots**.\n\n{\"answer\": 5.01704545, \"confidence\": 100}" }, { "taskId": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 1.6141304347826113, "answer": 1.6141304348, "confidence": 100, "grade": "exact", "relError": 1.0772832682032256e-11, "cost": 0.002098, "reply": "**Calculating and interpreting quotes**\n\nI need to calculate 1485/920, which comes out to approximately 1.614. Then I see that with a USD quote distance of 0.0092 per lot, it seems like I’m considering a short position. I’m making sure to keep all the currency values clear and accurate. This info should help in assessing any potential trades or decisions I'm looking at. I want to be as precise as possible in my calculations!\nThe stop distance is \\(0.76163-0.75243=0.00920\\), so each standard lot risks \\(0.00920\\times100000=920\\) USD.\n\nPosition size: \\(1485/920=1.6141304348\\) standard lots.\n\n{\"answer\": 1.6141304348, \"confidence\": 100}" }, { "taskId": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.023819567833442343, "answer": 0.023819568, "confidence": 100, "grade": "exact", "relError": 6.992471808068057e-9, "cost": 0.001812, "reply": "Using \\(z_{0.975}=1.959964\\),\n\n\\[\n\\mathrm{VaR}=-(-0.0003)+1.959964(0.012)=0.023819568.\n\\]\n\n{\"answer\": 0.023819568, \"confidence\": 100}" }, { "taskId": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.08208473980663146, "answer": 0.08208476, "confidence": 100, "grade": "exact", "relError": 2.4600636603490794e-7, "cost": 0.00196, "reply": "**Calculating values**\n\nI'm working on a computation: first, I multiply 1.6448536 by .0496, and then I add .0005. That gives me .08158473856. Next, I add another .0005, which brings the total to .08208473856. It seems straightforward, but I always want to double-check to make sure everything adds up correctly!\nUsing \\(z_{0.95}=1.644854\\):\n\n\\[\n\\mathrm{VaR}=-(-0.0005)+1.644854(0.0496)=0.08208476.\n\\]\n\n{\"answer\": 0.08208476, \"confidence\": 100}" }, { "taskId": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.10032876732331801, "answer": 0.1003287673, "confidence": 100, "grade": "exact", "relError": 2.324160338884447e-10, "cost": 0.00216, "reply": "**Calculating values**\n\nI need to perform a calculation starting with 2.326347874 multiplied by 0.0436, then subtract 0.0011. First, I find that 0.04 equals approximately 0.09305391496, and 0.0036 results in about 0.0083748523464. Adding these sums gives me roughly 0.1014287673064. Finally, subtracting 0.0011 from that leads to the result of about 0.1003287673. It's all adding up, but keeping track of the values is key!\nUsing \\(z_{0.99}=2.326347874\\),\n\n\\[\n\\mathrm{VaR}=-0.0011+2.326347874(0.0436)=0.1003287673.\n\\]\n\n{\"answer\": 0.1003287673, \"confidence\": 100}" }, { "taskId": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.07275151453115127, "answer": 0.0727515, "confidence": 100, "grade": "exact", "relError": 1.9973675272628119e-7, "cost": 0.00224, "reply": "**Calculating the expression**\n\nI need to compute the expression, starting with 2.326347874 multiplied by 0.0308 plus 0.0011. I'll check if it equals 0.071652. Next, I can also compute 2.326347874 multiplied by the sum of 0.03 and 0.0008, which goes to 0.06979043622. Then, I’ll add 0.0018610782992 to that, which gives me 0.0716515145192. Finally, adding 0.0011 gives me 0.0727515145192. It seems like there's a slight discrepancy!\nUsing \\(z_{0.99}=2.32635\\):\n\n\\[\n\\mathrm{VaR}=-(-0.0011)+2.32635(0.0308)\\approx0.0727515.\n\\]\n\n{\"answer\":0.0727515,\"confidence\":100}" }, { "taskId": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.11150152403343133, "answer": 0.111501524, "confidence": 100, "grade": "exact", "relError": 2.9982843230943654e-10, "cost": 0.001948, "reply": "**Calculating math expression**\n\nI need to compute the expression 2.326347874 multiplied by 0.0475, then add 0.001. That gives me 0.1105015240, and when I add 0.001, I get 0.111501524. It seems a bit short, so I'll double-check to make sure everything adds up correctly. Overall, I’m just working through the numbers to ensure my calculations are accurate.\nUsing \\(z_{0.99}=2.326347874\\),\n\n\\[\n\\mathrm{VaR}=-(-0.001)+2.326347874(0.0475)=0.111501524.\n\\]\n\n{\"answer\": 0.111501524, \"confidence\": 100}" }, { "taskId": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.04404168455292251, "answer": 0.04404168, "confidence": 100, "grade": "exact", "relError": 1.0337757416508617e-7, "cost": 0.00242, "reply": "**Calculating formula steps**\n\nI need to work through a short formula. I see it starting with z at 2.326347874 multiplied by phi, which is 0.0266521422. Then, I’ll divide that by 0.01. Next, I need to multiply 2.66521422 by 0.0172. After doing those calculations, I will subtract 0.0018. The goal is to reach a final result of 0.0440416846. Let’s get these numbers sorted!\nFor normally distributed returns, the 99% Expected Shortfall as a positive loss fraction is\n\\[\n\\mathrm{ES}_{0.99}=-\\mu+\\sigma\\frac{\\phi(\\Phi^{-1}(0.99))}{0.01}\n=-0.0018+0.0172(2.665214)\\approx 0.04404168.\n\\]\n\n{\"answer\": 0.04404168, \"confidence\": 100}" }, { "taskId": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.02233967069948375, "answer": 0.02233967, "confidence": 100, "grade": "exact", "relError": 3.131128303454666e-8, "cost": 0.00237, "reply": "**Calculating formulas**\n\nI need to compute a couple of formulas. The first one is: 0.0013 + (0.0102 * 0.10313564) / 0.05, which equals about 0.02233967. Then, I’ll compute 2.0627128 * 0.0102 and add that to 0.0013, getting approximately 0.02233967056. It seems straightforward, but I want to ensure accuracy in my calculations. Keeping things short and precise is key here!\nFor normally distributed returns, the 95% Expected Shortfall as a positive loss is\n\\[\nES_{0.95}=-\\mu+\\sigma\\frac{\\phi(1.64485)}{0.05}\n=0.0013+0.0102(2.06271)\n\\approx 0.02233967.\n\\]\n\n{\"answer\": 0.02233967, \"confidence\": 100}" }, { "taskId": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.09524684038587, "answer": 0.09524684, "confidence": 100, "grade": "exact", "relError": 4.051263024094849e-9, "cost": 0.00245, "reply": "**Calculating phi value**\n\nI need to compute the phi value using the formula provided. So, starting with 2.326347874 equals 0.0266521422 divided by 0.01, which gives me 2.66521422. Then I multiply that by 0.0366 and subtract 0.0023 from the result. The final step involves calculating 0.09754684045 minus 0.0023, yielding 0.09524684045. I should ensure I have that all laid out clearly in a compact formula.\nFor normally distributed returns, the 99% Expected Shortfall as a positive loss fraction is\n\\[\nES_{0.99}=-\\mu+\\sigma\\frac{\\phi(\\Phi^{-1}(0.99))}{0.01}\n=-0.0023+0.0366(2.665214)\n\\approx 0.09524684.\n\\]\n\n{\"answer\": 0.09524684, \"confidence\": 100}" }, { "taskId": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.12233732788143936, "answer": 0.1223373, "confidence": 100, "grade": "exact", "relError": 2.2790623147031247e-7, "cost": 0.002944, "reply": "**Computing a formula step-by-step**\n\nI’m working through a calculation involving z-scores and a few other ratios. First, I have the z-score z1.959963984, then I consider the ratio and some multipliers. I need to compute 2.337802792 times 0.05, which gives me a result to factor into the overall sum. Finally, I'll subtract 0.0018 from my total. Overall, it looks like a straightforward computation, but I want to be accurate with each step I take.\nFor normal returns, Expected Shortfall as a positive loss fraction is\n\\[\nES_{97.5\\%}=-\\mu+\\sigma\\frac{\\phi(\\Phi^{-1}(0.975))}{0.025}.\n\\]\nUsing \\(z=1.959964\\),\n\\[\nES=-0.0018+0.0531\\frac{0.0584451}{0.025}\n\\approx 0.1223373.\n\\]\n\n{\"answer\": 0.1223373, \"confidence\": 100}" }, { "taskId": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.1205670856534704, "answer": 0.1205671, "confidence": 100, "grade": "exact", "relError": 1.1899209079688076e-7, "cost": 0.002784, "reply": "**Calculating phi value**\n\nI need to compute phi using the formula phi1.959964/.025, which gives me 2.33780279. From there, I’ll multiply that by .0523 and subtract .0017. After working through this, I see that 2.33780279 multiplied by .05 equals .1168901395. Then, I multiply that by .0023 to get a smaller value. My calculations continue with adjustments to find the total, eventually landing at .120567085917 for the final result. It's a complex little formula!\nFor normally distributed returns, the expected shortfall as a positive loss is\n\\[\n\\mathrm{ES}_{97.5\\%}=-\\mu+\\sigma\\frac{\\phi(\\Phi^{-1}(0.975))}{0.025}\n=-0.0017+0.0523(2.337803)\n\\approx 0.1205671.\n\\]\n\n{\"answer\": 0.1205671, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 1.6781883884002948, "answer": 1.67819, "confidence": 100, "grade": "exact", "relError": 9.603210917092653e-7, "cost": 0.00533, "reply": "The mean monthly return is \\(0.02114167\\), and the sample standard deviation is \\(0.04364044\\).\n\nThus the annualized Sharpe ratio is\n\\[\n\\frac{0.02114167}{0.04364044}\\sqrt{12}\\approx 1.67819.\n\\]\n\n{\"answer\": 1.67819, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.48328609065675004, "answer": -0.483286, "confidence": 100, "grade": "exact", "relError": 1.875840248747503e-7, "cost": 0.00523, "reply": "The monthly mean is \\(-0.00545\\), and the sample standard deviation is\n\\[\ns=\\sqrt{\\frac{\\sum_{i=1}^{12}(r_i-\\bar r)^2}{11}}\n\\approx 0.03906455.\n\\]\nThus the annualized Sharpe ratio is\n\\[\n\\frac{-0.00545}{0.03906455}\\sqrt{12}\\approx -0.483286.\n\\]\n\n{\"answer\": -0.483286, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 2.241475231769422, "answer": 2.241475, "confidence": 100, "grade": "exact", "relError": 1.0340039409575755e-7, "cost": 0.006866, "reply": "The monthly mean is \\(0.02285\\), and the sample standard deviation is \\(0.03531367\\).\n\nThus, the annualized Sharpe ratio is\n\\[\n\\frac{0.02285}{0.03531367}\\sqrt{12}\\approx 2.241475.\n\\]\n\n{\"answer\": 2.241475, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 1.9830100176589147, "answer": 1.983, "confidence": 100, "grade": "exact", "relError": 0.000005051743977796405, "cost": 0.005644, "reply": "The monthly mean is \\(0.0259\\), and the sample standard deviation is\n\\[\ns=\\sqrt{\\frac{0.0305674-12(0.0259)^2}{11}}\\approx0.04524447.\n\\]\nThus the annualized Sharpe ratio is\n\\[\n\\frac{0.0259}{0.04524447}\\sqrt{12}\\approx1.9830.\n\\]\n\n{\"answer\":1.9830,\"confidence\":100}" }, { "taskId": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 1.3915400475486268, "answer": 1.39154, "confidence": 100, "grade": "exact", "relError": 3.416978684690232e-8, "cost": 0.00509, "reply": "The monthly mean is \\(0.014125\\), and the sample standard deviation is approximately \\(0.03516279\\).\n\nThus the annualized Sharpe ratio is\n\\[\n\\frac{0.014125}{0.03516279}\\sqrt{12}\\approx 1.39154.\n\\]\n\n{\"answer\": 1.39154, \"confidence\": 100}" }, { "taskId": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.4848484848484847, "answer": 0.4848484848, "confidence": 100, "grade": "exact", "relError": 9.999972724883402e-11, "cost": 0.002156, "reply": "**Calculating a formula**\n\nI need to work on this concise formula: f = (p - c) / (1 - c). I’m finding that when I plug in the numbers, it comes out to .16 divided by .33, which gives me approximately .4848. But I want to make sure I clarify what \"stake cash\" means in this context. Once I get that clear, I can confirm the output as .4848.\nFor a contract priced at \\(c\\), the full Kelly fraction is\n\\[\nf=\\frac{p-c}{1-c}=\\frac{0.83-0.67}{1-0.67}=0.484848\\ldots\n\\]\nSo you should stake about **48.48% of your bankroll**.\n\n{\"answer\": 0.4848484848, \"confidence\": 100}" }, { "taskId": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.28985507246376807, "answer": 0.2898550725, "confidence": 100, "grade": "exact", "relError": 1.2500007695592786e-10, "cost": 0.002346, "reply": "**Breaking down stake fractions**\n\nI need to distinguish the stake fraction. So, .2 divided by .69 equals about .2899. Then, I derive wealth for winning as 1 plus f, which is calculated by (.69 divided by .31). For losing, it’s 1 minus f. I define f as (p minus c) divided by (1 minus c). I think my final thoughts here need to be more concise, so I’ll wrap it up neatly!\nFor contract price \\(c=0.31\\) and probability \\(p=0.51\\), the full Kelly stake fraction is\n\\[\nf^*=\\frac{p-c}{1-c}=\\frac{0.51-0.31}{0.69}=0.289855.\n\\]\nSo stake about **28.99% of your bankroll**.\n\n{\"answer\": 0.2898550725, \"confidence\": 100}" }, { "taskId": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.09859154929577463, "answer": 0.09859155, "confidence": 100, "grade": "exact", "relError": 7.142857349246818e-9, "cost": 0.002486, "reply": "**Calculating a fraction**\n\nI need to derive the value of f using the formula (p-c)/(1-c)=.07/.71, which simplifies to approximately .09858955. When I express this as a percentage, it comes out to about 9.859%. I'll also represent this as a JSON number fraction, rounding it to .09859 for clarity. It's essential to ensure that all values are accurate and properly converted into their respective formats!\nIf \\(f\\) is the bankroll fraction spent on contracts, a win multiplies wealth by \\(1+f(1/0.29-1)\\), while a loss leaves \\(1-f\\).\n\nThe full Kelly fraction is\n\\[\nf=\\frac{p-c}{1-c}=\\frac{0.36-0.29}{1-0.29}=0.09859155,\n\\]\nso stake about **9.859% of your bankroll**.\n\n{\"answer\": 0.09859155, \"confidence\": 100}" }, { "taskId": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.8333333333333334, "answer": 0.8333333333, "confidence": 100, "grade": "exact", "relError": 4.000000330961484e-11, "cost": 0.002276, "reply": "**Calculating wealth ratios**\n\nI need to clarify the formula I'm working with: \\( f = \\frac{0.8333p - (1 - c)}{0.7} \\), which seems to involve some wealth comparison. Also, I need to relate wealth gains with a component of 1 + 5.25 and losses with 1 - f. Lastly, I plan to wrap this up in a short JSON format for clarity. I'll ensure I have a clean final output!\nFor a binary contract priced at \\(c\\), the full Kelly fraction is\n\\[\nf^*=\\frac{p-c}{1-c}=\\frac{0.86-0.16}{0.84}=0.833333\\ldots\n\\]\nSo you would stake about **83.33% of your bankroll**.\n\n{\"answer\": 0.8333333333, \"confidence\": 100}" }, { "taskId": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.576271186440678, "answer": 0.5762712, "confidence": 100, "grade": "exact", "relError": 2.352941170334061e-8, "cost": 0.002756, "reply": "Let \\(f\\) be the fraction of bankroll spent on contracts. Your bankroll becomes \\(1+f(1/0.41-1)\\) if the event occurs, and \\(1-f\\) otherwise.\n\nMaximizing expected log wealth gives\n\\[\nf^*=\\frac{p-c}{1-c}\n=\\frac{0.75-0.41}{1-0.41}\n=\\frac{0.34}{0.59}\n\\approx 0.5762712.\n\\]\n\nThus, full Kelly says to stake about **57.63% of your bankroll**.\n\n{\"answer\": 0.5762712, \"confidence\": 100}" }, { "taskId": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 24920.824742268043, "answer": 24920.82474, "confidence": 100, "grade": "exact", "relError": 9.10099621924795e-11, "cost": 0.002936, "reply": "**Calculating short PnL**\n\nI'm deriving a short PnL using the formula Q(1/P - 1/E). The margin can be calculated as Q/E multiplied by .04, and the maintenance is .01 times Q/E. I've calculated that 1/P equals .97/E, leading to P being approximately 24920.82. When I compute 24173.2 divided by .97, I confirm the result is 24920.82. Everything needs to be short and concise, so I’m streamlining the steps for clarity!\nFor USD contract size \\(Q\\), the short’s unrealized PnL is \\(Q(1/P-1/24173.2)\\) BTC.\n\nAt liquidation,\n\\[\n\\frac{Q}{25(24173.2)}+Q\\left(\\frac1P-\\frac1{24173.2}\\right)\n=\\frac{0.01Q}{24173.2}.\n\\]\nThus \\(1/P=0.97/24173.2\\), giving \\(P=24920.82474\\).\n\n{\"answer\":24920.82474,\"confidence\":100}" }, { "taskId": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 53342.9702970297, "answer": 53342.970297, "confidence": 100, "grade": "exact", "relError": 5.567829990665248e-13, "cost": 0.003306, "reply": "**Calculating short formula**\n\nI need a short formula to work with, focusing on some variables here. I’m looking at Q and E, and I'm setting it up with the margin as Q/(50E). For the profit and loss calculation, it seems like it’s Q times (1/E minus 1/P). Then, I’m maintaining the equation as 0.01Q/E. Okay, from that, I arrive at some final value for E, which looks to be around 53,342.97. That seems useful!\nLet \\(Q\\) be the USD contract size and \\(P\\) the liquidation price. For an inverse long, BTC PnL is \\(Q(1/53876.4-1/P)\\).\n\nThe liquidation condition is\n\\[\n\\frac{Q}{50(53876.4)}+Q\\left(\\frac1{53876.4}-\\frac1P\\right)\n=\\frac{0.01Q}{53876.4}.\n\\]\nCanceling \\(Q\\) gives \\(1/P=1.01/53876.4\\), so\n\\[\nP=\\frac{53876.4}{1.01}=53342.970297\\ldots\n\\]\n\n{\"answer\":53342.970297,\"confidence\":100}" }, { "taskId": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 53194.15525114154, "answer": 53194.15525, "confidence": 100, "grade": "exact", "relError": 2.1459867848544672e-11, "cost": 0.00302, "reply": "**Calculating entry point**\n\nI’m figuring out the entry point based on the equation where I need to divide by 1.095. So I start with 58247.6 divided by 1.095, and that gives me about 53200? When I multiply 53200 by 1.095, it results in 58254, so there’s a difference of -6.4. Dividing that by 1.095 gives me approximately -5.84, leading to an entry of about 53194. It feels slightly off. I need to double-check!\nFor USD contract size \\(Q\\), unrealized PnL is \\(Q(1/P_0-1/P)\\) BTC. Thus liquidation occurs when\n\\[\n\\frac{Q}{10P_0}+Q\\left(\\frac1{P_0}-\\frac1P\\right)=0.005\\frac{Q}{P_0}.\n\\]\nCanceling \\(Q\\) gives \\(P=P_0/1.095=58247.6/1.095\\).\n\n{\"answer\":53194.15525,\"confidence\":100}" }, { "taskId": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 170355.2216748768, "answer": 170355.2217, "confidence": 100, "grade": "exact", "relError": 1.474752749157994e-10, "cost": 0.003506, "reply": "For an inverse short with USD contract size \\(Q\\), unrealized PnL is \\(Q(1/P-1/P_0)\\). At liquidation,\n\\[\n\\frac{Q}{3P_0}+Q\\left(\\frac1P-\\frac1{P_0}\\right)=0.01\\frac{Q}{P_0}.\n\\]\nThus,\n\\[\nP=\\frac{P_0}{1-\\frac13+0.01}\n=\\frac{115273.7}{0.676666\\ldots}\n=170355.2217.\n\\]\n\n{\"answer\":170355.2217,\"confidence\":100}" }, { "taskId": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 115003.24607329845, "answer": 115003.2461, "confidence": 100, "grade": "exact", "relError": 2.3218089529464864e-10, "cost": 0.00354, "reply": "For an inverse short of \\(Q\\) USD contracts, BTC unrealized PnL is\n\\[\nQ\\left(\\frac{1}{P}-\\frac{1}{109828.1}\\right).\n\\]\n\nAt liquidation:\n\\[\n\\frac{Q}{20(109828.1)}\n+Q\\left(\\frac{1}{P}-\\frac{1}{109828.1}\\right)\n=\\frac{0.005Q}{109828.1}.\n\\]\n\nThus \\(1/P=0.955/109828.1\\), giving \\(P=109828.1/0.955\\).\n\n{\"answer\": 115003.2461, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 2398.0256188582507, "answer": 2398.026, "confidence": 100, "grade": "exact", "relError": 1.58939815376989e-7, "cost": 0.004948, "reply": "**Calculating harmonic weights**\n\nI need to compute a harmonic weighted average based on 86,000 divided by the sum, aiming for a manual accuracy of 1e-4 and using three terms. I’m checking if 7000 divided by 2309.36 gives me about 3.031142. After some manual calculations, it comes out to approximately that. \n\nI’ll repeat similar calculations for the other terms and sum them up to get the final quotient, which turns out to be around 2398.026.\nFor an inverse contract, the equivalent entry price is the USD-weighted **harmonic mean**, since ETH PnL is \\(Q(1/P_{\\text{entry}}-1/P_{\\text{exit}})\\).\n\n\\[\nP_{\\text{avg}}\n=\\frac{7000+34000+36000+9000}\n{\\frac{7000}{2309.36}+\\frac{34000}{3563.54}+\\frac{36000}{1854.34}+\\frac{9000}{2321.56}}\n\\approx 2398.026\n\\]\n\n{\"answer\": 2398.026, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 3712.5303509582454, "answer": 3712.53035, "confidence": 100, "grade": "exact", "relError": 2.581111388166906e-10, "cost": 0.004906, "reply": "For an inverse contract, the equivalent entry is the **USD-size-weighted harmonic mean**:\n\n\\[\nP_{\\text{entry}}\n=\\frac{34000+16000}{34000/3528.19+16000/4176.2}\n\\approx 3712.53035.\n\\]\n\nThis preserves the ETH PnL, \\(Q(1/P_{\\text{entry}}-1/P_{\\text{exit}})\\), at any exit price.\n\n{\"answer\":3712.53035,\"confidence\":100}" }, { "taskId": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 2495.29829080704, "answer": 2495.2982, "confidence": 100, "grade": "exact", "relError": 3.639125630933975e-8, "cost": 0.005838, "reply": "For an inverse contract, the equivalent entry is the USD-weighted harmonic mean:\n\n\\[\nP_{\\rm avg}=\n\\frac{41000+10000+27000+40000}\n{\\frac{41000}{3517.25}+\\frac{10000}{2976.86}+\\frac{27000}{4229.06}+\\frac{40000}{1545.09}}\n\\approx 2495.2982.\n\\]\n\nThis preserves the ETH PnL at any exit price.\n\n{\"answer\":2495.2982,\"confidence\":100}" }, { "taskId": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 2610.72093572179, "answer": 2610.721, "confidence": 100, "grade": "exact", "relError": 2.462086594375535e-8, "cost": 0.005178, "reply": "**Calculating harmonic sums**\n\nI need to compute values with a precision of .01% based on a total of 59,000. First, I calculate several ratios, starting with 10,000 divided by 3282.44. Each calculation yields a number I want to validate. For instance, some seem to match close enough to the expected result, while others need reevaluation. Ultimately, I'm checking my totals to ensure they add up accurately. I think I’ll finalize with the output and verify the last steps result in enough precision.\nFor an inverse contract, the equivalent entry is the USD-size-weighted **harmonic mean**:\n\n\\[\nP_{\\rm entry}\n=\\frac{\\sum_i Q_i}{\\sum_i Q_i/P_i}\n=\\frac{59000}{\n10000/3282.44+14000/2147.97+12000/1744.23+23000/3736.8}\n\\approx 2610.721.\n\\]\n\nThis preserves the combined ETH PnL, \\(\\sum_i Q_i(1/P_i-1/P_{\\rm exit})\\), at any exit price.\n\n{\"answer\": 2610.721, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 3424.594098647905, "answer": 3424.5941, "confidence": 100, "grade": "exact", "relError": 3.9481900700475597e-10, "cost": 0.005428, "reply": "**Calculating harmonic sum and price**\n\nI need to compute the harmonic sum accurately, starting with the values for ETH. The sum of the reciprocals yields 32.7046, and 3424.6 should provide good accuracy with at least four significant figures. I’m calculating various products and verifying residuals along the way, which all contribute to my final sum. Ultimately, I confirm that 3424.6 multiplied by the harmonic result gives me a total of approximately 111195.66, rounding off to 3424.59 for the final price.\nFor an inverse contract, the equivalent entry is the **USD-size-weighted harmonic mean**, since ETH PnL is \\(\\sum_i Q_i(1/P_i-1/P_{\\rm exit})\\).\n\n\\[\nP_{\\rm avg}\n=\\frac{\\sum_i Q_i}{\\sum_i Q_i/P_i}\n=\\frac{112000}{31000/3315.48+36000/3410.8+40000/3481.71+5000/3813.27}\n\\approx 3424.5941.\n\\]\n\n{\"answer\":3424.5941,\"confidence\":100}" }, { "taskId": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.02840777213808475, "answer": 0.02840777214, "confidence": 100, "grade": "exact", "relError": 6.741990184651402e-11, "cost": 0.004348, "reply": "For an inverse short, subtract both BTC-denominated fees from the gross PnL:\n\n\\[\n\\text{Net PnL}\n=44000\\left(\\frac{1}{41686.7}-\\frac{1}{42887.5}\\right)\n-0.00055(44000)\\left(\\frac{1}{42887.5}+\\frac{1}{41686.7}\\right)\n=0.02840777214\\text{ BTC}.\n\\]\n\n{\"answer\": 0.02840777214, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.06525146382766053, "answer": -0.06525146, "confidence": 100, "grade": "exact", "relError": 5.866014815941095e-8, "cost": 0.003954, "reply": "For an inverse long, net PnL in BTC is\n\\[\n45000\\left(\\frac1{49446}-\\frac1{46182.6}\\right)\n-0.0005(45000)\\left(\\frac1{49446}+\\frac1{46182.6}\\right)\n\\approx -0.06525146.\n\\]\n\n{\"answer\": -0.06525146, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.22922018359225318, "answer": 0.2292201836, "confidence": 100, "grade": "exact", "relError": 3.379633973554137e-11, "cost": 0.005408, "reply": "Net PnL after both fees is\n\\[\n189000\\left(\\frac{1}{91422.1}-\\frac{1}{102932.1}\\right)\n-0.0005(189000)\\left(\\frac{1}{91422.1}+\\frac{1}{102932.1}\\right)\n=0.2292201836\\text{ BTC}.\n\\]\n\n{\"answer\": 0.2292201836, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.026151408867684533, "answer": -0.02615141, "confidence": 100, "grade": "exact", "relError": 4.329844991575699e-8, "cost": 0.004868, "reply": "**Verifying complex calculations**\n\nI need to make sure my arithmetic is spot-on. The formula is complicated, involving a ratio and some specific constants. I'm calculating the differences and products carefully, aiming for accuracy. My intermediary calculations lead me through various products and ratios, and I wonder if I might have miscalculated any steps along the way. Ultimately, I believe my answer should be close to -0.02615141, but I want to verify each part thoroughly to be sure.\nFor an inverse long, net PnL in BTC is\n\\[\n14000\\left(\\frac{1}{89747.2}-\\frac{1}{76940.3}\\right)\n-0.00055(14000)\\left(\\frac{1}{89747.2}+\\frac{1}{76940.3}\\right)\n\\approx -0.02615141.\n\\]\n\n{\"answer\": -0.02615141, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.011586282591352507, "answer": -0.01158628, "confidence": 100, "grade": "exact", "relError": 2.2365693975986653e-7, "cost": 0.005358, "reply": "For an inverse short, net PnL after fees is\n\\[\n44000\\left(\\frac{1}{105593.7}-\\frac{1}{102777.6}\\right)\n-0.0002(44000)\\left(\\frac{1}{102777.6}+\\frac{1}{105593.7}\\right)\n= -0.01158628\\text{ BTC}.\n\\]\n\n{\"answer\": -0.01158628, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 6057.334255752001, "answer": 6057.334255752, "confidence": 100, "grade": "exact", "relError": 1.5014768268884594e-16, "cost": 0.002944, "reply": "Gross PnL: \\(1.703(26380.89-22814.19)=6074.0901\\)\n\nTotal fees: \\(0.0002 \\times 1.703(26380.89+22814.19)=16.755844248\\)\n\nNet PnL: \\(6074.0901-16.755844248=6057.334255752\\)\n\n{\"answer\":6057.334255752,\"confidence\":100}" }, { "taskId": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 2844.8890875599996, "answer": 2844.88908756, "confidence": 100, "grade": "exact", "relError": 1.5984712826765812e-16, "cost": 0.002684, "reply": "**Calculating totals and fees**\n\nI need to work through some arithmetic here. First, I’m taking 2485.82 multiplied by 1.164, which gives me 2893.49448 after adding 407.67448. I’ve confirmed that part. Then, when I sum 75922.20 with the same multiplier, I get 88373.4408. After applying a fee of 0.00055, I arrive at a net value of 2844.88908756. It’s all adding up correctly!\nNet PnL = size × price change − opening and closing fees:\n\n\\[\n1.164(39204.01-36718.19)-0.00055(1.164)(36718.19+39204.01)\n=2844.88908756\n\\]\n\n{\"answer\":2844.88908756,\"confidence\":100}" }, { "taskId": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -3783.556934676003, "answer": -3783.556934676, "confidence": 100, "grade": "exact", "relError": 8.413330395615781e-16, "cost": 0.002494, "reply": "**Calculating arithmetic results**\n\nI need to perform some arithmetic calculations. First, I find the difference: \\(-6121.88\\) multiplied by \\(0.612\\) gives \\(-3746.59056\\). Then I check another sum: \\(109822.86\\) multiplied by \\(0.612\\) results in \\(67211.59032\\), and multiplying that by \\(0.00055\\) gives me \\(36.966374676\\). Finally, I calculate the total, which is \\(-3783.556934676\\). I need these numbers to be clear and concise.\nNet PnL = 0.612 × (51850.49 − 57972.37) − 0.00055 × 0.612 × (57972.37 + 51850.49) = −3783.556934676.\n\n{\"answer\": -3783.556934676, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -1108.5578196799993, "answer": -1108.55781968, "confidence": 100, "grade": "exact", "relError": 6.15322912544697e-16, "cost": 0.003534, "reply": "Net PnL = size × (close − open) − opening and closing fees:\n\n\\(2.648(4440.76-4857.54)-0.0002(2.648)(4857.54+4440.76)=-1108.55781968\\)\n\n{\"answer\": -1108.55781968, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 3351.620043262996, "answer": 3351.620043263, "confidence": 100, "grade": "exact", "relError": 1.085438910178473e-15, "cost": 0.002994, "reply": "Net PnL = 1.949 × (28983.12 − 27232.54) − 0.00055 × 1.949 × (28983.12 + 27232.54) = **3351.620043263 USDT**.\n\n{\"answer\":3351.620043263,\"confidence\":100}" }, { "taskId": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.11472881355932601, "answer": 0.1147288, "confidence": 100, "grade": "exact", "relError": 1.1818588189447672e-7, "cost": 0.002916, "reply": "**Calculating risk and lot sizes**\n\nI need to compute a few values related to risk and lot sizes. For example, I've got 50 for risk and 59 pips, with a standard of 10. So, I’m calculating CAD divided by 1.3538, which gives me about 7.3866. From there, I’ll figure out how those relate to lot sizes as well. I'll clarify that 50 divided by (59 times 10 divided by 1.3538) is part of this. It feels a bit involved!\n- Risk amount: $5,000 × 1% = **$50**\n- Stop distance: 1.3538 − 1.3479 = **59 pips**\n- Pip value per standard lot: CAD 10 ÷ 1.3538 = **USD 7.3866**\n- Position size: 50 ÷ (59 × 7.3866) = **0.11473 standard lots**\n\nIf your broker requires 0.01-lot increments, round down to **0.11 lots** to stay within your risk limit.\n\n{\"answer\": 0.1147288, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.5939090909090843, "answer": 0.5939091, "confidence": 100, "grade": "exact", "relError": 1.5306914519508853e-8, "cost": 0.003036, "reply": "**Calculating values**\n\nI need to calculate: 250 divided by (55 times 10 divided by 1.3066). That gives me about 326.65 divided by 550, resulting in approximately 0.59390909. After that, I have to articulate the pip in USD, which is 7.65345 for one lot. Finally, I’ll prepare this all in JSON format. I want to make sure I get this right, so I'll double-check my numbers before finalizing!\nRisk amount: $25,000 × 1% = **$250** \nStop distance: 1.3066 − 1.3011 = **55 pips**\n\nFor one standard lot (100,000 USD), each pip is worth CAD 10. Converted at entry:\n\\[\n\\text{Pip value}=\\frac{10}{1.3066}=\\$7.65345\n\\]\n\n\\[\n\\text{Standard lots}=\\frac{250}{55\\times7.65345}=0.593909\n\\]\n\nTrade **0.5939 standard lots**, or round down to **0.59 lots** if your broker uses 0.01-lot increments.\n\n{\"answer\": 0.5939091, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 1.1982352941177008, "answer": 1.198235294, "confidence": 100, "grade": "exact", "relError": 9.822840190291062e-11, "cost": 0.002836, "reply": "**Calculating financial metrics**\n\nI need to compute a value: 250 multiplied by 0.8148, then divided by 170. After that, I find 203.7 divided by 170, which brings me to about 1.198. I also want to explain the 17 pip. For the pip standard, it's 10 CHF divided by 0.8148 USD, which gives me 12.273 lots. I’ll make sure to format this as final JSON numbers. Let's get this calculation squared away!\n- Risk budget: $25,000 × 1% = **$250**\n- Stop distance: 0.8148 − 0.8131 = **17 pips**\n- Pip value per standard lot: CHF 10 ÷ 0.8148 = **$12.27295**\n- Position size: $250 ÷ (17 × $12.27295) = **1.198235 standard lots**\n\nIf your broker requires 0.01-lot increments, round down to **1.19 lots** to stay within your risk limit.\n\n{\"answer\": 1.198235294, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 1.0178846153846508, "answer": 1.017884615, "confidence": 100, "grade": "exact", "relError": 3.778922972975077e-10, "cost": 0.002982, "reply": "**Calculating financial values**\n\nI need to calculate 250 divided by (39,000 divided by 158.79), which gives me a value of approximately 1.0179. To explain further, 39 pips in trading translates to a movement of 1,000 JPY, which is equivalent to about 6.2976 USD. So, I've got this lot at 1.0179, which reflects the value I'm analyzing. It's all about understanding these numbers in context, right?\n- Risk amount: $25,000 × 1% = **$250**\n- Stop distance: 158.79 − 158.40 = **39 pips**\n- Pip value per standard lot (100,000 USD), converted at entry: ¥1,000 ÷ 158.79 = **$6.2976**\n- Position size: $250 ÷ (39 × $6.2976) = **1.0179 standard lots**\n\nIf your broker requires 0.01-lot increments, round down to **1.01 lots** to stay within your risk limit, excluding fees and slippage.\n\n{\"answer\": 1.017884615, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-6.1-sol", "expected": 7.182727272727309, "answer": 7.1827, "confidence": 100, "grade": "exact", "relError": 0.000003796987728178971, "cost": 0.002892, "reply": "**Calculating risk per lot**\n\nI'm focusing on determining the risk for a position, specifically with a distance of 0.22 and a standard of 100k. I’m working through the equation using the pip amounts and the exchange rate of 158.02. After calculating, I arrive at 7.182727 as the concise final number. So, I’ll format this into a JSON structure for clarity, ensuring my confidence level is at 100%.\nRisk budget: $100,000 × 1% = **$1,000**.\n\nStop distance: 158.02 − 157.80 = **22 pips**.\n\nFor one standard lot (100,000 USD), pip value at entry:\n1,000 JPY ÷ 158.02 = **$6.3283 per pip**.\n\nPosition size: $1,000 ÷ (22 × $6.3283) = **7.1827 standard lots**.\n\n{\"answer\": 7.1827, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.23573231643213824, "answer": -0.2357323164, "confidence": 100, "grade": "exact", "relError": 1.3633360500973055e-10, "cost": 0.00525, "reply": "For an inverse short, realized PnL after both taker fees is\n\\[\n103000\\left(\\frac{1}{55827.2}-\\frac{1}{49553.7}\\right)\n-0.00055(103000)\\left(\\frac{1}{49553.7}+\\frac{1}{55827.2}\\right)\n= -0.2357323164\\ \\text{BTC}.\n\\]\n\n{\"answer\": -0.2357323164, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.0669895883004742, "answer": -0.06698959, "confidence": 100, "grade": "exact", "relError": 2.5369999144468978e-8, "cost": 0.00434, "reply": "For an inverse perpetual, the realized PnL after fees is\n\\[\n78000\\left(\\frac{1}{95195.8}-\\frac{1}{88094.2}\\right)\n-78000(0.00055)\\left(\\frac{1}{95195.8}+\\frac{1}{88094.2}\\right).\n\\]\n\nTrading PnL is approximately −0.06605196 BTC, and fees total 0.00093763 BTC.\n\n{\"answer\": -0.06698959, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.0024428835303150374, "answer": 0.002442884, "confidence": 100, "grade": "exact", "relError": 1.9226662137968772e-7, "cost": 0.00563, "reply": "**Calculating precision in finance**\n\nI need to ensure precise arithmetic here for a financial calculation involving pnl. It starts with the formula and combines the numerator and product, resulting in a numerator value of about 33,696,886.23. The product requires calculating specific products, which I verified, leading to a complex ratio around .002442. I’ll need to review all steps carefully and provide a clear formula alongside the final calculated value when I’m done.\nFor an inverse short, subtract fees in BTC at each fill price:\n\n\\[\n\\text{Net PnL}\n=42000\\left(\\frac1{116982.6}-\\frac1{117914.1}\\right)\n-42000(0.00055)\\left(\\frac1{117914.1}+\\frac1{116982.6}\\right)\n\\approx 0.002442884\\ \\text{BTC}.\n\\]\n\n{\"answer\": 0.002442884, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.001886499688812526, "answer": 0.001886498034, "confidence": 100, "grade": "exact", "relError": 8.77186747368351e-7, "cost": 0.004016, "reply": "For an inverse perpetual, net PnL in BTC is:\n\n\\[\n192000\\left(\\frac{1}{20434}-\\frac{1}{20460.6}\\right)\n-192000(0.00055)\\left(\\frac{1}{20434}+\\frac{1}{20460.6}\\right)\n=\\boxed{0.001886498034\\text{ BTC}}\n\\]\n\n{\"answer\": 0.001886498034, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.0962309957129481, "answer": -0.09623126645, "confidence": 100, "grade": "exact", "relError": 0.0000028134079866519854, "cost": 0.00433, "reply": "For an inverse short, PnL and fees are both settled in BTC:\n\n\\[\n\\text{Net PnL}\n=139000\\left(\\frac1{62000.9}-\\frac1{59513.2}\\right)\n-139000(0.00055)\\left(\\frac1{59513.2}+\\frac1{62000.9}\\right)\n=-0.09623126645\\ \\text{BTC}.\n\\]\n\n{\"answer\": -0.09623126645, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 70867.19713259465, "answer": 70867.1971, "confidence": 100, "grade": "exact", "relError": 4.5993981791230407e-10, "cost": 0.00505, "reply": "**Calculating harmonic average**\n\nI need to compute the harmonic average with some calculations. First, I’m looking at the formula involving specific products and denominators. I’ll start with manual calculations for clarity. The process includes adding various products and getting to the average base. I wonder if there’s a better way to manage the long division and ensure it’s normalized. I’m getting some large numbers here, but I think I’ll be able to round this to a sensible result. Let’s keep going!\nFor an inverse perpetual, the average entry is the **contract-weighted harmonic mean**:\n\n\\[\nP_{\\text{avg}}=\\frac{76000+55000}{76000/62771.7+55000/86235.1}\n=70867.1971\n\\]\n\nThe exchange will show approximately **70,867.2 USD**, depending on display precision.\n\n{\"answer\":70867.1971,\"confidence\":100}" }, { "taskId": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 34209.14079495059, "answer": 34209.3252, "confidence": 100, "grade": "exact", "relError": 0.000005390519759508746, "cost": 0.004844, "reply": "For an inverse contract, the average entry is the **contract-weighted harmonic mean**:\n\n\\[\nP_{\\text{avg}}=\n\\frac{76000+73000+42000}\n{\\frac{76000}{33536.1}+\\frac{73000}{24929.9}+\\frac{42000}{108003.1}}\n\\approx 34209.3252\n\\]\n\nThe exchange will show approximately **34,209.33**, subject to display rounding.\n\n{\"answer\":34209.3252,\"confidence\":100}" }, { "taskId": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 88678.94192491287, "answer": 88678.94, "confidence": 100, "grade": "exact", "relError": 2.1706538502484576e-8, "cost": 0.00508, "reply": "Bybit’s inverse contract uses a **contract-weighted harmonic mean**, not an arithmetic average:\n\n\\[\nP_{\\text{avg}}=\n\\frac{59000+34000+62000}\n{\\frac{59000}{82124.2}+\\frac{34000}{81534}+\\frac{62000}{101232.7}}\n\\approx 88678.94\n\\]\n\nYour displayed average entry price should be approximately **88,678.94 USD** (subject to display rounding).\n\n{\"answer\":88678.94,\"confidence\":100}" }, { "taskId": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 46924.5904555392, "answer": 46924.59, "confidence": 100, "grade": "exact", "relError": 9.707899338219995e-9, "cost": 0.00444, "reply": "**Calculating harmonic numeric**\n\nI’m trying to compute a harmonic numeric value using a specific formula. I start with the division and multiplication to get the needed denominator, ending up with a few intermediate products. After calculating several parts, I iterate through the full process, checking my results to ensure they're accurate, especially with final multiplications. By the end, I confirm my calculations, achieving the desired outcome while feeling good about the accuracy. Everything seems correct.\nFor an inverse contract, the average entry is weighted by BTC value, not the arithmetic average:\n\n\\[\nP_{\\text{avg}}=\\frac{42000+67000}{42000/30128.4+67000/72132.8}\n\\approx 46924.59.\n\\]\n\n{\"answer\":46924.59,\"confidence\":100}" }, { "taskId": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-6.1-sol", "expected": 37529.651947786224, "answer": 37529.65, "confidence": 100, "grade": "exact", "relError": 5.189992769330071e-8, "cost": 0.00441, "reply": "For an inverse perpetual, the average entry is weighted by the BTC value of each fill, not the arithmetic average:\n\n\\[\nP_{\\text{avg}}=\\frac{36000+61000}{36000/23930.3+61000/56468.2}\n\\approx 37529.65\n\\]\n\n{\"answer\":37529.65,\"confidence\":100}" }, { "taskId": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 4.917273424034056, "answer": 4.9173, "confidence": 100, "grade": "exact", "relError": 0.000005404614234884395, "cost": 0.011676, "reply": "Using the sample mean and sample standard deviation:\n\\[\n\\bar r=-0.25475\\%,\\qquad s=2.83461\\%.\n\\]\n\nThe 95% normal VaR, expressed as a positive loss, is\n\\[\n\\mathrm{VaR}_{95\\%}=1.644854s-\\bar r\n=4.9173\\%.\n\\]\n\n{\"answer\": 4.9173, \"confidence\": 100}" }, { "taskId": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 4.2934299989968965, "answer": 4.29343, "confidence": 100, "grade": "exact", "relError": 2.3363683077939204e-10, "cost": 0.009176, "reply": "Using the sample statistics:\n- Mean: \\(0.4610\\%\\)\n- Standard deviation: \\(2.89049\\%\\)\n\nThe 1-day 95% normal VaR (reported as a positive loss) is\n\\[\n\\mathrm{VaR}_{95\\%}=1.644854\\,s-\\bar r\n=1.644854(2.89049)-0.4610\n\\approx 4.29343\\%.\n\\]\n\n{\"answer\": 4.29343, \"confidence\": 100}" }, { "taskId": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 5.2422646771168795, "answer": 5.2423, "confidence": 100, "grade": "exact", "relError": 0.000006738096089440247, "cost": 0.007926, "reply": "**Calculating sums and squares**\n\nI need a manual for summing squares. I'm working with a group of 10 integers in the hundredths. The first sum I calculated was: 81 - 334 - 180 - 290 - 397 + 294 + 288 + 234 - 255 - 293, which totals -852. Then for the squares: 6561 + 111556 + 32400 + 84100 + 157609 + 86436 + 82944 + 54756 + 65025 + 85849, I got 767656? After correction, it’s actually 767236. For the second part, I’m checking if 420436159383451460102 - 334382343 sums to 2802?\nThe sample mean is \\(0.0750\\%\\), and the sample standard deviation is \\(3.2327\\%\\).\n\nThe 95% normal VaR (reported as a positive loss) is\n\\[\n\\mathrm{VaR}_{95\\%}=1.644854s-\\bar r\n=1.644854(3.2327)-0.0750\n\\approx 5.2423\\%.\n\\]\n\n{\"answer\": 5.2423, \"confidence\": 100}" }, { "taskId": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 7.037483251314367, "answer": 7.03748, "confidence": 100, "grade": "exact", "relError": 4.619995884998247e-7, "cost": 0.010796, "reply": "Using the sample standard deviation (denominator \\(n-1=39\\)):\n\n\\[\n\\bar r=-0.47075\\%,\\qquad s=2.822765\\%.\n\\]\n\nThe 99% normal VaR, expressed as a positive loss, is\n\\[\n\\mathrm{VaR}_{99\\%}=-\\bar r+2.326348s\n=7.03748\\%.\n\\]\n\n{\"answer\":7.03748,\"confidence\":100}" }, { "taskId": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 6.7877378350475945, "answer": 6.9351, "confidence": 100, "grade": "wrong", "relError": 0.02171005547555483, "cost": 0.007076, "reply": "Using the sample mean and sample standard deviation:\n\\[\n\\bar r=-0.03025\\%,\\qquad s=2.968108\\%.\n\\]\n\nThe 99% normal VaR (expressed as a positive loss) is\n\\[\n\\mathrm{VaR}_{99\\%}=2.326348s-\\bar r\\approx 6.9351\\%.\n\\]\n\n{\"answer\": 6.9351, \"confidence\": 100}" }, { "taskId": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.40527021190035356, "answer": 0.40527, "confidence": 100, "grade": "exact", "relError": 5.228619013177913e-7, "cost": 0.007188, "reply": "The mean daily return is **0.0476667%**, and the sample standard deviation is **2.24707%**.\n\n\\[\n\\text{Annualized Sharpe}=\\frac{0.0476667}{2.24707}\\sqrt{365}\\approx 0.40527\n\\]\n\n{\"answer\": 0.40527, \"confidence\": 100}" }, { "taskId": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.37258355219979333, "answer": 0.3726, "confidence": 99, "grade": "exact", "relError": 0.00004414526650342787, "cost": 0.008918, "reply": "The mean daily return is **0.045%**, and the sample standard deviation is **2.3075%**.\n\n\\[\n\\text{Annualized Sharpe}=\\frac{0.045}{2.3075}\\sqrt{365}\\approx 0.3726\n\\]\n\n{\"answer\": 0.3726, \"confidence\": 99}" }, { "taskId": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": -0.6855095501276544, "answer": -0.68551, "confidence": 100, "grade": "exact", "relError": 6.562597786046147e-7, "cost": 0.006188, "reply": "The annualized Sharpe ratio is\n\\[\n\\frac{\\bar r}{s}\\sqrt{365}\n=\\frac{-0.0833333}{2.322478}\\sqrt{365}\n\\approx -0.68551,\n\\]\nusing the sample standard deviation of daily returns.\n\n{\"answer\": -0.68551, \"confidence\": 100}" }, { "taskId": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.5941064255855656, "answer": 0.5941, "confidence": 100, "grade": "exact", "relError": 0.000010815546321174527, "cost": 0.007188, "reply": "The mean daily return is **0.0726667%**, and the sample standard deviation is **2.33678%**.\n\n\\[\n\\text{Annualized Sharpe}=\\frac{0.0726667}{2.33678}\\sqrt{365}\\approx 0.5941\n\\]\n\n{\"answer\": 0.5941, \"confidence\": 100}" }, { "taskId": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-6.1-sol", "expected": 6.721949210975506, "answer": 6.7128, "confidence": 100, "grade": "correct", "relError": 0.0013610949277283358, "cost": 0.006638, "reply": "The mean daily return is **0.858%**, and the sample standard deviation is **2.4419%**.\n\n\\[\n\\text{Annualized Sharpe}=\\sqrt{365}\\,\\frac{0.858}{2.4419}\\approx 6.7128\n\\]\n\n{\"answer\": 6.7128, \"confidence\": 100}" }, { "taskId": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.44697422776459944, "answer": 0.446974, "confidence": 99, "grade": "exact", "relError": 5.095698707232088e-7, "cost": 0.009822, "reply": "Writing \\(w=\\sigma\\sqrt{T}\\), the Black–Scholes price is\n\\[\nC=S\\Phi(d_1)-K\\Phi(d_1-w),\\qquad\nd_1=\\frac{\\ln(S/K)}{w}+\\frac{w}{2}.\n\\]\nMatching the quoted price gives \\(w\\approx0.157895\\), hence \\(d_1\\approx-0.133310\\). The call delta is \\(\\Phi(d_1)\\approx0.446974\\).\n\n{\"answer\":0.446974,\"confidence\":99}" }, { "taskId": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.25544432164825037, "answer": 0.2554, "confidence": 99, "grade": "exact", "relError": 0.00017350805828985117, "cost": 0.007222, "reply": "Writing total volatility as \\(w=\\sigma\\sqrt{T}\\), solve\n\\[\n5.0781=357.63\\,\\Phi(d_1)-383\\,\\Phi(d_1-w),\n\\qquad\nd_1=\\frac{\\ln(357.63/383)}{w}+\\frac{w}{2}.\n\\]\nThis gives \\(w\\approx0.09708\\), so \\(d_1\\approx-0.6575\\). The call delta is therefore \\(\\Phi(d_1)\\approx0.2554\\).\n\n{\"answer\": 0.2554, \"confidence\": 99}" }, { "taskId": "r_iv_then_delta-3", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.4262987878297406, "answer": 0.426299, "confidence": 100, "grade": "exact", "relError": 4.977031730297528e-7, "cost": 0.009132, "reply": "Writing \\(v=\\sigma\\sqrt{T}\\), the Black–Scholes price is\n\\[\n8.6404=152.85\\,N(d_1)-162\\,N(d_1-v),\n\\qquad\nd_1=\\frac{\\ln(152.85/162)}{v}+\\frac v2.\n\\]\nSolving gives \\(v\\approx0.202527\\), hence \\(d_1\\approx-0.185805\\). The call delta is \\(N(d_1)\\approx0.426299\\).\n\n{\"answer\": 0.426299, \"confidence\": 100}" }, { "taskId": "r_iv_then_delta-4", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.6063214899790841, "answer": 0.60632, "confidence": 99, "grade": "exact", "relError": 0.000002457407676848376, "cost": 0.006612, "reply": "Writing \\(v=\\sigma\\sqrt{T}\\), the Black–Scholes price equation is\n\\[\n28.2965=198.88\\,N(d_1)-192\\,N(d_1-v),\n\\qquad\nd_1=\\frac{\\ln(198.88/192)}{v}+\\frac v2.\n\\]\nSolving gives \\(v\\approx0.31821\\), so \\(d_1\\approx0.26974\\). Thus the call delta is \\(N(d_1)\\approx0.60632\\).\n\n{\"answer\": 0.60632, \"confidence\": 99}" }, { "taskId": "r_iv_then_delta-5", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 0.6017546889052645, "answer": 0.6018, "confidence": 99, "grade": "exact", "relError": 0.00007529828279015368, "cost": 0.006092, "reply": "Write \\(v=\\sigma\\sqrt{T}\\). Solving\n\\[\n4.3243=108.83\\,\\Phi(d_1)-107\\,\\Phi(d_2),\n\\quad d_1=\\frac{\\ln(108.83/107)}{v}+\\frac v2,\\quad d_2=d_1-v\n\\]\ngives \\(v\\approx0.07729\\). Thus \\(d_1\\approx0.2581\\), and the call delta is \\(\\Phi(d_1)\\approx0.6018\\).\n\n{\"answer\": 0.6018, \"confidence\": 99}" }, { "taskId": "r_deribit_straddle_breakeven-1", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 113713.12767394731, "answer": 113713.13, "confidence": 100, "grade": "exact", "relError": 2.04554455734024e-8, "cost": 0.002672, "reply": "**Calculating payoff formula**\n\nI need to use the concise formula for payoff: \\((S - K) / S\\). With premiums set at .1118, I'll plug in \\(S = 101000 / .8882\\) and calculate. That gives me \\(113713.12767\\). Now, I’ll check: \\(0.8882 \\times 113713\\) should result in about \\(100999.8866\\). So, it looks like \\(113713.127674\\) is my final output. Let’s ensure everything checks out!\nTotal premium paid: \\(0.054+0.0578=0.1118\\) BTC.\n\nAbove the strike, the BTC-settled payoff is \\((S-101000)/S\\). Break-even requires\n\\[\n\\frac{S-101000}{S}=0.1118\n\\quad\\Rightarrow\\quad\nS=\\frac{101000}{0.8882}\\approx113713.13.\n\\]\n\nYou profit in BTC above this expiry price.\n\n{\"answer\":113713.13,\"confidence\":100}" }, { "taskId": "r_deribit_straddle_breakeven-2", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 93608.77985797288, "answer": 93608.78, "confidence": 100, "grade": "exact", "relError": 1.5172413832323368e-9, "cost": 0.002902, "reply": "**Calculating breakeven point**\n\nI need to derive the equation x - K / x = 0.0706 and see what I get. I think I’ll compute 87000 divided by 0.9294 to find something around 93608.7809. I’ll also check if 0.9294 multiplied by 93600 equals about 86991.84. It seems like there's an 8.16 difference here, and dividing it by 0.9294 gives me 8.780934. Alright, this helps confirm the upper breakeven beyond that point!\nTotal premium paid is \\(0.015+0.0556=0.0706\\) BTC.\n\nAbove the strike, the call pays \\((S-87000)/S\\) BTC and the put expires worthless. Thus break-even requires\n\\[\n\\frac{S-87000}{S}=0.0706\n\\quad\\Rightarrow\\quad\nS=\\frac{87000}{0.9294}\\approx93608.78.\n\\]\n\nIgnoring fees, your straddle is profitable in BTC above **$93,608.78** at expiry.\n\n{\"answer\":93608.78,\"confidence\":100}" }, { "taskId": "r_deribit_straddle_breakeven-3", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 64193.23251006419, "answer": 64193.23, "confidence": 100, "grade": "exact", "relError": 3.910169481716397e-8, "cost": 0.002954, "reply": "**Calculating financial values**\n\nI need to figure out the calculation for 59000 divided by (1 - 0.0809), which simplifies to 59000 divided by 0.9191. When I compute that, I get approximately 64193.2325. Then, I check my math by doing 0.9191 times 64000, which gives me about 58822.4, leaving a remainder of 177.6. Dividing that by 0.9191 results in around 193.23251. So, yes, I confirm that 64193.23251 is correct. Now, I need to explain the payoff formula.\nTotal premium paid: \\(0.0196+0.0613=0.0809\\) BTC.\n\nAbove the strike, the put expires worthless and the call pays \\((S-59000)/S\\) BTC. Break-even therefore requires\n\\[\n\\frac{S-59000}{S}=0.0809\n\\quad\\Rightarrow\\quad\nS=\\frac{59000}{0.9191}\\approx64193.23.\n\\]\n\nYour BTC profit is positive above **$64,193.23**, excluding fees.\n\n{\"answer\":64193.23,\"confidence\":100}" }, { "taskId": "r_deribit_straddle_breakeven-4", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 102706.75082914304, "answer": 102706.7508, "confidence": 100, "grade": "exact", "relError": 2.837500712757901e-10, "cost": 0.002822, "reply": "**Calculating BTC payoff**\n\nI need to get the payoff for BTC, using the formula (S-K)/S, where the premium is .0653. I'm looking to break a value of 96000 divided by .9347. Then I'll calculate long division for 102706.7508. When I do .9347 times 102700, I get approximately 95993.69, with a remainder of about 6.31. Dividing that remainder by .9347 gives me roughly 6.750829. I need to explain this calculation clearly.\nThe total premium is \\(0.022+0.0433=0.0653\\) BTC.\n\nAbove the strike, the put expires worthless and the call pays \\((S-96000)/S\\) BTC. Thus break-even is:\n\\[\n\\frac{S-96000}{S}=0.0653\n\\quad\\Rightarrow\\quad\nS=\\frac{96000}{0.9347}\\approx102706.7508.\n\\]\n\nYour straddle is profitable in BTC above this price.\n\n{\"answer\":102706.7508,\"confidence\":100}" }, { "taskId": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-6.1-sol", "expected": 73994.63806970509, "answer": 73994.6381, "confidence": 100, "grade": "exact", "relError": 4.0942027028548813e-10, "cost": 0.003014, "reply": "**Calculating premium payoff**\n\nI need to derive a short premium for .0675 based on the formula (S - 69000)/S, which equals .0675. So, that leads me to compute 69000 divided by .9325. When I do that calculation, it gives me around 27,602,000 after working through the numbers. I think I’ve got it figured out, confirming that 74000 aligns with my calculations as well. Everything seems to check out!\nTotal premium paid: \\(0.0412+0.0263=0.0675\\) BTC.\n\nAbove the strike, the BTC-settled payoff is \\((S-69000)/S\\). Thus break-even requires\n\\[\n\\frac{S-69000}{S}=0.0675\n\\quad\\Rightarrow\\quad\nS=\\frac{69000}{0.9325}\\approx73994.6381.\n\\]\n\nThe straddle is profitable in BTC above this price, ignoring fees.\n\n{\"answer\":73994.6381,\"confidence\":100}" }, { "taskId": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 40.790551141452084, "answer": 40.9, "confidence": 94, "grade": "correct", "relError": 0.0026831914618748605, "cost": 0.003575, "reply": "**Calculating Black-Scholes Call Option**\n\nI need to compute the Black-Scholes call option manually with the given values: S=382.5, K=350, r=0.0597, sigma=0.607, T=0.04931507. I have to find d1 and d2 using the formulas. After some calculations, I determine N(d1) as approximately 0.7728 and N(d2) around 0.7303. The calculated call option price difference comes out to about 40.9. I’ll focus on ensuring precision in the results and adjusting my values for accuracy when needed.\nUsing Black-Scholes:\n\n- \\(T = 18/365 \\approx 0.049315\\)\n- \\(d_1 \\approx 0.748\\)\n- \\(d_2 \\approx 0.613\\)\n\nThen\n\n\\[\nC = S N(d_1) - K e^{-rT} N(d_2)\n\\]\n\n\\[\nC \\approx 382.5(0.773) - 350 e^{-0.0597\\cdot 0.049315}(0.730) \\approx 40.9\n\\]\n\n{\"answer\": 40.9, \"confidence\": 94}" }, { "taskId": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 3.2317412561362406, "answer": 3.234, "confidence": 96, "grade": "exact", "relError": 0.0006989247234662054, "cost": 0.004039, "reply": "**Calculating Black-Scholes Option Pricing**\n\nI need to compute the Black-Scholes formula using given values: S=22.89, K=23, T=171/365, sigma=0.518, r=0.0126, and q=0. I’ll calculate d1 and d2 accurately, starting with the logarithm for S/K. After computing to the correct precision, I find values for N(d1) and N(d2). The discount factor and final call option price will follow from these calculations. I really want to make sure everything is accurate in my final computation!\n**Calculating discount factors**\n\nI’m working through some financial calculations. I first subtract a value from another, then calculate terms like d1 and d2, which give me some values for N(d1) and N(d2). I'm checking my math on S*N(d1) which seems to add up around 13.09. I also want to evaluate the discount factor using e to the power of a small number. The difference I need to find is around 3.23. Each step seems important to ensure I get the right result!\n3.234\n{\"answer\":3.234,\"confidence\":96}" }, { "taskId": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 16.773153687059427, "answer": 16.82, "confidence": 99, "grade": "correct", "relError": 0.0027929341025901147, "cost": 0.003616, "reply": "**Calculating option pricing**\n\nI need to compute the Black-Scholes call option price. Starting with values like S=89.61, K=95, r=0.0577, sigma=0.788, and T=152/365. I’ll calculate d1 and d2 using the natural logarithm. After some iterations, I find d1 is about 0.1867 and d2 is around -0.3218. I’ll use the cumulative distribution functions N(d1) and N(d2) for final calculations. The estimated call price comes out to about 16.82. I'll refine this further for accuracy!\nUsing Black-Scholes:\n\n- \\(T = 152/365 \\approx 0.41644\\)\n- \\(d_1 \\approx 0.1867\\)\n- \\(d_2 \\approx -0.3218\\)\n\nThen\n\n\\[\nC = S N(d_1) - K e^{-rT} N(d_2)\n\\]\n\n\\[\nC \\approx 89.61(0.5741) - 95e^{-0.0577(0.41644)}(0.3738)\n\\approx 16.82\n\\]\n\n{\"answer\":16.82,\"confidence\":99}" }, { "taskId": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.6945968823131317, "answer": 0.7, "confidence": 96, "grade": "correct", "relError": 0.007778781944535873, "cost": 0.006383, "reply": "**Calculating option pricing**\n\nI need to compute some values for a Black-Scholes call option. Starting with the given parameters: S is 384.5, K is 427, and T is about 0.0246575. I need to calculate d1 and d2, starting with the natural logarithm and the numerator and denominator formulas. \n\nMy calculations are going everywhere from approximations to exact values, and I'm trying to ensure everything is precise, particularly with the ratios and square roots involved. I want to confirm everything matches up properly.\n**Calculating values accurately**\n\nI'm going through a systematic calculation where I start with a ratio of 3146 to 427, which gives me 0.000737. After multiplying 427 by 0.9, I get 384.3 and find a remainder. I dive into logarithmic calculations with ln values, adjusting using approximations. I'm trying to calculate d1 and d2 values more accurately, using z-scores and cumulative distribution functions. Each step feels meticulous, and I’m looking for precision in the final outcome.\n**Calculating price parameters**\n\nI’m working through some calculations: starting with values like -1.591 and using a cdf of 0.0559. Then, I’m trying to find the price using the formula: 384.5 multiplied by 0.0639, minus 427 multiplied by 0.999379 and then by 0.0559. The results are looking like 24.57 and 23.87, which gives me a final value of about 0.70. Let’s finalize that!\nUsing Black-Scholes:\n\n- \\(T = 9/365 \\approx 0.0246575\\)\n- \\(d_1 \\approx -1.525\\)\n- \\(d_2 \\approx -1.591\\)\n\nThen\n\n\\[\nC = S\\,N(d_1) - Ke^{-rT}N(d_2)\n\\approx 384.5(0.0639) - 427e^{-0.0252\\cdot 9/365}(0.0559)\n\\approx 0.70\n\\]\n\nSo the call option price is about **0.70**.\n\n{\"answer\":0.70,\"confidence\":96}" }, { "taskId": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 23.323917391138792, "answer": 23.3, "confidence": 91, "grade": "correct", "relError": 0.0010254448572124604, "cost": 0.003287, "reply": "**Calculating option pricing**\n\nI’m working through some calculations for option pricing. I need a few values: S is 314.3, K is 294, T is around 0.0548, with sigma at 0.355 and r at 0.0006. After some calculations, I find d1 at approximately 0.8415 and d2 at 0.7584. The N(d1) and N(d2) values come out around 0.799 and 0.776. After refining my values further, the estimated price ends up being about 23.3.\nUsing Black-Scholes:\n\n- \\(T = 20/365 \\approx 0.05479\\)\n- \\(d_1 \\approx 0.842\\)\n- \\(d_2 \\approx 0.759\\)\n\nSo the call price is about:\n\n\\[\nC = S N(d_1) - K e^{-rT} N(d_2) \\approx 23.3\n\\]\n\n{\"answer\": 23.3, \"confidence\": 91}" }, { "taskId": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.280639388201964, "answer": -0.2807, "confidence": 97, "grade": "exact", "relError": 0.00021597751628646894, "cost": 0.002706, "reply": "-0.2807\n{\"answer\": -0.2807, \"confidence\": 97}" }, { "taskId": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.31093085044630486, "answer": -0.311, "confidence": 96, "grade": "exact", "relError": 0.00022239528048078783, "cost": 0.003525, "reply": "**Calculating with Precision**\n\nI'm working on some calculations. First, I find T as 41 divided by 365, which equals roughly 0.1123. From there, I estimate square root T at about 0.335, and then I square it to confirm that I’m accurate. Next, I'm calculating some other values involving sigma, which seems to be coming together nicely. I also need to compute the natural logarithm of a ratio to finalize my results. Each step is confirming my process to reach an end value around -0.311, which I think is correct.\n{\"answer\": -0.3110, \"confidence\": 96}" }, { "taskId": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.5641608283248785, "answer": -0.5651, "confidence": 97, "grade": "correct", "relError": 0.001664723298691616, "cost": 0.002697, "reply": "**Calculating put delta**\n\nI need to compute the put delta using the equation: delta = N(d1) - 1 = -N(-d1). First, I’ll calculate d1 with the formula. For T, I have 65/365. S/K is approximately 0.9342, and the natural logarithm is around -0.0680. My volatility σ is 0.528, leading to σ² of 0.278784 and half being 0.139392. So, I’ll add these together carefully to refine my delta further.\n**Calculating logarithm and ratios**\n\nI'm computing more accurately now. First, I'm looking at ln(215.92/231), which gives a ratio of about 0.9342. I think the ln might be roughly -0.06801. To refine it, I try ln(1-x) with x being 0.0658, leading to an approximation of -0.06806. Then, using r+0.5σ², I arrive at about 0.176992. I’ll multiply this by T to get around 0.031512, resulting in a final calculation. It looks like my answer might be around -0.5651.\n-0.5651\n{\"answer\": -0.5651, \"confidence\": 97}" }, { "taskId": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.3853788948381438, "answer": -0.3854, "confidence": 99, "grade": "exact", "relError": 0.00005476470595277504, "cost": 0.002373, "reply": "Using Black-Scholes for a non-dividend-paying asset:\n\n- \\( T = 113/365 \\approx 0.30959 \\)\n- \\( d_1 = \\dfrac{\\ln(291.06/286) + (0.0594 + 0.5 \\cdot 0.729^2)T}{0.729\\sqrt{T}} \\approx 0.2913 \\)\n- Put delta \\(= N(d_1) - 1 \\approx 0.6146 - 1 = -0.3854 \\)\n\n{\"answer\": -0.3854, \"confidence\": 99}" }, { "taskId": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.4080759766740296, "answer": -0.4081, "confidence": 98, "grade": "exact", "relError": 0.000058869738341865825, "cost": 0.0035205, "reply": "Using \\( \\Delta_{\\text{put}} = N(d_1) - 1 \\):\n\n- \\(T = 162/365 \\approx 0.44384\\)\n- \\(d_1 \\approx 0.2325\\)\n- \\(N(d_1) \\approx 0.5919\\)\n\nSo the put delta is about:\n\n-0.4081\n\n{\"answer\": -0.4081, \"confidence\": 98}" }, { "taskId": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.007721610843754648, "answer": 0.00771, "confidence": 99, "grade": "correct", "relError": 0.0015036815490435144, "cost": 0.002684, "reply": "**Calculating gamma for Black-Scholes**\n\nI need to compute gamma for the Black-Scholes model using the formula φ(d1)/(S σ sqrt(T)). I've got T at 34/365, which is approximately 0.0932, with σ at 0.844. \n\nI’ll first find sqrt(T), then calculate d1, before using it to compute gamma. After working through the steps, I find my final gamma value is about 0.00771, but it may round to 0.00772. Let's keep it concise!\nUsing Black-Scholes,\n\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt{T}}\n\\]\n\nwith \\(T=34/365\\), \\(S=168.77\\), \\(K=150\\), \\(r=0.003\\), \\(\\sigma=0.844\\).\n\nThis gives \\(d_1 \\approx 0.588\\), \\(\\phi(d_1)\\approx 0.335\\), and\n\n\\[\n\\Gamma \\approx \\frac{0.335}{168.77 \\cdot 0.844 \\cdot \\sqrt{34/365}} \\approx 0.00771\n\\]\n\n{\"answer\": 0.00771, \"confidence\": 99}" }, { "taskId": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.011719035620003572, "answer": 0.0117, "confidence": 99, "grade": "correct", "relError": 0.0016243333172466141, "cost": 0.003026, "reply": "**Calculating gamma value**\n\nI'm working on computing the gamma using the Black-Scholes formula. For the parameters, T is 71 days converted to years, which is about 0.1945. I've got sigma as 0.161, r as 0.0448, S as 350.55, and K as 375. \n\nAs I work through the calculations for d1, I've estimated values for the natural logarithm and the numerator/denominator splits—calculating everything step-by-step. The final gamma appears to round to about 0.0117, but I want to double-check to ensure accuracy!\nFor a non-dividend-paying European option, Black-Scholes gamma is\n\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\,\\sigma\\sqrt{T}}\n\\]\n\nwith\n\n\\[\nd_1=\\frac{\\ln(S/K)+(r+\\tfrac12\\sigma^2)T}{\\sigma\\sqrt{T}}\n\\]\n\nUsing the given values:\n\n- \\(S=350.55\\)\n- \\(K=375\\)\n- \\(T=71/365 \\approx 0.19452\\)\n- \\(\\sigma=0.161\\)\n- \\(r=0.0448\\)\n\ngives\n\n\\[\nd_1 \\approx -0.7915,\\quad \\phi(d_1)\\approx 0.2912\n\\]\n\nand therefore\n\n\\[\n\\Gamma \\approx \\frac{0.2912}{350.55 \\cdot 0.161 \\cdot \\sqrt{0.19452}}\n\\approx 0.01170\n\\]\n\n{\"answer\": 0.01170, \"confidence\": 99}" }, { "taskId": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.00332788170697725, "answer": 0.00333, "confidence": 99, "grade": "exact", "relError": 0.0006365289422123021, "cost": 0.002801, "reply": "**Calculating gamma for options**\n\nI need to compute gamma using the formula gamma = pdf(d1) / (S sigma sqrt(T)). First, I've got T = 107/365, so that’s approximately 0.29315. The variables are sigma = 0.464, S = 471.11, K = 474, and r = 0.0503. I’ll calculate d1 as well by plugging in values. I find ln(471.11/474) and calculate the rest. Alright, the final calculation suggests gamma is around 0.00333. Just double-checking accuracy to be sure.\nUsing Black-Scholes,\n\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\,\\sigma \\sqrt{T}}\n\\]\n\nwith\n\n- \\(T = 107/365 \\approx 0.29315\\)\n- \\(\\sigma = 0.464\\)\n- \\(S = 471.11\\), \\(K = 474\\), \\(r = 0.0503\\)\n\ngives \\(d_1 \\approx 0.160\\), so \\(\\phi(d_1) \\approx 0.394\\).\n\nThus,\n\n\\[\n\\Gamma \\approx \\frac{0.394}{471.11 \\cdot 0.464 \\cdot \\sqrt{0.29315}} \\approx 0.00333\n\\]\n\n{\"answer\": 0.00333, \"confidence\": 99}" }, { "taskId": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.015589164355628841, "answer": 0.01559, "confidence": 97, "grade": "exact", "relError": 0.00005360417993518107, "cost": 0.002909, "reply": "0.01559\n{\"answer\":0.01559,\"confidence\":97}" }, { "taskId": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.00596435584251129, "answer": 0.00595, "confidence": 99, "grade": "correct", "relError": 0.0024069393058287294, "cost": 0.003047, "reply": "**Calculating gamma formula**\n\nI need to compute the gamma formula, which is N'(d1) divided by (S times sigma times the square root of T). With q set to 0, T is 103 divided by 365, and r is 0.0003. I want to approach this step-by-step, making sure to keep everything straight, especially with variable values. It’s important to be precise in my calculations to ensure accuracy! Let’s get going on this computation.\nUsing Black-Scholes gamma for a non-dividend-paying asset:\n\n\\[\n\\Gamma=\\frac{\\phi(d_1)}{S\\sigma\\sqrt{T}}\n\\]\n\nWith:\n- \\(S=335.65\\)\n- \\(K=300\\)\n- \\(T=103/365\\approx 0.28219\\)\n- \\(\\sigma=0.24\\)\n- \\(r=0.0003\\)\n\nCompute \\(d_1\\):\n\n\\[\nd_1=\\frac{\\ln(S/K)+(r+\\tfrac12\\sigma^2)T}{\\sigma\\sqrt{T}}\n\\approx \\frac{0.11241+0.00821}{0.12749}\n\\approx 0.946\n\\]\n\nThen:\n\n\\[\n\\phi(d_1)\\approx 0.255\n\\]\n\nSo:\n\n\\[\n\\Gamma \\approx \\frac{0.255}{335.65 \\cdot 0.24 \\cdot \\sqrt{103/365}}\n\\approx 0.00595\n\\]\n\n{\"answer\": 0.00595, \"confidence\": 99}" }, { "taskId": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.646446181061521, "answer": 0.6454, "confidence": 96, "grade": "correct", "relError": 0.0016183575557721904, "cost": 0.003321, "reply": "**Calculating vega for options**\n\nI'm working on computing vega using the Black-Scholes formula. I know I need to calculate d1, so I'll use the formula involving S, K, r, and sigma, adjusting for time. To refine my calculations, I'll ensure I compute the natural logarithm of the ratio accurately and incorporate the precise value for sigma. My current estimate for vega is around 64.5, which translates to a change of about 0.645 for a 1% increase in volatility. Let's finalize this number carefully.\nUsing Black-Scholes, the vega per 1.00 change in volatility is about 64.54, so for a 1 percentage point increase in volatility it is about 0.6454.\n\n{\"answer\": 0.6454, \"confidence\": 96}" }, { "taskId": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.26781898170848095, "answer": 0.268, "confidence": 98, "grade": "exact", "relError": 0.0006758979156903114, "cost": 0.002855, "reply": "Using Black-Scholes, the vega is\n\n\\[\n\\text{Vega} = S \\phi(d_1)\\sqrt{T}\n\\]\n\nwith \\(T = 104/365 \\approx 0.28493\\), giving \\(d_1 \\approx -0.559\\). Then \\(\\phi(d_1) \\approx 0.341\\), so\n\n\\[\n\\text{Vega} \\approx 26.8\n\\]\n\nper 1.00 change in volatility, or about\n\n\\[\n0.268\n\\]\n\nfor a 1 percentage point increase in volatility.\n\n{\"answer\": 0.2680, \"confidence\": 98}" }, { "taskId": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.45587809695630876, "answer": 0.4557, "confidence": 98, "grade": "exact", "relError": 0.00039066793842003697, "cost": 0.002666, "reply": "**Calculating vega for options pricing**\n\nI need to compute the vega using the Black-Scholes formula: vega = S * phi(d1) * sqrt(T). For a 1 percentage point increase in volatility, I should multiply vega by 0.01. I have T = 0.1288, sigma = 0.254, r = 0.0139, S = 469.67, and K = 512. After calculating, I find d1, then phi(d1), and finally the vega. This gives me a vega of approximately 45.5, translating to about 0.456 for a 1% point increase.\nUsing Black-Scholes, the vega is\n\n\\[\n\\text{Vega} = S \\phi(d_1)\\sqrt{T}\n\\]\n\nwith \\(T = 47/365 \\approx 0.12877\\).\n\nThis gives a vega of about \\(45.57\\) per 1.00 change in volatility, so for a 1 percentage point increase in volatility (\\(0.01\\)):\n\n\\[\n45.57 \\times 0.01 \\approx 0.4557\n\\]\n\nSo the option price changes by about **0.4557**.\n\n{\"answer\": 0.4557, \"confidence\": 98}" }, { "taskId": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.6097213395050691, "answer": 0.6095, "confidence": 95, "grade": "exact", "relError": 0.00036301748147564146, "cost": 0.00298, "reply": "**Calculating Black-Scholes Vega**\n\nI need to compute the Black-Scholes vega for a 1% volatility point. The formula is vega = S * phi(d1) * sqrt(T). If volatility changes from 72.6% to 73.6%, that's a change of 0.01 in sigma, so the price change would be approximately vega * 0.01.\n\nNow, I'm plugging in the details: T = 61/365, sigma = 0.726, r = 0.0203, S = 383.12, K = 429, and calculating d1. I’ll break it down and refine my calculations for accuracy. Let's keep going!\nUsing Black-Scholes, the vega is about 60.95 per 1.00 change in volatility, so for a 1 percentage point increase in volatility it is about 0.6095.\n\n{\"answer\": 0.6095, \"confidence\": 95}" }, { "taskId": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.2403754082675673, "answer": 0.2404, "confidence": 95, "grade": "exact", "relError": 0.00010230552538611792, "cost": 0.0033, "reply": "**Calculating Black-Scholes Vega**\n\nI need to compute the Black-Scholes vega. For non-dividend stocks, vega is calculated as S * sqrt(T) * phi(d1). I want to find out the price change for a 1% increase in volatility. So I'll start determining T as 65/365, which gives around 0.178. Then I'll calculate d1 using the given parameters. I also need to refine my calculations for accuracy, ultimately getting a standard vega value. Let's explore this together!\n**Calculating option price change**\n\nI'm manually computing some values. First, I find T as 65/365, which gives approximately 0.1781. After some square root calculations, I refine it to around 0.4220. Using that, I calculate sigma and other variables. Then, I derive the ratio and compute the natural logarithm. All these calculations lead to a final vega of about 0.2404, which reflects the change in option price for a one percentage point increase in volatility. I need to format this as a single JSON line.\n{\"answer\": 0.2404, \"confidence\": 95}" }, { "taskId": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 61.799447576373446, "answer": 61.76, "confidence": 95, "grade": "exact", "relError": 0.0006383160031438348, "cost": 0.006446, "reply": "{\"answer\": 61.76, \"confidence\": 95}" }, { "taskId": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 20.900010973832025, "answer": 20.89, "confidence": 96, "grade": "exact", "relError": 0.0004789937117525202, "cost": 0.005321, "reply": "**Calculating option pricing**\n\nI'm focusing on inverting the Black-Scholes formula for a put option. First, I need to compute the implied volatility (sigma), starting around 29%. Given the variables like stock price and strike price, I'm deriving values step-by-step. I realized my initial calculations led to prices too high, so I need to lower the volatility. Trying sigma equal to .20 got me closer, but I'm still fine-tuning it to find the right figure. It’s a bit tricky, but I’m working it out!\nThe implied volatility is about 20.89% per year.\n\n{\"answer\":20.89,\"confidence\":96}" }, { "taskId": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 81.50015130720656, "answer": 81.7, "confidence": 97, "grade": "correct", "relError": 0.0024521266474724517, "cost": 0.005582, "reply": "**Calculating implied volatility**\n\nI need to solve for implied volatility using the put option formula. Given specific values, I’m estimating the volatility around 20% initially. I compute values for the option components and notice that the intrinsic value appears low. As I try various volatility figures like 0.4, 0.6, and then 0.8, I adjust my calculations based on d1 and d2 outcomes. Each iteration shows I'm still too low, indicating I need to continue adjusting the volatility higher.\nUsing Black-Scholes for a European put and solving numerically for the volatility that matches the observed price gives an implied volatility of about **81.7** percent per year.\n\n{\"answer\":81.7,\"confidence\":97}" }, { "taskId": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 87.70003216471197, "answer": 87.9, "confidence": 97, "grade": "correct", "relError": 0.0022801341157146875, "cost": 0.009632, "reply": "**Calculating Implied Volatility**\n\nI'm focused on solving for implied volatility using the Black-Scholes put formula. I have the needed variables: S, K, r, and T. Time value seems significant, likely hinting at a volatility around 25-30%. Computing further, I determine the forward discount factor and call price. My initial estimates are too low for the put price, indicating that I might need to raise the volatility assumption. Adjusting to a higher sigma, I recalculate to refine my results.\n**Recalculating options carefully**\n\nI noticed a discrepancy while calculating derivatives for a financial model and need to double-check my formulas. Using a more precise approach seems essential, especially for terms like d1 and d2. After recalculating, I want to ensure everything is systematic and accurate. I’ll carefully calculate things like natural logarithms and parameters to ensure I’m not mixing up square roots or affecting my results. Getting these sigma values right feels crucial for the final outcome!\nImplied volatility is about 87.9% per year.\n\n{\"answer\": 87.9, \"confidence\": 97}" }, { "taskId": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 33.39992865638508, "answer": 33.46, "confidence": 94, "grade": "correct", "relError": 0.0017985470637656045, "cost": 0.006792, "reply": "33.46\n{\"answer\":33.46,\"confidence\":94}" }, { "taskId": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 6.438461538461515, "answer": 6.4385, "confidence": 99, "grade": "exact", "relError": 0.0000059737156548775595, "cost": 0.00173, "reply": "**Calculating pip difference**\n\nI need to compute the pip difference: 0.94593 minus 0.94203 equals 0.00390. Since I'm dealing with AUD/USD where the base currency is AUD, and a standard lot is 100k AUD, I need to find the value per unit in USD at entry. For a long position, the loss per AUD is 0.00390 USD, meaning for 100k AUD, the loss is 390 USD per lot. To risk 2511 USD, I calculate lots as 2511 divided by 390, which is about 6.43. Let's ensure it's exact by checking with stop distance times lot size.\nThe stop distance is:\n\n0.94593 − 0.94203 = 0.00390\n\nFor 1 standard lot (100,000 AUD), the loss would be:\n\n100000 × 0.00390 = 390 USD per lot\n\nSo the position size is:\n\n2511 / 390 = 6.438461538...\n\nSo the trader should use about **6.4385 standard lots**.\n\n{\"answer\": 6.4385, \"confidence\": 99}" }, { "taskId": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 12.3258666666662, "answer": 12.3259, "confidence": 100, "grade": "exact", "relError": 0.0000027043399627036095, "cost": 0.001609, "reply": "**Calculating position size**\n\nI need to compute the position size for EUR/JPY to determine pip values. The exact stop difference is 154.513 - 154.363, giving me 0.150 JPY per EUR. This means for one EUR, I’d experience a loss of 0.15 JPY. With one standard lot being 100,000 EUR, the total loss would be 15,000 JPY. To manage a risk of 184,888 JPY, I calculate the lots needed as 184,888 divided by 15,000, which equals approximately 12.33 lots.\nThe stop distance is:\n\n154.513 - 154.363 = 0.150 JPY per EUR\n\nFor 1 standard lot of 100,000 EUR, the loss at the stop is:\n\n100000 × 0.150 = 15000 JPY\n\nSo the position size is:\n\n184888 / 15000 = 12.3258666667\n\n{\"answer\": 12.3259, \"confidence\": 100}" }, { "taskId": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.4756567164179193, "answer": 0.4757, "confidence": 99, "grade": "exact", "relError": 0.00009099752108344542, "cost": 0.001654, "reply": "**Calculating position size**\n\nI need to compute the position size in lots based on the JPY risk and pip for EUR/JPY. With a long entry at 191.077 and a stop at 190.407, the difference is 0.670 JPY per EUR. A standard lot is 100k EUR, so if the stop is hit, the profit or loss would be 100k * 0.670, which equals 67,000 JPY. To find the size, I divide 31,869 by 67,000, giving me approximately 0.476 lots. Let's ensure we get this right!\nRisk per 1 standard lot if stopped out:\n\n- Price move = 191.077 − 190.407 = 0.670\n- Loss per lot = 100000 × 0.670 = 67000 JPY\n\nPosition size:\n\n- 31869 / 67000 = 0.4756567164\n\nSo the position size is **0.4757 standard lots**.\n\n{\"answer\":0.4757,\"confidence\":99}" }, { "taskId": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 5.01704545454548, "answer": 5.0161, "confidence": 100, "grade": "exact", "relError": 0.00018844847112627395, "cost": 0.001649, "reply": "**Calculating risk in JPY**\n\nI need to compute the risk per unit in JPY for a long position on USDJPY. My entry is at 142.806, and the stop is at 142.366, meaning a difference of 0.440 JPY per USD. For a position of 100,000 USD, the loss would be 44,000 JPY. To determine the number of lots, I'll divide the risk of 220,750 JPY by the loss of 44,000 JPY, which calculates to about 5.016 lots. Let's make sure that's precise!\nThe stop distance is:\n\n- \\(142.806 - 142.366 = 0.440\\) JPY per USD\n\nRisk per 1 standard lot (100,000 USD):\n\n- \\(0.440 \\times 100000 = 44000\\) JPY\n\nPosition size to risk exactly 220750 JPY:\n\n- \\(220750 / 44000 = 5.01613636\\)\n\nSo the position size is **5.0161** standard lots.\n\n{\"answer\": 5.0161, \"confidence\": 100}" }, { "taskId": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 1.6141304347826113, "answer": 1.61413, "confidence": 99, "grade": "exact", "relError": 2.693602709410174e-7, "cost": 0.001757, "reply": "**Calculating pip value for EUR/USD**\n\nOkay, I need to find the pip value in USD for the EUR/USD pair. To make this computation, I know the P/L in USD is calculated by units times the price change. The change is from 0.76163 to 0.75243, giving a delta of 0.00920. This means the loss per euro is 0.00920 USD. If I set N as 1485 euros, I get N divided by the loss per euro which equals about 161,413.043478, or roughly 1.614 when converted to lots.\nThe stop distance is:\n\n0.76163 - 0.75243 = 0.00920\n\nFor EUR/USD, each EUR loses 0.00920 USD if stopped out, so:\n\nPosition size in EUR = 1485 / 0.00920 = 161413.0435\n\nConvert to standard lots:\n\n161413.0435 / 100000 = 1.614130435\n\n{\"answer\": 1.61413, \"confidence\": 99}" }, { "taskId": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.023819567833442343, "answer": 0.02382, "confidence": 99, "grade": "exact", "relError": 0.0000181433416710188, "cost": 0.001282, "reply": "**Calculating VaR for confidence level**\n\nI'm trying to compute the z-value for a 97.5% confidence level in a one-sided Value at Risk (VaR) formula. I recall that the z-value is approximately 1.96 for this level. Using the formula, I calculate: 0.0003 (mean) + 1.96 (z) * 0.012 (standard deviation). That gives me 0.02382 as the result. I should also make sure that this reflects a positive loss fraction!\n0.02382\n{\"answer\":0.02382,\"confidence\":99}" }, { "taskId": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.08208473980663146, "answer": 0.08208, "confidence": 99, "grade": "exact", "relError": 0.000057742847728156604, "cost": 0.001979, "reply": "**Calculating VaR**\n\nI need to compute the Value at Risk (VaR) using the formula: -mean + z*stdev, with z set for the 95% confidence interval. For a parametric VaR at 95%, z is usually around 1.645. The mean and standard deviation of daily returns are -0.0005 and 0.0496, respectively. I’m making detailed calculations, comparing values using both approximations and precise constants to achieve clarity. Ultimately, I want to present the final answer with at least four significant digits.\n0.08208\n{\"answer\":0.08208,\"confidence\":99}" }, { "taskId": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.10032876732331801, "answer": 0.1003, "confidence": 99, "grade": "exact", "relError": 0.00028673055680336515, "cost": 0.001538, "reply": "**Calculating VaR**\n\nI need to compute the Value at Risk (VaR) for a positive loss fraction using the formula: -mean + z*stdev for a 99% confidence level. The z-score for this is 2.32635. When I calculate 2.32635 multiplied by the standard deviation 0.0436, I'm getting about 0.10142886. After subtracting the mean of 0.0011, I end up with approximately 0.10032886. I want to ensure that this number has at least four significant digits.\n0.1003\n{\"answer\":0.1003,\"confidence\":99}" }, { "taskId": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.07275151453115127, "answer": 0.07275, "confidence": 99, "grade": "exact", "relError": 0.00002081786421946403, "cost": 0.001903, "reply": "**Calculating Value at Risk**\n\nI'm computing the z-value for a 99% one-sided VaR. I know that to find it, I use the formula: VaR = -mean + z*stdev. For 99% confidence, z equals about 2.326. My mean is -0.0011, so -mean becomes 0.0011. Multiplying gives me approximately 0.07166. I double-check it by recalculating: 2.326 multiplied by 0.0308 gives around 0.071698. Finally, I add 0.0011 to get about 0.0727515145, which is the plain number I need!\n0.07275\n{\"answer\":0.07275,\"confidence\":99}" }, { "taskId": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.11150152403343133, "answer": 0.1115, "confidence": 99, "grade": "exact", "relError": 0.000013668274443252881, "cost": 0.002204, "reply": "**Calculating VaR precisely**\n\nI’m working on calculating Value at Risk (VaR). The formula is VaR = -mean + z*stdev. For a confidence level of 99%, z is around 2.326. Plugging in the numbers, I compute the result to be about 0.1115. I realize I made some scaling issues at one point, but after recalculating and adjusting the values carefully, I confirm that the approximation appears accurate. Alright, I’ll stick with 0.1115 as my final answer.\n0.1115\n{\"answer\": 0.1115, \"confidence\": 99}" }, { "taskId": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.04404168455292251, "answer": 0.044, "confidence": 99, "grade": "exact", "relError": 0.000946479530600684, "cost": 0.003002, "reply": "**Calculating CVaR/ES**\n\nI'm focusing on computing the Conditional Value at Risk (CVaR) for normally distributed returns. The daily return follows a normal distribution with given parameters. At a 99% confidence level, I need to establish the expected loss using the appropriate formulas for truncated returns. \n\nI'm using specific z-scores to compute relevant values and then plugging numbers in. I must ensure that I'm correctly applying the formulas for ES and validating the calculations for accuracy at this confidence level.\n**Clarifying loss distribution**\n\nI need to determine the mean of the return distribution, ensuring that the positive loss I calculated is indeed 0.0440. It’s important for me to express this clearly and concisely. I want to make sure all aspects of this positive loss are well defined and presented accurately. Let’s refine this to ensure the communication is straightforward. I’m keeping my focus on clarity to support any findings effectively!\nFor returns \\(R \\sim N(\\mu,\\sigma)\\) with \\(\\mu=0.0018\\), \\(\\sigma=0.0172\\), the 99% Expected Shortfall for a **loss** is\n\n\\[\n\\text{ES}_{99\\%} = -\\mu + \\sigma \\frac{\\phi(z_{0.01})}{0.01},\n\\]\n\nwhere \\(z_{0.01} \\approx -2.3263\\) and \\(\\phi(z_{0.01}) \\approx 0.02665\\).\n\nSo,\n\n\\[\n\\text{ES}_{99\\%} \\approx -0.0018 + 0.0172 \\cdot \\frac{0.02665}{0.01}\n\\approx 0.0440.\n\\]\n\n{\"answer\": 0.0440, \"confidence\": 99}" }, { "taskId": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.02233967069948375, "answer": 0.02234, "confidence": 99, "grade": "exact", "relError": 0.000014740616398397254, "cost": 0.00265, "reply": "**Calculating expected shortfall**\n\nI need to compute the expected shortfall (ES) for normal loss. It seems that the returns need to show a mean negative value, giving us a positive loss fraction. For 95% confidence, I'm looking at the expected loss, conditional on exceeding the Value at Risk (VaR). The formula for the ES of loss involves using quantiles and normal distribution properties, which leads to a pretty specific calculation of values. My calculations indicate the ES comes to approximately 0.02234.\nFor returns \\(R \\sim N(\\mu,\\sigma)\\), the 95% one-day Expected Shortfall as a positive loss fraction is\n\n\\[\n\\text{ES}_{95} = -\\mu + \\sigma \\frac{\\varphi(z_{0.05})}{0.05}\n\\]\n\nwith \\(z_{0.05} \\approx -1.64485\\) and \\(\\varphi(z_{0.05}) \\approx 0.10314\\).\n\nSo,\n\n\\[\n\\text{ES}_{95} = 0.0013 + 0.0102 \\times \\frac{0.10314}{0.05}\n\\approx 0.02234\n\\]\n\n{\"answer\": 0.02234, \"confidence\": 99}" }, { "taskId": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.09524684038587, "answer": 0.0953, "confidence": 99, "grade": "exact", "relError": 0.0005581246990937479, "cost": 0.002952, "reply": "**Calculating expected shortfall**\n\nI'm working on computing the expected shortfall for normal returns. Given that my parameters are a mean of 0.0023 and a standard deviation of 0.0366, at a 99% confidence level, I need to derive the parametric ES. I realize the expected shortfall of losses is calculated using the mean and standard deviation, and I’m aiming for the positive loss fraction. After some calculations, it looks like I arrive at a positive loss fraction of approximately 0.0952.\nFor a normal return \\(R \\sim N(\\mu,\\sigma)\\), the 99% Expected Shortfall for losses is\n\n\\[\n\\text{ES}_{0.99} = \\sigma \\frac{\\phi(z_{0.01})}{0.01} - \\mu\n\\]\n\nwith \\(z_{0.01} \\approx -2.3263\\) and \\(\\phi(z_{0.01}) \\approx 0.02665\\).\n\nSo:\n\n\\[\n\\text{ES}_{0.99} \\approx 0.0366 \\cdot \\frac{0.02665}{0.01} - 0.0023 \\approx 0.0953\n\\]\n\n{\"answer\": 0.0953, \"confidence\": 99}" }, { "taskId": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.12233732788143936, "answer": 0.1223, "confidence": 98, "grade": "exact", "relError": 0.00030512258266365974, "cost": 0.002814, "reply": "For normally distributed daily returns, the 97.5% 1-day Expected Shortfall as a positive loss fraction is\n\n\\[\n\\text{ES}_{0.975} = -\\mu + \\sigma \\frac{\\phi(z_{0.975})}{1-0.975}\n\\]\n\nUsing:\n- \\(\\mu = 0.0018\\)\n- \\(\\sigma = 0.0531\\)\n- \\(z_{0.975} \\approx 1.96\\)\n- \\(\\phi(1.96) \\approx 0.05844\\)\n\n\\[\n\\text{ES}_{0.975} \\approx -0.0018 + 0.0531 \\cdot \\frac{0.05844}{0.025}\n\\approx 0.1223\n\\]\n\n{\"answer\": 0.1223, \"confidence\": 98}" }, { "taskId": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.1205670856534704, "answer": 0.1206, "confidence": 99, "grade": "exact", "relError": 0.0002729961195562676, "cost": 0.002944, "reply": "**Calculating ES for Normal Returns**\n\nI need to compute the expected shortfall (ES) for normal returns, which involves using a positive loss fraction. So, for a return R with mean mu and standard deviation sigma, I’m considering the loss as L=-R. The VaR at a 97.5% confidence level corresponds to the 2.5% left tail of returns. This means I should derive the formula for ES when losses exceed this confidence level, focusing on conditional expectations carefully. Let’s calculate this precisely and ensure I’m aligned with the proper formulas!\n**Clarifying CVaR Calculation**\n\nI need to confirm how to express the 1-day parametric CVaR at a 97.5% confidence level as a positive loss fraction of the portfolio value. The calculated value seems to be 0.1206. Now, I'll make sure to communicate this clearly and concisely. It’s important that I present this information accurately, as clarity will help in understanding the implications for the portfolio. Let's keep it straightforward and focused!\nFor normally distributed daily returns, the 97.5% one-day Expected Shortfall as a positive loss fraction is\n\n\\[\n\\text{ES} = -\\mu + \\sigma \\frac{\\phi(z_{0.025})}{0.025}\n\\]\n\nwith \\(\\mu=0.0017\\), \\(\\sigma=0.0523\\), and \\(z_{0.025}\\approx -1.96\\).\n\nThis gives:\n\n\\[\n\\text{ES} \\approx -0.0017 + 0.0523 \\times \\frac{0.05844}{0.025} \\approx 0.1206\n\\]\n\n{\"answer\": 0.1206, \"confidence\": 99}" }, { "taskId": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 1.6781883884002948, "answer": 1.678, "confidence": 98, "grade": "exact", "relError": 0.00011225700380066998, "cost": 0.0059205, "reply": "**Calculating mean and std**\n\nI need to compute the mean and sample standard deviation for 12 returns. To obtain the Sharpe ratio, I’ll calculate the mean over the sample standard deviation since the risk-free rate is zero. Then, I can annualize the result by multiplying by the square root of 12. \n\nI've started adding the returns: the sum will help me find the mean. I’ll also compute the deviations and squares for the standard deviation calculation using the variance formula.\nAnnualized Sharpe ratio = monthly Sharpe × sqrt(12)\n\nMonthly mean return:\n0.2537 / 12 = 0.0211416667\n\nSample standard deviation:\n≈ 0.04364\n\nMonthly Sharpe:\n0.0211416667 / 0.04364 ≈ 0.4847\n\nAnnualized Sharpe:\n0.4847 × sqrt(12) ≈ 1.678\n\n{\"answer\": 1.678, \"confidence\": 98}" }, { "taskId": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.48328609065675004, "answer": -0.4831, "confidence": 99, "grade": "exact", "relError": 0.0003850527882918944, "cost": 0.005736, "reply": "**Calculating mean and sample standard deviation**\n\nI need to compute the mean and sample standard deviation from a list of values. First, I calculate the sum of the values, which is -0.0654. Then, to find the mean, I divide by 12, which results in -0.00545. \n\nNext, I compute the deviations from the mean for each value and square them. I’ll sum these squared deviations to help find the standard deviation later. This is a careful process, and I'm aiming for accuracy!\n**Calculating sums of squares**\n\nI'm computing the sum of squares for a series of values. Each individual squared value is adding up, and it's becoming clearer as I progress. I've calculated each square and am progressively summing them—starting from 0.00224676 up to 0.01714386 for the overall sum. Currently, I've noted the mean square as 0.0000297025, and when I multiply by n, I end up with 0.00035643. Everything seems to be aligning correctly!\n**Determining Sharpe ratio values**\n\nI'm calculating something involving the standard deviation, which is resulting in a value of about 0.03907. I’ve found the Sharpe ratio to be around -0.13949, and when I annualize the results, it comes out to -0.4831. So I think the answer to my question is -0.4831. It feels a bit negative overall, but that’s what the numbers are showing!\n{\"answer\": -0.4831, \"confidence\": 99}" }, { "taskId": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 2.241475231769422, "answer": 2.241, "confidence": 98, "grade": "exact", "relError": 0.0002120174083061153, "cost": 0.006756, "reply": "**Calculating mean and standard deviation**\n\nI need to carefully compute the mean and sample standard deviation. I’m summing a list of values: starting with 0.0236 and adding 0.0642 to get 0.0878, then continuing with various additions and subtractions. Eventually, I find the sum to be 0.2742. Dividing that by 12 gives me the mean, which is 0.02285. Now, I still need to find the sample standard deviation from the same dataset.\n**Calculating statistical metrics**\n\nI’m working through some calculations for sample variance and standard deviation based on given values. I see I need to compute the sample variance accurately, dividing the sum of squared deviations by 11. I end up with a standard deviation of about 0.035313. Then, I'm looking to find the monthly Sharpe ratio using that value. I’ll refine my calculations since accuracy is vital, especially when confirming each step along the way. Let’s keep verifying to ensure I’m correct!\n2.241\n{\"answer\":2.241,\"confidence\":98}" }, { "taskId": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 1.9830100176589147, "answer": 1.983, "confidence": 99, "grade": "exact", "relError": 0.000005051743977796405, "cost": 0.005797, "reply": "**Calculating mean and standard deviation**\n\nI need to compute the mean and sample standard deviation accurately. First, I calculate the total sum of the given numbers and then divide by 12 to find the mean, which comes out to be approximately 0.0259. Next, I determine the deviations from this mean and calculate their squares. Adding these squared deviations gives me a total which I divide by 11 to find the sample variance and take the square root for the standard deviation. The value I get is around 0.045244.\n**Calculating Variance and Standard Deviation**\n\nI’m calculating the Sharpe ratio and need to refine my standard deviation. I estimate the standard deviation around 0.045244 and calculate the variance using the formula with my values. I'll find the sum of squares first, breaking down the calculations for each squared difference. I need to verify these steps carefully to ensure accuracy. To finalize, I’ll approximate the square root of my variance and ascertain my delta for precision in the calculations.\n**Calculating Sharpe Ratio**\n\nI’m working through some math here. I calculated the Sharpe ratio as 0.0259 divided by 0.04524447, which gives me around 0.572608. Then, by taking the square root of 12, I get about 3.464101615, and when I multiply that by 0.572608, I arrive at approximately 1.983. I just need to format the final answer correctly in JSON. Alright, that’s the plan!\n{\"answer\": 1.983, \"confidence\": 99}" }, { "taskId": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 1.3915400475486268, "answer": 1.391, "confidence": 99, "grade": "exact", "relError": 0.00038809342898766044, "cost": 0.005394, "reply": "**Calculating mean and variance**\n\nI need to compute the mean and sample standard deviation carefully. The returned values are a series of numbers, and after summing them up, I get a total. The mean is calculated by dividing this total by 12. \n\nNow, for sample variance, I'll compute the deviations from the mean and their squares, which may involve approximations for calculations. It seems the sum of squares of returns will be useful in simplifying this process!\n**Calculating financial metrics**\n\nI’m working on calculating values for a financial analysis with different inputs. First, I'm verifying some squared calculations, and then moving on to compute the standard deviation and sample variance. The Sharpe ratio is next, specifically looking at the monthly and annualized values. I want to be accurate in my ratios and formulas. I’m refining the calculations to avoid rounding issues and ensuring I present it all concisely. Let's get this right!\nMonthly mean return = 0.014125.\n\nSample standard deviation:\n- variance = 0.0136006425 / 11 = 0.00123642295\n- std dev = 0.03516\n\nMonthly Sharpe = 0.014125 / 0.03516 ≈ 0.4017\n\nAnnualized Sharpe = 0.4017 × sqrt(12) ≈ 1.391\n\n{\"answer\": 1.391, \"confidence\": 99}" }, { "taskId": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.4848484848484847, "answer": 0.4848, "confidence": 99, "grade": "exact", "relError": 0.00009999999999966636, "cost": 0.004058, "reply": "**Calculating Kelly Fraction**\n\nI'm working on computing the Kelly fraction for a binary bet with a price of 0.67. This involves determining how much of my bankroll to allocate. If I set the probability of winning at q = 0.83 and losing at p = 0.17, I can model how much to invest. Based on this, I get an expected wealth formula for both scenarios of winning and losing, which should lead to deriving the optimal fraction of the bankroll to bet.\n**Verifying Betting Fraction**\n\nI'm examining the formula for buying at price c with a payoff of 1, calculating the edge as q - c. My guess for the fraction to bet, f, is (q - c) / (1 - c). If I invest fraction f of my bankroll in a contract costing c, I want to verify the wealth outcomes for winning and losing. I think there's a standard formula for net odds b that I'm trying to relate to, and I’m looking to derive the optimal fraction for my bet from there.\nThe full Kelly fraction is\n\n\\[\nf^*=\\frac{q-c}{1-c}\n\\]\n\nwith \\(q=0.83\\) and \\(c=0.67\\):\n\n\\[\nf^*=\\frac{0.83-0.67}{1-0.67}=\\frac{0.16}{0.33}\\approx 0.4848\n\\]\n\nSo you should stake about **48.48%** of your bankroll.\n\n{\"answer\": 0.4848, \"confidence\": 99}" }, { "taskId": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.28985507246376807, "answer": 0.289855, "confidence": 99, "grade": "exact", "relError": 2.4999999993502446e-7, "cost": 0.004112, "reply": "**Applying Kelly criterion**\n\nI'm looking at the math for determining how much of my bankroll to stake when betting on a contract. So, I'm calculating the expected value based on win and loss probabilities. After working it out, I found that with probabilities p = 0.51 and c = 0.31, I get a fraction of my bankroll to stake of about 28.99%. I want to be sure I've applied the Kelly criterion correctly here!\nFor a contract priced at \\(c=0.31\\) with your estimated probability \\(p=0.51\\), the full Kelly fraction is\n\n\\[\nf=\\frac{p-c}{1-c}=\\frac{0.51-0.31}{0.69}=\\frac{0.20}{0.69}\\approx 0.289855\n\\]\n\nSo you should stake about 29.0% of your bankroll.\n\n{\"answer\": 0.289855, \"confidence\": 99}" }, { "taskId": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.09859154929577463, "answer": 0.09859, "confidence": 99, "grade": "exact", "relError": 0.000015714285714109126, "cost": 0.00278, "reply": "**Calculating Kelly fraction for options**\n\nI need to compute the full Kelly fraction for a binary option purchase. The price is c=0.29, with a payoff of 1 if I win and 0 otherwise. The betting fraction f will determine how much of my bankroll I should invest. Each contract costs 0.29, yielding a net profit of 0.71 upon winning. After some calculations, I arrive at a Kelly fraction of around 0.09859, which I believe is the optimal fraction to bet given the probabilities involved.\nThe full Kelly fraction is:\n\n\\[\nf^*=\\frac{p-c}{1-c}=\\frac{0.36-0.29}{0.71}\\approx 0.09859\n\\]\n\nSo you should stake about **0.0986** of your bankroll, i.e. about **9.86%**.\n\n{\"answer\": 0.09859, \"confidence\": 99}" }, { "taskId": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.8333333333333334, "answer": 0.8333333333, "confidence": 99, "grade": "exact", "relError": 4.000000330961484e-11, "cost": 0.003135, "reply": "**Calculating Kelly Fraction for Betting**\n\nI need to compute the Kelly fraction for a binary bet where the cost is $0.16 and the payout is $1 if I win. If I win, the net profit per contract is $0.84, and if I lose, I lose $0.16. I want to derive the standard Kelly fraction based on this. If I wager a fraction of my bankroll at that cost, I need to analyze my expected outcomes mathematically. I'll also set up the log to maximize for expected value.\n**Calculating Standard Kelly Formula**\n\nI want to apply the standard Kelly formula to find the optimal fraction of my bankroll to wager. The net odds, or gain divided by stake, is calculated as 0.84 divided by 0.16, which equals 5.25. Using the formula \\( f^* = (bp - q) / b \\), I calculate the fraction as 0.8333, which implies I should stake about 83.33% of my bankroll since I have a high probability of success at 0.86. This gives me a substantial edge!\nFor a binary bet with purchase price \\(0.16\\) and payout \\(1\\), the net odds are\n\n\\[\nb=\\frac{1-0.16}{0.16}=5.25\n\\]\n\nWith estimated win probability \\(p=0.86\\) and \\(q=1-p=0.14\\), the Kelly fraction is\n\n\\[\nf^*=\\frac{bp-q}{b}\n=\\frac{5.25(0.86)-0.14}{5.25}\n=0.833333\\ldots\n\\]\n\nSo you should stake about \\(0.8333\\) of your bankroll, i.e. about 83.33%.\n\n{\"answer\": 0.8333333333, \"confidence\": 99}" }, { "taskId": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.576271186440678, "answer": 0.5763, "confidence": 99, "grade": "exact", "relError": 0.00005000000000002715, "cost": 0.003288, "reply": "**Calculating bankroll fraction**\n\nI'm working through a formula involving max p ln(1+af)+(1-p)ln(1-f). Taking the derivative gives me conditions to solve for p. After manipulating the terms, I've arrived at a workable expression for f. I've computed that f equals 57.63% of the bankroll. Oh, I should mention the full Kelly stake fraction on the bankroll to give a clearer picture of the bets.\nFull Kelly fraction:\n\n\\[\nf=\\frac{p-c}{1-c}=\\frac{0.75-0.41}{0.59}\\approx 0.5763\n\\]\n\nSo you should stake about **0.5763** of your bankroll, i.e. **57.63%**.\n\n{\"answer\": 0.5763, \"confidence\": 99}" }, { "taskId": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 24920.824742268043, "answer": 23469.1262, "confidence": 99, "grade": "wrong", "relError": 0.05825242772988281, "cost": 0.002543, "reply": "**Calculating liquidation details**\n\nI’m looking at a situation where the price is above entry but my PnL is negative. That certainly feels confusing at first, but when I think it through, it makes sense. The liquidation price is 3% above my entry point. So, it seems like I need to run some calculations to clarify the numbers and understand how this all adds up. Let’s dive into those computations.\nLiquidation price = 24173.2 / 1.03 = 23469.1262\n\n{\"answer\":23469.1262,\"confidence\":99}" }, { "taskId": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 53342.9702970297, "answer": 53343.9604, "confidence": 99, "grade": "exact", "relError": 0.000018561076835243294, "cost": 0.00401, "reply": "**Calculating margin values**\n\nI need to figure out the maintenance margin based on the entry value in BTC. The initial margin should equal the position value in BTC at entry divided by the leverage. Given that the maintenance margin is 1% of the position value in BTC at entry, it seems like I'm on the right track with my calculations. Just need to double-check everything to ensure it's accurate!\nFor an inverse long:\n\n- Initial margin in BTC: \\( \\frac{N}{P_0 \\cdot 50} \\)\n- Unrealized PnL in BTC: \\( N\\left(\\frac{1}{P_0} - \\frac{1}{P}\\right) \\)\n- Maintenance margin: \\( 0.01 \\cdot \\frac{N}{P_0} \\)\n\nSet margin + PnL = maintenance:\n\n\\[\n\\frac{N}{50P_0} + N\\left(\\frac{1}{P_0} - \\frac{1}{P_{liq}}\\right) = 0.01\\frac{N}{P_0}\n\\]\n\nCancel \\(N\\):\n\n\\[\n\\frac{1}{50P_0} + \\frac{1}{P_0} - \\frac{1}{P_{liq}} = \\frac{0.01}{P_0}\n\\]\n\n\\[\n\\frac{1}{P_{liq}} = \\frac{1 + 0.02 - 0.01}{P_0} = \\frac{1.01}{P_0}\n\\]\n\n\\[\nP_{liq} = \\frac{P_0}{1.01} = \\frac{53876.4}{1.01} \\approx 53343.9604\n\\]\n\n{\"answer\": 53343.9604, \"confidence\": 99}" }, { "taskId": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 53194.15525114154, "answer": 53194.2, "confidence": 99, "grade": "exact", "relError": 8.412363772456572e-7, "cost": 0.0040515, "reply": "Using inverse-contract BTC PnL:\n\n- Entry BTC value = \\(N / P_0\\)\n- Initial margin = \\(\\frac{1}{10}\\frac{N}{P_0}\\)\n- Maintenance margin = \\(0.005\\frac{N}{P_0}\\)\n- Unrealized PnL for a long inverse position = \\(N\\left(\\frac{1}{P_0}-\\frac{1}{P}\\right)\\)\n\nLiquidation when:\n\n\\[\n\\frac{1}{10}\\frac{N}{P_0} + N\\left(\\frac{1}{P_0}-\\frac{1}{P}\\right)=0.005\\frac{N}{P_0}\n\\]\n\nCancel \\(N\\) and solve:\n\n\\[\n\\frac{1}{P}=\\frac{1.095}{P_0}\n\\quad\\Rightarrow\\quad\nP=\\frac{P_0}{1.095}\n\\]\n\nWith \\(P_0=58247.6\\):\n\n\\[\nP \\approx \\frac{58247.6}{1.095} \\approx 53194.2\n\\]\n\n{\"answer\": 53194.2, \"confidence\": 99}" }, { "taskId": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 170355.2216748768, "answer": 170355.2, "confidence": 98, "grade": "exact", "relError": 1.2723341607440415e-7, "cost": 0.003451, "reply": "**Calculating pressure values**\n\nI’m working on calculating the pressure, starting with P = P0 / 0.6766666667. Using the numbers, I compute 115273.7 divided by this value to get about 170374, but I want to check it accurately. My calculations lead me to a series of fractions, and I need to ensure I’m precise in my methodology. Ultimately, it's a bit tricky, but I think the total is somewhere around 170,355. It seems I need to verify that division for accuracy!\n**Calculating total value**\n\nI'm working through some calculations here: I have 34,510,000 and a remainder of 72,110. Then I multiply 203 by 355, which gives me 72,065 with a remainder of 45. That leads to about 0.2217. So, the total is approximately 170,355.2217. Let me verify this with my calculator to make sure it checks out. Yep, it all adds up to about 170355.22!\n170355.2\n{\"answer\":170355.2,\"confidence\":98}" }, { "taskId": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 115003.24607329845, "answer": 114981.2, "confidence": 99, "grade": "exact", "relError": 0.00019169957415288025, "cost": 0.003372, "reply": "**Calculating short liquidation price**\n\nI need to derive the inverse perpetual short liquidation price. Starting with the notional value of N USD contracts and an entry price of Pe=109828.1, I’m considering the margin and PnL in BTC. For a short position, unrealized PnL is N*(1/P - 1/Pe). I should carefully derive the margins as well: initial margin IM= (N/Pe)/20 and maintenance margin MM = 0.5%*(N/Pe). Liquidation occurs when IM + PnL = MM, leading to the calculated price of P = Pe/0.955, which gives P=114981.255.\n**Calculating with precision**\n\nI need to compute 109828.1 divided by 0.955 to get an accurate result. This calculation seems straightforward, but I want to ensure I don't make any mistakes in the process. Let me go ahead and do that math carefully to make sure I get the right answer. After all, accuracy is key, especially when it comes to financial computations. Alright, let's see what that gives us!\nFor an inverse BTCUSD short:\n\n- Initial margin: \\( \\frac{N}{P_e \\cdot 20} \\)\n- Unrealized PnL in BTC for a short: \\( N\\left(\\frac{1}{P} - \\frac{1}{P_e}\\right) \\)\n- Maintenance margin: \\( 0.005 \\cdot \\frac{N}{P_e} \\)\n\nSet liquidation condition:\n\\[\n\\frac{N}{20P_e} + N\\left(\\frac{1}{P} - \\frac{1}{P_e}\\right) = 0.005\\frac{N}{P_e}\n\\]\n\nCancel \\(N\\) and solve:\n\\[\n\\frac{1}{P} = \\frac{1}{P_e}\\left(1 - \\frac{1}{20} + 0.005\\right) = \\frac{0.955}{P_e}\n\\]\n\\[\nP = \\frac{P_e}{0.955} = \\frac{109828.1}{0.955} \\approx 114981.2\n\\]\n\n{\"answer\":114981.2,\"confidence\":99}" }, { "taskId": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 2398.0256188582507, "answer": 2395.8, "confidence": 93, "grade": "exact", "relError": 0.0009281047044485664, "cost": 0.004837, "reply": "**Calculating average entry price**\n\nI need to compute the inverse and weighted average entry price based on contract value. For the profit and loss calculation, I consider the notional values in USD contracts. To find the combined average price, I can derive the formula and calculate it accurately. I have specific quantities and prices for computing, so I’ll sum them up for precision, verifying each calculation step. It’s crucial to be meticulous about the calculations. Let’s get to it!\n**Improving precision in calculations**\n\nI’m working on better precision for my calculations. First, I calculated the reciprocal of certain values, like 1 divided by 2309.36 and multiplied that by 7000, which gave me about 3.031. Then, I worked through additional values like dividing 34000 by 3563.54 and 36000 by 1854.34 for accuracy. However, since I want more exact results, I think I should explore using long division or more precise formulas. Let's dig into that to get it right!\n2395.8\n{\"answer\":2395.8,\"confidence\":93}" }, { "taskId": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 3712.5303509582454, "answer": 3712.53, "confidence": 99, "grade": "exact", "relError": 9.453343462758414e-8, "cost": 0.009623, "reply": "**Calculating average price**\n\nI need to compute the inverse contract weighted average instead of a simple one. For coin-margined inverse linear contracts, the calculation involves using the price in the denominator. For a USD contract size Q, purchased at different prices p_i, the ETH cost is Q divided by each p_i. The total ETH spent will be the sum of these values. I want to determine the average entry price, which satisfies the relationship between total ETH and contract size. Let’s calculate that!\n**Calculating sequentially**\n\nI need to break down these calculations carefully: starting with 34000 divided by 3528.19, which I estimate to be around 9.637. Then I work on refining that by examining the inverse of 3528.19. I find that using long division might help, zeroing in on the decimal. Through various checks, I ensure the precision of each step, double-checking multipliers and potential rounding errors to arrive at the most accurate result possible.\n**Calculating values**\n\nI’m working through some calculations. Starting with a value of approximately 0.0002834315 and multiplying by 34,000 gives me around 9.636. I’m also calculating a second value of 16,000 divided by 4,176.2 to find another component. After a few checks, I get results that lead to a sum of around 13.467905. Finally, I want to calculate the average price by dividing 50,000 by this sum to arrive at a price around 3,712.53.\n**Refining calculations**\n\nI'm refining my calculations for two terms. First, to find Term1, I divide 34,000 by 3528.19, which gives me around 9.63667. Then I explore the multiplication to confirm the result. Using some breakdowns, I estimate components of the product and resolve the total to 33,999. \n\nNext, for Term2, I divide 16,000 by 4176.2, estimating about 3.83123. I’ll run through some smaller components of the multiplication to reach 16,000 accurately. That’s my process to get to the exact numbers!\nThe combined inverse-position average entry price is the harmonic-size-weighted average:\n\n\\[\nP_{\\text{avg}}=\\frac{34000+16000}{34000/3528.19+16000/4176.2}\\approx 3712.53\n\\]\n\n{\"answer\":3712.53,\"confidence\":99}" }, { "taskId": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 2495.29829080704, "answer": 2496.9, "confidence": 96, "grade": "exact", "relError": 0.0006418908708674711, "cost": 0.006686, "reply": "**Calculating values accurately**\n\nI need to compute the sum to be precise. Starting with 118,000 as the total, I’ll divide each number: for 41,000, it's about 11.658, and I confirmed that with extra calculations. For 10,000, it's roughly 3.359, while 27,000 gives around 6.384. The value for 40,000 is approximately 25.888. Adding these results together, I approximate to 47.291. Finally, calculating the average leads me to around 2,496.5, but I want to verify these results thoroughly.\n**Calculating approximations**\n\nI’m working on some calculations using the formula N/P and trying to get precise results within 1%. I want to compute values more carefully, starting with 41000 divided by 3517.25, which leads me to the remaining value of about 2310.25. I’ll also look at other divisions like 10000 over 2976.86 and 27000 over 4229.06, checking each step carefully. After running these calculations, it seems I'm arriving at an approximate harmonic mean close to 2497.2.\nThe combined inverse-average entry price is the USD-weighted harmonic mean:\n\n\\[\nP_{\\text{avg}}=\\frac{\\sum N_i}{\\sum (N_i/P_i)}\n\\]\n\nUsing the fills:\n\n- Total notional: \\(41000+10000+27000+40000=118000\\)\n- \\(\\sum (N_i/P_i) \\approx 47.2829\\)\n\nSo:\n\n\\[\nP_{\\text{avg}} \\approx \\frac{118000}{47.2829} \\approx 2496.9\n\\]\n\n{\"answer\": 2496.9, \"confidence\": 96}" }, { "taskId": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 2610.72093572179, "answer": 2610.69, "confidence": 99, "grade": "exact", "relError": 0.000011849493895149411, "cost": 0.006592, "reply": "**Calculating inverse profit and loss**\n\nI need to derive the average entry price for an inverse perpetual contract based on contract size in USD and profit and loss in ETH. For each fill, I can find the ETH position as USD divided by price. The average entry price ensures that the total ETH quantity equals the sum of USD amounts divided by their respective prices. For calculating PnL in ETH at exit, there's a clear formula. I should compute everything carefully to get accurate results.\n**Refining calculations**\n\nI'm working on some precise arithmetic and summing various divisions. For example, calculating 10000 divided by 3282.44 gives me about 3.046. I want to verify this, so I'm checking by multiplying back and adjusting my approach to ensure accuracy. For the next number, 14000 divided by 2147.97, I’m estimating around 6.516 but need to refine that too, using multiplication to get back to the original numbers and fine-tuning my results.\nFor an inverse ETHUSD perp, the combined long position has total USD notional\n\n\\[\n10000+14000+12000+23000=59000\n\\]\n\nand total ETH exposure\n\n\\[\n\\sum \\frac{\\text{USD}_i}{\\text{price}_i}\n= \\frac{10000}{3282.44}+\\frac{14000}{2147.97}+\\frac{12000}{1744.23}+\\frac{23000}{3736.8}\n\\approx 22.599219\n\\]\n\nSo the single average entry price is\n\n\\[\nP_{\\text{avg}}=\\frac{59000}{22.599219}\\approx 2610.69\n\\]\n\n{\"answer\":2610.69,\"confidence\":99}" }, { "taskId": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 3424.594098647905, "answer": 3425, "confidence": 98, "grade": "exact", "relError": 0.00011852539028064685, "cost": 0.008972, "reply": "**Calculating averages and refining sums**\n\nI'm working on computing sums and averages here, starting with a sum of approximately 42.7110. To find the average, I'm dividing 112000 by that sum, trying different calculations to get a more precise figure. For example, I also want to refine with a weighted harmonic mean and check the division using pairwise calculations. It’s like juggling numbers to get closer to accurate averages—making sure I double-check each step!\n**Calculating precise averages**\n\nI'm working through some calculations involving averages and trying to get accurate numbers. I started with the sum and discovered my earlier estimate was off. The correct sum seems to be around 32.7, so I'm recalculating it to find a more precise average price, which is estimated to be around 3419. I’ll double-check all the computations with careful arithmetic to ensure everything aligns, especially since earlier results led to some confusion.\n**Calculating harmonic mean**\n\nI'm working on a calculation where I divide 5000 by 3813.27, getting around 1.311. I can break it down more simply, finding additional values. After my calculations, I reach a sum of about 32.7059, which leads me to compute 112000 divided by that sum to get approximately 3424.9. I'll round it and finalize with about 3425. My confidence in this answer feels pretty high considering it fits within 1%.\nFor an inverse USD-margined, ETH-settled perp, the equivalent single entry price is the weighted harmonic mean:\n\n\\[\nP_{avg}=\\frac{\\sum Q_i}{\\sum (Q_i/P_i)}\n\\]\n\nUsing the fills:\n\n- Total size = 31000 + 36000 + 40000 + 5000 = 112000\n- \\(\\sum (Q_i/P_i) \\approx 32.7059\\)\n\nSo:\n\n\\[\nP_{avg} \\approx \\frac{112000}{32.7059} \\approx 3425\n\\]\n\n{\"answer\":3425,\"confidence\":98}" }, { "taskId": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.02840777213808475, "answer": 0.02841, "confidence": 96, "grade": "exact", "relError": 0.0000784243799345561, "cost": 0.008783, "reply": "**Calculating inverse PnL for BTC**\n\nI need to calculate the inverse PnL for a BTC position using the formula: size times (1/entry - 1/exit). For a short profit if the price falls, I'm looking at an example involving the values 44000, 41686.7, and 42887.5. I should also factor in fees of 0.055% for each side based on the position value in BTC. \n\nTo get accurate results, I need to be careful with the calculations of the reciprocals, which involves playing with approximations and checking to ensure I have enough precision.\n**Calculating a ratio for BTC**\n\nI want to calculate a ratio relative to 1/43000, which gives me about 0.000023255814. Since 42887.5 is 112.5 less than 43000, its reciprocal will be slightly larger. I can find this factor by dividing 43000 by 42887.5, which results in about 1.002622. \n\nSo, when I multiply 0.0000232558 by this factor, I get approximately 0.0000233168. That looks good!\n**Calculating BTC margins**\n\nI'm working through some calculations involving Bitcoin values. So, 86.7 equals 0.997915, but I realize the denominator is larger. When I calculate the reciprocal and adjust for some small differences, I find it's around 0.0000239885. \n\nThen I need to compute potential fees and net values based on specific BTC entries. I'm trying to get accurate numbers within 1%. \n\nOverall, I want to ensure clarity while working through these computations to determine the gross PnL correctly. It seems important to get all these figures just right!\n**Calculating ratios and differences**\n\nI'm working through some calculations for gross ratios. First, I sum some values and determine a total product, then calculate gross based on a formula, but I realize my previous estimate was off. I need to refine my calculations, assessing the differences carefully. It seems the ratios and divisions are slightly off, so I'm adjusting those. I want to ensure accuracy as I finalize the computations for fees and values. I'll keep checking my calculations against my targets.\n**Calculating precise values**\n\nI'm trying to figure out a specific value here. I know that 42887.5 plus some amount equals 44000. So, I’ll calculate to find that exact amount. Dividing 44000 by 42887.5 gives me approximately 1.025970. I realize this is slightly off since the denominator is shorter by 1112.5 from 44000. It’s interesting to see how these figures fit together as I refine my calculations. I want to ensure everything aligns perfectly!\n0.02841\n{\"answer\": 0.02841, \"confidence\": 96}" }, { "taskId": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.06525146382766053, "answer": -0.06523, "confidence": 99, "grade": "exact", "relError": 0.00032894017086307895, "cost": 0.006162, "reply": "**Calculating PnL for BTCUSD**\n\nI’m trying to compute the inverse perpetual PnL for a long position with a notional value of 45,000 USD in BTCUSD. My approach involves determining profits in BTC based on the formula, factoring in fees of 0.05% on the position value for both opening and closing trades. I need to accurately calculate the size and its impacts. I’m also working through reciprocal approximations carefully to arrive at the right figures for my calculations. Let's proceed step by step and double-check the numbers!\n**Refining PnL Calculations**\n\nI'm trying to calculate the PnL more precisely using reciprocal series and aiming for accuracy within 1%. For instance, 1/46182.6 gives me around 2.1653e-5. This difference leads to a negative result when multiplied by the position size of 45,000. \n\nNow, I'll check the BTC value for fees: dividing 45,000 by the fill price gives me about 0.910 BTC. I need to confirm this calculation and compute the opening fee, which comes out to 0.00045503 BTC. Let's keep checking these numbers!\nNet PnL in BTC:\n\n\\[\n45000\\left(\\frac{1}{49446}-\\frac{1}{46182.6}\\right) - 0.0005\\left(\\frac{45000}{49446}+\\frac{45000}{46182.6}\\right)\n\\]\n\n\\[\n\\approx -0.06523\n\\]\n\n{\"answer\": -0.06523, \"confidence\": 99}" }, { "taskId": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.22922018359225318, "answer": 0.2288, "confidence": 92, "grade": "correct", "relError": 0.0018331003215695107, "cost": 0.007545, "reply": "**Calculating inverse PnL**\n\nI need to compute the inverse perpetual profit and loss (PnL). The position value is 189,000 USD long, and I’m using the formula for inverse coin margined BTCUSD PnL in BTC. I'll carefully consider the entry and exit prices of 91,422.1 and 102,932.1 respectively. I want to make sure I calculate fees accurately too, based on the position value. Ultimately, I’ll determine the net PnL by subtracting the fees from the gross PnL. Let’s do this step-by-step!\n**Calculating fees and PnL**\n\nI need to subtract fees for both the opening and closing when computing the PnL. I’ll ensure I calculate accurately by determining the opening fee based on the position value and fill price. I'm working with the values: size = 189,000, entry price (a) = 91,422.1, and exit price (b) = 102,932.1. I’ll start calculating the reciprocal of a, which is approximately 1 divided by 91,422.1. Let’s finish this step by step!\n**Calculating reciprocals and estimates**\n\nI’m calculating the product of 91422.1 and 938e-9, which simplifies down to 0.085754. It seems to slightly undercount, so I need to add 2.5e-5 to the numbers to refine my estimate. The reciprocal should be around 1.093827e-5. I'll also check my multiplication to ensure accuracy for 91422.1 with the new reciprocal values. Moving on to the next number, 102932.1, I’m estimating its reciprocal while monitoring for slight adjustments. I'll need to clarify my fees and net gain to ensure the numbers align logically.\nNet PnL is about 0.2288 BTC.\n\n{\"answer\": 0.2288, \"confidence\": 92}" }, { "taskId": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.026151408867684533, "answer": -0.02615, "confidence": 96, "grade": "exact", "relError": 0.00005387349078061194, "cost": 0.0061905, "reply": "**Calculating PnL and fees**\n\nI need to compute the inverse PnL in Bitcoin (BTC) for a long position with a notional value of $14,000. The formula for PnL is size multiplied by the difference between the reciprocals of the open and close prices. After calculating the fees at 0.055% each, I’ll subtract them from the PnL. The calculations involve finding the precise values for open and close fees based on their respective prices in USD. I'll proceed with these calculations carefully, ensuring accuracy throughout.\n**Refining calculations**\n\nI've found the net PnL to be approximately -0.02617 BTC, but I want to ensure that I'm calculating everything as accurately as possible. To do this, I will calculate 1/open using long division for better precision. Specifically, for an open price of 89,747.2, I want to see how close I can get to a precise value. This will help me in getting a clearer understanding of the overall results. Let's proceed with that calculation!\n**Calculating reciprocals and fractions**\n\nI’m working through the math involving various approximations and calculations. First, I’m considering whether 14 equals 0.999 by looking into some arithmetic and using a calculator-like approach for better accuracy. I’m trying different methods to find reciprocals and to simplify my calculations. For example, I’m calculating using fractions and breaking down larger numbers into manageable pieces to get to a final value. It's a little complex!\n**Calculating fees and values**\n\nI'm working through some calculations. First, I'm confirming that 905e9 multiplied by 0.02597 gives me around 179.3 million. Next, I’m computing fees using the numbers 7.7 and 89747.2. I find that 7.7 divided by 89747.2 leads to about 8.578e-5, which checks out. Then I calculate another fraction, summing up both values gives approximately 0.00018566. Finally, my net result is around -0.0261547. I'll represent this as -0.02615 for clarity.\n-0.02615\n{\"answer\": -0.02615, \"confidence\": 96}" }, { "taskId": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.011586282591352507, "answer": -0.01158, "confidence": 99, "grade": "exact", "relError": 0.0005422439253463194, "cost": 0.005934, "reply": "**Calculating PnL for BTC trades**\n\nI'm working on the profit and loss (PnL) calculation for Bitcoin (BTC) trades. For a long position, the formula is size times the difference in reciprocals of entry and exit prices. If the price increases, the exit price should be higher than the entry price, yielding a positive PnL.\n\nFor a short position, it's the negative of that. I'm trying to compute the specific values using the size and prices I have to arrive at an accurate result. Let's refine those calculations.\n**Calculating BTC fees**\n\nI'm working on calculating the closing fee for BTC transactions, starting with the formula 8.8 divided by 105593.7, which gives me a certain value. I’m realizing I need more precise figures here, aiming for about four significant figures. The division of 44000 by 102777.6 gives me 0.428 as well. I need to compute fees accurately and I’m breaking it down through long division. The final calculations need checking to ensure accuracy in my results.\n**Calculating final values**\n\nI’m calculating the ratio of 2816.1 divided by 10852667061.12, which results in approximately 2.5946e-7. When I multiply that by 44000, I get about 0.011416. However, it’s negative after I subtract fees, ending up at around -0.0115849, so I’ll round that to -0.01158 BTC. It seems I need just the final JSON line and maybe a brief explanation, so I should keep it succinct!\nNet PnL ≈ -0.01158 BTC\n\n{\"answer\": -0.01158, \"confidence\": 99}" }, { "taskId": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 6057.334255752001, "answer": 6067.173271752, "confidence": 99, "grade": "correct", "relError": 0.0016243145226228738, "cost": 0.00214, "reply": "{\"answer\": 6067.173271752, \"confidence\": 99}" }, { "taskId": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 2844.8890875599996, "answer": 2844.8891, "confidence": 99, "grade": "exact", "relError": 4.372754063820816e-9, "cost": 0.002504, "reply": "{\"answer\": 2844.8891, \"confidence\": 99}" }, { "taskId": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -3783.556934676003, "answer": -3783.556934676, "confidence": 100, "grade": "exact", "relError": 8.413330395615781e-16, "cost": 0.00218, "reply": "**Calculating pnl for BTC trade**\n\nI need to compute my profit and loss (PnL) for a BTC trade. I went long on 0.612 BTC, entering at 57972.37 and exiting at 51850.49, which gives me a gross loss. I have to account for fees of 0.055% on both sides. After calculating the PnL, considering the fees, it leads to a net loss of approximately -3783.557. I want to make sure I've captured everything accurately.\n-3783.556934676\n{\"answer\": -3783.556934676, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -1108.5578196799993, "answer": -1108.558, "confidence": 99, "grade": "exact", "relError": 1.6266179130010635e-7, "cost": 0.002329, "reply": "-1108.558\n{\"answer\": -1108.558, \"confidence\": 99}" }, { "taskId": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 3351.620043262996, "answer": 3351.620043, "confidence": 100, "grade": "exact", "relError": 7.846841401475839e-11, "cost": 0.002671, "reply": "**Calculating PNL for Short Position**\n\nI need to compute the short profit and loss (PNL) for a trade using the formula: profit = (entry - exit) * quantity. I'm considering a quantity of 1.949 BTC. There are fees for opening and closing that amount to 0.055% of the notional at respective fills. After calculating gross PNL, I compute the net by subtracting fees. It turns out the net profit is approximately $3351.62, which is positive, confirming that closing the short lower indeed yields a profit.\nNet PnL = 3411.88042 - 60.260377287 = 3351.620042713\n\n{\"answer\": 3351.620043, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.11472881355932601, "answer": 0.1149, "confidence": 99, "grade": "correct", "relError": 0.0014920963214307714, "cost": 0.0024915, "reply": "**Calculating risk amount**\n\nI need to compute the risk amount of 50 USD in the USD/CAD pair. The stop distance between 1.3538 and 1.3479 is 0.0059, which equals 59 pips since the pip value is 0.0001. For a standard lot of 100,000 USD, the pip in CAD is worth 10 CAD. I convert that to USD with the entry price. Actually, doing the math gives a risk per lot of 435.03 USD, leading to a position size of about 0.11494 lots.\nYour risk amount is:\n\n- 5000 × 1% = 50 USD\n\nStop distance:\n\n- 1.3538 − 1.3479 = 0.0059\n- = 59 pips\n\nPip value for 1 standard lot of USD/CAD:\n\n- 10 CAD per pip\n- Convert to USD at 1.3538:\n- 10 / 1.3538 = 7.387 USD per pip\n\nRisk per standard lot:\n\n- 59 × 7.387 = 435.0 USD\n\nPosition size:\n\n- 50 / 435.0 = 0.1149 standard lots\n\n{\"answer\": 0.1149, \"confidence\": 99}" }, { "taskId": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.5939090909090843, "answer": 0.5948, "confidence": 99, "grade": "correct", "relError": 0.0015000765345282367, "cost": 0.002262, "reply": "**Calculating risk amount**\n\nI'm figuring out the risk amount of 250 for USD/CAD. The pip value for a standard lot (100,000 USD) is 10 CAD per pip. When I convert that to USD at the entry rate of 1.3066, it comes out to about 7.654 USD per pip. With a stop distance of 55 pips, the total risk per lot is approximately 420.39 USD. Dividing my risk amount of 250 by that gives around 0.5947 lots. Let’s keep refining these calculations!\nRisk amount = 1% of 25000 = 250 USD.\n\nStop distance = 1.3066 − 1.3011 = 0.0055 = 55 pips.\n\nFor 1 standard lot of USD/CAD:\n- Pip value = 10 CAD\n- Converted to USD at 1.3066 = 10 / 1.3066 ≈ 7.6535 USD per pip\n\nRisk per standard lot:\n- 55 × 7.6535 ≈ 420.44 USD\n\nPosition size:\n- 250 / 420.44 ≈ 0.5948 standard lots\n\n{\"answer\": 0.5948, \"confidence\": 99}" }, { "taskId": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 1.1982352941177008, "answer": 1.199, "confidence": 99, "grade": "exact", "relError": 0.0006381934216537524, "cost": 0.002996, "reply": "**Computing position size**\n\nI'm figuring out how to manage the position size for USD/CHF. A standard lot is 100k USD, which has a pip value of 10 CHF. To express this in USD, I need to convert using the entry price of 0.8148 CHF/USD, so the pip value in USD would be approximately 12.27. With a stop distance of about 17 pips, risking 1% of $25,000, I’ll find the position size by dividing my risk by the pip value times the stop distance. I’ll recalculate this precisely.\n**Calculating trading values**\n\nI’m working through some calculations involving a trading scenario. First, I’m dividing 10 by 0.8148 and using that to find a total with some remainders. I’m also computing pip values in USD based on an entry price. Once I’ve got all the numbers, I’ll finalize it in a JSON format. It’s essential that I wrap it up neatly, including an explanation at the end. Let's finalize the JSON line!\nRisk amount = 250 USD.\n\nStop distance = 0.8148 - 0.8131 = 0.0017 = 17 pips.\n\nFor USD/CHF, 1 pip on 1 standard lot = 10 CHF. \nConverted to USD at 0.8148:\n\n10 / 0.8148 = 12.2730 USD per pip per standard lot\n\nRisk per standard lot at 17 pips:\n\n17 × 12.2730 = 208.64 USD\n\nPosition size:\n\n250 / 208.64 = 1.1987 standard lots\n\n{\"answer\": 1.199, \"confidence\": 99}" }, { "taskId": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 1.0178846153846508, "answer": 1.018, "confidence": 99, "grade": "exact", "relError": 0.00011335726427656952, "cost": 0.002355, "reply": "**Calculating lot size**\n\nI'm figuring out how to compute the lot size given a maximum risk of $250. For the USD/JPY pair, the pip size stands at 0.01, and with an entry point of 158.79 and a stop at 158.4, that's a difference of 0.39, which translates to 39 pips. A standard lot is 100k USD, with a pip value of 1000 JPY per pip, which I convert to USD at the entry price. I'm aiming for precision in my calculations as I finalize the lot size.\nA standard lot of USD/JPY has a pip value of:\n\n- \\(100{,}000 \\times 0.01 = 1000\\) JPY per pip\n- Converted to USD at 158.79: \\(1000 / 158.79 \\approx 6.2977\\) USD per pip\n\nYour stop is:\n\n- \\(158.79 - 158.40 = 0.39\\) JPY = 39 pips\n\nRisk per 1 standard lot:\n\n- \\(39 \\times 6.2977 \\approx 245.61\\) USD\n\nWith a $25,000 account and 1% risk:\n\n- Risk amount = \\(250\\) USD\n- Position size = \\(250 / 245.61 \\approx 1.018\\) standard lots\n\n{\"answer\": 1.018, \"confidence\": 99}" }, { "taskId": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "chat_gpt:gpt-5.4-mini", "expected": 7.182727272727309, "answer": 7.1825, "confidence": 99, "grade": "exact", "relError": 0.00003164156436398281, "cost": 0.002517, "reply": "**Calculating pip value and risk**\n\nI’m computing the risk amount for a position with a lot size of 100,000. Given that 1% of that amount is $1,000, I'm tracking pip values for USD/JPY. The stop distance suggests 22 pips is involved. \n\nI calculate the pip value in yen and then convert to USD at the entry price. My final calculations indicate that I can leverage around 7.18 standard lots in this scenario. I need to format this result as JSON for clarity.\nYou can risk $1000 total.\n\n- Stop distance: 158.02 − 157.80 = 0.22 JPY = 22 pips\n- Pip value for 1 standard lot (100,000 USD) = 1000 JPY per pip\n- Converted to USD at 158.02: 1000 / 158.02 = 6.3283 USD per pip\n- Risk per standard lot = 22 × 6.3283 = 139.22 USD\n\nSo position size = 1000 / 139.22 = 7.1825 standard lots\n\n{\"answer\": 7.1825, \"confidence\": 99}" }, { "taskId": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.23573231643213824, "answer": -0.23574, "confidence": 97, "grade": "exact", "relError": 0.000032594461285828227, "cost": 0.006074, "reply": "**Calculating inverse perpetual PnL**\n\nI need to compute the inverse perpetual short profit and loss (PnL) in BTC. The formula for long PnL is contracts multiplied by (1/entry - 1/exit) to signify profit. For shorts, profit occurs as price decreases, so for realized PnL, I’ll use contracts multiplied by (1/exit - 1/entry). Each contract is worth 1 USD notional. I need to ensure the fees are calculated correctly in BTC based on the taker fee and execution price too.\n**Calculating short trade fees and PnL**\n\nFor my open short, I need to calculate the fee based on the notional amount of $103,000 at an entry price of $49,553.7. The fee in BTC will be $103,000 multiplied by the fee rate (0.00055) divided by the entry price. I’ll do a similar calculation for the closing fee at the exit price of $55,827.2. Then, I can find the realized PnL after subtracting these fees from the trading PnL. I’ll work through the calculations and approximations to get accurate results.\n**Calculating BTC fees and corrections**\n\nI see that I need to compute some fees for BTC transactions. First, I’ll calculate the entry fee as 56.65 divided by 49553.7 and get an exit fee similarly. It seems I’m trying to achieve precision. I’ll focus on reciprocals, checking them carefully. I think I need to make a linear correction to deal with any discrepancies. I’ll adjust my calculations to ensure they're as accurate as possible to get the final numbers right.\n{\"answer\": -0.23574, \"confidence\": 97}" }, { "taskId": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.0669895883004742, "answer": -0.06698, "confidence": 99, "grade": "exact", "relError": 0.00014313120467614543, "cost": 0.004355, "reply": "**Calculating inverse PnL for BTC**\n\nI'm working on computing the inverse PnL for BTC with a contract size of 1 USD per contract. For the long position, with 78,000 contracts, I need to calculate the PnL in BTC, which involves the exit and entry prices. Considering fees of 0.055% on the notional, I’ll make sure to factor those in as well. The entry is 95,195.8 and the exit is 88,094.2. I’ll be careful with the calculations to maintain a 1% accuracy.\n**Refining calculations**\n\nI'm aiming for more accurate calculations. I start with the product of 95,195.8 and 88,094.2. First, I calculate 95,195.8 multiplied by 88,000, which gets me around 8,377,230,400. Then for 95,195.8 and 94.2, I break it down into parts and do the math step by step to ensure precision. By combining results carefully, I get a total of approximately 8,386,196,844.36. I’m trying to make sure everything is spot-on!\n**Calculating fees and net profit**\n\nI'm examining a difference in values, and first computing it gives me -7101.6. I divide to get -8.467e-7, which I multiply by 78000 to find -0.066041, but recalculating gives 0.0660426; so that's good. Next, I check fees based on prices—entry and exit. The entry fee computes as 0.0004506 and exit as 0.0004871, totaling about 0.0009377. My net result shows a loss around -0.0669793 BTC, refining this formula feels necessary.\nYour realized PnL in BTC after fees is approximately:\n\n-0.06698\n\n{\"answer\": -0.06698, \"confidence\": 99}" }, { "taskId": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.0024428835303150374, "answer": 0.0024424, "confidence": 99, "grade": "exact", "relError": 0.00019793424820998062, "cost": 0.005435, "reply": "**Calculating inverse perpetual PnL**\n\nI need to compute the inverse perpetual PnL in BTC. For BTCUSD, the contract value is 1 USD each. If I enter short at 117914.1 and exit at 116982.6, the PnL in BTC is calculated as contracts multiplied by (1/entry - 1/exit). For a short position, profit occurs when prices drop. I'll also need to consider fees based on the notional value, which is 42000 USD for each fill, and compute total fees accordingly.\n**Calculating fees and PnL**\n\nI’m working on entry and exit fees using the formulas involving prices and volumes. For my calculations, I’m starting with the price difference and total fees based on the current rates. The gross profit and loss are established by finding the ratio of the price difference. I'm stepping through the computations carefully to ensure precision, but I feel like I might have mixed up terms or made approximations at times. I’ll finalize the numbers to form a solid conclusion.\n**Refining fee calculations**\n\nI'm working on refining the computation for entry and exit fees. The entry fee is calculated as 23.1 divided by 117914.1, giving me about 0.000195928. The exit fee is computed from 23.1 divided by 116982.6, which equals around 0.000197487. Adding these gives a total of 0.000393415. My gross figure looks like 0.0028356, and net is 0.0024422. To do it clearly, the formula for short profits when prices fall is indeed Q*(1/exit - 1/entry). Great!\nYour gross PnL is:\n\n\\[\n42000\\left(\\frac{1}{116982.6}-\\frac{1}{117914.1}\\right)\\approx 0.0028358\\ \\text{BTC}\n\\]\n\nFees paid on both fills:\n\n\\[\n42000 \\times 0.00055 \\times \\frac{1}{117914.1}\n+\n42000 \\times 0.00055 \\times \\frac{1}{116982.6}\n\\approx 0.0003934\\ \\text{BTC}\n\\]\n\nNet realized PnL:\n\n\\[\n0.0028358 - 0.0003934 \\approx 0.0024424\\ \\text{BTC}\n\\]\n\n{\"answer\": 0.0024424, \"confidence\": 99}" }, { "taskId": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.001886499688812526, "answer": 0.001885, "confidence": 99, "grade": "exact", "relError": 0.0007949584203058974, "cost": 0.007823, "reply": "**Calculating BTC PnL**\n\nI'm calculating the inverse perpetual profit and loss (PnL) in BTC for a long position of 192,000 contracts at a size of 1 USD each. The formula involves contracts multiplied by the difference of the inverse prices at entry and exit, and fees charged based on the notional value in BTC. The fees here are 0.055% on the trade value, so I need to consider them carefully. My initial calculations show a gross PnL of around 0.0122208 BTC, but I want to ensure every detail is correct!\n**Calculating fees**\n\nI’m calculating the entry and exit fees for my BTC position. For the entry fee, I take 192,000 divided by the entry price of 20,434 and multiply by the fee rate of 0.00055, which gives me about 0.00517 BTC. For the exit fee, dividing 192,000 by 20,460.6 gives me around 0.00516 BTC. So, the total fees amount to roughly 0.01033 BTC. This leaves me with a net of about 0.00189 BTC, but I want to make sure those figures are exact!\n**Calculating fees and values**\n\nI'm trying to compute various values related to fees and gross amounts. I'm dividing values and checking approximations, such as dividing 192,000 by 20,434 and 20,460.6. I want that to be precise, so I'm also exploring different methods to calculate effectively, including using series approximations. My goal is to find the exact formula for the fees charged on orders in BTC and making sure all calculations are accurate and within the proper range.\n**Calculating net and fees**\n\nI'm working on calculating the net value by subtracting fees from the gross amount. The gross is 0.012220, and I'm determining the total fees, which I think adds up to around 0.010327. To refine my calculations, I want to use long division to get a more accurate fee from 105.6 divided by 20,434. I'm double-checking my multiplications to ensure everything adds up correctly. It's important to have precise values for clarity!\n**Calculating fees**\n\nI'm working through a multiplication: 20434 times 0.005167 equals 20434 times 5167 over 1e6. The breakdown includes various steps and calculations: first, 20434 times 5000 gives 102,170,000; then times 167, I try to find the total and fine-tune the numbers. I also check 0.005168 which is close. Ultimately, my calculations suggest a fee of around 0.00516114. I need to ensure I present this in a clear, concise formula format.\nRealized PnL in BTC:\n\n- Gross PnL: \n \\(192000 \\times \\left(\\frac{1}{20434} - \\frac{1}{20460.6}\\right)\\)\n- Entry fee: \n \\(192000 \\times \\frac{0.00055}{20434}\\)\n- Exit fee: \n \\(192000 \\times \\frac{0.00055}{20460.6}\\)\n\nNet:\n\\[\n192000\\left(\\frac{1}{20434} - \\frac{1}{20460.6}\\right)\n-192000\\cdot 0.00055\\left(\\frac{1}{20434}+\\frac{1}{20460.6}\\right)\n\\approx 0.001885\n\\]\n\n{\"answer\": 0.001885, \"confidence\": 99}" }, { "taskId": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.0962309957129481, "answer": -0.096206, "confidence": 99, "grade": "exact", "relError": 0.0002597470052441739, "cost": 0.00773, "reply": "**Calculating values**\n\nI'm diving into some calculations here. First, I'm working on dividing the gross amount to find ratios, but I see that my result seems high, so I readjust for accuracy. I'm making sure to check my math carefully, especially when determining fees from larger figures. The goal is to get a final, precise number, so I need to be diligent and ensure all my calculations align properly, which can be challenging! Let's aim for clarity and accuracy.\nYour realized PnL in BTC is approximately:\n\n- Gross short PnL: about **-0.093688 BTC**\n- Entry taker fee: about **0.0012846 BTC**\n- Exit taker fee: about **0.0012330 BTC**\n\nNet realized PnL after fees: **-0.096206 BTC**\n\n{\"answer\": -0.096206, \"confidence\": 99}" }, { "taskId": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 70867.19713259465, "answer": 70869.4, "confidence": 97, "grade": "exact", "relError": 0.000031084443783234955, "cost": 0.011246, "reply": "**Calculating with precision**\n\nI'm trying to figure out some calculations. First, I compute 1.8482691 times 70860, but it seems off. When I use 70000, I get 129378.837, then add 860 to get 1589, leading me to a total of 130968, possibly 70863. I realize I need to check my precision. Using a formula with values like a = 76000/62771.7, I calculate it and derive ratios to find accurate averages for total USD notional against BTC amounts.\n**Calculating precise divisions**\n\nI'm working through some calculations here. First, I’m trying to figure out if 76000 divided by 62771.7 gives me the right number. I double-check my math using multiplication to confirm the accuracy but find I'm low by 2.36. I adjust my quotient accordingly.\n\nNext, I’m looking at another calculation with 55000 divided by 86235.1 and breaking that down into smaller pieces to ensure accuracy. I’m also averaging out some values to refine my results further. Keeping everything precise is my goal!\n**Calculating averages**\n\nI'm working through calculations involving values around 62800. I want to be accurate, so let's use a weighted harmonic mean with weights of 76000 and 55000. After computing, I find that a is approximately 1.2107. The next step involves dividing 131000 by the sum from the harmonic mean calculation. I've encountered some small arithmetic issues when checking my intermediate results, but I'm refining the calculations to settle on an average around 70869.7. Let's verify these findings!\n**Calculating totals**\n\nI'm calculating some values based on 62771.7. First, I multiply by 0.2, which gives me 12554.34. Then, I check smaller multiples: 0.01 results in 627.717, and multiplying by 0.0006 gives me 37.663. Finally, I sum these values together for a total of 13219.72. It's a simple calculation, but I want to make sure I've captured everything accurately!\n**Calculating numbers precisely**\n\nI’m working on some complex math here. First, I estimated a value and added a little to it, getting 1.2107367. Next, I calculated a second number involving multiplication and some smaller divisions to reach a total. Then, I needed to find an average using two values divided by their respective totals.\n\nI’m realizing I'm digging into higher accuracy calculations now, thinking maybe I can use fractions or reciprocal methods for precision. Let me keep working through this!\n**Calculating average price**\n\nI'm working through a calculation here. It looks like I need to settle on a final number around 70,869.3. I'm using consistent division by letting x represent the average price as N divided by the sum of n/p values. With p1 at 62,771.7 and p2 at 86,235.1, I'm aiming for x to be between those two values but closer to the lower price, so around 70,800. I estimate a weighted harmonic average around 70,300, landing finally on 70,870.\nFor an inverse BTCUSD contract, the exchange’s average entry is the contract-weighted harmonic average:\n\n\\[\nP_{\\text{avg}}=\\frac{76000+55000}{76000/62771.7+55000/86235.1}\n\\]\n\nThis comes out to about **70869.4**.\n\n{\"answer\": 70869.4, \"confidence\": 97}" }, { "taskId": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 34209.14079495059, "answer": 34237.3, "confidence": 91, "grade": "exact", "relError": 0.0008231485619063256, "cost": 0.013676, "reply": "**Calculating weighted average**\n\nI need to find the weighted average for an inverse perpetual contract valued at 1 USD each. Typically, the average entry price is calculated as total contract size divided by the total BTC position size. Since each contract represents 1 USD, I can find the BTC amount per fill by dividing the number of contracts by the price. It seems I should compute the average entry as total contracts over total BTC, which points to a weighted harmonic mean.\n**Calculating BTC Contracts**\n\nI need to calculate some contracts precisely. I've got 76,000 at 33,536.1, 73,000 at 24,929.9, and 42,000 at 108,003.1. Total contracts add up to 191,000. \n\nI’ll compute each ratio carefully. So for the first one, 76,000 divided by 33,536.1 gives me around 2.266. The second ratio is 73,000 divided by 24,929.9, which is about 2.926. Finally, dividing 42,000 by 108,003.1 yields roughly 0.3889. \n\nSumming these ratios gives me around 5.58161, so now I can check how it relates to total contracts!\n**Refining Average Price**\n\nI’m estimating around 34,229, but I want to refine that further. It seems I should aim for an average price closer to 34,225. To get a more precise figure, I’ll use a formula with a proper denominator. \n\nOkay, let’s calculate it to ensure it’s accurate! I really need to nail down this number, so precision is key here. Time to work through the math a bit more clearly!\n**Calculating precise values**\n\nI'm diving into some calculations! First, I'm looking at the division: 76000 divided by 33536.1 to find 'a', which is about 2.266306 after checking my multiplication. For 'b', I’m calculating 73000 over 24929.9, which ends up around 2.926185. Lastly, for 'c', it's 42000 over 108003.1, giving a value close to 0.388889. Altogether, I’m summing these values, finally computing an approximate total to get to around 34225! Let’s verify this for accuracy!\n**Calculating values and refining results**\n\nI’m working through some calculations, beginning with values like 33536.1 multiplied by 0.266, which simplifies to around 8920.6026 when I adjust for decimals. Then I move to another calculation with 24929.9 times 2.9 and make necessary adjustments for remainders. I also realize I need to correct a previous mistake in my calculations regarding 0.388 versus 0.3888, ending up with totals that sum to approximately 5.5785503. Finally, I check price connections, aiming for accuracy.\n**Calculating BTC Values**\n\nI'm working through some calculations here. I divided 0.797 by 108003.1 and got 0.000137. Then, I checked that 0.388137 multiplied by 108003.1 equals about 42000.0. That looks correct! I realized that if I use 0.3889, it gives a significant difference. This all leads me to figure out BTC's total value and price calculations, refining things along the way. I'm determined to get these numbers right!\n34237.3\n{\"answer\":34237.3,\"confidence\":91}" }, { "taskId": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 88678.94192491287, "answer": 88679, "confidence": 97, "grade": "exact", "relError": 6.548915206903676e-7, "cost": 0.006528, "reply": "**Calculating harmonic for BTCUSD**\n\nI’m trying to compute the weighted harmonic average for the BTCUSD inverse perpetual contract priced at 1 USD. I need the position size and entry value calculations. So, I add up the contract values and use them to find averages. The first part of the math involves a summation for denominators, and I’m trying to be super precise with my division. It seems like I’m a bit off, so I need to improve my accuracy as I calculate!\n**Calculating approximation**\n\nI'm working on a calculation and starting with an approximate number of around 88,688. I keep trying different multiplications to find the right product. I think using the reciprocal would simplify things. After some adjustments, I estimate around 88,679 might be close to the result I want. I should check if this gives me the product with a minimal difference, so I'm looking at the number 88,678.95. Let me check this with weighted harmonic calculations for a better estimate!\nThe exchange should show an average entry price of about 88679.0\n\n{\"answer\":88679.0,\"confidence\":97}" }, { "taskId": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 46924.5904555392, "answer": 46924.55, "confidence": 97, "grade": "exact", "relError": 8.621394198966562e-7, "cost": 0.007461, "reply": "**Calculating precision**\n\nI’m working on computations. For 1/p1, I have 30128.4, and I'm approximating its reciprocal as around 0.00003319083. I need to verify with simple multiplications: 30128.4 times that reciprocal should equal 1. \n\nNext, for 42000 divided by 30128.4, I calculate about 1.394008 and confirm by multiplying back to check if it’s close to 42000. Just realized my estimate might've been too high, finding the correction. It's all about getting those numbers right!\n**Refining calculations**\n\nI'm working through some calculations with values like 67000 and 72132.8. First, I'm checking that 72132.8 multiplied by 0.928 gives me a number around 66966. I need 67000, so I'm finding the difference is 60.8 and then dividing. So, after some tweaks, I refine my value to about 0.928842. Then I start calculating the average price more accurately, considering weighted means and using some division. I want to ensure these calculations are precise.\n**Solving approximate calculations**\n\nI'm aiming to solve this with an approximation. I start with 30128.4 multiplied by 1.3939, which I can break down into simpler terms. So, I think: 30128.4 times (1 plus 0.3939) gives me 30128.4 plus something like 11859. I’ve got to make sure my calculations hold up as I keep refining this to get the right numbers. I'll double-check to ensure everything adds up accurately as I continue!\n**Calculating values and averages**\n\nI’m working through some computations. Starting with 30128.4 times 0.4 gives 12051.36. Then, subtracting another calculation results in a total of 11867.58. From there, I’m summing to 41995.98 and making adjustments to get btc1 around 1.394033. For btc2, I go through multiplications and additions, resulting in 66996.74624. \n\nMy final computations lead me to an average that’s a little off, so I'm fine-tuning to get accurate totals and need the answer as a plain number.\n46924.55\n{\"answer\":46924.55,\"confidence\":97}" }, { "taskId": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "chat_gpt:gpt-5.4-mini", "expected": 37529.651947786224, "answer": 37529.65, "confidence": 99, "grade": "exact", "relError": 5.189992769330071e-8, "cost": 0.004869, "reply": "**Calculating average entry price**\n\nI need to compute the average entry price for inverse perpetual contracts, specifically for the inverse BTCUSD contract. It seems that the average entry price is determined by the total contract value divided by the total Bitcoin (BTC). I'll sum the total contracts and BTC to find the weighted average entry price. For instance, with values N1=36000 at P1=23930.3 and N2=61000 at P2=56468.2, I can calculate it as a ratio. Let’s refine these calculations for more accuracy!\n**Refining calculations**\n\nI’m refining my calculations starting with 23930.3 multiplied by 0.004, which equals 95.7212. Adding 1.5 gives me a total of 35895.45. However, let me double-check: when I multiply 23930.3 by 1.504, it seems to break down like this: 23930.3 plus 11965.15 plus 95.7212, which gives me 35991.1712. The remainder is 8.8288, and dividing that by 23930.3 results in about 0.000369. Then onto the next calculation for 61000 divided by 56468.2.\nFor an inverse BTCUSD perp, the displayed average entry is:\n\n\\[\n\\text{Avg Price}=\\frac{\\text{total contracts}}{\\sum ( \\text{contracts} / \\text{entry price})}\n\\]\n\nSo:\n\n- \\(36000/23930.3 \\approx 1.504369\\)\n- \\(61000/56468.2 \\approx 1.080254\\)\n\nTotal BTC amount \\(\\approx 2.584623\\), and total contracts \\(=97000\\).\n\n\\[\n\\text{Avg Price} \\approx \\frac{97000}{2.584623} \\approx 37529.65\n\\]\n\n{\"answer\": 37529.65, \"confidence\": 99}" }, { "taskId": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 4.917273424034056, "answer": 4.916, "confidence": 99, "grade": "exact", "relError": 0.00025896953946697235, "cost": 0.008985, "reply": "**Calculating VaR**\n\nI need to compute the mean and sample standard deviation of returns in percentage, then determine the parametric VaR at the 95% confidence level. The formula for one-day likely loss quantile becomes - (mean + z_{0.05}*std). Since I'm asked for returns in percentage, the VaR will be represented as a positive number. I should calculate the sample mean and standard deviation from 40 values carefully, ensuring accuracy in my computations. Let's get to it!\n**Calculating statistics from returns**\n\nI have a list of returns, and I first need to sum them up, yielding a total of -10.19. From this, I can calculate the mean, which comes out to approximately -0.25475%. Next, I need to work on the sample standard deviation. To do that, I’ll need to find the sum of squares of the returns. This involves calculating how each return deviates from the mean. Let’s proceed step-by-step to ensure I get this right!\n**Calculating sample variance**\n\nI’m working on calculating the sample variance. I need to start with the formula that involves deviations and the sum of squares. First, I have m as -0.25475, so I’ll find deviations by adding 0.25475 to x. Then I need the sum of squares using x^2. I’ll compute that to get the values I need for my variance calculation. Let's get started!\nUsing the sample mean and sample standard deviation:\n\n- Sample mean ≈ -0.2548%\n- Sample standard deviation ≈ 2.8346%\n\n1-day 95% parametric VaR (normal), as a positive loss percentage:\n\n\\[\n\\text{VaR} = -(\\mu + z_{0.05}\\sigma) = -(\\mu - 1.64485\\,\\sigma) \\approx 4.916\n\\]\n\n{\"answer\": 4.916, \"confidence\": 99}" }, { "taskId": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 4.2934299989968965, "answer": 4.294, "confidence": 99, "grade": "exact", "relError": 0.00013276121963937663, "cost": 0.007896, "reply": "**Calculating VaR**\n\nI need to compute the mean and sample standard deviation of 40 returns expressed in percentage. The one-day 95% parametric VaR can be calculated using the formula that incorporates mean and standard deviation. \n\nI’ll use the cumulative returns: summing them gives me a total of 18.44, and dividing by 40 confirms the mean is 0.461. Now, I’ll proceed to determine the value at risk by applying the proper statistical methods. It’s important to be precise here!\n**Calculating variance steps**\n\nI'm working on calculating variance by computing squared deviations from the mean. First, I need to sum the squares of the deviations. For example, I'll take each number, square it, and keep a running total. So far, I've calculated several squared values: for instance, 4.59 squared is about 21.0681, and -3.68 squared is around 13.5424. I'll continue this until I've squared and summed all necessary deviations!\nSample mean = 0.4610% \nSample standard deviation = 2.8906%\n\n1-day 95% parametric normal VaR:\n\\( \\text{VaR} = 1.64485 \\times 2.8906 - 0.4610 \\approx 4.294 \\%\\)\n\n{\"answer\": 4.294, \"confidence\": 99}" }, { "taskId": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 5.2422646771168795, "answer": 5.24, "confidence": 99, "grade": "exact", "relError": 0.0004320035817277343, "cost": 0.007487, "reply": "**Calculating VaR for returns**\n\nI need to compute the mean and sample standard deviation from 40 returns. The 1-day Value at Risk (VaR) at 95% with a parametric normal distribution is usually calculated as the negative of the mean plus z_{0.05} times the standard deviation. Since the returns are given in percentages, I'm expecting a positive number for the loss percentage. I should be careful with the calculations, starting with the sum and sum of squares, doing it incrementally.\n**Summing returns incrementally**\n\nI’m working through a list of returns and calculating the incremental sums. The first return is 0.81, making the sum -2.53 after the second return of -3.34. It’s a chain reaction as I keep adding each return. By the time I reach the 10th value, the sum is -8.52. I’ll keep going through each value, noting how the sum changes with each addition, which helps in tracking the overall trend.\n**Finalizing the sum**\n\nI’m nearing the end of my calculations and have just added the last few returns. The 34th return gives me 1.89, then the 39th return adds 3.52 for a total of 5.41. After that, the 40th return is -2.41, which brings the final sum down to 3.00. So, I have my final sum tracked down! It's important to ensure that these calculations are accurate as they help evaluate overall performance.\n**Calculating standard deviation**\n\nI see that the mean is given as 0.075%. Now, I need to compute the sample standard deviation, which involves calculating the sum of squares. It's crucial that I do this carefully and use decimals for accurate results. I'll ensure I'm applying the right formulas and methods to get a precise outcome. This is where I really want to double-check my calculations to make sure everything adds up correctly!\n**Summing calculated values**\n\nI’m working through a series of calculations that provide a series of values leading to cumulative sums. Each step has a starting value and a new resultant total. I need to keep track of each of these values carefully. Precision is key here as I want to ensure that my final numbers are accurate. It looks like I’m moving through a lot of calculations, so I’ll double-check everything to maintain consistency.\n**Calculating VaR and Variance**\n\nI'm working through the sample variance calculation using the formula: variance = (sum of squares - n * mean^2) / (n - 1). With a given mean, I find the variance and then standard deviation. I'm confirming that the VaR calculation is correct by applying the normal distribution formula. After going through a few steps, it looks like the VaR is about 5.24%. Now, I’ll craft a more concise summary of these findings.\nUsing the sample mean and sample standard deviation:\n\n- Mean return = 0.075%\n- Sample standard deviation = 3.233%\n\n1-day 95% parametric VaR (normal) = \\(-\\mu + 1.64485\\sigma\\) ≈ 5.24%\n\n{\"answer\": 5.24, \"confidence\": 99}" }, { "taskId": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 7.037483251314367, "answer": 7.039, "confidence": 97, "grade": "exact", "relError": 0.00021552430485001015, "cost": 0.007023, "reply": "**Calculating VaR for returns**\n\nI need to compute the mean and sample standard deviation from 40 percent returns. To find the 99% one-day Value at Risk (VaR) using the parametric normal approach, I apply the formula: VaR = -(mu + z_alpha * sigma), where z = -2.326, which translates to a positive loss of mu + 2.326 * sigma. Since the returns are in percent, the VaR will also be in percent as a positive number. I'll carefully sum the returns and move forward.\n**Summing returns**\n\nI’m going through a series of returns, calculating cumulative totals. I track the values from negative to positive adjustments step by step. Starting with -1.04 and adjusting further with various returns, I notice a mix of losses and gains. For instance, after some ups and downs, I'm at -19.43 at the end of my list. I’m being careful to sum everything accurately. This will be important for understanding the overall percentage returns later!\n**Calculating sample statistics**\n\nI’m computing the sample standard deviation by first finding the mean, which is -0.47075%. Then, I'll list the returns and compute the sum of squares based on their deviations from the mean. I've begun calculating those squared deviations, and it seems I have a detailed list coming together. I want to ensure that everything is accurate as I go through this lengthy process to get the sample standard deviation right!\n**Calculating Value at Risk (VaR)**\n\nI'm computing the sample variance and standard deviation based on some data. The mean squared calculation gives me insights. Variance led to a standard deviation of 2.8235. For the 99% Value at Risk (VaR), I'm finding it with the formula incorporating the z-score. After simplifying my calculations, I determine that the VaR is approximately 7.039. It's really a matter of ensuring accuracy and expressing the final number, which I lean towards presenting plain.\n7.039\n{\"answer\":7.039,\"confidence\":97}" }, { "taskId": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 6.7877378350475945, "answer": 6.78, "confidence": 99, "grade": "correct", "relError": 0.0011399725852169764, "cost": 0.00754, "reply": "**Calculating VaR parameters**\n\nI need to compute the sample mean and standard deviation from 40 return percentages. For a 99% parametric normal Value at Risk (VaR), the usual loss quantile formula is \\(-(\\mu + z \\cdot \\sigma)\\), which I need to apply in terms of returns. Since the VaR is expressed in percentage, it likely represents a positive number for the loss threshold. I'll be careful in summing up these values for accurate calculations. Let's do this!\n**Computing statistics**\n\nI'm working through a series of return values to calculate the sample mean, which I've found to be approximately 0.04475%, and the sum is 1.79%. Now I need to compute the sample standard deviation, which may be easier by first finding the sum of squares. I'll square each value carefully and try to keep my calculations accurate. For example, squaring 2.50 gives me 6.25. I'll continue from here!\n**Calculating variance and VaR**\n\nI'm calculating the sample variance with the given values. After some squaring and summing, I compute the mean and its square. The numerator for variance comes out from the total minus the product of n and mean squared, which leads to a variance I can divide by n-1. Then I find the standard deviation by taking the square root. For the 99% Value at Risk (VaR), I'm determining the loss threshold. The VaR comes to approximately 6.78%. I need to be concise now!\nUsing the sample mean and sample standard deviation:\n\n- Mean = 0.04475\n- Sample SD ≈ 2.9370\n- 1% normal quantile factor = 2.3263\n\n99% 1-day parametric VaR ≈ 0.04475 + 2.3263 × 2.9370 = 6.78\n\n{\"answer\": 6.78, \"confidence\": 99}" }, { "taskId": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.40527021190035356, "answer": 0.4055, "confidence": 99, "grade": "exact", "relError": 0.0005669997273398564, "cost": 0.00646, "reply": "**Calculating mean and standard deviation**\n\nI need to compute the mean and sample standard deviation of daily returns as a percentage. The annualized Sharpe ratio is calculated using the daily returns as well, which makes sense to do carefully. I have a list of 30 values to sum sequentially, leading me to a mean of 1.43% daily after calculations. Now, I need to compute the sample standard deviation, using the deviations from the mean while keeping my calculations precise. Let's proceed with those computations!\n**Calculating squares and variance**\n\nI’m listing the squares of a series of numbers, making sure to track the calculations carefully. Each result seems to cascade into the next, and I can see how they build up. After getting all the squares, I’m also looking to find the sample variance. The formula requires me to sum up the squared differences from the mean, which I believe is about 0.0476667 squared. This process is critical for achieving accurate results.\n**Calculating Sharpe Ratio**\n\nI'm working through some calculations involving Sharpe ratios and daily returns. I multiply times, subtract from the mean, and divide, aiming for accuracy. I find the standard deviation and determine the annualized Sharpe ratio. It seems I need to ensure my calculations are precise, using the exact mean. I'm fluctuating between values like 0.4054 and 0.40545, aiming for a final answer around 0.4055. My confidence in this number feels really high, perhaps 99%.\nThe annualized Sharpe ratio is about 0.4055.\n\n{\"answer\":0.4055,\"confidence\":99}" }, { "taskId": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.37258355219979333, "answer": 0.373, "confidence": 99, "grade": "correct", "relError": 0.0011177299635152713, "cost": 0.006631, "reply": "**Calculating mean and standard deviation**\n\nI need to compute the mean and sample standard deviation of daily return percentages. The formula for annualized Sharpe ratio is sqrt(365) times mean/std since I’m using a risk-free rate of zero. I’ll start by listing 30 return values and carefully sum them up to find the mean, which turns out to be 1.35%. Then, I’ll calculate the sample standard deviation by computing the squared deviations from the mean to get an accurate result.\nThe annualized Sharpe ratio is about 0.373.\n\n{\"answer\": 0.373, \"confidence\": 99}" }, { "taskId": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": -0.6855095501276544, "answer": -0.686, "confidence": 99, "grade": "exact", "relError": 0.0007154530119301313, "cost": 0.005888, "reply": "**Calculating returns and variance**\n\nI need to compute the mean and sample standard deviation of daily return percentages, then annualize the Sharpe ratio. I have a list of 30 values, and the sum is -2.50%, giving me a mean of -0.0833%. Now, I’ll calculate the sample standard deviation by determining deviations from this mean. I plan to use the formula for variance involving the sums of the squares of the deviations. Next, I'll compute that sum of squares in percentage units.\n**Calculating precise values**\n\nI'm looking to compute some precise values using the mean of -0.0833 and a standard deviation of 2.322636. It seems the ratio is -0.035883, and the annual figure is about -0.6857. My goal is to get the answer as a plain number, perhaps rounding it to -0.686. I want to double-check that I'm using returns in decimal form and keeping the same ratio. Let's get that output!\n-0.686\n{\"answer\":-0.686,\"confidence\":99}" }, { "taskId": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.5941064255855656, "answer": 0.5945, "confidence": 97, "grade": "exact", "relError": 0.0006624644970747832, "cost": 0.007454, "reply": "**Calculating Sharpe Ratio**\n\nI need to compute the mean and sample standard deviation of daily returns in percent. For the annualized Sharpe ratio, it’s mean divided by standard deviation multiplied by the square root of 365. Since the risk-free rate is zero, I’ll just focus on those percent values; the same units will cancel out. I want to ensure accuracy, so I’ll sum the returns carefully as I go.\n**Calculating statistics step-by-step**\n\nI've got a list of numbers, and I'm working through them to find the sum incrementally. I’ve added and subtracted each value, updating the total along the way. After summing up all the values, I arrive at a mean of approximately 0.0727%. Now, I need to compute the sample standard deviation, which means figuring out the sum of squares. I’m considering using the sample variance formula to assist with this next step. Time to keep going!\n0.5945\n{\"answer\": 0.5945, \"confidence\": 97}" }, { "taskId": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "chat_gpt:gpt-5.4-mini", "expected": 6.721949210975506, "answer": 6.72, "confidence": 98, "grade": "exact", "relError": 0.0002899770459918843, "cost": 0.005312, "reply": "**Calculating mean and standard deviation**\n\nI need to compute the mean and sample standard deviation of daily returns in percent. For the annualized Sharpe ratio, I'll use the formula: sqrt(365) * mean/std, ensuring returns are in the same units. I’ll carefully sum the returns, pairing values together for accuracy.\n\nThe mean comes out to 0.858 after dividing the total by 30. Now, I need to compute the sample standard deviation using squared deviations for the variance calculation. Let's work on that!\nUsing the 30 daily returns:\n\n- Mean daily return = 0.858%\n- Sample daily standard deviation = 2.439%\n- Daily Sharpe = 0.858 / 2.439 ≈ 0.3518\n- Annualized Sharpe = 0.3518 × √365 ≈ 6.72\n\n{\"answer\": 6.72, \"confidence\": 98}" }, { "taskId": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.44697422776459944, "answer": 0.445, "confidence": 95, "grade": "correct", "relError": 0.004416871582222787, "cost": 0.003557, "reply": "**Calculating implied volatility**\n\nI need to infer the implied volatility (IV) from the option price and then calculate the delta. Using the Black-Scholes call option model, I have to solve for sigma, with given values: S=447.74, K=463, and T=23/365. The option price is around 21.6661. I'll start iterating to find a suitable volatility, beginning with 30%. Using the formula, I compute values for d1 and d2, adjusting sigma until the calculated price is close to the market price. It seems to converge around 0.6206 for sigma.\n**Calculating values for accuracy**\n\nI'm computing some values starting with x, which is 0.6206 multiplied by 0.251, resulting in 0.1558. Next, I’m working with 0.5 times x squared, giving me 0.0121. Then, I calculate d1 using the formula, leading to d1 equaling -0.1387. Finally, I'm identifying N, which is approximately 0.445. I think it's important to ensure the confidence level is high, so I'll provide that value.\n0.4450\n{\"answer\":0.4450,\"confidence\":95}" }, { "taskId": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.25544432164825037, "answer": 0.2595, "confidence": 93, "grade": "wrong", "relError": 0.015876956377736007, "cost": 0.00547, "reply": "**Calculating implied volatility**\n\nI need to find the implied volatility from the call price and then determine the delta, using the Black-Scholes formula with certain parameters. I’ll start by estimating with S=357.63, K=383, T adjusted for days, and the call price of 5.0781 since it's out of the money, indicating possibly low volatility. I’ll compute various values for sigma and refine it step by step until I get close to the target price. It seems like I might settle around 0.206 or 0.212 as I continue testing.\n**Calculating implied volatility**\n\nI'm working through some calculations for implied volatility based on specific prices. I've got a few values close to each other, but the delta around 0.259 seems essential for accuracy. As I refine my estimates, I'm checking the prices that come from different sigma values. The interpolation method gives me insights on adjusting this further down to about 0.2136. I want to ensure that my approximations are exact enough for reliable trading data.\n**Computing d1 for accuracy**\n\nI'm calculating d1 at a sigma of 0.2136, so I first square that value, resulting in 0.045624. Then I need to work through the rest of the formula, looking at my numerator and denominator to find d1. Right now, I'm getting a value of -0.6445 for d1 and a corresponding N value of 0.2595, which gives me a delta of 0.2595. It seems like I need to round this off and push for a more exact answer—I might target zero for my final goal.\n0.2595\n{\"answer\":0.2595,\"confidence\":93}" }, { "taskId": "r_iv_then_delta-3", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.4262987878297406, "answer": 0.4264, "confidence": 97, "grade": "exact", "relError": 0.00023742073200499564, "cost": 0.004075, "reply": "The implied volatility is about 68.4%, which gives a Black-Scholes delta of about 0.4264.\n\n{\"answer\": 0.4264, \"confidence\": 97}" }, { "taskId": "r_iv_then_delta-4", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.6063214899790841, "answer": 0.6064, "confidence": 92, "grade": "exact", "relError": 0.0001294857962541878, "cost": 0.00448, "reply": "**Estimating implied volatility**\n\nI need to infer the implied volatility from the option price and delta. Given the numbers I have, I'll use the Black-Scholes formula. My initial estimates for volatility seem high, around 0.45. I’ll calculate d1 and d2 to see how that affects the option price. After some rough iterative guesses, I’m narrowing in on a sigma of around 0.59, which gives me a delta between 0.606 and 0.607. Now I want to refine that more precisely!\nThe implied volatility is about 58.4%, which gives a Black-Scholes delta of about 0.6064.\n\n{\"answer\": 0.6064, \"confidence\": 92}" }, { "taskId": "r_iv_then_delta-5", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 0.6017546889052645, "answer": 0.6016, "confidence": 92, "grade": "exact", "relError": 0.0002570630659245954, "cost": 0.00295, "reply": "**Calculating implied volatility**\n\nI need to infer implied volatility from the prices, starting with the Black-Scholes formula where r=0 and q=0, and T is around 0.038356. The call price formula involves S, K, and the normal distribution functions N(d1) and N(d2). My calculations are hinting that the volatility should be around 30-35%. After trying a few volatility values, I refine it down to about 0.3958 for a more accurate delta of approximately 0.6016.\n{\"answer\": 0.6016, \"confidence\": 92}" }, { "taskId": "r_deribit_straddle_breakeven-1", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 113713.12767394731, "answer": 113746.5, "confidence": 99, "grade": "exact", "relError": 0.00029347821782177007, "cost": 0.003551, "reply": "**Calculating break-even points**\n\nI’m trying to compute various figures starting from 101,000 divided by 0.8882, which leads to around 113,770. I keep refining that calculation using the reciprocal and multiplying to ensure precision. I find that a more accurate division gives me about 113,746.5 as my adjustable answer. I'm also noting how this connects to break-even considerations based on premiums and potential payouts, especially regarding the upper break-even for BTC/USD at expiry. It's a lot of back-and-forth!\n**Finalizing the answer**\n\nI've calculated the answer to be approximately 113,746.5. Now, I just want to make sure that my formula is correct. I should aim to present this as a plain number. It's important to me that I get everything right. Double-checking helps to ensure that I'm providing the user with accurate information. Alright, I’ll finalize this as just 113,746.5 to keep it straightforward!\n113746.5\n{\"answer\":113746.5,\"confidence\":99}" }, { "taskId": "r_deribit_straddle_breakeven-2", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 93608.77985797288, "answer": 93608.8, "confidence": 99, "grade": "exact", "relError": 2.1517241384055273e-7, "cost": 0.003718, "reply": "**Computing break-even for options**\n\nI need to compute the break-even point for a long straddle on BTC-settled options since the profit is in BTC. For call payoffs, the formula is max(S-K,0)/S, and I realize the BTC options on Deribit settle differently. The intrinsic break-even point is calculated by determining where the payoff minus the premium of 0.0706 BTC equals zero. \n\nI've worked out that if K is 87000, S comes out to approximately 93593 using my calculations. Now, let's confirm the accuracy!\n**Finalizing break-even calculations**\n\nI need to find the exact break-even point. Starting with 8.16, I divide by 0.9294 which gives me 8.78, leading to about 93608.78. Checking back, I confirm that 87000 divided by 0.9294 is indeed around 93609. \n\nFor more accuracy, I calculate 1/0.9294, and that gives me a factor of about 1.07593. Multiplying it by 87000 gives around 93556, and I'm using a precise approach here to maintain accuracy: S equals K divided by (1 - premium), confirming all calculations yield approximately 93608.777.\nFor a BTC-settled call, the BTC payoff at expiry is:\n\n\\[\n\\frac{S-K}{S}\n\\]\n\nSo the straddle breaks even on the upside when:\n\n\\[\n\\frac{S-K}{S} = 0.015 + 0.0556 = 0.0706\n\\]\n\n\\[\n1 - \\frac{K}{S} = 0.0706\n\\Rightarrow \\frac{K}{S} = 0.9294\n\\Rightarrow S = \\frac{87000}{0.9294} \\approx 93608.8\n\\]\n\n{\"answer\": 93608.8, \"confidence\": 99}" }, { "taskId": "r_deribit_straddle_breakeven-3", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 64193.23251006419, "answer": 64193.25, "confidence": 99, "grade": "exact", "relError": 2.724576271668627e-7, "cost": 0.003516, "reply": "**Calculating break-even for straddle**\n\nI need to determine the break-even price for a long straddle in BTC where the premium is 0.0809 BTC. The payoff structure for calls and puts will differ based on if it's BTC or inverse options. For a long straddle, the break-even occurs when the call payoff equals the premium. Setting the equation equal to 0.0809, I can rearrange it to find the underlying price S. After calculating, it seems my resulting price is approximately 64,193.2, but I’ll need to check that again.\n**Verifying calculations**\n\nI'm working through my calculations for the straddle break-even price. So, I’m multiplying 9191 by 64193 and breaking it down into parts, first with 64000 and then adding 9191 multiplied by 193. After totaling everything, I find it’s slightly under the expected value by about 2,137, adjusting my final answer to around 64193.23. To double-check, I evaluate 59000 divided by 0.9191, which gives me 64193.25. That seems accurate!\nFor a BTC-settled Deribit straddle, the BTC payoff above the strike is\n\n\\[\n\\frac{S-K}{S}\n\\]\n\nSet that equal to your total premium:\n\n\\[\n0.0196 + 0.0613 = 0.0809\n\\]\n\nSo:\n\n\\[\n\\frac{S-59000}{S} = 0.0809\n\\quad\\Rightarrow\\quad\n1-\\frac{59000}{S}=0.0809\n\\quad\\Rightarrow\\quad\nS=\\frac{59000}{0.9191}\\approx 64193.25\n\\]\n\nSo the straddle breaks even above about **64193** USD/BTC.\n\n{\"answer\": 64193.25, \"confidence\": 99}" }, { "taskId": "r_deribit_straddle_breakeven-4", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 102706.75082914304, "answer": 102707, "confidence": 99, "grade": "exact", "relError": 0.0000024260416666504776, "cost": 0.006035, "reply": "**Calculating break-even for straddle**\n\nI need to figure out the break-even point for a straddle option with a premium in BTC. Since options are settled in BTC, I realize the payoff is influenced by intrinsic value measured in USD divided by the settlement price. The settlement value for a call option is calculated using a formula that considers the difference between settlement and strike prices. I'll also need to compute the upper break-even in relation to the strike price. My calculations lead me to an approximate break-even point of about 102,709 USD.\n**Refining the calculations**\n\nI want to compute the division of 96,000 by 0.9347 more accurately. First, I'll multiply 0.9347 by 102,700, which gives me around 96,019.69 — slightly higher than needed. To find the difference, I divide by 0.9347, which gives me approximately 21.07. So, I'll subtract this from 102,700, leading me to around 102,678.9. Checking again, I need to ensure my multiplication with the adjustment is correct, as my earlier arithmetic seems off. I’ve got to revisit my calculations carefully!\n**Refining calculations**\n\nI’m working through some calculations: it seems I need to find the right number I’m looking for is around 102704. So I tried using a mental calculator approach and approximated it to about 102705. \n\nAfter some verification and adjustments, I found that I need to aim slightly higher, landing around 102706.75. I had to make sure my final checks lined up correctly, which they seem to! It feels satisfying to fine-tune these calculations.\n**Clarifying break-even**\n\nI need to confirm the formula for calculating the break-even point, considering both call and put premiums along with any profits in Bitcoin. I realize there's a lower break-even point below the strike price, which might complicate things.\n\nBut the user is asking about the price above which they want to know the break-even. So I’ll focus on making sure I provide clear information based on those parameters. It’s important to get this right!\nFor BTC-settled Deribit options, the straddle’s BTC payoff at expiry is\n\n\\[\n\\frac{|S-K|}{S}\n\\]\n\nwith \\(K=96000\\) and total premium\n\n\\[\n0.022+0.0433=0.0653 \\text{ BTC}.\n\\]\n\nUpper break-even solves\n\n\\[\n\\frac{S-96000}{S}=0.0653\n\\quad\\Rightarrow\\quad\nS=\\frac{96000}{1-0.0653}=\\frac{96000}{0.9347}\\approx 102707.\n\\]\n\n{\"answer\": 102707, \"confidence\": 99}" }, { "taskId": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "chat_gpt:gpt-5.4-mini", "expected": 73994.63806970509, "answer": 73994.65, "confidence": 98, "grade": "exact", "relError": 1.612318840072502e-7, "cost": 0.002621, "reply": "**Evaluating straddle breakeven price**\n\nI'm calculating a straddle breakeven price for BTC/USD. I found that slightly under 74000 divided by 50k over 9325 equals about 5.36, giving me a figure of 73994.64. It seems I need to take a closer look at the downside breakeven point as well. Since the question focuses on the price at expiry, I think I'll conclude that the breakeven price is around 73995.\n73994.65\n{\"answer\":73994.65,\"confidence\":98}" }, { "taskId": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 40.790551141452084, "answer": 40.79, "confidence": 88, "grade": "exact", "relError": 0.000013511498047038315, "cost": 0.018728, "reply": "I'm computing d1 for the Black-Scholes formula, working through the time to expiration, volatility term, and drift components to get the numerator and dividing by σ√T, arriving at roughly 0.748.\n\nd2 comes out to about 0.613. I'm approximating the normal CDF at both points using interpolation around known values, getting N(d1)≈0.77276 and N(d2)≈0.73013. Then I discount the strike price by the risk-free rate, multiply both terms by the respective probabilities, and subtract to get the call price around $40.79.\n\n\nThe call price is about **40.79**, using the standard Black-Scholes call formula. Normal CDF values were interpolated from table values, so the last digit may be off by a few cents.\n\n**Inputs**\n- T = 18/365 = 0.049315\n- √T = 0.22207\n- σ√T = 0.607 × 0.22207 = 0.134796\n\n**d₁ and d₂**\n- ln(S/K) = ln(382.5/350) = 0.088796\n- rT = 0.0597 × 0.049315 = 0.002944\n- σ²T/2 = 0.134796² / 2 = 0.009085\n- d₁ = (0.088796 + 0.002944 + 0.009085) / 0.134796 = 0.74798\n- d₂ = d₁ − σ√T = 0.74798 − 0.134796 = 0.61319\n\n**Normal probabilities**\n- N(d₁) = N(0.74798) ≈ 0.77276\n- N(d₂) = N(0.61319) ≈ 0.73013\n\n**Discounted strike**\n- K·e^(−rT) = 350 × e^(−0.002944) ≈ 348.971\n\n**Call price**\n- C = S·N(d₁) − K·e^(−rT)·N(d₂)\n- S·N(d₁) = 382.5 × 0.77276 = 295.581\n- K·e^(−rT)·N(d₂) = 348.971 × 0.73013 = 254.794\n- C = 295.581 − 254.794 ≈ 40.79\n\n{\"answer\": 40.79, \"confidence\": 88}" }, { "taskId": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 3.2317412561362406, "answer": 3.2317, "confidence": 88, "grade": "exact", "relError": 0.00001276591563829087, "cost": 0.021268, "reply": "I'm computing the Black-Scholes d1 term, working through time to expiry, log moneyness, risk-free rate adjustment, and volatility squared to get the numerator before dividing by σ√T.\n\nI get d1≈0.180401 and d2≈-0.174152, then approximate N(d1)≈0.571581 and N(d2)≈0.430873 using a polynomial expansion. Discounting the strike gives K·e^(-rT)≈22.86463, so the call price comes out to C=22.89×0.571581-22.86463×0.430873≈3.2317.\n\n\n**Call price ≈ 3.232**\n\n**Inputs**\n- T = 171/365 = 0.468493\n- σ√T = 0.518 × 0.684466 = 0.354553\n- ln(S/K) = ln(22.89/23) = −0.004795\n- rT = 0.0126 × 0.468493 = 0.005903\n- ½σ²T = 0.062854\n\n**d₁ and d₂**\n- d₁ = (−0.004795 + 0.005903 + 0.062854) / 0.354553 = 0.180401\n- d₂ = d₁ − σ√T = −0.174152\n\n**Normal probabilities** (series approximation near zero)\n- N(d₁) ≈ 0.571581\n- N(d₂) ≈ 0.430873\n\n**Discounted strike**\n- K·e^(−rT) = 23 × e^(−0.005903) = 22.86463\n\n**Call price**\n- C = 22.89 × 0.571581 − 22.86463 × 0.430873\n- C = 13.08349 − 9.85175 = 3.2317\n\n{\"answer\": 3.2317, \"confidence\": 88}" }, { "taskId": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 16.773153687059427, "answer": 16.772, "confidence": 92, "grade": "exact", "relError": 0.00006878176167422208, "cost": 0.021308, "reply": "I'm computing d1 for the Black-Scholes formula: with T≈0.416 years, σ√T comes out to about 0.5085, and after working out the log moneyness, risk-free drift term, and variance adjustment, d1 lands at roughly 0.0949.\n\nNow I'm finishing d1=0.1866 and getting d2 by subtracting σ√T, giving -0.3219. Then I'm approximating N(d1)≈0.574 and N(d2)≈0.3738 using Taylor series for the normal CDF, discounting the strike with e^(-rT)=92.74, and combining everything to get a call price of about 16.77.\n\n\n**Call price ≈ 16.77**, using the standard Black-Scholes call formula with intermediate values rounded to about 6 significant figures.\n\n**Inputs**\n- Spot S = 89.61, strike K = 95\n- Time to expiry T = 152/365 = 0.416438\n- Volatility σ = 0.788, risk-free rate r = 0.0577\n\n**Intermediate quantities**\n- σ√T = 0.788 × 0.645320 = 0.508512\n- ln(S/K) = ln(0.943263) = −0.058411\n- rT = 0.024028\n- σ²T/2 = 0.129292\n\n**d₁ and d₂**\n- d₁ = (−0.058411 + 0.024028 + 0.129292) / 0.508512 = 0.186641\n- d₂ = d₁ − σ√T = −0.321871\n\n**Normal probabilities**\n- N(d₁) ≈ 0.574021\n- N(d₂) ≈ 0.373775\n\n**Price**\n- Discounted strike: K·e^(−rT) = 95 × 0.976258 = 92.7445\n- C = S·N(d₁) − K·e^(−rT)·N(d₂)\n- C = 89.61 × 0.574021 − 92.7445 × 0.373775\n- C = 51.4380 − 34.6656 = 16.7724\n\n{\"answer\": 16.772, \"confidence\": 92}" }, { "taskId": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.6945968823131317, "answer": 0.6939, "confidence": 80, "grade": "correct", "relError": 0.0010032902981236462, "cost": 0.020328, "reply": "I'm computing the time to expiration, volatility term, log-moneyness, and risk-free drift term needed to build d1 for the Black-Scholes calculation, with T≈0.0247 years, σ√T≈0.0669, and ln(S/K)≈-0.1048.\n\nContinuing, I get d1≈-1.5246 and d2≈-1.5914, then work out the normal CDF tail values using a small-x approximation, N(-1.5246)≈0.0637 and N(-1.5914)≈0.0558, and plug these into the call price formula to get C≈384.5(0.0637) - 427(0.9994)(0.0558) ≈ 24.49 - 23.79 ≈ 0.69, noting minor rounding sensitivity but accepting the result as close enough.\n\n\n**Call price ≈ 0.694**\n\n1. Time to expiry: T = 9/365 = 0.024658, so √T = 0.15703.\n2. Volatility over the period: σ√T = 0.426 × 0.15703 = 0.066894.\n3. Log-moneyness: ln(S/K) = ln(384.5/427) = −0.104841.\n4. Discount term: rT = 0.0252 × 0.024658 = 0.000621.\n5. d₁ = (−0.104841 + 0.000621 + 0.5 × 0.066894²) / 0.066894 = −0.101983 / 0.066894 ≈ −1.52455.\n6. d₂ = d₁ − σ√T ≈ −1.59144.\n7. Normal probabilities, from table values with a linear correction (density × step): N(d₁) ≈ 0.063684 and N(d₂) ≈ 0.055755.\n8. Discount factor: K·e^(−rT) = 427 × 0.999379 = 426.735.\n9. Call price: C = 384.5 × 0.063684 − 426.735 × 0.055755 = 24.4865 − 23.7926 ≈ 0.6939.\n\nThe result is a small difference of two large terms, so table rounding in the normal probabilities could move the last digit.\n\n{\"answer\": 0.6939, \"confidence\": 80}" }, { "taskId": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 23.323917391138792, "answer": 23.327, "confidence": 88, "grade": "exact", "relError": 0.00013216514230926843, "cost": 0.018928, "reply": "I'm working through the Black-Scholes d1 calculation, computing the time to expiration, log of the price ratio, and combining the drift and volatility terms to get the numerator before dividing by σ√T.\n\nNow I'm finishing d1 and d2, looking up their normal CDF values via interpolation, then plugging everything into the call price formula: stock price times N(d1) minus discounted strike times N(d2), arriving at a call value around 23.33.\n\n\n**Call price ≈ 23.33**\n\n**Inputs**\n- T = 20/365 = 0.054795\n- √T = 0.234082\n- σ√T = 0.355 × 0.234082 = 0.083099\n- ln(S/K) = ln(314.3/294) = ln(1.069048) = 0.066770\n- rT = 0.0006 × 0.054795 = 0.0000329\n- (σ²/2)·T = 0.0630125 × 0.054795 = 0.0034527\n\n**d1 and d2**\n- d1 = (0.066770 + 0.0000329 + 0.0034527) / 0.083099 = 0.0702556 / 0.083099 ≈ 0.84545\n- d2 = d1 − σ√T ≈ 0.76235\n\n**Normal CDF values** (from table values with linear interpolation)\n- N(d1) ≈ 0.80108\n- N(d2) ≈ 0.77707\n\n**Call price**\n- First term: S·N(d1) = 314.3 × 0.80108 ≈ 251.778\n- Discounted strike: K·e^(−rT) = 294 × e^(−0.0000329) ≈ 293.990\n- Second term: 293.990 × 0.77707 ≈ 228.452\n- C = 251.778 − 228.452 ≈ 23.327\n\nThe interpolated normal values could shift the last digit slightly.\n\n{\"answer\": 23.327, \"confidence\": 88}" }, { "taskId": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": -0.280639388201964, "answer": -0.2806, "confidence": 90, "grade": "exact", "relError": 0.00014035165276097544, "cost": 0.015104, "reply": "I'm computing the Black-Scholes d1 term, working through the volatility-adjusted time factor, the log of the price ratio, and the drift component before combining them.\n\nd1 comes out to roughly 0.581, so I'm finding N(d1) using the normal distribution approximation, landing around 0.7194 and giving a put delta of about -0.2806.\n\n\nThe put delta is about **−0.2806**. Under Black-Scholes it equals N(d₁) − 1, and d₁ = [ln(S/K) + (r + σ²/2)T] / (σ√T).\n\n1. **Time to expiry:** T = 50/365 = 0.136986, so √T = 0.370117.\n2. **Volatility over the period:** σ√T = 0.466 × 0.370117 = 0.172475.\n3. **Log-moneyness:** ln(114.52/106) = ln(1.080377) ≈ 0.077310.\n4. **Drift term:** (0.0585 + 0.466²/2) × T = (0.0585 + 0.108578) × 0.136986 ≈ 0.022887.\n5. **d₁:** (0.077310 + 0.022887) / 0.172475 = 0.100197 / 0.172475 ≈ 0.58094.\n6. **N(d₁):** Using N(0.58) = 0.71904 and a linear step with density ≈ 0.3372, N(0.58094) ≈ 0.71904 + 0.00094 × 0.3372 ≈ 0.71936.\n7. **Put delta:** N(d₁) − 1 = 0.71936 − 1 ≈ −0.2806.\n\n{\"answer\": -0.2806, \"confidence\": 90}" }, { "taskId": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": -0.31093085044630486, "answer": -0.3109, "confidence": 90, "grade": "exact", "relError": 0.00009921963761579435, "cost": 0.013684, "reply": "I'm computing the Black-Scholes d1 term, working through the time-to-maturity, volatility scaling, and log-moneyness pieces, then looking up the corresponding normal CDF value near 0.49.\n\nSo N(d1) comes out to about 0.68907, giving a put delta of roughly -0.31093.\n\n\nPut delta ≈ **−0.3109**, computed from the Black-Scholes formula Δ_put = N(d₁) − 1.\n\n**Inputs**\n- T = 41/365 = 0.112329, so √T = 0.335155.\n- σ√T = 0.727 × 0.335155 = 0.243658.\n\n**Compute d₁ = [ln(S/K) + (r + σ²/2)T] / (σ√T)**\n- ln(46.9/43) = ln(1.090698) = 0.086818.\n- (r + σ²/2)T = (0.0327 + 0.264265) × 0.112329 = 0.033358.\n- d₁ = (0.086818 + 0.033358) / 0.243658 = 0.120176 / 0.243658 = 0.49322.\n\n**Normal CDF**\n- N(0.49) = 0.68793, and the density there is about 0.3538.\n- By linear interpolation, N(0.49322) ≈ 0.68793 + 0.00322 × 0.3538 = 0.68907.\n\n**Put delta**\n- Δ_put = N(d₁) − 1 = 0.68907 − 1 = −0.31093.\n\n{\"answer\": -0.3109, \"confidence\": 90}" }, { "taskId": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": -0.5641608283248785, "answer": -0.5642, "confidence": 90, "grade": "exact", "relError": 0.00006943352525534937, "cost": 0.015824, "reply": "Computing the Black-Scholes d1 term: I'm calculating σ√T as 0.2228, the log moneyness as -0.06751, and now multiplying the drift adjustment (r+σ²/2) by T to get the numerator pieces for d1.\n\nThat gives d1 = -0.1615, and evaluating N(-0.1615) through a normal approximation gives roughly 0.4358, so the put delta works out to about -0.5642.\n\n\n**Put delta = N(d₁) − 1 ≈ −0.5642**\n\n**Inputs**\n- Time to expiry: T = 65/365 = 0.178082, so √T = 0.42200.\n- Volatility over the period: σ√T = 0.528 × 0.42200 = 0.222816.\n\n**Step 1 – compute d₁ = [ln(S/K) + (r + σ²/2)T] / (σ√T)**\n- ln(S/K) = ln(215.92/231) = ln(0.934719) ≈ −0.067510.\n- Drift term: (r + σ²/2)T = (0.0376 + 0.139392) × 0.178082 = 0.031519.\n- d₁ = (−0.067510 + 0.031519) / 0.222816 = −0.035991 / 0.222816 ≈ −0.16153.\n\n**Step 2 – evaluate N(d₁)**\n- Using a short series for the normal CDF near zero: N(−0.16153) ≈ 0.5 − 0.064156 = 0.435844.\n\n**Step 3 – put delta**\n- Put delta = N(d₁) − 1 = 0.435844 − 1 ≈ −0.5642.\n\n{\"answer\": -0.5642, \"confidence\": 90}" }, { "taskId": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": -0.3853788948381438, "answer": -0.3854, "confidence": 90, "grade": "exact", "relError": 0.00005476470595277504, "cost": 0.009204, "reply": "T = 113/365 = 0.309589.\nln(S/K) = ln(291.06/286) = ln(1.017692) ≈ 0.017537.\nσ²/2 = 0.729²/2 = 0.531441/2 = 0.2657205; adding r gives 0.2657205 + 0.0594 = 0.3251205.\nMultiplying by T: 0.3251205 × 0.309589 ≈ 0.100654.\nThe numerator is 0.017537 + 0.100654 = 0.118191.\nσ√T: √0.309589 ≈ 0.556407, so σ√T = 0.729 × 0.556407 ≈ 0.405621.\nd1 = 0.118191 / 0.405621 ≈ 0.291383.\n\nN(0.291383): φ(0.29) = 0.38251, and the table value is N(0.29) = 0.61409.\nAdding 0.001383 × 0.3825 ≈ 0.000529 gives N(d1) ≈ 0.614619.\n\nPut delta = N(d1) − 1 ≈ −0.38538.\n\n{\"answer\": -0.3854, \"confidence\": 90}" }, { "taskId": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": -0.4080759766740296, "answer": -0.40808, "confidence": 92, "grade": "exact", "relError": 0.000009859257100045587, "cost": 0.011544, "reply": "Put delta is N(d1) − 1, so I need d1.\n\n**Inputs**\n- T = 162/365 = 0.443836\n- σ = 0.799, so σ² = 0.638401\n- √T = 0.666211, so σ√T = 0.532303\n\n**Numerator of d1:** ln(S/K) + (r + σ²/2)T\n- ln(146.31/152) = ln(0.962566) = −0.038153\n- r + σ²/2 = 0.0456 + 0.3192005 = 0.3648005\n- (r + σ²/2)·T = 0.3648005 × 0.443836 = 0.161912\n- Sum = −0.038153 + 0.161912 = 0.123759\n\n**d1** = 0.123759 / 0.532303 = 0.232497\n\n**N(d1)**\n- φ(0.2325) = 0.39894 × e^(−0.027027) ≈ 0.388300\n- N(0.2325) ≈ 0.5 + 0.2325 × (0.398942 − 0.2325² × 0.398942/6) = 0.5 + 0.2325 × 0.395348 = 0.591918\n- The linear correction from 0.2325 to 0.232497 (−0.000003 × 0.3883 ≈ −0.0000012) is negligible.\n- N(d1) ≈ 0.59192\n\n**Put delta** = N(d1) − 1 = 0.59192 − 1 = −0.40808\n\n{\"answer\": -0.40808, \"confidence\": 92}" }, { "taskId": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.007721610843754648, "answer": 0.007722, "confidence": 92, "grade": "exact", "relError": 0.000050398324031885966, "cost": 0.016684, "reply": "Computing the Black-Scholes d1 term: working out T as 34/365, getting σ√T around 0.2576, then calculating the log of the price ratio and the drift term before combining them into the numerator of d1.\n\nd1 comes out to about 0.5876, then I compute φ(d1) using the normal density formula, getting roughly 0.3357. Dividing that by the stock price times σ√T gives Gamma ≈ 0.00772.\n\n\nGamma ≈ **0.007722** per $1 change in spot. It is computed with the Black-Scholes formula Γ = φ(d₁) / (S·σ·√T), which is the same for calls and puts.\n\n**Inputs**\n- S = 168.77, K = 150\n- σ = 0.844, r = 0.003\n- T = 34/365 = 0.093151\n- √T = 0.305207\n- σ√T = 0.844 × 0.305207 = 0.257595\n\n**d₁**\n- ln(S/K) = ln(1.125133) ≈ 0.117901\n- (r + σ²/2)·T = (0.003 + 0.356168) × 0.093151 ≈ 0.033457\n- d₁ = (0.117901 + 0.033457) / 0.257595 ≈ 0.58758\n\n**Standard normal density at d₁**\n- φ(d₁) = 0.398942 × e^(−0.58758²/2) = 0.398942 × e^(−0.172625)\n- e^(−0.172625) ≈ 0.841453, so φ(d₁) ≈ 0.335693\n\n**Gamma**\n- Γ = 0.335693 / (168.77 × 0.257595) = 0.335693 / 43.4743 ≈ 0.0077217\n\n{\"answer\": 0.007722, \"confidence\": 92}" }, { "taskId": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.011719035620003572, "answer": 0.011719, "confidence": 90, "grade": "exact", "relError": 0.000003039499556679529, "cost": 0.016684, "reply": "Computing the Black-Scholes d1 term: with time to expiry of about 0.1945 years, I'm working out ln(S/K) as roughly -0.0674, then combining it with the drift term (r+σ²/2)T to build toward d1.\n\nd1 comes out to -0.7913, and using the standard normal density I get φ(d1)≈0.2917. Dividing by S·σ√T gives a gamma of approximately 0.01172.\n\n\n**Gamma ≈ 0.01172**\n\nGamma is the same for calls and puts:\n\nΓ = φ(d₁) / (S·σ·√T)\n\nHere φ is the standard normal density.\n\n1. **Time to expiry:** T = 71/365 = 0.194521, so √T = 0.441045.\n2. **Volatility over the period:** σ√T = 0.161 × 0.441045 = 0.071008.\n3. **Log-moneyness:** ln(S/K) = ln(350.55/375) = ln(0.934800) ≈ −0.067423.\n4. **Drift term:** (r + σ²/2)·T = (0.0448 + 0.0129605) × 0.194521 ≈ 0.011236.\n5. **d₁:** d₁ = (−0.067423 + 0.011236) / 0.071008 ≈ −0.79128.\n6. **Density at d₁:** φ(d₁) = 0.398942 × e^(−0.313062) ≈ 0.398942 × 0.731205 ≈ 0.291707.\n7. **Gamma:** Γ = 0.291707 / (350.55 × 0.071008) = 0.291707 / 24.8919 ≈ 0.011719.\n\n{\"answer\": 0.011719, \"confidence\": 90}" }, { "taskId": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.00332788170697725, "answer": 0.0033279, "confidence": 92, "grade": "exact", "relError": 0.0000054968969333978925, "cost": 0.014704, "reply": "I'm computing the Black-Scholes d1 term using time to expiration of about 0.293 years, getting d1 ≈ 0.160, and now working out the normal density φ(d1) for the option pricing formula.\n\nDividing φ(d1) by S times σ√T gives gamma ≈ 0.003328.\n\n\n**Gamma ≈ 0.003328**\n\nFor a non-dividend-paying asset, Black-Scholes gamma is Γ = φ(d₁) / (S·σ·√T), where φ is the standard normal density.\n\n1. Time to expiry: T = 107/365 = 0.293151, so √T = 0.541434.\n2. σ√T = 0.464 × 0.541434 = 0.251225.\n3. ln(S/K) = ln(471.11/474) = −0.006116.\n4. (r + σ²/2)·T = (0.0503 + 0.107648) × 0.293151 = 0.046303.\n5. d₁ = (−0.006116 + 0.046303) / 0.251225 = 0.15996.\n6. φ(d₁) = 0.398942 × e^(−0.15996²/2) = 0.398942 × 0.987287 = 0.393870.\n7. S·σ·√T = 471.11 × 0.251225 = 118.3546.\n8. Γ = 0.393870 / 118.3546 = 0.0033279.\n\n{\"answer\": 0.0033279, \"confidence\": 92}" }, { "taskId": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.015589164355628841, "answer": 0.015589, "confidence": 92, "grade": "exact", "relError": 0.00001054294028157329, "cost": 0.016604, "reply": "I'm computing the Black-Scholes components: time to expiry of 0.4027 years, its square root, the volatility-adjusted term σ√T, the log moneyness ln(87.11/100), and now working out the drift term r+σ²/2 for the d1 calculation.\n\nPlugging these together, d1 comes out to about -0.27290. From there I'm computing the standard normal density φ(d1)≈0.38436, and finally deriving gamma as 0.38436/(87.11×0.28304)≈0.015589.\n\n\n**Gamma ≈ 0.01559**\n\nGamma is the same for a call and a put. I used the formula Γ = φ(d₁) / (S·σ·√T), where φ is the standard normal density.\n\n**Inputs:** S = 87.11, K = 100, σ = 0.446, r = 0.0514, T = 147/365 = 0.402740.\n\n1. Compute the volatility term: √T = 0.634618, so σ√T = 0.446 × 0.634618 = 0.283040.\n\n2. Compute the log-moneyness: ln(S/K) = ln(0.8711) = −0.137998.\n\n3. Compute the drift term: (r + σ²/2)·T = (0.0514 + 0.099458) × 0.402740 = 0.060757.\n\n4. Compute d₁ = (ln(S/K) + (r + σ²/2)T) / (σ√T) = (−0.137998 + 0.060757) / 0.283040 = −0.27290.\n\n5. Compute the density: φ(d₁) = (1/√(2π))·e^(−d₁²/2) = 0.398942 × e^(−0.037237) = 0.384360.\n\n6. Compute the denominator: S·σ√T = 87.11 × 0.283040 = 24.6556.\n\n7. So Γ = 0.384360 / 24.6556 ≈ 0.015589.\n\n{\"answer\": 0.015589, \"confidence\": 92}" }, { "taskId": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.00596435584251129, "answer": 0.0059643, "confidence": 92, "grade": "exact", "relError": 0.000009362706177324332, "cost": 0.015016, "reply": "Computing d1 for the Black-Scholes formula: with T≈0.282 years, σ√T≈0.1275, and ln(335.65/300)≈0.1123, I add the drift term (r+σ²/2)T≈0.00821 to get a numerator of about 0.1205, then divide by σ√T to find d1.\n\nNow I'm computing the standard normal density φ(d1)≈0.2552, then dividing by (S·σ√T) to get gamma≈0.00596.\n\n\n**Gamma ≈ 0.005964**, from Γ = φ(d₁) / (S·σ·√T).\n\n**Inputs**\n- T = 103/365 = 0.282192\n- √T = 0.531217\n- σ√T = 0.24 × 0.531217 = 0.127492\n\n**d₁**\n- ln(S/K) = ln(335.65/300) = ln(1.118833) ≈ 0.112287\n- (r + σ²/2)·T = (0.0003 + 0.0288) × 0.282192 = 0.008212\n- d₁ = (0.112287 + 0.008212) / 0.127492 ≈ 0.945149\n\n**Standard normal density at d₁**\n- φ(d₁) = (1/√(2π)) · e^(−d₁²/2) = 0.398942 × e^(−0.446653) ≈ 0.398942 × 0.639766 ≈ 0.255228\n\n**Gamma**\n- Γ = 0.255228 / (335.65 × 0.127492) = 0.255228 / 42.7927 ≈ 0.0059643\n\nGamma is the same for a call and a put.\n\n{\"answer\": 0.0059643, \"confidence\": 92}" }, { "taskId": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.646446181061521, "answer": 0.6464, "confidence": 88, "grade": "exact", "relError": 0.00007143837008234106, "cost": 0.00928, "reply": "Vega per 1 percentage point is S·φ(d1)·√T / 100.\n\n**Inputs**\n- T = 74/365 = 0.202740, so √T = 0.450267.\n- σ√T = 0.55 × 0.450267 = 0.247647.\n- ln(424.81/380) = ln(1.117921) = 0.111474.\n- (r + σ²/2)T = (0.0025 + 0.15125) × 0.202740 = 0.031172.\n\n**d1**\n\nd1 = (0.111474 + 0.031172) / 0.247647 = 0.142646 / 0.247647 = 0.57601.\n\n**φ(d1)**\n- d1²/2 = 0.165894.\n- e^(−0.165894) = 0.847135.\n- φ(d1) = 0.398942 × 0.847135 = 0.337958.\n\n**Vega**\n- S·√T = 424.81 × 0.450267 = 191.278.\n- Full vega = 191.278 × 0.337958 = 64.644.\n- Per 1 percentage point: 64.644 / 100 = 0.6464.\n\n{\"answer\": 0.6464, \"confidence\": 88}" }, { "taskId": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.26781898170848095, "answer": 0.2678, "confidence": 92, "grade": "exact", "relError": 0.00007087514245585115, "cost": 0.017408, "reply": "I'm computing the Black-Scholes inputs: time to expiry of 0.284932 years, volatility term σ√T of 0.087008, and the log-moneyness ln(S/K) of approximately -0.053060, now working toward the drift term (r+σ²/2)T for the d1 calculation.\n\nd1 comes out to roughly -0.55847, and I'm using that to find the normal density φ(d1)≈0.34134 via the exponential term exp(-0.155944)≈0.855607. From there I calculate vega as 146.99×0.34134×0.53379/100, giving approximately 0.26782.\n\n\nVega ≈ **0.2678**: the option price rises by about 0.268 for a 1 percentage point increase in volatility (16.3% → 17.3%). It is the same for a call or a put.\n\nBlack-Scholes vega is S·φ(d₁)·√T. I divide it by 100 to express it per 1 percentage point of volatility.\n\n**Inputs**\n- T = 104/365 = 0.284932, so √T = 0.533790\n- σ√T = 0.163 × 0.533790 = 0.087008\n\n**d₁**\n- ln(S/K) = ln(146.99/155) = −0.053060\n- (r + σ²/2)T = (0.0024 + 0.0132845) × 0.284932 = 0.004469\n- d₁ = (−0.053060 + 0.004469) / 0.087008 = −0.55847\n\n**Normal density at d₁**\n- φ(d₁) = 0.398942 × exp(−0.155944) = 0.398942 × 0.855607 = 0.341340\n\n**Vega**\n- S·φ(d₁)·√T = 146.99 × 0.341340 × 0.533790 = 26.782 per unit of volatility\n- Per 1 percentage point: 26.782 / 100 = 0.26782\n\n{\"answer\": 0.2678, \"confidence\": 92}" }, { "taskId": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.45587809695630876, "answer": 0.4559, "confidence": 90, "grade": "exact", "relError": 0.000048045834703398345, "cost": 0.011268, "reply": "Vega per unit volatility is S·φ(d1)·√T; dividing by 100 gives the change for a 1 percentage-point move.\n\n**Inputs**\n- T = 47/365 = 0.128767\n- √T = 0.358841\n- σ√T = 0.254 × 0.358841 = 0.091146\n\n**d1**\n- ln(S/K) = ln(469.67/512) = ln(0.917324) = −0.086296\n- (r + σ²/2)T = (0.0139 + 0.032258) × 0.128767 = 0.046158 × 0.128767 = 0.0059436\n- Numerator = −0.086296 + 0.0059436 = −0.080352\n- d1 = −0.080352 / 0.091146 = −0.88158\n\n**φ(d1)**\n- d1² = 0.777183, so d1²/2 = 0.388591\n- e^(−0.388591) ≈ 0.67803\n- φ(d1) = 0.398942 × 0.67803 = 0.270495\n\n**Vega**\n- S·√T = 469.67 × 0.358841 = 168.537\n- 168.537 × 0.270495 = 45.5884 per unit volatility\n- Per 1 percentage point: 45.5884 / 100 ≈ 0.4559\n\n**Answer: about 0.4559** — the option price rises by roughly 0.456 for a 1-point increase in volatility.\n\n{\"answer\": 0.4559, \"confidence\": 90}" }, { "taskId": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.6097213395050691, "answer": 0.6097, "confidence": 92, "grade": "exact", "relError": 0.00003499878335639152, "cost": 0.011948, "reply": "Use the Black-Scholes vega formula: vega = S·φ(d1)·√T, then divide by 100 to get the change for a 1 percentage-point move in volatility.\n\n**Inputs**\n- S = 383.12, K = 429, σ = 0.726, r = 0.0203\n- T = 61/365 = 0.167123\n- √T = 0.408807\n\n**Compute d1**\n- ln(S/K) = ln(0.893054) = −0.113105\n- (r + σ²/2)·T = (0.0203 + 0.263538) × 0.167123 = 0.283838 × 0.167123 = 0.047436\n- Numerator = −0.113105 + 0.047436 = −0.065669\n- σ√T = 0.726 × 0.408807 = 0.296794\n- d1 = −0.065669 / 0.296794 = −0.221261\n\n**Standard normal density**\n- d1²/2 = 0.048956/2 = 0.024478\n- e^(−0.024478) = 0.975819\n- φ(d1) = 0.398942 × 0.975819 = 0.389295\n\n**Vega**\n- S·√T = 383.12 × 0.408807 = 156.622\n- Vega = 156.622 × 0.389295 = 60.972 per unit of volatility\n- Per 1 percentage point: 60.972 / 100 = 0.60972\n\nThe vega is about **0.6097**, meaning the option price rises by roughly 0.61 for a 1 percentage-point increase in volatility.\n\n{\"answer\": 0.6097, \"confidence\": 92}" }, { "taskId": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.2403754082675673, "answer": 0.2404, "confidence": 92, "grade": "exact", "relError": 0.00010230552538611792, "cost": 0.017048, "reply": "I'm computing the Black-Scholes d1 term: time to expiry is 0.178082 years, so sqrt(T) is 0.421998 and volatility times sqrt(T) gives 0.238007. I'm also finding the log of the price ratio (158.91/148) which comes to 0.071127, and multiplying the drift term (r + σ²/2) by T to get 0.218948×0.178082.\n\nd1 works out to (0.071126+0.038991)/0.238007 = 0.462662. From there I calculate the standard normal density at d1, getting 0.358451, and multiply by S and sqrt(T) to get vega — 158.91×0.358451×0.421998 ≈ 24.0376, so vega per 1% is 0.240377.\n\n\nVega ≈ **0.2404** per 1 percentage-point rise in volatility (S·φ(d₁)·√T ≈ 24.038, divided by 100). Steps:\n\n**Inputs**\n- T = 65/365 = 0.178082, so √T = 0.421998\n- σ√T = 0.564 × 0.421998 = 0.238007\n\n**d₁**\n- ln(S/K) = ln(158.91/148) = ln(1.073716) ≈ 0.071126\n- (r + σ²/2)T = (0.0599 + 0.159048) × 0.178082 ≈ 0.038991\n- d₁ = (0.071126 + 0.038991) / 0.238007 ≈ 0.462662\n\n**Standard normal density at d₁**\n- φ(d₁) = 0.398942 × e^(−0.214056/2) ≈ 0.398942 × 0.898503 ≈ 0.358451\n\n**Vega**\n- Vega = S · φ(d₁) · √T = 158.91 × 0.358451 × 0.421998 ≈ 24.0376 per unit of volatility.\n- Dividing by 100 gives about 0.2404 per percentage point.\n\n{\"answer\": 0.2404, \"confidence\": 92}" }, { "taskId": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 61.799447576373446, "answer": 61.8, "confidence": 88, "grade": "exact", "relError": 0.00000893897353805309, "cost": 0.0234, "reply": "I'm computing the present value of the strike and testing a volatility of 0.6 to estimate d1, given the call price of 3.5195 suggests high implied volatility with T=136/365, S=27.09, K=29, r=0.054.\n\nGetting C=3.4010 against the target, I calculate vega to apply Newton's method, bumping sigma to 0.618 and finding C=3.5197, just 0.0002 above target, so I converge on implied volatility near 0.61797.\n\n\n**Implied volatility ≈ 61.80% per year** (σ ≈ 0.6180). I found it with two Newton (vega) steps on the Black-Scholes call formula. The interpolation step is a linear estimate, so it is good to about ±0.01 percentage points.\n\n**Fixed inputs**\n- T = 136/365 = 0.37260\n- ln(S/K) = ln(27.09/29) = −0.068131\n- rT = 0.020120\n- PV(K) = 29·e^(−0.020120) = 28.4223\n\n**Trial σ = 0.60**\n- σ√T = 0.36625\n- d1 = (−0.068131 + 0.020120 + 0.067068)/0.36625 = 0.05203\n- d2 = 0.05203 − 0.36625 = −0.31422\n- N(d1) = 0.52075, N(d2) = 0.37668\n- C = 27.09·0.52075 − 28.4223·0.37668 = 3.4010, which is 0.1185 below the market price 3.5195.\n- Vega = S·φ(d1)·√T = 27.09·0.3984·0.61041 ≈ 6.588\n- Newton step: Δσ = 0.1185/6.588 ≈ 0.0180, giving σ ≈ 0.618.\n\n**Check σ = 0.618**\n- σ√T = 0.37724\n- d1 = 0.061346, d2 = −0.31589\n- N(d1) = 0.52446, N(d2) = 0.37604\n- C = 14.2076 − 10.6879 = 3.5197\n\nThis is 0.0002 above the target. Correcting by 0.0002/vega moves σ down slightly, to about 0.61797.\n\n{\"answer\": 61.80, \"confidence\": 88}" }, { "taskId": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 20.900010973832025, "answer": 20.9, "confidence": 88, "grade": "exact", "relError": 5.250634576190053e-7, "cost": 0.02624, "reply": "Computing the discount factor for 53 days and deriving the put-call parity relationship, I get the call value around 17.47 by adding the put price to the forward difference, putting the average straddle value near 15.58.\n\nTesting σ=0.21 in Black-Scholes, I compute d1 and d2 using the cumulative normal approximation, getting a call price of 17.547, which is slightly high by 0.073. Using vega to adjust, I refine the volatility estimate down to about 0.20901.\n\n\n**Implied volatility ≈ 20.90% per year.** I found it from a closed-form first guess plus one Newton step on volatility; I did not run a second iteration to confirm convergence.\n\n**Setup**\n- T = 53/365 = 0.145205, so √T = 0.381057.\n- rT = 0.0508 × 0.145205 = 0.0073764.\n- Discounted strike: K·e^(−rT) = 490 × 0.992651 = 486.399.\n\n**Convert the put to a call price with put–call parity**\n- C = P + S − K·e^(−rT) = 13.6926 + 490.18 − 486.399 = 17.474.\n- Working with the call gives the same implied volatility as the put.\n\n**Initial guess**\nThe option is nearly at the money. The average of the call and put prices is about 15.58. Using the at-the-money approximation, this is roughly 0.3989 × S × σ√T, so σ√T ≈ 0.080 and σ ≈ 0.21.\n\n**Price the call at σ = 0.21**\n- σ√T = 0.080022.\n- ln(S/K) = ln(490.18/490) = 0.000367.\n- d₁ = (0.000367 + 0.0073764 + 0.5 × 0.0064035) / 0.080022 = 0.13678.\n- d₂ = d₁ − σ√T = 0.05676.\n- N(d₁) ≈ 0.554397 and N(d₂) ≈ 0.522632.\n- C = 490.18 × 0.554397 − 486.399 × 0.522632 = 271.755 − 254.208 = 17.547.\n\n**Newton correction**\n- The model price is too high by 17.547 − 17.474 = 0.073.\n- Vega = S·φ(d₁)·√T = 490.18 × 0.3952 × 0.381057 ≈ 73.82.\n- Δσ = −0.073 / 73.82 ≈ −0.00099, giving σ ≈ 0.20901.\n\n{\"answer\": 20.90, \"confidence\": 88}" }, { "taskId": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 81.50015130720656, "answer": 81.5, "confidence": 88, "grade": "exact", "relError": 0.0000018565266952024242, "cost": 0.026936, "reply": "I'm setting up the Black-Scholes put pricing equation with S=79.84, K=86, T≈0.441, and trying an initial volatility guess of 0.8 to start iterating toward the given price of 20.922.\n\nTesting sigma=0.8 gives a put value of 20.607, which is short of target, so I use vega to estimate a correction and bump sigma up to roughly 0.815. Recomputing with sigma=0.815 yields a put price of 20.922, matching the target almost exactly — so the implied volatility is about 81.5%.\n\n\nThe implied volatility is about **81.5% per year**. I found it with Newton–Raphson on the Black-Scholes put formula, using hand-rounded normal CDF values.\n\n**Inputs**\n- T = 161/365 = 0.44110, so √T = 0.66415.\n- r = 0.0004, so the discounted strike is K·e^(−rT) = 86 × e^(−0.000176) = 85.9848.\n- ln(S/K) = ln(79.84/86) = −0.074322. Adding rT = 0.000176 gives −0.074146.\n\nThe price is P = K·e^(−rT)·N(−d₂) − S·N(−d₁), where d₁ = [ln(S/K) + rT]/(σ√T) + σ√T/2 and d₂ = d₁ − σ√T.\n\n**First guess, σ = 0.80**\n- σ√T = 0.53132.\n- d₁ = 0.12611 and d₂ = −0.40521.\n- N(−d₂) = 0.65733 and N(−d₁) = 0.44982.\n- P = 85.9848 × 0.65733 − 79.84 × 0.44982 = 56.521 − 35.914 = 20.607.\n\nThis is 0.315 below the market price of 20.922.\n\n**Newton step**\n- Vega = S·φ(d₁)·√T = 79.84 × 0.3958 × 0.66415 ≈ 20.99.\n- Δσ ≈ 0.315 / 20.99 ≈ 0.0150, so the next guess is σ ≈ 0.815.\n\n**Check at σ = 0.815**\n- σ√T = 0.54128.\n- d₁ = 0.13366 and d₂ = −0.40762.\n- N(−d₂) = 0.65822 and N(−d₁) = 0.44683.\n- P = 85.9848 × 0.65822 − 79.84 × 0.44683 = 56.597 − 35.675 = 20.922.\n\nThis matches the market price, so σ ≈ 81.5%.\n\n{\"answer\": 81.5, \"confidence\": 88}" }, { "taskId": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 87.70003216471197, "answer": 87.7, "confidence": 88, "grade": "exact", "relError": 3.6675826876828525e-7, "cost": 0.02808, "reply": "I'm computing the time to expiry and discounted strike, then testing a volatility of 0.8 to work out d1 using the Black-Scholes formula for the put price.\n\nGetting a put value of 70.29 against a target of 75.88, I use vega to adjust the volatility up to roughly 0.877, then refine further and converge on about 87.70 as the implied volatility.\n\n\nImplied volatility ≈ **87.7% per year** (σ ≈ 0.877). I found it by solving the Black-Scholes put price for σ, using one Newton step and then a hand-computed check.\n\n**Inputs**\n- T = 105/365 = 0.287671, so √T = 0.53635.\n- Discounted strike: K·e^(−rT) = 362 · e^(−0.0030493) = 360.898.\n- ln(S / K·e^(−rT)) = ln(340.53 / 360.898) = −0.05810.\n\n**Pricing formula**\n\nP = K·e^(−rT)·N(−d₂) − S·N(−d₁), where\n- d₁ = ln(S / K·e^(−rT)) / (σ√T) + σ√T/2\n- d₂ = d₁ − σ√T\n\n**First guess: σ = 0.80**\n- σ√T = 0.42908, so d₁ = 0.07914 and d₂ = −0.34994.\n- N(0.34994) = 0.63681 and N(−0.07914) = 0.46846.\n- P = 360.898 × 0.63681 − 340.53 × 0.46846 = 229.82 − 159.53 = 70.29.\n- This is below the target 75.8825.\n\n**Newton step**\n- Vega = S·φ(d₁)·√T = 340.53 × 0.3977 × 0.53635 ≈ 72.64.\n- Δσ = (75.88 − 70.29) / 72.64 ≈ 0.077, giving σ ≈ 0.877.\n\n**Check at σ = 0.877**\n- σ√T = 0.47038, so d₁ = 0.11167 and d₂ = −0.35871.\n- N(0.35871) = 0.64009 and N(−0.11167) = 0.45555.\n- P = 360.898 × 0.64009 − 340.53 × 0.45555 = 231.007 − 155.128 = 75.879.\n- The remaining gap of about 0.0035 needs Δσ ≈ +0.00005, which is negligible.\n\nThe normal CDF values were computed by hand, so the last decimal place is approximate.\n\n{\"answer\": 87.70, \"confidence\": 88}" }, { "taskId": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 33.39992865638508, "answer": 33.4, "confidence": 88, "grade": "exact", "relError": 0.0000021360409375049508, "cost": 0.027116, "reply": "I'm computing the time to expiry and discount factor to get the forward value, then using a Brenner-Subrahmanyam approximation to estimate implied volatility from the option price relative to intrinsic value.\n\nThen I refine with a Newton-Raphson iteration: plugging σ=0.335 into Black-Scholes to compute d1, d2, N(d1), N(d2), and the resulting call price of 11.946, comparing it against the market price to get a vega-based correction bringing σ down to about 0.334.\n\n\nImplied volatility ≈ **33.40% per year** (σ ≈ 0.3340). I found it by pricing the call at a trial volatility and making one vega-based (Newton) correction.\n\n**Inputs**\n- T = 161/365 = 0.441096, so √T = 0.66415.\n- Discounted strike: K·e^(−rT) = 143·e^(−0.019×0.441096) = 143·e^(−0.008381) ≈ 141.806.\n\n**Initial guess**\n\nThe Brenner–Subrahmanyam approximation gives 0.4·S·σ√T ≈ 11.908 + 0.573. Here 0.573 is half the gap between the discounted strike and spot, a correction for the option being slightly out of the money. This gives σ√T ≈ 0.2218, so σ ≈ 0.334. I start at σ = 0.335.\n\n**Price at σ = 0.335**\n- σ√T = 0.22249.\n- ln(S/K) = ln(140.66/143) = −0.016499. Adding rT gives −0.008118.\n- d₁ = (−0.008118 + ½·0.22249²)/0.22249 = (−0.008118 + 0.024751)/0.22249 = 0.07476.\n- d₂ = d₁ − σ√T = −0.14773.\n- N(d₁) ≈ 0.529797 and N(d₂) ≈ 0.441277.\n- C = 140.66·0.529797 − 141.806·0.441277 = 74.5223 − 62.5763 ≈ 11.946.\n\n**Newton step**\n- The model price is 11.946 − 11.908 = 0.038 too high.\n- Vega = S·φ(d₁)·√T = 140.66·0.3978·0.66415 ≈ 37.16 per unit of σ.\n- σ ≈ 0.335 − 0.038/37.16 ≈ 0.335 − 0.00102 = 0.33398.\n\nI did not reprice at 0.33398 to confirm. The step is about 0.1 volatility point and vega barely changes over that range, so the remaining error should be well under 1%.\n\n{\"answer\": 33.40, \"confidence\": 88}" }, { "taskId": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 6.438461538461515, "answer": 6.4385, "confidence": 92, "grade": "exact", "relError": 0.0000059737156548775595, "cost": 0.005956, "reply": "With a stop distance of 0.0039, each lot risks 390 USD, so dividing 2511 by 390 gives about 6.44 lots.\n\n\n**Position size ≈ 6.438 standard lots**\n\n1. Stop distance = 0.94593 − 0.94203 = 0.00390, which is 39 pips.\n2. AUD/USD is quoted in USD, so the loss per standard lot at the stop is 100,000 × 0.00390 = 390 USD.\n3. Position size = 2511 ÷ 390 ≈ 6.4385 standard lots.\n\n{\"answer\": 6.4385, \"confidence\": 92}" }, { "taskId": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 12.3258666666662, "answer": 12.32587, "confidence": 95, "grade": "exact", "relError": 2.704340304114844e-7, "cost": 0.006756, "reply": "With a stop distance of 0.150 JPY per unit, dividing 184888 by that gives about 1,232,587 units, which works out to roughly 12.33 lots.\n\n\nThe position size is about **12.33 standard lots** (12.32587).\n\n1. The stop distance is 154.513 − 154.363 = 0.150 JPY per unit of EUR.\n2. Since EUR/JPY is quoted in JPY, each unit loses 0.150 JPY if the stop is hit. To risk 184,888 JPY you need 184,888 ÷ 0.150 ≈ 1,232,586.67 units.\n3. At 100,000 units per standard lot, that is 1,232,586.67 ÷ 100,000 ≈ 12.3259 lots.\n\n{\"answer\": 12.32587, \"confidence\": 95}" }, { "taskId": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 0.4756567164179193, "answer": 0.475657, "confidence": 92, "grade": "exact", "relError": 5.961906368996154e-7, "cost": 0.007136, "reply": "With a stop distance of 0.670 JPY per unit, dividing the risk amount of 31869 gives about 47565.67 units, which converts to roughly 0.4757 lots.\n\n\nThe position size is about **0.4757 standard lots** (≈ 47,566 EUR).\n\n1. **Stop distance per unit:** 191.077 − 190.407 = 0.670 JPY per EUR unit.\n2. **Units:** 31,869 JPY ÷ 0.670 JPY/unit ≈ 47,565.67 units.\n3. **Lots:** 47,565.67 ÷ 100,000 ≈ 0.475657 standard lots.\n\n{\"answer\": 0.475657, \"confidence\": 92}" }, { "taskId": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 5.01704545454548, "answer": 5.01705, "confidence": 92, "grade": "exact", "relError": 9.060022599983643e-7, "cost": 0.006252, "reply": "With a stop distance of 0.44 JPY per unit, dividing 220750 by 0.44 gives roughly 501,705 units, or about 5.02 lots.\n\n\n**Position size ≈ 5.017 standard lots** (about 501,705 units). Because the risk is stated in JPY and USD/JPY is quoted in JPY, no currency conversion is needed.\n\n1. Stop distance per unit: 142.806 − 142.366 = 0.440 JPY per USD.\n2. Units needed: 220,750 ÷ 0.440 = 501,704.545… units.\n3. Standard lots: 501,704.545 ÷ 100,000 = **5.01705 lots**.\n\n{\"answer\": 5.01705, \"confidence\": 92}" }, { "taskId": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 1.6141304347826113, "answer": 1.6141, "confidence": 92, "grade": "exact", "relError": 0.00001885521885678377, "cost": 0.006456, "reply": "Since EUR/USD is quoted in USD, P&L equals units times the price difference of 0.0092. Solving for units gives 1485/0.0092 ≈ 161,413, which converts to about 1.6141 lots.\n\n\nWith EUR/USD, the quote currency is USD, so the profit or loss in USD equals the number of units times the price move. No currency conversion is needed.\n\n1. Stop distance: 0.76163 − 0.75243 = 0.00920 (92 pips).\n2. Units needed for 1485 USD of risk: 1485 / 0.00920 ≈ 161,413 units.\n3. In standard lots: 161,413 / 100,000 ≈ 1.6141 lots.\n\n{\"answer\": 1.6141, \"confidence\": 92}" }, { "taskId": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.023819567833442343, "answer": 0.0238196, "confidence": 92, "grade": "exact", "relError": 0.0000013504257458398619, "cost": 0.004584, "reply": "With a normal distribution, the 97.5% one-tailed critical value is z = 1.959964.\n\nVaR = −mean + z × stdev\n= 0.0003 + 1.959964 × 0.012\n= 0.0003 + 0.023519568\n= 0.023819568\n\nThe 1-day 97.5% VaR is about 0.02382, a loss of roughly 2.38% of portfolio value.\n\n{\"answer\": 0.0238196, \"confidence\": 92}" }, { "taskId": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.08208473980663146, "answer": 0.08208, "confidence": 92, "grade": "exact", "relError": 0.000057742847728156604, "cost": 0.00452, "reply": "The parametric VaR formula given is VaR = −mean + z × stdev.\n\nAt 95% confidence, z = 1.644854.\n\n- z × stdev = 1.644854 × 0.0496 = 0.0815848\n- −mean = 0.0005\n\nVaR = 0.0005 + 0.0815848 = 0.0820848, which is a loss of about 8.21% of portfolio value.\n\n{\"answer\": 0.08208, \"confidence\": 92}" }, { "taskId": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.10032876732331801, "answer": 0.10033, "confidence": 90, "grade": "exact", "relError": 0.000012286373239487742, "cost": 0.00524, "reply": "Under normal returns, the 1-day 99% VaR is\n\nVaR = −μ + z·σ\n\nWith z₀.₉₉ = 2.326348, μ = 0.0011 and σ = 0.0436:\n\n- z·σ = 2.326348 × 0.0436 = 0.1014288\n- VaR = 0.1014288 − 0.0011 = 0.1003288\n\nThe 1-day 99% VaR is about **0.10033**, a loss of roughly 10.03% of portfolio value.\n\n{\"answer\": 0.10033, \"confidence\": 90}" }, { "taskId": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.07275151453115127, "answer": 0.0727515, "confidence": 88, "grade": "exact", "relError": 1.9973675272628119e-7, "cost": 0.00442, "reply": "VaR = -μ + z·σ, with z = 2.326348 for 99% confidence.\n\nz·σ = 2.326348 × 0.0308 = 0.0716515\n\nVaR = 0.0011 + 0.0716515 = 0.0727515\n\nThe 1-day 99% VaR is about 0.07275 of portfolio value (roughly a 7.28% loss).\n\n{\"answer\": 0.0727515, \"confidence\": 88}" }, { "taskId": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.11150152403343133, "answer": 0.1115015, "confidence": 92, "grade": "exact", "relError": 2.1554352316467348e-7, "cost": 0.004736, "reply": "Using the given formula, VaR = −mean + z × stdev.\n\nThe one-sided 99% z-value for a normal distribution is z = 2.326348.\n\nz × stdev = 2.326348 × 0.0475 = 0.11050153\n\nVaR = 0.001 + 0.11050153 = 0.11150153\n\nThe 1-day 99% VaR is a loss of about 0.1115 of portfolio value (11.15%).\n\n{\"answer\": 0.1115015, \"confidence\": 92}" }, { "taskId": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.04404168455292251, "answer": 0.04404, "confidence": 88, "grade": "exact", "relError": 0.00003824905744656407, "cost": 0.012896, "reply": "Working through the expected shortfall formula, I compute the normal density at the z-score, divide by the tail probability, and scale by sigma to get roughly 0.0458, then subtract the mean term to land on about 0.044.\n\n\nThe 1-day 99% parametric Expected Shortfall is about **0.04404**, a loss of roughly 4.40% of portfolio value. This assumes the standard normal-distribution ES formula and takes the mean's offset into account.\n\n**Formula.** For returns distributed N(μ, σ²) at confidence level c:\n\nES = −μ + σ · φ(z_c) / (1 − c)\n\nHere z_c is the standard normal quantile and φ is the standard normal density.\n\n**Values.**\n- μ = 0.0018, σ = 0.0172, c = 0.99, so 1 − c = 0.01.\n- z₀.₉₉ = 2.326348.\n\n**Density at the quantile.**\nφ(2.326348) = e^(−2.326348²/2) / √(2π) = e^(−2.705947) / 2.506628 = 0.066807 / 2.506628 ≈ 0.026652\n\n**Tail factor.** φ(z)/(1 − c) = 0.026652 / 0.01 = 2.6652\n\n**Result.**\nES = −0.0018 + 0.0172 × 2.6652 = −0.0018 + 0.045842 ≈ 0.044042\n\n{\"answer\": 0.04404, \"confidence\": 88}" }, { "taskId": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.02233967069948375, "answer": 0.02234, "confidence": 88, "grade": "exact", "relError": 0.000014740616398397254, "cost": 0.010196, "reply": "Computing expected shortfall: with phi(1.6449)≈0.10314 divided by 0.05 gives about 2.06271, multiplying by sigma yields roughly 0.0210396, then adding the mean term 0.0013 gives a final ES of about 0.0223396.\n\n\n**Answer: the 1-day 95% CVaR is about 0.02234 (2.234% of portfolio value).**\n\nFor normally distributed returns, the parametric Expected Shortfall at confidence α is the expected loss given that the loss exceeds the VaR:\n\nES = −μ + σ · φ(z_α) / (1 − α)\n\nHere φ is the standard normal density and z_α is the α-quantile.\n\n1. For α = 0.95, z = 1.6449, so φ(1.6449) ≈ 0.10314.\n2. Dividing by the tail probability: 0.10314 / 0.05 ≈ 2.0627.\n3. Scaling by volatility: σ × 2.0627 = 0.0102 × 2.0627 ≈ 0.021040.\n4. Adjusting for the mean: because μ = −0.0013 is negative, −μ = +0.0013. This increases the loss, giving ES ≈ 0.021040 + 0.0013 ≈ 0.022340.\n\n{\"answer\": 0.02234, \"confidence\": 88}" }, { "taskId": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.09524684038587, "answer": 0.09525, "confidence": 90, "grade": "exact", "relError": 0.000033172902294700856, "cost": 0.013276, "reply": "Computing the expected shortfall: with z=2.326348, I find phi(z)≈0.066807, divide by 2.506628 to get the normal density value, then divide by (1-c)=0.01, giving ES/sigma ≈ 2.6652.\n\nMultiplying by sigma=0.0366 gives 0.097547, then subtracting the mean term 0.0023 gives a final ES of approximately 0.095247.\n\n\nThe 1-day 99% Expected Shortfall is about **0.09525**, i.e. an expected loss of roughly **9.52%** of portfolio value on the worst 1% of days.\n\nFor normally distributed returns, the 99% Expected Shortfall as a positive loss is:\n\nES = −μ + σ · φ(z₀.₉₉) / (1 − 0.99)\n\n**Step 1 — Find the quantile and its density.**\n- The 99% standard normal quantile is z = 2.32635.\n- The normal density there is φ(z) = e^(−z²/2) / √(2π).\n- With z²/2 ≈ 2.70595, this gives φ(z) = 0.066807 / 2.506628 ≈ 0.026652.\n\n**Step 2 — Scale by the tail probability.**\n- φ(z) / 0.01 ≈ 2.6652.\n\n**Step 3 — Plug in μ = 0.0023 and σ = 0.0366.**\n- ES = −0.0023 + 0.0366 × 2.6652\n- ES = −0.0023 + 0.097547\n- ES ≈ 0.09525\n\n{\"answer\": 0.09525, \"confidence\": 90}" }, { "taskId": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.12233732788143936, "answer": 0.12234, "confidence": 88, "grade": "exact", "relError": 0.000021842217799887472, "cost": 0.014964, "reply": "I'm computing the expected shortfall formula using the normal density at the 97.5% quantile, getting phi(z)/(1-alpha) ≈ 2.3378, then multiplying by sigma and subtracting mu to arrive at roughly 0.1223. I'm double-checking the precision of phi(1.96) by accounting for the small adjustment at z=1.959964.\n\n\nTreat the loss as \\(L = -R\\), where returns \\(R\\) are normal with mean \\(\\mu\\) and standard deviation \\(\\sigma\\). The parametric Expected Shortfall at confidence \\(\\alpha\\) is\n\n$$\\text{ES}_\\alpha = -\\mu + \\sigma\\,\\frac{\\phi(z_\\alpha)}{1-\\alpha}$$\n\nHere \\(\\phi\\) is the standard normal density and \\(z_\\alpha\\) is the standard normal quantile at \\(\\alpha\\).\n\n**Inputs**\n- \\(\\alpha = 0.975\\), so \\(z_{0.975} = 1.959964\\).\n- The density at that point is \\(\\phi(1.959964) = \\frac{1}{\\sqrt{2\\pi}}e^{-z^2/2} \\approx 0.058445\\).\n- Dividing by the tail probability gives \\(\\phi(z)/(1-\\alpha) = 0.058445 / 0.025 \\approx 2.33780\\).\n\n**Calculation**\n- The volatility term is \\(\\sigma \\times 2.33780 = 0.0531 \\times 2.33780 \\approx 0.124137\\).\n- Subtracting the mean gives \\(\\text{ES} = 0.124137 - 0.0018 \\approx 0.122337\\).\n\nThe 1-day 97.5% Expected Shortfall is about **0.1223** of portfolio value, or roughly 12.23%.\n\n{\"answer\": 0.12234, \"confidence\": 88}" }, { "taskId": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.1205670856534704, "answer": 0.120565, "confidence": 88, "grade": "exact", "relError": 0.000017298696896284908, "cost": 0.007904, "reply": "ES = -μ + σ·φ(z)/(1−α), with z = 1.959964 for α = 0.975.\n\n1. φ(1.959964) = 0.398942 · e^(−1.920730) = 0.398942 · 0.146504 ≈ 0.058445.\n (e^(−1.92073): e^(−1.9) = 0.149569, e^(−0.02073) = 0.979483, so the product is 0.146500.)\n Recomputing gives φ ≈ 0.0584440.\n2. Divide by the tail probability: 0.058444 / 0.025 = 2.337760.\n3. Scale by σ: 2.33776 × 0.0523 = 0.122265.\n4. Subtract the mean: 0.122265 − 0.0017 = 0.120565.\n\nThe 1-day 97.5% parametric Expected Shortfall is about 0.1206 of portfolio value.\n\n{\"answer\": 0.120565, \"confidence\": 88}" }, { "taskId": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 1.6781883884002948, "answer": 1.6782, "confidence": 92, "grade": "exact", "relError": 0.00000691912766489489, "cost": 0.019144, "reply": "I'm adding up the deviations to get a total of 0.2537, giving a mean of about 0.0211, and now I'm starting to compute the squared deviations for variance.\n\nWorking through the sum of squares, I get 263.1301 (×1e-4), subtracting out the mean correction factor to get a sample variance of roughly 19.045e-4, so the standard deviation comes out to about 0.04364. Dividing the mean by this sd gives a monthly Sharpe ratio of about 0.4845, which annualizes to roughly 1.678 when multiplied by the square root of 12.\n\n\nAnnualized Sharpe ratio ≈ **1.678**\n\n1. **Mean monthly return.** The 12 returns sum to 0.2537, so the mean is 0.2537 / 12 = 0.0211417.\n\n2. **Sample standard deviation.**\n - The sum of the squared returns is 0.02631301.\n - Subtract n·mean², which is 0.2537² / 12 = 0.00536364.\n - That leaves a sum of squared deviations of 0.02094937.\n - Divide by n − 1 = 11 to get a variance of 0.00190449.\n - The standard deviation is √0.00190449 = 0.0436404.\n\n3. **Monthly Sharpe ratio.** With a risk-free rate of 0, it is 0.0211417 / 0.0436404 = 0.484453.\n\n4. **Annualize.** Multiply by √12 = 3.464102: 0.484453 × 3.464102 = 1.67822.\n\n{\"answer\": 1.6782, \"confidence\": 92}" }, { "taskId": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": -0.48328609065675004, "answer": -0.4833, "confidence": 88, "grade": "exact", "relError": 0.00002878076468342148, "cost": 0.020604, "reply": "I'm adding up these values sequentially and the running total ends up very close to zero, giving an average of about -0.00.\n\nComputing the squared deviations gives a sum around 171.43, and after subtracting out the mean contribution I get a sample variance leading to a standard deviation near 0.0391. Dividing the mean by this standard deviation gives a monthly Sharpe ratio of about -0.1395, which annualizes to roughly -0.483.\n\n\nThe annualized Sharpe ratio is about **-0.4833**. It is the monthly Sharpe ratio (mean return over sample standard deviation, risk-free rate 0) multiplied by √12.\n\n**Mean monthly return**\n- The twelve returns sum to -0.0654.\n- The mean is -0.0654 / 12 = -0.00545.\n\n**Sample standard deviation**\n- The sum of squared returns is 0.01714286.\n- Subtracting n·mean² = 12 × 0.00545² = 0.00035643 gives a sum of squared deviations of 0.01678643.\n- The sample variance is 0.01678643 / 11 = 0.001526039.\n- The standard deviation is √0.001526039 ≈ 0.039065.\n\n**Sharpe ratio**\n- Monthly Sharpe = -0.00545 / 0.039065 ≈ -0.13951.\n- Annualized Sharpe = -0.13951 × √12 = -0.13951 × 3.4641 ≈ -0.4833.\n\n{\"answer\": -0.4833, \"confidence\": 88}" }, { "taskId": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 2.241475231769422, "answer": 2.2415, "confidence": 88, "grade": "exact", "relError": 0.0000110499684433581, "cost": 0.026816, "reply": "I'm summing up the values to get a total of 0.2742, then computing the mean at 0.02285, and now starting to work out each deviation from that mean.\n\nContinuing through the squared deviations, I get a sum of 0.01371746, giving a variance of 0.00124704 and standard deviation of about 0.0353135. That makes the monthly Sharpe ratio 0.64706, which annualizes to 2.2415.\n\n\nThe annualized Sharpe ratio is about **2.2415**. Hand arithmetic (no tools) might move the 4th–5th digit slightly, but not by anywhere near 1%.\n\n**Mean monthly return**\n\nThe 12 returns sum to 0.2742, so the mean is 0.2742 / 12 = 0.02285.\n\n**Sample standard deviation**\n\nThe squared deviations from the mean are:\n\n| Return | Deviation | Squared |\n|---|---|---|\n| 0.0236 | 0.00075 | 0.00000056 |\n| 0.0642 | 0.04135 | 0.0017098 |\n| 0.0200 | −0.00285 | 0.0000081 |\n| −0.0290 | −0.05185 | 0.0026884 |\n| 0.0274 | 0.00455 | 0.0000207 |\n| 0.0497 | 0.02685 | 0.0007209 |\n| −0.0065 | −0.02935 | 0.0008614 |\n| 0.0325 | 0.00965 | 0.0000931 |\n| 0.0352 | 0.01235 | 0.0001525 |\n| 0.0197 | −0.00315 | 0.0000099 |\n| 0.0796 | 0.05675 | 0.0032206 |\n| −0.0422 | −0.06505 | 0.0042315 |\n\n- The squared deviations sum to 0.0137175.\n- The sample variance is 0.0137175 / 11 = 0.00124704.\n- The standard deviation is √0.00124704 ≈ 0.035314.\n\n**Sharpe ratio**\n\n- Monthly Sharpe ratio: 0.02285 / 0.035314 ≈ 0.64706.\n- Annualized: 0.64706 × √12 = 0.64706 × 3.4641 ≈ 2.2415.\n\n{\"answer\": 2.2415, \"confidence\": 88}" }, { "taskId": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 1.9830100176589147, "answer": 1.983, "confidence": 92, "grade": "exact", "relError": 0.000005051743977796405, "cost": 0.018512, "reply": "I'm summing up the values to get a total of about 0.311, giving a mean around 0.0259, and now I'm computing each term's deviation from that mean.\n\nAdding up the squared deviations gives roughly 0.02252, which divided by 11 is about 0.002047, so the standard deviation comes to roughly 0.04524. Dividing the mean by this standard deviation gives a monthly Sharpe ratio of about 0.572, which annualizes to approximately 1.983.\n\n\nThe annualized Sharpe ratio is about **1.983**.\n\n**Mean monthly return**\nThe 12 returns sum to 0.3108, so the mean is 0.3108 / 12 = 0.0259.\n\n**Sample standard deviation**\nThe squared deviations from the mean sum to 0.02251768.\nDividing by n − 1 = 11 gives a variance of 0.00204706.\nThe standard deviation is √0.00204706 ≈ 0.045245.\n\n**Monthly Sharpe ratio** (risk-free rate = 0)\n0.0259 / 0.045245 ≈ 0.57245\n\n**Annualized Sharpe ratio**\n0.57245 × √12 = 0.57245 × 3.46410 ≈ 1.983\n\n{\"answer\": 1.983, \"confidence\": 92}" }, { "taskId": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 1.3915400475486268, "answer": 1.3915, "confidence": 88, "grade": "exact", "relError": 0.000028779300098047822, "cost": 0.020624, "reply": "I'm summing up the deviations to get a running total of 0.1695, then computing the mean at 0.014125 before moving on to calculate the sum of squares for variance.\n\nWorking through the squared deviations, I get a sum of about 159.9483, then subtract the n*mean² correction of 23.941875 to get a sum of squares near 136.006, dividing by 11 degrees of freedom gives a variance around 12.364 and standard deviation about 3.5163. From there I compute the monthly Sharpe ratio as 0.4017 and annualize it by multiplying by √12 to get roughly 1.3915.\n\n\n**Annualized Sharpe ratio ≈ 1.3915**\n\nWith a risk-free rate of 0, the Sharpe ratio is the mean return divided by its standard deviation.\n\n1. **Mean.** The 12 returns sum to 0.1695, so the mean monthly return is 0.1695 / 12 = 0.014125.\n\n2. **Sample standard deviation (n − 1).** Working in percent to keep the numbers readable:\n - The sum of squared returns is 159.9483.\n - The correction term is n·mean² = 16.95² / 12 = 23.9419.\n - The sum of squared deviations is 159.9483 − 23.9419 = 136.0064.\n - The variance is 136.0064 / 11 = 12.3642.\n - The standard deviation is √12.3642 ≈ 3.5163%, which is 0.035163 as a decimal.\n\n3. **Monthly Sharpe ratio.** 0.014125 / 0.035163 ≈ 0.40170.\n\n4. **Annualize.** 0.40170 × √12 = 0.40170 × 3.4641 ≈ 1.3915.\n\n{\"answer\": 1.3915, \"confidence\": 88}" }, { "taskId": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-opus-5-5", "expected": 0.4848484848484847, "answer": 0.4848, "confidence": 92, "grade": "exact", "relError": 0.00009999999999966636, "cost": 0.0081, "reply": "Calculating the Kelly fraction for the binary contract gives about 0.485, taking the edge between my estimated probability and the price divided by the payoff denominator.\n\n\nFull Kelly says to stake **16/33 ≈ 0.4848 (about 48.5%)** of your bankroll.\n\n**Net odds.** Each contract costs $0.67 and pays $1, so a win nets $0.33 per $0.67 staked:\n\nb = 0.33 / 0.67\n\n**Kelly formula.** With p = 0.83 and q = 0.17:\n\nf* = (b·p − q) / b = p − q/b = 0.83 − 0.17 × (0.67 / 0.33)\n\nThis simplifies to the standard form for a binary contract bought at price c:\n\nf* = (p − c) / (1 − c) = (0.83 − 0.67) / 0.33 = 0.16 / 0.33 = 16/33 ≈ 0.4848\n\n{\"answer\": 0.4848, \"confidence\": 92}" }, { "taskId": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-opus-5-5", "expected": 0.28985507246376807, "answer": 0.2899, "confidence": 90, "grade": "exact", "relError": 0.0001550000000001245, "cost": 0.00762, "reply": "Applying the Kelly formula for this binary contract, I get f = (0.51-0.31)/0.69 ≈ 0.29, so roughly 29% of bankroll.\n\n\nFull Kelly says to stake about **0.2899** of your bankroll (roughly 29%).\n\n**Odds.** Buying a contract at price c = $0.31 risks $0.31 to win $0.69 net. The net odds are therefore b = 0.69 / 0.31.\n\n**Kelly formula.** With p = 0.51 and q = 1 − p = 0.49, the Kelly fraction is\n\nf* = (bp − q) / b.\n\nFor a binary contract this simplifies to\n\nf* = (p − c) / (1 − c).\n\n**Calculation.** f* = (0.51 − 0.31) / (1 − 0.31) = 0.20 / 0.69 ≈ 0.2899.\n\n{\"answer\": 0.2899, \"confidence\": 90}" }, { "taskId": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-opus-5-5", "expected": 0.09859154929577463, "answer": 0.09859, "confidence": 92, "grade": "exact", "relError": 0.000015714285714109126, "cost": 0.00746, "reply": "Applying Kelly's formula for this binary contract, I get roughly 9.86% of bankroll: (0.36-0.29)/(1-0.29).\n\n\nFull Kelly says to stake about **9.859%** of your bankroll (≈ 0.09859).\n\n**Setting up the bet.**\n- Buying one contract at $0.29 risks $0.29 to win $0.71.\n- The net odds are b = 0.71 / 0.29.\n- With p = 0.36 and q = 0.64, the Kelly formula is f* = p − q/b.\n\n**Simplifying for a binary contract.** Substituting b, this reduces to\n\nf* = (p − price) / (1 − price)\n\n**Computing it.**\n\nf* = (0.36 − 0.29) / 0.71 = 0.07 / 0.71 ≈ 0.098592\n\n{\"answer\": 0.09859, \"confidence\": 92}" }, { "taskId": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-opus-5-5", "expected": 0.8333333333333334, "answer": 0.8333, "confidence": 88, "grade": "exact", "relError": 0.00003999999999999559, "cost": 0.01008, "reply": "Computing Kelly fraction for this binary contract: (0.86-0.16)/0.84 comes out to about 0.833.\n\n\nFor a binary contract bought at price $c$ that pays $1 if the event happens, the full Kelly fraction is\n\n$$f^* = \\frac{p - c}{1 - c}$$\n\nwhere $p$ is your estimated probability of the event.\n\n**Why this formula holds:**\n- If you win, each $0.16 staked returns $1, so the net odds are $b = \\frac{1-c}{c} = \\frac{0.84}{0.16} = 5.25$.\n- The Kelly fraction is $f^* = p - \\frac{q}{b}$, with $q = 1 - p$.\n- Substituting $b$ gives $f^* = p - \\frac{(1-p)\\,c}{1-c} = \\frac{p - c}{1 - c}$.\n\n**Plugging in $p = 0.86$, $c = 0.16$:**\n\n$$f^* = \\frac{0.86 - 0.16}{1 - 0.16} = \\frac{0.70}{0.84} \\approx 0.8333$$\n\nFull Kelly says to stake about **83.33% of your bankroll** on this contract.\n\n{\"answer\": 0.8333, \"confidence\": 88}" }, { "taskId": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-opus-5-5", "expected": 0.576271186440678, "answer": 0.5763, "confidence": 92, "grade": "exact", "relError": 0.00005000000000002715, "cost": 0.0073, "reply": "Applying the Kelly formula for this binary contract, I get f = (0.75-0.41)/0.59 ≈ 0.576, suggesting a bet size of about 57.6% of the bankroll.\n\n\n**Answer: about 0.5763 of your bankroll (≈ 57.6%).**\n\nUse the Kelly formula for a binary contract.\n\nBuying at price $c$ risks $c$ to win $1 - c$, so the net odds are $b = (1-c)/c$.\n\nThe Kelly fraction is\n\n$$f^* = \\frac{bp - q}{b} = \\frac{p - c}{1 - c}.$$\n\nWith $p = 0.75$ and $c = 0.41$:\n\n$$f^* = \\frac{0.75 - 0.41}{1 - 0.41} = \\frac{0.34}{0.59} \\approx 0.5763$$\n\n{\"answer\": 0.5763, \"confidence\": 92}" }, { "taskId": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 24920.824742268043, "answer": 24920.82, "confidence": 88, "grade": "exact", "relError": 1.9029338285041182e-7, "cost": 0.012044, "reply": "Let Q be the position size in USD, E = 24173.2 the entry price and P the liquidation price.\n\n**Quantities in BTC**\n- Position value at entry = Q/E.\n- Initial margin = Q/(25E) = 0.04·Q/E.\n- Maintenance margin = 0.01·Q/E.\n- Unrealized PnL of the short = Q(1/E − 1/P).\n\n**Liquidation condition**\n\nMargin + PnL = maintenance margin:\n\n0.04·Q/E + Q/E − Q/P = 0.01·Q/E\n\nDividing by Q and collecting terms:\n\n1.03/E = 1/P, so P = E/1.03.\n\n**Calculation**\n\nP = 24173.2 / 1.03 ≈ 23469.13 (check: 1.03 × 23469.13 ≈ 24173.20).\n\nThe result is below the entry price, which cannot be right for a short: a short loses money as the price rises. I made a sign error in the PnL.\n\n**Corrected PnL**\n\nThe short's PnL in BTC is Q(1/P − 1/E). This is negative when P > E, as it should be.\n\nThe condition becomes:\n\n0.04/E + 1/P − 1/E = 0.01/E\n\n1/P = 0.97/E, so P = E/0.97.\n\n**Calculation**\n\nP = 24173.2 / 0.97 ≈ 24920.82 (check: 0.97 × 24920.82 = 24173.20).\n\nThe liquidation price is about **24920.82**.\n\n{\"answer\": 24920.82, \"confidence\": 88}" }, { "taskId": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 53342.9702970297, "answer": 53342.97, "confidence": 88, "grade": "exact", "relError": 5.5683007055427045e-9, "cost": 0.011344, "reply": "For the inverse long position, I'm working out PnL and margin formulas, then solving for the liquidation price P where the loss condition equals 1% margin. Setting up 1/P = 1.01/E gives P = E/1.01 ≈ 53343.0.\n\n\n**Liquidation price ≈ 53342.97**\n\nLet Q be the position size in USD, E = 53876.4 the entry price, and P the liquidation price.\n\n1. Position value in BTC at entry is Q/E.\n2. Initial margin is (Q/E)/50 = 0.02·Q/E.\n3. For an inverse long, unrealized PnL in BTC is Q(1/E − 1/P).\n4. Maintenance margin is 0.01·Q/E.\n\nLiquidation happens when margin plus PnL equals maintenance margin:\n\n0.02·Q/E + Q/E − Q/P = 0.01·Q/E\n\nDivide through by Q and rearrange:\n\n1/P = (1 + 0.02 − 0.01)/E = 1.01/E\n\nSo P = E/1.01 = 53876.4/1.01 ≈ 53342.97.\n\n{\"answer\": 53342.97, \"confidence\": 88}" }, { "taskId": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 53194.15525114154, "answer": 53194.16, "confidence": 88, "grade": "exact", "relError": 8.927406475571944e-8, "cost": 0.012132, "reply": "For the inverse short position, I set margin as Q/(10E) and compute PnL as Q(1/E - 1/P), then solve for the liquidation condition where margin plus PnL equals 0.5% of position value. This gives 1/P = 1.095/E, so P equals E/1.095, which works out to roughly 53,194.\n\n\n**Liquidation price ≈ 53194.16**\n\n**Setup.** Let Q be the position size in USD contracts and E = 58247.6 the entry price.\n\n- Position value at entry, in BTC: Q/E\n- Initial margin: Q/(10E)\n- Maintenance margin: 0.005·Q/E\n- Unrealized PnL of an inverse long at price P, in BTC: Q·(1/E − 1/P)\n\n**Liquidation condition.** Liquidation happens when margin plus unrealized PnL equals the maintenance margin:\n\nQ/(10E) + Q(1/E − 1/P) = 0.005·Q/E\n\nDivide through by Q:\n\n0.1/E + 1/E − 1/P = 0.005/E\n\nSo 1/P = (1 + 0.1 − 0.005)/E = 1.095/E.\n\n**Result.**\n\nP = E/1.095 = 58247.6/1.095 ≈ 53194.155\n\n{\"answer\": 53194.16, \"confidence\": 88}" }, { "taskId": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 170355.2216748768, "answer": 170355.22, "confidence": 88, "grade": "exact", "relError": 9.831672864603095e-9, "cost": 0.014904, "reply": "Working out the inverse contract math: for a short position sized Q in USD with entry price E, position value is Q/E and margin is a third of that. For PnL, I need to get the sign right—longs gain when price falls below entry in the 1/E - 1/P sense, so shorts should profit the opposite way, giving PnL = Q(1/P - 1/E).\n\nSetting margin plus PnL equal to 1% of the position value and solving for the liquidation price P: I get 1/P ≈ 0.676667/E, so P ≈ E/0.676667. Plugging in E = 115273.7 gives P ≈ 170355.22.\n\n\nThe liquidation price is about **170355.22**.\n\n**Setup.** Let Q be the position size in USD, E = 115273.7 the entry price, and P the liquidation price.\n\n- Position value at entry, in BTC: Q/E\n- Initial margin: Q/(3E)\n- Maintenance margin: 0.01 · Q/E\n- Unrealized PnL of an inverse short: Q·(1/P − 1/E). This is negative when the price rises above E.\n\n**Liquidation condition.** Margin plus unrealized PnL equals maintenance margin:\n\nQ/(3E) + Q·(1/P − 1/E) = 0.01·Q/E\n\nDivide by Q and solve for 1/P:\n\n1/P = (1 − 1/3 + 0.01)/E = 0.676667/E\n\n**Liquidation price.**\n\nP = E / 0.676667 = 3E/2.03 = 345821.1/2.03 ≈ 170355.22\n\n{\"answer\": 170355.22, \"confidence\": 88}" }, { "taskId": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 115003.24607329845, "answer": 115003.25, "confidence": 88, "grade": "exact", "relError": 3.414426711223029e-8, "cost": 0.014992, "reply": "I'm working out the inverse contract math: position value at entry equals Q/E, margin is 5% of that, and the short's PnL in BTC terms comes from the difference between 1/P and 1/E, which turns positive as price falls below entry.\n\nSetting the margin plus PnL equal to the maintenance threshold, I solve for liquidation price: 1/P works out to 0.955/E, giving P ≈ 109828.1/0.955 ≈ 115003.25.\n\n\n**Liquidation price ≈ 115003.25**\n\nLet Q be the position size in USD, E = 109828.1 the entry price, and P the liquidation price. For an inverse contract, the position value in BTC is Q/price.\n\n- Position value at entry: Q/E BTC.\n- Initial margin: (Q/E) / 20 = 0.05·Q/E.\n- Maintenance margin: 0.005·Q/E.\n- A short makes money when the price falls, so its unrealized PnL in BTC is Q(1/P − 1/E).\n\nLiquidation happens when margin plus PnL equals maintenance margin:\n\n0.05·Q/E + Q/P − Q/E = 0.005·Q/E\n\nDivide through by Q:\n\n1/P = (1 − 0.05 + 0.005)/E = 0.955/E\n\nSo P = E / 0.955 = 109828.1 / 0.955 ≈ 115003.25.\n\n{\"answer\": 115003.25, \"confidence\": 88}" }, { "taskId": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 2398.0256188582507, "answer": 2398.03, "confidence": 92, "grade": "exact", "relError": 0.0000018269787090834882, "cost": 0.014692, "reply": "I'm computing the weighted harmonic mean by dividing total USD by the sum of each USD amount divided by its price, working through the division for each of the three price tiers in turn.\n\nSumming those quotients to about 35.863, then dividing the total USD of 86000 by that sum to get a final weighted price of roughly 2398.03.\n\n\n**Average entry ≈ 2398.03**\n\nOn an inverse contract, a position of Q USD entered at price P has ETH PnL = Q·(1/P − 1/P_exit). So the combined entry price is the USD-weighted harmonic mean of the fill prices:\n\nP_avg = ΣQ / Σ(Q/P)\n\n**ETH value of each fill (Q/P):**\n- 7000 / 2309.36 ≈ 3.031143\n- 34000 / 3563.54 ≈ 9.541074\n- 36000 / 1854.34 ≈ 19.413915\n- 9000 / 2321.56 ≈ 3.876704\n\n**Totals:**\n- Σ(Q/P) ≈ 35.862836 ETH\n- ΣQ = 86000 USD\n\n**Result:**\nP_avg = 86000 / 35.862836 ≈ 2398.03\n\n{\"answer\": 2398.03, \"confidence\": 92}" }, { "taskId": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 3712.5303509582454, "answer": 3712.53, "confidence": 92, "grade": "exact", "relError": 9.453343462758414e-8, "cost": 0.013316, "reply": "I'm computing the harmonic mean by dividing total volume by the sum of each segment's ratio, working through the division for both portions to get roughly 9.64 and 3.83.\n\nAdding those together gives about 13.47, and dividing 50000 by that sum yields a harmonic mean of roughly 3712.53.\n\n\n**Average entry ≈ 3712.53**\n\nOn an inverse contract, the ETH PnL of a long of $Q$ USD entered at price $P$ is $Q\\left(\\frac{1}{P} - \\frac{1}{\\text{exit}}\\right)$. For a single entry price to reproduce the combined PnL at every exit price, the $\\frac{1}{P}$ terms must match. So the average entry is the USD-weighted harmonic mean of the fill prices:\n\n$$P_{avg} = \\frac{\\sum Q_i}{\\sum Q_i / P_i}$$\n\n**Working:**\n\n- 34000 / 3528.19 ≈ 9.636669 ETH\n- 16000 / 4176.2 ≈ 3.831234 ETH\n- Sum ≈ 13.467903 ETH\n- Total size = 50000 USD\n\n$$P_{avg} = \\frac{50000}{13.467903} \\approx 3712.53$$\n\n{\"answer\": 3712.53, \"confidence\": 92}" }, { "taskId": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 2495.29829080704, "answer": 2495.3, "confidence": 92, "grade": "exact", "relError": 6.849653873569959e-7, "cost": 0.016972, "reply": "I'm calculating the harmonic mean weighted by USD amounts, dividing each USD value by its price and summing those ratios to get the denominator.\n\nSumming the ratios gives 47.288936, and dividing the total 118000 by that sum yields approximately 2495.298 as the weighted harmonic mean.\n\n\nThe average entry price is about **2495.30**.\n\nOn an inverse contract, the ETH PnL of a fill of size Q (in USD) bought at price P and closed at exit X is:\n\nPnL = Q × (1/P − 1/X)\n\nSumming over all fills, the combined PnL is:\n\nΣ(Q/P) − (ΣQ)/X\n\nA single entry price P_avg reproduces this at every X only if ΣQ / P_avg = Σ(Q/P). So the entry price is the USD-weighted harmonic mean:\n\nP_avg = ΣQ / Σ(Q/P)\n\n**ETH value of each fill (Q/P):**\n\n| Fill | Q (USD) | Price | Q/P (ETH) |\n|---|---|---|---|\n| 1 | 41000 | 3517.25 | 11.656835 |\n| 2 | 10000 | 2976.86 | 3.359244 |\n| 3 | 27000 | 4229.06 | 6.384398 |\n| 4 | 40000 | 1545.09 | 25.888459 |\n| **Total** | **118000** | | **47.288936** |\n\n**Average entry:**\n\nP_avg = 118000 / 47.288936 ≈ 2495.30\n\n{\"answer\": 2495.30, \"confidence\": 92}" }, { "taskId": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 2610.72093572179, "answer": 2610.72, "confidence": 90, "grade": "exact", "relError": 3.584150941701721e-7, "cost": 0.016312, "reply": "I'm calculating the harmonic mean weighted by dividing total USD by the sum of each allocation's USD-to-price ratio, working through each term's division individually.\n\nAfter computing the fourth term (6.154999), I sum all four ratios to get 22.599123, then divide the total 59000 by that sum, arriving at a weighted harmonic mean of approximately 2610.72.\n\n\nFor an inverse contract, the ETH PnL of a fill of size Q USD entered at price P and exited at X is\n\nQ · (1/P − 1/X)\n\nTo give the same PnL at every exit price X, the combined entry price P̄ must satisfy\n\nΣ Q/P̄ = Σ Q/Pᵢ, so P̄ = ΣQ / Σ(Qᵢ/Pᵢ).\n\nThis is the USD-weighted harmonic mean of the fill prices.\n\n**ETH value of each fill (Qᵢ/Pᵢ):**\n\n| Size (USD) | Price | ETH value |\n|---|---|---|\n| 10,000 | 3282.44 | 3.046514 |\n| 14,000 | 2147.97 | 6.517782 |\n| 12,000 | 1744.23 | 6.879828 |\n| 23,000 | 3736.80 | 6.154999 |\n\nThe ETH values sum to 22.599123. The total size is 59,000 USD.\n\n**Average entry:**\n\nP̄ = 59,000 / 22.599123 ≈ 2610.72\n\n{\"answer\": 2610.72, \"confidence\": 90}" }, { "taskId": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 3424.594098647905, "answer": 3424.59, "confidence": 92, "grade": "exact", "relError": 0.000001196827357271257, "cost": 0.017092, "reply": "I'm computing the weighted harmonic mean by dividing each USD amount by its price and summing those ratios, working through each term one at a time.\n\nAdding them all up gives roughly 32.7046, then dividing the total USD of 112000 by that sum yields a weighted average price of approximately 3424.59.\n\n\n**Average entry ≈ 3424.59**\n\nOn an inverse contract, PnL in ETH for a long of size Q (USD) is Q·(1/entry − 1/exit). Adding the fills gives Σ Qᵢ/Pᵢ − (ΣQᵢ)/exit.\n\nFor one entry price to give the same PnL at every exit, it must satisfy ΣQᵢ / P_avg = Σ Qᵢ/Pᵢ. So the average entry is the USD-weighted harmonic mean of the fill prices:\n\nP_avg = ΣQᵢ / Σ(Qᵢ/Pᵢ)\n\n**ETH value of each fill (Qᵢ/Pᵢ):**\n\n| Fill (USD) | Price | ETH |\n|---|---|---|\n| 31000 | 3315.48 | 9.350079 |\n| 36000 | 3410.8 | 10.554709 |\n| 40000 | 3481.71 | 11.488607 |\n| 5000 | 3813.27 | 1.311211 |\n| **112000** | | **32.704606** |\n\n**Average entry:**\n\nP_avg = 112000 / 32.704606 ≈ 3424.59\n\n{\"answer\": 3424.59, \"confidence\": 92}" }, { "taskId": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 0.02840777213808475, "answer": 0.0284078, "confidence": 90, "grade": "exact", "relError": 9.807849455850231e-7, "cost": 0.024148, "reply": "I'm computing the inverse short PnL using the formula size times the difference of reciprocals of close and open prices, working through the precise decimal approximation of 1/41686.7.\n\nTaking the difference of reciprocals and multiplying by 44000 gives a raw PnL of about 0.0295526, then subtracting fees calculated as 0.00055 times 44000 times the sum of inverses (roughly 0.00114479) yields a net PnL of approximately 0.0284078.\n\n\n**Net PnL ≈ 0.028408 BTC** (a profit). This is the gross short PnL of 0.0295526 BTC minus total fees of 0.0011448 BTC.\n\n**Gross PnL (short, inverse contract)**\n\nPnL = size × (1/close − 1/open)\n= 44000 × (1/41686.7 − 1/42887.5)\n= 44000 × (42887.5 − 41686.7) / (41686.7 × 42887.5)\n\n- Price difference: 42887.5 − 41686.7 = 1200.8\n- Price product: 41686.7 × 42887.5 = 1,787,838,346.25\n- 1200.8 / 1,787,838,346.25 ≈ 6.71650 × 10⁻⁷\n- Gross PnL ≈ 44000 × 6.71650 × 10⁻⁷ ≈ 0.0295526 BTC\n\n**Fees (0.055% on each fill's BTC value)**\n\n- Open value: 44000 / 42887.5 ≈ 1.025932 BTC\n- Close value: 44000 / 41686.7 ≈ 1.055492 BTC\n- Combined value: 44000 × 4.73053 × 10⁻⁵ ≈ 2.081433 BTC\n- Fees: 0.00055 × 2.081433 ≈ 0.0011448 BTC\n\n**Net PnL**\n\n0.0295526 − 0.0011448 ≈ 0.0284078 BTC\n\n{\"answer\": 0.0284078, \"confidence\": 90}" }, { "taskId": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -0.06525146382766053, "answer": -0.0652515, "confidence": 88, "grade": "exact", "relError": 5.543529194454429e-7, "cost": 0.01452, "reply": "I'm computing the PnL by dividing 45000 by both price levels, getting roughly 0.9101 and 0.9743, to then find their difference.\n\nPnL comes out to about -0.0643, and after subtracting fees of roughly 0.00094, the net result is approximately -0.0653.\n\n\nNet PnL ≈ **−0.06525 BTC**: about −0.06431 BTC from the price move, minus 0.000942 BTC in fees.\n\nFor an inverse contract, PnL in BTC is size × (1/entry − 1/exit) for a long.\n\n**Position values in BTC**\n- At entry: 45000 / 49446 ≈ 0.9100837 BTC\n- At exit: 45000 / 46182.6 ≈ 0.9743930 BTC\n\n**Gross PnL**\n0.9100837 − 0.9743930 ≈ −0.0643093 BTC\n\n**Fees**\n0.0005 × (0.9100837 + 0.9743930) = 0.0005 × 1.8844767 ≈ 0.0009422 BTC\n\n**Net PnL**\n−0.0643093 − 0.0009422 ≈ −0.0652515 BTC\n\n{\"answer\": -0.0652515, \"confidence\": 88}" }, { "taskId": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 0.22922018359225318, "answer": 0.22922, "confidence": 92, "grade": "exact", "relError": 8.009427891588274e-7, "cost": 0.013188, "reply": "I'm computing the PnL by multiplying 189000 by the difference in inverse prices, working through the division 189000/91422.1 to get roughly 2.0673, then continuing to divide 189000 by 102932.\n\nSo the PnL comes out to about 0.2312, and after subtracting fees of about 0.00195 (0.05% applied to the sum of both notional ratios), the net result is approximately 0.2292.\n\n\nNet PnL ≈ **0.229220 BTC** (a profit, after fees).\n\nFor an inverse contract, PnL in BTC = size × (1/entry − 1/exit). Each fee is 0.05% of that fill's position value in BTC.\n\n**Position value at each fill**\n- At opening: 189000 / 91422.1 ≈ 2.0673338 BTC\n- At closing: 189000 / 102932.1 ≈ 1.8361619 BTC\n\n**Gross PnL**\n2.0673338 − 1.8361619 ≈ 0.2311720 BTC\n\n**Fees**\n0.0005 × (2.0673338 + 1.8361619) = 0.0005 × 3.9034957 ≈ 0.0019517 BTC\n\n**Net PnL**\n0.2311720 − 0.0019517 ≈ 0.2292202 BTC\n\n{\"answer\": 0.22922, \"confidence\": 92}" }, { "taskId": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -0.026151408867684533, "answer": -0.0261514, "confidence": 90, "grade": "exact", "relError": 3.390901261185902e-7, "cost": 0.020188, "reply": "I'm computing the inverse contract PnL for the short-size position, working out 1/entry and 1/exit price precisely to get an accurate reciprocal before multiplying by the position size.\n\nComputing the difference gives roughly -0.02597 before fees. Factoring in the fee rate applied to both entry and exit notional, fees add about 0.000186, bringing net PnL to approximately -0.02615.\n\n\n**Net PnL ≈ −0.026151 BTC**\n\n**1. Gross PnL**\n\nFor a long position in an inverse contract, PnL in BTC is size × (1/entry − 1/exit).\n\n- 1/89747.2 ≈ 1.11424089 × 10⁻⁵\n- 1/76940.3 ≈ 1.29970897 × 10⁻⁵\n- The difference is ≈ −1.8546808 × 10⁻⁶.\n\nGross PnL = 14000 × (−1.8546808 × 10⁻⁶) ≈ **−0.0259655 BTC**\n\n**2. Fees**\n\nEach fee is 0.055% of that fill's value in BTC (size / fill price).\n\n- Opening value is 14000 / 89747.2 ≈ 0.155994 BTC.\n- Closing value is 14000 / 76940.3 ≈ 0.181959 BTC.\n- Total fees = 0.00055 × (0.155994 + 0.181959) ≈ **0.000185874 BTC**\n\n**3. Net PnL**\n\nNet = −0.0259655 − 0.000185874 ≈ **−0.0261514 BTC**\n\n{\"answer\": -0.0261514, \"confidence\": 90}" }, { "taskId": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -0.011586282591352507, "answer": -0.011586, "confidence": 88, "grade": "exact", "relError": 0.00002439016572207775, "cost": 0.019608, "reply": "I'm working out the short position PnL formula using inverse price calculations, since shorts profit as price falls, meaning the reciprocal of exit price exceeding the reciprocal of entry price gives a positive result. Now I'm computing the precise reciprocal of 105593.7 to use in that calculation.\n\n\n**Net PnL ≈ −0.011586 BTC.** The short loses because the price rose. Inverse contracts settle in BTC, so PnL and fees are both computed on 1/price.\n\n**1. Reciprocals of the fill prices**\n- 1/102,777.6 ≈ 9.7297465 × 10⁻⁶\n- 1/105,593.7 ≈ 9.4702620 × 10⁻⁶\n\n**2. Gross PnL**\n\nFor a short inverse position, PnL = size × (1/exit − 1/entry).\n\nPnL = 44,000 × (9.4702620 − 9.7297465) × 10⁻⁶\n = 44,000 × (−0.2594845 × 10⁻⁶)\n ≈ −0.0114173 BTC\n\n**3. Fees**\n\nEach fee is 0.02% of that fill's position value in BTC (size / fill price).\n- Opening fee: 0.0002 × 44,000 / 102,777.6 ≈ 0.0000856 BTC\n- Closing fee: 0.0002 × 44,000 / 105,593.7 ≈ 0.0000833 BTC\n- Total fees ≈ 0.000169 BTC\n\n**4. Net PnL**\n\nNet = −0.0114173 − 0.0001690 ≈ **−0.011586 BTC**\n\n{\"answer\": -0.011586, \"confidence\": 88}" }, { "taskId": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 6057.334255752001, "answer": 6057.334256, "confidence": 97, "grade": "exact", "relError": 4.094196982022914e-11, "cost": 0.011396, "reply": "Calculating the short position's profit at roughly 6074.09, then working out fees as 0.02% of total notional across entry and exit, which comes to about 16.76.\n\n\n**Net PnL ≈ +6057.33 USDT** (a profit), worked out as gross PnL from the price drop minus the two fees.\n\n**Gross PnL**\n\nA short position gains when the price falls:\n\n(26380.89 − 22814.19) × 1.703 = 3566.70 × 1.703 = 6074.0901 USDT\n\n**Fees**\n\nEach fee is 0.02% of that fill's notional, so the total is 0.0002 × 1.703 × (open price + close price).\n\n- Sum of fill prices: 26380.89 + 22814.19 = 49195.08\n- Combined notional: 1.703 × 49195.08 = 83779.22124 USDT\n- Total fees: 0.0002 × 83779.22124 = 16.75584 USDT\n\n**Net PnL**\n\n6074.0901 − 16.75584 = 6057.33426 USDT\n\n{\"answer\": 6057.334256, \"confidence\": 97}" }, { "taskId": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 2844.8890875599996, "answer": 2844.889, "confidence": 97, "grade": "exact", "relError": 3.0778001086810425e-8, "cost": 0.009816, "reply": "Computing the price difference and scaling by position size gives about 2893.49, then subtracting fees of roughly 48.61 based on the combined notional, leaving a net profit near 2844.89.\n\n\n**Net PnL ≈ 2844.889 USDT** (a profit).\n\nFor a linear USDT-margined perpetual, PnL is size × (exit price − entry price). The fee is charged on each fill's notional value (size × fill price).\n\n**Gross PnL**\n- Price change: 39204.01 − 36718.19 = 2485.82\n- Gross PnL: 1.164 × 2485.82 = 2893.49448 USDT\n\n**Fees**\n- Sum of the two fill prices: 36718.19 + 39204.01 = 75922.20\n- Combined notional: 1.164 × 75922.20 = 88373.4408 USDT\n- Total fees: 88373.4408 × 0.00055 = 48.60539 USDT\n\n**Net PnL**\n- 2893.49448 − 48.60539 = 2844.88909 USDT\n\n{\"answer\": 2844.889, \"confidence\": 97}" }, { "taskId": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -3783.556934676003, "answer": -3783.557, "confidence": 97, "grade": "exact", "relError": 1.726523424342638e-8, "cost": 0.009516, "reply": "I'm calculating the price difference times position size for a loss of about -3746.59, then adding fees of roughly 36.97 on the combined notional, giving a net result near -3783.56.\n\n\n**Net PnL ≈ −3783.557 USDT** (a loss).\n\nFor a linear USDT-margined contract, PnL is size × (exit price − entry price). Fees are 0.055% of each fill's notional.\n\n**1. Price PnL**\n- Price change: 51850.49 − 57972.37 = −6121.88\n- Price PnL: 0.612 × (−6121.88) = −3746.59056 USDT\n\n**2. Fees**\n- Combined notional of both fills: 0.612 × (57972.37 + 51850.49) = 0.612 × 109822.86 = 67211.59032 USDT\n- Total fees: 67211.59032 × 0.00055 = 36.96637 USDT\n\n**3. Net PnL**\n- Net PnL: −3746.59056 − 36.96637 = −3783.55693 USDT\n\n{\"answer\": -3783.557, \"confidence\": 97}" }, { "taskId": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -1108.5578196799993, "answer": -1108.55782, "confidence": 97, "grade": "exact", "relError": 2.886639766704435e-10, "cost": 0.012056, "reply": "I'm computing the price difference multiplied by position size to get the P&L, getting about -1103.63, then starting to calculate the opening notional value for fees based on the entry price and position size.\n\n\n**Net PnL ≈ −1108.56 USDT**: a gross loss of 1103.63 USDT plus 4.92 USDT in fees.\n\n**Gross PnL**\n- Price change: 4440.76 − 4857.54 = −416.78 per BTC.\n- Gross PnL: 2.648 × (−416.78) = −1103.63344 USDT.\n\n**Fees**\n- Opening notional: 2.648 × 4857.54 = 12,862.76592 USDT.\n- Closing notional: 2.648 × 4440.76 = 11,759.13248 USDT.\n- Total fees: (12,862.76592 + 11,759.13248) × 0.0002 = 4.92437968 USDT.\n\n**Net PnL**\n- Net PnL: −1103.63344 − 4.92437968 = −1108.55782 USDT.\n\n{\"answer\": -1108.55782, \"confidence\": 97}" }, { "taskId": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 3351.620043262996, "answer": 3351.62, "confidence": 97, "grade": "exact", "relError": 1.2908084972596765e-8, "cost": 0.011016, "reply": "I'm calculating the gross PnL as 1.949 times the price difference (28983.12-27232.54), getting about 3411.88, then working out fees using the 0.00055 rate applied to the position size times the sum of entry and exit prices.\n\nFees come to about 60.26, so subtracting that from the gross PnL gives a net result of roughly 3351.62.\n\n\n**Net PnL ≈ +3351.62 USDT** (a profit). For a short position, gross PnL = size × (entry price − exit price), and each fee is 0.055% of that fill's notional.\n\n**Gross PnL**\n1.949 × (28983.12 − 27232.54) = 1.949 × 1750.58 = 3411.88042 USDT\n\n**Fees**\n- The opening notional plus the closing notional is 1.949 × (28983.12 + 27232.54) = 1.949 × 56215.66 = 109564.32134 USDT.\n- Total fees = 0.00055 × 109564.32134 = 60.26038 USDT.\n\n**Net PnL**\n3411.88042 − 60.26038 = 3351.62004 USDT\n\n{\"answer\": 3351.62, \"confidence\": 97}" }, { "taskId": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 0.11472881355932601, "answer": 0.1147, "confidence": 88, "grade": "exact", "relError": 0.0002511449254299066, "cost": 0.009176, "reply": "Work out the dollar risk, the stop distance, and the USD value of one pip per standard lot, then divide.\n\n1. **Dollar risk:** 1% of 5,000 USD = 50 USD.\n2. **Stop distance:** 1.3538 − 1.3479 = 0.0059, which is 59 pips.\n3. **Pip value per standard lot:** 100,000 units × 0.0001 = 10 CAD. Converted to USD at the entry price: 10 / 1.3538 = 7.38661 USD per pip.\n4. **Risk per standard lot at the stop:** 59 × 7.38661 = 435.81 USD.\n5. **Lot size:** 50 / 435.81 = 0.114729 standard lots.\n\nAs a cross-check, the loss per unit is 0.0059 CAD, or 0.0059 / 1.3538 = 0.00435810 USD. Then 50 / 0.0043581 = 11,472.9 units, which is 0.11473 lots.\n\nYou should trade about **0.1147 standard lots** (roughly 0.11 if your broker only allows two decimals).\n\n{\"answer\": 0.1147, \"confidence\": 88}" }, { "taskId": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 0.5939090909090843, "answer": 0.5939, "confidence": 88, "grade": "exact", "relError": 0.000015306903402306825, "cost": 0.010136, "reply": "With a $250 risk and 55 pip stop, I calculate the pip value for USD/CAD at the given rate (about 7.65 USD per pip per standard lot), giving a position size of roughly 0.594 lots.\n\n\nYou should trade about **0.5939 standard lots**.\n\n1. The amount at risk is 1% of $25,000, which is **$250**.\n2. The stop distance is 1.3066 − 1.3011 = 0.0055, which is **55 pips**.\n3. One standard lot is 100,000 units, so a pip is worth 10 CAD. Converted at the entry price, that is 10 / 1.3066 ≈ **$7.6534 per pip**.\n4. Divide the risk by the risk per lot:\n - Risk per lot = 55 × $7.6534 ≈ $420.94.\n - Lots = $250 / $420.94 ≈ **0.5939 standard lots**.\n\nIn practice you would round down to 0.59 lots (59,000 units) so you don't exceed 1% risk.\n\n{\"answer\": 0.5939, \"confidence\": 88}" }, { "taskId": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 1.1982352941177008, "answer": 1.1982, "confidence": 88, "grade": "exact", "relError": 0.000029455081046378537, "cost": 0.009636, "reply": "With $250 risk and a 17 pip stop, converting CHF pip value to USD gives roughly 1.198 lots.\n\n\nYou should trade about **1.198 standard lots**. In practice that rounds down to 1.19 lots, so you don't exceed your 1% risk.\n\n1. **Amount at risk:** 1% × 25,000 USD = 250 USD.\n2. **Stop distance:** 0.8148 − 0.8131 = 0.0017, which is 17 pips.\n3. **Pip value for one standard lot (100,000 USD):** 100,000 × 0.0001 = 10 CHF. Converting at the entry price gives 10 ÷ 0.8148 ≈ 12.273 USD per pip.\n4. **Risk per lot:** 17 pips × 12.273 USD ≈ 208.64 USD.\n5. **Lot size:** 250 ÷ 208.64 ≈ 1.198 lots.\n\n{\"answer\": 1.1982, \"confidence\": 88}" }, { "taskId": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 1.0178846153846508, "answer": 1.0179, "confidence": 88, "grade": "exact", "relError": 0.000015114301873409756, "cost": 0.010848, "reply": "Calculating the pip value for a standard lot at this price gives about $6.30, so for a 39-pip stop with $250 risk, the position size comes out to roughly 1.02 lots.\n\n\nYou should trade about **1.0179 standard lots**. In practice that rounds down to **1.01 lots**, which keeps the risk slightly under $250.\n\n1. **Risk amount:** 1% of $25,000 = **$250**.\n2. **Stop distance:** 158.79 − 158.40 = 0.39. For USD/JPY one pip is 0.01, so this is **39 pips**.\n3. **Pip value per standard lot:** 100,000 × 0.01 = ¥1,000 per pip. At the entry price, ¥1,000 ÷ 158.79 ≈ **$6.2976 per pip**.\n4. **Risk per lot:** 39 × $6.2976 ≈ **$245.61**.\n5. **Position size:** $250 ÷ $245.61 ≈ **1.0179 lots**.\n\n{\"answer\": 1.0179, \"confidence\": 88}" }, { "taskId": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-opus-5-5", "expected": 7.182727272727309, "answer": 7.1827, "confidence": 88, "grade": "exact", "relError": 0.000003796987728178971, "cost": 0.009948, "reply": "I'm calculating position size: with a $1000 risk and a 22 pip stop, I convert the JPY pip value to USD, getting about 6.33 USD per pip per lot, then divide risk by (pips × pip value) to land on roughly 7.18 lots.\n\n\nYou should trade about **7.1827 standard lots** (in practice, round down to 7.18).\n\n1. **Dollar risk:** 1% of $100,000 is **$1,000**.\n2. **Stop distance:** 158.02 − 157.80 = 0.22, which is **22 pips** (one JPY pip is 0.01).\n3. **Pip value per standard lot:** 100,000 × 0.01 = 1,000 JPY. Converted at the entry price, 1,000 ÷ 158.02 ≈ **$6.3283 per pip**.\n4. **Position size:** $1,000 ÷ (22 pips × $6.3283) = $1,000 ÷ $139.22 ≈ **7.1827 lots**.\n\n{\"answer\": 7.1827, \"confidence\": 88}" }, { "taskId": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -0.23573231643213824, "answer": -0.235732, "confidence": 88, "grade": "exact", "relError": 0.0000013423366937182556, "cost": 0.022396, "reply": "I'm working out the inverse PnL formula for a short position: since the price rose, this is a loss, calculated as Q times the difference between the reciprocal of exit and entry prices. I'm now computing the reciprocal of the entry price, 55827.2, to plug into the formula.\n\n\nYour realized PnL is about **−0.23573 BTC**. The price rose after you went short, so the trade lost about 0.23357 BTC, plus about 0.00216 BTC in fees.\n\nFor the BTCUSD inverse contract, both PnL and fees are paid in BTC, and each is based on the USD notional divided by price.\n\n**1. Trade PnL for a short**\n\nPnL = Q × (1/exit − 1/entry), with Q = 103,000 USD.\n\n- 1/55,827.2 ≈ 1.7912416 × 10⁻⁵\n- 1/49,553.7 ≈ 2.0180128 × 10⁻⁵\n- Difference = −2.267712 × 10⁻⁶\n- PnL = 103,000 × (−2.267712 × 10⁻⁶) ≈ −0.233574 BTC\n\n**2. Taker fees on both fills**\n\nEach fill's fee = 0.055% × (Q / fill price).\n\n- Fees = 0.00055 × 103,000 × (1/49,553.7 + 1/55,827.2)\n- = 56.65 × 3.8092544 × 10⁻⁵\n- ≈ 0.002158 BTC\n\n**3. Net**\n\n−0.233574 − 0.002158 ≈ −0.235732 BTC\n\nThis ignores any funding payments you paid or received while the position was open.\n\n{\"answer\": -0.235732, \"confidence\": 88}" }, { "taskId": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -0.0669895883004742, "answer": -0.06699, "confidence": 88, "grade": "exact", "relError": 0.0000061457240779507516, "cost": 0.023696, "reply": "I'm computing the inverse PnL formula using reciprocals of entry and exit prices, since this is a long position that lost value as price fell, carrying out the division precisely.\n\nWorking out the difference of reciprocals times quantity gives a price-based loss of about -0.06605, then I calculate the trading fees using the sum of reciprocals times the fee rate, getting about 0.0009376, and subtracting that from the loss gives a total PnL of roughly -0.06699.\n\n\n**Realized PnL ≈ −0.06699 BTC after fees.** The trade loses about 0.06605 BTC on price, and the two taker fees add about 0.00094 BTC.\n\nFor an inverse contract, each contract is worth 1 USD, so PnL in BTC is Q × (1/entry − 1/exit), with Q = 78,000 contracts.\n\n**1. Price PnL (long)**\n\nPnL = Q × (1/entry − 1/exit) = 78,000 × (1/95,195.8 − 1/88,094.2)\n\nThe difference of reciprocals equals (88,094.2 − 95,195.8) / (95,195.8 × 88,094.2):\n- Numerator: −7,101.6\n- Denominator: 8,386,197,844.36\n- Result: −8.468203 × 10⁻⁷\n\nMultiplying by 78,000 gives a price PnL of ≈ −0.06605198 BTC. The price fell, so the long loses.\n\n**2. Taker fees (0.055% per fill)**\n\nOn an inverse contract, each fill's fee in BTC is Q × rate ÷ fill price.\n- Entry fee: 42.9 / 95,195.8 ≈ 0.00045064 BTC\n- Exit fee: 42.9 / 88,094.2 ≈ 0.00048699 BTC\n- Total fees: ≈ 0.00093763 BTC\n\n**3. Net**\n\n−0.06605198 − 0.00093763 ≈ **−0.06698961 BTC**\n\n{\"answer\": -0.06699, \"confidence\": 88}" }, { "taskId": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 0.0024428835303150374, "answer": 0.0024429, "confidence": 88, "grade": "exact", "relError": 0.000006741903475231194, "cost": 0.021556, "reply": "I'm computing the inverse contract PnL formula for the short position, working out 1/116982.6 precisely by adjusting from the 1/117000 approximation.\n\nThen I subtract the two reciprocals to get the net BTC difference, multiply by position size to get 0.002836255 BTC gross profit. I calculate fees on both entry and exit fills (0.055% of 42000 USD converted to BTC via each price), getting a total fee of about 0.000393371 BTC, and subtract to land on a net profit of roughly 0.002442884 BTC.\n\n\n**Realized PnL ≈ +0.0024429 BTC after fees.**\n\nFor an inverse contract, PnL and fees are settled in BTC.\n\n**Gross PnL on the short**\n\nPnL = Q × (1/exit − 1/entry) = 42,000 × (1/116,982.6 − 1/117,914.1)\n\nThe difference of reciprocals equals the price change divided by the product of the prices:\n\n931.5 / (117,914.1 × 116,982.6) ≈ 6.75299 × 10⁻⁸\n\nGross PnL = 42,000 × 6.75299 × 10⁻⁸ ≈ **0.0028363 BTC**\n\n**Fees**\n\nEach fill's fee is 0.055% of the 42,000 USD notional, which is 23.1 USD. Each fee is converted to BTC at that fill's price.\n\n- Entry fee: 23.1 / 117,914.1 ≈ 0.00019591 BTC\n- Exit fee: 23.1 / 116,982.6 ≈ 0.00019747 BTC\n- Total fees ≈ 0.00039337 BTC\n\n**Net**\n\n0.0028363 − 0.00039337 ≈ **0.0024429 BTC**\n\n{\"answer\": 0.0024429, \"confidence\": 88}" }, { "taskId": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 0.001886499688812526, "answer": 0.0018865, "confidence": 88, "grade": "exact", "relError": 1.6495495644927526e-7, "cost": 0.018728, "reply": "I'm computing the inverse PnL formula using quantity 192000 and the price difference between entry 20434 and exit 20460.6, working through the multiplication to get the denominator product of roughly 418 million before dividing by the 26.6 price gap.\n\nThat gives a gross PnL ratio of about 0.012216. Then I'm calculating fees on both legs at 0.055% of notional divided by each price, summing them to about 0.010329 total fees, and subtracting to get a net result of roughly 0.0018865.\n\n\nRealized PnL after fees is **≈ +0.0018865 BTC**. The gross gain of about 0.0122155 BTC is reduced by roughly 0.0103290 BTC in taker fees.\n\nFor an inverse contract, PnL in BTC is the number of contracts × (1/entry − 1/exit). Each fee is the number of contracts × fee rate ÷ the fill price.\n\n**Gross PnL**\n192,000 × (1/20434 − 1/20460.6)\n= 192,000 × 26.6 / (20434 × 20460.6)\n= 5,107,200 / 418,091,900.4\n≈ 0.0122155 BTC\n\n**Fees (0.055% on each fill)**\n- Entry: 192,000 × 0.00055 / 20434 = 105.6 / 20434 ≈ 0.0051679 BTC\n- Exit: 105.6 / 20460.6 ≈ 0.0051611 BTC\n- Total ≈ 0.0103290 BTC\n\n**Net**\n0.0122155 − 0.0103290 ≈ **0.0018865 BTC**\n\n{\"answer\": 0.0018865, \"confidence\": 88}" }, { "taskId": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": -0.0962309957129481, "answer": -0.096231, "confidence": 88, "grade": "exact", "relError": 4.4549595089137153e-8, "cost": 0.021956, "reply": "I'm working out the inverse contract PnL formula for a short position, reasoning that it should be Q*(1/exit - 1/entry) by symmetry with the long formula Q*(1/entry - 1/exit).\n\nComputing the reciprocals of the entry and exit prices carefully, I find the difference comes to about -6.742e-7, which scaled by the position size of 139000 gives roughly -0.0937. Adding in fees of about 0.00252 brings the total loss to approximately -0.0962.\n\n\n**Realized PnL ≈ −0.096231 BTC** after fees. You shorted and the price rose, so this is a loss: −0.093713 BTC on the trade plus 0.002518 BTC in fees.\n\nOn an inverse contract, PnL and fees are paid in BTC. Each contract is worth 1 USD, so the BTC value of a fill is quantity ÷ price.\n\n**Trading PnL for a short**\n\nPnL = Q × (1/entry − 1/exit)\n\n- 1/59,513.2 ≈ 1.6802995 × 10⁻⁵\n- 1/62,000.9 ≈ 1.6128798 × 10⁻⁵\n- Difference = 1.6802995 × 10⁻⁵ − 1.6128798 × 10⁻⁵ ≈ 6.74197 × 10⁻⁷\n- 139,000 × 6.74197 × 10⁻⁷ ≈ **0.0937134 BTC lost** (−0.0937134)\n\n**Fees (0.055% taker on each fill, charged on the BTC value)**\n\n- Entry fee: 139,000 × 0.00055 ÷ 59,513.2 ≈ 0.0012846 BTC\n- Exit fee: 139,000 × 0.00055 ÷ 62,000.9 ≈ 0.0012331 BTC\n- Total fees ≈ **0.0025176 BTC**\n\n**Net**\n\n−0.0937134 − 0.0025176 ≈ **−0.096231 BTC**\n\n{\"answer\": -0.096231, \"confidence\": 88}" }, { "taskId": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 70867.19713259465, "answer": 70867.2, "confidence": 88, "grade": "exact", "relError": 4.046167285427324e-8, "cost": 0.013328, "reply": "I'm computing the weighted average entry price using the inverse contract formula, dividing total contracts by the sum of each position's contracts over its price to get roughly 1.2107 for one leg and about 0.6378 for the other.\n\nSumming those gives 1.848528, and dividing 131000 by that sum yields approximately 70867.2, which I verify by checking that 1.848528 times 867.2 lands close to the remainder of 1603.04.\n\n\nThe exchange should show an average entry of about **70,867.2**.\n\nOn an inverse contract the margin is in BTC, so the average entry is the harmonic mean of the fill prices, weighted by contracts:\n\nAverage entry = total contracts ÷ Σ(contracts / price)\n\n1. 76,000 / 62,771.7 ≈ 1.2107367 BTC\n2. 55,000 / 86,235.1 ≈ 0.6377913 BTC\n3. Total ≈ 1.8485280 BTC\n4. 131,000 / 1.8485280 ≈ **70,867.2**\n\nThis is lower than the simple contract-weighted average of about 72,617. The lower-priced fill buys more BTC per contract, so it carries more weight.\n\n{\"answer\": 70867.2, \"confidence\": 88}" }, { "taskId": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 34209.14079495059, "answer": 34209.14, "confidence": 92, "grade": "exact", "relError": 2.3237958347514953e-8, "cost": 0.013848, "reply": "I'm computing weighted averages by dividing total contracts by the sum of contracts-over-price ratios, getting roughly 2.266 and 2.928 for two groups so far.\n\nContinuing the sum, I get 0.3889 for the third group, bringing the total to about 5.5833. Dividing 191000 by that sum gives a weighted average price of approximately 34209.14.\n\n\n**Average entry price ≈ 34209.14**\n\nOn an inverse contract each contract is worth 1 USD, so the BTC value of a fill is contracts ÷ price. Bybit therefore shows the harmonic (contract-weighted) average:\n\nAverage entry = total contracts ÷ Σ(contracts ÷ price)\n\n**BTC value of each fill:**\n- 76,000 ÷ 33,536.1 ≈ 2.2662147 BTC\n- 73,000 ÷ 24,929.9 ≈ 2.9282106 BTC\n- 42,000 ÷ 108,003.1 ≈ 0.3888778 BTC\n\nThe sum is about 5.5833031 BTC.\n\n**Average entry:** 191,000 ÷ 5.5833031 ≈ **34,209.14**\n\nThis is well below the simple contract-weighted average of about 51,682, because inverse averaging weights the lower-priced fills more heavily.\n\n{\"answer\": 34209.14, \"confidence\": 92}" }, { "taskId": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 88678.94192491287, "answer": 88678.94, "confidence": 90, "grade": "exact", "relError": 2.1706538502484576e-8, "cost": 0.01552, "reply": "I'm computing the inverse-contract-weighted average entry price by dividing total contracts by the sum of each position's contracts-over-price ratio, working through each term's division carefully.\n\nAdding these three ratios gives about 1.74788, then dividing 155000 by that sum yields an average entry price around 88678.94.\n\n\n**Average entry ≈ 88,678.94 USD**\n\nFor an inverse contract, each contract is worth a fixed 1 USD, so the value held in BTC is contracts ÷ price. The exchange therefore shows a harmonic (contract-weighted) average rather than a simple price average:\n\nAverage entry = total contracts ÷ Σ(contractsᵢ / priceᵢ)\n\n**BTC value of each fill:**\n- 59,000 / 82,124.2 ≈ 0.71842405 BTC\n- 34,000 / 81,534 ≈ 0.41700395 BTC\n- 62,000 / 101,232.7 ≈ 0.61245032 BTC\n\n**Totals:**\n- Total BTC value ≈ 1.74787832 BTC\n- Total contracts = 155,000\n\n**Average entry:** 155,000 / 1.74787832 ≈ **88,678.94**\n\nThese figures were worked by hand, so the last digit or two may differ slightly from what the exchange displays.\n\n{\"answer\": 88678.94, \"confidence\": 90}" }, { "taskId": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 46924.5904555392, "answer": 46924.6, "confidence": 88, "grade": "exact", "relError": 2.0339998087891724e-7, "cost": 0.016728, "reply": "I'm computing the inverse-weighted average entry price for each position by dividing total contracts by the sum of contracts/price, getting approximately 1.3940 and 0.9288 for the two calculations, then moving to sum these results.\n\nDividing 109000 by the sum 2.3228759 gives roughly 46924.6, which I double-checked by refining each division more precisely and confirming the result holds steady at about 46924.6.\n\n\n**The exchange will show an average entry price of about 46,924.6.**\n\nOn an inverse contract each contract is worth a fixed 1 USD, so your BTC exposure is contracts ÷ price. The average entry is therefore the contract-weighted harmonic mean of the fill prices, not the simple weighted average:\n\nAverage entry = total contracts ÷ Σ(contracts / price)\n\n**Step 1 – BTC value of each fill**\n- 42,000 / 30,128.4 ≈ 1.3940335 BTC\n- 67,000 / 72,132.8 ≈ 0.9288424 BTC\n- Total ≈ 2.3228759 BTC\n\n**Step 2 – Divide total contracts by total BTC**\n- Total contracts = 42,000 + 67,000 = 109,000\n- 109,000 / 2.3228759 ≈ 46,924.6\n\nThe display may differ by a few cents due to rounding in the hand calculation.\n\n{\"answer\": 46924.6, \"confidence\": 88}" }, { "taskId": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-opus-5-5", "expected": 37529.651947786224, "answer": 37529.65, "confidence": 88, "grade": "exact", "relError": 5.189992769330071e-8, "cost": 0.011768, "reply": "I'm computing the weighted average entry prices by dividing total contracts by the sum of contracts-over-price, getting roughly 1.5044 and 1.0803 for the two positions.\n\n\nThe exchange will show an average entry of about **37,529.65**.\n\nOn an inverse contract each contract is worth 1 USD, so the position's value in BTC is contracts ÷ price. The average entry is therefore the harmonic (contract-weighted) average, not the simple weighted average of prices:\n\nAverage entry = total contracts ÷ Σ(contracts / price)\n\n1. BTC value of each fill:\n - 36,000 / 23,930.3 ≈ 1.5043689 BTC\n - 61,000 / 56,468.2 ≈ 1.0802540 BTC\n2. Total BTC value ≈ 2.5846229 BTC\n3. Average entry = 97,000 / 2.5846229 ≈ **37,529.65**\n\nA simple weighted average of the prices would give about 44,392, which is not what an inverse contract uses.\n\n{\"answer\": 37529.65, \"confidence\": 88}" }, { "taskId": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 4.917273424034056, "answer": 4.9173, "confidence": 88, "grade": "exact", "relError": 0.000005404614234884395, "cost": 0.029232, "reply": "I'm adding up the data values one by one to get the running total before computing the mean and standard deviation.\n\nContinuing the final total comes out to -10.19, giving a mean of -0.25475. Now I'm squaring each deviation from the mean and summing these squared differences to work toward the variance.\n\nSubtracting n times the mean squared gives a sum of squares around 313.37, which divided by 39 degrees of freedom gives a variance near 8.035 and a standard deviation of about 2.8346. Using the 95th percentile z-score, the VaR estimate comes out to approximately 4.6625.\n\n\n**1-day 95% parametric VaR ≈ 4.917%.** In other words, a one-day loss of about 4.92% should be exceeded on only about 5% of days, assuming returns are normally distributed.\n\n**Inputs**\n- Number of returns: n = 40.\n- Sum of returns: −10.19, so the sample mean is μ = −10.19 / 40 = −0.25475%.\n- Sum of squared returns: Σr² = 315.9621.\n\n**Sample standard deviation**\n- Sum of squared deviations: Σ(r − μ)² = Σr² − (Σr)²/n = 315.9621 − 103.8361/40 = 315.9621 − 2.5959 = 313.3662.\n- Sample variance (dividing by n − 1 = 39): 313.3662 / 39 = 8.03503.\n- Sample standard deviation: σ = √8.03503 ≈ 2.83461%.\n\n**VaR**\nThe 95% one-sided normal critical value is z = 1.644854.\n\nVaR = −(μ − zσ) = zσ − μ = 1.644854 × 2.83461 + 0.25475 = 4.66252 + 0.25475 ≈ **4.917%**\n\nThe negative mean adds to the VaR rather than reducing it.\n\n{\"answer\": 4.9173, \"confidence\": 88}" }, { "taskId": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 4.2934299989968965, "answer": 4.2934, "confidence": 88, "grade": "exact", "relError": 0.000006987186679038246, "cost": 0.028412, "reply": "I'm adding up the data values one by one to get the running sum needed for computing the mean and sample standard deviation.\n\nFinishing the sum of squared deviations, I get 325.84196, giving a variance of 8.35492 for 39 degrees of freedom, and taking the square root yields a standard deviation of about 2.8905.\n\nNow computing VaR: 1.644854 times the standard deviation gives roughly 4.7544, and subtracting the mean adjustment of 0.461 yields a VaR of about 4.2934.\n\n\n**1-day 95% VaR ≈ 4.293% (a loss)**\n\nFormula: VaR = −(μ − z·σ) = z·σ − μ, with z = 1.644854 (the 95% one-sided normal quantile).\n\n1. **Mean.** The 40 returns sum to 18.44, so μ = 18.44 / 40 = 0.461%.\n2. **Sample standard deviation.**\n - The sum of squared returns is 334.3428.\n - Subtract n·μ² = 18.44² / 40 = 8.50084, giving a sum of squared deviations of 325.84196.\n - Divide by n − 1 = 39 to get a variance of 8.35492.\n - So σ ≈ 2.89049%.\n3. **VaR.** 1.644854 × 2.89049 − 0.461 = 4.75443 − 0.461 ≈ 4.2934%.\n\n{\"answer\": 4.2934, \"confidence\": 88}" }, { "taskId": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 5.2422646771168795, "answer": 5.2423, "confidence": 88, "grade": "exact", "relError": 0.000006738096089440247, "cost": 0.027712, "reply": "I'm running a cumulative sum across the data points to compute the mean, tracking the running total as I add each value.\n\nNow I'm squaring the deviations from the mean for each data point and summing them to get the variance components.\n\nFinishing the sum of squared deviations gives 407.78, so dividing by 39 degrees of freedom yields a variance around 10.45, giving a standard deviation of roughly 3.233. Using the 95% z-score of 1.6449, I compute VaR as 1.6449 times the standard deviation minus the mean, landing at approximately 5.2.\n\n\n**1-day 95% parametric VaR ≈ 5.242%**\n\nI used VaR = −(μ − z·σ), with the sample mean μ, the sample standard deviation σ (n − 1 divisor) and z₀.₉₅ = 1.644854.\n\n1. **Mean.** The 40 returns sum to 3.00, so μ = 3.00 / 40 = 0.075%.\n2. **Sum of squares.** The sum of the squared returns is 407.7804.\n3. **Squared deviations.** Σ(rᵢ − μ)² = 407.7804 − 40 × 0.075² = 407.7804 − 0.225 = 407.5554.\n4. **Standard deviation.** σ² = 407.5554 / 39 = 10.45014, so σ ≈ 3.23267%.\n5. **VaR.** VaR = 1.644854 × 3.23267 − 0.075 = 5.31727 − 0.075 ≈ 5.2423%.\n\nSo on 95% of days, the loss should not exceed about 5.24%.\n\n{\"answer\": 5.2423, \"confidence\": 88}" }, { "taskId": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 7.037483251314367, "answer": 7.0375, "confidence": 88, "grade": "exact", "relError": 0.0000023799254697953426, "cost": 0.031532, "reply": "I'm adding up the data values to get the running sum needed to calculate the mean.\n\nContinuing total: -18.83, giving a mean of about -0.47075. Now I'm squaring each deviation from the mean to prepare for the variance calculation.\n\nDividing gives variance 7.968, so standard deviation is about 2.8228. Computing the VaR estimate: 2.326348 times the sd plus the mean gives roughly 6.5667 + 0.4708 ≈ 7.0375.\n\n\n**1-day 99% parametric VaR ≈ 7.037%**, meaning you would expect a daily loss larger than about 7.04% on roughly 1 day in 100.\n\nParametric VaR under a normal assumption is VaR = −(μ − z₀.₉₉·σ) = z₀.₉₉·σ − μ, with z₀.₉₉ = 2.326348.\n\n**1. Sample mean**\n- The 40 returns sum to −18.83.\n- μ = −18.83 / 40 = −0.47075%.\n\n**2. Sample standard deviation**\n- The sum of squared returns is 319.6163.\n- Sum of squared deviations = 319.6163 − (−18.83)²/40 = 319.6163 − 8.8642 = 310.7521.\n- Sample variance = 310.7521 / 39 = 7.9680.\n- σ ≈ 2.82276%.\n\n**3. VaR**\n- z·σ = 2.326348 × 2.82276 = 6.5667.\n- VaR = 6.5667 − (−0.47075) = 7.0375%.\n\nThe negative sample mean adds to the VaR, on top of the volatility term.\n\n{\"answer\": 7.0375, \"confidence\": 88}" }, { "taskId": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 6.7877378350475945, "answer": 6.788, "confidence": 88, "grade": "exact", "relError": 0.00003862331733734097, "cost": 0.029012, "reply": "I'm adding up the deviations sequentially to get a running total, working toward computing the mean and standard deviation.\n\nNow I'm squaring each deviation from the mean and summing those squares to get the variance.\n\nSo VaR works out to roughly 6.7877 after subtracting the mean adjustment.\n\n\n**1-day 99% parametric VaR ≈ 6.788%**\n\nApproach: fit a normal distribution with the sample mean and sample standard deviation, then VaR₉₉ = z₀.₉₉·σ − μ, reported as a positive loss.\n\n1. **Mean.** The 40 returns sum to 1.79%, so μ = 1.79 / 40 = 0.04475%.\n\n2. **Sample standard deviation (n − 1 = 39).**\n - Sum of squared returns: Σr² = 336.4933.\n - Sum of squared deviations: Σ(r − μ)² = Σr² − n·μ² = 336.4933 − 40 × 0.0020026 = 336.4132.\n - Variance: s² = 336.4132 / 39 = 8.62598.\n - Standard deviation: s = 2.93700%.\n\n3. **VaR at 99%.** The one-sided 99% normal quantile is z = 2.32635.\n - VaR = 2.32635 × 2.93700 − 0.04475\n - = 6.83250 − 0.04475\n - = 6.78775%.\n\nSo with 99% confidence, the 1-day loss should not exceed about 6.79%.\n\n{\"answer\": 6.788, \"confidence\": 88}" }, { "taskId": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.40527021190035356, "answer": 0.4053, "confidence": 88, "grade": "exact", "relError": 0.00007350182365181334, "cost": 0.02588, "reply": "I'm running a cumulative sum through this sequence of values, tracking the running total as it fluctuates between negative and positive.\n\nNow I'm computing squared deviations from the mean for each data point and summing them up, working toward a variance calculation.\n\nI get sum of squares = 146.4985, subtract the correction factor to get 146.4303, divide by 29 degrees of freedom for variance of 5.04933, then take the square root for a standard deviation of 2.24707. Dividing the mean by this gives a daily Sharpe ratio of 0.021213, which annualizes by multiplying by the square root of 365 to about 0.40528.\n\n\n**Annualized Sharpe ratio ≈ 0.405**\n\n1. **Mean daily return:** the 30 returns sum to 1.43%, so the mean is 1.43 / 30 = 0.04767% per day.\n2. **Sample standard deviation:**\n - The sum of squared returns is 146.4985.\n - Subtract the mean correction 1.43² / 30 = 0.0682, which gives 146.4303.\n - The sample variance is 146.4303 / 29 = 5.0493.\n - The daily standard deviation is √5.0493 = 2.2471%.\n3. **Daily Sharpe:** 0.04767 / 2.2471 = 0.02121.\n4. **Annualized Sharpe:** 0.02121 × √365 = 0.02121 × 19.105 ≈ 0.4053.\n\nThirty days is a very short sample, so this estimate is statistically noisy.\n\n{\"answer\": 0.4053, \"confidence\": 88}" }, { "taskId": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.37258355219979333, "answer": 0.3726, "confidence": 88, "grade": "exact", "relError": 0.00004414526650342787, "cost": 0.02654, "reply": "I'm adding up the values one by one to get the running total before computing the mean and sample standard deviation.\n\nMean comes out to 0.045, so now I'm squaring each deviation from the mean and summing them up to find the variance.\n\nVariance works out to about 5.324, giving a standard deviation near 2.307. Dividing the mean by that gives a daily Sharpe ratio of roughly 0.0195, which annualizes to about 0.373 after multiplying by the square root of 365.\n\n\n**Annualized Sharpe ratio ≈ 0.3726**\n\n1. **Mean daily return.** The 30 returns sum to 1.35%, so the mean is 1.35 / 30 = 0.045%.\n\n2. **Sample standard deviation.**\n - The sum of squared returns is 154.4683.\n - Subtracting n × mean² gives the sum of squared deviations: 154.4683 − 30 × 0.045² = 154.4076.\n - The sample variance is 154.4076 / 29 = 5.3244.\n - The standard deviation is √5.3244 ≈ 2.3075%.\n\n3. **Daily Sharpe ratio.** With a risk-free rate of 0, it is 0.045 / 2.3075 ≈ 0.019502.\n\n4. **Annualize.** Multiply by √365 ≈ 19.105: 0.019502 × 19.105 ≈ 0.3726.\n\n{\"answer\": 0.3726, \"confidence\": 88}" }, { "taskId": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": -0.6855095501276544, "answer": -0.6855, "confidence": 88, "grade": "exact", "relError": 0.000013931429040744886, "cost": 0.02686, "reply": "I'm adding up the values sequentially to get the running total, working toward the mean and sample standard deviation.\n\nMean comes out to -0.083333.\n\nNow I'm computing the squared deviations from the mean for each data point and keeping a running sum of squares, which I'll need to get the variance.\n\nWith the sum of squares totaled, I divide by 29 degrees of freedom to get variance, then take the square root to get the standard deviation (about 2.3225). Finally I compute the daily Sharpe ratio (mean over std), which comes out to about -0.0359, then annualize by multiplying by sqrt(365), giving an annualized Sharpe of roughly -0.686.\n\n\n**Annualized Sharpe ratio ≈ −0.6855**\n\nArithmetic was done by hand, so treat the last digit as approximate.\n\n**1. Mean daily return.**\nThe 30 returns sum to −2.50%, so the mean is −2.50 / 30 = −0.08333% per day.\n\n**2. Sample standard deviation.**\n- The sum of the squared returns is 156.6316.\n- The sum of squared deviations from the mean is 156.6316 − (−2.50)²/30 = 156.6316 − 0.2083 = 156.4233.\n- The sample variance is 156.4233 / 29 = 5.3939.\n- The standard deviation is √5.3939 ≈ 2.3225% per day.\n\n**3. Daily Sharpe ratio.**\nWith a risk-free rate of 0, the daily Sharpe is −0.08333 / 2.3225 = −0.03588.\n\n**4. Annualize with 365 trading days.**\n−0.03588 × √365 = −0.03588 × 19.105 ≈ −0.6855.\n\nThe negative value reflects that the bot lost money on average over this period.\n\n{\"answer\": -0.6855, \"confidence\": 88}" }, { "taskId": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 0.5941064255855656, "answer": 0.5941, "confidence": 88, "grade": "exact", "relError": 0.000010815546321174527, "cost": 0.02624, "reply": "I'm running a cumulative sum through the data values to find the total, which I'll need to compute the mean and sample standard deviation.\n\nFinishing the sum of squares gives 158.5138, so the corrected sum of squares is 158.3554, variance 5.46053, and standard deviation 2.33678. Dividing the mean return by this sd gives a daily Sharpe of 0.031097, which annualizes (×√365) to about 0.594.\n\n\n**Annualized Sharpe ≈ 0.594**\n\n1. **Mean daily return:** the 30 returns sum to 2.18%, so the mean is 2.18 / 30 = 0.07267%.\n\n2. **Sample standard deviation:**\n - The sum of squared returns is 158.5138.\n - The sum of squared deviations is 158.5138 − 2.18² / 30 = 158.3554.\n - The sample variance is 158.3554 / 29 = 5.4605.\n - The standard deviation is √5.4605 ≈ 2.3368%.\n\n3. **Daily Sharpe:** 0.07267 / 2.3368 ≈ 0.03110.\n\n4. **Annualized Sharpe:** 0.03110 × √365 = 0.03110 × 19.105 ≈ **0.594**.\n\nWith only 30 days of data, this estimate is statistically very noisy.\n\n{\"answer\": 0.5941, \"confidence\": 88}" }, { "taskId": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-opus-5-5", "expected": 6.721949210975506, "answer": 6.722, "confidence": 92, "grade": "exact", "relError": 0.000007555698935003174, "cost": 0.02696, "reply": "I'm adding up the data points sequentially to get the running total, working toward computing the mean and sample standard deviation.\n\nFinishing the sum of squares gives 194.5396, so the corrected sum of squares is 172.455, variance 5.947, and standard deviation about 2.4386. That makes the daily Sharpe ratio 0.858/2.4386 ≈ 0.3518, which annualizes to roughly 6.72 after multiplying by √365.\n\n\n**Annualized Sharpe ratio ≈ 6.722**\n\n1. **Mean daily return.** The 30 returns sum to 25.74%, so the mean is 25.74 / 30 = 0.858%.\n\n2. **Sample standard deviation.**\n - Sum of squared returns = 194.5396.\n - Sum of squared deviations = 194.5396 − 25.74²/30 = 194.5396 − 22.0849 = 172.4547.\n - Sample variance = 172.4547 / 29 = 5.9467.\n - Standard deviation = √5.9467 ≈ 2.4386%.\n\n3. **Daily Sharpe ratio.** 0.858 / 2.4386 ≈ 0.35185.\n\n4. **Annualize.** Multiply by √365 ≈ 19.105: 0.35185 × 19.105 ≈ 6.722.\n\nThis assumes daily returns are independent, which is what the √365 scaling relies on. With only 30 observations, the estimate carries a lot of statistical uncertainty.\n\n{\"answer\": 6.722, \"confidence\": 92}" }, { "taskId": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.44697422776459944, "answer": 0.447, "confidence": 88, "grade": "exact", "relError": 0.000057659332014418574, "cost": 0.029872, "reply": "I need to back out implied volatility for this option. With S=447.74, K=463, T≈0.063 years, the call is trading at 21.6661 with no intrinsic value, so I'm testing trial volatilities like 0.40 and computing d1 using the standard Black-Scholes formula to see how close it gets to the market price.\n\nTrying σ=0.40 gives a theoretical price of only about 11.62, far too low, so I push up to σ=0.75, which overshoots to 27.07. Using vega to interpolate, I narrow in on σ≈0.63, computing d1 and d2 again and getting a price of 21.71, very close to target, so I'm fine-tuning the vega-based correction to converge on the precise implied volatility.\n\n\nDelta ≈ **0.4470**, at an implied volatility of about **62.9%**. The volatility and N(d₁) were found by hand iteration, so the 4th digit could be off by about 1.\n\n**Inputs**\n- S = 447.74, K = 463.\n- T = 23/365 = 0.063014, so √T = 0.25103.\n- ln(S/K) = ln(0.96704) = −0.033515.\n\n**Step 1 – Bracket the implied volatility.** I priced the call at three trial volatilities:\n\n| σ | d₁ | d₂ | Call price |\n|---|---|---|---|\n| 0.40 | −0.2836 | −0.3840 | 11.62 (too low) |\n| 0.75 | −0.0839 | −0.2722 | 27.07 (too high) |\n| 0.63 | −0.13285 | −0.29100 | 21.71 (0.044 too high) |\n\nAt σ = 0.63, the price is 200.21 − 178.50 = 21.71, using N(d₁) = 0.44716 and N(d₂) = 0.38552.\n\n**Step 2 – Refine with vega.** Vega is about S·φ(d₁)·√T ≈ 44.4 per unit of volatility. Removing the 0.044 excess gives σ ≈ 0.629.\n\n**Step 3 – Delta.** For a call, delta = N(d₁).\n- d₁ moves by about 0.462 per unit of σ, so at σ ≈ 0.629, d₁ ≈ −0.13331.\n- N(0.13331) ≈ 0.5 + 0.3989 × (0.13331 − 0.13331³/6) = 0.553021.\n- So N(−0.13331) ≈ 0.44698.\n\n{\"answer\": 0.4470, \"confidence\": 88}" }, { "taskId": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.25544432164825037, "answer": 0.255, "confidence": 80, "grade": "correct", "relError": 0.001739407027658275, "cost": 0.025632, "reply": "I'm solving for implied volatility using Black-Scholes with S=357.63, K=383, T=0.2137, trying sigma=0.25 and computing d1, d2 and their normal CDF values to compare against the market price of 5.0781.\n\nThe result at sigma=0.25 gives a call price of 7.27, too high, so I try sigma=0.20, getting 4.54, too low. Using vega to interpolate, I estimate sigma around 0.2104, recompute d1, d2, and the option price comes to 5.12, slightly overshooting, so I nudge sigma down to about 0.2096 and refine d1 accordingly, converging toward the correct implied volatility.\n\n\nThe call's delta is about **0.255** (N(d₁) ≈ 0.2550), at an implied volatility of about 20.96%. I found the volatility by hand iteration, so the third decimal of delta is approximate.\n\n**Setup**\n- T = 78/365 = 0.21370 years, so √T = 0.46228.\n- ln(S/K) = ln(357.63/383) = −0.06854.\n- With zero rates and no dividends:\n - d₁ = ln(S/K)/(σ√T) + σ√T/2\n - d₂ = d₁ − σ√T\n - C = S·N(d₁) − K·N(d₂)\n\n**Solving for implied volatility**\n1. σ = 0.25: d₁ = −0.5353, d₂ = −0.6509, which gives C ≈ 7.27. This is too high.\n2. σ = 0.20: d₁ = −0.6951, d₂ = −0.7876, which gives C ≈ 4.54. This is too low.\n3. Vega at σ = 0.20 is about 51.85 per unit of vol. Closing the 0.54 gap needs about +0.0104, so try σ ≈ 0.2104.\n4. σ = 0.2104: d₁ = −0.6560, d₂ = −0.7533, which gives C ≈ 5.12. This is 0.04 above 5.0781, so lower σ by about 0.0008. That puts the implied volatility at about 0.2096.\n\n**Delta**\n- The sensitivity of d₁ to σ is about 3.58. The −0.0008 adjustment therefore moves d₁ by about −0.0029, giving d₁ ≈ −0.6589.\n- Delta = N(d₁) = N(−0.6589) ≈ 0.2550.\n\n{\"answer\": 0.255, \"confidence\": 80}" }, { "taskId": "r_iv_then_delta-3", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.4262987878297406, "answer": 0.4263, "confidence": 88, "grade": "exact", "relError": 0.0000028434757357895263, "cost": 0.025732, "reply": "I need to find the implied volatility given the call price of 8.6404, with the stock trading well out of the money. I'm trying sigma = 0.6 and working through the Black-Scholes d1 calculation to see how close it gets.\n\nUsing vega to refine my guess, I iterate up to sigma = 0.684, computing N(d1) and N(d2) carefully through series expansion. That gives a call price of 8.640, matching the target almost exactly, with delta landing around 0.4263.\n\n\nThe call's delta is **≈ 0.4263**, using an implied volatility of about 68.4%. The vol was found by hand iteration and the normal CDF values are series approximations, so the last digit is approximate.\n\n**Setup.** S = 152.85, K = 162, T = 32/365 = 0.08767 years, √T = 0.29610, r = 0. ln(S/K) = −0.05814.\n\n**Step 1: find the implied volatility.**\n\nFirst trial, σ = 0.60:\n- σ√T = 0.17766\n- d1 = (−0.05814 + 0.01578)/0.17766 = −0.2384\n- d2 = −0.4161\n- N(d1) = 0.4058 and N(d2) = 0.3387\n- Price C = 152.85 × 0.4058 − 162 × 0.3387 ≈ 7.16, which is too low against 8.6404.\n\nNewton step:\n- Vega = S·φ(d1)·√T ≈ 152.85 × 0.3877 × 0.2961 ≈ 17.55 per unit of vol.\n- The price is short by 1.48, so σ ≈ 0.60 + 1.48/17.55 ≈ 0.684.\n\nSecond trial, σ = 0.684:\n- σ√T = 0.20253\n- d1 = (−0.05814 + 0.02051)/0.20253 = −0.1858\n- d2 = −0.3883\n- N(d1) ≈ 0.42631 and N(d2) ≈ 0.34890\n- Price C = 152.85 × 0.42631 − 162 × 0.34890 = 65.162 − 56.522 = 8.640, which matches the market price.\n\n**Step 2: delta.** With zero rates and no dividends, call delta = N(d1) = 0.4263.\n\n{\"answer\": 0.4263, \"confidence\": 88}" }, { "taskId": "r_iv_then_delta-4", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.6063214899790841, "answer": 0.6063, "confidence": 88, "grade": "exact", "relError": 0.00003544320865960742, "cost": 0.021612, "reply": "I need to back out implied volatility from the option price. With S=198.88, K=192, T=0.296 years, intrinsic value of 6.88 against the 28.30 price suggests substantial time value, pointing to high volatility. I'm setting up the Black-Scholes d1 term and testing a trial volatility around 35% to see how close it gets to matching the observed price.\n\nTesting σ=0.35 gives a call price of about 30.73 — too high. Dropping to σ=0.32 gives 28.43, which is close but still slightly above target, so I'm using vega to estimate the adjustment needed, landing on σ≈0.3182 as a refined guess and recalculating d1 for that volatility.\n\n\n**Delta ≈ 0.6063**, at an implied volatility of about 58.5%.\n\n**Inputs:** S = 198.88, K = 192, T = 108/365 = 0.29589, r = 0, no dividends.\n\nWith zero rates, C = S·N(d₁) − K·N(d₂), where\n- v = σ√T,\n- d₁ = ln(S/K)/v + v/2,\n- d₂ = d₁ − v,\n- ln(S/K) = ln(1.035833) = 0.035206.\n\n**Solving for implied volatility**\n\n| v | d₁ | d₂ | N(d₁) | N(d₂) | Model price |\n|---|---|---|---|---|---|\n| 0.35 | 0.27559 | −0.07441 | 0.60857 | 0.47034 | 30.727 (too high) |\n| 0.32 | 0.27002 | −0.04998 | 0.60643 | 0.48007 | 28.434 (slightly high) |\n\nAt v = 0.32 the price is 0.1375 above the market price of 28.2965. Vega in terms of v is S·φ(d₁) ≈ 198.88 × 0.3846 ≈ 76.5. The correction is therefore Δv ≈ −0.1375/76.5 ≈ −0.0018, giving v ≈ 0.3182.\n\nThis corresponds to σ = 0.3182/√0.29589 ≈ 0.585.\n\n**Delta**\n\nAt v = 0.3182:\n- d₁ = 0.035206/0.3182 + 0.3182/2 = 0.11064 + 0.1591 = 0.26974.\n- Delta = N(d₁) = N(0.26974) ≈ 0.6063.\n\n{\"answer\": 0.6063, \"confidence\": 88}" }, { "taskId": "r_iv_then_delta-5", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 0.6017546889052645, "answer": 0.6018, "confidence": 88, "grade": "exact", "relError": 0.00007529828279015368, "cost": 0.027372, "reply": "I need to find implied volatility by solving for the option's time value numerically, computing d1 using the Black-Scholes formula with the given strike, spot, and time to expiration, trying sigma=0.35 as an initial guess.\n\nChecking the resulting call price against the target shows it's too low, so I use vega to estimate a better sigma of about 0.3953, then recompute d1 and d2 with this refined volatility. The new call price comes out to 4.327, very close to the target of 4.3243, so I make a tiny adjustment and converge on a delta of approximately 0.6018.\n\n\n**Delta ≈ 0.6018** (implied volatility ≈ 39.5%). I solved for the vol by hand with Newton steps, so the last digit is approximate.\n\n**Inputs:** S = 108.83, K = 107, T = 14/365 = 0.038356 years, r = q = 0.\n\n**1. Implied volatility**\n\n- √T = 0.19585 and ln(S/K) = ln(1.017103) = 0.016958.\n- First guess σ = 0.35:\n - d1 = 0.016958/0.06855 + 0.06855/2 = 0.2817 and d2 = 0.2131.\n - N(d1) = 0.61089 and N(d2) = 0.58438.\n - The model price is 3.954, which is 0.370 below the market price.\n - Vega is S·φ(d1)·√T ≈ 8.17, so the Newton step gives σ ≈ 0.3953.\n- At σ = 0.3953:\n - σ√T = 0.07742, so d1 = 0.21904 + 0.03871 = 0.2578 and d2 = 0.1803.\n - N(d1) = 0.60170 and N(d2) = 0.57155.\n - The model price is 108.83 × 0.60170 − 107 × 0.57155 = 4.327, about 0.003 above 4.3243.\n - So σ comes down by roughly 0.0003–0.0004, giving implied vol ≈ 0.395.\n\n**2. Delta**\n\nWith zero rates and no dividends, delta = N(d1).\n\n- At σ = 0.3953, N(0.2578) ≈ 0.6017.\n- Lowering σ by about 0.0004 raises d1 by about 0.0002. This is because ∂d1/∂σ = −ln(S/K)/(σ²√T) + √T/2 ≈ −0.456.\n- That change adds about 0.0001 to delta, giving 0.6018.\n\n{\"answer\": 0.6018, \"confidence\": 88}" }, { "taskId": "r_deribit_straddle_breakeven-1", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 113713.12767394731, "answer": 113713.13, "confidence": 88, "grade": "exact", "relError": 2.04554455734024e-8, "cost": 0.009196, "reply": "For inverse options, the payoff in BTC is max(S-K,0)/S, so the upper breakeven solves (S-K)/S = total premium of 0.1118 BTC, giving S = K/(1-0.1118) = 101000/0.8882 ≈ 113713.13.\n\n\nThe straddle breaks even in BTC terms above **≈ 113,713 USD**.\n\nDeribit BTC options are inverse: they settle in BTC. A call at strike K pays max(S − K, 0)/S BTC at expiry, where S is the BTC/USD settlement price.\n\n- Total premium paid = 0.054 + 0.0578 = 0.1118 BTC.\n- Above the strike the put expires worthless, so the straddle breaks even when the call payoff equals the premium:\n\n(S − 101000)/S = 0.1118\n\nS = 101000 / (1 − 0.1118) = 101000 / 0.8882 ≈ 113,713.13\n\n{\"answer\": 113713.13, \"confidence\": 88}" }, { "taskId": "r_deribit_straddle_breakeven-2", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 93608.77985797288, "answer": 93608.78, "confidence": 85, "grade": "exact", "relError": 1.5172413832323368e-9, "cost": 0.009756, "reply": "For inverse options, the payoff in BTC is max(S-K,0)/S, so with total premium of 0.0706 BTC, the upper breakeven is S = K/(1-P) = 87000/0.9294 ≈ 93608.78.\n\n\n**The straddle breaks even above about 93,608.8 USD (BTC-denominated P&L).**\n\nBecause Deribit options settle in BTC, the payoff is the USD intrinsic value divided by the expiry price S. So the breakeven is not simply the strike plus the premium.\n\n1. The total premium is P = 0.015 + 0.0556 = 0.0706 BTC.\n2. Above the strike the put expires worthless, so only the call pays. Its BTC payoff is (S − 87000)/S.\n3. Set that payoff equal to the premium: (S − 87000)/S = 0.0706.\n4. Rearranging gives S = 87000 / (1 − 0.0706) = 87000 / 0.9294 ≈ 93,608.8.\n\n{\"answer\": 93608.78, \"confidence\": 85}" }, { "taskId": "r_deribit_straddle_breakeven-3", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 64193.23251006419, "answer": 64193.23, "confidence": 85, "grade": "exact", "relError": 3.910169481716397e-8, "cost": 0.01022, "reply": "For inverse options, payoff in BTC is max(S-K,0)/S for a call, with total premium of 0.0809 BTC. Solving for the upper breakeven, I get S = K/(1-P) = 59000/0.9191, which works out to roughly 64193.23.\n\n\nAbove about **64193.23 USD**, taking breakeven in BTC terms.\n\n**Total premium:** 0.0196 + 0.0613 = 0.0809 BTC.\n\n**Payoff above the strike:** Deribit options are inverse, so they pay out in BTC. Above the strike, the put expires worthless. The call pays (S − K)/S BTC, where S is the settlement price and K = 59000.\n\n**Breakeven condition:** set the payoff equal to the premium.\n\n(S − 59000)/S = 0.0809\n\nS = 59000 / (1 − 0.0809) = 59000 / 0.9191 ≈ 64193.23\n\n{\"answer\": 64193.23, \"confidence\": 85}" }, { "taskId": "r_deribit_straddle_breakeven-4", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 102706.75082914304, "answer": 102706.75, "confidence": 85, "grade": "exact", "relError": 8.072916682855671e-9, "cost": 0.009416, "reply": "For inverse options, the payoff in BTC is max(S-K,0)/S for a call, so with total premium of 0.0653 BTC, the upper breakeven solves (S-K)/S = 0.0653, giving S = 96000/0.9347 ≈ 102706.75.\n\n\nThe upper breakeven is about **102706.75 USD**.\n\nDeribit BTC options are inverse: they are paid in BTC, so the payoff in BTC is the USD intrinsic value divided by the settlement price S.\n\n1. Total premium paid is 0.022 + 0.0433 = 0.0653 BTC.\n\n2. Above the 96000 strike the put expires worthless. The call pays (S − 96000)/S BTC.\n\n3. Breakeven in BTC terms requires the payoff to equal the premium:\n (S − 96000)/S = 0.0653\n 1 − 96000/S = 0.0653\n S = 96000 / 0.9347 ≈ 102706.75\n\n{\"answer\": 102706.75, \"confidence\": 85}" }, { "taskId": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "claude:claude-opus-5-5", "expected": 73994.63806970509, "answer": 73994.64, "confidence": 88, "grade": "exact", "relError": 2.6086956541816556e-8, "cost": 0.01014, "reply": "For inverse options, the payoff is in BTC terms, normalized by spot price. With total premium at 0.0675 BTC, I'm solving for the upper breakeven where (S-K)/S equals that premium, giving S around 73994.64.\n\n\n**Break-even ≈ 73,994.64 USD.**\n\nOn Deribit, BTC options are inverse: they pay out in BTC. A call with strike K pays (S − K)/S BTC at expiry, where S is the BTC/USD settlement price.\n\n**Total premium paid**\n0.0412 + 0.0263 = 0.0675 BTC\n\n**Upper break-even**\nAbove the strike, only the call pays, so break-even is where the call payout equals the premium:\n\n(S − 69000)/S = 0.0675\n\n1 − 69000/S = 0.0675\n\nS = 69000 / 0.9325 ≈ 73,994.64\n\nAt this price the payout in BTC exactly covers the 0.0675 BTC paid, so the profit in BTC is positive above it.\n\n{\"answer\": 73994.64, \"confidence\": 88}" }, { "taskId": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 40.790551141452084, "answer": 40.69, "confidence": 55, "grade": "correct", "relError": 0.0024650596434306303, "cost": 0.000957, "reply": "I need to compute the Black-Scholes call price.\n\nInputs: S = 382.5, K = 350, T = 18/365 = 0.049315, σ = 0.607, r = 0.0597.\n\nsqrt(T) = sqrt(0.049315) ≈ 0.222072\nσ√T = 0.607 × 0.222072 ≈ 0.134790\n\nln(S/K) = ln(1.092857) ≈ 0.088773\n(r + σ²/2)T = (0.0597 + 0.184245) × 0.049315 ≈ 0.012019\n\nd1 = (0.088773 + 0.012019) / 0.134790 ≈ 0.100792 / 0.134790 ≈ 0.747829\nd2 = d1 − σ√T ≈ 0.613039\n\nN(d1) ≈ 0.772, more precisely N(0.747829) ≈ 0.77262\nN(d2) ≈ N(0.613039) ≈ 0.72996\n\nDiscount factor e^(−rT) = e^(−0.0597 × 0.049315) = e^(−0.0029443) ≈ 0.997060\n\nCall = S·N(d1) − K·e^(−rT)·N(d2)\n= 382.5 × 0.77262 − 350 × 0.997060 × 0.72996\n= 295.52 − 254.8\n\nComputing: 382.5 × 0.77262 ≈ 295.5 (295.5002); 350 × 0.99706 = 349.0710; × 0.72996 ≈ 254.7 (349.0710 × 0.72996 ≈ 254.7 — 349.071 × 0.73 = 254.82, minus 349.071 × 0.00004 = 0.014, giving 254.81).\n\nCall ≈ 295.50 − 254.81 ≈ 40.69\n\nGiven the rounding in the normal CDF values, my estimate is roughly 40.7, but I can't guarantee four significant digits by hand.\n\n{\"answer\": 40.69, \"confidence\": 55}" }, { "taskId": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 3.2317412561362406, "answer": 3.231, "confidence": 72, "grade": "exact", "relError": 0.00022936741449622883, "cost": 0.000896, "reply": "I need to compute the Black-Scholes call price.\n\nInputs: S = 22.89, K = 23, T = 171/365 = 0.468493, σ = 0.518, r = 0.0126.\n\nsqrt(T) = 0.684468.\nσ√T = 0.518 × 0.684468 = 0.354555.\n\nln(S/K) = ln(22.89/23) = ln(0.995217) = -0.004794.\n\n(r + σ²/2)T = (0.0126 + 0.134162) × 0.468493 = 0.146762 × 0.468493 = 0.068764.\n\nd1 = (-0.004794 + 0.068764)/0.354555 = 0.063970/0.354555 = 0.180438.\nd2 = d1 − 0.354555 = -0.174117.\n\nN(d1): N(0.180438) ≈ 0.571569.\nN(d2): N(-0.174117) ≈ 0.430912.\n\ne^(-rT) = e^(-0.0126 × 0.468493) = e^(-0.005903) ≈ 0.994114.\n\nCall = S·N(d1) − K·e^(-rT)·N(d2)\n= 22.89 × 0.571569 − 23 × 0.994114 × 0.430912\n= 13.08421 − 9.85303 (23 × 0.994114 = 22.86462; × 0.430912 = 9.85294)\n≈ 13.08421 − 9.85294\n≈ 3.23127\n\nRounding uncertainty in the normal CDF values is on the order of 0.0001 to 0.001, so the result is around 3.231.\n\n{\"answer\": 3.231, \"confidence\": 72}" }, { "taskId": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 16.773153687059427, "answer": 16.79, "confidence": 60, "grade": "correct", "relError": 0.0010043616874249044, "cost": 0.000875, "reply": "I need to compute the Black-Scholes call price.\n\nS = 89.61, K = 95, T = 152/365 = 0.416438, σ = 0.788, r = 0.0577.\n\nsqrt(T) = 0.645314\nσ√T = 0.788 × 0.645314 = 0.508507\n\nln(S/K) = ln(89.61/95) = ln(0.943263) = -0.058370\n\n(r + σ²/2)T = (0.0577 + 0.310472) × 0.416438 = 0.368172 × 0.416438 = 0.153318\n\nd1 = (-0.058370 + 0.153318) / 0.508507 = 0.094948 / 0.508507 = 0.186729\nd2 = d1 − σ√T = 0.186729 − 0.508507 = −0.321778\n\nN(d1): N(0.186729) ≈ 0.57406\nN(d2): N(−0.321778) ≈ 0.37369\n\ne^(−rT) = e^(−0.0577 × 0.416438) = e^(−0.024030) ≈ 0.976256\n\nCall = S·N(d1) − K·e^(−rT)·N(d2)\n= 89.61 × 0.57406 − 95 × 0.976256 × 0.37369\n= 51.4422 − 92.7443 × 0.37369\n= 51.4422 − 34.6539\n= 16.788\n\nRounding precision on the normal CDF values limits accuracy to roughly ±0.01.\n\n{\"answer\": 16.79, \"confidence\": 60}" }, { "taskId": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.6945968823131317, "answer": 0.6954, "confidence": 60, "grade": "correct", "relError": 0.0011562356631861555, "cost": 0.000849, "reply": "I need to compute the Black-Scholes call price with S=384.5, K=427, T=9/365, σ=0.426, r=0.0252.\n\nT = 9/365 = 0.024658\n√T = 0.157028\nσ√T = 0.426 × 0.157028 = 0.066894\n\nln(S/K) = ln(384.5/427) = ln(0.900468) = -0.104923\n\n(r + σ²/2)T = (0.0252 + 0.090738) × 0.024658 = 0.115938 × 0.024658 = 0.002859\n\nd1 = (-0.104923 + 0.002859)/0.066894 = -0.102064/0.066894 = -1.52583\nd2 = d1 - σ√T = -1.52583 - 0.066894 = -1.59272\n\nN(d1): N(-1.52583) ≈ 0.063486\nN(d2): N(-1.59272) ≈ 0.055600\n\nDiscount factor e^(-rT) = e^(-0.000621) = 0.999379\n\nCall = S·N(d1) - K·e^(-rT)·N(d2)\n= 384.5 × 0.063486 - 427 × 0.999379 × 0.055600\n= 24.4103 - 23.7149\n= 0.6954\n\nRounding carefully, the call price is about 0.6954.\n\n{\"answer\": 0.6954, \"confidence\": 60}" }, { "taskId": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 23.323917391138792, "answer": 23.31, "confidence": 60, "grade": "exact", "relError": 0.0005967004129452551, "cost": 0.000898, "reply": "I need to compute the Black-Scholes call price.\n\nInputs: S = 314.3, K = 294, T = 20/365 = 0.054795, σ = 0.355, r = 0.0006.\n\nsqrt(T) = 0.234080\nσ√T = 0.355 × 0.234080 = 0.083098\n\nln(S/K) = ln(314.3/294) = ln(1.069048) ≈ 0.066798\n\n(r + σ²/2)T = (0.0006 + 0.063013) × 0.054795 = 0.063613 × 0.054795 ≈ 0.003486\n\nd1 = (0.066798 + 0.003486)/0.083098 = 0.070284/0.083098 ≈ 0.84582\nd2 = d1 − σ√T = 0.84582 − 0.083098 = 0.76272\n\nN(d1): N(0.84582) ≈ 0.80118\nN(d2): N(0.76272) ≈ 0.77721\n\ne^(−rT) = e^(−0.0000329) ≈ 0.999967\n\nCall = S·N(d1) − K·e^(−rT)·N(d2)\n= 314.3 × 0.80118 − 294 × 0.999967 × 0.77721\n= 251.8109 − 228.4 (computed: 294 × 0.999967 = 293.9903; × 0.77721 = 228.5 approx)\n\nRefining: 293.9903 × 0.77721 ≈ 228.5 (293.9903 × 0.77721 = 228.4968)\n\nCall ≈ 251.8109 − 228.4968 ≈ 23.314\n\n{\"answer\": 23.31, \"confidence\": 60}" }, { "taskId": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": -0.280639388201964, "answer": -0.2807, "confidence": 85, "grade": "exact", "relError": 0.00021597751628646894, "cost": 0.00079, "reply": "Put delta = N(d1) − 1.\n\nT = 50/365 = 0.136986.\nσ√T = 0.466 × √0.136986 = 0.466 × 0.370130 = 0.172481.\n\nln(S/K) = ln(114.52/106) = ln(1.080377) ≈ 0.077288.\n\n(r + σ²/2)T = (0.0585 + 0.108578) × 0.136986 = 0.167078 × 0.136986 ≈ 0.022887.\n\nd1 = (0.077288 + 0.022887)/0.172481 = 0.100175/0.172481 ≈ 0.580765.\n\nN(0.580765): N(0.58) ≈ 0.719043, density φ(0.58) ≈ 0.337468; adding 0.000765×0.3375 ≈ 0.000258 gives N ≈ 0.719301.\n\nPut delta = 0.719301 − 1 = −0.280699.\n\n{\"answer\": -0.2807, \"confidence\": 85}" }, { "taskId": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": -0.31093085044630486, "answer": -0.3109, "confidence": 80, "grade": "exact", "relError": 0.00009921963761579435, "cost": 0.000829, "reply": "Setup: S=46.9, K=43, T=41/365=0.112329, σ=0.727, r=0.0327.\n\nd1 = [ln(S/K) + (r + σ²/2)T] / (σ√T)\n\nln(46.9/43) = ln(1.090698) ≈ 0.086860\n\nσ² /2 = 0.528529/2 = 0.264265; r+σ²/2 = 0.296965; times T = 0.033359\n\nNumerator = 0.086860 + 0.033359 = 0.120219\n\nσ√T = 0.727 × √0.112329 = 0.727 × 0.335156 = 0.243658\n\nd1 = 0.120219 / 0.243658 ≈ 0.49331\n\nd2 = d1 − 0.243658 ≈ 0.24965\n\nN(d1) for 0.49331: N(0.49)=0.68793, N(0.50)=0.69146; interpolate at 0.49331: 0.68793 + 0.331×0.00353 ≈ 0.68910.\n\nPut delta = N(d1) − 1 = −0.31090\n\nCheck the precision: d1 uncertainty is small enough that the delta is about −0.3109.\n\n{\"answer\": -0.3109, \"confidence\": 80}" }, { "taskId": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": -0.5641608283248785, "answer": -0.5643, "confidence": 85, "grade": "exact", "relError": 0.0002466879445260238, "cost": 0.0008215, "reply": "Put delta = N(d1) − 1.\n\nT = 65/365 = 0.178082\nσ√T = 0.528 × √0.178082 = 0.528 × 0.422001 = 0.222817\n\nln(S/K) = ln(215.92/231) = ln(0.934718) = −0.067607\n(r + σ²/2)T = (0.0376 + 0.139392) × 0.178082 = 0.176992 × 0.178082 = 0.031524\n\nd1 = (−0.067607 + 0.031524)/0.222817 = −0.036083/0.222817 = −0.161948\n\nN(−0.161948): N(0.161948) ≈ 0.5 + φ(0)·(0.161948 − 0.161948³/6·…) ≈ 0.5 + 0.398942·(0.161948 − 0.000708) ≈ 0.5 + 0.398942·0.161240 ≈ 0.564327 (small higher-order correction ≈ +0.0000, negligible at this precision). So N(−0.161948) ≈ 0.435673.\n\nPut delta = 0.435673 − 1 = −0.564327\n\n{\"answer\": -0.5643, \"confidence\": 85}" }, { "taskId": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": -0.3853788948381438, "answer": -0.3854, "confidence": 80, "grade": "exact", "relError": 0.00005476470595277504, "cost": 0.000843, "reply": "I need d1 for the put delta: Δ_put = N(d1) − 1.\n\nInputs: S = 291.06, K = 286, T = 113/365 = 0.309589, σ = 0.729, r = 0.0594.\n\nln(S/K) = ln(1.017692) ≈ 0.017540.\n\nσ²/2 = 0.265641 (0.729² = 0.531441, halved).\n(r + σ²/2)T = (0.0594 + 0.265641) × 0.309589 = 0.325041 × 0.309589 ≈ 0.100630.\n\nNumerator: 0.017540 + 0.100630 = 0.118170.\n\nσ√T: √0.309589 ≈ 0.556409; × 0.729 ≈ 0.405662.\n\nd1 ≈ 0.118170 / 0.405662 ≈ 0.291280.\n\nN(0.291280): N(0.29) ≈ 0.614092, with density φ(0.29) ≈ 0.38triangle 0.3825, so N(0.29128) ≈ 0.614092 + 0.00128 × 0.3824 ≈ 0.614581.\n\nΔ_put = 0.614581 − 1 ≈ −0.385419.\n\nDelta ≈ −0.3854\n\n{\"answer\": -0.3854, \"confidence\": 80}" }, { "taskId": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": -0.4080759766740296, "answer": -0.408, "confidence": 82, "grade": "exact", "relError": 0.00018618266786709934, "cost": 0.000835, "reply": "Setup: S=146.31, K=152, T=162/365=0.443836, σ=0.799, r=0.0456.\n\nd1 = [ln(S/K) + (r + σ²/2)T] / (σ√T)\n\nln(146.31/152) = ln(0.962566) ≈ -0.038103\n\nσ² /2 = 0.638401/2 = 0.319201; r+σ²/2 = 0.364801; times T: 0.364801×0.443836 ≈ 0.161911\n\nNumerator: -0.038103 + 0.161911 = 0.123808\n\nσ√T = 0.799×√0.443836 = 0.799×0.666210 ≈ 0.532282\n\nd1 ≈ 0.123808/0.532282 ≈ 0.232590\n\nPut delta = N(d1) − 1 = N(0.232590) − 1.\n\nN(0.232590): φ(0.23259)... N(0.23)=0.59095, N(0.24)=0.59483. Interpolating at 0.23259: 0.59095 + 0.259×0.00388 ≈ 0.59095 + 0.001005 = 0.591955.\n\nPut delta ≈ 0.591955 − 1 = −0.408045\n\n{\"answer\": -0.4080, \"confidence\": 82}" }, { "taskId": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.007721610843754648, "answer": 0.007723, "confidence": 80, "grade": "exact", "relError": 0.00017990498012152137, "cost": 0.000808, "reply": "Gamma = φ(d1) / (S σ √T)\n\nInputs: S = 168.77, K = 150, T = 34/365 = 0.093151, σ = 0.844, r = 0.003.\n\nσ√T = 0.844 × √0.093151 = 0.844 × 0.305207 = 0.257595\n\nln(S/K) = ln(1.125133) = 0.117832\n\n(r + σ²/2)T = (0.003 + 0.356168) × 0.093151 = 0.359168 × 0.093151 = 0.033458\n\nd1 = (0.117832 + 0.033458) / 0.257595 = 0.151290 / 0.257595 = 0.587331\n\nφ(d1) = (1/√(2π)) exp(−d1²/2) = 0.398942 × exp(−0.172499) = 0.398942 × 0.841564 = 0.335737\n\nDenominator: S σ √T = 168.77 × 0.257595 = 43.4704\n\nGamma = 0.335737 / 43.4704 = 0.0077230\n\n{\"answer\": 0.007723, \"confidence\": 80}" }, { "taskId": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.011719035620003572, "answer": 0.011697, "confidence": 80, "grade": "correct", "relError": 0.0018803270779344594, "cost": 0.000823, "reply": "Gamma = φ(d1) / (S σ √T), where d1 = [ln(S/K) + (r + σ²/2)T] / (σ√T).\n\nT = 71/365 = 0.194521\nσ√T = 0.161 × √0.194521 = 0.161 × 0.441043 = 0.071008\n\nln(S/K) = ln(350.55/375) = ln(0.934800) = −0.067604\n\n(r + σ²/2)T = (0.0448 + 0.0129610) × 0.194521 = 0.0577610 × 0.194521 = 0.011236\n\nd1 = (−0.067604 + 0.011236) / 0.071008 = −0.056368 / 0.071008 = −0.79384\n\nφ(d1) = (1/√(2π)) exp(−d1²/2); d1² = 0.630183, half = 0.315092, exp(−0.315092) = 0.729835\nφ(d1) = 0.398942 × 0.729835 = 0.291172\n\nS σ √T = 350.55 × 0.071008 = 24.8926\n\nGamma = 0.291172 / 24.8926 = 0.011697\n\n{\"answer\": 0.011697, \"confidence\": 80}" }, { "taskId": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.00332788170697725, "answer": 0.003328, "confidence": 85, "grade": "exact", "relError": 0.0000355460419466294, "cost": 0.000826, "reply": "Gamma = φ(d1) / (S σ √T), with d1 = [ln(S/K) + (r + σ²/2)T] / (σ√T).\n\nT = 107/365 = 0.293151\n√T = 0.541437\nσ√T = 0.464 × 0.541437 = 0.251227\n\nln(S/K) = ln(471.11/474) = ln(0.993903) = −0.006116\n\nσ²/2 = 0.215296/2 = 0.107648\nr + σ²/2 = 0.05030 + 0.107648 = 0.157948\n(r + σ²/2)T = 0.157948 × 0.293151 = 0.046300\n\nNumerator = −0.006116 + 0.046300 = 0.040184\nd1 = 0.040184 / 0.251227 = 0.159955\n\nφ(d1) = (1/√(2π)) exp(−d1²/2) = 0.398942 × exp(−0.012792) = 0.398942 × 0.987289 = 0.393873\n\nS σ √T = 471.11 × 0.251227 = 118.3543\n\nGamma = 0.393873 / 118.3543 = 0.0033278\n\n{\"answer\": 0.003328, \"confidence\": 85}" }, { "taskId": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.015589164355628841, "answer": 0.01559, "confidence": 80, "grade": "exact", "relError": 0.00005360417993518107, "cost": 0.000796, "reply": "Gamma = φ(d1) / (S σ √T)\n\nT = 147/365 = 0.402740\nσ√T = 0.446 × √0.402740 = 0.446 × 0.634611 = 0.283057\n\nln(S/K) = ln(0.8711) = -0.138 (ln 0.8711 ≈ -0.138180)\n(r + σ²/2)T = (0.0514 + 0.099458) × 0.402740 = 0.150858 × 0.402740 = 0.060762\n\nd1 = (-0.138180 + 0.060762) / 0.283057 = -0.077418 / 0.283057 = -0.273489\n\nφ(d1) = (1/√(2π)) exp(-d1²/2) = 0.398942 × exp(-0.037398) = 0.398942 × 0.963290 = 0.384307\n\nGamma = 0.384307 / (87.11 × 0.283057) = 0.384307 / 24.6585 = 0.015586\n\n{\"answer\": 0.01559, \"confidence\": 80}" }, { "taskId": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.00596435584251129, "answer": 0.005966, "confidence": 80, "grade": "exact", "relError": 0.0002756638825924912, "cost": 0.000843, "reply": "Gamma = φ(d1) / (S σ √T), where d1 = [ln(S/K) + (r + σ²/2)T] / (σ√T).\n\nT = 103/365 = 0.282192\n√T = 0.531219\nσ√T = 0.24 × 0.531219 = 0.127493\n\nln(S/K) = ln(335.65/300) = ln(1.118833) = 0.112231\n\n(r + σ²/2)T = (0.0003 + 0.0288) × 0.282192 = 0.0291 × 0.282192 = 0.008212\n\nNumerator = 0.112231 + 0.008212 = 0.120443\nd1 = 0.120443 / 0.127493 = 0.944700\n\nφ(d1) = (1/√(2π)) exp(-d1²/2)\nd1² = 0.892460, half = 0.446230\nexp(-0.446230) = 0.640000 (approx: e^-0.44623 ≈ 0.63998)\nφ(d1) = 0.398942 × 0.63998 = 0.255308\n\nDenominator = S σ √T = 335.65 × 0.127493 = 42.7937\n\nGamma = 0.255308 / 42.7937 = 0.0059656\n\nGamma ≈ 0.005966\n\n{\"answer\": 0.005966, \"confidence\": 80}" }, { "taskId": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.646446181061521, "answer": 0.6463, "confidence": 80, "grade": "exact", "relError": 0.00022613028865130883, "cost": 0.0008724, "reply": "I need the Black-Scholes vega for a call (assuming a European call, since no type is specified; vega is the same for call and put).\n\nInputs: S = 424.81, K = 380, T = 74/365 = 0.202740, σ = 0.55, r = 0.0025.\n\nd1 = [ln(S/K) + (r + σ²/2)T] / (σ√T)\n\nln(S/K) = ln(1.117921) ≈ 0.111583\nσ²/2 = 0.15125; r + σ²/2 = 0.15375\n(0.15375)(0.202740) ≈ 0.031171\nNumerator ≈ 0.142754\nσ√T = 0.55 × √0.202740 = 0.55 × 0.450266 ≈ 0.247646\nd1 ≈ 0.576471\nd2 = d1 − σ√T ≈ 0.328825\n\nStandard normal density at d1:\nφ(d1) = exp(−d1²/2)/√(2π) = exp(−0.166199)/2.506628 ≈ 0.846915/2.506628 ≈ 0.337888\n\nVega (per unit σ) = S φ(d1) √T = 424.81 × 0.337888 × 0.450266 ≈ 64.6268\n\nPer 1 percentage point (0.01) change in volatility: 0.646268\n\n**Answer: approximately 0.6463** (change in option price per 1 vol point, for a call or put)\n\n{\"answer\": 0.6463, \"confidence\": 80}" }, { "taskId": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.26781898170848095, "answer": 0.2679, "confidence": 80, "grade": "exact", "relError": 0.00030251138661733373, "cost": 0.000899, "reply": "I need the vega of a European option (call presumably, but vega is the same for call and put).\n\nInputs: S = 146.99, K = 155, T = 104/365 = 0.284932, σ = 0.163, r = 0.0024.\n\nsqrt(T) = sqrt(0.284932) ≈ 0.533791.\nσ√T = 0.163 × 0.533791 ≈ 0.087008.\n\nln(S/K) = ln(146.99/155) = ln(0.948323) ≈ -0.052990.\n\n(r + σ²/2)T = (0.0024 + 0.0132845) × 0.284932 = 0.0156845 × 0.284932 ≈ 0.0044694.\n\nd1 = (-0.052990 + 0.0044694)/0.087008 = -0.0485206/0.087008 ≈ -0.557622.\nd2 = d1 − σ√T = -0.557622 − 0.087008 ≈ -0.644630.\n\nφ(d1) = (1/√(2π)) exp(-d1²/2). d1² = 0.310962; half = 0.155481; exp(-0.155481) ≈ 0.855946; times 0.398942 ≈ 0.341465.\n\nVega (per unit vol) = S φ(d1) √T = 146.99 × 0.341465 × 0.533791.\n146.99 × 0.341465 ≈ 50.1958.\n× 0.533791 ≈ 26.7928.\n\nFor a 1 percentage point change: 26.7928 × 0.01 ≈ 0.267928.\n\nVega ≈ 0.2679\n\n{\"answer\": 0.2679, \"confidence\": 80}" }, { "taskId": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.45587809695630876, "answer": 0.4558, "confidence": 72, "grade": "exact", "relError": 0.0001713110518583802, "cost": 0.000939, "reply": "Setting up: S = 469.67, K = 512, T = 47/365 = 0.128767, σ = 0.254, r = 0.0139.\n\nVega (per unit vol) = S·φ(d1)·√T.\n\nd1 = [ln(S/K) + (r + σ²/2)T] / (σ√T)\n\nln(S/K) = ln(469.67/512) = ln(0.917324) ≈ -0.086, more precisely: ln(0.917324) = -0.086 (compute: ln(0.917324) ≈ -0.0863 ... let's refine: ln(0.92) = -0.083382, and 0.917324/0.92 = 0.997091, ln = -0.002913; total = -0.086295).\n\nσ² /2 = 0.032258/2 = 0.032258... (0.254² = 0.064516; /2 = 0.032258)\nr + σ²/2 = 0.0139 + 0.032258 = 0.046158\ntimes T: 0.046158 × 0.128767 = 0.0059436\n\nNumerator: -0.086295 + 0.005944 = -0.080351\n\nσ√T: √0.128767 = 0.358843; × 0.254 = 0.091136\n\nd1 = -0.080351 / 0.091136 = -0.88167\n\nφ(d1) = (1/√(2π)) exp(-d1²/2): d1² = 0.77734; half = 0.38867; exp(-0.38867) = 0.67796; × 0.398942 = 0.27046\n\nVega = S·φ(d1)·√T = 469.67 × 0.27046 × 0.358843\n469.67 × 0.27046 = 127.03\n× 0.358843 = 45.583\n\nPer 1 percentage point (0.01 change in σ): 45.583 × 0.01 = 0.45583\n\nAnswer ≈ 0.4558\n\n{\"answer\": 0.4558, \"confidence\": 72}" }, { "taskId": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.6097213395050691, "answer": 0.6097, "confidence": 80, "grade": "exact", "relError": 0.00003499878335639152, "cost": 0.000839, "reply": "I need vega = S·φ(d1)·√T / 100 (per 1 vol point). Vega is the same for call and put.\n\nT = 61/365 = 0.167123\n√T = 0.408809\nσ√T = 0.726 × 0.408809 = 0.296805\n\nln(S/K) = ln(383.12/429) = ln(0.893051) = -0.113198\n(r + σ²/2)T = (0.0203 + 0.263538) × 0.167123 = 0.283838 × 0.167123 = 0.047432\n\nd1 = (-0.113198 + 0.047432)/0.296805 = -0.065766/0.296805 = -0.221586\n\nφ(d1) = (1/√(2π)) exp(-d1²/2)\nd1² = 0.049100, half = 0.024550, exp(-0.024550) = 0.975749\nφ(d1) = 0.398942 × 0.975749 = 0.389268\n\nVega (per unit vol) = S·φ(d1)·√T = 383.12 × 0.389268 × 0.408809\n383.12 × 0.389268 = 149.1420\n× 0.408809 = 60.9703\n\nPer 1 percentage point: 60.9703/100 = 0.609703\n\n{\"answer\": 0.6097, \"confidence\": 80}" }, { "taskId": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.2403754082675673, "answer": 0.2403, "confidence": 80, "grade": "exact", "relError": 0.00031371040869261586, "cost": 0.000903, "reply": "Setting up: S = 158.91, K = 148, T = 65/365 = 0.178082, σ = 0.564, r = 0.0599.\n\nVega (per unit vol) = S·φ(d1)·√T, and the requested quantity is vega/100.\n\nd1 = [ln(S/K) + (r + σ²/2)T] / (σ√T)\n\nln(158.91/148) = ln(1.073716) ≈ 0.071179\nσ²/2 = 0.159048; r + σ²/2 = 0.218948; times T = 0.038991\nNumerator = 0.110170\nσ√T = 0.564 × 0.422000 = 0.238008 (√0.178082 ≈ 0.421996, so ≈ 0.237806)\n\nRecompute: σ√T = 0.564 × 0.421996 ≈ 0.238006\nd1 ≈ 0.110170 / 0.238006 ≈ 0.462860\nd2 = d1 − 0.238006 ≈ 0.224854\n\nφ(d1) = (1/√(2π)) exp(−d1²/2); d1² = 0.214237, half = 0.107118, exp(−0.107118) ≈ 0.898399\nφ(d1) ≈ 0.398942 × 0.898399 ≈ 0.358385\n\nVega = S·φ(d1)·√T = 158.91 × 0.358385 × 0.421996 ≈ 158.91 × 0.151240 ≈ 24.0319\n\nPer 1 percentage point: 24.0319 / 100 ≈ 0.24032\n\nAnswer ≈ 0.2403\n\n{\"answer\": 0.2403, \"confidence\": 80}" }, { "taskId": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 61.799447576373446, "answer": 61.77, "confidence": 60, "grade": "exact", "relError": 0.0004765022589732766, "cost": 0.00119, "reply": "I need to solve for sigma such that the Black-Scholes call price equals 3.5195, with S=27.09, K=29, T=136/365≈0.372603, r=0.054.\n\nSetup: d1 = [ln(S/K) + (r+σ²/2)T]/(σ√T), d2 = d1 − σ√T.\n\nln(27.09/29) = ln(0.934138) ≈ −0.068 (precisely −0.067 9...; compute: ln(0.934138) = −0.067 9 approximately: ln(0.934138) ≈ −0.06794).\n\n√T = √0.372603 ≈ 0.610412.\n\nTry σ = 0.30: σ√T = 0.183124; σ²/2·T = 0.045·0.372603=0.016767; rT = 0.054·0.372603=0.020121.\nd1 = (−0.06794 + 0.020121 + 0.016767)/0.183124 = (−0.031052)/0.183124 = −0.16957.\nd2 = −0.35269.\nN(d1)=0.43267, N(d2)=0.36213.\nCall = 27.09·0.43267 − 29·e^{−0.020121}·0.36213\n= 11.7204 − 29·0.980083·0.36213 = 11.7204 − 10.2947 = 1.4257? Recheck: 29·0.980083=28.4224; ×0.36213=10.2930. Call ≈ 11.7204−10.2930=1.4274.\n\nTarget 3.5195 is higher, so σ larger. Try σ = 0.45: σ√T=0.274686; σ²/2·T=0.10125·0.372603=0.037731; d1=(−0.06794+0.020121+0.037731)/0.274686=(−0.010088)/0.274686=−0.03673; d2=−0.31142.\nN(d1)=0.48535, N(d2)=0.37779.\nCall = 27.09·0.48535 − 28.4224·0.37779 = 13.1482 − 10.7380 = 2.4102.\n\nTry σ = 0.55: σ√T=0.335727; σ²/2·T=0.15125·0.372603=0.056356; d1=(−0.06794+0.020121+0.056356)/0.335727=0.008537/0.335727=0.02543; d2=−0.31030.\nN(d1)=0.51014, N(d2)=0.37818.\nCall = 27.09·0.51014 − 28.4224·0.37818 = 13.8189 − 10.7477 = 3.0712.\n\nTry σ = 0.62: σ√T=0.378455; σ²/2·T=0.1922·0.372603=0.071617; d1=(−0.06794+0.020121+0.071617)/0.378455=0.023798/0.378455=0.06289; d2=−0.31557.\nN(d1)=0.52507, N(d2)=0.37613.\nCall = 27.09·0.52507 − 28.4224·0.37613 = 14.2254 − 10.6905 = 3.5349.\n\nSlightly above target 3.5195. Vega ≈ S√T φ(d1) ≈ 27.09·0.6104·0.3981 ≈ 6.58. Difference −0.0154 → Δσ ≈ −0.0023. σ ≈ 0.6177.\n\nSo implied volatility ≈ 61.8%.\n\n{\"answer\": 61.77, \"confidence\": 60}" }, { "taskId": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 20.900010973832025, "answer": 20.89, "confidence": 60, "grade": "exact", "relError": 0.0004789937117525202, "cost": 0.001131, "reply": "I need to solve for sigma such that the Black-Scholes put price equals 13.6926.\n\nSetup: S=490.18, K=490, T=53/365=0.145205, r=0.0508.\n\nPut-call parity: P = C - S + K e^{-rT}. Compute K e^{-rT}: rT = 0.0508×0.145205 = 0.0073764, e^{-0.0073764} ≈ 0.992651, so K e^{-rT} ≈ 486.3990. Discount of strike ≈ 486.3990.\n\nATM-ish: S≈K, so at-the-money forward roughly. Put price ≈ C - S + PV(K). Call price C = P + S - PV(K) = 13.6926 + 490.18 - 486.3990 = 17.4736.\n\nApproximate ATM call: C ≈ 0.4 S σ √T + small correction. Let's use the more precise form: with d1 = (ln(S/K)+(r+σ²/2)T)/(σ√T), and S/K = 1.000367, ln ≈ 0.000367.\n\nTry σ = 0.20: σ√T = 0.2×0.381048=0.0762096. (r+σ²/2)T = (0.0508+0.02)×0.145205=0.0103 (0.0708×0.145205=0.010280). d1 = (0.000367+0.010280)/0.0762096 = 0.13975; d2 = 0.06331.\nN(d1)=0.55557, N(d2)=0.52524.\nC = 490.18×0.55557 − 490×0.992651×0.52524 = 272.3 − 255.5… compute: 490.18×0.55557=272.3 (272.3036); 486.399×0.52524=255.4 (255.4772). C ≈ 16.826.\n\nNeed 17.4736, so σ higher. Vega ≈ S√T φ(d1) = 490.18×0.381048×0.39556 ≈ 73.90. Δσ ≈ (17.4736−16.826)/73.9 ≈ 0.00877. σ ≈ 0.2088.\n\nCheck σ = 0.2088: σ√T = 0.079563; σ²/2=0.021797; (0.0508+0.021797)×0.145205=0.010547; d1 = (0.000367+0.010547)/0.079563 = 0.13719; d2 = 0.05763.\nN(d1)=0.554568, N(d2)=0.522977.\nC = 490.18×0.554568 − 486.399×0.522977 = 271.8 (271.8402) − 254.3 (254.4 compute: 486.399×0.522977 = 254.3… 486.399×0.5=243.1995; ×0.022977=11.1765; total 254.376). C = 271.8402−254.376 = 17.464.\n\nNeed 17.4736, difference 0.0096; vega ≈ 73.7 → Δσ ≈ 0.00013. σ ≈ 0.2089.\n\nImplied volatility ≈ 20.89%.\n\n{\"answer\": 20.89, \"confidence\": 60}" }, { "taskId": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 81.50015130720656, "answer": 81.48, "confidence": 60, "grade": "exact", "relError": 0.00024725484411191554, "cost": 0.001216, "reply": "I need the volatility σ such that the Black-Scholes put price equals 20.922.\n\nSetup: S=79.84, K=86, T=161/365≈0.44110, r=0.0004.\n\nPut price P = K e^{-rT} N(-d2) - S N(-d1).\n\nDiscounting: e^{-rT} ≈ 0.999824, so K e^{-rT} ≈ 85.9849.\n\nIntrinsic value (forward-ish): K e^{-rT} - S ≈ 6.145, so the put must carry substantial time value; price 20.922 implies high volatility.\n\nTry σ=0.40: σ√T = 0.40×0.66416 = 0.26566.\nln(S/K) = ln(0.92842) = -0.074704.\nd1 = (-0.074704 + (0.0004+0.08)×0.44110)/0.26566 = (-0.074704+0.035444)/0.26566 = -0.14779.\nd2 = -0.14779-0.26566 = -0.41345.\nN(-d1)=N(0.14779)≈0.55875; N(-d2)=N(0.41345)≈0.66024.\nP = 85.9849×0.66024 - 79.84×0.55875 = 56.7742 - 44.6064 = 12.168. Too low.\n\nTry σ=0.55: σ√T=0.36511. Drift term (0.0004+0.15125)×0.44110=0.066912. d1=(-0.074704+0.066912)/0.36511=-0.021346. d2=-0.38646.\nN(-d1)=N(0.021346)≈0.50851; N(-d2)=N(0.38646)≈0.65049.\nP = 85.9849×0.65049 - 79.84×0.50851 = 55.9337 - 40.5995 = 15.334.\n\nTry σ=0.70: σ√T=0.46491. Drift (0.0004+0.245)×0.44110=0.108268. d1=(0.033564)/0.46491=0.072200. d2=-0.39271.\nN(-d1)=N(-0.0722)=0.47123; N(-d2)=N(0.39271)=0.65271.\nP = 85.9849×0.65271 - 79.84×0.47123 = 56.1278 - 37.6230 = 18.505.\n\nTry σ=0.80: σ√T=0.53133. Drift (0.0004+0.32)×0.44110=0.141427. d1=0.066723/0.53133=0.125577. d2=-0.40575.\nN(-d1)=N(-0.125577)=0.45003; N(-d2)=N(0.40575)=0.65747.\nP = 85.9849×0.65747 - 79.84×0.45003 = 56.5383 - 35.9302 = 20.608.\n\nTry σ=0.82: σ√T=0.54461. Drift (0.0004+0.3362)×0.44110=0.148497. d1=0.073793/0.54461=0.135495. d2=-0.40912.\nN(-d1)=N(-0.135495)=0.44610; N(-d2)=N(0.40912)=0.65878.\nP = 85.9849×0.65878 - 79.84×0.44610 = 56.6479 - 35.6163 = 21.032.\n\nInterpolate between 0.80 (20.608) and 0.82 (21.032): slope ≈ 21.2 per unit. Need 20.922-20.608=0.314 → Δσ≈0.0148. σ≈0.8148.\n\nEstimate ≈ 81.5%.\n\n{\"answer\": 81.48, \"confidence\": 60}" }, { "taskId": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 87.70003216471197, "answer": 87.7, "confidence": 55, "grade": "exact", "relError": 3.6675826876828525e-7, "cost": 0.001253, "reply": "I need the implied volatility σ such that the Black-Scholes put price equals 75.8825.\n\nSetup: S = 340.53, K = 362, T = 105/365 = 0.287671, r = 0.0106.\n\nDiscounted strike: K e^{-rT} = 362 × e^{-0.0030486} ≈ 362 × 0.996957 ≈ 360.9 (362 × 0.996957 = 360.9)\n\nPut price P = K e^{-rT} N(-d2) − S N(-d1).\n\nCheck intrinsic: K − S = 21.47, and the put price 75.88 is well above that, so volatility must be substantial. Since put ≈ 75.88 is large relative to spot, try σ near 0.5.\n\nTry σ = 0.50:\nσ√T = 0.5 × 0.536349 = 0.268174\nd1 = [ln(340.53/362) + (0.0106 + 0.125)×0.287671]/0.268174\nln(0.940690) = −0.061033\ndrift term = 0.1356 × 0.287671 = 0.039012\nd1 = (−0.061033 + 0.039012)/0.268174 = −0.082117\nd2 = −0.350291\nN(−d1) = N(0.082117) ≈ 0.532730\nN(−d2) = N(0.350291) ≈ 0.636806\nP = 360.9 × 0.636806 − 340.53 × 0.532730 ≈ 229.84 − 181.42 = 48.42\n\nToo low, so increase σ.\n\nTry σ = 0.65:\nσ√T = 0.348727\nd1 = (−0.061033 + (0.0106 + 0.21125)×0.287671)/0.348727\ndrift = 0.22185 × 0.287671 = 0.063819\nd1 = 0.002786/0.348727 = 0.007989\nd2 = −0.340738\nN(−d1) = N(−0.007989) ≈ 0.496814\nN(−d2) = N(0.340738) ≈ 0.633314\nP = 360.9 × 0.633314 − 340.53 × 0.496814 ≈ 228.57 − 169.19 = 59.38\n\nTry σ = 0.80:\nσ√T = 0.429079\ndrift = (0.0106 + 0.32)×0.287671 = 0.095 (0.3306×0.287671 = 0.095105)\nd1 = 0.034072/0.429079 = 0.079410\nd2 = −0.349669\nN(−d1) = N(−0.07941) ≈ 0.468358\nN(−d2) = N(0.349669) ≈ 0.636601\nP = 360.9 × 0.636601 − 340.53 × 0.468358 ≈ 229.75 − 159.50 = 70.25\n\nTry σ = 0.90:\nσ√T = 0.482714\ndrift = (0.0106 + 0.405)×0.287671 = 0.119555\nd1 = 0.058522/0.482714 = 0.121236\nd2 = −0.361478\nN(−d1) = N(−0.121236) ≈ 0.451756\nN(−d2) = N(0.361478) ≈ 0.641108\nP = 360.9 × 0.641108 − 340.53 × 0.451756 ≈ 231.37 − 153.84 = 77.53\n\nInterpolate between 0.80 (70.25) and 0.90 (77.53): slope ≈ 72.8 per unit σ. Need 75.88 − 70.25 = 5.63 → Δσ ≈ 0.0773, giving σ ≈ 0.877.\n\nCheck σ = 0.877 approximately: the curvature is small, so the estimate is close to 0.876–0.878.\n\nImplied volatility ≈ 87.7% per year.\n\n{\"answer\": 87.7, \"confidence\": 55}" }, { "taskId": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 33.39992865638508, "answer": 33.4, "confidence": 72, "grade": "exact", "relError": 0.0000021360409375049508, "cost": 0.001119, "reply": "I need to solve for sigma such that BS call price = 11.908, with S=140.66, K=143, T=161/365=0.44110, r=0.019.\n\nRough check: at-the-money-ish forward. Forward F = 140.66·e^{0.019·0.4411} = 140.66·1.008410 ≈ 141.844.\n\nTry sigma = 0.30: sqrt(T)=0.66415. d1 = [ln(140.66/143) + (0.019+0.045)·0.4411]/(0.30·0.66415)\nln(140.66/143)=ln(0.983636)= -0.016496.\n(0.064)·0.4411=0.028230. Numerator = 0.011734. Denominator = 0.199245. d1=0.058896. d2 = d1 - 0.199245 = -0.140349.\nN(d1)=0.52348, N(d2)=0.44419.\nCall = 140.66·0.52348 - 143·e^{-0.008381}·0.44419\ne^{-0.008381}=0.991654; 143·0.991654=141.8066; ×0.44419=62.9827.\n140.66·0.52348=73.6310. Call = 73.6310 - 62.9827 = 10.6483. Too low; need higher sigma.\n\nTry sigma = 0.33: sigma·sqrtT = 0.219172. σ²/2=0.05445; (0.019+0.05445)·0.4411=0.032421. Numerator = -0.016496+0.032421=0.015925. d1=0.072652. d2=-0.146520.\nN(d1)=0.52896, N(d2)=0.44173.\n140.66·0.52896=74.4 (140.66·0.52896=74.4 approx: 140.66·0.5=70.33, ·0.02896=4.0735; total 74.4035).\n143·0.991654=141.8066·0.44173=62.6374. Call=74.4035-62.6374=11.7661. Still slightly low.\n\nSlope ≈ (11.7661-10.6483)/0.03 = 37.26 per unit sigma. Need 11.908-11.7661=0.1419 → Δσ≈0.00381. σ≈0.33381.\n\nCheck vega-based refinement at σ=0.3338: sigma·sqrtT=0.221682; σ²/2=0.055720; (0.074720)(0.4411)=0.032960; numerator=0.016464; d1=0.074271; d2=-0.147411.\nN(d1)=0.529597; N(d2)=0.441399.\n140.66·0.529597=74.4917. 141.8066·0.441399=62.5902. Call=11.9015. Slightly below 11.908 by 0.0065; vega ≈ S·φ(d1)·√T ≈ 140.66·0.3982·0.66415 ≈ 37.2. Δσ≈0.000175. σ≈0.33398.\n\nSo σ ≈ 33.40%.\n\n{\"answer\": 33.40, \"confidence\": 72}" }, { "taskId": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 6.438461538461515, "answer": 6.4385, "confidence": 85, "grade": "exact", "relError": 0.0000059737156548775595, "cost": 0.000698, "reply": "Stop distance = 0.94593 − 0.94203 = 0.00390 USD per AUD.\n\nFor AUD/USD, the quote is USD per AUD, so loss per unit = 0.00390 USD.\n\nUnits needed = 2511 / 0.00390 = 643,846.15 AUD.\n\nLots = 643,846.15 / 100,000 = 6.4385 standard lots.\n\n{\"answer\": 6.4385, \"confidence\": 85}" }, { "taskId": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 12.3258666666662, "answer": 12.3259, "confidence": 85, "grade": "exact", "relError": 0.0000027043399627036095, "cost": 0.000679, "reply": "Risk per unit = 154.513 − 154.363 = 0.150 JPY per EUR.\n\nUnits needed = 184888 / 0.150 = 1,232,586.67 EUR.\n\nLots = 1,232,586.67 / 100,000 = 12.3258667.\n\n{\"answer\": 12.3259, \"confidence\": 85}" }, { "taskId": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 0.4756567164179193, "answer": 0.4757, "confidence": 85, "grade": "exact", "relError": 0.00009099752108344542, "cost": 0.000687, "reply": "Risk per unit (in JPY, since EUR/JPY quotes JPY per EUR): 191.077 − 190.407 = 0.670 JPY per EUR.\n\nUnits needed: 31869 / 0.670 = 47565.67 EUR.\n\nLots: 47565.67 / 100000 = 0.4756567.\n\n{\"answer\": 0.4757, \"confidence\": 85}" }, { "taskId": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 5.01704545454548, "answer": 5.017, "confidence": 85, "grade": "exact", "relError": 0.000009060022655040646, "cost": 0.000694, "reply": "Stop distance in price: 142.806 − 142.366 = 0.440 JPY per USD.\n\nRisk per unit (in JPY): 0.440 JPY per 1 USD unit.\n\nUnits needed: 220750 / 0.440 = 501,704.5 units of USD.\n\nLots: 501,704.5 / 100,000 = 5.017045 standard lots.\n\n{\"answer\": 5.0170, \"confidence\": 85}" }, { "taskId": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 1.6141304347826113, "answer": 1.6141, "confidence": 85, "grade": "exact", "relError": 0.00001885521885678377, "cost": 0.0006788, "reply": "Stop distance = 0.76163 − 0.75243 = 0.00920 USD per EUR.\n\nRisk per lot = 100,000 × 0.00920 = 920 USD per standard lot.\n\nPosition size = 1485 / 920 = 1.61413 standard lots.\n\n{\"answer\": 1.6141, \"confidence\": 85}" }, { "taskId": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.023819567833442343, "answer": 0.02382, "confidence": 85, "grade": "exact", "relError": 0.0000181433416710188, "cost": 0.000681, "reply": "VaR = -mean + z * stdev, with z for 97.5% confidence = 1.959964.\n\nz * stdev = 1.959964 × 0.012 = 0.02351957\n-mean = 0.0003\n\nVaR = 0.02351957 + 0.0003 = 0.02381957\n\n{\"answer\": 0.02382, \"confidence\": 85}" }, { "taskId": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.08208473980663146, "answer": 0.08209, "confidence": 80, "grade": "exact", "relError": 0.0000640824759989248, "cost": 0.000671, "reply": "VaR = -mean + z * stdev, with z = 1.645 (95% one-tailed).\n\n-(-0.0005) + 1.645 × 0.0496 = 0.0005 + 0.081592 = 0.082092\n\n{\"answer\": 0.08209, \"confidence\": 80}" }, { "taskId": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.10032876732331801, "answer": 0.10033, "confidence": 88, "grade": "exact", "relError": 0.000012286373239487742, "cost": 0.000712, "reply": "VaR = -mean + z * stdev, with z for 99% confidence = 2.326348.\n\nz * stdev = 2.326348 * 0.0436 = 0.10142... let's compute: 2.326348 * 0.0436 = 0.101432 (2.326348*0.04 = 0.09305392; 2.326348*0.0036 = 0.00837485; sum = 0.10142877).\n\nVaR = -0.0011 + 0.10142877 = 0.10032877.\n\n{\"answer\": 0.10033, \"confidence\": 88}" }, { "taskId": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.07275151453115127, "answer": 0.07275, "confidence": 85, "grade": "exact", "relError": 0.00002081786421946403, "cost": 0.000701, "reply": "VaR = -mean + z·σ, with z = 2.326348 for 99% confidence.\n\nz·σ = 2.326348 × 0.0308 = 0.07165 (2.326348 × 0.03 = 0.06979044; 2.326348 × 0.0008 = 0.00186108; sum = 0.07165152)\n\nVaR = 0.0011 + 0.07165152 = 0.07275152\n\n{\"answer\": 0.07275, \"confidence\": 85}" }, { "taskId": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.11150152403343133, "answer": 0.1115, "confidence": 80, "grade": "exact", "relError": 0.000013668274443252881, "cost": 0.000674, "reply": "VaR = -mean + z·σ, with z for 99% confidence = 2.326348.\n\nz·σ = 2.326348 × 0.0475 = 0.1105016\n\nVaR = 0.001 + 0.1105016 = 0.1115016\n\n{\"answer\": 0.1115, \"confidence\": 80}" }, { "taskId": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.04404168455292251, "answer": 0.04398, "confidence": 80, "grade": "correct", "relError": 0.0014005947671776652, "cost": 0.000826, "reply": "For a normal distribution with mean μ and standard deviation σ, the parametric ES at confidence level α is:\n\nES = -μ + σ · φ(z_α) / (1 − α)\n\nwhere z_α = Φ⁻¹(α) and φ is the standard normal density.\n\n**Step 1: z at 99%.** z = 2.326348 (Φ⁻¹(0.99)).\n\n**Step 2: density at z.** φ(z) = exp(−z²/2)/√(2π). z² = 5.412258, so z²/2 = 2.706129. exp(−2.706129) ≈ 0.066726. Dividing by √(2π) = 2.506628 gives φ(z) ≈ 0.026618.\n\n**Step 3: ES of standard normal.** φ(z)/(1−α) = 0.026618 / 0.01 = 2.6618.\n\n**Step 4: scale and shift.** σ · 2.6618 = 0.0172 × 2.6618 ≈ 0.045783. Subtract the mean: −0.0018 + 0.045783 = 0.043983.\n\nSo the 1-day 99% ES as a positive loss fraction is about 0.04398.\n\n{\"answer\": 0.04398, \"confidence\": 80}" }, { "taskId": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.02233967069948375, "answer": 0.02234, "confidence": 82, "grade": "exact", "relError": 0.000014740616398397254, "cost": 0.000776, "reply": "ES for normal: ES = -μ + σ·φ(z)/(1-α), where z = Φ^{-1}(0.05) = -1.644854, φ(z) = exp(-z²/2)/√(2π).\n\nz² = 2.705540, half = 1.352770, exp(-1.352770) ≈ 0.258717... let's compute: exp(-1.35277) = exp(-1.35)·exp(-0.00277) = 0.259240·0.997234 = 0.258523.\nφ = 0.258523 / 2.506628 = 0.103146.\n\nES quantile term = φ/(0.05) = 2.06292.\nσ·2.06292 = 0.0102·2.06292 = 0.0210418.\nES loss = -μ + that = 0.0013 + 0.0210418 = 0.0223418.\n\nAnswer ≈ 0.02234\n\n{\"answer\": 0.02234, \"confidence\": 82}" }, { "taskId": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.09524684038587, "answer": 0.09525, "confidence": 85, "grade": "exact", "relError": 0.000033172902294700856, "cost": 0.000779, "reply": "ES for normal returns at level α: loss ES = -μ + σ·φ(z_α)/α, where z_α = Φ^{-1}(0.01) = -2.326348, φ(z) = exp(-z²/2)/√(2π).\n\nz² = 5.411620; z²/2 = 2.705810; exp(-2.705810) ≈ 0.066726. Let me compute more carefully: exp(-2.705810) = exp(-2.7)·exp(-0.005810) = 0.0672055 × 0.994206 = 0.0668158.\nφ = 0.0668158 / 2.506628 = 0.0266567.\nφ/α = 0.0266567/0.01 = 2.66567.\nσ·φ/α = 0.0366 × 2.66567 = 0.0975535.\nES loss = -0.0023 + 0.0975535 = 0.0952535.\n\n{\"answer\": 0.09525, \"confidence\": 85}" }, { "taskId": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.12233732788143936, "answer": 0.12234, "confidence": 80, "grade": "exact", "relError": 0.000021842217799887472, "cost": 0.000864, "reply": "ES for normal: loss L = -R, mean of L = -0.0018, sd = 0.0531.\nES = -μ + σ·φ(z)/(1-α), where z = Φ^{-1}(0.975) = 1.959964, φ(z) = 0.058441.\n\nφ(1.959964): φ = exp(-z²/2)/√(2π). z² = 3.841459, half = 1.920729, exp(-1.920729) ≈ 0.146456. Divide by 2.506628: 0.058434.\n\nLet me refine: exp(-1.920729): exp(-1.92)=0.146607, times exp(-0.000729)=0.999271 → 0.146500. /2.506628 = 0.058444.\n\nCheck: standard value φ(1.96)=0.058441, and at 1.959964 it's slightly higher: derivative of φ is -zφ ≈ -0.1146, times (-0.000036) gives +0.0000041 → 0.058445. Good, φ ≈ 0.058445.\n\nES term: σ·φ/(0.025) = 0.0531 × 0.058445 / 0.025 = 0.0031034 / 0.025 = 0.124136.\n\nCompute 0.0531×0.058445 = 0.00310344. /0.025 = 0.1241376.\n\nES = -0.0018 + 0.124138 = 0.122338.\n\n**Answer: 0.12234**\n\n{\"answer\": 0.12234, \"confidence\": 80}" }, { "taskId": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.1205670856534704, "answer": 0.12056, "confidence": 80, "grade": "exact", "relError": 0.000058769384960984604, "cost": 0.000795, "reply": "For a normal return with mean μ and standard deviation σ, the Expected Shortfall at confidence level α (loss tail) is:\n\nES = −μ + σ · φ(z_α) / (1 − α)\n\nwhere z_α = Φ⁻¹(α) = Φ⁻¹(0.975) = 1.959964, and φ(z) = exp(−z²/2)/√(2π).\n\nCompute φ(1.959964):\n- z² = 3.841459, so z²/2 = 1.920729\n- exp(−1.920729) ≈ 0.146525\n- divide by √(2π) = 2.506628: φ ≈ 0.058441\n\nThen:\n- σ · φ / (1 − α) = 0.0523 × 0.058441 / 0.025 = 0.0030565 / 0.025 ≈ 0.122261\n- ES = −0.0017 + 0.122261 = 0.120561\n\nExpressed as a positive loss fraction: **≈ 0.12056**\n\n{\"answer\": 0.12056, \"confidence\": 80}" }, { "taskId": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 1.6781883884002948, "answer": 1.406, "confidence": 80, "grade": "wrong", "relError": 0.16219179579698675, "cost": 0.000979, "reply": "I need mean and sample standard deviation of the 12 returns.\n\nSum: -0.0353 + 0.0402 = 0.0049; +0.0555 = 0.0604; +0.0622 = 0.1226; +0.0058 = 0.1284; +0.0795 = 0.2079; +0.0049 = 0.2128; +0.0435 = 0.2563; +0.0415 = 0.2978; +0.0469 = 0.3447; -0.0363 = 0.3084; -0.0547 = 0.2537.\n\nMean = 0.2537/12 = 0.0211417.\n\nSum of squares:\n- (-0.0353)^2 = 0.00124609\n- 0.0402^2 = 0.00161604\n- 0.0555^2 = 0.00308025\n- 0.0622^2 = 0.00386884\n- 0.0058^2 = 0.00003364\n- 0.0795^2 = 0.00632025\n- 0.0049^2 = 0.00002401\n- 0.0435^2 = 0.00189225\n- 0.0415^2 = 0.00172225\n- 0.0469^2 = 0.00219961\n- (-0.0363)^2 = 0.00131769\n- (-0.0547)^2 = 0.00299209\n\nTotal = 0.03521106.\n\nn·mean² = 12 × (0.0211417)^2 = 12 × 0.00044697 = 0.00536366. (Precisely: 0.2537²/12 = 0.06436369/12 = 0.00536364.)\n\nSum of squared deviations = 0.03521106 − 0.00536364 = 0.02984742.\n\nSample variance = 0.02984742/11 = 0.00271340. Std dev = sqrt(0.00271340) ≈ 0.0520894.\n\nMonthly Sharpe = 0.0211417/0.0520894 ≈ 0.405868.\n\nAnnualized = 0.405868 × sqrt(12) = 0.405868 × 3.4641016 ≈ 1.40600.\n\nAnswer ≈ 1.406\n\n{\"answer\": 1.406, \"confidence\": 80}" }, { "taskId": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": -0.48328609065675004, "answer": -0.4833, "confidence": 72, "grade": "exact", "relError": 0.00002878076468342148, "cost": 0.000966, "reply": "Sum of returns:\n-0.0474 + 0.0376 = -0.0098\n+0.0074 = -0.0024\n-0.0481 = -0.0505\n-0.0542 = -0.1047\n+0.0097 = -0.0950\n-0.0123 = -0.1073\n+0.0621 = -0.0452\n-0.0496 = -0.0948\n+0.0357 = -0.0591\n+0.0095 = -0.0496\n-0.0158 = -0.0654\n\nMean = -0.0654/12 = -0.00545\n\nSquared deviations (x - mean), mean = -0.00545:\n1. -0.0474: d = -0.04195, d² = 0.0017598\n2. 0.0376: d = 0.04305, d² = 0.0018533\n3. 0.0074: d = 0.01285, d² = 0.0001651\n4. -0.0481: d = -0.04265, d² = 0.0018190\n5. -0.0542: d = -0.04875, d² = 0.0023766\n6. 0.0097: d = 0.01515, d² = 0.0002295\n7. -0.0123: d = -0.00685, d² = 0.0000469\n8. 0.0621: d = 0.06755, d² = 0.0045630\n9. -0.0496: d = -0.04415, d² = 0.0019492\n10. 0.0357: d = 0.04115, d² = 0.0016933\n11. 0.0095: d = 0.01495, d² = 0.0002235\n12. -0.0158: d = -0.01035, d² = 0.0001071\n\nSum of squares ≈ 0.0167853\n\nSample variance = 0.0167853/11 = 0.00152594\nStd dev = sqrt(0.00152594) ≈ 0.039063\n\nMonthly Sharpe = -0.00545/0.039063 ≈ -0.13951\n\nAnnualized = -0.13951 × sqrt(12) = -0.13951 × 3.46410 ≈ -0.48330\n\n{\"answer\": -0.4833, \"confidence\": 72}" }, { "taskId": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 2.241475231769422, "answer": 2.2404, "confidence": 78, "grade": "exact", "relError": 0.0004796982604056022, "cost": 0.000973, "reply": "Sum of returns: 0.0236+0.0642=0.0878; +0.02=0.1078; -0.029=0.0788; +0.0274=0.1062; +0.0497=0.1559; -0.0065=0.1494; +0.0325=0.1819; +0.0352=0.2171; +0.0197=0.2368; +0.0796=0.3164; -0.0422=0.2742.\n\nMean = 0.2742/12 = 0.02285.\n\nDeviations squared (x - 0.02285):\n1. 0.00135 → 1.8225e-6\n2. 0.04135 → 1.71012e-3\n3. -0.00285 → 8.1225e-6\n4. -0.05185 → 2.68858e-3\n5. 0.00455 → 2.0703e-5\n6. 0.02685 → 7.2092e-4\n7. -0.02935 → 8.6142e-4\n8. 0.01035 → 1.0712e-4\n9. 0.01235 → 1.5252e-4\n10. -0.00315 → 9.923e-6\n11. 0.05675 → 3.22056e-3\n12. -0.06505 → 4.23150e-3\n\nSum ≈ 0.0000018225+0.00171012=0.00171194; +0.0000081225=0.00172006; +0.00268858=0.00440864; +0.0000207=0.00442934; +0.00072092=0.00515026; +0.00086142=0.00601168; +0.00010712=0.0061188; +0.00015252=0.00627132; +0.00000992=0.00628124; +0.00322056=0.0095018; +0.0042315=0.0137333.\n\nSample variance = 0.0137333/11 = 0.00124848. Std = sqrt(0.00124848) ≈ 0.035335.\n\nMonthly Sharpe = 0.02285/0.035335 ≈ 0.64673.\n\nAnnualized = 0.64673 × sqrt(12) = 0.64673 × 3.46410 ≈ 2.2404.\n\n{\"answer\": 2.2404, \"confidence\": 78}" }, { "taskId": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 1.9830100176589147, "answer": 1.983, "confidence": 80, "grade": "exact", "relError": 0.000005051743977796405, "cost": 0.000982, "reply": "Sum of returns:\n0.05 + 0.044 = 0.094\n+0.0617 = 0.1557\n+0.0753 = 0.231\n-0.0527 = 0.1783\n-0.0366 = 0.1417\n+0.0167 = 0.1584\n+0.046 = 0.2044\n-0.0225 = 0.1819\n-0.0093 = 0.1726\n+0.0695 = 0.2421\n+0.0687 = 0.3108\n\nMean = 0.3108/12 = 0.0259\n\nDeviations (r - 0.0259) and squares:\n1. 0.0241 → 0.00058081\n2. 0.0181 → 0.00032761\n3. 0.0358 → 0.00128164\n4. 0.0494 → 0.00244036\n5. -0.0786 → 0.00617796\n6. -0.0625 → 0.00390625\n7. -0.0092 → 0.00008464\n8. 0.0201 → 0.00040401\n9. -0.0484 → 0.00234256\n10. -0.0352 → 0.00123904\n11. 0.0436 → 0.00190096\n12. 0.0428 → 0.00183184\n\nSum of squares:\n0.00058081 + 0.00032761 = 0.00090842\n+0.00128164 = 0.00219006\n+0.00244036 = 0.00463042\n+0.00617796 = 0.01080838\n+0.00390625 = 0.01471463\n+0.00008464 = 0.01479927\n+0.00040401 = 0.01520328\n+0.00234256 = 0.01754584\n+0.00123904 = 0.01878488\n+0.00190096 = 0.02068584\n+0.00183184 = 0.02251768\n\nSample variance = 0.02251768/11 = 0.00204706\nStd dev = sqrt(0.00204706) ≈ 0.045244\n\nMonthly Sharpe = 0.0259/0.045244 ≈ 0.57246\nAnnualized = 0.57246 × sqrt(12) = 0.57246 × 3.46410 ≈ 1.98312\n\n{\"answer\": 1.983, \"confidence\": 80}" }, { "taskId": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 1.3915400475486268, "answer": 1.3937, "confidence": 80, "grade": "correct", "relError": 0.0015522028670164072, "cost": 0.000962, "reply": "I need the mean and sample standard deviation of the 12 returns.\n\nSum of returns:\n0.0603 - 0.0347 = 0.0256\n+0.0392 = 0.0648\n+0.0484 = 0.1132\n-0.0114 = 0.1018\n-0.0247 = 0.0771\n+0.0144 = 0.0915\n+0.0301 = 0.1216\n+0.0712 = 0.1928\n+0.0025 = 0.1953\n-0.0155 = 0.1798\n-0.0103 = 0.1695\n\nMean = 0.1695 / 12 = 0.0141250\n\nDeviations from mean (r - 0.014125), squared:\n1. 0.0603 → 0.046175 → 0.00213212\n2. -0.0347 → -0.048825 → 0.00238390\n3. 0.0392 → 0.025075 → 0.00062876\n4. 0.0484 → 0.034275 → 0.00117476\n5. -0.0114 → -0.025525 → 0.00065153\n6. -0.0247 → -0.038825 → 0.00150739\n7. 0.0144 → 0.000275 → 0.00000008\n8. 0.0301 → 0.015975 → 0.00025520\n9. 0.0712 → 0.057075 → 0.00325756\n10. 0.0025 → -0.011625 → 0.00013514\n11. -0.0155 → -0.029625 → 0.00087764\n12. -0.0103 → -0.024425 → 0.00059648\n\nSum of squares ≈ 0.01356111\n\nSample variance = 0.01356111 / 11 = 0.00123283\nSample std dev = sqrt(0.00123283) ≈ 0.035112\n\nMonthly Sharpe = 0.014125 / 0.035112 ≈ 0.40230\n\nAnnualized = 0.40230 × sqrt(12) = 0.40230 × 3.464102 ≈ 1.39366\n\n{\"answer\": 1.3937, \"confidence\": 80}" }, { "taskId": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-haiku-5-5", "expected": 0.4848484848484847, "answer": 0.4848, "confidence": 92, "grade": "exact", "relError": 0.00009999999999966636, "cost": 0.000763, "reply": "Kelly for a binary contract priced at c, with win probability p, net odds b = (1-c)/c.\n\nHere c = 0.67, so b = 0.33/0.67 = 0.492537...\np = 0.83, q = 0.17.\n\nKelly fraction f* = (bp - q)/b = p - q/b = 0.83 - 0.17/0.492537\n0.17/0.492537 = 0.17 × 0.67/0.33 = 0.1139/0.33 = 0.345152...\nf* = 0.83 - 0.345152 = 0.484848...\n\nCheck: (p - c)/(1 - c) = (0.83 - 0.67)/0.33 = 0.16/0.33 = 0.484848...\n\nSo the full Kelly stake is about 0.4848 of bankroll.\n\n{\"answer\": 0.4848, \"confidence\": 92}" }, { "taskId": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-haiku-5-5", "expected": 0.28985507246376807, "answer": 0.2899, "confidence": 85, "grade": "exact", "relError": 0.0001550000000001245, "cost": 0.000734, "reply": "Kelly for a binary bet: buying at price c pays net odds b = (1-c)/c per dollar risked. Kelly fraction f* = (p·b − q)/b = p − q/b = p − (1−p)·c/(1−c).\n\nHere p = 0.51, c = 0.31:\n- b = 0.69/0.31 = 2.225806...\n- q/b = 0.49 / 2.225806 = 0.22017...\n- f* = 0.51 − 0.22017 = 0.28983...\n\nEquivalently f* = (p − c)/(1 − c) = 0.20/0.69 = 0.289855...\n\n{\"answer\": 0.2899, \"confidence\": 85}" }, { "taskId": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-haiku-5-5", "expected": 0.09859154929577463, "answer": 0.09859, "confidence": 88, "grade": "exact", "relError": 0.000015714285714109126, "cost": 0.000736, "reply": "Kelly for a binary contract priced at c with win probability p: buying at price c pays net odds b = (1-c)/c per dollar staked.\n\nHere c = 0.29, so b = 0.71/0.29 = 2.448276.\np = 0.36, q = 0.64.\n\nKelly fraction f* = (bp - q)/b = (2.448276×0.36 - 0.64)/2.448276\n= (0.881379 - 0.64)/2.448276\n= 0.241379/2.448276\n= 0.098592\n\nEquivalently f* = (p - c)/(1 - c) = 0.07/0.71 = 0.098592.\n\n{\"answer\": 0.09859, \"confidence\": 88}" }, { "taskId": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-haiku-5-5", "expected": 0.8333333333333334, "answer": 0.8333, "confidence": 85, "grade": "exact", "relError": 0.00003999999999999559, "cost": 0.000731, "reply": "Buying at price c = 0.16 pays b = (1−c)/c = 0.84/0.16 = 5.25 per unit staked.\n\nKelly fraction f* = (bp − q)/b, with p = 0.86, q = 0.14:\nbp − q = 5.25×0.86 − 0.14 = 4.515 − 0.14 = 4.375\nf* = 4.375 / 5.25 = 0.833333...\n\nEquivalently, f* = (p − c)/(1 − c) = (0.86 − 0.16)/0.84 = 0.70/0.84 = 0.8333.\n\n{\"answer\": 0.8333, \"confidence\": 85}" }, { "taskId": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "claude:claude-haiku-5-5", "expected": 0.576271186440678, "answer": 0.5763, "confidence": 85, "grade": "exact", "relError": 0.00005000000000002715, "cost": 0.000696, "reply": "Kelly for a binary contract priced at c, with win probability p: net odds b = (1 - c)/c, fraction f = (p(b+1) - 1)/b = (p - c)/(1 - c).\n\nHere p = 0.75, c = 0.41:\nf = (0.75 - 0.41)/(1 - 0.41) = 0.34/0.59 ≈ 0.576271\n\n{\"answer\": 0.5763, \"confidence\": 85}" }, { "taskId": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 24920.824742268043, "answer": 24919.9, "confidence": 80, "grade": "exact", "relError": 0.000037107209637144675, "cost": 0.000907, "reply": "Setup: Let contract size be N USD, entry price P0 = 24173.2.\n\nPosition value in BTC at entry = N / P0. Initial margin IM = (N/P0)/25 = 0.04·N/P0.\n\nShort inverse PnL in BTC: PnL = N(1/P0 − 1/P) (short gains when price falls; for a short, PnL = N(1/P − 1/P0)·(−1)... let's be careful).\n\nFor a short: profit when P < P0. PnL_BTC = N(1/P − 1/P0)·(+1)? Check: short gains when P falls, so 1/P rises, giving positive PnL when PnL = N(1/P − 1/P0). Yes.\n\nLiquidation condition: IM + PnL = MM, with MM = 0.01·N/P0.\n\n0.04·N/P0 + N(1/P − 1/P0) = 0.01·N/P0\n\nDivide by N: 0.04/P0 + 1/P − 1/P0 = 0.01/P0\n1/P = 1/P0 − 0.04/P0 + 0.01/P0 = (1 − 0.03)/P0 = 0.97/P0\n\nSo P = P0/0.97.\n\nNote this is the liquidation for a short where price rises: wait, the margin decreases when price rises for a short. Check direction: PnL = N(1/P − 1/P0); as P rises, 1/P falls, PnL becomes negative, margin falls toward MM. Consistent.\n\nP = 24173.2 / 0.97 = 24919.8969...\n\nCompute: 24173.2 / 0.97 = 24919.8969.\n\n{\"answer\": 24919.90, \"confidence\": 80}" }, { "taskId": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 53342.9702970297, "answer": 53344.06, "confidence": 85, "grade": "exact", "relError": 0.000020428239451788548, "cost": 0.000814, "reply": "Setup: inverse contract, size N USD. Position value in BTC at entry = N/P0. Initial margin IM = N/(P0·L). Maintenance margin MM = 0.01·N/P0.\n\nFor a long inverse position, PnL in BTC = N(1/P0 − 1/P).\n\nLiquidation when IM + PnL = MM:\nN/(P0·L) + N(1/P0 − 1/P) = 0.01·N/P0\n\nDivide by N and rearrange:\n1/(P0·L) + 1/P0 − 1/P = 0.01/P0\n\n1/P = 1/P0 + 1/(P0·L) − 0.01/P0 = (1 + 1/L − 0.01)/P0\n\nWith L = 50: 1 + 0.02 − 0.01 = 1.01, so 1/P = 1.01/P0\n\nP = P0/1.01 = 53876.4/1.01 = 53344.0594...\n\nCheck: 53876.4/1.01 ≈ 53344.06.\n\nLiquidation price ≈ 53344.06\n\n{\"answer\": 53344.06, \"confidence\": 85}" }, { "taskId": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 53194.15525114154, "answer": 53194.15, "confidence": 85, "grade": "exact", "relError": 9.871651343515556e-8, "cost": 0.000835, "reply": "Setup: contract size N (USD), entry price P0 = 58247.6, leverage 10.\n\nFor an inverse long, BTC position value at entry = N/P0. Initial margin M = N/(10·P0).\n\nUnrealized PnL in BTC at price P: N/P0 − N/P (long profits when P rises; loss when P falls, PnL = N/P0 − N/P).\n\nLiquidation when margin + PnL = maintenance = 0.005·N/P0:\n\nN/(10P0) + N/P0 − N/P = 0.005·N/P0\n\nDivide by N/P0:\n0.1 + 1 − P0/P = 0.005\nP0/P = 1.095\nP = 58247.6 / 1.095 = 53193.24...\n\nCompute: 58247.6 / 1.095. 1.095 × 53000 = 58035; remainder 212.6; 212.6/1.095 = 194.15. So P ≈ 53194.15.\n\nCheck: 1.095 × 53194.15 = 58247.6 (53194.15 + 0.095×53194.15 = 53194.15 + 5053.44 = 58247.59). Good.\n\nLiquidation price ≈ 53194.15\n\n{\"answer\": 53194.15, \"confidence\": 85}" }, { "taskId": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 170355.2216748768, "answer": 87108.6, "confidence": 85, "grade": "wrong", "relError": 0.48866492530386363, "cost": 0.000997, "reply": "Setup: contract size N (USD), entry price P0 = 115273.7.\n\nPosition value in BTC at entry: N/P0. Initial margin M = N/(3·P0).\n\nShort inverse PnL in BTC when price moves to P: PnL = N/P - N/P0 ... careful: for a short, PnL = N(1/P0 - 1/P).\n\nLiquidation when M + PnL = 0.01·N/P0:\nN/(3P0) + N(1/P0 - 1/P) = 0.01 N/P0\nDivide by N: 1/(3P0) + 1/P0 - 1/P = 0.01/P0\n1/P = (1/3 + 1 - 0.01)/P0 = (1.323333…)/P0\nP = P0/1.323333… = 115273.7 × 0.755668...\n\nCompute: 1/1.3233333 = 0.7556675 (since 1.3233333 × 0.7556675 ≈ 1.0000). Check: 1.3233333 × 0.75 = 0.99250; remainder 0.0075/1.3233333 = 0.005668; total 0.755668.\n\nP = 115273.7 × 0.755668 ≈ 87,100.\n\nMore precisely: 115273.7 × 0.755668 = 115273.7×0.75 = 86455.275; 115273.7×0.005668 = 653.3; total ≈ 87108.6.\n\nRefine 1/1.323333: exact = 1/(119/90) = 90/119 = 0.7563025. Let me recheck: 1/3 + 1 - 0.01 = 1.32333…, and 1.323333 = 3970/3000 = 397/300. So 1/P = 300/397 /P0, i.e., P = 115273.7 × 397/300 = 115273.7 × 1.323333 ... wait, P = P0 × 300/397.\n\nP = 115273.7 × 300/397 = 34582110/397 ≈ 87109.6.\n\nCompute: 397 × 87100 = 34578700; remainder 3410; 3410/397 ≈ 8.59. So P ≈ 87108.6.\n\n**Liquidation price ≈ 87108.6**\n\n{\"answer\": 87108.6, \"confidence\": 85}" }, { "taskId": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 115003.24607329845, "answer": 115003.2, "confidence": 82, "grade": "exact", "relError": 4.006260694662902e-7, "cost": 0.00108, "reply": "Setup: short inverse perpetual, position size N USD. Entry price P0 = 109828.1.\n\nPosition value in BTC at entry: N/P0. Initial margin M = N/(P0·20).\n\nShort PnL in BTC at price P: N/P − N/P0 (short gains when price falls; in BTC terms, PnL = N(1/P0 − 1/P)... let's be careful).\n\nFor a short inverse contract: PnL (BTC) = N·(1/P0 − 1/P). Equity = M + N(1/P0 − 1/P).\n\nLiquidation when equity = MM = 0.005·N/P0.\n\nSo: N/(20P0) + N/P0 − N/P = 0.005N/P0.\n\nDivide by N: 1/(20P0) + 1/P0 − 1/P = 0.005/P0.\n\nMultiply by P0: 1/20 + 1 − P0/P = 0.005 → P0/P = 1.05 − 0.005 = 1.045... wait recompute: 0.05 + 1 − 0.005 = 1.045.\n\nSo P = P0/1.045 ... that gives a lower price, which is wrong for a short. Recheck sign: for a short, equity = M − N(1/P − 1/P0) = M + N(1/P0 − 1/P). Setting equal to MM:\n\n1/(20P0) + 1/P0 − 1/P = 0.005/P0 ⇒ 1/P = (1.05 − 0.005)/P0 = 1.045/P0 ⇒ P = P0/1.045.\n\nCheck with direction: short liquidation should be above entry, so the sign must be the other way: losses occur when price rises. PnL for short in BTC = N(1/P − 1/P0)·(−1)... Short loses when P rises: PnL = N(1/P0 − 1/P) is positive when P > P0? 1/P0 − 1/P > 0 when P > P0. That's a gain for rising price, wrong. So the correct short PnL is N(1/P − 1/P0).\n\nEquity = M + N(1/P − 1/P0) = MM ⇒ 1/(20P0) + 1/P − 1/P0 = 0.005/P0 ⇒ 1/P = (0.005 + 1 − 0.05)/P0 = 0.955/P0 ⇒ P = P0/0.955.\n\nP = 109828.1 / 0.955 = 115001.15...\n\nCompute: 0.955 × 115000 = 109825; remainder 3.1 → 3.1/0.955 ≈ 3.246. So P ≈ 115003.2.\n\nCheck: 0.955 × 115003.246 = 109828.1. Good.\n\n{\"answer\": 115003.2, \"confidence\": 82}" }, { "taskId": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 2398.0256188582507, "answer": 2398, "confidence": 85, "grade": "exact", "relError": 0.00001068331299267227, "cost": 0.000811, "reply": "For inverse contracts, the ETH PnL depends on the total of contracts divided by entry price (the \"coin quantity\" 1/P). The equivalent single entry price is the harmonic-weighted average: total USD / sum(USD_i / P_i).\n\nTotal USD = 7000 + 34000 + 36000 + 9000 = 86000.\n\nCompute each USD_i / P_i:\n- 7000 / 2309.36 ≈ 3.031 (7000/2309.36 = 3.03109)\n- 34000 / 3563.54 ≈ 9.54149\n- 36000 / 1854.34 ≈ 19.41414\n- 9000 / 2321.56 ≈ 3.87672\n\nSum ≈ 3.03109 + 9.54149 + 19.41414 + 3.87672 = 35.86344\n\nAverage entry = 86000 / 35.86344 ≈ 2398.04\n\nCheck: 35.86344 × 2398 ≈ 85,999.2, so 86000/35.86344 ≈ 2398.04.\n\n{\"answer\": 2398.0, \"confidence\": 85}" }, { "taskId": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 3712.5303509582454, "answer": 3712.55, "confidence": 85, "grade": "exact", "relError": 0.000005292627910684943, "cost": 0.000861, "reply": "For inverse contracts, the ETH PnL depends on 1/price, so the correct average entry is the harmonic-type weighted average: the price P such that total contracts / P = sum of (contracts_i / P_i).\n\nTotal contracts = 34000 + 16000 = 50000 USD.\n\nSum of contracts_i / P_i:\n- 34000 / 3528.19 = 9.63790...\n Compute: 3528.19 × 9.6 = 33870.62; remainder 129.38; 129.38/3528.19 = 0.036672; so 9.636672.\n- 16000 / 4176.2 = 3.831 ... compute: 4176.2 × 3.83 = 15994.85; remainder 5.15; 5.15/4176.2 = 0.001233; so 3.831233.\n\nSum = 9.636672 + 3.831233 = 13.467905.\n\nAverage entry P = 50000 / 13.467905 = 3712.6...\nCheck: 13.467905 × 3712 = 49992.6; remainder 7.4; 7.4/13.467905 = 0.5495; so P ≈ 3712.55.\n\nRefine: 13.467905 × 3712.55 = 49992.6 + 13.467905×0.55 = 49992.6+7.4073 ≈ 50000.0 (approx). Good.\n\nAverage entry price ≈ 3712.55\n\n{\"answer\": 3712.55, \"confidence\": 85}" }, { "taskId": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 2495.29829080704, "answer": 2495.5, "confidence": 80, "grade": "exact", "relError": 0.00008083570357230291, "cost": 0.000912, "reply": "For inverse contracts, the ETH PnL on a long of size N (USD) bought at price P is N/P − N/exit. So the position's equivalent entry is the harmonic-weighted average: total USD / Σ(N_i / P_i).\n\nTotal notional: 41000 + 10000 + 27000 + 40000 = 118000.\n\nCompute each N_i/P_i:\n- 41000 / 3517.25 ≈ 11.65 (41000/3517.25 = 11.65 approx; more precisely 11.65 × 3517.25 = 40975.9, remainder 24.1/3517.25 ≈ 0.00685, so 11.65685)\n- 10000 / 2976.86 ≈ 3.35912 (2976.86 × 3.359 = 9999.7; remainder 0.3/2976.86 ≈ 0.0001, so 3.35910)\n- 27000 / 4229.06 ≈ 6.38397 (4229.06 × 6.384 = 27000.2; slightly less: 6.38395)\n- 40000 / 1545.09 ≈ 25.88 (1545.09 × 25.89 = 40000.9; 25.89 − 0.9/1545.09 ≈ 25.88942)\n\nSum ≈ 11.65685 + 3.35910 + 6.38395 + 25.88942 = 47.28932\n\nAverage entry = 118000 / 47.28932 ≈ 2495.5\n\nCheck: 47.28932 × 2495 = 117,976.7; remainder 23.3/47.28932 ≈ 0.493; so ≈ 2495.49.\n\nAverage entry price ≈ 2495.5\n\n{\"answer\": 2495.5, \"confidence\": 80}" }, { "taskId": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 2610.72093572179, "answer": 2611.1, "confidence": 85, "grade": "exact", "relError": 0.00014519524971951087, "cost": 0.0007687, "reply": "For inverse contracts, the ETH PnL of a long position with notional N (USD) entered at price P is N(1/P_entry − 1/P_exit) in ETH. The equivalent single entry price P* satisfies:\n\nTotal notional / P* = Σ N_i / P_i\n\nTotal notional = 10000 + 14000 + 12000 + 23000 = 59000.\n\nCompute Σ N_i/P_i:\n- 10000/3282.44 = 3.046311\n- 14000/2147.97 = 6.517607\n- 12000/1744.23 = 6.879622\n- 23000/3736.8 = 6.154570\n\nSum = 22.598110\n\nP* = 59000 / 22.598110 = 2611.1\n\nCheck: 59000/22.598110 ≈ 2611.09.\n\n{\"answer\": 2611.1, \"confidence\": 85}" }, { "taskId": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 3424.594098647905, "answer": 3424.58, "confidence": 80, "grade": "exact", "relError": 0.000004116881446065742, "cost": 0.000949, "reply": "For inverse contracts, the ETH quantity bought in each fill is size/price. The equivalent single entry price is total USD notional divided by total ETH quantity (harmonic-weighted average).\n\nTotal USD: 31000 + 36000 + 40000 + 5000 = 112000\n\nETH quantities:\n- 31000/3315.48 = 9.35063...\n- 36000/3410.8 = 10.55475...\n- 40000/3481.71 = 11.48845...\n- 5000/3813.27 = 1.31136...\n\nCompute more carefully:\n- 31000/3315.48: 3315.48 × 9.35 = 30999.74; remainder 0.26/3315.48 = 0.0000784 → 9.350078\n- 36000/3410.8: 3410.8 × 10.55 = 35983.94; remainder 16.06/3410.8 = 0.004709 → 10.554709\n- 40000/3481.71: 3481.71 × 11.49 = 40001.85; overshoot −1.85/3481.71 = −0.000531 → 11.489469\n- 5000/3813.27: 3813.27 × 1.311 = 4999.19; remainder 0.81/3813.27 = 0.000212 → 1.311212\n\nTotal ETH = 9.350078 + 10.554709 + 11.489469 + 1.311212 = 32.705468\n\nAverage entry = 112000 / 32.705468 ≈ 3424.4\n\nCheck: 32.705468 × 3424 = 111,980.9; remainder 19.1/32.705 = 0.584 → 3424.58\n\nRefining: 32.705468 × 3424.58 ≈ 111980.9 + 32.705468×0.58 = 111980.9 + 18.969 = 111999.87; remainder 0.13/32.7 = 0.004 → 3424.584.\n\nAverage entry price ≈ 3424.58\n\n{\"answer\": 3424.58, \"confidence\": 80}" }, { "taskId": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 0.02840777213808475, "answer": 0.028296, "confidence": 80, "grade": "correct", "relError": 0.0039345619058560825, "cost": 0.001, "reply": "Setup: short 44000 USD notional, inverse contract, so BTC quantity fixed.\n\nGross PnL for a short in inverse contract: size × (1/P_entry − 1/P_exit).\n\n1/42887.5 = 0.0233185...\nCompute: 1/42887.5 ≈ 2.331545e-5 × 1000? Let's do carefully.\n1/42887.5: 42887.5 × 0.0000233 = 0.99927; remainder 0.00073/42887.5 ≈ 1.702e-8. So ≈ 0.00002331702.\nCheck: 42887.5 × 0.00002331702 = 0.999999... good enough: 0.00002331702.\n\n1/41686.7: 41686.7 × 0.000024 = 1.0004808; too big. 0.0000239 → 0.99641; remainder 0.00359/41686.7 = 8.612e-8. So ≈ 0.00002398612.\nCheck: 41686.7 × 0.00002398612 ≈ 0.99641 + 0.0000... fine.\n\nDifference: 0.00002331702 − 0.00002398612 = −0.00000066910.\n\nGross PnL = 44000 × (−6.6910e-7) = −0.0294404 BTC? Careful: short PnL = size × (1/P_exit − 1/P_entry)... Let's define: short profits when price falls. Short PnL = size × (1/P_exit − 1/P_entry) in BTC for inverse: 1/41686.7 − 1/42887.5 = +6.6910e-7. Times 44000 = +0.0294404 BTC.\n\nFees: opening fee = 0.00055 × (44000/42887.5) = 0.00055 × 1.025937 = 0.00056427 BTC.\nClosing fee = 0.00055 × (44000/41686.7) = 0.00055 × 1.055337 = 0.00058044 BTC.\nTotal fees = 0.00114471 BTC.\n\nNet PnL = 0.0294404 − 0.0011447 = 0.0282957 BTC.\n\nRefine gross: 44000 × 6.6910e-7 = 0.02944040. Net ≈ 0.028296.\n\n{\"answer\": 0.028296, \"confidence\": 80}" }, { "taskId": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -0.06525146382766053, "answer": -0.06527, "confidence": 80, "grade": "exact", "relError": 0.00028407289663904964, "cost": 0.000925, "reply": "Inverse contract PnL for a long: size × (1/entry − 1/exit).\n\n1/49446 = 0.0000202238...\nCompute precisely: 1/49446 ≈ 2.022381e-5.\n1/46182.6 ≈ 2.165400e-5.\n\nLet me refine:\n1/49446: 49446 × 2.0224e-5 = 1.000026... so slightly less: 2.02235e-5 gives 49446×2.02235e-5 = 1.0000... check: 49446×2.02235 = 100,000.0? 49446×2 = 98892; 49446×0.02235 = 1105.12; total 99997.12 → need 100000, difference 2.88 → add 2.88/49446 = 0.0000582 → 2.0223582e-5.\n\n1/46182.6: 46182.6×2.1654 = 100,000.1? 46182.6×2 = 92365.2; ×0.1654 = 7638.59; total 100003.79. Too high by 3.79 → subtract 3.79/46182.6 = 0.0000821 → 2.1653179e-5.\n\nDifference: 2.0223582e-5 − 2.1653179e-5 = −1.429597e-6.\n\nGross PnL = 45000 × (−1.429597e-6) = −0.0643319 BTC.\n\nFees:\nOpening position value in BTC = 45000/49446 = 0.910132 BTC; fee = 0.0005 × 0.910132 = 0.000455066.\nClosing position value = 45000/46182.6 = 0.974377 BTC; fee = 0.000487188.\nTotal fees = 0.000942254 BTC.\n\nNet PnL = −0.0643319 − 0.0009423 = −0.0652742 BTC.\n\n{\"answer\": -0.06527, \"confidence\": 80}" }, { "taskId": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 0.22922018359225318, "answer": 0.2292, "confidence": 72, "grade": "exact", "relError": 0.00008805329415973827, "cost": 0.00089, "reply": "Inverse perpetual: PnL in BTC = size × (1/entry − 1/exit).\n\n1/91422.1 = 0.000010938...\nCompute precisely: 1/91422.1 ≈ 1.093804e-5 (since 91422.1 × 1.0938e-5 ≈ 0.99999...). Let's refine: 91422.1 × 1.0938e-5 = 0.999972; remainder 0.000028/91422.1 ≈ 3.06e-10, so 1/91422.1 ≈ 1.093831e-5.\n\n1/102932.1: 102932.1 × 9.7e-6 = 0.998441; remainder 0.001559/102932.1 ≈ 1.5145e-8, so ≈ 9.715145e-6.\n\nDifference: 1.093831e-5 − 0.9715145e-5 = 1.2231655e-6... let me recompute: 10.93831e-6 − 9.715145e-6 = 1.223165e-6.\n\nGross PnL = 189000 × 1.223165e-6 = 0.2311782 BTC.\n\nFees:\n- Opening: 189000 / 91422.1 = 2.067666 BTC × 0.0005 = 0.00103383 BTC.\n- Closing: 189000 / 102932.1 = 1.836371 BTC × 0.0005 = 0.00091819 BTC.\n- Total fees = 0.00195202 BTC.\n\nNet PnL = 0.2311782 − 0.0019520 = 0.2292262 BTC.\n\n{\"answer\": 0.2292, \"confidence\": 72}" }, { "taskId": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -0.026151408867684533, "answer": -0.02615, "confidence": 72, "grade": "exact", "relError": 0.00005387349078061194, "cost": 0.001008, "reply": "Inverse contract PnL in BTC for a long: size × (1/entry − 1/exit).\n\n1/89747.2 = 0.0000111442...\nCompute: 1/89747.2 ≈ 1.114269e-5 (since 89747.2 × 1.114269e-5 ≈ 1.00000).\n1/76940.3 ≈ 1.299726e-5 (76940.3 × 1.299726e-5 ≈ 1.000000).\n\nDifference: 1.114269e-5 − 1.299726e-5 = −1.85457e-6.\n\nGross PnL = 14000 × (−1.85457e-6) = −0.0259640 BTC.\n\nFees:\n- Opening: 14000/89747.2 = 0.156006 BTC × 0.00055 = 0.0000858033 BTC\n- Closing: 14000/76940.3 = 0.181960 BTC × 0.00055 = 0.000100078 BTC\n- Total fees = 0.000185881 BTC\n\nNet PnL = −0.0259640 − 0.000185881 = −0.0261499 BTC\n\nCheck the gross more precisely: 14000/76940.3 = 0.1819600...; 14000/89747.2 = 0.1560067...; difference = −0.0259533, which gives gross −0.0259533 (the earlier multiplication was slightly off). Net = −0.0259533 − 0.0001859 = −0.0261392.\n\nRe-verify: 14000/89747.2: 89747.2 × 0.156 = 14000.5, so 0.156 − 0.5/89747.2 ≈ 0.156 − 0.00000557 = 0.1559944. Recompute fee: 0.1559944 × 0.00055 = 0.0000857969.\n14000/76940.3: 76940.3 × 0.182 = 14003.1, so 0.182 − 3.1/76940.3 ≈ 0.182 − 0.0000403 = 0.1819597. Fee: 0.1819597 × 0.00055 = 0.0001000778.\nGross: 0.1559944 − 0.1819597 = −0.0259653.\nTotal fees: 0.0001858747.\nNet: −0.0259653 − 0.0001859 = −0.0261512.\n\n{\"answer\": -0.02615, \"confidence\": 72}" }, { "taskId": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -0.011586282591352507, "answer": 0.011179, "confidence": 72, "grade": "wrong", "relError": 1.9648478631397714, "cost": 0.000915, "reply": "Short inverse perpetual: PnL in BTC = size × (1/entry − 1/exit) for a short (profit when price falls).\n\nEntry: 1/102777.6 = 9.72985e-6 BTC per USD... compute precisely:\n1/102777.6 ≈ 0.0000097287 (check: 102777.6 × 9.7287e-6 ≈ 1.000003; adjust to 9.72867e-6 → 102777.6 × 9.72867e-6 ≈ 1.000000). So 1/entry ≈ 9.72867e-6.\n\nExit: 1/105593.7 ≈ 9.47077e-6 (check: 105593.7 × 9.47077e-6 ≈ 1.000000).\n\nDifference (entry − exit): 9.72867e-6 − 9.47077e-6 = 0.25790e-6 = 2.5790e-7 BTC per USD.\n\nGross PnL = 44000 × 2.5790e-7 ≈ 0.0113476 BTC.\n\nFees:\n- Opening: 0.02% × (44000 / 102777.6) = 0.0002 × 0.428067 = 0.0000856134 BTC.\n- Closing: 0.02% × (44000 / 105593.7) = 0.0002 × 0.416667 ≈ 0.0000833334 BTC. (44000/105593.7 ≈ 0.416685; ×0.0002 = 0.0000833370.)\n\nTotal fees ≈ 0.0000856134 + 0.0000833370 = 0.0001689504 BTC.\n\nNet PnL ≈ 0.0113476 − 0.0001690 = 0.0111786 BTC.\n\n{\"answer\": 0.011179, \"confidence\": 72}" }, { "taskId": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 6057.334255752001, "answer": 6057.33, "confidence": 88, "grade": "exact", "relError": 7.0257836554604e-7, "cost": 0.000829, "reply": "Short position: profit = size × (entry − exit) = 1.703 × (26380.89 − 22814.19) = 1.703 × 3566.70.\n\n1.703 × 3566.70 = 6074.2101 (3566.70 × 1.7 = 6063.39; 3566.70 × 0.003 = 10.7001; total 6074.0901). Let me recompute: 6063.39 + 10.7001 = 6074.0901.\n\nFees:\n- Opening notional: 1.703 × 26380.89 = 44926.85 (26380.89 × 1.7 = 44847.513; × 0.003 = 79.14267; total 44926.6557). Fee = 0.0002 × 44926.6557 = 8.98533.\n- Closing notional: 1.703 × 22814.19 = 38852.6 (22814.19 × 1.7 = 38784.123; × 0.003 = 68.44257; total 38852.5656). Fee = 0.0002 × 38852.5656 = 7.77051.\n\nTotal fees = 8.98533 + 7.77051 = 16.75584.\n\nNet PnL = 6074.0901 − 16.75584 = 6057.3343.\n\n{\"answer\": 6057.33, \"confidence\": 88}" }, { "taskId": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 2844.8890875599996, "answer": 2844.89, "confidence": 88, "grade": "exact", "relError": 3.2072955121873763e-7, "cost": 0.000824, "reply": "Gross PnL = 1.164 × (39204.01 − 36718.19) = 1.164 × 2485.82 = 2893.494...\n\nCompute: 2485.82 × 1.164 = 2485.82 + 2485.82×0.164 = 2485.82 + 407.6745 = 2893.4945.\n\nFees:\n- Opening notional = 1.164 × 36718.19 = 42739.53...\n 1.164 × 36718.19 = 36718.19 + 36718.19×0.164 = 36718.19 + 6021.79 = 42739.98 (36718.19×0.164 = 6021.7832), so 42739.9732.\n Fee = 0.00055 × 42739.9732 = 23.5070\n- Closing notional = 1.164 × 39204.01 = 39204.01 + 39204.01×0.164 = 39204.01 + 6429.4576 = 45633.4676\n Fee = 0.00055 × 45633.4676 = 25.0984\n\nTotal fees = 23.5070 + 25.0984 = 48.6054\n\nNet PnL = 2893.4945 − 48.6054 = 2844.8891\n\n{\"answer\": 2844.89, \"confidence\": 88}" }, { "taskId": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -3783.556934676003, "answer": -3783.557, "confidence": 85, "grade": "exact", "relError": 1.726523424342638e-8, "cost": 0.00078, "reply": "Gross PnL = 0.612 × (51850.49 − 57972.37) = 0.612 × (−6121.88) = −3746.99 (approximately: 6121.88 × 0.612 = 3746.79856... let me compute precisely: 6121.88 × 0.6 = 3673.128; 6121.88 × 0.012 = 73.46256; total 3746.59056). So gross PnL = −3746.59056.\n\nFees:\n- Opening notional = 0.612 × 57972.37 = 35479.17444; fee = 0.00055 × 35479.17444 = 19.51354594\n- Closing notional = 0.612 × 51850.49 = 31731.89988; fee = 0.00055 × 31731.89988 = 17.45254493\n\nTotal fees = 36.96609087\n\nNet PnL = −3746.59056 − 36.96609 = −3783.55665\n\n{\"answer\": -3783.557, \"confidence\": 85}" }, { "taskId": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -1108.5578196799993, "answer": -1108.582, "confidence": 85, "grade": "exact", "relError": 0.000021812412101159933, "cost": 0.000819, "reply": "Gross PnL = 2.648 × (4440.76 − 4857.54) = 2.648 × (−416.78) = −1103.56 (2.648 × 416.78 = 1103.5... compute precisely: 416.78 × 2.648 = 416.78×2 = 833.56; 416.78×0.648 = 270.09744; total = 1103.65744). So gross = −1103.65744.\n\nFees:\n- Opening notional = 2.648 × 4857.54 = 12862.19... compute: 4857.54×2 = 9715.08; 4857.54×0.648 = 3148.04592; total = 12863.12592. Fee = 0.0002 × 12863.12592 = 2.572625184.\n- Closing notional = 2.648 × 4440.76 = 4440.76×2 = 8881.52; 4440.76×0.648 = 2877.41248; total = 11758.93248. Fee = 0.0002 × 11758.93248 = 2.351786496.\n\nTotal fees = 4.92441168.\n\nNet PnL = −1103.65744 − 4.92441168 = −1108.58185168.\n\n{\"answer\": -1108.582, \"confidence\": 85}" }, { "taskId": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 3351.620043262996, "answer": 3351.61, "confidence": 88, "grade": "exact", "relError": 0.0000029965398423563937, "cost": 0.000845, "reply": "Short position: profit = size × (entry − exit).\n\nGross PnL = 1.949 × (28983.12 − 27232.54) = 1.949 × 1750.58 = 3412.0814 (approx).\n\nLet me compute: 1750.58 × 1.949 = 1750.58×2 − 1750.58×0.051 = 3501.16 − 89.28958 = 3411.87042.\n\nFees:\n- Opening notional: 1.949 × 28983.12 = 56487.80 (1.949×28983.12 = 57966.24 − 1478.12 ... let's do directly: 28983.12×1.949 = 28983.12×2 − 28983.12×0.051 = 57966.24 − 1478.14 = 56488.10). Check 28983.12×0.051 = 1478.139... = 1478.1391. So 57966.24 − 1478.1391 = 56488.1009.\n- Closing notional: 1.949 × 27232.54 = 54465.08 − 1388.8595 = 53076.2205.\n- Total notional = 109564.3214; fee at 0.055% = 60.26037677.\n\nNet PnL = 3411.87042 − 60.26038 = 3351.61004.\n\nNet PnL ≈ 3351.61 USDT (profit).\n\n{\"answer\": 3351.61, \"confidence\": 88}" }, { "taskId": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 0.11472881355932601, "answer": 0.1147, "confidence": 85, "grade": "exact", "relError": 0.0002511449254299066, "cost": 0.000764, "reply": "Risk amount: 1% of 5,000 = 50 USD.\n\nStop distance: 1.3538 − 1.3479 = 0.0059 = 59 pips (4-decimal pair, pip = 0.0001).\n\nUSD/CAD is quoted in CAD, so pip value per standard lot (100,000 units) in CAD = 100,000 × 0.0001 = 10 CAD.\n\nConvert to USD at entry: pip value in USD = 10 / 1.3538 ≈ 7.3872 USD per pip per lot.\n\nRisk per lot at the stop = 59 × 7.3872 ≈ 435.84 USD.\n\nLots = 50 / 435.84 ≈ 0.11473 standard lots.\n\nSo about 0.1147 standard lots (roughly 11,473 units).\n\n{\"answer\": 0.1147, \"confidence\": 85}" }, { "taskId": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 0.5939090909090843, "answer": 0.5939, "confidence": 85, "grade": "exact", "relError": 0.000015306903402306825, "cost": 0.000742, "reply": "Risk amount: 1% of 25,000 = 250 USD.\n\nStop distance: 1.3066 − 1.3011 = 0.0055 = 55 pips (4-decimal pair, pip = 0.0001).\n\nPip value for a standard lot (100,000 units) in USD/CAD: 0.0001 × 100,000 = 10 CAD per pip. Convert to USD at entry price 1.3066: 10 / 1.3066 = 7.6537 USD per pip per lot.\n\nRisk per lot at the stop: 55 × 7.6537 = 420.95 USD.\n\nLots: 250 / 420.95 = 0.5939 standard lots.\n\n{\"answer\": 0.5939, \"confidence\": 85}" }, { "taskId": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 1.1982352941177008, "answer": 1.1983, "confidence": 80, "grade": "exact", "relError": 0.000054000981791114666, "cost": 0.000776, "reply": "Risk amount: 1% of 25,000 = 250 USD.\n\nStop distance: 0.8148 − 0.8131 = 0.0017 = 17 pips (4-decimal pair, pip = 0.0001).\n\nUSD/CHF is quoted in CHF, so pip value for 1 standard lot (100,000 units) is 100,000 × 0.0001 = 10 CHF per pip. Convert to USD at the entry price: 1 CHF = 1/0.8148 USD, so 10 CHF = 10/0.8148 = 12.2730 USD per pip per lot.\n\nRisk per lot = 17 × 12.2730 = 208.64 USD.\n\nLots = 250 / 208.64 = 1.1983 standard lots.\n\nRounded to two decimals, that is about 1.20 lots, though 1.19 keeps the risk at or under 1%.\n\n{\"answer\": 1.1983, \"confidence\": 80}" }, { "taskId": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 1.0178846153846508, "answer": 1.018, "confidence": 80, "grade": "exact", "relError": 0.00011335726427656952, "cost": 0.000756, "reply": "Risk amount: 1% of 25,000 = 250 USD.\n\nStop distance: 158.79 − 158.40 = 0.39 yen = 39 pips (JPY pairs use 0.01 pip size).\n\nPip value for 1 standard lot (100,000 units): a pip is 0.01 JPY, so 100,000 × 0.01 = 1,000 JPY per pip. Convert to USD at entry: 1,000 / 158.79 = 6.2972 USD per pip per lot.\n\nRisk per lot: 39 × 6.2972 = 245.59 USD.\n\nLots: 250 / 245.59 = 1.01795 lots.\n\nAnswer: about 1.018 standard lots (round down to 1.01 lots to stay within risk).\n\n{\"answer\": 1.018, \"confidence\": 80}" }, { "taskId": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "claude:claude-haiku-5-5", "expected": 7.182727272727309, "answer": 7.182, "confidence": 85, "grade": "exact", "relError": 0.00010125300595361607, "cost": 0.000782, "reply": "Risk amount: 1% of 100,000 = 1,000 USD.\n\nStop distance: 158.02 − 157.80 = 0.22 yen = 22 pips (pip = 0.01 for USD/JPY).\n\nStandard lot = 100,000 units. Pip value in JPY per lot = 0.01 × 100,000 = 1,000 JPY per pip.\n\nConvert to USD at entry price 158.02: 1,000 / 158.02 = 6.3284 USD per pip per lot.\n\nRisk per lot = 22 pips × 6.3284 = 139.22 USD.\n\nLots = 1,000 / 139.22 = 7.182 standard lots.\n\nEquivalently, lots = 1,000 / (22 × 6.3284) ≈ 7.18.\n\nPosition size: about 7.18 standard lots (roughly 7.18 lots, or 7.2 if your broker only allows one decimal).\n\n{\"answer\": 7.182, \"confidence\": 85}" }, { "taskId": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -0.23573231643213824, "answer": 0.2315, "confidence": 72, "grade": "wrong", "relError": 1.9820460915321443, "cost": 0.000888, "reply": "Inverse contract: PnL in BTC = N × (1/entry − 1/exit) for a short, where N = 103,000 USD.\n\n1/49553.7 = 2.018106...e-5\nLet me compute more precisely: 1/49553.7 ≈ 0.0000201810\n1/55827.2 ≈ 0.0000179123\n\nDifference (short gain): 0.0000201810 − 0.0000179123 = 0.0000022687 BTC per USD\n\nGross PnL = 103,000 × 0.0000022687 ≈ 0.23368 BTC\n\nFees: 0.055% on each fill, based on notional in BTC at each fill.\n- Entry fee: 103,000 × 0.00055 × (1/49553.7) = 56.65 / 49553.7 ≈ 0.0011433 BTC\n- Exit fee: 103,000 × 0.00055 × (1/55827.2) = 56.65 / 55827.2 ≈ 0.0010148 BTC\n\nTotal fees ≈ 0.0021581 BTC\n\nNet PnL ≈ 0.23368 − 0.00216 ≈ 0.23152 BTC\n\nLet me refine the gross: 103,000 × 0.0000022687 = 0.233676; more precisely, 1/49553.7 = 0.00002018106, 1/55827.2 = 0.00001791243, difference = 0.00000226863, times 103,000 = 0.233669.\n\nNet ≈ 0.233669 − 0.002158 ≈ 0.231511 BTC\n\n{\"answer\": 0.2315, \"confidence\": 72}" }, { "taskId": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -0.0669895883004742, "answer": -0.06691, "confidence": 62, "grade": "correct", "relError": 0.001188069705955233, "cost": 0.001043, "reply": "Inverse contract: PnL in BTC = N × (1/entry − 1/exit) for a long.\n\n1/95195.8 = 0.000010504...\nCompute more precisely: 1/95195.8 ≈ 1.050557e-5\n1/88094.2 ≈ 1.135009e-5\n\nDifference (exit − entry in inverse terms): 1/88094.2 − 1/95195.8 = 1.135009e-5 − 1.050557e-5 = 8.4452e-7 — but for a long, loss when price falls: PnL = N(1/exit − 1/entry)? Let's check sign: long profits when price rises. PnL = N(1/entry − 1/exit) gives 1/95195.8 − 1/88094.2 = 1.050557e-5 − 1.135009e-5 = −8.4452e-7 BTC, a loss. Correct.\n\nGross PnL = 78,000 × (−8.4452e-7) = −0.065873 BTC.\n\nLet me refine the reciprocals:\n1/95195.8: 95195.8 × 1.0505e-5 = 0.99998... ; more precisely 1/95195.8 = 1.0505e-5 × (1 + 0.0000... ) ≈ 1.050570e-5. Check: 95195.8 × 1.05057e-5 = 1.000003... slightly high, so 1/95195.8 ≈ 1.050567e-5.\n1/88094.2: 88094.2 × 1.135e-5 = 0.99987; remainder 0.00013/88094.2 ≈ 1.476e-9, so ≈ 1.135148e-5.\n\nDifference: 1.050567e-5 − 1.135148e-5 = −8.4581e-7.\nGross PnL = 78,000 × (−8.4581e-7) = −0.065973 BTC.\n\nFees: taker 0.055% on notional in BTC for each fill.\nEntry fee = 0.00055 × 78,000 / 95195.8 = 0.00055 × 0.819 = 0.00045045 BTC (78000×1.050567e-5 = 0.819442; ×0.00055 = 0.00045069).\nExit fee = 0.00055 × 78000 × 1.135148e-5 = 0.00055 × 0.885415 = 0.00048698 BTC.\nTotal fees ≈ 0.00093767 BTC.\n\nNet PnL = −0.065973 − 0.000938 = −0.066911 BTC.\n\n{\"answer\": -0.06691, \"confidence\": 62}" }, { "taskId": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 0.0024428835303150374, "answer": 0.002451, "confidence": 45, "grade": "correct", "relError": 0.0033224955607752365, "cost": 0.001, "reply": "Inverse perpetual: PnL in BTC = contracts × (1/entry − 1/exit) for a short, with sign: short profits when price falls.\n\nShort PnL (BTC) = N × (1/P_entry − 1/P_exit)\n= 42,000 × (1/117914.1 − 1/116982.6)\n\n1/117914.1 = 8.48143e-6\n1/116982.6 = 8.54936e-6\nDifference (entry − exit) = −6.793e-8 per contract... let me compute more precisely:\n\n1/117914.1 ≈ 0.00000848100 (check: 117914.1 × 8.481e-6 ≈ 1.000014, so adjust: 8.481e-6 × (1 − 0.000014) ≈ 8.48088e-6)\n1/116982.6 ≈ 0.00000854814 (check: 116982.6 × 8.548e-6 ≈ 0.99993, so 8.548e-6 × 1.00007 ≈ 8.54860e-6)\n\nGross PnL = 42,000 × (8.48088e-6 − 8.54860e-6) = 42,000 × (−6.772e-8) ≈ −0.002844... that sign is wrong for a short, so recheck: for a short, PnL = N × (1/P_exit − 1/P_entry) = 42,000 × (8.54860e-6 − 8.48088e-6) = 42,000 × 6.772e-8 ≈ 0.0028442 BTC.\n\nFees: taker 0.055% on each fill, paid in BTC on notional converted to BTC.\n- Entry fee = 0.00055 × 42,000 / 117914.1 = 0.00055 × 0.356200 ≈ 0.00019591 BTC\n- Exit fee = 0.00055 × 42,000 / 116982.6 = 0.00055 × 0.359074 ≈ 0.00019749 BTC\n- Total fees ≈ 0.00039340 BTC\n\nNet PnL ≈ 0.0028442 − 0.0003934 ≈ 0.0024508 BTC\n\nPrecision on the gross figure is limited by my hand arithmetic, so I'm not highly confident in the 4th significant digit.\n\n{\"answer\": 0.002451, \"confidence\": 45}" }, { "taskId": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 0.001886499688812526, "answer": 0.001887, "confidence": 62, "grade": "exact", "relError": 0.00026520608004393647, "cost": 0.000877, "reply": "Inverse contract: PnL in BTC = N × (1/entry − 1/exit) for a long, with N = 192,000 USD.\n\n1/20434 = 0.0000489386...\nLet me compute more precisely: 1/20434 ≈ 4.893613e-5\n1/20460.6 ≈ 4.887250e-5\n\nDifference: 4.893613e-5 − 4.887250e-5 = 6.363e-8 (approximately)\n\nGross PnL = 192,000 × 6.363e-8 ≈ 0.012217 BTC\n\nFees: taker fee 0.055% on each fill, charged in BTC on notional value in BTC.\nEntry fee = 192,000 × 0.00055 / 20434 ≈ 105.6 / 20434 ≈ 0.0051685 BTC\nExit fee = 192,000 × 0.00055 / 20460.6 ≈ 105.6 / 20460.6 ≈ 0.0051615 BTC\nTotal fees ≈ 0.0103300 BTC\n\nNet PnL ≈ 0.012217 − 0.010330 ≈ 0.001887 BTC\n\nLet me refine the gross figure: 192,000 × 6.363e-8 = 0.01221696. Recomputing the difference more precisely: 1/20434 = 0.00004893613, 1/20460.6 = 0.00004887250, difference = 0.00000006363, consistent.\n\nNet ≈ 0.001887 BTC\n\n{\"answer\": 0.001887, \"confidence\": 62}" }, { "taskId": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": -0.0962309957129481, "answer": 0.09136, "confidence": 72, "grade": "wrong", "relError": 1.9493822580047075, "cost": 0.000992, "reply": "Inverse contract: PnL in BTC = N × (1/entry − 1/exit) for a short, where N = 139,000 USD.\n\nEntry: 1/59513.2 = 0.0000168037...\nCompute precisely: 1/59513.2 ≈ 1.680272e-5 (since 59513.2 × 1.68e-5 = 0.99982, remainder adds 0.00018/59513.2 ≈ 3.0e-9, so ≈ 1.68030e-5). Let me refine: 59513.2 × 1.6803e-5 = 1.000000... check: 59513.2×1.6803 = 100,000.8 (59513.2×1.68=99982.18; ×0.0003=17.85; total 100000.03). So 1.6803e-5 gives 1.0000003, adjust: 1/59513.2 ≈ 1.680295e-5.\n\nExit: 1/62000.9 ≈ 1.612754e-5 (62000.9×1.6127e-5 = 0.99998..., adjust: 62000.9×1.61275=99999.5 → need +0.5/62000.9 ≈ 8.1e-6 relative... let's directly: 1.61275e-5 × 62000.9 = 0.999995; remainder 5e-6/62000.9 ≈ 8.1e-11, so 1/62000.9 ≈ 1.612758e-5).\n\nGross BTC PnL = 139000 × (1.680295e-5 − 1.612758e-5) = 139000 × 6.7537e-7 ≈ 0.093876 BTC.\n\nFees: taker 0.055% on each fill, notional in BTC terms:\n- Entry fee: 139000 × 1.680295e-5 × 0.00055 = 2.33561 × 0.00055 ≈ 0.00128459 BTC\n- Exit fee: 139000 × 1.612758e-5 × 0.00055 = 2.24185 × 0.00055 ≈ 0.00123302 BTC\nTotal fees ≈ 0.00251761 BTC\n\nNet PnL ≈ 0.093876 − 0.002518 ≈ 0.091358 BTC\n\n{\"answer\": 0.09136, \"confidence\": 72}" }, { "taskId": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 70867.19713259465, "answer": 70862, "confidence": 72, "grade": "exact", "relError": 0.00007333622331537425, "cost": 0.000781, "reply": "For inverse contracts, the average entry price is the weighted average in terms of contract count when using the standard approach. Bybit's inverse perpetual calculates entry price as the position-weighted average, which for inverse contracts is the harmonic-type weighting based on the cost in BTC. The BTC cost of each fill is contracts / price.\n\nTotal contracts: 76,000 + 55,000 = 131,000.\n\nBTC cost:\n- 76,000 / 62771.7 = 1.210742...\n- 55,000 / 86235.1 = 0.637829...\n- Total = 1.848571...\n\nAverage entry = 131,000 / 1.848571 = 70,859.\n\nLet me refine: 76000/62771.7 = 1.2107432; 55000/86235.1 = 0.6378222. Sum = 1.8485654. 131000/1.8485654 = 70,861.9.\n\n**Answer: approximately 70,862**\n\n{\"answer\": 70862, \"confidence\": 72}" }, { "taskId": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 34209.14079495059, "answer": 34213, "confidence": 85, "grade": "exact", "relError": 0.00011281210108560574, "cost": 0.000881, "reply": "For inverse contracts, the average entry price is the total contracts divided by the sum of contracts/price (harmonic weighting).\n\nTotal contracts = 76,000 + 73,000 + 42,000 = 191,000\n\nSum of contracts/price:\n- 76,000 / 33536.1 ≈ 2.266 (76000/33536.1 = 2.26611...)\n- 73,000 / 24929.9 ≈ 2.92834...\n- 42,000 / 108003.1 ≈ 0.388884...\n\nCompute more precisely:\n- 76000/33536.1: 33536.1 × 2.266 = 75992.2; remainder 7.8 → 7.8/33536.1 ≈ 0.000233; total 2.266233\n- 73000/24929.9: 24929.9 × 2.928 = 73002.8; slightly over, so 2.928 − 2.8/24929.9 ≈ 2.928 − 0.000112 = 2.927888\n- 42000/108003.1: 108003.1 × 0.3889 = 42000.4; so 0.3889 − 0.4/108003.1 ≈ 0.3889 − 0.0000037 = 0.388896\n\nSum ≈ 2.266233 + 2.927888 + 0.388896 = 5.583017\n\nAverage entry = 191,000 / 5.583017 ≈ 34,213.0\n\nCheck: 5.583017 × 34,213 ≈ 191,000.0, consistent.\n\nAverage entry price ≈ 34213.\n\n{\"answer\": 34213, \"confidence\": 85}" }, { "taskId": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 88678.94192491287, "answer": 88680, "confidence": 80, "grade": "exact", "relError": 0.000011931525841008828, "cost": 0.0007284, "reply": "For inverse contracts, the average entry price is the weighted harmonic mean: total contracts divided by the sum of (contracts / price).\n\nTotal contracts = 59,000 + 34,000 + 62,000 = 155,000.\n\nSum of contracts/price:\n- 59,000 / 82124.2 ≈ 0.718487\n- 34,000 / 81534 ≈ 0.417025\n- 62,000 / 101232.7 ≈ 0.612459\n\nTotal ≈ 1.747971\n\nAverage entry = 155,000 / 1.747971 ≈ 88,679.6\n\n{\"answer\": 88680, \"confidence\": 80}" }, { "taskId": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 46924.5904555392, "answer": 46926.1, "confidence": 72, "grade": "exact", "relError": 0.000032169582006937134, "cost": 0.000871, "reply": "For an inverse perpetual, the average entry price is the weighted harmonic mean of entry prices, weighted by contract count, since each contract's P&L depends on 1/price.\n\nTotal contracts: 42,000 + 67,000 = 109,000.\n\nAverage entry = total contracts / sum(contracts_i / price_i).\n\nCompute sum:\n- 42,000 / 30128.4 ≈ 1.394 (42000/30128.4 = 1.39390...)\n- 67,000 / 72132.8 ≈ 0.92886 (67000/72132.8 ≈ 0.928829...)\n\nLet me be more precise:\n- 42000/30128.4: 30128.4 × 1.3939 = 41,998.0; remainder 2.0/30128.4 ≈ 0.0000664; so ≈ 1.3939664.\n- 67000/72132.8: 72132.8 × 0.9288 = 66,998.9; remainder 1.1/72132.8 ≈ 0.0000152; so ≈ 0.9288152.\n\nSum ≈ 1.3939664 + 0.9288152 = 2.3227816.\n\nAverage entry = 109000 / 2.3227816 ≈ 46,926.\n\nCheck: 2.3227816 × 46926 ≈ 109,000.0 (2.3227816 × 46926 = 108,999.8), so refine slightly: 46926 + 0.2/2.3228 ≈ 46926.1.\n\n{\"answer\": 46926.1, \"confidence\": 72}" }, { "taskId": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "claude:claude-haiku-5-5", "expected": 37529.651947786224, "answer": 37530.4, "confidence": 72, "grade": "exact", "relError": 0.000019932298194992473, "cost": 0.000876, "reply": "For an inverse contract, the average entry is the volume-weighted harmonic mean of the entry prices, since each contract's P&L depends on 1/price.\n\nTotal contracts: 36,000 + 61,000 = 97,000.\n\nAverage entry = total contracts / (sum of contracts_i / price_i)\n\nSum of contracts_i / price_i:\n- 36,000 / 23930.3 ≈ 1.504 (36000/23930.3 = 1.504 approximately; more precisely 1.504...)\n\nLet me compute more carefully:\n- 36000 / 23930.3: 23930.3 × 1.5 = 35895.45; remainder 104.55; 104.55 / 23930.3 ≈ 0.0043690. So ≈ 1.5043690.\n- 61000 / 56468.2: 56468.2 × 1.08 = 60985.656; remainder 14.344; 14.344 / 56468.2 ≈ 0.00025403. So ≈ 1.08025403.\n\nSum = 1.5043690 + 1.0802540 = 2.5846230\n\nAverage entry = 97000 / 2.5846230 ≈ 37530.2\n\nCheck: 2.5846230 × 37530 = 96,999.0; remainder ~1.0 → 1.0/2.5846 ≈ 0.39, so ≈ 37530.4.\n\nRefining: 2.5846230 × 37530.4 ≈ 97,000.0. So the average entry is about 37530.4.\n\n{\"answer\": 37530.4, \"confidence\": 72}" }, { "taskId": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 4.917273424034056, "answer": 5.07, "confidence": 60, "grade": "wrong", "relError": 0.03105919943753099, "cost": 0.001226, "reply": "I need the sample mean and sample standard deviation of the 40 returns.\n\nSum of returns:\nPositives: 3.56+2.19+1.35+4.41+0.39+3.69+3.17+0.28+4.97+1.30+3.09+0.45+3.89+3.33+1.07+3.17+0.65 = 41.97\nNegatives: -0.91-3.67-0.89-0.92-1.11-1.75-3.56-2.60-3.03-4.20-3.23-3.99-2.38-4.46-3.26-2.96-0.70-4.41-2.16-1.49-1.47-1.30 = -51.57... let me recount carefully.\n\nNegatives list: -0.91, -3.67, -0.89, -0.92, -1.11, -1.75, -3.56, -2.60, -3.03, -4.20, -3.23, -3.99, -2.38, -4.46, -3.26, -2.96, -0.70, -4.41, -2.16, -1.49, -1.47, -1.30 (22 values).\nSum: 0.91+3.67=4.58; +0.89=5.47; +0.92=6.39; +1.11=7.50; +1.75=9.25; +3.56=12.81; +2.60=15.41; +3.03=18.44; +4.20=22.64; +3.23=25.87; +3.99=29.86; +2.38=32.24; +4.46=36.70; +3.26=39.96; +2.96=42.92; +0.70=43.62; +4.41=48.03; +2.16=50.19; +1.49=51.68; +1.47=53.15; +1.30=54.45. So negatives sum = -54.45.\n\nPositives count: 40-22 = 18 values. My positive list had 17 entries; I missed 3.30 (listed as 3.30 after 3.33? The list has 3.30 at position 17). Adding 3.30: 41.97+3.30 = 45.27.\n\nTotal sum = 45.27 - 54.45 = -9.18, mean = -0.2295.\n\nSum of squares:\nPositives: 3.56²=12.67, 2.19²=4.80, 1.35²=1.82, 4.41²=19.45, 0.39²=0.15, 3.69²=13.62, 3.17²=10.05, 0.28²=0.08, 4.97²=24.70, 1.30²=1.69, 3.09²=9.55, 0.45²=0.20, 3.89²=15.13, 3.33²=11.09, 1.07²=1.14, 3.17²=10.05, 0.65²=0.42, 3.30²=10.89. Sum ≈ 150.50.\nNegatives: 0.83, 13.47, 0.79, 0.85, 1.23, 3.06, 12.67, 6.76, 9.18, 17.64, 10.43, 15.92, 5.66, 19.89, 10.63, 8.76, 0.49, 19.45, 4.67, 2.22, 2.16, 1.69. Sum ≈ 188.76.\n\nTotal ≈ 339.26. Sample variance = (339.26 - 40·0.2295²)/39 = (339.26 - 2.107)/39 = 337.15/39 ≈ 8.645. SD ≈ 2.940.\n\nVaR (95%, one-tailed z=1.645): loss = -mean + 1.645·SD = 0.2295 + 4.836 ≈ 5.07%.\n\nVaR ≈ 5.07% (loss).\n\n{\"answer\": 5.07, \"confidence\": 60}" }, { "taskId": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 4.2934299989968965, "answer": 4.294, "confidence": 80, "grade": "exact", "relError": 0.00013276121963937663, "cost": 0.001367, "reply": "I need the sample mean and sample standard deviation of the 40 returns, then VaR = -(mean - 1.645·sd).\n\nSum of returns:\n4.59-3.68=0.91; +2.15=3.06; -0.27=2.79; +3.28=6.07; +4.83=10.90; +2.46=13.36; -1.45=11.91; +3.40=15.31; +4.44=19.75; -0.87=18.88; -0.24=18.64; -1.92=16.72; +3.13=19.85; -0.12=19.73; +4.82=24.55; +0.68=25.23; -3.67=21.56; +0.20=21.76; +2.82=24.58; -0.87=23.71; -4.38=19.33; +0.18=19.51; +3.37=22.88; -3.79=19.09; +2.97=22.06; -1.94=20.12; +3.40=23.52; -1.52=22.00; -4.07=17.93; +4.19=22.12; +1.52=23.64; -3.76=19.88; -4.57=15.31; +0.37=15.68; -1.28=14.40; +0.73=15.13; -1.77=13.36; +2.77=16.13; +2.31=18.44.\n\nMean = 18.44/40 = 0.461.\n\nSum of squares:\n4.59²=21.0681; 3.68²=13.5424; 2.15²=4.6225; 0.27²=0.0729; 3.28²=10.7584; 4.83²=23.3289; 2.46²=6.0516; 1.45²=2.1025; 3.40²=11.56; 4.44²=19.7136; 0.87²=0.7569; 0.24²=0.0576; 1.92²=3.6864; 3.13²=9.7969; 0.12²=0.0144; 4.82²=23.2324; 0.68²=0.4624; 3.67²=13.4689; 0.20²=0.04; 2.82²=7.9524; 0.87²=0.7569; 4.38²=19.1844; 0.18²=0.0324; 3.37²=11.3569; 3.79²=14.3641; 2.97²=8.8209; 1.94²=3.7636; 3.40²=11.56; 1.52²=2.3104; 4.07²=16.5649; 4.19²=17.5561; 1.52²=2.3104; 3.76²=14.1376; 4.57²=20.8849; 0.37²=0.1369; 1.28²=1.6384; 0.73²=0.5329; 1.77²=3.1329; 2.77²=7.6729; 2.31²=5.3361.\n\nRunning total: 21.0681+13.5424=34.6105; +4.6225=39.233; +0.0729=39.3059; +10.7584=50.0643; +23.3289=73.3932; +6.0516=79.4448; +2.1025=81.5473; +11.56=93.1073; +19.7136=112.8209; +0.7569=113.5778; +0.0576=113.6354; +3.6864=117.3218; +9.7969=127.1187; +0.0144=127.1331; +23.2324=150.3655; +0.4624=150.8279; +13.4689=164.2968; +0.04=164.3368; +7.9524=172.2892; +0.7569=173.0461; +19.1844=192.2305; +0.0324=192.2629; +11.3569=203.6198; +14.3641=217.9839; +8.8209=226.8048; +3.7636=230.5684; +11.56=242.1284; +2.3104=244.4388; +16.5649=261.0037; +17.5561=278.5598; +2.3104=280.8702; +14.1376=295.0078; +20.8849=315.8927; +0.1369=316.0296; +1.6384=317.668; +0.5329=318.2009; +3.1329=321.3338; +7.6729=329.0067; +5.3361=334.3428.\n\nSample variance = (334.3428 - 40·0.461²)/39 = (334.3428 - 8.4942)/39 = 325.8486/39 = 8.35507.\nSD = √8.35507 ≈ 2.89051.\n\nVaR = -(0.461 - 1.645·2.89051) = -(0.461 - 4.75489) = 4.2939.\n\n**VaR ≈ 4.294%**\n\n{\"answer\": 4.294, \"confidence\": 80}" }, { "taskId": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 5.2422646771168795, "answer": 5.242, "confidence": 72, "grade": "exact", "relError": 0.000050489079278054144, "cost": 0.001226, "reply": "I need the sample mean and sample standard deviation of the 40 returns.\n\nSum of returns:\n0.81 - 3.34 = -2.53\n-1.80 → -4.33\n-2.90 → -7.23\n-3.97 → -11.20\n+2.94 → -8.26\n+2.88 → -5.38\n+2.34 → -3.04\n-2.55 → -5.59\n-2.93 → -8.52\n+4.20 → -4.32\n+4.36 → 0.04\n+1.59 → 1.63\n+3.83 → 5.46\n+4.51 → 9.97\n+4.60 → 14.57\n+1.02 → 15.59\n-3.34 → 12.25\n+3.82 → 16.07\n+3.43 → 19.50\n-4.88 → 14.62\n-4.91 → 9.71\n+0.19 → 9.90\n-2.96 → 6.94\n-1.54 → 5.40\n-3.61 → 1.79\n-2.21 → -0.42\n-3.84 → -4.26\n-2.12 → -6.38\n+2.49 → -3.89\n+4.47 → 0.58\n+1.13 → 1.71\n+4.34 → 6.05\n+1.00 → 7.05\n-3.76 → 3.29\n-1.06 → 2.23\n+3.00 → 5.23\n-3.34 → 1.89\n+3.52 → 5.41\n-2.41 → 3.00\n\nSum = 3.00, so mean = 0.075.\n\nSum of squares:\n0.6561, 11.1556, 3.24, 8.41, 15.7609, 8.6436, 8.2944, 5.4756, 6.5025, 8.5849, 17.64, 19.0096, 2.5281, 14.6689, 20.3401, 21.16, 1.0404, 11.1556, 14.5924, 11.7649, 23.8144, 24.1081, 0.0361, 8.7616, 2.3716, 13.0321, 4.8841, 14.7456, 4.4944, 6.2001, 19.9809, 1.2769, 18.8356, 1.00, 14.1376, 1.1236, 9.00, 11.1556, 12.3904, 5.8081\n\nRunning total:\n0.6561, 11.8117, 15.0517, 23.4617, 39.2226, 47.8662, 56.1606, 61.6362, 68.1387, 76.7236, 94.3636, 113.3732, 115.9013, 130.5702, 150.9103, 172.0703, 173.1107, 184.2663, 198.8587, 210.6236, 234.4380, 258.5461, 258.5822, 267.3438, 269.7154, 282.7475, 287.6316, 302.3772, 306.8716, 313.0717, 333.0526, 334.3295, 353.1651, 354.1651, 368.3027, 369.4263, 378.4263, 389.5819, 401.9723, 407.7804\n\nSum of squares ≈ 407.7804.\n\nSample variance = (407.7804 - 40·0.075²)/39 = (407.7804 - 0.225)/39 = 407.5554/39 ≈ 10.44757.\n\nStandard deviation ≈ 3.2324.\n\nVaR (95%, parametric): VaR = -(mean - 1.645·σ) = -(0.075 - 1.645·3.2324) = -(0.075 - 5.3173) = 5.2423.\n\nSo the 1-day 95% parametric VaR is about 5.242%.\n\n{\"answer\": 5.242, \"confidence\": 72}" }, { "taskId": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 7.037483251314367, "answer": 7.038, "confidence": 80, "grade": "exact", "relError": 0.00007342805192995108, "cost": 0.001352, "reply": "I need the sample mean and sample standard deviation of the 40 returns.\n\nSum of returns:\nPositives: 3.83+1.02+4.44+3.60+2.25+1.56+3.91+1.55+1.01+2.20+2.12+2.02+1.40+4.86+2.98+1.89+0.93 = \n3.83+1.02=4.85; +4.44=9.29; +3.60=12.89; +2.25=15.14; +1.56=16.70; +3.91=20.61; +1.55=22.16; +1.01=23.17; +2.20=25.37; +2.12=27.49; +2.02=29.51; +1.40=30.91; +4.86=35.77; +2.98=38.75; +1.89=40.64; +0.93=41.57.\n\nNegatives: 1.04+4.23+1.35+2.72+1.23+3.35+2.10+3.18+3.34+3.57+4.06+2.55+1.96+2.03+3.66+0.72+3.08+3.86+3.83+4.40+1.20+2.94 =\n1.04+4.23=5.27; +1.35=6.62; +2.72=9.34; +1.23=10.57; +3.35=13.92; +2.10=16.02; +3.18=19.20; +3.34=22.54; +3.57=26.11; +4.06=30.17; +2.55=32.72; +1.96=34.68; +2.03=36.71; +3.66=40.37; +0.72=41.09; +3.08=44.17; +3.86=48.03; +3.83=51.86; +4.40=56.26; +1.20=57.46; +2.94=60.40.\n\nNet sum = 41.57 − 60.40 = −18.83; mean = −0.47075.\n\nSum of squares:\nSquares of all values:\n1.0816, 17.8929, 14.6689, 1.0404, 1.8225, 7.3984, 1.5129, 19.7136, 11.2225, 12.9600, 4.4100, 10.1124, 5.0625, 11.1556, 2.4336, 12.7449, 15.2881, 16.4836, 6.5025, 2.4025, 3.8416, 4.1209, 13.3956, 1.0201, 0.5184, 4.8400, 9.4864, 14.8996, 4.4944, 14.6689, 4.0804, 1.9600, 23.6196, 8.8804, 3.5721, 19.3600, 1.4400, 8.6436, 0.8649, 0.0000\n\nSum: running:\n1.0816+17.8929=18.9745; +14.6689=33.6434; +1.0404=34.6838; +1.8225=36.5063; +7.3984=43.9047; +1.5129=45.4176; +19.7136=65.1312; +11.2225=76.3537; +12.9600=89.3137; +4.4100=93.7237; +10.1124=103.8361; +5.0625=108.8986; +11.1556=120.0542; +2.4336=122.4878; +12.7449=135.2327; +15.2881=150.5208; +16.4836=167.0044; +6.5025=173.5069; +2.4025=175.9094; +3.8416=179.7510; +4.1209=183.8719; +13.3956=197.2675; +1.0201=198.2876; +0.5184=198.8060; +4.8400=203.6460; +9.4864=213.1324; +14.8996=228.0320; +4.4944=232.5264; +14.6689=247.1953; +4.0804=251.2757; +1.9600=253.2357; +23.6196=276.8553; +8.8804=285.7357; +3.5721=289.3078; +19.3600=308.6678; +1.4400=310.1078; +8.6436=318.7514; +0.8649=319.6163; +0=319.6163.\n\nSample variance = (319.6163 − 40·0.47075²)/39 = (319.6163 − 8.8642)/39 = 310.7521/39 = 7.96800.\nStd dev = √7.968 ≈ 2.82277.\n\nVaR (99%, parametric): −(mean − 2.326348·σ) = −(−0.47075 − 6.56716) = 7.0379 ≈ 7.038.\n\n{\"answer\": 7.038, \"confidence\": 80}" }, { "taskId": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 6.7877378350475945, "answer": 6.788, "confidence": 72, "grade": "exact", "relError": 0.00003862331733734097, "cost": 0.001354, "reply": "I need the sample mean and sample standard deviation of the 40 returns.\n\nSum of returns:\n-2.50 + 4.47 = 1.97\n1.97 - 1.71 = 0.26\n0.26 + 0.32 = 0.58\n0.58 - 3.74 = -3.16\n-3.16 + 0.12 = -3.04\n-3.04 + 2.74 = -0.30\n-0.30 + 2.80 = 2.50\n2.50 - 0.95 = 1.55\n1.55 - 3.34 = -1.79\n-1.79 - 0.21 = -2.00\n-2.00 + 3.76 = 1.76\n1.76 - 0.86 = 0.90\n0.90 - 0.92 = -0.02\n-0.02 - 2.76 = -2.78\n-2.78 + 4.54 = 1.76\n1.76 - 4.23 = -2.47\n-2.47 + 1.16 = -1.31\n-1.31 + 0.50 = -0.81\n-0.81 + 4.48 = 3.67\n3.67 + 4.64 = 8.31\n8.31 + 0.58 = 8.89\n8.89 - 4.75 = 4.14\n4.14 - 3.14 = 1.00\n1.00 + 4.66 = 5.66\n5.66 + 3.24 = 8.90\n8.90 + 2.86 = 11.76\n11.76 - 2.04 = 9.72\n9.72 - 1.49 = 8.23\n8.23 - 4.19 = 4.04\n4.04 + 1.99 = 6.03\n6.03 - 0.03 = 6.00\n6.00 - 3.00 = 3.00\n3.00 - 3.51 = -0.51\n-0.51 + 2.36 = 1.85\n1.85 + 1.91 = 3.76\n3.76 + 2.20 = 5.96\n5.96 + 1.18 = 7.14\n7.14 - 4.75 = 2.39\n2.39 - 0.60 = 1.79\n\nMean = 1.79/40 = 0.04475\n\nSum of squares (approximate, computing x²):\n6.25, 19.9809, 2.9241, 0.1024, 13.9876, 0.0144, 7.5076, 7.84, 0.9025, 11.1556, 0.0441, 14.1376, 0.7396, 0.8464, 7.6176, 20.6116, 17.8929, 1.3456, 0.25, 20.0704, 21.5296, 0.3364, 22.5625, 9.8596, 21.7156, 10.4976, 8.1796, 4.1616, 2.2201, 17.5561, 3.9601, 0.0009, 9.0000, 12.3201, 5.5696, 3.6481, 4.8400, 1.3924, 22.5625, 0.3600\n\nRunning total:\n6.25 → 26.2309 → 29.1550 → 29.2574 → 43.2450 → 43.2594 → 50.7670 → 58.6070 → 59.5095 → 70.6651 → 70.7092 → 84.8468 → 85.5864 → 86.4328 → 94.0504 → 114.6620 → 132.5549 → 133.9005 → 134.1505 → 154.2209 → 175.7505 → 176.0869 → 198.6494 → 208.5090 → 230.2246 → 240.7222 → 248.9018 → 253.0634 → 255.2835 → 272.8396 → 276.7997 → 276.8006 → 285.8006 → 298.1207 → 303.6903 → 307.3384 → 312.1784 → 313.5708 → 336.1333 → 336.4933\n\nSum of squares ≈ 336.4933\n\nSample variance = (336.4933 - 40·0.04475²)/39 = (336.4933 - 0.08022)/39 = 336.4131/39 ≈ 8.62597\nSample std ≈ 2.93702\n\nVaR (99%, normal) = -(mean - 2.326348·σ) = 2.326348·2.93702 - 0.04475 ≈ 6.83253 - 0.04475 ≈ 6.7878\n\nSo the 1-day 99% parametric VaR ≈ 6.788% (loss).\n\n{\"answer\": 6.788, \"confidence\": 72}" }, { "taskId": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.40527021190035356, "answer": 0.4053, "confidence": 72, "grade": "exact", "relError": 0.00007350182365181334, "cost": 0.001168, "reply": "I need mean and sample standard deviation of the 30 returns.\n\nSum: let me add them.\n-0.03-0.14 = -0.17\n-2.24 → -2.41\n-0.42 → -2.83\n-2.93 → -5.76\n+2.82 → -2.94\n+3.50 → 0.56\n-3.33 → -2.77\n-2.35 → -5.12\n-0.77 → -5.89\n+1.20 → -4.69\n+0.38 → -4.31\n-0.96 → -5.27\n+1.19 → -4.08\n-3.90 → -7.98\n-2.28 → -10.26\n+2.86 → -7.40\n-0.20 → -7.60\n+4.21 → -3.39\n+2.17 → -1.22\n-0.68 → -1.90\n+2.42 → 0.52\n+3.83 → 4.35\n+0.96 → 5.31\n-0.03 → 5.28\n+2.59 → 7.87\n-1.03 → 6.84\n-1.20 → 5.64\n-2.51 → 3.13\n-1.70 → 1.43\n\nMean = 1.43/30 = 0.04767\n\nSum of squares:\n0.0009, 0.0196, 5.0176, 0.1764, 8.5849, 7.9524, 12.25, 11.0889, 5.5225, 0.5929, 1.44, 0.1444, 0.9216, 1.4161, 15.21, 5.1984, 8.1796, 0.04, 17.7241, 4.7089, 0.4624, 5.8564, 14.6689, 0.9216, 0.0009, 6.7081, 1.0609, 1.44, 6.3001, 2.89\n\nRunning total: 0.0009+0.0196=0.0205; +5.0176=5.0381; +0.1764=5.2145; +8.5849=13.7994; +7.9524=21.7518; +12.25=34.0018; +11.0889=45.0907; +5.5225=50.6132; +0.5929=51.2061; +1.44=52.6461; +0.1444=52.7905; +0.9216=53.7121; +1.4161=55.1282; +15.21=70.3382; +5.1984=75.5366; +8.1796=83.7162; +0.04=83.7562; +17.7241=101.4803; +4.7089=106.1892; +0.4624=106.6516; +5.8564=112.508; +14.6689=127.1769; +0.9216=128.0985; +0.0009=128.0994; +6.7081=134.8075; +1.0609=135.8684; +1.44=137.3084; +6.3001=143.6085; +2.89=146.4985\n\nSample variance = (146.4985 - 30·0.04767²)/29 = (146.4985 - 0.06823)/29 = 146.4303/29 = 5.04932\nStd dev = √5.04932 ≈ 2.24708\n\nSharpe (daily) = 0.04767/2.24708 ≈ 0.021216\nAnnualized = 0.021216 × √365 (√365 ≈ 19.1050) ≈ 0.40533\n\n**Annualized Sharpe ratio ≈ 0.4053**\n\n{\"answer\": 0.4053, \"confidence\": 72}" }, { "taskId": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.37258355219979333, "answer": 0.3726, "confidence": 80, "grade": "exact", "relError": 0.00004414526650342787, "cost": 0.001208, "reply": "I need the mean and sample standard deviation of the 30 returns.\n\nSum: 0.49+3.50=3.99; +1.12=5.11; +2.59=7.70; -3.20=4.50; -0.05=4.45; -0.34=4.11; -2.32=1.79; -0.09=1.70; +1.99=3.69; -3.04=0.65; +2.82=3.47; +3.56=7.03; -2.59=4.44; +1.21=5.65; -3.72=1.93; -1.96=-0.03; +0.68=0.65; -0.86=-0.21; +2.46=2.25; -1.70=0.55; -2.02=-1.47; +2.95=1.48; +0.85=2.33; -0.52=1.81; -1.58=0.23; -3.05=-2.82; -1.92=-4.74; +2.48=-2.26; +3.61=1.35.\n\nMean = 1.35/30 = 0.045.\n\nSum of squares:\n0.49²=0.2401; 3.50²=12.25; 1.12²=1.2544; 2.59²=6.7081; 3.20²=10.24; 0.05²=0.0025; 0.34²=0.1156; 2.32²=5.3824; 0.09²=0.0081; 1.99²=3.9601; 3.04²=9.2416; 2.82²=7.9524; 3.56²=12.6736; 2.59²=6.7081; 1.21²=1.4641; 3.72²=13.8384; 1.96²=3.8416; 0.68²=0.4624; 0.86²=0.7396; 2.46²=6.0516; 1.70²=2.89; 2.02²=4.0804; 2.95²=8.7025; 0.85²=0.7225; 0.52²=0.2704; 1.58²=2.4964; 3.05²=9.3025; 1.92²=3.6864; 2.48²=6.1504; 3.61²=13.0321.\n\nRunning total: 0.2401+12.25=12.4901; +1.2544=13.7445; +6.7081=20.4526; +10.24=30.6926; +0.0025=30.6951; +0.1156=30.8107; +5.3824=36.1931; +0.0081=36.2012; +3.9601=40.1613; +9.2416=49.4029; +7.9524=57.3553; +12.6736=70.0289; +6.7081=76.737; +1.4641=78.2011; +13.8384=92.0395; +3.8416=95.8811; +0.4624=96.3435; +0.7396=97.0831; +6.0516=103.1347; +2.89=106.0247; +4.0804=110.1051; +8.7025=118.8076; +0.7225=119.5301; +0.2704=119.8005; +2.4964=122.2969; +9.3025=131.5994; +3.6864=135.2858; +6.1504=141.4362; +13.0321=154.4683.\n\nSample variance = (154.4683 − 30·0.045²)/29 = (154.4683 − 0.06075)/29 = 154.4076/29 = 5.32440.\nStd dev = √5.32440 ≈ 2.30748.\n\nSharpe (daily) = 0.045/2.30748 ≈ 0.019502.\nAnnualized = 0.019502 × √365 (√365 ≈ 19.10497) ≈ 0.37259.\n\n{\"answer\": 0.3726, \"confidence\": 80}" }, { "taskId": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": -0.6855095501276544, "answer": -0.6856, "confidence": 80, "grade": "exact", "relError": 0.00013194545915339796, "cost": 0.001212, "reply": "I need mean and sample standard deviation of the 30 returns.\n\nSum: \n1.48 - 2.22 = -0.74\n+0.55 = -0.19\n+4.03 = 3.84\n-0.27 = 3.57\n+3.13 = 6.70\n-1.18 = 5.52\n+0.65 = 6.17\n+0.44 = 6.61\n-1.96 = 4.65\n+1.71 = 6.36\n-3.20 = 3.16\n+2.07 = 5.23\n+3.86 = 9.09\n+3.92 = 13.01\n-1.40 = 11.61\n-1.88 = 9.73\n-3.08 = 6.65\n-0.74 = 5.91\n-1.95 = 3.96\n-2.32 = 1.64\n+2.23 = 3.87\n-0.71 = 3.16\n+2.02 = 5.18\n-3.28 = 1.90\n+0.61 = 2.51\n+1.45 = 3.96\n-2.96 = 1.00\n+0.14 = 1.14\n-3.64 = -2.50\n\nMean = -2.50/30 = -0.08333\n\nSum of squares:\n1.48²=2.1904\n2.22²=4.9284\n0.55²=0.3025\n4.03²=16.2409\n0.27²=0.0729\n3.13²=9.7969\n1.18²=1.3924\n0.65²=0.4225\n0.44²=0.1936\n1.96²=3.8416\n1.71²=2.9241\n3.20²=10.24\n2.07²=4.2849\n3.86²=14.8996\n3.92²=15.3664\n1.40²=1.96\n1.88²=3.5344\n3.08²=9.4864\n0.74²=0.5476\n1.95²=3.8025\n2.32²=5.3824\n2.23²=4.9729\n0.71²=0.5041\n2.02²=4.0804\n3.28²=10.7584\n0.61²=0.3721\n1.45²=2.1025\n2.96²=8.7616\n0.14²=0.0196\n3.64²=13.2496\n\nTotal: let me add sequentially:\n2.1904+4.9284=7.1188\n+0.3025=7.4213\n+16.2409=23.6622\n+0.0729=23.7351\n+9.7969=33.5320\n+1.3924=34.9244\n+0.4225=35.3469\n+0.1936=35.5405\n+3.8416=39.3821\n+2.9241=42.3062\n+10.24=52.5462\n+4.2849=56.8311\n+14.8996=71.7307\n+15.3664=87.0971\n+1.96=89.0571\n+3.5344=92.5915\n+9.4864=102.0779\n+0.5476=102.6255\n+3.8025=106.4280\n+5.3824=111.8104\n+4.9729=116.7833\n+0.5041=117.2874\n+4.0804=121.3678\n+10.7584=132.1262\n+0.3721=132.4983\n+2.1025=134.6008\n+8.7616=143.3624\n+0.0196=143.3820\n+13.2496=156.6316\n\nSample variance = (156.6316 - 30·(0.08333)²)/29 = (156.6316 - 0.2083)/29 = 156.4233/29 = 5.39390\nStd dev = 2.32247\n\nSharpe (daily) = -0.08333/2.32247 = -0.035883\nAnnualized = -0.035883 × √365 (19.10497) = -0.68557\n\n{\"answer\": -0.6856, \"confidence\": 80}" }, { "taskId": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 0.5941064255855656, "answer": 0.5942, "confidence": 62, "grade": "exact", "relError": 0.0001575044645277682, "cost": 0.001154, "reply": "I need mean and sample standard deviation of the 30 returns.\n\nSum: let me add.\n-2.72 -1.97 = -4.69\n-3.54 → -8.23\n+0.68 → -7.55\n-3.46 → -11.01\n+4.40 → -6.61\n+0.18 → -6.43\n+0.01 → -6.42\n+3.68 → -2.74\n+0.26 → -2.48\n+3.31 → 0.83\n+2.79 → 3.62\n+0.26 → 3.88\n-1.31 → 2.57\n-3.29 → -0.72\n-0.19 → -0.91\n-1.70 → -2.61\n-1.24 → -3.85\n+0.43 → -3.42\n-2.23 → -5.65\n-0.57 → -6.22\n+0.39 → -5.83\n+0.29 → -5.54\n+2.92 → -2.62\n-3.54 → -6.16\n+2.00 → -4.16\n+0.24 → -3.92\n-0.21 → -4.13\n+2.00 → -2.13\n+4.31 → 2.18\n\nMean = 2.18/30 = 0.07267\n\nSquares of returns:\n7.3984, 3.8809, 12.5316, 0.4624, 11.9716, 19.3600, 0.0324, 0.0001, 13.5424, 0.0676, 10.9561, 7.7841, 0.0676, 1.7161, 10.8241, 0.0361, 2.8900, 1.5376, 0.1849, 4.9729, 0.3249, 0.1521, 0.0841, 8.5264, 12.5316, 4.0000, 0.0576, 0.0441, 4.0000, 18.5761\n\nSum: \n7.3984+3.8809=11.2793\n+12.5316=23.8109\n+0.4624=24.2733\n+11.9716=36.2449\n+19.36=55.6049\n+0.0324=55.6373\n+0.0001=55.6374\n+13.5424=69.1798\n+0.0676=69.2474\n+10.9561=80.2035\n+7.7841=87.9876\n+0.0676=88.0552\n+1.7161=89.7713\n+10.8241=100.5954\n+0.0361=100.6315\n+2.89=103.5215\n+1.5376=105.0591\n+0.1849=105.244\n+4.9729=110.2169\n+0.3249=110.5418\n+0.1521=110.6939\n+0.0841=110.778\n+8.5264=119.3044\n+12.5316=131.836\n+4=135.836\n+0.0576=135.8936\n+0.0441=135.9377\n+4=139.9377\n+18.5761=158.5138\n\nSum of squares minus n·mean²: 158.5138 − 30·(0.07267)² = 158.5138 − 0.1584 = 158.3554\nSample variance = 158.3554/29 = 5.4602; std = 2.3366\n\nSharpe daily = 0.07267/2.3366 = 0.03110\nAnnualized = 0.03110 × √365 = 0.03110 × 19.105 = 0.5942\n\n{\"answer\": 0.5942, \"confidence\": 62}" }, { "taskId": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "claude:claude-haiku-5-5", "expected": 6.721949210975506, "answer": 6.722, "confidence": 80, "grade": "exact", "relError": 0.000007555698935003174, "cost": 0.001206, "reply": "I need the mean and sample standard deviation of the 30 returns.\n\nSum: \n2.83+1.74=4.57; -2.40→2.17; +3.16→5.33; -2.74→2.59; -1.88→0.71; +0.21→0.92; +3.34→4.26; -0.56→3.70; +4.01→7.71; -0.04→7.67; -0.90→6.77; +4.15→10.92; +0.88→11.80; +3.74→15.54; +2.93→18.47; +1.64→20.11; -1.87→18.24; +2.92→21.16; -0.05→21.11; +1.95→23.06; -3.52→19.54; -0.10→19.44; +0.47→19.91; +1.98→21.89; -2.33→19.56; +3.66→23.22; -3.52→19.70; +1.75→21.45; +4.29→25.74.\n\nMean = 25.74/30 = 0.858.\n\nSum of squares:\n2.83²=8.0089; 1.74²=3.0276; 2.40²=5.76; 3.16²=9.9856; 2.74²=7.5076; 1.88²=3.5344; 0.21²=0.0441; 3.34²=11.1556; 0.56²=0.3136; 4.01²=16.0801; 0.04²=0.0016; 0.90²=0.81; 4.15²=17.2225; 0.88²=0.7744; 3.74²=13.9876; 2.93²=8.5849; 1.64²=2.6896; 1.87²=3.4969; 2.92²=8.5264; 0.05²=0.0025; 1.95²=3.8025; 3.52²=12.3904; 0.10²=0.01; 0.47²=0.2209; 1.98²=3.9204; 2.33²=5.4289; 3.66²=13.3956; 3.52²=12.3904; 1.75²=3.0625; 4.29²=18.4041.\n\nTotal: running sum ≈ 8.0089+3.0276=11.0365; +5.76=16.7965; +9.9856=26.7821; +7.5076=34.2897; +3.5344=37.8241; +0.0441=37.8682; +11.1556=49.0238; +0.3136=49.3374; +16.0801=65.4175; +0.0016=65.4191; +0.81=66.2291; +17.2225=83.4516; +0.7744=84.226; +13.9876=98.2136; +8.5849=106.7985; +2.6896=109.4881; +3.4969=112.985; +8.5264=121.5114; +0.0025=121.5139; +3.8025=125.3164; +12.3904=137.7068; +0.01=137.7168; +0.2209=137.9377; +3.9204=141.8581; +5.4289=147.287; +13.3956=160.6826; +12.3904=173.073; +3.0625=176.1355; +18.4041=194.5396.\n\nSample variance = (194.5396 − 30·0.858²)/29 = (194.5396 − 22.0993)/29 = 172.4403/29 = 5.94622.\nStd dev = √5.94622 ≈ 2.43849.\n\nDaily Sharpe = 0.858/2.43849 ≈ 0.35184.\nAnnualized = 0.35184 × √365 (√365 ≈ 19.1050) ≈ 6.7220.\n\n{\"answer\": 6.722, \"confidence\": 80}" }, { "taskId": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.44697422776459944, "answer": 0.4597, "confidence": 60, "grade": "wrong", "relError": 0.028470930637420618, "cost": 0.001162, "reply": "I need implied vol first, then delta = N(d1).\n\nSetup: S=447.74, K=463, T=23/365=0.063014, r=0.\n\nIntrinsic value is 0, so price 21.6661 is pure time value. Try σ such that call = 21.6661.\n\nATM-ish approximation: C ≈ 0.4σ√T·S... Let's compute directly.\n\nσ√T: try σ=0.40: σ√T=0.40·0.25100=0.10040. d1 = [ln(447.74/463) + 0.5·0.10040²]/0.10040. ln(0.96705)= -0.033501. 0.5·0.010080=0.005040. d1=(-0.028461)/0.10040=-0.28347. d2=-0.38387.\nN(d1)=0.38853, N(d2)=0.35046.\nC = 447.74·0.38853 − 463·0.35046 = 173.97 − 162.26 = 11.71. Too low; need higher vol.\n\nTry σ=0.70: σ√T=0.17570. d1=(-0.028461+0.5·0.030870)/0.17570 = (-0.028461+0.015435)/0.17570 = -0.013026/0.17570 = -0.07415. d2=-0.24985.\nN(d1)=0.47043, N(d2)=0.40136.\nC=447.74·0.47043 − 463·0.40136 = 210.65 − 185.83 = 24.82. Slightly high.\n\nTry σ=0.65: σ√T=0.16315. σ²T/2=0.5·0.026616=0.013308. d1=(-0.028461+0.013308)/0.16315=-0.015153/0.16315=-0.09288. d2=-0.25603.\nN(d1)=0.46300, N(d2)=0.39893.\nC=447.74·0.46300 − 463·0.39893 = 207.28 − 184.71 = 22.57. Slightly high.\n\nTry σ=0.63: σ√T=0.15813. σ²T/2=0.5·0.025005=0.012503. d1=(-0.015958)/0.15813=-0.1091. d2=-0.26704.\nN(d1)=0.45657, N(d2)=0.39471.\nC=447.74·0.45657 − 463·0.39471 = 204.43 − 182.75 = 21.68. Close to 21.6661.\n\nVega ≈ S·φ(d1)·√T ≈ 447.74·0.3968·0.2510 ≈ 44.6 per unit vol. Difference needed: (21.6661−21.68)/44.6 ≈ −0.00031. σ≈0.6297.\n\nDelta at σ=0.6297: σ√T=0.15808, σ²T/2≈0.012479. d1=(-0.028461+0.012479)/0.15808=-0.015982/0.15808=-0.10110.\nN(-0.10110)=0.5-φ(0.1011)·0.1011·... compute: N(0.1011)≈0.5+0.1011·0.39695·(1-0.1011²/6·...)≈0.5+0.040131−small. More precisely N(0.1011)≈0.54027, so N(-0.1011)≈0.45973.\n\n{\"answer\": 0.4597, \"confidence\": 60}" }, { "taskId": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 0.25544432164825037, "answer": 0.255, "confidence": 60, "grade": "correct", "relError": 0.001739407027658275, "cost": 0.001024, "reply": "I need the implied volatility first, then delta = N(d1).\n\nSetup: S=357.63, K=383, T=78/365=0.21370, r=0.\n\nIntrinsic value is zero (S K).\nSet (S − 96000)/S = 0.0653 (total premium in BTC).\n\nSolve: S − 96000 = 0.0653S → 0.9347S = 96000 → S = 96000 / 0.9347 ≈ 102,704.0\n\nCheck the lower breakeven as well for context: put payoff (96000 − S)/S = 0.0653 gives S = 96000 / 1.0653 ≈ 90,118.0, but the question asks for the upper breakeven.\n\nAbove roughly 102,704 USD per BTC at expiry, the straddle is profitable in BTC terms.\n\n{\"answer\": 102704, \"confidence\": 72}" }, { "taskId": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "claude:claude-haiku-5-5", "expected": 73994.63806970509, "answer": 73995, "confidence": 62, "grade": "exact", "relError": 0.000004891304347854031, "cost": 0.000797, "reply": "Total premium paid = 0.0412 + 0.0263 = 0.0675 BTC.\n\nFor a long straddle, breakeven points are K ± total premium, but since the options are settled in BTC and profit is counted in BTC, the premium cost in USD depends on the underlying price. The upper breakeven (where the call's intrinsic value in BTC equals the total premium) is found by:\n\nCall payoff in BTC = (S − K)/S = 0.0675\nSolve: S − K = 0.0675·S → S(1 − 0.0675) = K → S = 69000 / 0.9325 ≈ 73994.6\n\nLower breakeven (put payoff in BTC = (K − S)/S = 0.0675):\n(K − S)/S = 0.0675 → K/S = 1.0675 → S = 69000 / 1.0675 ≈ 64637.7\n\nThe question asks for the price above which the straddle breaks even: the upper breakeven.\n\nUpper breakeven ≈ 73994.6 USD\n\n{\"answer\": 73995, \"confidence\": 62}" }, { "taskId": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 40.790551141452084, "answer": 40.79, "confidence": 99, "grade": "exact", "relError": 0.000013511498047038315, "cost": 0.008754, "reply": "To find the call option price using the Black-Scholes model, we use the formula:\n$$C = S N(d_1) - K e^{-rT} N(d_2)$$\n\n**1. Parameters:**\n* Spot price ($S$) = 382.5\n* Strike price ($K$) = 350\n* Time to expiry ($T$) = $18 / 365 \\approx 0.049315$ years\n* Volatility ($\\sigma$) = 0.607\n* Risk-free rate ($r$) = 0.0597\n\n**2. Intermediate Calculations:**\n* $\\ln(S / K) = \\ln(382.5 / 350) = \\ln(1.092857) \\approx 0.088795$\n* $rT = 0.0597 \\times 0.049315 \\approx 0.002944$\n* $\\sigma \\sqrt{T} = 0.607 \\times \\sqrt{0.049315} \\approx 0.607 \\times 0.222070 \\approx 0.134796$\n* $\\frac{1}{2}\\sigma^2 T = 0.5 \\times (0.134796)^2 \\approx 0.009085$\n\n**3. $d_1$ and $d_2$:**\n* $d_1 = \\frac{\\ln(S / K) + rT + \\frac{1}{2}\\sigma^2 T}{\\sigma \\sqrt{T}} = \\frac{0.088795 + 0.002944 + 0.009085}{0.134796} = \\frac{0.100824}{0.134796} \\approx 0.74798$\n* $d_2 = d_1 - \\sigma \\sqrt{T} = 0.74798 - 0.13480 \\approx 0.61318$\n\n**4. Cumulative Normal Distribution Values:**\n* $N(d_1) = N(0.7480) \\approx 0.77277$\n* $N(d_2) = N(0.6132) \\approx 0.73013$\n\n**5. Call Option Price ($C$):**\n* $S N(d_1) = 382.5 \\times 0.77277 \\approx 295.58$\n* $K e^{-rT} = 350 \\times e^{-0.002944} \\approx 350 \\times 0.997060 \\approx 348.97$\n* $K e^{-rT} N(d_2) = 348.97 \\times 0.73013 \\approx 254.79$\n* $C = 295.58 - 254.79 = 40.79$\n\n{\"answer\": 40.79, \"confidence\": 99}" }, { "taskId": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 3.2317412561362406, "answer": 3.2316, "confidence": 100, "grade": "exact", "relError": 0.00004370898690376947, "cost": 0.010414, "reply": "To find the call option price using the Black-Scholes model, we use the following parameters:\n* Spot price ($S$) = 22.89\n* Strike price ($K$) = 23.00\n* Time to expiry ($T$) = $171 / 365 \\approx 0.468493$ years\n* Volatility ($\\sigma$) = 0.518\n* Risk-free rate ($r$) = 0.0126\n\nFirst, calculate $\\sigma\\sqrt{T}$:\n$$\\sqrt{T} = \\sqrt{0.468493} \\approx 0.684466$$\n$$\\sigma\\sqrt{T} = 0.518 \\times 0.684466 \\approx 0.354553$$\n\nNext, calculate $d_1$ and $d_2$:\n$$\\ln(S/K) = \\ln(22.89 / 23) = \\ln(0.995217) \\approx -0.004794$$\n$$(r + \\frac{\\sigma^2}{2})T = (0.0126 + \\frac{0.518^2}{2}) \\times 0.468493 = (0.0126 + 0.134162) \\times 0.468493 = 0.146762 \\times 0.468493 \\approx 0.068757$$\n\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2/2)T}{\\sigma\\sqrt{T}} = \\frac{-0.004794 + 0.068757}{0.354553} = \\frac{0.063963}{0.354553} \\approx 0.180405$$\n$$d_2 = d_1 - \\sigma\\sqrt{T} = 0.180405 - 0.354553 \\approx -0.174148$$\n\nNow, find the cumulative standard normal distribution values $N(d_1)$ and $N(d_2)$:\n* For $x = 0.180405$, $N(d_1) \\approx 0.5 + \\frac{1}{\\sqrt{2\\pi}}(x - \\frac{x^3}{6}) \\approx 0.5 + 0.398942 \\times (0.180405 - 0.000979) \\approx 0.57158$\n* For $x = -0.174148$, $N(d_2) \\approx 0.5 - 0.398942 \\times (0.174148 - 0.000880) \\approx 0.43088$\n\nCalculate the discounted strike price $K e^{-rT}$:\n$$rT = 0.0126 \\times 0.468493 \\approx 0.005903$$\n$$K e^{-rT} = 23 \\times e^{-0.005903} \\approx 23 \\times 0.994114 \\approx 22.8646$$\n\nFinally, calculate the call price $C$:\n$$C = S N(d_1) - K e^{-rT} N(d_2)$$\n$$C = (22.89 \\times 0.57158) - (22.8646 \\times 0.43088) \\approx 13.0835 - 9.8519 = 3.2316$$\n\n{\"answer\": 3.2316, \"confidence\": 100}" }, { "taskId": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 16.773153687059427, "answer": 16.77, "confidence": 99, "grade": "exact", "relError": 0.00018801992268516554, "cost": 0.008554, "reply": "To find the price of the European call option using the Black-Scholes model, we use the following parameters:\n* Spot price ($S$) = 89.61\n* Strike price ($K$) = 95\n* Time to expiry ($T$) = $152 / 365 \\approx 0.416438$ years\n* Volatility ($\\sigma$) = 0.788\n* Risk-free rate ($r$) = 0.0577\n\n1. **Calculate $\\sigma \\sqrt{T}$:**\n $$\\sqrt{T} = \\sqrt{0.416438} \\approx 0.64532$$\n $$\\sigma \\sqrt{T} = 0.788 \\times 0.64532 \\approx 0.50851$$\n\n2. **Calculate $d_1$ and $d_2$:**\n $$\\ln(S / K) = \\ln(89.61 / 95) = \\ln(0.943263) \\approx -0.05841$$\n $$\\left(r + \\frac{\\sigma^2}{2}\\right)T = \\left(0.0577 + \\frac{0.788^2}{2}\\right) \\times 0.416438 = (0.0577 + 0.310472) \\times 0.416438 \\approx 0.15332$$\n $$d_1 = \\frac{-0.05841 + 0.15332}{0.50851} = \\frac{0.09491}{0.50851} \\approx 0.18664$$\n $$d_2 = d_1 - \\sigma \\sqrt{T} = 0.18664 - 0.50851 = -0.32187$$\n\n3. **Find the cumulative normal distribution values $N(d_1)$ and $N(d_2)$:**\n $$N(0.18664) \\approx 0.57403$$\n $$N(-0.32187) \\approx 0.37378$$\n\n4. **Calculate the Call Option Price ($C$):**\n $$K e^{-rT} = 95 \\times e^{-0.0577 \\times 0.416438} = 95 \\times e^{-0.024028} \\approx 95 \\times 0.976258 \\approx 92.7445$$\n $$C = S N(d_1) - K e^{-rT} N(d_2)$$\n $$C = (89.61 \\times 0.57403) - (92.7445 \\times 0.37378)$$\n $$C \\approx 51.4388 - 34.6660 = 16.7728$$\n\n{\"answer\": 16.77, \"confidence\": 99}" }, { "taskId": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.6945968823131317, "answer": 0.6945, "confidence": 95, "grade": "exact", "relError": 0.0001394799135996935, "cost": 0.008572, "reply": "Given parameters:\n$S = 384.5$\n$K = 427$\n$T = 9 / 365 \\approx 0.0246575$\n$\\sigma = 0.426$\n$r = 0.0252$\n\nFirst, calculate $\\sigma \\sqrt{T}$:\n$\\sqrt{T} = \\sqrt{9 / 365} = 3 / \\sqrt{365} \\approx 3 / 19.104973 \\approx 0.157027$\n$\\sigma \\sqrt{T} = 0.426 \\times 0.157027 \\approx 0.066894$\n\nNext, calculate $\\ln(S/K)$:\n$S/K = 384.5 / 427 \\approx 0.9004684$\n$\\ln(S/K) \\approx -0.1048403$\n\nNow calculate $d_1$ and $d_2$:\n$(r + \\sigma^2 / 2)T = (0.0252 + 0.426^2 / 2) \\times 0.0246575 = (0.0252 + 0.090738) \\times 0.0246575 \\approx 0.115938 \\times 0.0246575 \\approx 0.0028588$\n$d_1 = \\frac{-0.1048403 + 0.0028588}{0.066894} = \\frac{-0.1019815}{0.066894} \\approx -1.52453$\n$d_2 = d_1 - \\sigma \\sqrt{T} = -1.52453 - 0.06689 \\approx -1.59142$\n\nFind the cumulative normal distribution values $N(d_1)$ and $N(d_2)$:\n$N(-1.52453) \\approx 0.06369$\n$N(-1.59142) \\approx 0.05576$\n\nFinally, calculate the call price $C$:\n$K e^{-rT} = 427 \\times e^{-0.0252 \\times 0.0246575} \\approx 427 \\times e^{-0.0006214} \\approx 426.7348$\n$C = S N(d_1) - K e^{-rT} N(d_2)$\n$C \\approx 384.5 \\times 0.063689 - 426.7348 \\times 0.055758$\n$C \\approx 24.4884 - 23.7939 = 0.6945$\n\n{\"answer\": 0.6945, \"confidence\": 95}" }, { "taskId": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 23.323917391138792, "answer": 23.33, "confidence": 99, "grade": "exact", "relError": 0.0002607884755893082, "cost": 0.007494, "reply": "To find the call option price using the Black-Scholes model, we use the following parameters:\n- Spot price ($S$) = 314.3\n- Strike price ($K$) = 294\n- Time to expiry ($T$) = 20 / 365 $\\approx$ 0.0547945 years\n- Volatility ($\\sigma$) = 0.355\n- Risk-free rate ($r$) = 0.0006\n\nFirst, we calculate $\\ln(S/K)$ and $\\sigma\\sqrt{T}$:\n$$\\ln(314.3 / 294) = \\ln(1.0690476) \\approx 0.066768$$\n$$\\sigma\\sqrt{T} = 0.355 \\times \\sqrt{0.0547945} \\approx 0.355 \\times 0.234082 \\approx 0.083099$$\n\nNext, we compute $d_1$ and $d_2$:\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma\\sqrt{T}} = \\frac{0.066768 + (0.0006 + 0.0630125) \\times 0.0547945}{0.083099} = \\frac{0.070254}{0.083099} \\approx 0.8454$$\n$$d_2 = d_1 - \\sigma\\sqrt{T} = 0.8454 - 0.0831 = 0.7623$$\n\nUsing the standard normal cumulative distribution function:\n$$N(d_1) = N(0.8454) \\approx 0.8011$$\n$$N(d_2) = N(0.7623) \\approx 0.7771$$\n\nFinally, we calculate the call price $C$:\n$$C = S \\cdot N(d_1) - K \\cdot e^{-rT} \\cdot N(d_2)$$\n$$K \\cdot e^{-rT} = 294 \\times e^{-0.0000329} \\approx 293.9903$$\n$$C = (314.3 \\times 0.8011) - (293.9903 \\times 0.7771) \\approx 251.786 - 228.460 = 23.33$$\n\n{\"answer\": 23.33, \"confidence\": 99}" }, { "taskId": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": -0.280639388201964, "answer": -0.2806, "confidence": 99, "grade": "exact", "relError": 0.00014035165276097544, "cost": 0.007518, "reply": "To find the delta of the European put option using the Black-Scholes model, we first calculate $d_1$:\n\n$$d_1 = \\frac{\\ln(S / K) + \\left(r + \\frac{\\sigma^2}{2}\\right)T}{\\sigma \\sqrt{T}}$$\n\nGiven:\n* $S = 114.52$\n* $K = 106$\n* $T = \\frac{50}{365} \\approx 0.136986$\n* $\\sigma = 0.466$\n* $r = 0.0585$\n\n1. **Calculate $\\ln(S / K)$:**\n $$\\frac{S}{K} = \\frac{114.52}{106} \\approx 1.080377$$\n $$\\ln(1.080377) \\approx 0.077310$$\n\n2. **Calculate the drift term:**\n $$\\frac{\\sigma^2}{2} = \\frac{0.466^2}{2} = \\frac{0.217156}{2} = 0.108578$$\n $$\\left(r + \\frac{\\sigma^2}{2}\\right)T = (0.0585 + 0.108578) \\times 0.136986 = 0.167078 \\times 0.136986 \\approx 0.022887$$\n\n3. **Calculate the denominator:**\n $$\\sqrt{T} = \\sqrt{0.136986} \\approx 0.370117$$\n $$\\sigma \\sqrt{T} = 0.466 \\times 0.370117 \\approx 0.172474$$\n\n4. **Compute $d_1$:**\n $$d_1 = \\frac{0.077310 + 0.022887}{0.172474} = \\frac{0.100197}{0.172474} \\approx 0.58094$$\n\n5. **Calculate Put Delta:**\n $$\\Delta_{\\text{put}} = N(d_1) - 1 = -N(-d_1)$$\n Using the standard normal cumulative distribution function:\n $$N(0.58094) \\approx 0.71936$$\n $$\\Delta_{\\text{put}} = 0.71936 - 1 = -0.28064$$\n\n{\"answer\": -0.2806, \"confidence\": 99}" }, { "taskId": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": -0.31093085044630486, "answer": -0.3109, "confidence": 99, "grade": "exact", "relError": 0.00009921963761579435, "cost": 0.008872, "reply": "To find the delta of the European put option using the Black-Scholes model, we use the formula:\n$$\\Delta_{\\text{put}} = N(d_1) - 1$$\n\nwhere:\n$$d_1 = \\frac{\\ln(S / K) + \\left(r + \\frac{\\sigma^2}{2}\\right)T}{\\sigma \\sqrt{T}}$$\n\nGiven:\n* Spot price $S = 46.9$\n* Strike price $K = 43$\n* Time to expiry $T = \\frac{41}{365} \\approx 0.112329\\text{ years}$\n* Volatility $\\sigma = 0.727$\n* Risk-free rate $r = 0.0327$\n\n**Step 1: Calculate the components of $d_1$**\n* $\\frac{S}{K} = \\frac{46.9}{43} \\approx 1.090698$\n* $\\ln(S / K) = \\ln(1.090698) \\approx 0.086818$\n* $\\frac{\\sigma^2}{2} = \\frac{0.727^2}{2} = \\frac{0.528529}{2} \\approx 0.264265$\n* $\\left(r + \\frac{\\sigma^2}{2}\\right)T = (0.0327 + 0.264265) \\times 0.112329 = 0.296965 \\times 0.112329 \\approx 0.033358$\n* $\\sqrt{T} = \\sqrt{0.112329} \\approx 0.335155$\n* $\\sigma \\sqrt{T} = 0.727 \\times 0.335155 \\approx 0.243658$\n\n**Step 2: Calculate $d_1$**\n$$d_1 = \\frac{0.086818 + 0.033358}{0.243658} = \\frac{0.120176}{0.243658} \\approx 0.493216$$\n\n**Step 3: Calculate Put Delta**\n$$\\Delta_{\\text{put}} = N(0.493216) - 1 = -N(-0.493216)$$\n\nUsing standard normal cumulative distribution values:\n* $N(0.49) \\approx 0.68793$\n* $N(0.50) \\approx 0.69146$\n\nInterpolating for $d_1 = 0.493216$:\n$$N(0.493216) \\approx 0.68793 + 0.3216 \\times (0.69146 - 0.68793) \\approx 0.68907$$\n\nTherefore:\n$$\\Delta_{\\text{put}} = 0.68907 - 1 = -0.31093$$\n\n{\"answer\": -0.3109, \"confidence\": 99}" }, { "taskId": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": -0.5641608283248785, "answer": -0.5642, "confidence": 99, "grade": "exact", "relError": 0.00006943352525534937, "cost": 0.009188, "reply": "To find the delta of the European put option using the Black-Scholes model, we use the formula:\n$$\\Delta_{\\text{put}} = N(d_1) - 1$$\n\nGiven parameters:\n* Spot price ($S$) = 215.92\n* Strike price ($K$) = 231\n* Time to expiry ($T$) = $65 / 365 \\approx 0.178082$ years\n* Volatility ($\\sigma$) = 0.528\n* Risk-free rate ($r$) = 0.0376\n\n**Step 1: Calculate $d_1$**\n$$d_1 = \\frac{\\ln(S / K) + (r + \\frac{\\sigma^2}{2})T}{\\sigma \\sqrt{T}}$$\n\n1. $\\ln(215.92 / 231) = \\ln(0.9347186) \\approx -0.067510$\n2. $\\frac{\\sigma^2}{2} = \\frac{0.528^2}{2} = \\frac{0.278784}{2} = 0.139392$\n3. $(r + \\frac{\\sigma^2}{2})T = (0.0376 + 0.139392) \\times 0.178082 = 0.176992 \\times 0.178082 \\approx 0.031519$\n4. Numerator = $-0.067510 + 0.031519 = -0.035991$\n5. $\\sqrt{T} = \\sqrt{0.178082} \\approx 0.42200$\n6. Denominator = $\\sigma \\sqrt{T} = 0.528 \\times 0.42200 \\approx 0.222815$\n7. $d_1 = \\frac{-0.035991}{0.222815} \\approx -0.16153$\n\n**Step 2: Calculate $N(d_1)$ and Put Delta**\nUsing the standard normal cumulative distribution function:\n$$N(-0.16153) = 0.5 - \\frac{1}{\\sqrt{2\\pi}} \\int_0^{0.16153} e^{-x^2/2} dx$$\n$$\\frac{1}{\\sqrt{2\\pi}} \\approx 0.398942$$\n$$\\int_0^{0.16153} e^{-x^2/2} dx \\approx 0.16153 - \\frac{(0.16153)^3}{6} \\approx 0.16153 - 0.00070 = 0.16083$$\n$$N(-0.16153) \\approx 0.5 - (0.398942 \\times 0.16083) = 0.5 - 0.06416 = 0.43584$$\n\nFinally, the put delta is:\n$$\\Delta_{\\text{put}} = 0.43584 - 1 = -0.56416$$\n\n{\"answer\": -0.5642, \"confidence\": 99}" }, { "taskId": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": -0.3853788948381438, "answer": -0.3854, "confidence": 99, "grade": "exact", "relError": 0.00005476470595277504, "cost": 0.00816, "reply": "To find the delta of the European put option using the Black-Scholes model, we use the formula:\n$$\\Delta_{\\text{put}} = N(d_1) - 1$$\n\nGiven parameters:\n* Spot price ($S$) = 291.06\n* Strike price ($K$) = 286\n* Time to expiry ($T$) = $113 / 365 \\approx 0.309589$ years\n* Volatility ($\\sigma$) = 0.729\n* Risk-free rate ($r$) = 0.0594\n\nFirst, calculate $\\ln(S/K)$:\n$$\\ln(291.06 / 286) = \\ln(1.0176923) \\approx 0.017538$$\n\nNext, calculate the components of $d_1$:\n$$\\sigma^2 / 2 = (0.729)^2 / 2 = 0.531441 / 2 = 0.2657205$$\n$$(r + \\sigma^2 / 2)T = (0.0594 + 0.2657205) \\times 0.309589 = 0.3251205 \\times 0.309589 \\approx 0.100654$$\n$$\\sigma\\sqrt{T} = 0.729 \\times \\sqrt{0.309589} \\approx 0.729 \\times 0.556407 \\approx 0.405621$$\n\nNow, compute $d_1$:\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma\\sqrt{T}} = \\frac{0.017538 + 0.100654}{0.405621} = \\frac{0.118192}{0.405621} \\approx 0.291385$$\n\nUsing the standard normal cumulative distribution function $N(d_1)$:\nFor $z = 0.291385$:\n* $N(0.29) \\approx 0.61409$\n* $N(0.30) \\approx 0.61791$\nInterpolating for $0.291385$:\n$$N(0.291385) \\approx 0.61409 + 0.1385 \\times (0.61791 - 0.61409) \\approx 0.61462$$\n\nFinally, calculate the put delta:\n$$\\Delta_{\\text{put}} = 0.61462 - 1 = -0.38538$$\n\n{\"answer\": -0.3854, \"confidence\": 99}" }, { "taskId": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": -0.4080759766740296, "answer": -0.4081, "confidence": 99, "grade": "exact", "relError": 0.000058869738341865825, "cost": 0.00744, "reply": "To find the delta of the European put option using the Black-Scholes model, we first calculate the parameter $d_1$:\n\n$$d_1 = \\frac{\\ln(S / K) + (r + \\sigma^2 / 2)T}{\\sigma \\sqrt{T}}$$\n\nGiven:\n* Spot price ($S$) = 146.31\n* Strike price ($K$) = 152\n* Time to expiry ($T$) = $162 / 365 \\approx 0.443836$ years\n* Volatility ($\\sigma$) = 0.799\n* Risk-free rate ($r$) = 0.0456\n\nStep-by-step calculation:\n1. $\\ln(S / K) = \\ln(146.31 / 152) = \\ln(0.962566) \\approx -0.038153$\n2. $\\sigma^2 / 2 = 0.799^2 / 2 = 0.638401 / 2 \\approx 0.3192005$\n3. $(r + \\sigma^2 / 2)T = (0.0456 + 0.3192005) \\times 0.443836 = 0.3648005 \\times 0.443836 \\approx 0.161911$\n4. Numerator = $-0.038153 + 0.161911 = 0.123758$\n5. $\\sqrt{T} = \\sqrt{0.443836} \\approx 0.666210$\n6. Denominator = $\\sigma \\sqrt{T} = 0.799 \\times 0.666210 \\approx 0.532302$\n7. $d_1 = 0.123758 / 0.532302 \\approx 0.232496$\n\nThe delta of a European put option on a non-dividend-paying asset is given by $\\Delta_{\\text{put}} = N(d_1) - 1 = -N(-d_1)$.\n\nUsing the standard normal cumulative distribution function:\n* $N(0.2325) \\approx 0.591926$\n* $\\Delta_{\\text{put}} = 0.591926 - 1 = -0.408074$\n\n{\"answer\": -0.4081, \"confidence\": 99}" }, { "taskId": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.007721610843754648, "answer": 0.007722, "confidence": 100, "grade": "exact", "relError": 0.000050398324031885966, "cost": 0.008448, "reply": "To find the option's gamma ($\\Gamma$) using the Black-Scholes model, we use the formula:\n$$\\Gamma = \\frac{N'(d_1)}{S \\sigma \\sqrt{T}}$$\nwhere $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$ and $d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma \\sqrt{T}}$.\n\n**1. Given parameters:**\n* Spot price ($S$) = $168.77$\n* Strike price ($K$) = $150$\n* Time to expiry ($T$) = $\\frac{34}{365} \\approx 0.0931507\\text{ years}$\n* Volatility ($\\sigma$) = $0.844$\n* Risk-free rate ($r$) = $0.003$\n\n**2. Intermediate calculations:**\n* $\\sqrt{T} = \\sqrt{0.0931507} \\approx 0.305206$\n* $\\sigma \\sqrt{T} = 0.844 \\times 0.305206 \\approx 0.257594$\n* $\\ln(S/K) = \\ln(168.77 / 150) = \\ln(1.125133) \\approx 0.117901$\n* $(r + \\sigma^2 / 2)T = \\left(0.003 + \\frac{0.844^2}{2}\\right) \\times 0.0931507 = (0.003 + 0.356168) \\times 0.0931507 \\approx 0.033457$\n\n**3. Calculate $d_1$ and $N'(d_1)$:**\n* $d_1 = \\frac{0.117901 + 0.033457}{0.257594} = \\frac{0.151358}{0.257594} \\approx 0.587584$\n* $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-0.587584^2 / 2} \\approx 0.398942 \\times e^{-0.172627} \\approx 0.398942 \\times 0.841451 \\approx 0.335690$\n\n**4. Calculate Gamma ($\\Gamma$):**\n* $S \\sigma \\sqrt{T} = 168.77 \\times 0.257594 \\approx 43.4741$\n* $\\Gamma = \\frac{0.335690}{43.4741} \\approx 0.0077216$\n\n{\"answer\": 0.007722, \"confidence\": 100}" }, { "taskId": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.011719035620003572, "answer": 0.01172, "confidence": 99, "grade": "exact", "relError": 0.00008229175400588628, "cost": 0.0089, "reply": "To find the option gamma using the Black-Scholes model, we use the formula:\n$$\\Gamma = \\frac{N'(d_1)}{S \\sigma \\sqrt{T}}$$\nwhere $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$ and\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma \\sqrt{T}}$$\n\n**1. Identify the given parameters:**\n* Spot price ($S$) = 350.55\n* Strike price ($K$) = 375\n* Time to expiry ($T$) = $71 / 365 \\approx 0.1945205$ years\n* Volatility ($\\sigma$) = 0.161\n* Risk-free rate ($r$) = 0.0448\n\n**2. Calculate intermediate values:**\n* $\\sqrt{T} = \\sqrt{0.1945205} \\approx 0.4410448$\n* $\\sigma \\sqrt{T} = 0.161 \\times 0.4410448 \\approx 0.0710082$\n* $\\ln(S / K) = \\ln(350.55 / 375) = \\ln(0.9348) \\approx -0.0674227$\n* $(r + \\sigma^2 / 2)T = (0.0448 + 0.161^2 / 2) \\times 0.1945205 = (0.0448 + 0.0129605) \\times 0.1945205 = 0.0577605 \\times 0.1945205 \\approx 0.0112356$\n\n**3. Calculate $d_1$ and $N'(d_1)$:**\n* $d_1 = \\frac{-0.0674227 + 0.0112356}{0.0710082} = \\frac{-0.0561871}{0.0710082} \\approx -0.791276$\n* $d_1^2 / 2 = (-0.791276)^2 / 2 \\approx 0.626118 / 2 \\approx 0.313059$\n* $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-0.313059} \\approx 0.3989423 \\times 0.731207 \\approx 0.291709$\n\n**4. Calculate Gamma ($\\Gamma$):**\n* $S \\sigma \\sqrt{T} = 350.55 \\times 0.0710082 \\approx 24.89193$\n* $\\Gamma = \\frac{0.291709}{24.89193} \\approx 0.011719$\n\n{\"answer\": 0.01172, \"confidence\": 99}" }, { "taskId": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.00332788170697725, "answer": 0.003328, "confidence": 99, "grade": "exact", "relError": 0.0000355460419466294, "cost": 0.008212, "reply": "To find the option gamma using the Black-Scholes model, we use the formula:\n$$\\Gamma = \\frac{N'(d_1)}{S \\sigma \\sqrt{T}}$$\nwhere $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$ and $d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma \\sqrt{T}}$.\n\n**1. Parameters:**\n* $S = 471.11$\n* $K = 474$\n* $T = \\frac{107}{365} \\approx 0.2931507$\n* $\\sigma = 0.464$\n* $r = 0.0503$\n\n**2. Intermediate Calculations:**\n* $\\sqrt{T} = \\sqrt{0.2931507} \\approx 0.5414339$\n* $\\sigma \\sqrt{T} = 0.464 \\times 0.5414339 \\approx 0.2512253$\n* $\\ln(S/K) = \\ln(471.11 / 474) \\approx -0.0061158$\n* $(r + \\sigma^2 / 2)T = (0.0503 + 0.464^2 / 2) \\times 0.2931507 = (0.0503 + 0.107648) \\times 0.2931507 \\approx 0.0463026$\n\n**3. Calculate $d_1$ and $N'(d_1)$:**\n* $d_1 = \\frac{-0.0061158 + 0.0463026}{0.2512253} = \\frac{0.0401868}{0.2512253} \\approx 0.159963$\n* $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-(0.159963)^2 / 2} \\approx 0.3989423 \\times e^{-0.012794} \\approx 0.3989423 \\times 0.9872875 \\approx 0.393871$\n\n**4. Calculate Gamma ($\\Gamma$):**\n* $S \\sigma \\sqrt{T} = 471.11 \\times 0.2512253 \\approx 118.35476$\n* $\\Gamma = \\frac{0.393871}{118.35476} \\approx 0.0033279$\n\n{\"answer\": 0.003328, \"confidence\": 99}" }, { "taskId": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.015589164355628841, "answer": 0.015589, "confidence": 99, "grade": "exact", "relError": 0.00001054294028157329, "cost": 0.00835, "reply": "To find the option gamma under the Black-Scholes model, we use the formula:\n$$\\Gamma = \\frac{N'(d_1)}{S \\sigma \\sqrt{T}}$$\nwhere $N'(x) = \\frac{1}{\\sqrt{2\\pi}} e^{-x^2/2}$ is the standard normal probability density function, and:\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2/2)T}{\\sigma\\sqrt{T}}$$\n\n**1. Given parameters:**\n* Spot price $S = 87.11$\n* Strike price $K = 100$\n* Time to expiry $T = \\frac{147}{365} \\approx 0.402740$ years\n* Volatility $\\sigma = 0.446$\n* Risk-free rate $r = 0.0514$\n\n**2. Intermediate calculations:**\n* $\\sqrt{T} = \\sqrt{0.402740} \\approx 0.634618$\n* $\\sigma\\sqrt{T} = 0.446 \\times 0.634618 \\approx 0.283040$\n* $\\ln(S/K) = \\ln(0.8711) \\approx -0.137999$\n* $r + \\frac{\\sigma^2}{2} = 0.0514 + \\frac{0.446^2}{2} = 0.0514 + 0.099458 = 0.150858$\n* $(r + \\frac{\\sigma^2}{2})T = 0.150858 \\times 0.402740 \\approx 0.060756$\n\n**3. Calculate $d_1$ and $N'(d_1)$:**\n* $d_1 = \\frac{-0.137999 + 0.060756}{0.283040} = \\frac{-0.077243}{0.283040} \\approx -0.272904$\n* $N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-(-0.272904)^2 / 2} \\approx 0.398942 \\times e^{-0.037238} \\approx 0.398942 \\times 0.963447 \\approx 0.384360$\n\n**4. Calculate Gamma:**\n* $S \\sigma \\sqrt{T} = 87.11 \\times 0.283040 \\approx 24.65557$\n* $\\Gamma = \\frac{0.384360}{24.65557} \\approx 0.015589$\n\n{\"answer\": 0.015589, \"confidence\": 99}" }, { "taskId": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.00596435584251129, "answer": 0.005964, "confidence": 99, "grade": "exact", "relError": 0.000059661515960215854, "cost": 0.008448, "reply": "To find the option's gamma using the Black-Scholes model, we use the formula:\n$$\\Gamma = \\frac{N'(d_1)}{S \\sigma \\sqrt{T}}$$\n\n**1. Calculate the inputs:**\n* Spot price $S = 335.65$\n* Strike price $K = 300$\n* Time to expiry $T = \\frac{103}{365} \\approx 0.282192$\n* Volatility $\\sigma = 0.24$\n* Risk-free rate $r = 0.0003$\n\n**2. Calculate $\\sigma \\sqrt{T}$ and $S \\sigma \\sqrt{T}$:**\n* $\\sqrt{T} = \\sqrt{0.282192} \\approx 0.531217$\n* $\\sigma \\sqrt{T} = 0.24 \\times 0.531217 \\approx 0.127492$\n* $S \\sigma \\sqrt{T} = 335.65 \\times 0.127492 \\approx 42.79274$\n\n**3. Calculate $d_1$:**\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma \\sqrt{T}}$$\n* $\\ln(335.65 / 300) = \\ln(1.118833) \\approx 0.112286$\n* $(r + \\sigma^2 / 2)T = (0.0003 + 0.0576 / 2) \\times 0.282192 = 0.0291 \\times 0.282192 \\approx 0.008212$\n* $d_1 = \\frac{0.112286 + 0.008212}{0.127492} = \\frac{0.120498}{0.127492} \\approx 0.94514$\n\n**4. Calculate $N'(d_1)$:**\n$$N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$$\n* $d_1^2 / 2 = (0.94514)^2 / 2 \\approx 0.89329 / 2 \\approx 0.446646$\n* $e^{-0.446646} \\approx 0.63977$\n* $N'(d_1) = 0.398942 \\times 0.63977 \\approx 0.25523$\n\n**5. Calculate Gamma ($\\Gamma$):**\n$$\\Gamma = \\frac{0.25523}{42.79274} \\approx 0.005964$$\n\n{\"answer\": 0.005964, \"confidence\": 99}" }, { "taskId": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.646446181061521, "answer": 0.6464, "confidence": 99, "grade": "exact", "relError": 0.00007143837008234106, "cost": 0.008696, "reply": "To find the Black-Scholes vega, we first identify the given parameters:\n* Spot price, $S = 424.81$\n* Strike price, $K = 380$\n* Time to expiry, $T = 74 / 365 \\approx 0.2027397$ years\n* Volatility, $\\sigma = 0.55$\n* Risk-free rate, $r = 0.0025$\n\nFirst, calculate $\\sqrt{T}$ and $\\sigma \\sqrt{T}$:\n$$\\sqrt{T} = \\sqrt{0.2027397} \\approx 0.450266$$\n$$\\sigma \\sqrt{T} = 0.55 \\times 0.450266 \\approx 0.247646$$\n\nNext, calculate $d_1$:\n$$d_1 = \\frac{\\ln(S/K) + (r + \\frac{1}{2}\\sigma^2)T}{\\sigma \\sqrt{T}}$$\n$$\\ln(424.81 / 380) = \\ln(1.117921) \\approx 0.111471$$\n$$(r + 0.5\\sigma^2)T = (0.0025 + 0.5 \\times 0.3025) \\times 0.2027397 = 0.15375 \\times 0.2027397 \\approx 0.031171$$\n$$d_1 = \\frac{0.111471 + 0.031171}{0.247646} = \\frac{0.142642}{0.247646} \\approx 0.575992$$\n\nNow, compute the standard normal probability density function at $d_1$, $N'(d_1)$:\n$$N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$$\n$$d_1^2 / 2 \\approx (0.575992)^2 / 2 \\approx 0.165883$$\n$$N'(d_1) \\approx 0.3989423 \\times e^{-0.165883} \\approx 0.3989423 \\times 0.847145 \\approx 0.337962$$\n\nFinally, calculate the vega for a 1 percentage point (0.01) change in volatility:\n$$\\text{Vega}_{1\\%} = \\frac{S \\sqrt{T} N'(d_1)}{100}$$\n$$\\text{Vega}_{1\\%} = \\frac{424.81 \\times 0.450266 \\times 0.337962}{100} \\approx \\frac{191.2775 \\times 0.337962}{100} \\approx 0.646445$$\n\n{\"answer\": 0.6464, \"confidence\": 99}" }, { "taskId": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.26781898170848095, "answer": 0.2678, "confidence": 99, "grade": "exact", "relError": 0.00007087514245585115, "cost": 0.008762, "reply": "To find the vega of the European option, we use the Black-Scholes formula for vega:\n$$\\mathcal{V} = S \\sqrt{T} \\phi(d_1)$$\nwhere $\\phi(x) = \\frac{1}{\\sqrt{2\\pi}} e^{-x^2/2}$ is the standard normal probability density function, and $d_1$ is given by:\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2/2)T}{\\sigma\\sqrt{T}}$$\n\n**1. Identify the given parameters:**\n* Spot price, $S = 146.99$\n* Strike price, $K = 155$\n* Time to expiry, $T = \\frac{104}{365} \\approx 0.28493$\n* Volatility, $\\sigma = 0.163$\n* Risk-free rate, $r = 0.0024$\n\n**2. Calculate intermediate values:**\n* $\\sqrt{T} = \\sqrt{0.28493} \\approx 0.53379$\n* $\\sigma\\sqrt{T} = 0.163 \\times 0.53379 \\approx 0.087008$\n* $\\ln(S/K) = \\ln(146.99 / 155) = \\ln(0.94832) \\approx -0.05306$\n* $(r + \\sigma^2/2)T = (0.0024 + 0.163^2 / 2) \\times 0.28493 = (0.0024 + 0.0132845) \\times 0.28493 \\approx 0.004469$\n\n**3. Calculate $d_1$ and $\\phi(d_1)$:**\n* $d_1 = \\frac{-0.05306 + 0.004469}{0.087008} = \\frac{-0.04859}{0.087008} \\approx -0.55846$\n* $\\phi(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-(-0.55846)^2 / 2} \\approx 0.39894 \\times e^{-0.15594} \\approx 0.39894 \\times 0.85561 \\approx 0.34134$\n\n**4. Calculate Vega:**\n* $\\mathcal{V} = 146.99 \\times 0.53379 \\times 0.34134 \\approx 78.4616 \\times 0.34134 \\approx 26.782$\n\nFor a 1 percentage point (0.01) increase in volatility, we divide by 100:\n$$\\text{Vega (per 1\\%)} = \\frac{26.782}{100} \\approx 0.2678$$\n\n{\"answer\": 0.2678, \"confidence\": 99}" }, { "taskId": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.45587809695630876, "answer": 0.4559, "confidence": 99, "grade": "exact", "relError": 0.000048045834703398345, "cost": 0.00862, "reply": "To find the vega of the option for a 1 percentage point increase in volatility, we use the Black-Scholes formula for vega:\n$$\\text{Vega} = S \\sqrt{T} N'(d_1) \\times 0.01$$\n\n**1. Parameters:**\n* Spot price, $S = 469.67$\n* Strike price, $K = 512$\n* Time to expiry, $T = \\frac{47}{365} \\approx 0.128767$\n* Volatility, $\\sigma = 0.254$\n* Risk-free rate, $r = 0.0139$\n\n**2. Intermediate values:**\n* $\\sqrt{T} = \\sqrt{0.128767} \\approx 0.358841$\n* $\\sigma \\sqrt{T} = 0.254 \\times 0.358841 \\approx 0.091146$\n* $\\ln(S/K) = \\ln(469.67 / 512) = \\ln(0.917324) \\approx -0.086294$\n* $(r + \\frac{\\sigma^2}{2})T = (0.0139 + \\frac{0.254^2}{2}) \\times 0.128767 = (0.0139 + 0.032258) \\times 0.128767 = 0.046158 \\times 0.128767 \\approx 0.005944$\n\n**3. Calculate $d_1$ and $N'(d_1)$:**\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2 / 2)T}{\\sigma \\sqrt{T}} = \\frac{-0.086294 + 0.005944}{0.091146} = \\frac{-0.080350}{0.091146} \\approx -0.88155$$\n\n$$N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2} \\approx 0.398942 \\times e^{-(-0.88155)^2 / 2} = 0.398942 \\times e^{-0.388565} \\approx 0.398942 \\times 0.678029 \\approx 0.270494$$\n\n**4. Calculate Vega (per 1% change):**\n$$\\text{Vega}_{1\\%} = 469.67 \\times 0.358841 \\times 0.270494 \\times 0.01 \\approx 168.5369 \\times 0.270494 \\times 0.01 \\approx 0.45588$$\n\n{\"answer\": 0.4559, \"confidence\": 99}" }, { "taskId": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.6097213395050691, "answer": 0.6097, "confidence": 99, "grade": "exact", "relError": 0.00003499878335639152, "cost": 0.00868, "reply": "To find the vega of the European option, we use the Black-Scholes formula for vega:\n$$\\text{Vega} = S \\sqrt{T} \\phi(d_1)$$\n\nGiven:\n- Spot price $S = 383.12$\n- Strike price $K = 429$\n- Time to expiry $T = \\frac{61}{365} \\approx 0.167123$\n- Volatility $\\sigma = 0.726$\n- Risk-free rate $r = 0.0203$\n\nFirst, calculate $\\sqrt{T}$ and $\\sigma \\sqrt{T}$:\n$$\\sqrt{T} = \\sqrt{0.167123} \\approx 0.408807$$\n$$\\sigma \\sqrt{T} = 0.726 \\times 0.408807 \\approx 0.296794$$\n\nNext, calculate $d_1$:\n$$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2/2)T}{\\sigma\\sqrt{T}}$$\n$$\\ln(383.12 / 429) = \\ln(0.8930536) \\approx -0.113109$$\n$$(r + \\sigma^2 / 2)T = (0.0203 + 0.726^2 / 2) \\times 0.167123 = (0.0203 + 0.263538) \\times 0.167123 \\approx 0.047436$$\n$$d_1 = \\frac{-0.113109 + 0.047436}{0.296794} = \\frac{-0.065673}{0.296794} \\approx -0.221275$$\n\nNow, compute the standard normal probability density function $\\phi(d_1)$:\n$$\\phi(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$$\n$$d_1^2 / 2 = (-0.221275)^2 / 2 \\approx 0.024481$$\n$$\\phi(d_1) \\approx 0.398942 \\times e^{-0.024481} \\approx 0.398942 \\times 0.975816 \\approx 0.389294$$\n\nFinally, calculate the unscaled vega and divide by 100 for a 1 percentage point increase in volatility:\n$$\\text{Vega}_{1\\%} = \\frac{S \\sqrt{T} \\phi(d_1)}{100} = \\frac{383.12 \\times 0.408807 \\times 0.389294}{100} \\approx \\frac{60.9718}{100} \\approx 0.6097$$\n\n{\"answer\": 0.6097, \"confidence\": 99}" }, { "taskId": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.2403754082675673, "answer": 0.2404, "confidence": 100, "grade": "exact", "relError": 0.00010230552538611792, "cost": 0.00866, "reply": "To find the vega of the option (change in price for a 1 percentage point increase in volatility), we use the Black-Scholes formula for vega:\n$$\\text{Vega}_{1\\%} = S \\sqrt{T} \\phi(d_1) \\times 0.01$$\n\n**1. Parameters:**\n* Spot price $S = 158.91$\n* Strike price $K = 148$\n* Time to expiry $T = \\frac{65}{365} \\approx 0.178082$\n* Volatility $\\sigma = 0.564$\n* Risk-free rate $r = 0.0599$\n\n**2. Intermediate Calculations:**\n* $\\sqrt{T} = \\sqrt{0.178082} \\approx 0.42200$\n* $\\sigma \\sqrt{T} = 0.564 \\times 0.42200 \\approx 0.23801$\n* $\\ln(S/K) = \\ln(158.91 / 148) = \\ln(1.073716) \\approx 0.071126$\n* $(r + \\frac{\\sigma^2}{2})T = (0.0599 + \\frac{0.564^2}{2}) \\times 0.178082 = (0.0599 + 0.159048) \\times 0.178082 \\approx 0.038991$\n\n**3. Calculate $d_1$ and $\\phi(d_1)$:**\n* $d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2/2)T}{\\sigma \\sqrt{T}} = \\frac{0.071126 + 0.038991}{0.23801} = \\frac{0.110117}{0.23801} \\approx 0.46266$\n* $d_1^2 / 2 = 0.46266^2 / 2 \\approx 0.10703$\n* $\\phi(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2/2} \\approx 0.398942 \\times e^{-0.10703} \\approx 0.398942 \\times 0.89850 \\approx 0.35845$\n\n**4. Calculate Vega:**\n* $\\text{Vega} = 158.91 \\times 0.42200 \\times 0.35845 \\approx 24.038$\n* For a 1 percentage point (0.01) increase in volatility:\n $$\\text{Vega}_{1\\%} = 24.038 \\times 0.01 \\approx 0.2404$$\n\n{\"answer\": 0.2404, \"confidence\": 100}" }, { "taskId": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 61.799447576373446, "answer": 61.81, "confidence": 99, "grade": "exact", "relError": 0.0001707527177086113, "cost": 0.010956, "reply": "To find the implied volatility, we use the Black-Scholes formula for a European call option:\n$$C = S N(d_1) - K e^{-rT} N(d_2)$$\n\nGiven parameters:\n- Spot price, $S = 27.09$\n- Strike price, $K = 29$\n- Time to expiry, $T = \\frac{136}{365} \\approx 0.372603$\n- Risk-free rate, $r = 0.054$\n- Call price, $C = 3.5195$\n\n1. **Present Value of Strike:**\n$$rT = 0.054 \\times 0.372603 \\approx 0.020121$$\n$$K e^{-rT} = 29 \\times e^{-0.020121} \\approx 29 \\times 0.98008 = 28.4223$$\n\n2. **Moneyness Ratio:**\n$$\\ln\\left(\\frac{S}{K e^{-rT}}\\right) = \\ln\\left(\\frac{27.09}{28.4223}\\right) \\approx \\ln(0.953125) \\approx -0.04801$$\n\n3. **Approximation for Total Volatility ($v = \\sigma\\sqrt{T}$):**\nFor near-at-the-money options, we can approximate $N(d) \\approx 0.5 + \\frac{1}{\\sqrt{2\\pi}} d - \\frac{1}{6\\sqrt{2\\pi}} d^3$.\nTesting $\\sigma = 60\\%$ per year ($0.60$):\n- $\\sqrt{T} = \\sqrt{0.372603} \\approx 0.610412$\n- $v = 0.60 \\times 0.610412 = 0.366247$\n- $d_1 = \\frac{-0.04801}{0.366247} + \\frac{0.366247}{2} = -0.13109 + 0.18312 = 0.05203$\n- $d_2 = d_1 - v = 0.05203 - 0.366247 = -0.31422$\n\n4. **Evaluate Normal CDFs:**\n- $N(0.05203) \\approx 0.5 + 0.39894 \\times 0.05203 \\approx 0.52075$\n- $N(-0.31422) \\approx 0.5 - 0.398942 \\times \\left(0.31422 - \\frac{0.31422^3}{6}\\right) \\approx 0.5 - 0.12330 = 0.37670$\n\n5. **Calculate Call Price:**\n$$C = 27.09 \\times 0.52075 - 28.4223 \\times 0.37670 = 14.1071 - 10.7067 = 3.4004$$\n\n6. **Newton-Raphson Step (using Vega):**\n$$\\text{Vega} = S \\sqrt{T} N'(d_1) \\approx 27.09 \\times 0.610412 \\times 0.39840 \\approx 6.588$$\n$$\\Delta \\sigma = \\frac{3.5195 - 3.4004}{6.588} = \\frac{0.1191}{6.588} \\approx 0.01808$$\n\n$$\\sigma \\approx 0.60 + 0.01808 = 0.61808 \\text{ (or } 61.81\\% \\text{)}$$\n\n{\"answer\": 61.81, \"confidence\": 99}" }, { "taskId": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 20.900010973832025, "answer": 20.9, "confidence": 100, "grade": "exact", "relError": 5.250634576190053e-7, "cost": 0.013852, "reply": "We are given:\n- Spot price $S = 490.18$\n- Strike price $K = 490$\n- Time to expiry $T = \\frac{53}{365} \\approx 0.1452055$ years\n- Risk-free rate $r = 0.0508$\n- Put price $P = 13.6926$\n\nFirst, calculate the discount factor and the forward price:\n$$rT = 0.0508 \\times \\frac{53}{365} \\approx 0.00737644$$\n$$e^{-rT} \\approx 1 - 0.00737644 + \\frac{(0.00737644)^2}{2} \\approx 0.9926508$$\n$$K e^{-rT} = 490 \\times 0.9926508 \\approx 486.3989$$\n\nThe intrinsic value adjustment (difference between discounted strike and spot):\n$$S - K e^{-rT} = 490.18 - 486.3989 = 3.7811$$\n\nUsing the Brenner-Subrahmanyam / Taylor approximation for an option near the forward money:\n$$P \\approx \\frac{S + K e^{-rT}}{2} \\frac{\\sigma \\sqrt{T}}{\\sqrt{2\\pi}} - \\frac{S - K e^{-rT}}{2}$$\n\nHere, $\\frac{S + K e^{-rT}}{2} \\approx 488.2895$ and $\\frac{S - K e^{-rT}}{2} \\approx 1.89055$.\n$$13.6926 + 1.89055 = 15.58315$$\n\nSince $\\frac{1}{\\sqrt{2\\pi}} \\approx 0.3989423$ and $\\sqrt{T} = \\sqrt{0.1452055} \\approx 0.3810584$:\n$$\\sigma \\sqrt{T} \\approx \\frac{15.58315}{488.2895 \\times 0.3989423} = \\frac{15.58315}{194.7993} \\approx 0.079996$$\n\nSolving for $\\sigma$:\n$$\\sigma \\approx \\frac{0.079996}{0.3810584} \\approx 0.20993 \\text{ (or } 21.0\\%)$$\n\nLet's verify $\\sigma = 0.21$:\n$$\\sigma \\sqrt{T} = 0.21 \\times 0.3810584 = 0.0800223$$\n$$\\ln(S/K) = \\ln(490.18/490) \\approx 0.0003673$$\n$$d_1 = \\frac{0.0003673 + 0.0073764 + 0.5(0.0800223)^2}{0.0800223} = \\frac{0.0109455}{0.0800223} \\approx 0.13678$$\n$$d_2 = 0.13678 - 0.0800223 = 0.05676$$\n\nUsing standard normal CDF values:\n$$N(-d_1) = N(-0.13678) \\approx 0.5 - 0.3989423 \\times \\left(0.13678 - \\frac{0.13678^3}{6}\\right) \\approx 0.445602$$\n$$N(-d_2) = N(-0.05676) \\approx 0.5 - 0.3989423 \\times \\left(0.05676 - \\frac{0.05676^3}{6}\\right) \\approx 0.477368$$\n\nThen the Black-Scholes put price is:\n$$P = 486.3989 \\times 0.477368 - 490.18 \\times 0.445602 = 232.1913 - 218.4252 = 13.7661$$\n\nSince $13.7661 > 13.6926$, we adjust using Vega ($\\mathcal{V} = S \\sqrt{T} N'(d_1) \\approx 490.18 \\times 0.381058 \\times 0.39523 \\approx 73.825$):\n$$\\Delta \\sigma = \\frac{13.6926 - 13.7661}{73.825} \\approx -0.0010$$\n$$\\sigma = 0.21 - 0.0010 = 0.2090 \\text{ (or } 20.90\\%)$$\n\n{\"answer\": 20.9, \"confidence\": 100}" }, { "taskId": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 81.50015130720656, "answer": 81.5, "confidence": 99, "grade": "exact", "relError": 0.0000018565266952024242, "cost": 0.010878, "reply": "To find the implied volatility $\\sigma$, we use the Black-Scholes formula for a European put option:\n$$P = K e^{-rT} N(-d_2) - S N(-d_1)$$\n\n**1. Given parameters:**\n* Spot price, $S = 79.84$\n* Strike price, $K = 86$\n* Time to expiry, $T = \\frac{161}{365} \\approx 0.441096\\text{ years}$\n* Risk-free rate, $r = 0.04\\% = 0.0004$\n* Put price, $P = 20.922$\n\n**2. Intermediate values:**\n* $\\sqrt{T} \\approx \\sqrt{0.441096} \\approx 0.6641505$\n* Discounted strike, $K' = K e^{-rT} = 86 \\times e^{-0.0004 \\times 0.441096} \\approx 85.9848$\n* $\\ln(S / K') = \\ln(79.84 / 85.9848) = \\ln(0.928536) \\approx -0.074146$\n\n**3. Solve for $\\sigma\\sqrt{T}$:**\nLet $x = \\sigma\\sqrt{T}$. Then:\n$$d_1 = \\frac{-0.074146}{x} + \\frac{x}{2}, \\quad d_2 = d_1 - x$$\n\nUsing the approximation $P \\approx \\frac{K'-S}{2} + \\sqrt{S K'} \\frac{x}{\\sqrt{2\\pi}}$:\n$$20.922 \\approx \\frac{85.9848 - 79.84}{2} + \\sqrt{79.84 \\times 85.9848} \\frac{x}{2.5066}$$\n$$20.922 \\approx 3.0724 + 33.0548 x \\implies x \\approx 0.540$$\n\nNow, refine $x$ using the exact Black-Scholes formula. \nTest $x = 0.548$:\n* $-d_1 = \\frac{0.074146}{0.548} - 0.274 = 0.13530 - 0.274 = -0.13870$\n* $-d_2 = -0.13870 + 0.548 = 0.40930$\n\nUsing standard normal cumulative distribution values:\n* $N(-d_1) = N(-0.13870) \\approx 0.44484$\n* $N(-d_2) = N(0.40930) \\approx 0.65884$\n\nCalculate the put price for $x = 0.548$:\n$$P = 85.9848(0.65884) - 79.84(0.44484) = 56.6502 - 35.5160 = 21.1342$$\n\nVega with respect to $x$ is $S N'(d_1) \\approx 79.84 \\times 0.3951 \\approx 31.545$.\nAdjust $x$:\n$$x \\approx 0.548 - \\frac{21.1342 - 20.922}{31.545} = 0.548 - 0.00673 = 0.54127$$\n\n**4. Convert to annualized implied volatility:**\n$$\\sigma = \\frac{x}{\\sqrt{T}} = \\frac{0.54127}{0.6641505} \\approx 0.8150 \\text{ or } 81.50\\%$$\n\n{\"answer\": 81.50, \"confidence\": 99}" }, { "taskId": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 87.70003216471197, "answer": 87.8, "confidence": 95, "grade": "correct", "relError": 0.0011398836787228786, "cost": 0.012344, "reply": "We have a European put option with the following parameters:\n- Spot price, $S = 340.53$\n- Strike price, $K = 362$\n- Time to expiry, $T = 105 / 365 \\approx 0.287671$ years\n- Risk-free rate, $r = 0.0106$\n- Put price, $P = 75.8825$\n\nFirst, calculate the discounted strike price:\n$$K e^{-rT} = 362 \\times e^{-0.0106 \\times 0.287671} = 362 \\times e^{-0.0030493} \\approx 362 \\times 0.996955 = 360.898$$\n\nThe intrinsic value (using the forward/discounted strike) is:\n$$K e^{-rT} - S = 360.898 - 340.53 = 20.368$$\n\nThe time value of the put (which equals the value of a call at the same strike) is:\n$$C = P - (K e^{-rT} - S) = 75.8825 - 20.368 = 55.5145$$\n\nNotice that the average of $S$ and $K e^{-rT}$ is approximately $\\frac{340.53 + 360.90}{2} = 350.715$.\nUsing the Brenner-Subrahmanyam approximation for a near-the-money option:\n$$C \\approx 0.3989 \\times F \\times \\sigma \\sqrt{T} - \\frac{K e^{-rT} - S}{2} \\implies \\frac{P + C}{2} \\approx 0.3989 \\times 350.715 \\times \\sigma \\sqrt{T}$$\n$$\\frac{75.8825 + 55.5145}{2} = 65.6985$$\n$$\\sigma \\sqrt{T} \\approx \\frac{65.6985}{0.3989 \\times 350.715} = \\frac{65.6985}{139.90} \\approx 0.4696$$\n\nLet's refine this using the exact Black-Scholes formula with $\\sigma\\sqrt{T} = 0.472$:\n$$\\ln(S / K e^{-rT}) = \\ln(340.53 / 360.898) \\approx -0.05809$$\n$$d_1 = \\frac{-0.05809}{0.472} + \\frac{0.472}{2} = -0.1231 + 0.2360 = 0.1129$$\n$$d_2 = d_1 - 0.472 = -0.3591$$\n\nUsing standard normal CDF values:\n$$N(-d_2) = N(0.3591) \\approx 0.5 + 0.3591 \\times 0.386 = 0.6401$$\n$$N(-d_1) = N(-0.1129) \\approx 0.5 - 0.1129 \\times 0.397 = 0.4551$$\n\nCalculate the put price for $\\sigma\\sqrt{T} = 0.472$:\n$$P = 360.898 \\times 0.6401 - 340.53 \\times 0.4551 = 231.01 - 154.98 = 76.03$$\n\nThis is slightly above $75.8825$. Adjusting by Vega ($\\approx S N'(d_1) = 340.53 \\times 0.3964 \\approx 135$):\n$$\\Delta(\\sigma\\sqrt{T}) = \\frac{75.8825 - 76.03}{135} \\approx -0.0011$$\nSo $\\sigma\\sqrt{T} \\approx 0.472 - 0.0011 = 0.4709$.\n\nSince $\\sqrt{T} = \\sqrt{105/365} \\approx 0.53635$:\n$$\\sigma = \\frac{0.4709}{0.53635} \\approx 0.8780 \\text{ or } 87.80\\%$$\n\n{\"answer\": 87.80, \"confidence\": 95}" }, { "taskId": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 33.39992865638508, "answer": 33.4, "confidence": 99, "grade": "exact", "relError": 0.0000021360409375049508, "cost": 0.01075, "reply": "We are given the following parameters for the European call option:\n- Spot price $S = 140.66$\n- Strike price $K = 143$\n- Time to expiry $T = \\frac{161}{365} \\approx 0.441096$ years\n- Risk-free rate $r = 1.9\\% = 0.019$\n- Call price $C = 11.908$\n\nFirst, we calculate the discounted strike price $K e^{-rT}$:\n$$rT = 0.019 \\times \\frac{161}{365} \\approx 0.0083808$$\n$$K e^{-rT} \\approx 143 \\times e^{-0.0083808} \\approx 143 \\times 0.991654 \\approx 141.8065$$\n\nThe forward moneyness is very close to 1:\n$$\\ln\\left(\\frac{S}{K e^{-rT}}\\right) = \\ln\\left(\\frac{140.66}{141.8065}\\right) \\approx -0.008118$$\n\nFor a near-the-money option, we can use the Brenner-Subrahmanyam approximation mixed with the intrinsic adjustment:\n$$C \\approx \\frac{S - K e^{-rT}}{2} + \\frac{\\sqrt{S \\cdot K e^{-rT}}}{\\sqrt{2\\pi}} \\sigma \\sqrt{T}$$\nHere, $\\frac{S - K e^{-rT}}{2} = \\frac{140.66 - 141.8065}{2} \\approx -0.5733$, and $\\sqrt{S \\cdot K e^{-rT}} \\approx 141.232$.\n$$11.908 \\approx -0.5733 + 141.232 \\times 0.39894 \\times \\sigma \\sqrt{T}$$\n$$12.4813 \\approx 56.343 \\times \\sigma \\sqrt{T} \\implies \\sigma \\sqrt{T} \\approx \\frac{12.4813}{56.343} \\approx 0.22152$$\n\nChecking with the exact Black-Scholes formula for $\\sigma \\sqrt{T} = 0.22195$:\n- $d_1 = \\frac{-0.008118}{0.22195} + \\frac{0.22195}{2} = -0.03658 + 0.11098 = 0.0744$\n- $d_2 = 0.0744 - 0.22195 = -0.14755$\n- $N(d_1) = N(0.0744) \\approx 0.52965$\n- $N(d_2) = N(-0.14755) \\approx 0.44135$\n- $C = 140.66(0.52965) - 141.8065(0.44135) = 74.501 - 62.586 = 11.915$ (very close to $11.908$)\n\nRefining $\\sigma \\sqrt{T}$ slightly down by $\\frac{0.007}{56.3} \\approx 0.00012$ gives $\\sigma \\sqrt{T} \\approx 0.22183$.\n\nSince $\\sqrt{T} = \\sqrt{0.441096} \\approx 0.66415$, we find $\\sigma$:\n$$\\sigma = \\frac{0.22183}{0.66415} \\approx 0.3340 = 33.40\\%$$\n\n{\"answer\": 33.40, \"confidence\": 99}" }, { "taskId": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 6.438461538461515, "answer": 6.43846, "confidence": 100, "grade": "exact", "relError": 2.3894862234357026e-7, "cost": 0.003212, "reply": "To find the position size in standard lots:\n\n1. **Calculate the stop-loss distance in price:**\n $$\\text{Stop Distance} = 0.94593 - 0.94203 = 0.00390 \\text{ USD per AUD}$$\n\n2. **Calculate the risk per standard lot (100,000 units of AUD):**\n $$\\text{Risk per Standard Lot} = 100,000 \\times 0.00390 = 390 \\text{ USD}$$\n\n3. **Calculate the number of standard lots for a 2511 USD risk:**\n $$\\text{Position Size} = \\frac{2511}{390} = \\frac{837}{130} \\approx 6.43846 \\text{ standard lots}$$\n\n{\"answer\": 6.43846, \"confidence\": 100}" }, { "taskId": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 12.3258666666662, "answer": 12.3259, "confidence": 100, "grade": "exact", "relError": 0.0000027043399627036095, "cost": 0.003416, "reply": "To find the position size in standard lots:\n\n1. **Calculate the risk per unit of the base currency (EUR):**\n $$\\text{Price Difference} = 154.513 - 154.363 = 0.150 \\text{ JPY}$$\n\n2. **Calculate the risk per standard lot (100,000 EUR):**\n $$\\text{Risk per Standard Lot} = 0.150 \\text{ JPY/EUR} \\times 100,000 \\text{ EUR} = 15,000 \\text{ JPY}$$\n\n3. **Calculate the number of standard lots for a total risk of 184,888 JPY:**\n $$\\text{Position Size} = \\frac{184,888 \\text{ JPY}}{15,000 \\text{ JPY/lot}} = 12.3258667 \\text{ lots}$$\n\n{\"answer\": 12.3259, \"confidence\": 100}" }, { "taskId": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 0.4756567164179193, "answer": 0.475657, "confidence": 100, "grade": "exact", "relError": 5.961906368996154e-7, "cost": 0.003434, "reply": "To find the position size in standard lots:\n\n1. **Calculate the risk per unit of the base currency (EUR):**\n $$\\text{Risk per EUR} = 191.077 - 190.407 = 0.670 \\text{ JPY}$$\n\n2. **Calculate the risk per standard lot (100,000 EUR):**\n $$\\text{Risk per standard lot} = 0.670 \\text{ JPY/EUR} \\times 100,000 \\text{ EUR} = 67,000 \\text{ JPY}$$\n\n3. **Calculate the position size in standard lots:**\n $$\\text{Position Size} = \\frac{\\text{Total Risk}}{\\text{Risk per standard lot}} = \\frac{31,869 \\text{ JPY}}{67,000 \\text{ JPY}} = 0.4756567... \\text{ lots}$$\n\n{\"answer\": 0.475657, \"confidence\": 100}" }, { "taskId": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 5.01704545454548, "answer": 5.01705, "confidence": 100, "grade": "exact", "relError": 9.060022599983643e-7, "cost": 0.003566, "reply": "To find the position size in standard lots:\n\n1. **Calculate the risk per unit of the base currency (USD):**\n $$\\text{Price Difference} = 142.806 - 142.366 = 0.440 \\text{ JPY}$$\n\n2. **Calculate the risk per standard lot (100,000 units of USD):**\n $$\\text{Risk per standard lot} = 0.440 \\text{ JPY/USD} \\times 100,000 \\text{ USD} = 44,000 \\text{ JPY}$$\n\n3. **Calculate the number of standard lots for a total risk of 220,750 JPY:**\n $$\\text{Position Size} = \\frac{220,750 \\text{ JPY}}{44,000 \\text{ JPY/lot}} = \\frac{220.75}{44} = 5.017045... \\text{ lots}$$\n\n{\"answer\": 5.01705, \"confidence\": 100}" }, { "taskId": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 1.6141304347826113, "answer": 1.61413, "confidence": 100, "grade": "exact", "relError": 2.693602709410174e-7, "cost": 0.003192, "reply": "To find the position size in standard lots, we first calculate the price movement from the entry to the stop-loss:\n\n$$\\text{Stop-loss distance} = 0.76163 - 0.75243 = 0.00920 \\text{ USD per EUR}$$\n\nNext, we determine the risk per standard lot (100,000 units of EUR):\n\n$$\\text{Risk per standard lot} = 100,000 \\times 0.00920 = 920 \\text{ USD}$$\n\nFinally, divide the total desired risk by the risk per standard lot to find the position size:\n\n$$\\text{Position size} = \\frac{1485 \\text{ USD}}{920 \\text{ USD/lot}} \\approx 1.61413 \\text{ lots}$$\n\n{\"answer\": 1.61413, \"confidence\": 100}" }, { "taskId": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.023819567833442343, "answer": 0.02382, "confidence": 100, "grade": "exact", "relError": 0.0000181433416710188, "cost": 0.003288, "reply": "To find the 1-day parametric Value at Risk (VaR) at 97.5% confidence, we use the given formula:\n\n$$\\text{VaR} = -\\mu + z_{0.975} \\times \\sigma$$\n\nGiven:\n* Mean ($\\mu$) = $-0.0003$\n* Standard deviation ($\\sigma$) = $0.012$\n* One-tailed critical value at 97.5% confidence ($z_{0.975}$) $\\approx 1.959964$\n\nCalculating VaR:\n$$\\text{VaR} = -(-0.0003) + 1.959964 \\times 0.012$$\n$$\\text{VaR} = 0.0003 + 0.02351957 = 0.02381957$$\n\n{\"answer\": 0.02382, \"confidence\": 100}" }, { "taskId": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.08208473980663146, "answer": 0.082085, "confidence": 100, "grade": "exact", "relError": 0.0000031698141354686314, "cost": 0.003216, "reply": "To find the 1-day parametric Value at Risk (VaR) at 95% confidence, we use the formula:\n$$\\text{VaR} = -\\mu + z_{0.95} \\cdot \\sigma$$\n\nGiven:\n- Mean return ($\\mu$) = -0.0005\n- Standard deviation ($\\sigma$) = 0.0496\n- 95% one-tailed standard normal critical value ($z_{0.95}$) $\\approx 1.644854$\n\nCalculation:\n$$\\text{VaR} = -(-0.0005) + 1.644854 \\times 0.0496$$\n$$\\text{VaR} = 0.0005 + 0.08158476 = 0.08208476$$\n\n{\"answer\": 0.082085, \"confidence\": 100}" }, { "taskId": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.10032876732331801, "answer": 0.10033, "confidence": 100, "grade": "exact", "relError": 0.000012286373239487742, "cost": 0.003166, "reply": "To find the 1-day parametric Value at Risk (VaR) at 99% confidence, we use the formula:\n\n$$\\text{VaR} = -\\mu + z \\cdot \\sigma$$\n\nGiven:\n* $\\mu = 0.0011$\n* $\\sigma = 0.0436$\n* $z_{0.99} \\approx 2.32635$ (standard normal quantile for 99% one-tailed confidence)\n\nNow, calculate the product $z \\cdot \\sigma$:\n$$2.32635 \\times 0.0436 \\approx 0.101429$$\n\nSubtract the mean:\n$$\\text{VaR} = -0.0011 + 0.101429 = 0.100329$$\n\n{\"answer\": 0.10033, \"confidence\": 100}" }, { "taskId": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.07275151453115127, "answer": 0.07275, "confidence": 100, "grade": "exact", "relError": 0.00002081786421946403, "cost": 0.003286, "reply": "To find the 1-day parametric Value at Risk (VaR) at the 99% confidence level, we use the given formula:\n\n$$\\text{VaR} = -\\mu + z \\cdot \\sigma$$\n\nGiven:\n* Mean ($\\mu$) = $-0.0011$\n* Standard deviation ($\\sigma$) = $0.0308$\n* For a 99% confidence level (one-tailed), the standard normal critical value $z \\approx 2.32635$\n\nSubstitute the values into the formula:\n$$\\text{VaR} = -(-0.0011) + 2.32635 \\times 0.0308$$\n$$\\text{VaR} = 0.0011 + 0.0716516$$\n$$\\text{VaR} = 0.0727516$$\n\n{\"answer\": 0.07275, \"confidence\": 100}" }, { "taskId": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.11150152403343133, "answer": 0.1115, "confidence": 100, "grade": "exact", "relError": 0.000013668274443252881, "cost": 0.003124, "reply": "To find the 1-day parametric Value at Risk (VaR) at 99% confidence, we use the given formula:\n$$\\text{VaR} = -\\mu + z \\cdot \\sigma$$\n\nGiven:\n- Mean daily return ($\\mu$) = -0.001\n- Standard deviation ($\\sigma$) = 0.0475\n- 99% one-tailed standard normal critical value ($z$) $\\approx 2.3263479$\n\nSubstitute the values into the formula:\n$$\\text{VaR} = -(-0.001) + 2.3263479 \\times 0.0475$$\n$$\\text{VaR} = 0.001 + 0.1105015 = 0.1115015$$\n\n{\"answer\": 0.1115, \"confidence\": 100}" }, { "taskId": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.04404168455292251, "answer": 0.04404, "confidence": 100, "grade": "exact", "relError": 0.00003824905744656407, "cost": 0.006196, "reply": "To find the 1-day parametric Conditional Value at Risk (CVaR, or Expected Shortfall) at a 99% confidence level for a normally distributed portfolio return, we use the formula for the expected loss in the worst $(1 - \\alpha)$ tail:\n\n$$\\text{CVaR}_{\\alpha} = -\\mu + \\sigma \\frac{\\phi(z_{\\alpha})}{1 - \\alpha}$$\n\nwhere:\n- $\\mu = 0.0018$ (mean daily return)\n- $\\sigma = 0.0172$ (daily standard deviation)\n- $\\alpha = 0.99$ (confidence level)\n- $1 - \\alpha = 0.01$ (tail probability)\n- $z_{0.99} \\approx 2.32635$ (the 99th percentile of the standard normal distribution)\n- $\\phi(z) = \\frac{1}{\\sqrt{2\\pi}} e^{-z^2 / 2}$ (the standard normal probability density function)\n\nFirst, calculate $\\phi(2.32635)$:\n$$\\phi(2.32635) = \\frac{1}{\\sqrt{2\\pi}} e^{-(2.32635)^2 / 2} \\approx 0.3989423 \\times e^{-2.70595} \\approx 0.3989423 \\times 0.066807 \\approx 0.026652$$\n\nNext, compute the tail expectation factor for the standard normal distribution:\n$$\\frac{\\phi(z_{0.99})}{0.01} \\approx \\frac{0.026652}{0.01} = 2.6652$$\n\nNow, substitute the values into the CVaR formula:\n$$\\text{CVaR}_{0.99} = -0.0018 + 0.0172 \\times 2.6652$$\n$$\\text{CVaR}_{0.99} = -0.0018 + 0.045841 = 0.044041$$\n\n{\"answer\": 0.04404, \"confidence\": 100}" }, { "taskId": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.02233967069948375, "answer": 0.02234, "confidence": 100, "grade": "exact", "relError": 0.000014740616398397254, "cost": 0.006126, "reply": "To find the 1-day parametric Conditional Value at Risk (CVaR), also known as Expected Shortfall (ES), at the 95% confidence level for a normally distributed return, we use the formula for the expected loss in the worst $\\alpha = 1 - 0.95 = 0.05$ tail:\n\n$$\\text{CVaR}_{0.95} = -\\mu + \\sigma \\frac{\\phi(z_{0.95})}{1 - 0.95}$$\n\nWhere:\n* $\\mu = -0.0013$ (mean daily return)\n* $\\sigma = 0.0102$ (daily standard deviation)\n* $z_{0.95} \\approx 1.644854$ (the 95th percentile of the standard normal distribution)\n* $\\phi(z)$ is the standard normal probability density function: $\\phi(z) = \\frac{1}{\\sqrt{2\\pi}} e^{-z^2/2}$\n\nFirst, calculate $\\phi(1.644854)$:\n$$\\phi(1.644854) = \\frac{1}{\\sqrt{2\\pi}} e^{-(1.644854)^2 / 2} \\approx 0.3989423 \\times e^{-1.35277} \\approx 0.1031356$$\n\nNext, divide by the tail probability $\\alpha = 0.05$:\n$$\\frac{\\phi(1.644854)}{0.05} \\approx \\frac{0.1031356}{0.05} \\approx 2.062713$$\n\nNow, compute the CVaR as a positive loss fraction:\n$$\\text{CVaR}_{0.95} = -(-0.0013) + 0.0102 \\times 2.062713$$\n$$\\text{CVaR}_{0.95} = 0.0013 + 0.0210397 = 0.0223397$$\n\n{\"answer\": 0.02234, \"confidence\": 100}" }, { "taskId": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.09524684038587, "answer": 0.09525, "confidence": 99, "grade": "exact", "relError": 0.000033172902294700856, "cost": 0.005946, "reply": "To find the 1-day parametric Conditional Value at Risk (Expected Shortfall) at 99% confidence for a normally distributed portfolio return, we use the formula for Expected Shortfall (ES) as a positive loss fraction:\n\n$$\\text{ES}_{\\alpha} = -\\mu + \\sigma \\frac{\\phi(z_{\\alpha})}{1 - \\alpha}$$\n\nwhere:\n* $\\mu = 0.0023$ (mean daily return)\n* $\\sigma = 0.0366$ (daily standard deviation)\n* $\\alpha = 0.99$ (confidence level)\n* $1 - \\alpha = 0.01$ (tail probability)\n* $z_{0.99} \\approx 2.32635$ (99th percentile of the standard normal distribution)\n* $\\phi(z)$ is the standard normal probability density function: $\\phi(z) = \\frac{1}{\\sqrt{2\\pi}} e^{-z^2/2}$\n\nFirst, calculate the expected value in the 1% tail of a standard normal distribution:\n$$\\phi(2.32635) = \\frac{1}{\\sqrt{2\\pi}} e^{-(2.32635)^2 / 2} \\approx 0.398942 \\times e^{-2.70595} \\approx 0.398942 \\times 0.066807 \\approx 0.026652$$\n\n$$\\frac{\\phi(z_{0.99})}{0.01} \\approx \\frac{0.026652}{0.01} = 2.6652$$\n\nNow, compute the Expected Shortfall:\n$$\\text{ES}_{0.99} = -0.0023 + 0.0366 \\times 2.6652$$\n$$\\text{ES}_{0.99} = -0.0023 + 0.097546 = 0.095246$$\n\n{\"answer\": 0.09525, \"confidence\": 99}" }, { "taskId": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.12233732788143936, "answer": 0.122337, "confidence": 100, "grade": "exact", "relError": 0.0000026801422349040934, "cost": 0.00634, "reply": "To find the 1-day parametric Conditional Value at Risk (CVaR) or Expected Shortfall (ES) for a normally distributed portfolio return at a confidence level $c = 97.5\\%$, we use the formula:\n\n$$\\text{ES} = -\\mu + \\sigma \\frac{\\phi(z_\\alpha)}{\\alpha}$$\n\nwhere:\n- $\\mu = 0.0018$ is the mean daily return.\n- $\\sigma = 0.0531$ is the daily standard deviation.\n- $\\alpha = 1 - c = 1 - 0.975 = 0.025$ is the tail probability.\n- $z_\\alpha = \\Phi^{-1}(0.975) \\approx 1.959964$ is the standard normal quantile.\n- $\\phi(z_\\alpha) = \\frac{1}{\\sqrt{2\\pi}} e^{-z_\\alpha^2 / 2}$ is the standard normal probability density function evaluated at $z_\\alpha$.\n\nFirst, we calculate the density $\\phi(1.959964)$:\n$$z_\\alpha^2 \\approx (1.959964)^2 \\approx 3.841459$$\n$$e^{-3.841459 / 2} = e^{-1.920729} \\approx 0.146500$$\n$$\\phi(1.959964) \\approx 0.3989423 \\times 0.146500 \\approx 0.058445$$\n\nNext, we find the expected tail loss factor for the standard normal distribution:\n$$\\frac{\\phi(z_\\alpha)}{\\alpha} = \\frac{0.058445}{0.025} \\approx 2.337803$$\n\nFinally, we substitute the values into the Expected Shortfall formula:\n$$\\text{ES} = -0.0018 + 0.0531 \\times 2.337803$$\n$$\\text{ES} = -0.0018 + 0.124137 = 0.122337$$\n\n{\"answer\": 0.122337, \"confidence\": 100}" }, { "taskId": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.1205670856534704, "answer": 0.12057, "confidence": 99, "grade": "exact", "relError": 0.000024171991168299684, "cost": 0.00606, "reply": "To find the 1-day parametric Conditional Value at Risk (CVaR, or Expected Shortfall) for a normal distribution, we use the formula for the expected loss in the worst $(1 - c)$ tail:\n\n$$\\text{CVaR}_c = -\\mu + \\sigma \\frac{\\phi(z_c)}{1 - c}$$\n\nGiven:\n* Mean return ($\\mu$) = $0.0017$\n* Standard deviation ($\\sigma$) = $0.0523$\n* Confidence level ($c$) = $97.5\\% = 0.975$\n* Tail probability ($1 - c$) = $0.025$\n\n1. **Find the critical $z$-score ($z_{0.975}$):**\n For a standard normal distribution, the 97.5th percentile is $z_{0.975} \\approx 1.95996$.\n\n2. **Evaluate the standard normal probability density function $\\phi(z_c)$:**\n $$\\phi(1.95996) = \\frac{1}{\\sqrt{2\\pi}} e^{-\\frac{1.95996^2}{2}} \\approx 0.398942 \\times e^{-1.92072} \\approx 0.398942 \\times 0.146501 \\approx 0.058445$$\n\n3. **Calculate the tail expectation factor for the standard normal distribution:**\n $$\\frac{\\phi(z_{0.975})}{0.025} \\approx \\frac{0.058445}{0.025} \\approx 2.3378$$\n\n4. **Compute the CVaR:**\n $$\\text{CVaR}_{0.975} = -0.0017 + 0.0523 \\times 2.3378$$\n $$\\text{CVaR}_{0.975} = -0.0017 + 0.122267 = 0.120567$$\n\n{\"answer\": 0.12057, \"confidence\": 99}" }, { "taskId": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 1.6781883884002948, "answer": 1.6782, "confidence": 100, "grade": "exact", "relError": 0.00000691912766489489, "cost": 0.009994, "reply": "To find the annualized Sharpe ratio, we first calculate the sample mean and sample standard deviation of the 12 monthly returns.\n\n**1. Calculate the Mean ($\\bar{r}$):**\nSum of returns:\n$$-0.0353 + 0.0402 + 0.0555 + 0.0622 + 0.0058 + 0.0795 + 0.0049 + 0.0435 + 0.0415 + 0.0469 - 0.0363 - 0.0547 = 0.2537$$\n\nMean monthly return:\n$$\\bar{r} = \\frac{0.2537}{12} \\approx 0.0211417$$\n\n**2. Calculate the Sample Standard Deviation ($s$):**\nSum of squared returns ($\\sum r_i^2$):\n$$(-0.0353)^2 + (0.0402)^2 + (0.0555)^2 + (0.0622)^2 + (0.0058)^2 + (0.0795)^2 + (0.0049)^2 + (0.0435)^2 + (0.0415)^2 + (0.0469)^2 + (-0.0363)^2 + (-0.0547)^2$$\n$$= 0.00124609 + 0.00161604 + 0.00308025 + 0.00386884 + 0.00003364 + 0.00632025 + 0.00002401 + 0.00189225 + 0.00172225 + 0.00219961 + 0.00131769 + 0.00299209 = 0.02631301$$\n\nUsing the variance formula $\\sum (r_i - \\bar{r})^2 = \\sum r_i^2 - \\frac{(\\sum r_i)^2}{n}$:\n$$\\sum (r_i - \\bar{r})^2 = 0.02631301 - \\frac{0.2537^2}{12} = 0.02631301 - \\frac{0.06436369}{12} = 0.02631301 - 0.00536364 = 0.02094937$$\n\nSample variance ($s^2$, dividing by $n - 1 = 11$):\n$$s^2 = \\frac{0.02094937}{11} \\approx 0.001904488$$\n\nSample standard deviation ($s$):\n$$s = \\sqrt{0.001904488} \\approx 0.04364044$$\n\n**3. Calculate the Annualized Sharpe Ratio:**\n$$\\text{Monthly Sharpe} = \\frac{0.0211417}{0.04364044} \\approx 0.484452$$\n\n$$\\text{Annualized Sharpe} = 0.484452 \\times \\sqrt{12} \\approx 0.484452 \\times 3.4641016 \\approx 1.67819$$\n\n{\"answer\": 1.6782, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": -0.48328609065675004, "answer": -0.4833, "confidence": 100, "grade": "exact", "relError": 0.00002878076468342148, "cost": 0.009704, "reply": "To find the annualized Sharpe ratio, we first calculate the sample mean and sample standard deviation of the 12 monthly returns. \n\nTo simplify the calculations, we scale the returns by $10,000$:\n$$x = [-474, 376, 74, -481, -542, 97, -123, 621, -496, 357, 95, -158]$$\n\n**1. Calculate the Sum and Mean:**\n$$\\sum x_i = (-474 - 481 - 542 - 123 - 496 - 158) + (376 + 74 + 97 + 621 + 357 + 95)$$\n$$\\sum x_i = -2274 + 1620 = -654$$\n\n$$\\bar{x} = \\frac{-654}{12} = -54.5 \\quad (\\text{or } -0.00545 \\text{ in decimal})$$\n\n**2. Calculate the Sample Standard Deviation:**\nFirst, find the sum of squared values ($\\sum x_i^2$):\n* $(-474)^2 = 224,676$\n* $376^2 = 141,376$\n* $74^2 = 5,476$\n* $(-481)^2 = 231,361$\n* $(-542)^2 = 293,764$\n* $97^2 = 9,409$\n* $(-123)^2 = 15,129$\n* $621^2 = 385,641$\n* $(-496)^2 = 246,016$\n* $357^2 = 127,449$\n* $95^2 = 9,025$\n* $(-158)^2 = 24,964$\n\n$$\\sum x_i^2 = 1,714,286$$\n\nUsing the variance formula $\\sum (x_i - \\bar{x})^2 = \\sum x_i^2 - n\\bar{x}^2$:\n$$\\sum (x_i - \\bar{x})^2 = 1,714,286 - 12(-54.5)^2 = 1,714,286 - 35,643 = 1,678,643$$\n\nSample variance ($s^2$):\n$$s^2 = \\frac{1,678,643}{11} \\approx 152,603.909$$\n\nSample standard deviation ($s$):\n$$s = \\sqrt{152,603.909} \\approx 390.6455 \\quad (\\text{or } 0.03906455 \\text{ in decimal})$$\n\n**3. Calculate the Annualized Sharpe Ratio:**\n$$\\text{Monthly Sharpe} = \\frac{\\bar{x}}{s} = \\frac{-54.5}{390.6455} \\approx -0.139513$$\n\n$$\\text{Annualized Sharpe} = \\text{Monthly Sharpe} \\times \\sqrt{12} = -0.139513 \\times 3.4641016 \\approx -0.483287$$\n\n{\"answer\": -0.4833, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 2.241475231769422, "answer": 2.2415, "confidence": 99, "grade": "exact", "relError": 0.0000110499684433581, "cost": 0.008788, "reply": "To find the annualized Sharpe ratio, we first scale the returns by 10,000 for easier calculation:\n$x = [236, 642, 200, -290, 274, 497, -65, 325, 352, 197, 796, -422]$\n\n1. **Sum and Mean:**\n$$\\sum x = 236 + 642 + 200 - 290 + 274 + 497 - 65 + 325 + 352 + 197 + 796 - 422 = 2742$$\n$$\\bar{x} = \\frac{2742}{12} = 228.5 \\quad (\\text{or } 0.02285)$$\n\n2. **Sample Variance and Standard Deviation:**\n$$\\sum x^2 = 236^2 + 642^2 + 200^2 + (-290)^2 + 274^2 + 497^2 + (-65)^2 + 325^2 + 352^2 + 197^2 + 796^2 + (-422)^2$$\n$$\\sum x^2 = 55696 + 412164 + 40000 + 84100 + 75076 + 247009 + 4225 + 105625 + 123904 + 38809 + 633616 + 178084 = 1,998,308$$\n\nUsing the variance formula $\\sum (x_i - \\bar{x})^2 = \\sum x_i^2 - n\\bar{x}^2$:\n$$n\\bar{x}^2 = 12 \\times (228.5)^2 = 12 \\times 52212.25 = 626,547$$\n$$\\sum (x_i - \\bar{x})^2 = 1,998,308 - 626,547 = 1,371,761$$\n\nSample variance ($s^2$):\n$$s^2 = \\frac{1,371,761}{11} \\approx 124,705.545$$\n$$s = \\sqrt{124,705.545} \\approx 353.1367 \\quad (\\text{or } 0.035314)$$\n\n3. **Annualized Sharpe Ratio:**\n$$\\text{Monthly Sharpe} = \\frac{228.5}{353.1367} \\approx 0.647058$$\n$$\\text{Annualized Sharpe} = 0.647058 \\times \\sqrt{12} \\approx 0.647058 \\times 3.464102 \\approx 2.2415$$\n\n{\"answer\": 2.2415, \"confidence\": 99}" }, { "taskId": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 1.9830100176589147, "answer": 1.983, "confidence": 100, "grade": "exact", "relError": 0.000005051743977796405, "cost": 0.011656, "reply": "To find the annualized Sharpe ratio, we first calculate the sample mean and sample standard deviation of the 12 monthly returns.\n\n**1. Sum and Mean of Returns:**\nSum of returns ($\\sum r_i$):\n$$0.05 + 0.044 + 0.0617 + 0.0753 - 0.0527 - 0.0366 + 0.0167 + 0.046 - 0.0225 - 0.0093 + 0.0695 + 0.0687 = 0.3108$$\n\nMean monthly return ($\\bar{r}$):\n$$\\bar{r} = \\frac{0.3108}{12} = 0.0259$$\n\n**2. Sample Standard Deviation:**\nDeviations from the mean ($r_i - 0.0259$):\n- $0.05 - 0.0259 = 0.0241 \\rightarrow (0.0241)^2 = 0.00058081$\n- $0.044 - 0.0259 = 0.0181 \\rightarrow (0.0181)^2 = 0.00032761$\n- $0.0617 - 0.0259 = 0.0358 \\rightarrow (0.0358)^2 = 0.00128164$\n- $0.0753 - 0.0259 = 0.0494 \\rightarrow (0.0494)^2 = 0.00244036$\n- $-0.0527 - 0.0259 = -0.0786 \\rightarrow (-0.0786)^2 = 0.00617796$\n- $-0.0366 - 0.0259 = -0.0625 \\rightarrow (-0.0625)^2 = 0.00390625$\n- $0.0167 - 0.0259 = -0.0092 \\rightarrow (-0.0092)^2 = 0.00008464$\n- $0.046 - 0.0259 = 0.0201 \\rightarrow (0.0201)^2 = 0.00040401$\n- $-0.0225 - 0.0259 = -0.0484 \\rightarrow (-0.0484)^2 = 0.00234256$\n- $-0.0093 - 0.0259 = -0.0352 \\rightarrow (-0.0352)^2 = 0.00123904$\n- $0.0695 - 0.0259 = 0.0436 \\rightarrow (0.0436)^2 = 0.00190096$\n- $0.0687 - 0.0259 = 0.0428 \\rightarrow (0.0428)^2 = 0.00183184$\n\nSum of squared deviations:\n$$\\sum (r_i - \\bar{r})^2 = 0.02251768$$\n\nSample variance ($s^2$):\n$$s^2 = \\frac{0.02251768}{11} \\approx 0.0020470618$$\n\nSample standard deviation ($s$):\n$$s = \\sqrt{0.0020470618} \\approx 0.04524447$$\n\n**3. Annualized Sharpe Ratio:**\n$$\\text{Sharpe}_{\\text{annual}} = \\frac{\\bar{r}}{s} \\times \\sqrt{12} = \\frac{0.0259}{0.04524447} \\times 3.4641016 \\approx 0.5724456 \\times 3.4641016 \\approx 1.9830$$\n\n{\"answer\": 1.983, \"confidence\": 100}" }, { "taskId": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 1.3915400475486268, "answer": 1.3915, "confidence": 100, "grade": "exact", "relError": 0.000028779300098047822, "cost": 0.008554, "reply": "To find the annualized Sharpe ratio, we first calculate the mean and sample standard deviation of the 12 monthly returns (multiplied by $10^4$ for simplicity):\n\nReturns ($x_i \\times 10^4$): $603, -347, 392, 484, -114, -247, 144, 301, 712, 25, -155, -103$\n\n1. **Sum of returns:**\n$$\\sum x_i = 603 - 347 + 392 + 484 - 114 - 247 + 144 + 301 + 712 + 25 - 155 - 103 = 1695$$\n$$\\text{Mean } (\\bar{x}) = \\frac{1695}{12} = 141.25 \\quad (\\text{or } 0.014125)$$\n\n2. **Sum of squared returns:**\n$$\\sum x_i^2 = 603^2 + (-347)^2 + 392^2 + 484^2 + (-114)^2 + (-247)^2 + 144^2 + 301^2 + 712^2 + 25^2 + (-155)^2 + (-103)^2$$\n$$\\sum x_i^2 = 363609 + 120409 + 153664 + 234256 + 12996 + 61009 + 20736 + 90601 + 506944 + 625 + 24025 + 10609 = 1599483$$\n\n3. **Sample variance and standard deviation:**\n$$\\sum (x_i - \\bar{x})^2 = \\sum x_i^2 - \\frac{(\\sum x_i)^2}{12} = 1599483 - \\frac{1695^2}{12} = 1599483 - 239418.75 = 1360064.25$$\n$$s = \\sqrt{\\frac{1360064.25}{11}} = \\sqrt{123642.2045} \\approx 351.628 \\quad (\\text{or } 0.0351628)$$\n\n4. **Annualized Sharpe ratio:**\n$$\\text{Sharpe}_{\\text{annual}} = \\frac{\\bar{x}}{s} \\times \\sqrt{12} = \\frac{141.25}{351.628} \\times 3.4641016 \\approx 0.401703 \\times 3.4641016 \\approx 1.3915$$\n\n{\"answer\": 1.3915, \"confidence\": 100}" }, { "taskId": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-4-argon", "expected": 0.4848484848484847, "answer": 0.484848, "confidence": 100, "grade": "exact", "relError": 9.999999996818112e-7, "cost": 0.00298, "reply": "For a binary contract that costs $c = 0.67$ and pays $1$ with estimated probability $p = 0.83$, the net odds received on the wager are $b = \\frac{1 - c}{c}$. \n\nThe full Kelly fraction $f^*$ of the bankroll to stake is given by:\n$$f^* = p - \\frac{1 - p}{b} = p - \\frac{(1 - p)c}{1 - c} = \\frac{p - c}{1 - c}$$\n\nSubstituting the given values:\n$$f^* = \\frac{0.83 - 0.67}{1 - 0.67} = \\frac{0.16}{0.33} \\approx 0.484848$$\n\n{\"answer\": 0.484848, \"confidence\": 100}" }, { "taskId": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-4-argon", "expected": 0.28985507246376807, "answer": 0.289855, "confidence": 100, "grade": "exact", "relError": 2.4999999993502446e-7, "cost": 0.00282, "reply": "For a binary contract costing $c$ that pays $1$ with probability $p$ and $0$ otherwise, the net fractional odds are $b = \\frac{1 - c}{c}$. \n\nThe Kelly criterion fraction $f^*$ is given by:\n$$f^* = \\frac{p b - (1 - p)}{b} = \\frac{p - c}{1 - c}$$\n\nGiven $p = 0.51$ and $c = 0.31$:\n$$f^* = \\frac{0.51 - 0.31}{1 - 0.31} = \\frac{0.20}{0.69} \\approx 0.289855$$\n\n{\"answer\": 0.289855, \"confidence\": 100}" }, { "taskId": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-4-argon", "expected": 0.09859154929577463, "answer": 0.09859, "confidence": 100, "grade": "exact", "relError": 0.000015714285714109126, "cost": 0.00375, "reply": "Here's my analysis of the situation:\n\n**Kelly Criterion Application**\n\nOkay, so I'm looking at a binary contract, and I've got a cost, let's call it $c$, of 0.29. The probability of success, which I'll denote as $p$, is 0.36. I need to figure out the optimal fraction of my capital to bet, which, of course, means I'm reaching for the Kelly fraction. The formula for the Kelly fraction in this binary scenario is $f = \\frac{p - c}{1 - c}$. Plugging in those values, I get $\\frac{0.36 - 0.29}{1 - 0.29}$. That's $\\frac{0.07}{0.71}$, which comes out to roughly 0.0985915. So, according to this, I should be betting about 9.86% of my capital on each contract, assuming I'm maximizing for long-term growth and accounting for the risk involved. It's a fairly standard Kelly calculation, nothing too exotic here.\n\n\n\nTo find the optimal fraction of your bankroll to stake using the Kelly criterion for a binary contract, we use the formula:\n\n$$f^* = \\frac{p - c}{1 - c}$$\n\nwhere:\n* $p = 0.36$ is your estimated probability of winning.\n* $c = 0.29$ is the cost of the contract.\n\nSubstituting the given values:\n\n$$f^* = \\frac{0.36 - 0.29}{1 - 0.29} = \\frac{0.07}{0.71} \\approx 0.0985915$$\n\n{\"answer\": 0.09859, \"confidence\": 100}" }, { "taskId": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-4-argon", "expected": 0.8333333333333334, "answer": 0.833333, "confidence": 100, "grade": "exact", "relError": 4.0000000005591114e-7, "cost": 0.0036, "reply": "Let $c = 0.16$ be the cost of the contract and $p = 0.86$ be the estimated probability of winning. \n\nIf you buy the contract, your net profit per dollar staked if the event occurs is:\n$$b = \\frac{1 - c}{c} = \\frac{0.84}{0.16} = 5.25$$\n\nThe Kelly criterion fraction $f^*$ of the bankroll to stake is given by:\n$$f^* = \\frac{p \\cdot b - (1 - p)}{b} = p - \\frac{1 - p}{b}$$\n\nAlternatively, for binary prediction markets paying $1, this simplifies directly to:\n$$f^* = \\frac{p - c}{1 - c}$$\n\nSubstituting the given values:\n$$f^* = \\frac{0.86 - 0.16}{1 - 0.16} = \\frac{0.70}{0.84} = \\frac{5}{6} \\approx 0.833333$$\n\n{\"answer\": 0.833333, \"confidence\": 100}" }, { "taskId": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-4-argon", "expected": 0.576271186440678, "answer": 0.576271, "confidence": 100, "grade": "exact", "relError": 3.23529411836051e-7, "cost": 0.0036, "reply": "Okay, let me break this down.\n\n**Kelly Fraction Calculation**\n\nAlright, so I'm looking at a Kelly fraction calculation, a classic in portfolio optimization and betting strategies. The variables I'm working with are:\n\n* `c = 0.41`: This seems to represent the cost or disadvantage associated with the bet, or perhaps the edge I am *losing* or *paying* to play the game.\n* `p = 0.75`: This is the probability of success, the likelihood of winning whatever I'm betting on, so a 75% chance.\n\nThe goal, as I see it, is to figure out the optimal fraction of my capital to wager on this event. I recall the Kelly criterion formula. It's:\n\n`f = (p - c) / (1 - c)`\n\nIn this case, plugging in the values:\n\n`f = (0.75 - 0.41) / (1 - 0.41)`\n\nWhich simplifies to:\n\n`f = 0.34 / 0.59`\n\nAnd the approximate result is:\n\n`f ≈ 0.57627`\n\nSo, my optimal Kelly fraction, given these parameters, is approximately 57.63%. This means I should bet roughly 57.63% of my available capital on this particular bet. This minimizes risk and maximizes long-term growth (when the parameters are held constant, which is a big assumption!). I'll need to carefully consider whether these probabilities and costs are accurate before I actually make any investment decisions.\n\n\n\nFor a binary contract costing $c = 0.41$ that pays $1$ with probability $p = 0.75$, the net odds received on a winning bet are $b = \\frac{1 - c}{c}$. \n\nThe full Kelly fraction $f^*$ is given by:\n$$f^* = \\frac{bp - (1 - p)}{b} = \\frac{p - c}{1 - c}$$\n\nSubstituting the given values:\n$$f^* = \\frac{0.75 - 0.41}{1 - 0.41} = \\frac{0.34}{0.59} \\approx 0.576271$$\n\n{\"answer\": 0.576271, \"confidence\": 100}" }, { "taskId": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 24920.824742268043, "answer": 24920.8247, "confidence": 100, "grade": "exact", "relError": 1.6960932370518094e-9, "cost": 0.005334, "reply": "Let $P_0 = 24173.2$ be the entry price, $L = 25$ be the leverage, and $Q$ be the position size in USD.\n\n1. **Position value at entry (in BTC):**\n $$V_0 = \\frac{Q}{P_0}$$\n\n2. **Initial Margin (IM) and Maintenance Margin (MM) in BTC:**\n $$\\text{IM} = \\frac{V_0}{L} = 0.04 V_0$$\n $$\\text{MM} = 0.01 V_0$$\n\n3. **Unrealized PnL for a short inverse position at liquidation price $P_{\\text{liq}}$:**\n $$\\text{uPnL} = Q \\left(\\frac{1}{P_{\\text{liq}}} - \\frac{1}{P_0}\\right) = V_0 \\left(\\frac{P_0}{P_{\\text{liq}}} - 1\\right)$$\n\n4. **Liquidation condition:**\n $$\\text{IM} + \\text{uPnL} = \\text{MM}$$\n $$0.04 V_0 + V_0 \\left(\\frac{P_0}{P_{\\text{liq}}} - 1\\right) = 0.01 V_0$$\n $$0.04 + \\frac{P_0}{P_{\\text{liq}}} - 1 = 0.01$$\n $$\\frac{P_0}{P_{\\text{liq}}} = 0.97$$\n\n5. **Solve for $P_{\\text{liq}}$:**\n $$P_{\\text{liq}} = \\frac{24173.2}{0.97} \\approx 24920.8247$$\n\n{\"answer\": 24920.8247, \"confidence\": 100}" }, { "taskId": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 53342.9702970297, "answer": 53342.97, "confidence": 100, "grade": "exact", "relError": 5.5683007055427045e-9, "cost": 0.004904, "reply": "For an inverse (coin-margined) long position, the values in BTC are given by:\n- **Entry Value in BTC**: $V_0 = \\frac{N}{P_0}$, where $N$ is the USD position size and $P_0 = 53876.4$.\n- **Initial Margin (IM)**: $\\frac{V_0}{L} = \\frac{V_0}{50} = 0.02 \\, V_0$\n- **Maintenance Margin (MM)**: $0.01 \\, V_0$\n- **Unrealized PnL in BTC** at price $P$: $\\text{PnL} = N \\left(\\frac{1}{P_0} - \\frac{1}{P}\\right) = V_0 \\left(1 - \\frac{P_0}{P}\\right)$\n\nLiquidation occurs when $\\text{IM} + \\text{PnL} = \\text{MM}$:\n$$0.02 \\, V_0 + V_0 \\left(1 - \\frac{P_0}{P_{\\text{liq}}}\\right) = 0.01 \\, V_0$$\n\nDivide the entire equation by $V_0$:\n$$0.02 + 1 - \\frac{P_0}{P_{\\text{liq}}} = 0.01$$\n$$1.01 = \\frac{P_0}{P_{\\text{liq}}}$$\n$$P_{\\text{liq}} = \\frac{53876.4}{1.01} \\approx 53342.9703$$\n\n{\"answer\": 53342.97, \"confidence\": 100}" }, { "taskId": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 53194.15525114154, "answer": 53194.16, "confidence": 100, "grade": "exact", "relError": 8.927406475571944e-8, "cost": 0.005698, "reply": "Let $P_0 = 58247.6$ be the entry price, $N$ be the position size in USD, and $L = 10$ be the leverage.\n\n1. **Position Value at Entry (in BTC):**\n $$V_0 = \\frac{N}{P_0}$$\n\n2. **Initial Margin (IM) and Maintenance Margin (MM) in BTC:**\n $$\\text{IM} = \\frac{V_0}{L} = 0.1 V_0$$\n $$\\text{MM} = 0.005 V_0$$\n\n3. **Unrealized PnL (in BTC) for a long inverse position at price $P$:**\n $$\\text{PnL} = N \\left(\\frac{1}{P_0} - \\frac{1}{P}\\right) = V_0 \\left(1 - \\frac{P_0}{P}\\right)$$\n\n4. **Liquidation Condition:**\n Liquidation occurs when $\\text{IM} + \\text{PnL} = \\text{MM}$:\n $$0.1 V_0 + V_0 \\left(1 - \\frac{P_0}{P_{\\text{liq}}}\\right) = 0.005 V_0$$\n Divide by $V_0$:\n $$0.1 + 1 - \\frac{P_0}{P_{\\text{liq}}} = 0.005$$\n $$1.095 = \\frac{P_0}{P_{\\text{liq}}}$$\n\n5. **Calculate $P_{\\text{liq}}$:**\n $$P_{\\text{liq}} = \\frac{58247.6}{1.095} = \\frac{58247600}{1095} = \\frac{11649520}{219} \\approx 53194.15525$$\n\n{\"answer\": 53194.16, \"confidence\": 100}" }, { "taskId": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 170355.2216748768, "answer": 170355.22, "confidence": 100, "grade": "exact", "relError": 9.831672864603095e-9, "cost": 0.006084, "reply": "For an inverse (coin-margined) short position:\nLet $N$ be the position size in USD, $P_0 = 115273.7$ be the entry price, and $L = 3$ be the leverage.\n\n1. **Position value in BTC at entry**:\n $$V_0 = \\frac{N}{P_0}$$\n\n2. **Initial Margin (IM) and Maintenance Margin (MM)**:\n $$\\text{IM} = \\frac{V_0}{3}$$\n $$\\text{MM} = 0.01 V_0$$\n\n3. **Unrealized PnL at price $P$**:\n $$\\text{uPnL} = N \\left(\\frac{1}{P} - \\frac{1}{P_0}\\right) = V_0 \\left(\\frac{P_0}{P} - 1\\right)$$\n\n4. **Liquidation Condition**:\n $$\\text{IM} + \\text{uPnL} = \\text{MM}$$\n $$\\frac{V_0}{3} + V_0 \\left(\\frac{P_0}{P_{\\text{liq}}} - 1\\right) = 0.01 V_0$$\n\n Dividing by $V_0$:\n $$\\frac{1}{3} + \\frac{P_0}{P_{\\text{liq}}} - 1 = 0.01$$\n $$\\frac{P_0}{P_{\\text{liq}}} = 1 - \\frac{1}{3} + 0.01 = \\frac{2}{3} + \\frac{1}{100} = \\frac{203}{300}$$\n\n5. **Liquidation Price ($P_{\\text{liq}}$)**:\n $$P_{\\text{liq}} = P_0 \\times \\frac{300}{203} = 115273.7 \\times \\frac{300}{203} = \\frac{34582110}{203} \\approx 170355.2217$$\n\n{\"answer\": 170355.22, \"confidence\": 100}" }, { "taskId": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 115003.24607329845, "answer": 115003.25, "confidence": 100, "grade": "exact", "relError": 3.414426711223029e-8, "cost": 0.00531, "reply": "Let the position size in USD be $Q$, the entry price be $P_0 = 109828.1$, and the leverage be $L = 20$.\n\nThe position value in BTC at entry is:\n$$V_0 = \\frac{Q}{P_0}$$\n\nThe initial margin (IM) and maintenance margin (MM) in BTC are:\n$$\\text{IM} = \\frac{V_0}{20} = 0.05 V_0$$\n$$\\text{MM} = 0.005 V_0$$\n\nFor an inverse short position, the unrealized PnL in BTC at price $P$ is:\n$$\\text{PnL} = Q \\left(\\frac{1}{P} - \\frac{1}{P_0}\\right) = V_0 \\left(\\frac{P_0}{P} - 1\\right)$$\n\nLiquidation occurs when $\\text{IM} + \\text{PnL} = \\text{MM}$:\n$$0.05 V_0 + V_0 \\left(\\frac{P_0}{P} - 1\\right) = 0.005 V_0$$\n\nDividing by $V_0$:\n$$0.05 + \\frac{P_0}{P} - 1 = 0.005$$\n$$\\frac{P_0}{P} = 1 - 0.05 + 0.005 = 0.955$$\n\nSolving for the liquidation price $P$:\n$$P = \\frac{109828.1}{0.955} = \\frac{109828100}{955} \\approx 115003.2461$$\n\n{\"answer\": 115003.25, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 2398.0256188582507, "answer": 2398.03, "confidence": 100, "grade": "exact", "relError": 0.0000018269787090834882, "cost": 0.005136, "reply": "For an inverse (coin-margined) contract, the average entry price is the harmonic weighted average of the fill prices:\n\n$$\\text{Average Entry Price} = \\frac{\\sum Q_i}{\\sum \\frac{Q_i}{P_i}}$$\n\n1. **Total USD Quantity ($\\sum Q_i$):**\n$$7000 + 34000 + 36000 + 9000 = 86000\\text{ USD}$$\n\n2. **Total ETH Value at Entry ($\\sum \\frac{Q_i}{P_i}$):**\n* $\\frac{7000}{2309.36} \\approx 3.031143\\text{ ETH}$\n* $\\frac{34000}{3563.54} \\approx 9.541074\\text{ ETH}$\n* $\\frac{36000}{1854.34} \\approx 19.413915\\text{ ETH}$\n* $\\frac{9000}{2321.56} \\approx 3.876704\\text{ ETH}$\n\n$$\\text{Total ETH} \\approx 3.031143 + 9.541074 + 19.413915 + 3.876704 = 35.862836\\text{ ETH}$$\n\n3. **Average Entry Price:**\n$$\\frac{86000}{35.862836} \\approx 2398.0256$$\n\n{\"answer\": 2398.03, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 3712.5303509582454, "answer": 3712.53, "confidence": 100, "grade": "exact", "relError": 9.453343462758414e-8, "cost": 0.003882, "reply": "For an inverse (coin-margined) contract, the average entry price is the harmonic weighted average of the fill prices:\n\n$$\\text{Average Entry Price} = \\frac{\\text{Total USD Contracts}}{\\text{Total ETH Value at Entry}}$$\n\n1. **Total USD Contracts:**\n $$34000 + 16000 = 50000\\text{ USD}$$\n\n2. **Total ETH Value at Entry:**\n $$\\frac{34000}{3528.19} \\approx 9.6366692\\text{ ETH}$$\n $$\\frac{16000}{4176.2} \\approx 3.8312341\\text{ ETH}$$\n $$\\text{Total ETH} \\approx 9.6366692 + 3.8312341 = 13.4679033\\text{ ETH}$$\n\n3. **Average Entry Price:**\n $$\\frac{50000}{13.4679033} \\approx 3712.53$$\n\n{\"answer\": 3712.53, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 2495.29829080704, "answer": 2495.3, "confidence": 100, "grade": "exact", "relError": 6.849653873569959e-7, "cost": 0.00547, "reply": "For an inverse (coin-margined) contract, the average entry price is the weighted harmonic mean of the entry prices, where the weights are the USD contract quantities.\n\n1. **Calculate the ETH value of each fill ($\\text{USD Quantity} / \\text{Price}$):**\n * Fill 1: $41000 / 3517.25 \\approx 11.656834\\text{ ETH}$\n * Fill 2: $10000 / 2976.86 \\approx 3.359244\\text{ ETH}$\n * Fill 3: $27000 / 4229.06 \\approx 6.384398\\text{ ETH}$\n * Fill 4: $40000 / 1545.09 \\approx 25.888460\\text{ ETH}$\n\n2. **Sum the total USD contracts and total ETH value:**\n * $\\text{Total USD} = 41000 + 10000 + 27000 + 40000 = 118000\\text{ USD}$\n * $\\text{Total ETH} = 11.656834 + 3.359244 + 6.384398 + 25.888460 = 47.288936\\text{ ETH}$\n\n3. **Calculate the average entry price:**\n $$\\text{Average Entry Price} = \\frac{\\text{Total USD}}{\\text{Total ETH}} = \\frac{118000}{47.288936} \\approx 2495.30$$\n\n{\"answer\": 2495.30, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 2610.72093572179, "answer": 2610.72, "confidence": 100, "grade": "exact", "relError": 3.584150941701721e-7, "cost": 0.004978, "reply": "For an inverse perpetual contract, the average entry price is the weighted harmonic mean of the fill prices, which equals the total USD value divided by the total ETH value of the fills.\n\n1. **Total USD value:**\n$$\\text{Total USD} = 10000 + 14000 + 12000 + 23000 = 59000$$\n\n2. **ETH value for each fill:**\n$$\\text{ETH}_1 = \\frac{10000}{3282.44} \\approx 3.046514$$\n$$\\text{ETH}_2 = \\frac{14000}{2147.97} \\approx 6.517782$$\n$$\\text{ETH}_3 = \\frac{12000}{1744.23} \\approx 6.879827$$\n$$\\text{ETH}_4 = \\frac{23000}{3736.8} \\approx 6.154999$$\n\n3. **Total ETH value:**\n$$\\text{Total ETH} \\approx 3.046514 + 6.517782 + 6.879827 + 6.154999 = 22.599122$$\n\n4. **Average Entry Price:**\n$$\\text{Average Entry Price} = \\frac{59000}{22.599122} \\approx 2610.72$$\n\n{\"answer\": 2610.72, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 3424.594098647905, "answer": 3424.59, "confidence": 100, "grade": "exact", "relError": 0.000001196827357271257, "cost": 0.005246, "reply": "To find the average entry price $P_{avg}$ for an inverse (coin-margined) contract, we use the weighted harmonic mean, as the ETH value of each fill is given by $\\frac{\\text{USD Contracts}}{\\text{Price}}$.\n\n1. **Calculate total USD contracts:**\n$$\\text{Total USD} = 31000 + 36000 + 40000 + 5000 = 112000\\text{ USD}$$\n\n2. **Calculate the ETH cost of each fill:**\n* $\\frac{31000}{3315.48} \\approx 9.35008\\text{ ETH}$\n* $\\frac{36000}{3410.80} \\approx 10.55471\\text{ ETH}$\n* $\\frac{40000}{3481.71} \\approx 11.48861\\text{ ETH}$\n* $\\frac{5000}{3813.27} \\approx 1.31121\\text{ ETH}$\n\n3. **Calculate total ETH cost:**\n$$\\text{Total ETH} = 9.35008 + 10.55471 + 11.48861 + 1.31121 = 32.70461\\text{ ETH}$$\n\n4. **Calculate the average entry price:**\n$$P_{avg} = \\frac{\\text{Total USD}}{\\text{Total ETH}} = \\frac{112000}{32.70461} \\approx 3424.59$$\n\n{\"answer\": 3424.59, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 0.02840777213808475, "answer": 0.028408, "confidence": 100, "grade": "exact", "relError": 0.000008021111762719801, "cost": 0.005604, "reply": "To find the net PnL in BTC for an inverse (coin-margined) short position, we calculate the BTC values at opening and closing:\n\n1. **Opening Value in BTC ($V_{\\text{open}}$):**\n $$V_{\\text{open}} = \\frac{44000}{42887.5} \\approx 1.02593996 \\text{ BTC}$$\n\n2. **Closing Value in BTC ($V_{\\text{close}}$):**\n $$V_{\\text{close}} = \\frac{44000}{41686.7} \\approx 1.05549252 \\text{ BTC}$$\n\n3. **Gross PnL in BTC:**\n For a short position in an inverse contract, Gross PnL = $V_{\\text{close}} - V_{\\text{open}}$:\n $$\\text{Gross PnL} = 1.05549252 - 1.02593996 = 0.02955256 \\text{ BTC}$$\n\n4. **Total Fees in BTC:**\n $$\\text{Total Fees} = 0.00055 \\times (V_{\\text{open}} + V_{\\text{close}})$$\n $$\\text{Total Fees} = 0.00055 \\times (1.02593996 + 1.05549252) = 0.00055 \\times 2.08143248 \\approx 0.00114479 \\text{ BTC}$$\n\n5. **Net PnL in BTC:**\n $$\\text{Net PnL} = 0.02955256 - 0.00114479 = 0.02840777 \\text{ BTC}$$\n\n{\"answer\": 0.028408, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -0.06525146382766053, "answer": -0.0652515, "confidence": 100, "grade": "exact", "relError": 5.543529194454429e-7, "cost": 0.005248, "reply": "To find the net PnL in BTC for an inverse (coin-margined) contract, we first calculate the BTC value of the position at the entry and exit prices:\n\n1. **Position Values in BTC:**\n * $\\text{Entry Value} = \\frac{45000}{49446} \\approx 0.91008373 \\text{ BTC}$\n * $\\text{Exit Value} = \\frac{45000}{46182.6} \\approx 0.97439295 \\text{ BTC}$\n\n2. **Gross PnL in BTC (Long Position):**\n $$\\text{Gross PnL} = 45000 \\times \\left(\\frac{1}{49446} - \\frac{1}{46182.6}\\right) = 0.91008373 - 0.97439295 = -0.06430922 \\text{ BTC}$$\n\n3. **Trading Fees (0.05% each on entry and exit):**\n $$\\text{Total Fees} = 0.0005 \\times (0.91008373 + 0.97439295) = 0.0005 \\times 1.88447668 \\approx 0.00094224 \\text{ BTC}$$\n\n4. **Net PnL in BTC:**\n $$\\text{Net PnL} = -0.06430922 - 0.00094224 = -0.06525146 \\text{ BTC}$$\n\n{\"answer\": -0.0652515, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 0.22922018359225318, "answer": 0.22922, "confidence": 100, "grade": "exact", "relError": 8.009427891588274e-7, "cost": 0.004986, "reply": "To find the net PnL in BTC for an inverse (coin-margined) BTCUSD perpetual contract:\n\n1. **Position Values in BTC:**\n * **Open Value:** $\\frac{189000}{91422.1} \\approx 2.0673338$ BTC\n * **Close Value:** $\\frac{189000}{102932.1} \\approx 1.8361619$ BTC\n\n2. **Gross PnL in BTC (Long Position):**\n $$\\text{Gross PnL} = \\text{Open Value} - \\text{Close Value} = 2.0673338 - 1.8361619 = 0.2311719 \\text{ BTC}$$\n\n3. **Total Fees (0.05% at open and close):**\n $$\\text{Total Fees} = 0.0005 \\times (\\text{Open Value} + \\text{Close Value})$$\n $$\\text{Total Fees} = 0.0005 \\times (2.0673338 + 1.8361619) = 0.0005 \\times 3.9034957 \\approx 0.0019517 \\text{ BTC}$$\n\n4. **Net PnL in BTC:**\n $$\\text{Net PnL} = 0.2311719 - 0.0019517 = 0.2292202 \\text{ BTC}$$\n\n{\"answer\": 0.22922, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -0.026151408867684533, "answer": -0.0261514, "confidence": 100, "grade": "exact", "relError": 3.390901261185902e-7, "cost": 0.005664, "reply": "To find the net PnL in BTC for an inverse (coin-margined) perpetual contract, we calculate the BTC values at the entry and exit prices, and then subtract the trading fees.\n\n**1. Position Value in BTC:**\n* **Entry Value:** $\\frac{14000}{89747.2} \\approx 0.15599372\\text{ BTC}$\n* **Exit Value:** $\\frac{14000}{76940.3} \\approx 0.18195926\\text{ BTC}$\n\n**2. Gross PnL in BTC (Long Position):**\n$$\\text{Gross PnL} = \\text{Entry Value} - \\text{Exit Value}$$\n$$\\text{Gross PnL} = 0.15599372 - 0.18195926 = -0.02596554\\text{ BTC}$$\n\n**3. Trading Fees (0.055% = 0.00055):**\n$$\\text{Total Fees} = 0.00055 \\times (\\text{Entry Value} + \\text{Exit Value})$$\n$$\\text{Total Fees} = 0.00055 \\times (0.15599372 + 0.18195926) = 0.00055 \\times 0.33795298 \\approx 0.00018587\\text{ BTC}$$\n\n**4. Net PnL in BTC:**\n$$\\text{Net PnL} = \\text{Gross PnL} - \\text{Total Fees}$$\n$$\\text{Net PnL} = -0.02596554 - 0.00018587 = -0.02615141\\text{ BTC}$$\n\n{\"answer\": -0.0261514, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -0.011586282591352507, "answer": -0.011586, "confidence": 100, "grade": "exact", "relError": 0.00002439016572207775, "cost": 0.007896, "reply": "To find the net PnL in BTC for the inverse perpetual short position, we calculate the BTC value at the entry and exit prices, and then subtract the trading fees.\n\n1. **Position Values in BTC:**\n * $\\text{Entry Value} = \\frac{44000}{102777.6} = \\frac{44000}{1427.4666... \\times 72} \\approx 0.42810885 \\text{ BTC}$ (Note: $102777.6 = 1.8 \\times 57098.66...$, more simply: $102777.6 / 105593.7 = 73 / 75$)\n * Using the exact ratio: $102777.6 = 73 \\times 1407.9123...$ and $105593.7 = 75 \\times 1407.916...$ (specifically, $105593.7 - 102777.6 = 2816.1$)\n * $\\text{Exit Value} = \\frac{44000}{105593.7} \\approx 0.41669153 \\text{ BTC}$\n\n2. **Gross PnL (Short Position):**\n $$\\text{Gross PnL} = \\text{Exit Value} - \\text{Entry Value} = 44000 \\times \\left(\\frac{1}{105593.7} - \\frac{1}{102777.6}\\right)$$\n $$\\text{Gross PnL} = 44000 \\times \\left(\\frac{-2816.1}{10852667067.12}\\right) \\approx -0.01141732 \\text{ BTC}$$\n\n3. **Fees (0.02% on each fill):**\n $$\\text{Total Fees} = 0.0002 \\times (\\text{Entry Value} + \\text{Exit Value})$$\n $$\\text{Total Fees} = 0.0002 \\times (0.42810885 + 0.41669153) = 0.0002 \\times 0.84480038 \\approx 0.00016896 \\text{ BTC}$$\n\n4. **Net PnL:**\n $$\\text{Net PnL} = \\text{Gross PnL} - \\text{Total Fees} = -0.01141732 - 0.00016896 = -0.01158628 \\text{ BTC}$$\n\n{\"answer\": -0.011586, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 6057.334255752001, "answer": 6057.3343, "confidence": 100, "grade": "exact", "relError": 7.3048634385547545e-9, "cost": 0.004404, "reply": "To find the net PnL in USDT after fees, we calculate the gross PnL and subtract the total fees.\n\n1. **Gross PnL:**\n - Price difference = $26380.89 - 22814.19 = 3566.70$\n - Gross PnL = $1.703 \\times 3566.70 = 6074.0901\\text{ USDT}$\n\n2. **Total Fees:**\n - Combined price for fee basis = $26380.89 + 22814.19 = 49195.08$\n - Total notional traded = $1.703 \\times 49195.08 = 83779.22124\\text{ USDT}$\n - Total fee (0.02% on each fill) = $83779.22124 \\times 0.0002 = 16.755844248\\text{ USDT}$\n\n3. **Net PnL:**\n - Net PnL = $6074.0901 - 16.755844248 = 6057.334255752\\text{ USDT}$\n\n{\"answer\": 6057.3343, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 2844.8890875599996, "answer": 2844.8891, "confidence": 100, "grade": "exact", "relError": 4.372754063820816e-9, "cost": 0.004616, "reply": "To find the net PnL in USDT after fees, we calculate the gross PnL and subtract the opening and closing fees.\n\n1. **Price Difference**: \n $$\\Delta P = 39204.01 - 36718.19 = 2485.82\\text{ USDT}$$\n\n2. **Gross PnL**:\n $$\\text{Gross PnL} = 1.164 \\times 2485.82 = 2893.49448\\text{ USDT}$$\n\n3. **Total Fees**:\n $$\\text{Sum of Prices} = 36718.19 + 39204.01 = 75922.20\\text{ USDT}$$\n $$\\text{Total Notional Traded} = 1.164 \\times 75922.20 = 88373.4408\\text{ USDT}$$\n $$\\text{Total Fees} = 88373.4408 \\times 0.00055 = 48.60539244\\text{ USDT}$$\n\n4. **Net PnL**:\n $$\\text{Net PnL} = 2893.49448 - 48.60539244 = 2844.88908756\\text{ USDT}$$\n\n{\"answer\": 2844.8891, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -3783.556934676003, "answer": -3783.56, "confidence": 100, "grade": "exact", "relError": 8.101699141391267e-7, "cost": 0.005096, "reply": "To find the net PnL in USDT after fees, we calculate the gross PnL and subtract the opening and closing fees.\n\n1. **Position Size:** 0.612 BTC\n2. **Entry Price:** 57972.37 USDT\n3. **Exit Price:** 51850.49 USDT\n4. **Fee Rate:** 0.055% = 0.00055\n\n**Gross PnL:**\n$$\\text{Price Difference} = 51850.49 - 57972.37 = -6121.88\\text{ USDT}$$\n$$\\text{Gross PnL} = 0.612 \\times (-6121.88) = -3746.59056\\text{ USDT}$$\n\n**Total Fees:**\n$$\\text{Combined Price} = 57972.37 + 51850.49 = 109822.86\\text{ USDT}$$\n$$\\text{Total Notional Traded} = 0.612 \\times 109822.86 = 67211.59032\\text{ USDT}$$\n$$\\text{Total Fees} = 67211.59032 \\times 0.00055 \\approx 36.96637\\text{ USDT}$$\n\n**Net PnL:**\n$$\\text{Net PnL} = -3746.59056 - 36.96637 = -3783.55693\\text{ USDT}$$\n\n{\"answer\": -3783.56, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -1108.5578196799993, "answer": -1108.5578, "confidence": 100, "grade": "exact", "relError": 1.775279456924296e-8, "cost": 0.00438, "reply": "To find the net PnL in USDT after fees, we calculate the raw PnL and subtract the opening and closing fees.\n\n1. **Raw PnL:**\n $$\\text{Price Difference} = 4440.76 - 4857.54 = -416.78$$\n $$\\text{Raw PnL} = 2.648 \\times (-416.78) = -1103.63344\\text{ USDT}$$\n\n2. **Total Fees:**\n $$\\text{Sum of Prices} = 4857.54 + 4440.76 = 9298.30$$\n $$\\text{Total Notional Traded} = 2.648 \\times 9298.30 = 24621.8984\\text{ USDT}$$\n $$\\text{Total Fee} = 24621.8984 \\times 0.0002 = 4.92437968\\text{ USDT}$$\n\n3. **Net PnL:**\n $$\\text{Net PnL} = -1103.63344 - 4.92437968 = -1108.55781968\\text{ USDT}$$\n\n{\"answer\": -1108.5578, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 3351.620043262996, "answer": 3351.62, "confidence": 100, "grade": "exact", "relError": 1.2908084972596765e-8, "cost": 0.004396, "reply": "To find the net PnL in USDT after fees:\n\n1. **Gross PnL (Short Position):**\n $$\\text{Price Difference} = 28983.12 - 27232.54 = 1750.58$$\n $$\\text{Gross PnL} = 1.949 \\times 1750.58 = 3411.88042 \\text{ USDT}$$\n\n2. **Total Fees:**\n $$\\text{Sum of Prices} = 28983.12 + 27232.54 = 56215.66$$\n $$\\text{Total Notional Traded} = 1.949 \\times 56215.66 = 109564.32134 \\text{ USDT}$$\n $$\\text{Total Fees} = 109564.32134 \\times 0.00055 = 60.260376737 \\text{ USDT}$$\n\n3. **Net PnL:**\n $$\\text{Net PnL} = 3411.88042 - 60.260376737 = 3351.620043 \\text{ USDT}$$\n\n{\"answer\": 3351.62, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 0.11472881355932601, "answer": 0.114729, "confidence": 100, "grade": "exact", "relError": 0.0000016250553648953176, "cost": 0.004268, "reply": "To find the number of standard lots to trade, follow these steps:\n\n1. **Calculate the risk amount in USD:**\n $$\\text{Risk} = \\$5,000 \\times 1\\% = \\$50$$\n\n2. **Calculate the stop loss in pips:**\n $$\\text{Stop Loss} = 1.3538 - 1.3479 = 0.0059 = 59\\text{ pips}$$\n\n3. **Calculate the pip value per standard lot (100,000 units) in USD at the entry price:**\n $$\\text{Pip Value (CAD)} = 100,000 \\times 0.0001 = 10\\text{ CAD}$$\n $$\\text{Pip Value (USD)} = \\frac{10}{1.3538} \\approx \\$7.386615$$\n\n4. **Calculate the number of standard lots:**\n $$\\text{Lots} = \\frac{\\text{Risk}}{\\text{Stop Loss (pips)} \\times \\text{Pip Value (USD)}} = \\frac{50}{59 \\times \\frac{10}{1.3538}} = \\frac{50 \\times 1.3538}{590} \\approx 0.114729$$\n\n{\"answer\": 0.114729, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 0.5939090909090843, "answer": 0.5939, "confidence": 100, "grade": "exact", "relError": 0.000015306903402306825, "cost": 0.00453, "reply": "To find the number of standard lots to trade, follow these steps:\n\n1. **Calculate the total risk in USD:**\n $$\\text{Risk} = 25,000 \\times 1\\% = 250 \\text{ USD}$$\n\n2. **Calculate the stop loss in pips:**\n $$\\text{Stop Loss} = 1.3066 - 1.3011 = 0.0055 = 55 \\text{ pips}$$\n\n3. **Calculate the pip value per standard lot (100,000 units) in USD at the entry price:**\n $$\\text{Pip Value in CAD} = 100,000 \\times 0.0001 = 10 \\text{ CAD}$$\n $$\\text{Pip Value in USD} = \\frac{10}{1.3066} \\approx 7.65345 \\text{ USD/pip}$$\n\n4. **Calculate the number of standard lots:**\n $$\\text{Lots} = \\frac{\\text{Risk}}{\\text{Stop Loss (pips)} \\times \\text{Pip Value (USD)}} = \\frac{250}{55 \\times \\frac{10}{1.3066}} = \\frac{250 \\times 1.3066}{550} = \\frac{326.65}{550} \\approx 0.593909$$\n\n{\"answer\": 0.5939, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 1.1982352941177008, "answer": 1.1982, "confidence": 100, "grade": "exact", "relError": 0.000029455081046378537, "cost": 0.0046, "reply": "To find the number of standard lots to trade, we follow these steps:\n\n1. **Calculate the dollar risk:**\n $$\\text{Risk} = 25,000 \\text{ USD} \\times 1\\% = 250 \\text{ USD}$$\n\n2. **Calculate the stop loss in pips:**\n $$\\text{Stop Loss} = 0.8148 - 0.8131 = 0.0017 = 17 \\text{ pips}$$\n\n3. **Calculate the pip value in USD per standard lot (100,000 units):**\n $$\\text{Pip Value} = \\frac{10 \\text{ CHF}}{0.8148} \\approx 12.27295 \\text{ USD per pip}$$\n\n4. **Calculate the risk per standard lot:**\n $$\\text{Risk per lot} = 17 \\text{ pips} \\times 12.27295 \\text{ USD/pip} \\approx 208.64016 \\text{ USD}$$\n\n5. **Calculate the position size in standard lots:**\n $$\\text{Standard Lots} = \\frac{250 \\text{ USD}}{208.64016 \\text{ USD/lot}} = \\frac{250 \\times 0.8148}{170} = \\frac{203.7}{170} \\approx 1.198235$$\n\n{\"answer\": 1.1982, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 1.0178846153846508, "answer": 1.0179, "confidence": 100, "grade": "exact", "relError": 0.000015114301873409756, "cost": 0.005308, "reply": "To find the number of standard lots to trade, we follow these steps:\n\n1. **Calculate the risk amount in USD:**\n $$\\text{Risk Amount} = \\$25,000 \\times 1\\% = \\$250$$\n\n2. **Calculate the stop loss in pips:**\n For USD/JPY, 1 pip is $0.01$.\n $$\\text{Stop Loss Distance} = 158.79 - 158.40 = 0.39$$\n $$\\text{Pips} = \\frac{0.39}{0.01} = 39\\text{ pips}$$\n\n3. **Calculate the pip value per standard lot in USD:**\n 1 standard lot is $100,000$ units of the base currency (USD). Thus, 1 pip ($0.01$ JPY) is worth $1,000$ JPY per standard lot.\n Converting this to USD at the entry price of $158.79$:\n $$\\text{Pip Value} = \\frac{1,000}{158.79} \\approx \\$6.297626\\text{ per pip}$$\n\n4. **Calculate the number of standard lots:**\n $$\\text{Risk per Standard Lot} = 39\\text{ pips} \\times \\$6.297626/\\text{pip} \\approx \\$245.6074$$\n $$\\text{Standard Lots} = \\frac{\\$250}{\\$245.6074} = \\frac{250 \\times 158.79}{39,000} = \\frac{39,697.5}{39,000} \\approx 1.01788$$\n\n{\"answer\": 1.0179, \"confidence\": 100}" }, { "taskId": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-4-argon", "expected": 7.182727272727309, "answer": 7.1827, "confidence": 100, "grade": "exact", "relError": 0.000003796987728178971, "cost": 0.00456, "reply": "To find the number of standard lots to trade, we follow these steps:\n\n1. **Calculate the dollar risk:**\n $$\\text{Risk} = \\$100,000 \\times 1\\% = \\$1,000$$\n\n2. **Calculate the stop loss in pips:**\n $$\\text{Stop Loss} = 158.02 - 157.80 = 0.22\\text{ JPY} = 22\\text{ pips}$$\n\n3. **Calculate the pip value per standard lot in USD (at the entry price):**\n One standard lot is 100,000 units of the base currency (USD). For USD/JPY, 1 pip (0.01) is worth $100,000 \\times 0.01 = 1,000\\text{ JPY}$.\n $$\\text{Pip Value in USD} = \\frac{1,000\\text{ JPY}}{158.02} \\approx \\$6.328313$$\n\n4. **Calculate the number of standard lots:**\n $$\\text{Lots} = \\frac{\\text{Risk}}{\\text{Stop Loss in Pips} \\times \\text{Pip Value}} = \\frac{\\$1,000}{22 \\times \\frac{1,000}{158.02}} = \\frac{158.02}{22} \\approx 7.182727$$\n\n{\"answer\": 7.1827, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -0.23573231643213824, "answer": -0.235732, "confidence": 100, "grade": "exact", "relError": 0.0000013423366937182556, "cost": 0.00558, "reply": "To calculate the realized PnL in BTC for an inverse perpetual contract, we use the following formulas:\n\n**1. Position Values in BTC:**\n* **Entry Value:** $\\frac{103,000}{49,553.7} \\approx 2.07855317\\text{ BTC}$\n* **Exit Value:** $\\frac{103,000}{55,827.2} \\approx 1.84497880\\text{ BTC}$\n\n**2. Gross PnL (Short Position):**\n$$\\text{Gross PnL} = \\text{ Contracts} \\times \\left(\\frac{1}{\\text{Exit Price}} - \\frac{1}{\\text{Entry Price}}\\right)$$\n$$\\text{Gross PnL} = 1.84497880 - 2.07855317 = -0.23357437\\text{ BTC}$$\n\n**3. Taker Fees (0.055% on both fills):**\n* **Open Fee:** $2.07855317 \\times 0.00055 = 0.00114320\\text{ BTC}$\n* **Close Fee:** $1.84497880 \\times 0.00055 = 0.00101474\\text{ BTC}$\n* **Total Fees:** $0.00114320 + 0.00101474 = 0.00215794\\text{ BTC}$\n\n**4. Realized PnL After Fees:**\n$$\\text{Realized PnL} = -0.23357437 - 0.00215794 = -0.23573231\\text{ BTC}$$\n\n{\"answer\": -0.235732, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -0.0669895883004742, "answer": -0.06698959, "confidence": 100, "grade": "exact", "relError": 2.5369999144468978e-8, "cost": 0.005608, "reply": "To find the realized PnL in BTC after fees for an inverse perpetual contract (where 1 contract = 1 USD):\n\n**1. Raw PnL (before fees):**\n$$\\text{Raw PnL} = \\text{Contracts} \\times \\left(\\frac{1}{\\text{Entry Price}} - \\frac{1}{\\text{Exit Price}}\\right)$$\n$$\\text{Entry Value in BTC} = \\frac{78,000}{95,195.8} \\approx 0.81936388\\text{ BTC}$$\n$$\\text{Exit Value in BTC} = \\frac{78,000}{88,094.2} \\approx 0.88541584\\text{ BTC}$$\n$$\\text{Raw PnL} = 0.81936388 - 0.88541584 = -0.06605196\\text{ BTC}$$\n\n**2. Trading Fees:**\n$$\\text{Fee Rate} = 0.055\\% = 0.00055$$\n$$\\text{Total Fees} = (\\text{Entry Value in BTC} + \\text{Exit Value in BTC}) \\times 0.00055$$\n$$\\text{Total Fees} = (0.81936388 + 0.88541584) \\times 0.00055 = 1.70477972 \\times 0.00055 \\approx 0.00093763\\text{ BTC}$$\n\n**3. Realized PnL (after fees):**\n$$\\text{Realized PnL} = -0.06605196 - 0.00093763 = -0.06698959\\text{ BTC}$$\n\n{\"answer\": -0.06698959, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 0.0024428835303150374, "answer": 0.00244288, "confidence": 100, "grade": "exact", "relError": 0.000001445142592127571, "cost": 0.005782, "reply": "To calculate the realized PnL in BTC for an inverse perpetual contract on Bybit:\n\n**1. Calculate Position Values in BTC:**\n* **Entry Value:** $\\frac{42,000}{117,914.1} \\approx 0.35619150 \\text{ BTC}$\n* **Exit Value:** $\\frac{42,000}{116,982.6} \\approx 0.35902775 \\text{ BTC}$\n\n**2. Gross PnL (Short Position):**\n$$\\text{Gross PnL} = 42,000 \\times \\left(\\frac{1}{116,982.6} - \\frac{1}{117,914.1}\\right)$$\n$$\\text{Gross PnL} = 42,000 \\times \\frac{931.5}{13,793,897,834.66} \\approx 0.0028362506 \\text{ BTC}$$\n\n**3. Calculate Taker Fees (0.055% on both fills):**\n* $$\\text{Total Fees} = 0.00055 \\times (\\text{Entry Value} + \\text{Exit Value})$$\n* $$\\text{Total Fees} = 0.00055 \\times (0.35619150 + 0.35902775) = 0.00055 \\times 0.71521925 \\approx 0.0003933706 \\text{ BTC}$$\n\n**4. Realized PnL After Fees:**\n$$\\text{Net PnL} = 0.0028362506 - 0.0003933706 = 0.00244288 \\text{ BTC}$$\n\n{\"answer\": 0.00244288, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 0.001886499688812526, "answer": 0.00188659, "confidence": 100, "grade": "exact", "relError": 0.00004787235747225214, "cost": 0.005876, "reply": "For a Bybit BTCUSD inverse perpetual contract (1 contract = 1 USD):\n\n* **Quantity ($Q$):** $192,000\\text{ USD}$\n* **Entry Price ($P_{\\text{entry}}$):** $20,434$\n* **Exit Price ($P_{\\text{exit}}$):** $20,460.6$\n* **Taker Fee Rate ($f$):** $0.055\\% = 0.00055$\n\n**1. Values in BTC:**\n* $\\text{Entry Value} = \\frac{192,000}{20,434} \\approx 9.39610453\\text{ BTC}$\n* $\\text{Exit Value} = \\frac{192,000}{20,460.6} \\approx 9.38388894\\text{ BTC}$\n\n**2. Gross PnL (Long):**\n$$\\text{Gross PnL} = 192,000 \\times \\left(\\frac{1}{20,434} - \\frac{1}{20,460.6}\\right) = \\frac{192,000 \\times 26.6}{418,091,900.4} \\approx 0.01221559\\text{ BTC}$$\n\n**3. Total Fees:**\n$$\\text{Fees} = 0.00055 \\times (\\text{Entry Value} + \\text{Exit Value}) = 0.00055 \\times 18.77999347 \\approx 0.01032900\\text{ BTC}$$\n\n**4. Realized PnL after fees:**\n$$\\text{Net PnL} = 0.01221559 - 0.01032900 = 0.00188659\\text{ BTC}$$\n\n{\"answer\": 0.00188659, \"confidence\": 100}" }, { "taskId": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": -0.0962309957129481, "answer": -0.096231, "confidence": 100, "grade": "exact", "relError": 4.4549595089137153e-8, "cost": 0.00539, "reply": "To find the realized PnL in BTC after fees for the Bybit BTCUSD inverse perpetual contract:\n\n**1. Calculate the Open and Close Values in BTC:**\n* $\\text{Open Value} = \\frac{139,000}{59,513.2} \\approx 2.33561630\\text{ BTC}$\n* $\\text{Close Value} = \\frac{139,000}{62,000.9} \\approx 2.24190294\\text{ BTC}$\n\n**2. Calculate the Gross PnL (Short Position):**\n$$\\text{Gross PnL} = 139,000 \\times \\left(\\frac{1}{62,000.9} - \\frac{1}{59,513.2}\\right)$$\n$$\\text{Gross PnL} = 2.24190294 - 2.33561630 = -0.09371336\\text{ BTC}$$\n\n**3. Calculate the Taker Fees (0.055% on both fills):**\n$$\\text{Total Fees} = 0.00055 \\times (2.33561630 + 2.24190294)$$\n$$\\text{Total Fees} = 0.00055 \\times 4.57751924 \\approx 0.00251764\\text{ BTC}$$\n\n**4. Calculate the Realized PnL after Fees:**\n$$\\text{Net PnL} = -0.09371336 - 0.00251764 = -0.09623100\\text{ BTC}$$\n\n{\"answer\": -0.096231, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 70867.19713259465, "answer": 70867.2, "confidence": 100, "grade": "exact", "relError": 4.046167285427324e-8, "cost": 0.00424, "reply": "For an inverse perpetual contract, the average entry price is the harmonic weighted mean of the entry prices, because contract values are denominated in the base currency (BTC).\n\n1. **Calculate the BTC value of each fill:**\n $$\\text{BTC}_1 = \\frac{76,000}{62,771.7} \\approx 1.2107366855\\text{ BTC}$$\n $$\\text{BTC}_2 = \\frac{55,000}{86,235.1} \\approx 0.6377913402\\text{ BTC}$$\n\n2. **Sum the total contracts and total BTC value:**\n $$\\text{Total Contracts} = 76,000 + 55,000 = 131,000$$\n $$\\text{Total BTC} \\approx 1.2107366855 + 0.6377913402 = 1.8485280257\\text{ BTC}$$\n\n3. **Calculate the average entry price:**\n $$\\text{Average Entry Price} = \\frac{131,000}{1.8485280257} \\approx 70,867.2$$\n\n{\"answer\": 70867.2, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 34209.14079495059, "answer": 34209.14, "confidence": 100, "grade": "exact", "relError": 2.3237958347514953e-8, "cost": 0.004528, "reply": "For an inverse perpetual contract, the average entry price is calculated as the harmonic mean weighted by the number of contracts:\n\n$$\\text{Average Entry Price} = \\frac{\\text{Total Contracts}}{\\text{Total BTC Value}}$$\n\n1. **Total Contracts:**\n$$76,000 + 73,000 + 42,000 = 191,000\\text{ USD}$$\n\n2. **Total Value in BTC:**\n* $\\frac{76,000}{33,536.1} \\approx 2.2662146\\text{ BTC}$\n* $\\frac{73,000}{24,929.9} \\approx 2.9282107\\text{ BTC}$\n* $\\frac{42,000}{108,003.1} \\approx 0.3888777\\text{ BTC}$\n\n$$\\text{Total BTC} = 2.2662146 + 2.9282107 + 0.3888777 = 5.5833030\\text{ BTC}$$\n\n3. **Average Entry Price:**\n$$\\text{Average Entry Price} = \\frac{191,000}{5.5833030} \\approx 34,209.14$$\n\n{\"answer\": 34209.14, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 88678.94192491287, "answer": 88678.94, "confidence": 100, "grade": "exact", "relError": 2.1706538502484576e-8, "cost": 0.005064, "reply": "For an inverse perpetual contract, the average entry price is calculated using the harmonic mean weighted by the contract quantities:\n\n$$\\text{Average Entry Price} = \\frac{\\text{Total Contracts}}{\\text{Total Value in BTC}}$$\n\n1. **Calculate the total number of contracts:**\n$$\\text{Total Contracts} = 59,000 + 34,000 + 62,000 = 155,000$$\n\n2. **Calculate the BTC value of each piece:**\n* $\\text{Piece 1} = \\frac{59,000}{82,124.2} \\approx 0.71842405\\text{ BTC}$\n* $\\text{Piece 2} = \\frac{34,000}{81,534.0} \\approx 0.41700395\\text{ BTC}$\n* $\\text{Piece 3} = \\frac{62,000}{101,232.7} \\approx 0.61245032\\text{ BTC}$\n\n3. **Sum the total BTC value:**\n$$\\text{Total BTC} = 0.71842405 + 0.41700395 + 0.61245032 = 1.74787832\\text{ BTC}$$\n\n4. **Calculate the average entry price:**\n$$\\text{Average Entry Price} = \\frac{155,000}{1.74787832} \\approx 88,678.94$$\n\n{\"answer\": 88678.94, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 46924.5904555392, "answer": 46924.59, "confidence": 100, "grade": "exact", "relError": 9.707899338219995e-9, "cost": 0.00423, "reply": "For an inverse perpetual contract, the average entry price is calculated as the harmonic mean weighted by the contract quantities (since value in BTC is $\\text{Contracts} / \\text{Price}$):\n\n$$\\text{Average Entry Price} = \\frac{\\text{Total Contracts}}{\\text{Total Cost in BTC}}$$\n\n1. **Total Contracts:**\n$$42,000 + 67,000 = 109,000\\text{ USD}$$\n\n2. **Total Cost in BTC:**\n$$\\text{Cost}_1 = \\frac{42,000}{30128.4} \\approx 1.39403354\\text{ BTC}$$\n$$\\text{Cost}_2 = \\frac{67,000}{72132.8} \\approx 0.92884236\\text{ BTC}$$\n$$\\text{Total Cost} = 1.39403354 + 0.92884236 = 2.32287590\\text{ BTC}$$\n\n3. **Average Entry Price:**\n$$\\text{Average Entry Price} = \\frac{109,000}{2.32287590} \\approx 46924.59$$\n\n{\"answer\": 46924.59, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-4-argon", "expected": 37529.651947786224, "answer": 37529.65, "confidence": 100, "grade": "exact", "relError": 5.189992769330071e-8, "cost": 0.00411, "reply": "For an inverse perpetual contract, the value is denominated in the quote currency (USD) and settled in the base currency (BTC). Therefore, the average entry price is calculated using the harmonic weighted mean (Total Contracts in USD / Total Contract Value in BTC).\n\n1. **Calculate the BTC value of each trade:**\n * First trade: $36,000 / 23930.3 \\approx 1.50436894 \\text{ BTC}$\n * Second trade: $61,000 / 56468.2 \\approx 1.08025402 \\text{ BTC}$\n\n2. **Calculate total USD contracts and total BTC value:**\n * Total Contracts = $36,000 + 61,000 = 97,000 \\text{ USD}$\n * Total BTC Value = $1.50436894 + 1.08025402 = 2.58462296 \\text{ BTC}$\n\n3. **Calculate the average entry price:**\n * Average Entry Price = $97,000 / 2.58462296 \\approx 37529.65$\n\n{\"answer\": 37529.65, \"confidence\": 100}" }, { "taskId": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 4.917273424034056, "answer": 4.8525, "confidence": 95, "grade": "wrong", "relError": 0.013172630124138286, "cost": 0.006946, "reply": "To find the 1-day 95% parametric Value at Risk (VaR), we first calculate the sample mean ($\\mu$) and sample standard deviation ($s$) of the 40 returns.\n\n1. **Sample Mean ($\\mu$):**\n Sum of positive returns (18 values) = $47.26$\n Sum of negative returns (22 values) = $-53.39$\n Total sum = $47.26 - 53.39 = -6.13$\n $$\\mu = \\frac{-6.13}{40} = -0.15325\\%$$\n\n2. **Sample Standard Deviation ($s$):**\n Sum of squared returns ($\\sum x_i^2$):\n $$\\sum x_{pos}^2 = 160.2994$$\n $$\\sum x_{neg}^2 = 158.9679$$\n $$\\sum x_i^2 = 160.2994 + 158.9679 = 319.2673$$\n\n Using the sample variance formula:\n $$s^2 = \\frac{\\sum x_i^2 - n\\mu^2}{n - 1} = \\frac{319.2673 - 40(-0.15325)^2}{39} = \\frac{319.2673 - 0.9394}{39} = \\frac{318.3279}{39} \\approx 8.16225$$\n $$s = \\sqrt{8.16225} \\approx 2.85697\\%$$\n\n3. **95% Parametric VaR:**\n Using the standard normal $z$-score for a one-tailed 95% confidence level ($z_{0.05} \\approx 1.64485$):\n $$\\text{VaR}_{95\\%} = -\\mu + z_{0.05} \\cdot s = -(-0.15325) + 1.64485 \\times 2.85697 = 0.15325 + 4.69929 = 4.85254\\%$$\n\n{\"answer\": 4.8525, \"confidence\": 95}" }, { "taskId": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 4.2934299989968965, "answer": 4.2546, "confidence": 98, "grade": "correct", "relError": 0.00904405079527759, "cost": 0.007056, "reply": "To find the 1-day 95% parametric (normal) Value at Risk (VaR), we calculate the sample mean ($\\mu$) and sample standard deviation ($s$) of the $N=40$ returns, and then apply the formula $\\text{VaR}_{95\\%} = -\\mu + z_{0.95} \\cdot s$, where $z_{0.95} \\approx 1.64485$.\n\n1. **Sum of returns ($\\sum x_i$):**\n - Sum of positive returns (22 values): $57.51$\n - Sum of negative returns (18 values): $-39.97$\n - Total sum: $57.51 - 39.97 = 17.54$\n - **Sample Mean ($\\mu$):** $\\frac{17.54}{40} = 0.4385\\%$\n\n2. **Sum of squared returns ($\\sum x_i^2$):**\n - Sum of squares of positive returns: $194.2895$\n - Sum of squares of negative returns: $130.8925$\n - Total sum of squares: $194.2895 + 130.8925 = 325.1820$\n\n3. **Sample Variance ($s^2$) and Standard Deviation ($s$):**\n - $s^2 = \\frac{1}{N-1} \\left(\\sum x_i^2 - N\\mu^2\\right) = \\frac{325.1820 - 40(0.4385)^2}{39} = \\frac{325.1820 - 7.6913}{39} = \\frac{317.4907}{39} \\approx 8.14079$\n - $s = \\sqrt{8.14079} \\approx 2.85321\\%$\n\n4. **95% Parametric VaR:**\n - $\\text{VaR}_{95\\%} = -0.4385 + 1.644854 \\times 2.85321 \\approx -0.4385 + 4.6931 = 4.2546\\%$\n\n{\"answer\": 4.2546, \"confidence\": 98}" }, { "taskId": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 5.2422646771168795, "answer": 5.0962, "confidence": 99, "grade": "wrong", "relError": 0.027862896307862876, "cost": 0.006436, "reply": "To find the 1-day 95% parametric Value at Risk (VaR), we first calculate the sample mean ($\\mu$) and sample standard deviation ($s$) of the 40 daily returns.\n\n**1. Sample Mean ($\\mu$):**\nSum of the 40 returns:\n$$\\sum x_i = 0.81 - 3.34 - 1.80 - \\dots - 2.41 = 9.97\\%$$\n$$\\mu = \\frac{9.97}{40} = 0.24925\\%$$\n\n**2. Sample Standard Deviation ($s$):**\nSum of squared returns:\n$$\\sum x_i^2 = (0.81)^2 + (-3.34)^2 + \\dots + (-2.41)^2 = 414.3775$$\nUsing the sample variance formula:\n$$s^2 = \\frac{\\sum x_i^2 - n\\mu^2}{n - 1} = \\frac{414.3775 - 40(0.24925)^2}{39} = \\frac{414.3775 - 2.4850}{39} = \\frac{411.8925}{39} \\approx 10.561346$$\n$$s = \\sqrt{10.561346} \\approx 3.24982\\%$$\n\n**3. 95% Parametric VaR:**\nUsing the standard normal quantile for 95% confidence ($z_{0.95} \\approx 1.644854$):\n$$\\text{VaR}_{95\\%} = -\\mu + z_{0.95} \\times s$$\n$$\\text{VaR}_{95\\%} = -0.24925 + (1.644854 \\times 3.24982) = -0.24925 + 5.34548 = 5.09623\\%$$\n\n{\"answer\": 5.0962, \"confidence\": 99}" }, { "taskId": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 7.037483251314367, "answer": 6.8048, "confidence": 95, "grade": "wrong", "relError": 0.033063418129046195, "cost": 0.006376, "reply": "To find the 1-day 99% parametric (normal) VaR, we first compute the sample mean ($\\mu$) and sample standard deviation ($s$) of the 40 returns.\n\n**1. Sample Mean ($\\mu$):**\nSum of returns:\n$$(-1.04 - 4.23 + 3.83 + \\dots + 0.93 + 0.00) = -9.09\\%$$\n$$\\mu = \\frac{-9.09}{40} = -0.22725\\%$$\n\n**2. Sample Standard Deviation ($s$):**\nSum of squared returns ($\\sum x_i^2$):\n$$1.0816 + 17.8929 + 14.6689 + \\dots + 0.8649 + 0.0000 = 313.8419$$\n\nUsing the sample variance formula:\n$$s^2 = \\frac{\\sum x_i^2 - n\\mu^2}{n - 1} = \\frac{313.8419 - 40(-0.22725)^2}{39} = \\frac{313.8419 - 2.0657}{39} = \\frac{311.7762}{39} \\approx 7.99426$$\n$$s = \\sqrt{7.99426} \\approx 2.82741\\%$$\n\n**3. 99% Parametric VaR:**\nUsing the 99% normal critical value $z_{0.99} \\approx 2.32635$:\n$$\\text{VaR}_{99\\%} = -\\mu + z_{0.99} \\cdot s = -(-0.22725) + 2.32635 \\times 2.82741 \\approx 0.22725 + 6.57755 = 6.8048\\%$$\n\n{\"answer\": 6.8048, \"confidence\": 95}" }, { "taskId": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 6.7877378350475945, "answer": 6.7416, "confidence": 95, "grade": "correct", "relError": 0.006797232917477725, "cost": 0.005996, "reply": "To find the 1-day 99% parametric (normal) Value at Risk (VaR), we calculate the sample mean ($\\mu$) and sample standard deviation ($s$) of the 40 daily returns.\n\n**1. Sample Mean ($\\mu$):**\nSum of the 40 returns:\n$\\sum x_i = -2.50 + 4.47 - 1.71 + \\dots - 0.60 = 5.76\\%$\n\n$\\mu = \\frac{5.76}{40} = 0.144\\%$\n\n**2. Sample Standard Deviation ($s$):**\nSum of squared returns:\n$\\sum x_i^2 = (-2.50)^2 + (4.47)^2 + \\dots + (-0.60)^2 = 342.4956$\n\nSample variance ($s^2$):\n$s^2 = \\frac{\\sum x_i^2 - n\\mu^2}{n - 1} = \\frac{342.4956 - 40(0.144)^2}{39} = \\frac{342.4956 - 0.82944}{39} = \\frac{341.66616}{39} \\approx 8.76067$\n\n$s = \\sqrt{8.76067} \\approx 2.95984\\%$\n\n**3. 99% Parametric VaR:**\nUsing the 99% normal critical value $z_{0.99} \\approx 2.32635$:\n$\\text{VaR}_{99\\%} = -\\mu + z_{0.99} \\cdot s = -0.144 + 2.32635 \\times 2.95984 \\approx 6.7416\\%$\n\n{\"answer\": 6.7416, \"confidence\": 95}" }, { "taskId": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.40527021190035356, "answer": 0.4009, "confidence": 99, "grade": "wrong", "relError": 0.010783452057483354, "cost": 0.007174, "reply": "To find the annualized Sharpe ratio, we calculate the mean and sample standard deviation of the 30 daily returns.\n\n**1. Sum of daily returns ($\\sum x_i$):**\n$$\\text{Positive returns} = 2.82 + 3.50 + 1.20 + 0.38 + 1.19 + 2.86 + 4.21 + 2.17 + 2.42 + 3.83 + 0.96 + 2.59 = 28.13$$\n$$\\text{Negative returns} = -0.03 - 0.14 - 2.24 - 0.42 - 2.93 - 3.33 - 2.35 - 0.77 - 0.96 - 3.90 - 2.28 - 0.20 - 0.68 - 0.03 - 1.03 - 1.20 - 2.51 - 1.70 = -26.70$$\n$$\\sum x_i = 28.13 - 26.70 = 1.43$$\n\n**2. Mean daily return ($\\bar{x}$):**\n$$\\bar{x} = \\frac{1.43}{30} \\approx 0.047667\\%$$\n\n**3. Sample standard deviation ($s$):**\n$$\\sum x_i^2 = (-0.03)^2 + (-0.14)^2 + \\dots + (-1.70)^2 = 149.7069$$\n$$s = \\sqrt{\\frac{\\sum x_i^2 - \\frac{(\\sum x_i)^2}{n}}{n - 1}} = \\sqrt{\\frac{149.7069 - \\frac{1.43^2}{30}}{29}} = \\sqrt{\\frac{149.6387}{29}} = \\sqrt{5.159956} \\approx 2.271554\\%$$\n\n**4. Annualized Sharpe Ratio:**\n$$\\text{Sharpe Ratio} = \\frac{\\bar{x}}{s} \\times \\sqrt{365} = \\frac{0.047667}{2.271554} \\times 19.104973 \\approx 0.4009$$\n\n{\"answer\": 0.4009, \"confidence\": 99}" }, { "taskId": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.37258355219979333, "answer": 0.3674, "confidence": 95, "grade": "wrong", "relError": 0.01391245579465009, "cost": 0.007614, "reply": "To find the annualized Sharpe ratio, we calculate the sample mean and sample standard deviation of the 30 daily returns, then multiply the daily Sharpe ratio by $\\sqrt{365}$.\n\n1. **Sum of returns ($\\sum r_i$):**\n $$\\text{Positive sum} = 0.49 + 3.50 + 1.12 + 2.59 + 1.99 + 2.82 + 3.56 + 1.21 + 0.68 + 2.46 + 2.95 + 0.85 + 2.48 + 3.61 = 30.31$$\n $$\\text{Negative sum} = -(3.20 + 0.05 + 0.34 + 2.32 + 0.09 + 3.04 + 2.59 + 3.72 + 1.96 + 0.86 + 1.70 + 2.02 + 0.52 + 1.58 + 3.05 + 1.92) = -28.96$$\n $$\\sum r_i = 30.31 - 28.96 = 1.35$$\n\n2. **Mean daily return ($\\bar{r}$):**\n $$\\bar{r} = \\frac{1.35}{30} = 0.045$$\n\n3. **Sum of squared returns ($\\sum r_i^2$):**\n $$\\sum r_i^2 = 0.49^2 + 3.50^2 + \\dots + 3.61^2 \\approx 158.8551$$\n\n4. **Sample standard deviation ($s$):**\n $$s = \\sqrt{\\frac{\\sum r_i^2 - 30\\bar{r}^2}{29}} = \\sqrt{\\frac{158.8551 - 30(0.045)^2}{29}} = \\sqrt{\\frac{158.79435}{29}} \\approx \\sqrt{5.475667} \\approx 2.3400$$\n\n5. **Annualized Sharpe Ratio:**\n $$\\text{Sharpe} = \\frac{\\bar{r}}{s} \\times \\sqrt{365} = \\frac{0.045}{2.3400} \\times 19.10497 \\approx 0.019231 \\times 19.10497 \\approx 0.3674$$\n\n{\"answer\": 0.3674, \"confidence\": 95}" }, { "taskId": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": -0.6855095501276544, "answer": -0.6855, "confidence": 99, "grade": "exact", "relError": 0.000013931429040744886, "cost": 0.012434, "reply": "To find the annualized Sharpe ratio, we calculate the sample mean and sample standard deviation of the 30 daily returns, then scale by $\\sqrt{365}$.\n\n**1. Sum of returns ($\\sum r_i$):**\nPositive returns: $1.48 + 0.55 + 4.03 + 3.13 + 0.65 + 0.44 + 1.71 + 2.07 + 3.86 + 3.92 + 2.23 + 2.02 + 0.61 + 1.45 + 0.14 = 28.29$\nNegative returns: $-2.22 - 0.27 - 1.18 - 1.96 - 3.20 - 1.40 - 1.88 - 3.08 - 0.74 - 1.95 - 2.32 - 0.71 - 3.28 - 2.96 - 3.64 = -30.79$\nTotal sum: $\\sum r_i = 28.29 - 30.79 = -2.50$\n\n**2. Mean daily return ($\\bar{r}$):**\n$$\\bar{r} = \\frac{-2.50}{30} = -0.083333\\%$$\n\n**3. Sum of squared returns ($\\sum r_i^2$):**\n$1.48^2 + (-2.22)^2 + 0.55^2 + 4.03^2 + (-0.27)^2 + 3.13^2 + (-1.18)^2 + 0.65^2 + 0.44^2 + (-1.96)^2 + 1.71^2 + (-3.20)^2 + 2.07^2 + 3.86^2 + 3.92^2 + (-1.40)^2 + (-1.88)^2 + (-3.08)^2 + (-0.74)^2 + (-1.95)^2 + (-2.32)^2 + 2.23^2 + (-0.71)^2 + 2.02^2 + (-3.28)^2 + 0.61^2 + 1.45^2 + (-2.96)^2 + 0.14^2 + (-3.64)^2$\n$= 2.1904 + 4.9284 + 0.3025 + 16.2409 + 0.0729 + 9.7969 + 1.3924 + 0.4225 + 0.1936 + 3.8416 + 2.9241 + 10.2400 + 4.2849 + 14.8996 + 15.3664 + 1.9600 + 3.5344 + 9.4864 + 0.5476 + 3.8025 + 5.3824 + 4.9729 + 0.5041 + 4.0804 + 10.7584 + 0.3721 + 2.1025 + 8.7616 + 0.0196 + 13.2496 = 156.6316$\n\n**4. Sample standard deviation ($s$):**\n$$s = \\sqrt{\\frac{\\sum r_i^2 - \\frac{(\\sum r_i)^2}{n}}{n - 1}} = \\sqrt{\\frac{156.6316 - \\frac{(-2.50)^2}{30}}{29}} = \\sqrt{\\frac{156.6316 - 0.208333}{29}} = \\sqrt{\\frac{156.423267}{29}} = \\sqrt{5.393906} \\approx 2.322478\\%$$\n\n**5. Annualized Sharpe Ratio:**\n$$\\text{Sharpe} = \\frac{\\bar{r}}{s} \\times \\sqrt{365} = \\frac{-0.083333}{2.322478} \\times 19.104973 \\approx -0.035881 \\times 19.104973 \\approx -0.6855$$\n\n{\"answer\": -0.6855, \"confidence\": 99}" }, { "taskId": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 0.5941064255855656, "answer": 0.5861, "confidence": 99, "grade": "wrong", "relError": 0.013476416414238088, "cost": 0.007754, "reply": "To find the annualized Sharpe ratio, we calculate the sample mean and sample standard deviation of the 30 daily returns, and then scale the ratio by $\\sqrt{365}$.\n\n1. **Sum of returns ($\\sum r_i$):**\n $$\\text{Positive returns} = 0.68 + 4.40 + 0.18 + 0.01 + 3.68 + 0.26 + 3.31 + 2.79 + 0.26 + 0.43 + 0.39 + 0.29 + 2.92 + 2.00 + 0.24 + 2.00 + 4.31 = 28.15$$\n $$\\text{Negative returns} = -2.72 - 1.97 - 3.54 - 3.46 - 1.31 - 3.29 - 0.19 - 1.70 - 1.24 - 2.23 - 0.57 - 3.54 - 0.21 = -25.97$$\n $$\\sum r_i = 28.15 - 25.97 = 2.18$$\n\n2. **Sample Mean ($\\bar{r}$):**\n $$\\bar{r} = \\frac{2.18}{30} \\approx 0.072667\\%$$\n\n3. **Sum of squared returns ($\\sum r_i^2$):**\n $$\\sum r_i^2 = (-2.72)^2 + (-1.97)^2 + (-3.54)^2 + \\dots + (4.31)^2 = 162.8856$$\n\n4. **Sample Variance ($s^2$) and Standard Deviation ($s$):**\n $$s^2 = \\frac{\\sum r_i^2 - \\frac{(\\sum r_i)^2}{30}}{29} = \\frac{162.8856 - \\frac{4.7524}{30}}{29} = \\frac{162.72719}{29} \\approx 5.61128$$\n $$s = \\sqrt{5.61128} \\approx 2.36881\\%$$\n\n5. **Annualized Sharpe Ratio:**\n $$\\text{Sharpe} = \\frac{\\bar{r}}{s} \\times \\sqrt{365} = \\frac{0.072667}{2.36881} \\times 19.10497 \\approx 0.58607$$\n\n{\"answer\": 0.5861, \"confidence\": 99}" }, { "taskId": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-4-argon", "expected": 6.721949210975506, "answer": 6.7219, "confidence": 99, "grade": "exact", "relError": 0.000007320938311433693, "cost": 0.012874, "reply": "Let the 30 daily returns be $r_i$.\n\n1. **Sum of returns ($\\sum r_i$):**\nPositive returns: $2.83 + 1.74 + 3.16 + 0.21 + 3.34 + 4.01 + 4.15 + 0.88 + 3.74 + 2.93 + 1.64 + 2.92 + 1.95 + 0.47 + 1.98 + 3.66 + 1.75 + 4.29 = 45.65$\nNegative returns: $-2.40 - 2.74 - 1.88 - 0.56 - 0.04 - 0.90 - 1.87 - 0.05 - 3.52 - 0.10 - 2.33 - 3.52 = -19.91$\nTotal sum: $\\sum r_i = 45.65 - 19.91 = 25.74$\n\n2. **Mean daily return ($\\bar{r}$):**\n$$\\bar{r} = \\frac{25.74}{30} = 0.858\\%$$\n\n3. **Sum of squared returns ($\\sum r_i^2$):**\n- Positives squared:\n$2.83^2 + 1.74^2 + 3.16^2 + 0.21^2 + 3.34^2 + 4.01^2 + 4.15^2 + 0.88^2 + 3.74^2 + 2.93^2 + 1.64^2 + 2.92^2 + 1.95^2 + 0.47^2 + 1.98^2 + 3.66^2 + 1.75^2 + 4.29^2$\n$= 8.0089 + 3.0276 + 9.9856 + 0.0441 + 11.1556 + 16.0801 + 17.2225 + 0.7744 + 13.9876 + 8.5849 + 2.6896 + 8.5264 + 3.8025 + 0.2209 + 3.9204 + 13.3956 + 3.0625 + 18.4041 = 142.8933$\n\n- Negatives squared:\n$(-2.40)^2 + (-2.74)^2 + (-1.88)^2 + (-0.56)^2 + (-0.04)^2 + (-0.90)^2 + (-1.87)^2 + (-0.05)^2 + (-3.52)^2 + (-0.10)^2 + (-2.33)^2 + (-3.52)^2$\n$= 5.7600 + 7.5076 + 3.5344 + 0.3136 + 0.0016 + 0.8100 + 3.4969 + 0.0025 + 12.3904 + 0.0100 + 5.4289 + 12.3904 = 51.6463$\n\nTotal sum of squares: $\\sum r_i^2 = 142.8933 + 51.6463 = 194.5396$\n\n4. **Sample standard deviation ($s$):**\n$$s^2 = \\frac{\\sum r_i^2 - 30\\bar{r}^2}{29} = \\frac{194.5396 - 30(0.858)^2}{29} = \\frac{194.5396 - 22.08492}{29} = \\frac{172.45468}{29} \\approx 5.946713$$\n$$s = \\sqrt{5.946713} \\approx 2.43859\\%$$\n\n5. **Annualized Sharpe Ratio:**\n$$\\text{Sharpe} = \\frac{\\bar{r}}{s} \\times \\sqrt{365} = \\frac{0.858}{2.43859} \\times 19.10497 \\approx 0.351843 \\times 19.10497 \\approx 6.7219$$\n\n{\"answer\": 6.7219, \"confidence\": 99}" }, { "taskId": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.44697422776459944, "answer": 0.447, "confidence": 99, "grade": "exact", "relError": 0.000057659332014418574, "cost": 0.014768, "reply": "To find the Black-Scholes delta of the call option, we use the standard formula with zero interest rates ($r = 0$) and no dividends ($q = 0$):\n$$C = S N(d_1) - K N(d_2)$$\nwhere $S = 447.74$, $K = 463$, $C = 21.6661$, and:\n$$d_1 = \\frac{\\ln(S/K)}{\\sigma\\sqrt{T}} + \\frac{1}{2}\\sigma\\sqrt{T}, \\quad d_2 = d_1 - \\sigma\\sqrt{T}$$\n\nLet $v = \\sigma\\sqrt{T}$ be the total implied volatility over the life of the option.\n1. Compute $\\ln(S/K)$:\n$$\\ln(447.74 / 463) = \\ln(0.967041) \\approx -0.033514$$\n\n2. Using the linear approximation for near-the-money options ($C \\approx \\frac{S-K}{2} + 0.3989 \\sqrt{SK} v$):\n$$21.6661 \\approx \\frac{447.74 - 463}{2} + 0.3989 (455.31) v$$\n$$21.6661 \\approx -7.63 + 181.62 v \\implies v \\approx \\frac{29.2961}{181.62} \\approx 0.1613$$\n\n3. Refining $v$ using exact values:\nFor $v = 0.16178$:\n$$d_1 = \\frac{-0.033514}{0.16178} + \\frac{0.16178}{2} = -0.20716 + 0.08089 = -0.12627$$\n$$d_2 = -0.12627 - 0.16178 = -0.28805$$\n\nUsing the standard normal cumulative distribution function $N(x)$:\n$$N(d_1) = N(-0.12627) \\approx 0.5 - 0.398942 \\times 0.12627 + \\frac{0.398942 \\times 0.12627^3}{6} \\approx 0.44976$$\n$$N(d_2) = N(-0.28805) \\approx 0.38665$$\n\nChecking the call price:\n$$C = 447.74(0.44976) - 463(0.38665) = 201.375 - 179.019 = 22.356 \\text{ (Wait: } 0.398942 \\times 0.28805 = 0.1149, \\text{ so } N(d_2) = 0.5 - 0.1149 + 0.0016 = 0.3867 \\text{)}$$\nLet's re-evaluate $N(-0.28805)$:\n$$\\phi(0) \\times (0.28805 - 0.28805^3 / 6) = 0.398942 \\times (0.28805 - 0.00398) = 0.398942 \\times 0.28407 = 0.11333$$\n$$N(-0.28805) = 0.5 - 0.11333 = 0.38667$$\n$$463 \\times 0.38667 = 179.028$$\n$$447.74 \\times 0.44976 = 201.375$$\n$$C = 201.375 - 179.028 = 22.347$$\n\nSince $22.347 > 21.6661$, we need a slightly smaller $v$. \nThe option vega with respect to $v$ is $S \\phi(d_1) \\approx 447.74 \\times 0.3958 \\approx 177.2$.\n$$\\Delta v \\approx \\frac{21.6661 - 22.347}{177.2} \\approx -0.00384 \\implies v \\approx 0.16178 - 0.00384 = 0.15794$$\n\n4. Recalculating $d_1$ with $v = 0.15794$:\n$$d_1 = \\frac{-0.033514}{0.15794} + 0.07897 = -0.21220 + 0.07897 = -0.13323$$\n\n5. Finding the delta, $\\Delta = N(d_1)$:\n$$N(-0.13323) \\approx 0.5 - 0.398942 \\times \\left(0.13323 - \\frac{0.13323^3}{6}\\right) = 0.5 - 0.398942 \\times 0.13284 = 0.5 - 0.052995 = 0.4470$$\n\n{\"answer\": 0.4470, \"confidence\": 99}" }, { "taskId": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.25544432164825037, "answer": 0.2555, "confidence": 98, "grade": "exact", "relError": 0.0002179666840522005, "cost": 0.011986, "reply": "To find the Black-Scholes delta of the call option at its implied volatility, we use the Black-Scholes formula with $r = 0$ and $q = 0$:\n\n$$C = S N(d_1) - K N(d_2)$$\nwhere\n$$d_1 = \\frac{\\ln(S/K)}{\\sigma\\sqrt{T}} + \\frac{1}{2}\\sigma\\sqrt{T}, \\quad d_2 = d_1 - \\sigma\\sqrt{T}$$\n\nGiven:\n- $S = 357.63$\n- $K = 383$\n- $C = 5.0781$\n\nFirst, calculate $\\ln(S/K)$:\n$$\\frac{S}{K} = \\frac{357.63}{383} \\approx 0.93376$$\n$$\\ln(S/K) \\approx \\ln(0.93376) \\approx -0.068536$$\n\nLet $v = \\sigma\\sqrt{T}$ be the total implied volatility over the life of the option. We can rewrite the call price normalized by the spot price:\n$$\\frac{C}{S} = \\frac{5.0781}{357.63} \\approx 0.014199$$\n$$\\frac{C}{S} = N(d_1) - \\frac{K}{S} N(d_2)$$\nSince $\\frac{K}{S} = \\frac{383}{357.63} \\approx 1.07094$, we need to find $v$ such that:\n$$N(d_1) - 1.07094 N(d_2) = 0.014199$$\n\nTesting $v = 0.12$:\n$$d_1 = \\frac{-0.068536}{0.12} + 0.06 = -0.57113 + 0.06 = -0.51113$$\n$$d_2 = -0.51113 - 0.12 = -0.63113$$\n\nUsing the standard normal cumulative distribution function $N(x)$:\n- $N(-0.51113) \\approx 0.30463$\n- $N(-0.63113) \\approx 0.26398$\n\nChecking the normalized call price:\n$$0.30463 - 1.07094 \\times 0.26398 = 0.30463 - 0.28271 = 0.02192 \\quad (\\text{too high})$$\n\nTesting $v = 0.10$:\n$$d_1 = \\frac{-0.068536}{0.10} + 0.05 = -0.63536$$\n$$d_2 = -0.73536$$\n- $N(-0.63536) \\approx 0.26260$\n- $N(-0.73536) \\approx 0.23106$\n$$0.26260 - 1.07094 \\times 0.23106 = 0.26260 - 0.24745 = 0.01515 \\quad (\\text{slightly high})$$\n\nTesting $v = 0.097$:\n$$d_1 = \\frac{-0.068536}{0.097} + 0.0485 = -0.70656 + 0.0485 = -0.65806$$\n$$d_2 = -0.75506$$\n- $N(-0.65806) \\approx 0.25525$\n- $N(-0.75506) \\approx 0.22511$\n$$0.25525 - 1.07094 \\times 0.22511 = 0.25525 - 0.24108 = 0.01417 \\quad (\\text{extremely close to } 0.01420)$$\n\nSince $r = 0$ and $q = 0$, the Black-Scholes delta of the call is simply $N(d_1)$:\n$$\\Delta = N(d_1) \\approx 0.2555$$\n\n{\"answer\": 0.2555, \"confidence\": 98}" }, { "taskId": "r_iv_then_delta-3", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.4262987878297406, "answer": 0.4263, "confidence": 99, "grade": "exact", "relError": 0.0000028434757357895263, "cost": 0.014016, "reply": "We are given:\n- Underlying price $S = 152.85$\n- Strike price $K = 162$\n- Call price $C = 8.6404$\n- Risk-free rate $r = 0$, dividend yield $q = 0$\n\nIn the Black-Scholes model with $r = q = 0$, the call price is given by:\n$$C = S N(d_1) - K N(d_2)$$\nwhere $d_1 = \\frac{\\ln(S/K) + \\frac{1}{2}v^2}{v}$, $d_2 = d_1 - v$, and $v = \\sigma \\sqrt{T}$ is the total implied volatility over the remaining life of the option. \n\nNote that $\\ln(S/K) = \\ln(152.85 / 162) = \\ln(0.9435185) \\approx -0.058139$.\n\nLet's find $v$ such that $152.85 N(d_1) - 162 N(d_2) = 8.6404$.\n\nUsing the linear approximation near at-the-money for initial estimation:\n$$C \\approx \\frac{S - K}{2} + 0.3989 \\sqrt{S K} v \\implies 8.6404 \\approx -4.575 + 62.77 v \\implies v \\approx \\frac{13.2154}{62.77} \\approx 0.2105$$\n\nLet's refine $v$ around $0.212$:\nFor $v = 0.2121$:\n- $\\frac{1}{2}v^2 = 0.02249$\n- $d_1 = \\frac{-0.058139 + 0.02249}{0.2121} = \\frac{-0.03565}{0.2121} \\approx -0.16808$\n- $d_2 = -0.16808 - 0.2121 = -0.38018$\n\nUsing the standard normal cumulative distribution function $N(x)$:\n- $N(-0.16808) \\approx 0.5 - 0.06674 = 0.43326$\n- $N(-0.38018) \\approx 0.5 - 0.14809 = 0.35191$\n\nCalculating the call price for $v = 0.2121$:\n$$C \\approx 152.85(0.43326) - 162(0.35191) = 66.2238 - 57.0094 = 9.2144$$\n\nThis is slightly too high, by $9.2144 - 8.6404 = 0.5740$.\n\nThe sensitivity of the call price to $v$ (total vega) is:\n$$\\frac{\\partial C}{\\partial v} = S N'(d_1) = 152.85 \\times \\frac{1}{\\sqrt{2\\pi}} e^{-(-0.168)^2/2} \\approx 152.85 \\times 0.39894 \\times 0.9860 \\approx 60.12$$\n\nAdjusting $v$:\n$$\\Delta v \\approx -\\frac{0.5740}{60.12} \\approx -0.00955 \\implies v \\approx 0.2121 - 0.00955 = 0.20255$$\n\nLet's recompute at $v = 0.2025$:\n- $\\frac{1}{2}v^2 = 0.02050$\n- $d_1 = \\frac{-0.058139 + 0.02050}{0.2025} = \\frac{-0.037639}{0.2025} \\approx -0.18587$\n- $d_2 = -0.18587 - 0.2025 = -0.38837$\n\nValues of $N(x)$:\n- $N(-0.18587) \\approx 0.42627$\n- $N(-0.38837) \\approx 0.34887$\n\nRecomputing $C$:\n$$C \\approx 152.85(0.42627) - 162(0.34887) = 65.1554 - 56.5169 = 8.6385$$\n\nThis is extremely close to $8.6404$ (off by $0.0019$, which corresponds to $\\Delta v \\approx +0.00003$).\nWith $v \\approx 0.20253$, $d_1 \\approx -0.18583$, and the Black-Scholes delta (since $q=0$) is:\n$$\\Delta = N(d_1) \\approx N(-0.18583) \\approx 0.4263$$\n\n{\"answer\": 0.4263, \"confidence\": 99}" }, { "taskId": "r_iv_then_delta-4", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.6063214899790841, "answer": 0.6063, "confidence": 99, "grade": "exact", "relError": 0.00003544320865960742, "cost": 0.01035, "reply": "To find the Black-Scholes delta of the call option at its implied volatility, we use the Black-Scholes formula with $r = 0$ and $q = 0$:\n\n$$C = S N(d_1) - K N(d_2)$$\n\nwhere:\n* $S = 198.88$\n* $K = 192$\n* $C = 28.2965$\n* $T = \\frac{108}{365} \\approx 0.29589$\n* $d_1 = \\frac{\\ln(S/K)}{\\sigma\\sqrt{T}} + \\frac{1}{2}\\sigma\\sqrt{T}$\n* $d_2 = d_1 - \\sigma\\sqrt{T}$\n* $\\Delta = N(d_1)$\n\nFirst, compute $\\ln(S/K)$:\n$$\\ln\\left(\\frac{198.88}{192}\\right) = \\ln(1.035833) \\approx 0.035206$$\n\nLet $v = \\sigma\\sqrt{T}$. Using the normal approximation for near-at-the-money options, the time value is roughly:\n$$C - (S - K) = 28.2965 - 6.88 = 21.4165$$\nUsing the Brenner-Subrahmanyam approximation to estimate $v$:\n$$v \\approx \\frac{C - 0.5(S - K)}{0.3989 \\sqrt{S K}} = \\frac{28.2965 - 3.44}{0.3989 \\times 195.41} \\approx \\frac{24.8565}{77.95} \\approx 0.3189$$\n\nRefining $v$ by plugging $v = 0.3199$ into $d_1$ and $d_2$:\n$$d_1 = \\frac{0.035206}{0.3199} + \\frac{0.3199}{2} \\approx 0.11005 + 0.15995 = 0.2700$$\n$$d_2 = 0.2700 - 0.3199 = -0.0499$$\n\nUsing the standard normal cumulative distribution function:\n* $N(0.2700) \\approx 0.60642$\n* $N(-0.0499) \\approx 0.48010$\n\nChecking the call price with these values:\n$$C = 198.88 \\times 0.60642 - 192 \\times 0.48010 = 120.6048 - 92.1792 = 28.4256$$\n\nSince this is slightly higher than $28.2965$, we adjust $v$ slightly downward by $\\Delta v \\approx \\frac{28.2965 - 28.4256}{S N'(d_1)} \\approx \\frac{-0.1291}{198.88 \\times 0.3847} \\approx -0.00169$:\n* $v \\approx 0.3199 - 0.00169 = 0.3182$\n* $d_1 = \\frac{0.035206}{0.3182} + \\frac{0.3182}{2} \\approx 0.11064 + 0.15910 = 0.26974$\n\nNow we evaluate the delta $\\Delta = N(d_1)$:\n$$\\Delta = N(0.26974) \\approx 0.6063$$\n\n{\"answer\": 0.6063, \"confidence\": 99}" }, { "taskId": "r_iv_then_delta-5", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 0.6017546889052645, "answer": 0.6018, "confidence": 99, "grade": "exact", "relError": 0.00007529828279015368, "cost": 0.012056, "reply": "To find the Black-Scholes delta of the call option at its implied volatility, we use the Black-Scholes formula with $S = 108.83$, $K = 107$, $r = 0$, $q = 0$, $T = 14/365$, and $C = 4.3243$.\n\nFor $r = 0$ and $q = 0$, the call price is:\n$$C = S N(d_1) - K N(d_2)$$\nwhere $d_1 = \\frac{\\ln(S/K) + 0.5\\sigma^2\\tau}{\\sigma\\sqrt{\\tau}}$ and $d_2 = d_1 - \\sigma\\sqrt{\\tau}$. The call's delta is simply $\\Delta = N(d_1)$.\n\n1. **Calculate intrinsic value and moneyness:**\n $$\\text{Intrinsic Value} = S - K = 108.83 - 107 = 1.83$$\n $$\\ln(S/K) = \\ln(108.83 / 107) = \\ln(1.0171028) \\approx 0.016958$$\n\n2. **Estimate total volatility $\\nu = \\sigma\\sqrt{\\tau}$:**\n Using the Brenner-Subrahmanyam-style expansion for near-the-money options, the time value is $4.3243 - 1.83 = 2.4943$.\n For an ATM option at $\\sqrt{S K} \\approx 107.91$, the price is roughly $\\frac{1}{\\sqrt{2\\pi}} \\sqrt{S K} \\nu \\approx 43.05 \\nu$.\n Adding half the intrinsic value gives a first-order approximation:\n $$C \\approx 0.5(S - K) + 43.05 \\nu \\implies 4.3243 \\approx 0.915 + 43.05 \\nu \\implies \\nu \\approx \\frac{3.4093}{43.05} \\approx 0.0792$$\n\n3. **Refine $\\nu$ and calculate $d_1$:**\n Let's test $\\nu = 0.0790$:\n $$d_1 = \\frac{0.016958}{0.0790} + 0.5(0.0790) = 0.21466 + 0.03950 = 0.25416$$\n $$d_2 = 0.25416 - 0.0790 = 0.17516$$\n\n Using standard normal CDF values:\n $$N(0.25416) \\approx 0.5 + \\frac{1}{\\sqrt{2\\pi}}\\left(0.25416 - \\frac{0.25416^3}{6}\\right) \\approx 0.60031$$\n $$N(0.17516) \\approx 0.5 + \\frac{1}{\\sqrt{2\\pi}}\\left(0.17516 - \\frac{0.17516^3}{6}\\right) \\approx 0.56952$$\n\n Check the call price:\n $$C = 108.83(0.60031) - 107(0.56952) = 65.3317 - 60.9386 = 4.3931$$\n\n Since $4.3931 > 4.3243$, we reduce $\\nu$. The sensitivity $\\frac{\\partial C}{\\partial \\nu} = S n(d_1) \\approx 108.83 \\times 0.3863 \\approx 42.04$.\n $$\\Delta \\nu \\approx \\frac{4.3243 - 4.3931}{42.04} = -0.00164 \\implies \\nu \\approx 0.07736$$\n\n4. **Calculate final $d_1$ and $\\Delta$:**\n $$d_1 = \\frac{0.016958}{0.07736} + 0.5(0.07736) = 0.21921 + 0.03868 = 0.25789$$\n $$\\Delta = N(0.25789) \\approx 0.60175$$\n\n{\"answer\": 0.6018, \"confidence\": 99}" }, { "taskId": "r_deribit_straddle_breakeven-1", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 113713.12767394731, "answer": 113713.13, "confidence": 100, "grade": "exact", "relError": 2.04554455734024e-8, "cost": 0.004108, "reply": "To find the upper break-even price for the straddle, we calculate the total BTC cost of the options and equate it to the BTC payoff of the call option at expiry.\n\n1. **Total BTC Cost ($C_{total}$):**\n $$C_{total} = 0.054 + 0.0578 = 0.1118\\text{ BTC}$$\n\n2. **Payoff at Expiry (in BTC):**\n For an inverse (BTC-settled) option on Deribit, the payoff of the call at expiry price $S > K$ (where $K = 101000$) is:\n $$\\text{Payoff} = \\frac{S - K}{S} = 1 - \\frac{K}{S}$$\n\n3. **Break-even Condition:**\n Set the payoff equal to the total cost:\n $$1 - \\frac{101000}{S} = 0.1118$$\n $$\\frac{101000}{S} = 1 - 0.1118 = 0.8882$$\n $$S = \\frac{101000}{0.8882} \\approx 113713.13$$\n\n{\"answer\": 113713.13, \"confidence\": 100}" }, { "taskId": "r_deribit_straddle_breakeven-2", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 93608.77985797288, "answer": 93608.78, "confidence": 100, "grade": "exact", "relError": 1.5172413832323368e-9, "cost": 0.004026, "reply": "To find the upper break-even price for the straddle in BTC terms on Deribit (where options are inverse and settled in BTC):\n\n1. **Total Premium Paid in BTC:**\n $$\\text{Total Cost} = 0.015 + 0.0556 = 0.0706\\text{ BTC}$$\n\n2. **Payoff at Expiry (above the strike $K = 87000$):**\n Above the strike price, the put expires worthless and the call is in-the-money. The payoff in BTC for an inverse call option at expiry price $S$ is:\n $$\\text{Payoff} = \\frac{S - K}{S} = 1 - \\frac{K}{S}$$\n\n3. **Break-Even Condition:**\n Set the payoff equal to the total BTC cost:\n $$1 - \\frac{87000}{S} = 0.0706$$\n $$\\frac{87000}{S} = 1 - 0.0706 = 0.9294$$\n $$S = \\frac{87000}{0.9294} \\approx 93608.78$$\n\n{\"answer\": 93608.78, \"confidence\": 100}" }, { "taskId": "r_deribit_straddle_breakeven-3", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 64193.23251006419, "answer": 64193.23, "confidence": 100, "grade": "exact", "relError": 3.910169481716397e-8, "cost": 0.004068, "reply": "To find the upper break-even price for the straddle in BTC terms, we look at the scenario where the BTC/USD price at expiry ($S$) is above the strike price ($K = 59000$).\n\n1. **Total Premium Paid in BTC:**\n $$\\text{Cost} = 0.0196 + 0.0613 = 0.0809 \\text{ BTC}$$\n\n2. **Payoff at Expiry (for $S > K$):**\n On Deribit, options are inverse (settled in BTC). The payoff of the in-the-money call in BTC is:\n $$\\text{Payoff} = \\frac{S - K}{S} = 1 - \\frac{K}{S}$$\n\n3. **Break-Even Condition:**\n Set the payoff equal to the total BTC cost:\n $$1 - \\frac{59000}{S} = 0.0809$$\n $$\\frac{59000}{S} = 1 - 0.0809 = 0.9191$$\n $$S = \\frac{59000}{0.9191} \\approx 64193.2325$$\n\n{\"answer\": 64193.23, \"confidence\": 100}" }, { "taskId": "r_deribit_straddle_breakeven-4", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 102706.75082914304, "answer": 102706.75, "confidence": 100, "grade": "exact", "relError": 8.072916682855671e-9, "cost": 0.004036, "reply": "To find the upper break-even price for the straddle in BTC terms, we look at the payoff of the call option at expiry when the BTC/USD price $S$ is above the strike price $K = 96000$. \n\n1. **Total BTC Cost (Premium):**\n $$\\text{Cost} = 0.022 + 0.0433 = 0.0653\\text{ BTC}$$\n\n2. **Payoff at Expiry (for $S > K$):**\n On Deribit, inverse options are settled in BTC. The payoff of the call in BTC is:\n $$\\text{Payoff} = \\frac{S - K}{S} = 1 - \\frac{K}{S}$$\n\n3. **Break-even Condition:**\n Set the payoff equal to the total cost:\n $$1 - \\frac{96000}{S} = 0.0653$$\n $$\\frac{96000}{S} = 1 - 0.0653 = 0.9347$$\n $$S = \\frac{96000}{0.9347} \\approx 102706.75$$\n\n{\"answer\": 102706.75, \"confidence\": 100}" }, { "taskId": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-4-argon", "expected": 73994.63806970509, "answer": 73994.64, "confidence": 100, "grade": "exact", "relError": 2.6086956541816556e-8, "cost": 0.003528, "reply": "For an inverse (BTC-settled) option on Deribit with strike $K = 69000$:\n\nThe total cost of the straddle in BTC is:\n$$\\text{Cost} = 0.0412 + 0.0263 = 0.0675\\text{ BTC}$$\n\nAbove the strike price ($S > K$), the put expires worthless and the call payoff in BTC is:\n$$\\text{Payoff} = \\frac{S - K}{S} = 1 - \\frac{K}{S}$$\n\nTo break even in BTC terms, the payoff must equal the total BTC cost:\n$$1 - \\frac{69000}{S} = 0.0675$$\n$$\\frac{69000}{S} = 1 - 0.0675 = 0.9325$$\n$$S = \\frac{69000}{0.9325} \\approx 73994.64$$\n\n{\"answer\": 73994.64, \"confidence\": 100}" }, { "taskId": "bs_call_price-1", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 40.790551141452084, "answer": 40.79, "confidence": 98, "grade": "exact", "relError": 0.000013511498047038315, "cost": 0.003598, "reply": "**Black-Scholes Call Option Valuation**\n\nOkay, let's break this down. I've got a Black-Scholes call option to value. The underlying asset (S) is at 382.5, the strike price (K) is 350, and the time to expiration (T) is about 18 days, which translates to roughly 0.049315 years. The volatility (sigma) is looking pretty high at 0.607, and the risk-free rate (r) is 0.0597.\n\nFirst, I need to calculate a few intermediate values. Sigma times the square root of time (sigma * sqrt(T)) is coming out to be 0.134796. The natural log of the ratio of the spot price to the strike price (ln(S/K)) is 0.08880.\n\nNow for d1 and d2, the core components of the Black-Scholes formula. I'll need to calculate a couple of intermediate terms. The risk-free rate plus half the volatility squared (r + 0.5 * sigma^2) is 0.2439245, which then gets multiplied by time (T) to get 0.012029.\n\nWith those values, I get d1 equal to 0.7480. d2 is simply d1 minus sigma * sqrt(T), which is 0.6132.\n\nNext, I look up the cumulative standard normal distribution functions for d1 and d2. I'm interpolating from the standard normal tables to find N(0.748) is 0.77277, and N(0.6132) is approximately 0.73013.\n\nOne last calculation: the discount factor, exp(-rT), which comes out to be 0.99706.\n\nFinally, I'm ready to plug everything into the Black-Scholes formula for the call option price (C). S * N(d1) is 295.5845. K * exp(-rT) * N(d2) is 254.794. Subtracting that, I've got a final call option value of 40.79.\n\n\n\n```json\n{\"answer\": 40.79, \"confidence\": 98}\n```" }, { "taskId": "bs_call_price-2", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 3.2317412561362406, "answer": 3.231, "confidence": 99, "grade": "exact", "relError": 0.00022936741449622883, "cost": 0.004581, "reply": "**Black-Scholes Call Option Pricing: Deep Dive**\n\nOkay, let's break this down step-by-step. I'm looking at a Black-Scholes formula, and it's time to price a call option. First, I've got my inputs: spot price (S) at 22.89, strike price (K) at 23, time to expiration (T) at approximately 0.468 years (171/365), implied volatility (σ) at 0.518, and risk-free rate (r) at 0.0126.\n\nI need to calculate d1 and d2. The formula's pretty straightforward, so I'll crank through these: S/K = 0.995, ln(S/K) is then -0.0048. Calculate the term (r + σ²/2)T which equates to approximately 0.0688. Then plug it into the numerator of d1: -0.0048 + 0.0688 = 0.064. Dividing that by σ√T = 0.3545 gives me d1 = 0.1804. So, d2 will be d1 - σ√T which is -0.1741.\n\nNow, I'll calculate the cumulative normal distribution values. Approximating using the error function, N(d1) which is N(0.1804) around 0.5716. For N(d2) which is N(-0.1741), I'll subtract N(0.1741) from 1, and I get 0.43089.\n\nThe discount factor: e^-rT = e^(-0.0059), which gets me about 0.994115. Multiplying this by the strike price gives me K*e^-rT = 22.8646.\n\nFinally, I plug everything into the Black-Scholes formula. I calculate S*N(d1) as 13.08347. Calculate K*e^-rT*N(d2) as 9.85213.\n\nThe call option price is the difference between those two: C = 13.08347 - 9.85213 = 3.2313. There's my option price.\n\n\n\n{\"answer\": 3.231, \"confidence\": 99}" }, { "taskId": "bs_call_price-3", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 16.773153687059427, "answer": 16.77, "confidence": 98, "grade": "exact", "relError": 0.00018801992268516554, "cost": 0.003816, "reply": "**Black-Scholes Call Option Price Calculation**\n\nOkay, let's break this down step-by-step. I'm looking at a Black-Scholes call option price calculation. The input parameters are: current stock price ($S$) of 89.61, a strike price ($K$) of 95, time to expiration ($T$) of approximately 0.416438 years (152/365), a volatility ($\\sigma$) of 0.788, and a risk-free interest rate ($r$) of 0.0577.\n\nFirst, I need to get the square root of the time to expiration, which is roughly 0.64532. Then, the product of volatility and the square root of time, sigma times root T, comes out to be approximately 0.508512. That is used for calculating the d1 and d2 values.\n\nNow, calculating the natural log of the ratio of the stock price to the strike price: ln(S/K) gives me -0.058414. Next, I calculate the term (r + 0.5 sigma squared), which is 0.368172, and then multiply that by T, which results in 0.153321.\n\nNow for the core of the Black-Scholes calculation, figuring out d1. The numerator is the sum of ln(S/K) and the (r + 0.5 sigma squared)*T term, which yields 0.094907. Dividing that by the product of sigma and root T I found earlier, the d1 value is approximately 0.186637. Then, to get d2, I subtract sigma times root T from d1 to get -0.321875.\n\nNext, I need the cumulative normal distribution functions for d1 and d2. I look these up or approximate these. $N(d_1)$ is roughly 0.57408, and $N(d_2)$ is roughly 0.3738.\n\nThen I work out the discounted strike price, $K e^{-rT}$, this comes out to about 92.7448.\n\nFinally, I put it all together using the Black-Scholes formula. I calculate $S N(d_1)$ as roughly 51.4433, and $K e^{-rT} N(d_2)$ as about 34.668. Now I just subtract to find the call option price, which is approximately 16.775.\n\n\n\n```json\n{\"answer\": 16.77, \"confidence\": 98}\n```" }, { "taskId": "bs_call_price-4", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.6945968823131317, "answer": 0.6953, "confidence": 98, "grade": "correct", "relError": 0.0010122672657655234, "cost": 0.00479, "reply": "**Black-Scholes Call Option Pricing: A Refined Calculation**\n\nOkay, let's break this down. I've got a set of parameters here: Spot price ($S$), Strike price ($K$), Time to maturity ($T$), Volatility ($\\sigma$), and risk-free rate ($r$). My goal, of course, is to calculate the price of a European call option.\n\nFirst, the basic inputs: $S$ is 384.5, $K$ is 427, $T$ is 9/365, which is roughly 0.0246575 years. Volatility, $\\sigma$, is 0.426, and the risk-free rate, $r$, is 0.0252.\n\nNow, let's get into the Black-Scholes calculations. I start by calculating $\\sigma \\sqrt{T}$ which gives me 0.0668936. Next, I need $\\ln(S/K)$, that gives me -0.10484. Then, I calculate $r + \\sigma^2/2$, which equals 0.115938. And multiplying by $T$, I get (r + \\sigma^2/2) * T = 0.002859.\n\nNow I can start to piece together the pieces for $d_1$ and $d_2$.\nThe numerator for $d_1$ is $\\ln(S/K) + (r + \\sigma^2/2)T$ which works out to -0.101981, and dividing that by $\\sigma \\sqrt{T}$ gets me $d_1$, which is about -1.5245. $d_2$ is simply $d_1 - \\sigma \\sqrt{T}$, and so is approximately -1.5914.\n\nI need the cumulative normal distribution values for $d_1$ and $d_2$. $N(d_1)$ is approximately 0.06369, and $N(d_2)$ is 0.05576. Now to put it together using the Black-Scholes formula for a call: $C = S N(d_1) - K e^{-rT} N(d_2)$. First, I'll calculate $K e^{-rT}$, which is roughly 426.735. Then, I compute $S N(d_1)$, which results in 24.4888, and $K e^{-rT} N(d_2)$, which gives me 23.7947. Subtracting, I get a call price of about 0.694.\n\nBut, I am not satisfied. I'm going to refine this a bit, by interpolating the Normal CDF values for $d_1$ and $d_2$. Looking up values for $N(-1.52)$, and $N(-1.53)$ and interpolating, I get a slightly more accurate value for $N(d_1)$, 0.063691. Similarly, I interpolate between $N(-1.59)$ and $N(-1.60)$ to get $N(d_2) = 0.055758$.\n\nLet's plug these new values into the call price formula. $S N(d_1)$ is now 24.4892, and $K e^{-rT} N(d_2)$ is 23.7939. Subtracting them, I get a refined call price, which is approximately 0.6953. This is the final refined estimate.\n\n\n\n{\"answer\": 0.6953, \"confidence\": 98}" }, { "taskId": "bs_call_price-5", "type": "bs_call_price", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 23.323917391138792, "answer": 23.33, "confidence": 99, "grade": "exact", "relError": 0.0002607884755893082, "cost": 0.004611, "reply": "**Black-Scholes Call Option Valuation - A Quick Run-Through**\n\nOkay, let's break this down. We're looking at a Black-Scholes valuation for a European call option. First, I've got the givens: Spot price (S) is 314.3, strike price (K) is 294, time to maturity (T) is about 0.0547945 years (20 days), volatility (σ) is 0.355, and the risk-free rate (r) is 0.0006.\n\nI'll start with the standard calculations to derive the inputs for the Black-Scholes formula. First, the square root of time: √T ≈ 0.234082. Then, volatility times the square root of time: σ√T ≈ 0.083099. That’s a key component.\n\nNow, on to calculating d1 and d2. First, d1: It involves the natural log of the spot price to strike price ratio, ln(S/K). So, ln(314.3 / 294) ≈ 0.066779. Then, I need to compute σ²/2 which is 0.0630125 and add that to the risk-free rate, r. The sum becomes 0.0636125. Multiplying this by T, gives 0.003486. Summing the ratio and the previously calculated result yields 0.070265. This result divided by σ√T, which gives me d1 ≈ 0.84556.\n\nFor d2, it’s just d1 minus σ√T, so d2 ≈ 0.76246. Straightforward.\n\nNow, the cumulative standard normal distributions, N(d1) and N(d2). I can either use a table or a built-in function; this time I used a table. For N(0.84556), I get approximately 0.8011. And for N(0.76246), it's roughly 0.7771.\n\nI need the present value factor, e^(-rT). e^(-0.0006 * 0.05479) is approximately 0.999967. Multiplying this by the strike, K, gives 293.990.\n\nFinally, the Black-Scholes call option formula: C = S * N(d1) - K * e^(-rT) * N(d2). Plugging in the numbers: 314.3 * 0.80110 - 293.990 * 0.77710 which results in 251.7857 - 228.4596. This produces a call value (C) of 23.326.\n\nI'll run a quick sanity check. The intrinsic value (S-K) is 20.3, so the time value is about 3.0. That's quite sensible. All checks out.\n\n\n\n```json\n{\"answer\": 23.33, \"confidence\": 99}\n```" }, { "taskId": "bs_put_delta-1", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": -0.280639388201964, "answer": -0.2806, "confidence": 98, "grade": "exact", "relError": 0.00014035165276097544, "cost": 0.00294, "reply": "**Black-Scholes Delta Calculation**\n\nAlright, let's break this down. I see we're calculating the put delta, which I know is the sensitivity of a put option's price to changes in the underlying asset's price. The inputs are as follows. The current asset price, *S*, is 114.52; the strike price, *K*, is 106; and the time to expiration, *days*, is 50. I'll need to convert those days into years, so *T* = 50/365, which equals approximately 0.137. Then, we have the volatility, *sigma*, at 0.466, and the risk-free rate, *r*, at 0.0585.\n\nNow, into the Black-Scholes formula. First, I have to calculate *d1*. That requires a few intermediate steps. I start by computing S/K = 1.080377, and taking the natural log of that, yielding 0.077310. Then I square the volatility, *sigma*, giving me 0.217156. Half of that gives me 0.108578. Adding the risk-free rate, *r*, to half the squared volatility results in 0.167078. Multiplying that by *T*, the time to expiration, gives me 0.022887. Adding the first calculated value ln(S/K) gives a numerator of 0.100197.\n\nFor the denominator, I multiply *sigma* by the square root of *T*, resulting in 0.172474. Dividing the numerator by the denominator, I get *d1* equal to 0.58094.\n\nNow, I'll need the cumulative standard normal distribution of *d1*, which is N(d1). I'll interpolate to get that. I know N(0.58) is 0.71904 and N(0.59) is 0.72240. Given d1 = 0.58094, the value I am looking for is a weighted average between the two values, N(0.58) + 0.094(N(0.59) - N(0.58)). I'll calculate it out, and the end result is 0.71936.\n\nFinally, I calculate the put delta, which is N(d1) - 1. So, 0.71936 - 1, and I end up with a put delta of -0.28064. That's a negative value as expected, with puts.\n\n\n\n```json\n{\"answer\": -0.2806, \"confidence\": 98}\n```" }, { "taskId": "bs_put_delta-2", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": -0.31093085044630486, "answer": -0.311, "confidence": 98, "grade": "exact", "relError": 0.00022239528048078783, "cost": 0.003324, "reply": "**Black-Scholes Put Delta Calculation**\n\nAlright, let's break this down. I'm starting with a Black-Scholes put delta calculation. The formula I'm using is `Delta_put = N(d1) - 1 = -N(-d1)`. We've got our inputs: a spot price (S) of 46.9, a strike price (K) of 43, a time to expiration (T) of approximately 0.1123, an implied volatility (sigma) of 0.727, and a risk-free rate (r) of 0.0327.\n\nFirst, I need to compute the components of `d1`. I've calculated the natural log of the ratio of the spot to strike, which is approximately 0.086817. The volatility multiplied by the square root of time gives us about 0.2436576. Then, I calculate `r + sigma^2 / 2`, which is 0.2969645, and multiply that by T, to arrive at approximately 0.0333576.\n\nNow, I can construct the numerator for d1, which is the sum of `ln(S/K)` and `(r + sigma^2 / 2)T`, resulting in 0.1201746. Finally, dividing the numerator by `sigma * sqrt(T)` gives us d1, which is about 0.49321.\n\nNext, I need to find the cumulative standard normal distribution of d1, which is `N(0.49321)`. Since I don't have a perfect normal distribution table, I'll need to approximate. I know that `N(0) = 0.5`. I'm going to approximate this using a standard normal table. I've found `N(0.49) = 0.6879` and `N(0.50) = 0.6915`. Using linear interpolation between these two values I can find an approximation for `N(0.4932)` of 0.68905.\n\nFinally, I plug this value back into the put delta formula: `N(d1) - 1`. This leads me to `-0.31095`. So, the Black-Scholes delta of this put option is approximately -0.31095.\n\n\n\n```json\n{\"answer\": -0.3110, \"confidence\": 98}\n```" }, { "taskId": "bs_put_delta-3", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": -0.5641608283248785, "answer": -0.5642, "confidence": 98, "grade": "exact", "relError": 0.00006943352525534937, "cost": 0.003844, "reply": "**Black-Scholes Delta Calculation for a Put Option**\n\nOkay, let's break this down. We're looking at the Black-Scholes delta for a put option, which is given by $\\Delta_{put} = N(d_1) - 1 = -N(-d_1)$. I know I have:\n\n* Spot price, $S = 215.92$\n* Strike price, $K = 231$\n* Time to expiry, $T = 65/365 \\approx 0.178082$ years\n* Volatility, $\\sigma = 0.528$\n* Risk-free rate, $r = 0.0376$\n\nMy first step is to calculate $d_1$. Let me run through the calculations:\n\n1. $\\ln(S/K) = \\ln(215.92/231) = \\ln(0.9347186) \\approx -0.067510$\n2. $\\sigma \\sqrt{T} = 0.528 \\times \\sqrt{0.178082} = 0.528 \\times 0.421998 \\approx 0.222815$\n3. $r + \\frac{1}{2}\\sigma^2 = 0.0376 + 0.5 \\times (0.528)^2 = 0.0376 + 0.5 \\times 0.278784 = 0.0376 + 0.139392 = 0.176992$\n4. $(r + \\frac{1}{2}\\sigma^2) T = 0.176992 \\times 0.178082 \\approx 0.031519$\n\nSo the numerator for $d_1$ is $-0.067510 + 0.031519 \\approx -0.035991$. Therefore, $d_1 = -0.035991 / 0.222815 \\approx -0.16153$.\n\nNow for the put delta: $\\Delta_{put} = N(d_1) - 1$. This means I need to find the cumulative standard normal distribution of $d_1$, or $N(-0.16153)$. I know it's probably best to use the error function, which is $\\text{erf}(x/\\sqrt{2})$.\n\nI also know I can approximate the cumulative normal by: $N(-0.16153) \\approx 0.5 - 0.16153 \\times \\frac{1}{\\sqrt{2\\pi}} \\times (1 - 0.16153^2 / 6)$ and that $1/\\sqrt{2\\pi} \\approx 0.39894$ so the result should be close to: $N(-0.16153) \\approx 0.5 - 0.06444 + \\dots \\approx 0.4358$.\n\nBut to be more precise, let's use the normal distribution table since $N(-x) = 1 - N(x)$. I calculate $-0.16153 / 1.41421 = -0.11422$ or $x=0.1615$ to use with a table or tool. From the table, I find $\\Phi(0.16) = 0.5636$ and $\\Phi(0.17) = 0.5675$. Interpolating, $\\Phi(0.1615) \\approx 0.5642$. Therefore, $N(-0.16153) = 1 - 0.56417 = 0.43583$.\n\nFinally, $\\Delta_{put} = 0.43583 - 1 = -0.56417 \\approx -0.5642$. Okay, delta is approximately -0.5642.\n\n\n\n```json\n{\"answer\": -0.5642, \"confidence\": 98}\n```" }, { "taskId": "bs_put_delta-4", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": -0.3853788948381438, "answer": -0.3854, "confidence": 99, "grade": "exact", "relError": 0.00005476470595277504, "cost": 0.003327, "reply": "**Black-Scholes Put Delta Calculation**\n\nAlright, let's break this down. I'm looking at calculating the Black-Scholes put delta, which we know is $\\Delta_P = N(d_1) - 1 = -N(-d_1)$. First, I need to get my parameters straight. I see the spot price (S) is 291.06, the strike price (K) is 286, time to maturity (T) is 113/365, which is roughly 0.309589. The volatility (sigma) is 0.729, and the risk-free rate (r) is 0.0594.\n\nNow, let's crunch the numbers. I've got $\\sqrt{T}$ as approximately 0.556407, and $\\sigma \\sqrt{T}$ comes out to around 0.405621. The natural log of the ratio of the spot and strike prices, $\\ln(S/K)$, is roughly 0.017538. Then I calculate $(r + \\sigma^2 / 2)$ as 0.3251205. Multiplying that by T gives me approximately 0.100654.\n\nNow, for the numerator of $d_1$: I take $\\ln(S/K) + (r + \\sigma^2 / 2)T$ which is 0.118192. Dividing that by $\\sigma \\sqrt{T}$, I get $d_1 \\approx 0.291385$.\n\nFinally, to get the put delta, I need $N(d_1)$. I know I can approximate this using a standard normal distribution. $N(0.2914)$ is roughly 0.61462, through a linear interpolation, which puts the put delta ($\\Delta_P$) at $0.61462 - 1 = -0.38538$. So, there we have it, the Black-Scholes put delta for those parameters is -0.38538. That's what I'd expect given the parameters.\n\n\n\n```json\n{\"answer\": -0.3854, \"confidence\": 99}\n```" }, { "taskId": "bs_put_delta-5", "type": "bs_put_delta", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": -0.4080759766740296, "answer": -0.4081, "confidence": 99, "grade": "exact", "relError": 0.000058869738341865825, "cost": 0.003245, "reply": "**My Black-Scholes Delta Calculation for a Put Option**\n\nOkay, let's break this down. I'm looking at the Black-Scholes delta for a put option, and I know the formula: $\\Delta_{put} = N(d_1) - 1 = -N(-d_1)$. My goal is to determine this delta given the parameters. I'll need to calculate $d_1$ first.\n\nThe provided data gives me: $S = 146.31$, $K = 152$, $T \\approx 0.4438$, $r = 0.0456$, and $\\sigma = 0.799$. I'll start working through the formula step-by-step.\n\nFirst, I quickly square root T and get $\\sqrt{T} \\approx 0.66621$. Then, I calculate the product of the volatility and the square root of time: $\\sigma \\sqrt{T} = 0.53230$. That's the denominator, I'll need later.\n\nNext, I calculate the natural log of the ratio of the spot price to the strike price: $\\ln(S/K) = \\ln(0.9625658) \\approx -0.038153$. Then, I compute the intermediate value: $r + \\frac{1}{2}\\sigma^2 = 0.0456 + 0.5 \\times (0.799)^2 = 0.3648005$. I multiply this result by time: $(r + \\frac{1}{2}\\sigma^2)T = 0.16191$.\n\nNow I'm ready for the numerator of $d_1$. I sum $\\ln(S/K)$ and $(r + \\frac{1}{2}\\sigma^2)T$ which is $-0.038153 + 0.16191 = 0.123757$. I'll divide this result by $\\sigma \\sqrt{T}$ to get $d_1 = 0.123757 / 0.53230 \\approx 0.2325$.\n\nNow I need to find the cumulative standard normal distribution $N(d_1)$ for $d_1 = 0.2325$. I'm interpolating, since I don't have a lookup table handy. I know $N(0.23) \\approx 0.59095$ and $N(0.24) \\approx 0.59483$. Interpolating between these values, I approximate $N(0.2325) \\approx 0.59192$.\n\nFinally, I can calculate the delta: $\\Delta_{put} = N(d_1) - 1 = 0.59192 - 1 = -0.40808$. Therefore, the delta for the put option is approximately -0.4081.\n\n\n\n```json\n{\"answer\": -0.4081, \"confidence\": 99}\n```" }, { "taskId": "bs_gamma-1", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.007721610843754648, "answer": 0.007722, "confidence": 98, "grade": "exact", "relError": 0.000050398324031885966, "cost": 0.003394, "reply": "**Gamma Calculation for a European Option**\n\nOkay, let's break this down. We're after the Gamma of a European option, using the Black-Scholes model. I see the formula for Gamma: Γ = N'(d₁) / (S * σ * √T). So, the plan is clear; get the pieces and plug them in.\n\nFirst, I note the given parameters: Spot price (S) is 168.77, Strike (K) is 150, Time to Maturity (T) is about 0.093 years (34 days), and implied volatility (σ) is a hefty 0.844. The risk-free rate (r) is a mere 0.003, which is negligible for this.\n\nNow, I'm calculating the inputs needed for d₁. I calculate S/K, which is 1.125, then its natural log (~0.118). I square the volatility to get σ² (~0.712), then calculate the risk-free rate plus half the variance (~0.359). Multiplying this by T gives me a figure around 0.033. The numerator for d₁ is the sum of ln(S/K) and (r + σ²/2)T, resulting in around 0.151.\n\nFor the denominator, it's just sigma multiplied by the square root of T (0.258). Therefore, d₁ is about 0.588.\n\nNext, I need N'(d₁). This is the normal probability density function, which I compute as e^(-d₁²/2) / √(2π). I square d₁ (0.345), divide by 2, get 0.173, and then e^(-0.173) (~0.841). Dividing by the square root of 2π (~2.507) gives me N'(d₁) (~0.336).\n\nFinally, I put it all together. Gamma is N'(d₁) divided by (S * σ * √T). That's 0.336 divided by (168.77 * 0.844 * √0.093), and the result is approximately 0.00772. There you have it – the Gamma.\n\n\n\n```json\n{\"answer\": 0.007722, \"confidence\": 98}\n```" }, { "taskId": "bs_gamma-2", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.011719035620003572, "answer": 0.01172, "confidence": 98, "grade": "exact", "relError": 0.00008229175400588628, "cost": 0.003402, "reply": "**Gamma Calculation for the Black-Scholes Model**\n\nOkay, let's break down this Gamma calculation. We're using the Black-Scholes formula, specifically the Gamma, which is the second derivative of the option price with respect to the underlying asset's price – a measure of how the delta changes.\n\nWe have our inputs: spot price ($S$) at 350.55, strike price ($K$) at 375, time to expiry ($T$) at roughly 0.1945205 (71 days out of 365), implied volatility ($\\sigma$) at 0.161, and the risk-free rate ($r$) at 0.0448.\n\nThe formula requires $d_1$, which is the trickier part. First, I calculated the natural log of the spot price divided by the strike price, getting approximately -0.067423. Then, I tackled the $(r + 0.5 \\sigma^2) T$ term. Calculating $\\sigma^2$ and multiplying by 0.5, then adding the risk-free rate and multiplying by $T$ gave me roughly 0.0112356. Adding these, I found $d_1$'s numerator, and the denominator ($\\sigma \\sqrt{T}$), and divided to find $d_1$ equals -0.79128.\n\nNow, to get $\\phi(d_1)$, which is the probability density function of the standard normal distribution, I squared $d_1$ and divided by 2. I was then able to calculate the value of the PDF, which came to approximately 0.291709.\n\nFinally, the denominator of the Gamma formula is the spot price times the product of the implied volatility and the square root of the time to maturity ($S \\sigma \\sqrt{T}$), which equals about 24.892. Therefore, Gamma comes out to approximately 0.011719. That's our Gamma.\n\n\n\n```json\n{\"answer\": 0.01172, \"confidence\": 98}\n```" }, { "taskId": "bs_gamma-3", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.00332788170697725, "answer": 0.003328, "confidence": 98, "grade": "exact", "relError": 0.0000355460419466294, "cost": 0.0038, "reply": "**Gamma Calculation for the Black-Scholes Model**\n\nOkay, let's break down this Gamma calculation. We're using the Black-Scholes formula, and my target is to calculate the Gamma for a specific set of parameters: current asset price (S), strike price (K), time to expiration (T, in years), volatility (σ), and risk-free interest rate (r).\n\nFirst, let's lay out the known values: S = 471.11, K = 474, days = 107 (so T = 107/365 ≈ 0.29315), volatility σ = 0.464, and the risk-free rate r = 0.0503. The square root of T, which we'll need, is approximately 0.5414339.\n\nNow, let's calculate d1, a crucial parameter. We start with ln(S/K) which is ln(471.11 / 474) ≈ -0.0061157. Then, we calculate the other part of the d1 numerator, (r + σ²/2)T. We have σ² = 0.215296 and σ²/2 = 0.107648. Adding the interest rate, r + σ²/2 = 0.157948. Multiplying by T gives us (r + σ²/2)T ≈ 0.04630256. Adding this to our first part of the numerator we get 0.04018686. The denominator for d1 is σ√T, or 0.464 * 0.5414339 ≈ 0.2512253. Therefore, d1 is approximately 0.04018686 / 0.2512253 ≈ 0.1599634.\n\nNext, we need the standard normal probability density function value at d1, or N'(d1). The formula is (1/√(2π)) * exp(-d1²/2). We have d1² ≈ 0.025588 and then d1²/2 ≈ 0.012794. The exponential of this, e^(-0.012794), is approximately 0.987287. Multiplying that by 1/√(2π) ≈ 0.3989423, we have N'(d1) ≈ 0.39387.\n\nFinally, we arrive at the Gamma calculation. Gamma = N'(d1) / (S * σ * √T). The denominator becomes 471.11 * 0.2512253 ≈ 118.3547. Therefore, Gamma = 0.39387 / 118.3547 ≈ 0.00332788.\n\nSo, for these specific parameters, the calculated Gamma is roughly 0.00332788.\n\n\n\n```json\n{\"answer\": 0.003328, \"confidence\": 98}\n```" }, { "taskId": "bs_gamma-4", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.015589164355628841, "answer": 0.01559, "confidence": 98, "grade": "exact", "relError": 0.00005360417993518107, "cost": 0.003552, "reply": "**Gamma Calculation for the Given Black-Scholes Parameters**\n\nOkay, let's break this down. My goal is to calculate Gamma for this option, and I have the standard Black-Scholes formula in mind: $\\Gamma = \\frac{N'(d_1)}{S \\sigma \\sqrt{T}}$. I've got the necessary parameters laid out: the underlying price $S$, strike price $K$, time to maturity $T$, volatility $\\sigma$, and risk-free rate $r$. Now, let's step through the calculation.\n\nFirst, the easy part: I'll calculate $\\sigma \\sqrt{T}$. T is approximately 0.40274, so its square root is around 0.634618. Multiplying that by the volatility of 0.446, I get approximately 0.28304. That's one part done.\n\nNext, I need to find $d_1$, which requires a few intermediate calculations. I'll take the natural log of $S/K$ which gives me ln(0.8711), around -0.137996. Now, I have to compute $(r + \\sigma^2 / 2) T$. First, I square sigma, getting roughly 0.198916, which divided by 2 is about 0.099458. Adding the risk-free rate, 0.0514, gives me 0.150858. Multiplying that by $T$, I get approximately 0.060757. Now, I can combine those pieces to get the numerator for $d_1$: -0.137996 + 0.060757 = -0.077239.\n\nNow I can compute d1; -0.077239 / 0.28304 gives me approximately -0.27289.\n\nThat was easy enough.\n\nAlright, moving on to $N'(d_1)$. This is the probability density function of the standard normal distribution evaluated at $d_1$. I know the formula is $\\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$. I'll square $d_1$ and divide by 2, giving me approximately 0.037235. The negative exponential of that value can be quickly approximated. This results in 0.963458. Now I just divide this value by the square root of 2π, which is about 2.506628, giving me a value for $N'(d_1)$ of about 0.384364.\n\nFinally, I can plug everything into the Gamma formula. I divide the value I calculated for $N'(d_1)$ (0.384364) by the product of $S$, $\\sigma$, and $\\sqrt{T}$ (87.11 * 0.28304). This comes out to approximately 0.015589. So, Gamma is around 0.015589.\n\n\n\n```json\n{\"answer\": 0.01559, \"confidence\": 98}\n```" }, { "taskId": "bs_gamma-5", "type": "bs_gamma", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.00596435584251129, "answer": 0.005964, "confidence": 98, "grade": "exact", "relError": 0.000059661515960215854, "cost": 0.003371, "reply": "**Gamma Calculation Walkthrough**\n\nAlright, let's break down this Gamma calculation using the Black-Scholes framework. My goal here is to get a handle on the sensitivity of this option's delta to changes in the underlying asset's price.\n\nFirst, I start with the Gamma formula: Γ = N'(d1) / (S * σ * √T). It's a standard formula, so I'm familiar with its components. I need to calculate d1 and the probability density function of a standard normal distribution, N'(d1).\n\nI have the necessary inputs: current asset price (S), strike price (K), time to expiry (T), volatility (σ), and risk-free rate (r). I make a quick calculation of σ * √T = 0.127492, as this is a recurrent term in the formula.\n\nNow, I'll calculate d1, the first input into the cumulative normal distribution function. I take the natural log of the ratio of the asset price to the strike price ( ln(S/K) ), giving me approximately 0.112286. I also calculate the term (r + σ²/2) * T = 0.008212. I plug these into the d1 formula, which simplifies to approximately 0.94514.\n\nNext, I tackle N'(d1). This is the probability density function. The formula is (1/√(2π)) * e^(-d1²/2). I square d1 and divide by two. I then calculate e to the power of that negative number to get about 0.63977. I calculate 1/√(2π), which is approximately 0.398942. Now I can calculate N'(d1) ≈ 0.25523.\n\nFinally, I plug everything back into the Gamma formula. I divide the N'(d1) value by the product of the asset price and σ * √T, which is 42.7927. The final result is Gamma ≈ 0.0059643. So that's the estimated sensitivity. It's a standard calculation but worth keeping fresh.\n\n\n\n```json\n{\"answer\": 0.005964, \"confidence\": 98}\n```" }, { "taskId": "bs_vega-1", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.646446181061521, "answer": 0.6464, "confidence": 98, "grade": "exact", "relError": 0.00007143837008234106, "cost": 0.003565, "reply": "**My Black-Scholes Vega Calculation**\n\nOkay, let's break this down. I'm looking at the Black-Scholes Vega calculation. First, I've got my inputs: $S = 424.81$, $K = 380$, $T = 74/365 \\approx 0.2027397$, $\\sigma = 0.55$, and $r = 0.0025$. Straightforward enough.\n\nI need to calculate $\\sqrt{T}$, which is roughly $0.450266$, and then $\\sigma \\sqrt{T}$, which gives me approximately $0.247646$. That's the first step.\n\nNow, for $d_1$: I need $\\ln(S/K)$, and that comes out to be about $0.111470$. I also calculate $r + \\sigma^2/2$, which is $0.15375$, and then multiply that by $T$, resulting in $0.031171$. Adding the log term and the time-dependent term in the numerator, that's $0.142641$. Dividing the numerator by $\\sigma \\sqrt{T}$, I get $d_1 \\approx 0.575988$.\n\nNext, I need $\\phi(d_1)$, the cumulative standard normal distribution. I'm going to calculate the parts one step at a time. I square $d_1$, divide by 2, and get $0.165881$. Then, I find $e^{-0.165881}$, which is about $0.847143$. Dividing this by $\\sqrt{2\\pi}$, which I know is about $2.506628$, I get $\\phi(d_1) \\approx 0.337961$.\n\nNow for the Vega. $Vega = S \\sqrt{T} \\phi(d_1)$. I multiply $S$ and $\\sqrt{T}$ which gives $191.2785$. Then, multiply this by $\\phi(d_1)$ to get $64.6447$. This is the Vega per 100% volatility.\n\nFinally, to get the Vega per 1 percentage point increase in volatility, I divide the Vega by 100. That gives me $Vega_{1\\%} \\approx 0.6464$. So, the option's value changes by roughly 0.6464 per 1% change in volatility.\n\n\n\n```json\n{\"answer\": 0.6464, \"confidence\": 98}\n```" }, { "taskId": "bs_vega-2", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.26781898170848095, "answer": 0.2678, "confidence": 98, "grade": "exact", "relError": 0.00007087514245585115, "cost": 0.003601, "reply": "**My Vega Calculation for This Option**\n\nOkay, let's break this down. First, I'm starting with the given parameters: the current asset price, *S*, at 146.99; the strike price, *K*, at 155; the time to expiration, *T*, which is approximately 0.2849315 years; the volatility, *σ*, at 0.163; and the risk-free rate, *r*, at 0.0024. I need to calculate the Vega of this option, specifically how the option's value changes with a 1% change in volatility.\n\nThe core of this is the *d1* value. Let's compute that. I'm taking the natural log of *S* divided by *K*, which is roughly -0.053066. Then, I calculate *r* plus *σ* squared divided by two, resulting in approximately 0.0156845. Multiplying this by *T* gives me roughly 0.004469. Now, I sum these two values and the numerator is about -0.048597. The denominator is just the product of volatility and the square root of time, or roughly 0.0870077. So, *d1* is the numerator divided by the denominator, which is about -0.558537.\n\nNow, to calculate the cumulative standard normal distribution function, phi(*d1*). I square *d1* and divide by 2, which gives me approximately 0.155982. Using this to solve for the standard normal distribution, I have 0.855576, which I divide by the square root of 2π which is 2.506628, giving me a final result of approximately 0.341325 for phi(*d1*).\n\nFinally, I can calculate Vega. The formula is *S* times the square root of *T* multiplied by phi(*d1*). This needs to be divided by 100 to get a 1% change in volatility for this formula. I have *S* multiplied by the square root of *T* which is roughly 78.4618. Multiplying that by phi(*d1*) I get 26.781. When I divide this by 100, the Vega result for a 1% change is about 0.26781.\n\n\n\n{\"answer\": 0.2678, \"confidence\": 98}" }, { "taskId": "bs_vega-3", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.45587809695630876, "answer": 0.4558, "confidence": 98, "grade": "exact", "relError": 0.0001713110518583802, "cost": 0.003807, "reply": "**My Vega Calculation for the Given Option**\n\nOkay, let me break this down. I need to calculate the Vega of this option, which is the sensitivity of its price to changes in volatility. The prompt specifically asks for the change in the option price for a *one percentage point* increase in volatility, so I need to remember to divide my final Vega value by 100.\n\nFirst, I recall the standard formula: $S \\sqrt{T} \\phi(d_1)$. Since I need it per 1%, I'll be dividing this total by 100, or I could multiply by 0.01 at the end. I'm going to take the division by 100 approach at the end.\n\nNow, I'm working with these inputs: S = 469.67, K = 512, days = 47 (implying T = 47/365 = 0.128767), sigma = 0.254, and r = 0.0139.\n\nLet's start computing:\n\n* I need d1. The formula for d1 is [ln(S/K) + (r + sigma^2 / 2) T] / (sigma \\* sqrt(T)).\n * ln(S/K) = ln(469.67 / 512) = ln(0.917324) = -0.086303\n * sigma^2 / 2 = 0.254^2 / 2 = 0.032258\n * r + sigma^2 / 2 = 0.0139 + 0.032258 = 0.046158\n * (r + sigma^2 / 2) \\* T = 0.046158 \\* 0.128767 = 0.005944\n * Numerator = -0.086303 + 0.005944 = -0.080359\n * Denominator = sigma \\* sqrt(T) = 0.254 \\* 0.35884 = 0.091145\n * d1 = -0.080359 / 0.091145 = -0.88166\n\n* Next, I need to calculate phi(d1), which is the standard normal cumulative distribution function.\n * phi(d1) = (1 / sqrt(2*pi)) \\* exp(-d1^2 / 2)\n * d1^2 / 2 = (-0.88166)^2 / 2 = 0.38866\n * exp(-0.38866) = 0.67796\n * 1 / sqrt(2\\*pi) = 0.398942\n * phi(d1) = 0.398942 \\* 0.67796 = 0.27047\n\n* Now, I can calculate Vega (nu):\n * nu = S \\* sqrt(T) \\* phi(d1) = 469.67 \\* 0.35884 \\* 0.27047\n * 469.67 \\* 0.35884 = 168.536\n * nu = 168.536 \\* 0.27047 = 45.584\n\n* Finally, to get the Vega for a 1 percentage point increase in volatility, I divide by 100:\n * Vega\\_1% = nu / 100 = 0.45584\n\nSo, my final answer is 0.45584. This means that for every 1 percentage point increase in the implied volatility of this option, its price would be expected to increase by approximately 0.45584.\n\n\n\n```json\n{\"answer\": 0.4558, \"confidence\": 98}\n```" }, { "taskId": "bs_vega-4", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.6097213395050691, "answer": 0.6097, "confidence": 99, "grade": "exact", "relError": 0.00003499878335639152, "cost": 0.003316, "reply": "**Black-Scholes Vega Calculation: A Quick Run-Through**\n\nAlright, let's see, I've got a Black-Scholes Vega calculation to run. First, I need to define the inputs. Let's see: spot price ($S$) is 383.12, strike ($K$) is 429, time to maturity ($T$) is 61 days, which I'll convert to approximately 0.167123 years. The square root of T is about 0.408807. Volatility ($\\sigma$) is 0.726, and the risk-free rate ($r$) is 0.0203.\n\nNow, let's calculate $d_1$. Remember, it's:\n\n$d_1 = \\frac{\\ln(S/K) + (r + \\sigma^2/2)T}{\\sigma \\sqrt{T}}$\n\nFirst the ratio of the spot to strike: $\\ln(383.12/429) \\approx -0.113108$. Then, $\\sigma^2/2 = 0.726^2/2 \\approx 0.263538$. Adding the risk-free rate gives me $r + \\sigma^2/2 \\approx 0.283838$. Multiplying that by $T$, I get $(r + \\sigma^2/2)T \\approx 0.047436$. Adding that to $\\ln(S/K)$ gives us $-0.065672$. Then, $\\sigma \\sqrt{T} = 0.726 \\times 0.408807 \\approx 0.296794$.\n\nSo, $d_1 \\approx -0.065672/0.296794 \\approx -0.22127$.\n\nNext, I need $N'(d_1)$, the normal PDF at $d_1$. Remember, it's:\n\n$N'(d_1) = \\frac{1}{\\sqrt{2\\pi}} e^{-d_1^2 / 2}$\n\nSo, $N'(d_1) = 0.398942 \\times e^{-0.02448} \\approx 0.389294$.\n\nFinally, the Vega itself:\n\n$\\nu = S \\sqrt{T} N'(d_1) = 383.12 \\times 0.408807 \\times 0.389294 \\approx 60.9725$.\n\nThat's the Vega per unit of volatility. So, for a 1% increase in volatility:\n\n$\\nu_{1\\%} = \\nu / 100 \\approx 0.6097$.\n\nGood, that checks out. Ready to move on to the next one.\n\n\n\n{\"answer\": 0.6097, \"confidence\": 99}" }, { "taskId": "bs_vega-5", "type": "bs_vega", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.2403754082675673, "answer": 0.2404, "confidence": 98, "grade": "exact", "relError": 0.00010230552538611792, "cost": 0.00359325, "reply": "**My Vega Calculation for the Option**\n\nOkay, let's see. I've been given the following parameters for this option: current asset price ($S$) is 158.91, the strike price ($K$) is 148, the time to expiration ($T$) is 65/365, which is roughly 0.178082 in years. The volatility ($\\sigma$) is 0.564, and the risk-free rate ($r$) is 0.0599. I want to calculate the option's Vega.\n\nFirst, I'll calculate the square root of time, which is approximately 0.421998. Then I multiply the volatility by the square root of time, and that gives me approximately 0.238007, which is a key part of the calculation.\n\nNow for $d_1$. I need to work through the formula. First I'll take the natural log of the ratio of the asset price to the strike, that’s $\\ln(S/K)$, which is ln(158.91/148), which gives me about 0.071132. Next, I need to compute the risk-free rate plus half the squared volatility. That's $r + \\sigma^2/2$, which evaluates to approximately 0.218948. Then, I multiply this by the time to expiration, resulting in about 0.038991. Now I sum the results for the numerator of $d_1$, so $0.071132 + 0.038991$, which gives me 0.110123.\n\nI then divide the numerator by the denominator, which I computed before, so $d_1 = 0.110123 / 0.238007$, to get about 0.462688.\n\nNext, I need to get the value of the standard normal cumulative distribution function $\\phi$ at $d_1$. So I calculate $d_1^2 / 2$, which is $(0.462688)^2 / 2$, which equals about 0.107040. I exponentiate the negative of this value to get $e^{-0.107040}$, and I find that that is approximately 0.89849. To get $\\phi(d_1)$ I then divide by the square root of $2\\pi$, which is about 0.398942. Multiplying, I get $\\phi(d_1) \\approx 0.358445$.\n\nFinally, I can calculate Vega: $\\text{Vega} = S \\sqrt{T} \\phi(d_1)$, which is $158.91 \\times 0.421998 \\times 0.358445$. This evaluates to approximately 24.037. The option's Vega is therefore 24.037.\n\nNow, I want to convert this to the change from a 1 percentage point increase in volatility, so I divide by 100: $\\text{Vega}_{1\\%} = \\text{Vega} / 100$, or 0.24037.\n\n\n\n```json\n{\"answer\": 0.2404, \"confidence\": 98}\n```" }, { "taskId": "implied_vol-1", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 61.799447576373446, "answer": 61.8, "confidence": 98, "grade": "exact", "relError": 0.00000893897353805309, "cost": 0.005703, "reply": "**Implied Volatility Calculation**\n\nOkay, let's break this down. We're given the Black-Scholes formula and a set of market parameters: the current stock price (S), the strike price (K), time to expiration (T), the risk-free rate (r), and the observed call option price (C). My immediate goal is to determine the implied volatility (σ) that makes the Black-Scholes model price match the observed market price.\n\nFirst, I've calculated some preliminary values. T is approximately 0.3726027 years, and its square root is about 0.61041. I compute the forward price, F, which is S compounded at the risk-free rate, which turns out to be about 27.6406. Since the option is out of the money (S < K), the intrinsic value is zero, of course.\n\nThe Black-Scholes formula rearranged, tells us that C * exp(rT) needs to be 3.59104. So, I need to find the sigma (σ) value that generates that.\n\nI'll start with a guess for sigma. Let's try 70%. If sigma is 0.70, then the value of sigma multiplied by the square root of time is around 0.4273. I calculate d1 and d2 using the standard formulas and the forward price. Using those, and the cumulative normal distribution N, I calculate the theoretical option price. 4.14 is too high.\n\nLet's adjust sigma to 60%. Repeating all of the calculations with this value, I get a theoretical value of 3.470. That's too low.\n\nNow, it's time to interpolate between the two values of sigma. Based on this range of 0.60 to 0.70, I estimate a value of sigma of 0.618, based on the observed call price.\n\nLet's test this estimate of sigma. Running the numbers, I find a result very close to 3.59104. That looks good! Therefore, based on these market parameters and this process, the implied volatility is around 61.8%.\n\n\n\n```json\n{\"answer\": 61.8, \"confidence\": 98}\n```" }, { "taskId": "implied_vol-2", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 20.900010973832025, "answer": 20.9, "confidence": 95, "grade": "exact", "relError": 5.250634576190053e-7, "cost": 0.006125, "reply": "**My Implied Volatility Calculation**\n\nOkay, let's break this down. I've been given some option pricing data, and I need to back out the implied volatility. Here's what I'm thinking:\n\nFirst, I have my known inputs: spot price ($S = 490.18$), strike ($K = 490$), time to maturity ($T \\approx 0.1452$), risk-free rate ($r = 0.0508$), and put option price ($P = 13.6926$).\n\nI'll start with the Black-Scholes put formula. I need to figure out the implied volatility ($\\sigma$) that, when plugged into the formula, produces the observed put price. Let's crunch some intermediate values: $e^{-rT} \\approx 0.9926507$ and $Ke^{-rT} = 486.3988$. Using put-call parity I know that $C = P + S - Ke^{-rT} = 17.4738$ .\n\nThe option is near-the-money. I can approximate the call price. Knowing the forward price $F = Se^{rT} \\approx 493.809$, I can try the at-the-money approximation to estimate $\\sigma$, giving $C \\approx 0.4 \\times S \\times \\sigma \\sqrt{T} \\approx 74.7 \\sigma$, from which, if $C \\approx 17.47$, it follows that $\\sigma \\approx 0.234$.\n\nAlright, that's my first guess. Now, let's iterate. I'll test a few different volatility values and then interpolate.\n\nFirst, let's plug in $\\sigma = 0.20$. That gives me $\\sigma \\sqrt{T} \\approx 0.0762117$. After calculating $d_1$ and $d_2$ and using the standard normal cumulative distribution function (CDF), I calculate a put price of around $13.03$. That's a bit too low (I need 13.6926).\n\nNext, I'll try $\\sigma = 0.22$. With that, I get a put price of approximately $14.508$. That's too high.\n\nNow for some interpolation. My calculated put prices are: at 0.20, it's 13.033; at 0.22, it's 14.508. The slope is roughly $1.475 / 0.02 = 73.75$. I'm aiming for 13.6926. The difference to hit my target is 0.6596. Therefore, the adjustment I need is $0.6596 / 73.75 \\approx 0.00894$.\n\nSo, adding that to my initial value means my implied volatility is approximately $0.20894$, or 20.90%.\n\n\n\n```json\n{\"answer\": 20.9, \"confidence\": 95}\n```" }, { "taskId": "implied_vol-3", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 81.50015130720656, "answer": 81.51, "confidence": 98, "grade": "exact", "relError": 0.00012084263201324131, "cost": 0.00648, "reply": "**Estimating Implied Volatility for a Put Option**\n\nOkay, let's take a look at this put option. I have the following parameters: a spot price ($S$) of 79.84, a strike price ($K$) of 86, a time to maturity ($T$) of approximately 0.4411 years, and a risk-free rate ($r$) of 0.0004. The observed market price ($P$) is 20.922. First thing, let's determine if this put is trading above its intrinsic value.\n\nThe intrinsic value, $K e^{-r T} - S$, comes out to roughly 6.1448. Hmm, the market price, 20.922, is significantly higher. We need to find the implied volatility, $\\sigma$, that makes the Black-Scholes put formula price match this market price.\n\nI know I can use the Black-Scholes put formula: $P = K e^{-rT} N(-d_2) - S N(-d_1)$. Where I have some useful equations for $d_1$ and $d_2$.\nI start by plugging in the values I know and calculate $d_1$ and $d_2$.\n\nFirst, I calculate a trial value for $\\sigma$. Let's start with $\\sigma = 1.0$ (100%). Plugging this into the formula gives me a price of around 24.77, which is too high. Next, I try $\\sigma = 0.8$ (80%). This time I get a price of about 20.61, which is a bit too low, but we're getting closer. So, the implied volatility lies between 0.8 and 0.85.\n\nI will perform an iterative approximation. Let's try $\\sigma = 0.82$. This gives me a price of approximately 21.025, which is close to the market price. The difference of $0.103$. Now, calculate the sensitivity of the put price to changes in volatility (Vega). The Vega comes out to be 20.96. Thus, I need to adjust $\\sigma$ by -0.00491. So, the implied volatility is approximately 0.8151 or 81.51%.\n\n\n\n{\"answer\": 81.51, \"confidence\": 98}" }, { "taskId": "implied_vol-4", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 87.70003216471197, "answer": 87.71, "confidence": 98, "grade": "exact", "relError": 0.00011365828543029919, "cost": 0.007401, "reply": "**Implied Volatility Calculation**\n\nOkay, here's what I'm thinking. We've got a put option with a price of 75.8825. Let's start by checking its intrinsic value: that's the difference between the strike price, discounted back to the present, and the asset price. In this case, it comes out to roughly 20.366. But the put price is way above this, meaning volatility has to be pretty high.\n\nNext, I'll calculate the corresponding call price using put-call parity to have something to compare with. This yields a call price of 55.5165.\n\nNow, my focus is on finding the implied volatility. I'll need to use the Black-Scholes model in reverse, effectively. I'll start with the standard Black-Scholes formula, and I'm going to start by calculating d1 and d2. I'll need to estimate the implied volatility sigma.\nLet's make some guesses and iterate. I'll use a volatility of 1.0 (100%) as the initial estimate. Plugging that into the formula, it gives me a put price of 84.74. That's too high.\n\nThen I'll try 0.8 (80%). This gives me a put price of 70.297. This is too low, so now I need to interpolate.\nI'll perform a linear interpolation between these two points to get a better approximation. I calculate the slope between the two volatilities and the corresponding option prices, and I estimate the volatility to be around 0.877 (87.7%) to get close to the target put price of 75.8825.\n\nI'll then plug this 88% into the calculations, and I get a put price of 76.095. This is a bit too high, but very close.\n\nFinally, I'll calculate Vega. Then, I estimate the change in sigma based on the difference between the theoretical put price and the calculated price. With that, I estimate my implied volatility at approximately 0.877064. That should be a good approximation.\n\n\n\n{\"answer\": 87.71, \"confidence\": 98}" }, { "taskId": "implied_vol-5", "type": "implied_vol", "tier": "spec", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 33.39992865638508, "answer": 33.41, "confidence": 95, "grade": "exact", "relError": 0.0003015378780754506, "cost": 0.005652, "reply": "**Analyzing Implied Volatility**\n\nOkay, here's what I'm thinking. I've got some option data: spot price *S* at 140.66, a strike *K* of 143, and a time to maturity *T* of roughly 0.441 years. The risk-free rate *r* is 0.019, and the observed call price *C* is 11.908. My goal is to determine the implied volatility, which I can then use for analysis and potential arbitrage opportunities.\n\nFirst, I calculate the forward price *F*, which comes out to about 141.8438. The discount factor *df* is approximately 0.9917. This gives me a discounted call price of about 12.0078. I'm going to look at Brenner-Subrahmanyam approximation which helps to infer the volatility.\n\nNow, let's play with it. I'm going to start by approximating *sigma* using Brenner-Subrahmanyam approximation. First I need F-K which is -1.1562, and m = F/K which is 0.9919. So I make a quick test using the approximation sigma * sqrt(T) ≈ sqrt(2*pi) * (C/df - (F-K)/2) / F?\n\nI decided to start by testing a *sigma* of 0.35. Calculating *sigma* * sqrt(*T*) gives me 0.23245. I then calculated *ln(S/K)* which equals -0.01650. I need to calculate d1 and d2 for Black-Scholes using this sigma value of 0.35.\n\nI found d1 = 0.08127, d2 = -0.15118, and calculate N(d1) and N(d2). Then calculate the theoretical call price using those values. I calculate a theoretical call price of 12.497. That's a bit too high, my target is 11.908.\n\nLet's test *sigma* = 0.33. That gives me *sigma* * sqrt(*T*) as 0.21917, d1 = 0.07255 and d2 = -0.14662. After calculating N(d1) and N(d2), my new call price is 11.758.\n\nNow, I'll interpolate between these two *sigma* values. I have 0.33 which produced 11.758 and 0.35 which gives 12.497. The difference is 0.739. The required difference is 0.150. Based on this, it seems the implied volatility is around 33.41%. I can use this as my implied volatility value.\n\n\n\n```json\n{\"answer\": 33.41, \"confidence\": 95}\n```" }, { "taskId": "fx_position_size-1", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 6.438461538461515, "answer": 6.438, "confidence": 100, "grade": "exact", "relError": 0.00007168458780997271, "cost": 0.00143, "reply": "**My Calculated Risk Exposure**\n\nAlright, let me break down this trade I'm considering. The AUD/USD entry point is set at 0.94593, and I've got a stop-loss order at 0.94203. That gives me a price difference, or the risk per AUD, of 0.00390 USD. Knowing that, and assuming I'm going in with a standard lot size of 100,000 AUD, the risk per lot works out to be 390 USD.\n\nNow, my total risk tolerance for this particular trade is 2511 USD. Given the risk per lot, I need to figure out my optimal position size. Dividing my total risk by the risk per lot (2511/390), I arrive at a position size of approximately 6.438 lots. This gives me the necessary exposure based on my overall risk parameters.\n\n\n\n```json\n{\"answer\": 6.438, \"confidence\": 100}\n```" }, { "taskId": "fx_position_size-2", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 12.3258666666662, "answer": 12.3259, "confidence": 100, "grade": "exact", "relError": 0.0000027043399627036095, "cost": 0.001526, "reply": "**Calculating My Position Size**\n\nOkay, let's break this down meticulously. My entry point for this trade is 154.513, and I've set a stop-loss at 154.363. This means that if the market moves against me, I'm facing a potential loss of 0.150 JPY per EUR. I need to be precise about my risk, which in this case is 184,888 JPY.\n\nNow, the crucial step: I need to figure out how many units of the base currency (EUR) I can trade while adhering to my risk parameters. Dividing my total risk in JPY (184,888) by the loss per EUR (0.150) gives me 1,232,586.6667 EUR.\n\nSince I'm dealing in standard lots of 100,000 units, I need to convert that EUR figure into lots. Dividing 1,232,586.6667 EUR by 100,000, I arrive at a position size of 12.32586667 lots. Therefore, based on the parameters I've set, I need to open a position of approximately 12.33 standard lots. This ensures I'm properly managing my risk at the specified stop-loss level.\n\n\n\n{\"answer\": 12.3259, \"confidence\": 100}" }, { "taskId": "fx_position_size-3", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 0.4756567164179193, "answer": 0.475657, "confidence": 100, "grade": "exact", "relError": 5.961906368996154e-7, "cost": 0.00185925, "reply": "**Calculating My Position Size**\n\nOkay, let's break this down meticulously. The current EUR/JPY rate is 191.077. I've decided on a stop-loss at 190.407. So, the risk I'm taking per EUR is the difference: 191.077 minus 190.407, which gives me 0.67 JPY. I'm aiming for a total risk of 31869 JPY on this trade.\n\nNow, to figure out how many EUR units I can buy, I need to divide the total risk by the risk per unit: 31869 divided by 0.67. This gives me a rough number of units of EUR. I then remember that one standard lot is 100,000 units. To get my position size in lots, I'll divide the number of units by 100,000. So it's 31869 divided by (0.67 * 100000), or 31869 divided by 67000. That gives me 0.4756567... lots.\n\nI'm a little hesitant, so I just want to double-check my math. 31869 divided by 67. Let's see... 67 goes into 318 six times (6 * 67 = 402, too big, so five times (5 * 67 = 335... still too big). So it goes into 318 four times (4 * 67 = 268) with 50 leftover. 67 goes into 506 seven times (7 * 67 = 469) with 37 leftover. 67 goes into 379... five times (5 * 67 = 335) with 44 leftover. 67 goes into 440... six times (6 * 67 = 402) with 38 leftover. 67 goes into 380... five times (5 * 67 = 335) with 45 leftover. So, the result is 475.6567...\nSo I'm confident my original calculation of 0.4756567... lots is correct. The correct position size will be approximately 0.475 lots.\n\n\n\n{\"answer\": 0.475657, \"confidence\": 100}" }, { "taskId": "fx_position_size-4", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 5.01704545454548, "answer": 5.017, "confidence": 100, "grade": "exact", "relError": 0.000009060022655040646, "cost": 0.001924, "reply": "**Calculating Position Size for a Forex Trade**\n\nOkay, here's the breakdown of my thinking. I've got an entry point at 142.806 JPY, and a stop-loss set at 142.366 JPY. That's a risk of 0.440 JPY per USD traded (142.806 - 142.366). My total risk for this trade is 220,750 JPY. Now, to determine the appropriate position size, I need to divide the total risk by the risk per USD.\n\nSo, the position size in USD is 220,750 JPY / 0.440 JPY. Knowing that one standard lot is 100,000 USD, I then need to divide the result by 100,000 to get the number of lots. Initially, I calculated (220750 / 0.440) / 100000 = 220.75 / 44 = 5.01704545... but let's reconfirm with a detailed calculation.\n\nDividing 220750 by 44000, 220.75 / 44 = 5.01704545... This confirms I am looking at approximately 5.017 lots. It's crucial to get this precise to manage risk effectively.\n\n\n\n```json\n{\"answer\": 5.017, \"confidence\": 100}\n```" }, { "taskId": "fx_position_size-5", "type": "fx_position_size", "tier": "spec", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 1.6141304347826113, "answer": 1.614, "confidence": 100, "grade": "exact", "relError": 0.0000808080808096388, "cost": 0.001817, "reply": "**Calculating the EUR/USD Position Size**\n\nOkay, so I'm looking at a long EUR/USD trade. My entry is at $0.76163, and I've set my stop-loss at $0.75243. I need to figure out my position size to manage my risk.\n\nFirst, I calculate the risk per unit: $0.76163 minus $0.75243, which gives me 0.00920 USD. The total risk I'm comfortable with is $1485. Since the quote currency is USD, the price difference is directly in USD, which makes things simpler.\n\nNow, I calculate the number of units: $1485 divided by 0.00920 USD per unit. This gives me the number of units I need to trade. A standard lot is 100,000 units. Therefore to calculate position size in standard lots I must adjust for standard lots. So I'll divide the number of units I calculated by 100,000. 1485 / (0.00920 * 100000)\n\nWhich gives me 1.61413043478 standard lots, approximately. I want to double-check my work, so I'll work backwards. If I take a look at the risk per lot, which will be 100,000 * 0.00920 = 920 USD. Then if I take the risk per lot and divide it by the total risk 1485 / 920 I get 1.61413 which is correct, good. So the final position size is 1.61413 lots.\n\n\n\n```json\n{\"answer\": 1.614, \"confidence\": 100}\n```" }, { "taskId": "var_parametric-1", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.023819567833442343, "answer": 0.02382, "confidence": 100, "grade": "exact", "relError": 0.0000181433416710188, "cost": 0.00194625, "reply": "**Calculating Value at Risk (VaR)**\n\nOkay, let's break this down. I'm looking at calculating the Value at Risk, or VaR, at a 97.5% confidence level. The data I've got gives me a mean (μ) of -0.0003 and a standard deviation (σ) of 0.012. I'll need to figure out the corresponding z-score for this confidence level – I know that at 97.5% it's 1.95996, though for quick calculations, 1.96 is usually fine.\n\nNow, applying the VaR formula, which is $\\text{VaR} = -\\mu + z \\cdot \\sigma$, the first thing I need to do is multiply the z-score by the standard deviation. So, that's 1.95996 (or 1.96) times 0.012. Doing the math, z * σ = 0.0235195678 (or 0.02352 using 1.96).\n\nNext, I plug that into the formula: - (-0.0003) + 0.0235195678 (or 0.02352). This simplifies to 0.0003 + 0.0235195678 (or 0.0003 + 0.02352). Therefore, the VaR at 97.5% confidence is 0.0238195678 (or 0.02382 if I use 1.96). So, I'm confident that the loss will not exceed 2.38% (approximately) 97.5% of the time.\n\n\n\n```json\n{\"answer\": 0.02382, \"confidence\": 100}\n```" }, { "taskId": "var_parametric-2", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.08208473980663146, "answer": 0.08208, "confidence": 99, "grade": "exact", "relError": 0.000057742847728156604, "cost": 0.002066, "reply": "**Calculating Value at Risk (VaR)**\n\nOkay, let's break this down. I've got a mean (μ) of -0.0005 and a standard deviation (σ) of 0.0496. We're aiming for a 95% confidence level, which, for a one-tailed test, means a z-score around 1.644853 (or we can round it to 1.645 – the precision isn't *that* critical, depending on the application, though, it's good practice to stick with the higher precision whenever possible).\n\nThe formula provided is VaR = -μ + z * σ. So, substituting in the values: VaR = -(-0.0005) + (1.644853 * 0.0496).\n\nLet's do the multiplication: 1.644853 * 0.0496 yields approximately 0.0815847. Adding the -μ, 0.0005, results in 0.0820847. A slight difference, due to the rounding of z.\n\nNow, let's consider using the rounded value of z = 1.645: 0.0496 * 1.645 gives us roughly 0.081592, which, when we add 0.0005, results in approximately 0.082092.\n\nGiven the inherent assumptions, and the rounding involved in the initial values, I'll go with 0.08208 as the final VaR. It's a negligible difference, but good practice to keep the maximum number of decimal places when appropriate.\n\n\n\n{\"answer\": 0.08208, \"confidence\": 99}" }, { "taskId": "var_parametric-3", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.10032876732331801, "answer": 0.1003, "confidence": 99, "grade": "exact", "relError": 0.00028673055680336515, "cost": 0.001743, "reply": "**My Value at Risk Calculation**\n\nOkay, let's break this down. I've got a mean of 0.0011 and a standard deviation of 0.0436. I'm aiming for a 99% confidence level for my Value at Risk (VaR) calculation, which is standard practice in my field. With a normal distribution assumed, I know I need the z-critical value for a 99% confidence interval. I recall that's approximately 2.326348 (or, more precisely, the standard normal quantile where the cumulative probability Φ(z) = 0.99, which gives me essentially the same value).\n\nNow, the standard VaR formula I'm using is VaR = -mean + z * stdev. So, I substitute my values: -0.0011 + 2.326348 * 0.0436. Let's do the multiplication: 2.326348 multiplied by 0.0436 gives me 0.10142877.\n\nFinally, I have VaR = -0.0011 + 0.10142877 = 0.10032877, which I can reasonably round to 0.1003. So, my Value at Risk at the 99% confidence level is approximately 0.1003. This indicates the potential loss I could incur.\n\n\n\n```json\n{\"answer\": 0.1003, \"confidence\": 99}\n```" }, { "taskId": "var_parametric-4", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.07275151453115127, "answer": 0.07275, "confidence": 98, "grade": "exact", "relError": 0.00002081786421946403, "cost": 0.002467, "reply": "**Calculating Value at Risk (VaR)**\n\nOkay, let's break this down. I've got the formula for Value at Risk: $\\text{VaR} = -\\mu + z \\cdot \\sigma$. The inputs are: a mean, $\\mu = -0.0011$, a standard deviation, $\\sigma = 0.0308$, and a confidence level of 99%. For that confidence level, I need the z-score. Looking it up, the standard z-score for a 99% one-tailed normal distribution is typically approximated as 2.326, 2.33, or more precisely 2.3263. I'll use 2.3263479 here for greater accuracy and see how much the results change with the common approximations.\n\nFirst, let's calculate $-\\mu$, which is just $-(-0.0011) = 0.0011$. Now the more complex term: $z \\cdot \\sigma$.\n\nWith $z=2.3263479$:\n$2.3263479 \\times 0.0308 = 0.07165158$.\n\nNow, let's calculate using 2.326 and 2.33 to see how much the result changes, as often the final VaR number is rounded:\n\nWith $z=2.326$:\n$2.326 \\times 0.0308 = 0.0716408$.\n\nWith $z=2.33$:\n$2.33 \\times 0.0308 = 0.071764$.\n\nSo, back to the full calculation using the accurate z score:\n$\\text{VaR} = 0.0011 + 0.07165158$. That's 0.07275158. Using 2.326 as z gives 0.0716408 + 0.0011 = 0.0727408, and with $z=2.33$, that's 0.071764 + 0.0011 = 0.072864. So rounding, the VaR is approximately 0.07275.\n\n\n\n```json\n{\"answer\": 0.07275, \"confidence\": 98}\n```" }, { "taskId": "var_parametric-5", "type": "var_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.11150152403343133, "answer": 0.1115, "confidence": 99, "grade": "exact", "relError": 0.000013668274443252881, "cost": 0.001769, "reply": "**Calculating Value at Risk (VaR)**\n\nOkay, let's break this VaR calculation down. The core formula we're working with is VaR = -mean + z * stdev. We're aiming for a 99% confidence level, which is standard. I know the corresponding z-score from my work with the standard normal distribution – it's 2.3263479..., but let's go with 2.32635 for the sake of efficiency, although I know I could get more precision if I wanted it.\n\nWe have the mean, which is -0.001, and a standard deviation of 0.0475. Plugging those into the formula is straightforward: VaR = -(-0.001) + 2.326348 * 0.0475. This becomes 0.001 + 0.11050153, which equals 0.11150153. Always good to double check, so I'll quickly verify that z-score. I'm confident in the initial result, but confirming that Phi(2.32635) = 0.99 is a good habit. Yep, 2.326347874 * 0.0475 = 0.1105015. Add the 0.001 back in, and it's 0.1115015. Alright, after rounding to more than four significant digits, that comes out to 0.1115. Confirmed.\n\n\n\n```json\n{\"answer\": 0.1115, \"confidence\": 99}\n```" }, { "taskId": "cvar_parametric-1", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.04404168455292251, "answer": 0.04404, "confidence": 99, "grade": "exact", "relError": 0.00003824905744656407, "cost": 0.003232, "reply": "**Calculating Parametric CVaR (Expected Shortfall) for a Normal Distribution**\n\nOkay, so I'm diving into a parametric CVaR calculation, specifically the Expected Shortfall (ES), for a normally distributed random variable, which I'll denote as R. I've got my parameters: R follows a normal distribution with a mean (μ) of 0.0018 and a standard deviation (σ) of 0.0172. The confidence level, α, is set to 0.99, or equivalently, a tail probability, p, of 0.01.\n\nI'm defining the Loss, L, as the negative of R (L = -R). Therefore L follows a normal distribution with mean -μ, which equals -0.0018.\n\nNow, the core of the ES calculation is this formula: $ES_{0.99}(L) = \\mu_L + \\sigma \\frac{\\phi(z_\\alpha)}{1 - \\alpha}$, where $z_\\alpha = \\Phi^{-1}(\\alpha)$.\n\nFirst, I need to find $z_{0.99}$, which is the inverse of the standard normal cumulative distribution function (CDF) at 0.99. That's approximately 2.3263479.\n\nNext, I need to determine $\\phi(z_\\alpha)$, which is the standard normal probability density function (PDF) evaluated at $z_\\alpha$. I know this is $\\phi(z) = \\frac{1}{\\sqrt{2\\pi}} e^{-z^2/2}$. I can approximate this.\n$\\phi(2.32635) \\approx \\frac{1}{\\sqrt{2\\pi}} e^{-2.32635^2/2} = 0.39894228 \\times e^{-2.70588} = 0.39894228 \\times 0.0668115 \\approx 0.026654$.\n\nSo, $\\frac{\\phi(z_\\alpha)}{1 - \\alpha}$ becomes $\\frac{0.026654}{0.01} = 2.6654$.\n\nI want more precision so let's tighten that up a little:\n$z = 2.32634787$\n$z^2 / 2 = 5.411894 / 2 = 2.705947$\n$\\exp(-2.705947) = 0.066807$\n$\\phi(z) = 0.3989423 \\times 0.066807 = 0.026652$\nRatio $= 2.66521$.\n\nNow I can put it all together:\n$ES = -\\mu + \\sigma \\times 2.66521 = -0.0018 + 0.0172 \\times 2.66521$\n$0.0172 \\times 2.665214 = 0.0458417$\n$ES = 0.0458417 - 0.0018 = 0.0440417$\nTherefore, the parametric CVaR at the 99% confidence level, is approximately 0.0440417.\n\n\n\n```json\n{\"answer\": 0.04404, \"confidence\": 99}\n```" }, { "taskId": "cvar_parametric-2", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.02233967069948375, "answer": 0.02234, "confidence": 99, "grade": "exact", "relError": 0.000014740616398397254, "cost": 0.002587, "reply": "**Calculating Conditional Value at Risk (CVaR)**\n\nOkay, here's the breakdown. I'm starting with a random variable, let's call it 'R', that's normally distributed with a mean (µ) of -0.0013 and a standard deviation (σ) of 0.0102. The confidence level, alpha, is set at 0.95.\n\nMy loss function is defined as L = -R, so the loss is also normally distributed, but with a mean of -µ (which is 0.0013) and the same standard deviation, σ, of 0.0102. My goal is to determine the CVaR at the 95% level of this loss distribution.\n\nCVaR is essentially the expected value of the loss given that the loss exceeds the Value at Risk (VaR). I know that for a standard normal distribution, the conditional expectation above a certain quantile can be calculated using the ratio of the probability density function (PDF) at that quantile to (1-alpha).\n\nFirst, I need to find the z-score corresponding to the 95th percentile, which I know to be approximately 1.6448536. Then I need to find the PDF value at that z-score. I can calculate this using the standard normal PDF formula. Doing that calculation, I find that the PDF value is approximately 0.1031356. Dividing that by (1-0.95), which is 0.05, gives me 2.062713.\n\nNow I can put it all together. The CVaR is equal to the mean loss plus the standard deviation of the loss multiplied by this ratio. The mean loss is 0.0013, the standard deviation is 0.0102, and the ratio is 2.062713. So, I multiply 0.0102 by 2.062713 and get roughly 0.0210397. Finally, I add the mean loss to this product, and that gives me my CVaR, which is approximately 0.0223397.\n\n\n\n```json\n{\"answer\": 0.02234, \"confidence\": 99}\n```" }, { "taskId": "cvar_parametric-3", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.09524684038587, "answer": 0.09525, "confidence": 99, "grade": "exact", "relError": 0.000033172902294700856, "cost": 0.003067, "reply": "**Calculating Expected Shortfall**\n\nOkay, here's what I'm thinking. I have a return variable, $R$, that's normally distributed, $R \\sim N(\\mu, \\sigma^2)$. The mean, $\\mu$, is 0.0023, and the standard deviation, $\\sigma$, is 0.0366. I'm working with a confidence level of 99%, so $\\alpha = 0.99$, which means the tail probability, $p$, is 0.01.\n\nNow, I need to calculate the Expected Shortfall, or ES. For a standard normal distribution, the formula for ES is: $ES_p(Z) = \\frac{\\phi(z_p)}{p}$, where $\\phi$ is the standard normal probability density function, and $z_p$ is the z-score corresponding to the (1-p) quantile. So, $z_p = \\Phi^{-1}(1-p) = \\Phi^{-1}(0.99)$, which I know is approximately 2.326348.\n\nNext, I need to calculate $\\phi(z_p)$. That's $\\frac{1}{\\sqrt{2\\pi}} e^{-z_p^2 / 2}$. Plugging in $z_p$ gives me: $e^{-(2.326348)^2 / 2}$, which is roughly $e^{-2.70597}$. That equals about 0.066805. Then $\\phi(z_p)$ becomes $0.026652$.\n\nTherefore, the ES for the standard normal distribution is $-\\frac{\\phi(z_p)}{p}$, and that's equal to $-\\frac{0.026652}{0.01}$, giving me -2.6652. This is, of course, for the standard normal case.\n\nNow, let's look at the loss, $L = -R$. Then the expected loss, conditional on the loss being greater than or equal to VaR, is $E[L \\mid L \\ge VaR] = -\\mu + \\sigma \\frac{\\phi(z_p)}{p}$.\n\nPlugging in the numbers gives $-\\mu + \\sigma * -(-2.6652) = -\\mu + \\sigma \\times 2.66521$. So the ratio is $2.66521$.\n\nNow, to calculate the expected shortfall in terms of the actual variables, I'll take the mean of the return and add the standard deviation times $2.66521$, $0.0975467$, which is $0.0952467$ or $0.09525$ when rounded to five significant digits.\n\n\n\n```json\n{\"answer\": 0.09525, \"confidence\": 99}\n```" }, { "taskId": "cvar_parametric-4", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.12233732788143936, "answer": 0.1223, "confidence": 99, "grade": "exact", "relError": 0.00030512258266365974, "cost": 0.002772, "reply": "**Calculating 1-Day Parametric CVaR (Expected Shortfall)**\n\nOkay, let's break this down. We're given a return distribution, *R*, which we're assuming follows a normal distribution, *N*(μ, σ²), with μ = 0.0018 and σ = 0.0531. The goal is to calculate the 1-day parametric CVaR (or Expected Shortfall) at a 97.5% confidence level, and express it as a positive loss fraction.\n\nFirst, I need to define the loss, *L*, as the negative of the return: *L* = -*R*. This means the loss distribution will also be normal, but with a mean and sign changed. Therefore, *L* ~ *N*(-μ, σ²). The mean of the loss, μₗ, becomes -0.0018, and the standard deviation remains 0.0531.\n\nNow, let's get to the CVaR calculation. The formula is: ESα(*L*) = μₗ + σ * (φ(*z*α) / (1 - α)), where *z*α is the inverse of the standard normal cumulative distribution function (CDF) at the confidence level α, and φ(*z*) is the standard normal probability density function (PDF).\n\nWith our confidence level α = 0.975, I've got to find *z*0.975 which is approximately 1.95996. Then, 1 - α is 0.025. I then compute the PDF for this value, φ(1.95996). It should be: φ(*z*) = (1 / √(2π)) * e^(-*z*²/2), so for 1.95996 I get approximately 0.0584409.\n\nNow the ratio φ(*z*α) / (1 - α) equals 0.0584409 / 0.025 which results in 2.337637.\n\nTherefore, the Expected Shortfall (ES) will be μₗ + σ * 2.337637 = -0.0018 + 0.0531 * 2.337637. 0.0531 * 2.337637 equals 0.1241285. Finally, ES = 0.1241285 - 0.0018 = 0.1223285. So the 1-day parametric CVaR at a 97.5% confidence level, expressed as a positive loss fraction, is 0.1223285.\n\n\n\n```json\n{\"answer\": 0.1223, \"confidence\": 99}\n```" }, { "taskId": "cvar_parametric-5", "type": "cvar_parametric", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.1205670856534704, "answer": 0.1206, "confidence": 99, "grade": "exact", "relError": 0.0002729961195562676, "cost": 0.002742, "reply": "**Calculating Expected Shortfall**\n\nOkay, here's the breakdown of how I'm thinking about this expected shortfall calculation. We're dealing with a return, *R*, that's normally distributed with a mean (µ) of 0.0017 and a standard deviation (σ) of 0.0523. We're interested in the loss, *L*, which is simply the negative of the return, so *L = -R*. Therefore, the loss *L* is also normally distributed, but with a mean of -µ = -0.0017 and the same standard deviation, σ = 0.0523.\n\nNow, we're targeting the Expected Shortfall (ES) at a confidence level of α = 0.975. This means we are focused on the tail. Mathematically, ES is the expected value of *L*, given that *L* is greater than the Value at Risk (VaR) at the same confidence level. In other words, ES is the average of the losses we see in the worst-case scenario.\n\nI know that for a standard normal distribution, *Z ~ N(0, 1)*, the ES has a convenient formula: ESα(Z) = φ(zα) / (1 - α), where φ(zα) is the probability density function (PDF) of the standard normal evaluated at the z-score (zα) corresponding to the confidence level α.\n\nWith α = 0.975, the z-score, zα, is approximately 1.959964. Plugging this value into the standard normal PDF, I find that φ(zα) is about 0.05844094. And, 1-α is 0.025. Therefore, ESα(Z) = 0.05844094 / 0.025 = 2.3376378.\n\nTo calculate ES(L) itself, I have to transform this standard normal result back to the actual loss distribution. I use the relationship: ESα(L) = µL + σL * ESα(Z), which is really just a scale and shift of the standard normal result. In our case, that's ESα(L) = -µ + σ * ESα(Z), where µL = -0.0017 and σ = 0.0523.\n\nSo, ESα(L) = -0.0017 + 0.0523 * 2.3376378 = -0.0017 + 0.12225846. That gives me an Expected Shortfall of roughly 0.12055846. Therefore, if we do have a \"loss\" that is greater than the 97.5% VaR, our average loss will be about 12.06%.\n\n\n\n```json\n{\"answer\": 0.1206, \"confidence\": 99}\n```" }, { "taskId": "sharpe_annualized-1", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 1.6781883884002948, "answer": 1.678, "confidence": 99, "grade": "exact", "relError": 0.00011225700380066998, "cost": 0.006366, "reply": "**Sharpe Ratio Calculation**\n\nOkay, here's what I'm seeing. First, I'm given a set of returns, `x`, with `n = 12`. I immediately recognize this as likely representing monthly returns, given the sample size. My immediate thought is to calculate the Sharpe Ratio, a classic measure of risk-adjusted return.\n\nFirst, I need the mean return. I'll sum the returns and divide by the number of observations (12). After summing all the values, I get a mean of 0.02114167, or around 2.1%.\n\nNext, I'll calculate the standard deviation. First, I'll calculate the deviations from the mean (x - mean). Then, I'll square each of those deviations and sum them to obtain the sum of squares. I recognize a shortcut here, I could calculate sum of squares by doing sum(x^2)-n*mean(x)^2 but to follow steps as given, I am doing things step by step. This gives me a sum of squares of 0.02631301.\n\nNow, I calculate the sample variance, which is the sum of squares divided by (n-1), in this case, 11, giving me a sample variance of 0.001904488. Then I take the square root of the sample variance to get the sample standard deviation, which is 0.04364044, or around 4.4%.\n\nWith the mean and standard deviation in hand, I can now calculate the monthly Sharpe Ratio. It's the mean return divided by the standard deviation. This gives me a monthly Sharpe Ratio of 0.484451.\n\nFinally, I need to annualize this figure to be useful. I'll multiply the monthly Sharpe by the square root of 12 (since there are 12 months in a year). The annualized Sharpe Ratio is approximately 1.6781. This is a reasonable Sharpe Ratio, indicating a good risk-adjusted return over the period. I'd need more context to evaluate this fully, but that's the bottom line from the numbers.\n\n\n\n```json\n{\"answer\": 1.678, \"confidence\": 99}\n```" }, { "taskId": "sharpe_annualized-2", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": -0.48328609065675004, "answer": -0.4833, "confidence": 98, "grade": "exact", "relError": 0.00002878076468342148, "cost": 0.004806, "reply": "**Analyzing the Performance Data**\n\nOkay, let's break this down. I've got a vector of returns, `x`, and I need to calculate the Sharpe ratio. First, I'm going to sum these returns to get a sense of the overall performance. Let me do that step-by-step to be careful; I see the intermediate sums calculated, so I'll follow those. The sum looks like it ends up being -0.0654. Simple enough.\n\nNow, to get the average, or mean monthly return, I'll divide that sum by the number of data points, 12, to get -0.00545. That gives me a basic measure of central tendency. Not looking great so far, but we need to quantify the risk too.\n\nTo calculate the standard deviation, which will be the denominator in the Sharpe ratio, I need to do a few more calculations. I see that the code moves into summing the squares of the returns, but with values multiplied by 10^8 - I wonder why that scale is applied. Okay, I'm noting the squares are calculated individually: I'm seeing the individual squared returns are given as 224676, 141376, 5476, 231361, 293764, 9409, 15129, 385641, 246016, 127449, 9025, and 24964. These are each summed by hand too, and now I'm able to get the total of 1714286, which, after the scale, gives us a total of 0.01714286.\n\nNext, I need to subtract the square of the sum of the returns divided by the number of returns from the sum of the squared returns. That gives me the sum of squares around the mean. Okay, let's see, I need to square -0.0654, which yields 0.00427716. I divide that by 12, giving me 0.00035643. Subtracting that from 0.01714286, I get a sum of squares of 0.01678643.\n\nNow I can get the sample variance, s^2, which is the sum of squares divided by n-1. So, that's 0.01678643 divided by 11, which is 0.001526039. Finally, the standard deviation, s, is the square root of that, which comes out to 0.03906455.\n\nNow I have the ingredients for the Sharpe ratio. The monthly Sharpe ratio is the mean monthly return (-0.00545) divided by the monthly standard deviation (0.03906455), which equals -0.1395127.\n\nFinally, to annualize this, I multiply the monthly Sharpe ratio by the square root of 12 (approximately 3.4641016), resulting in an annualized Sharpe ratio of -0.483286. So, the portfolio is underperforming, with negative performance and returns that are low relative to the risk.\n\n\n\n{\"answer\": -0.4833, \"confidence\": 98}" }, { "taskId": "sharpe_annualized-3", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 2.241475231769422, "answer": 2.2415, "confidence": 98, "grade": "exact", "relError": 0.0000110499684433581, "cost": 0.005861, "reply": "**Analyzing the Return Series**\n\nOkay, let's break this down. I've been given a series of returns, r1 through r12. My immediate goal is to understand their basic characteristics and derive some key metrics.\n\nFirst, I need to get a sense of the central tendency. I'll start by calculating the sum of the returns. I can break it down logically by summing the positive terms, then the negative ones, and then computing the total. The sum comes out to be 0.2742. Dividing that sum by the number of observations (12), I immediately find the mean monthly return which is 0.02285, or 2.285%.\n\nNow, I need to assess the risk, specifically the volatility. I'll calculate the standard deviation. First I need to compute the sum of squared returns. I meticulously go through squaring each return value and summing them. Next, I need to take that, calculate the square of the sum of the returns, divide that by 12, and then calculate the sum of squares, minus that value from the calculation, and finally, divide that by 11 to get the variance. The variance is roughly 0.001247. Taking the square root gives us the standard deviation, s = 0.03531367, or 3.53%.\n\nWith the mean and standard deviation in hand, I can now calculate the Sharpe Ratio. I'm going to start with the monthly Sharpe ratio, by dividing the monthly mean by the monthly standard deviation, which gives me 0.647058. Then, I need to annualize this, which I do by multiplying the monthly Sharpe ratio by the square root of 12. This yields an annualized Sharpe ratio of 2.2415. The portfolio seems to have performed well considering the Sharpe ratio is greater than 1.\n\n\n\n{\"answer\": 2.2415, \"confidence\": 98}" }, { "taskId": "sharpe_annualized-4", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 1.9830100176589147, "answer": 1.983, "confidence": 99, "grade": "exact", "relError": 0.000005051743977796405, "cost": 0.004961, "reply": "**Sharpe Ratio Calculation**\n\nOkay, let's break this down step-by-step. I'm given a set of 12 monthly returns, which I'll call `x`. The first thing to do is calculate the sum and the mean. Summing them up gives a total of 0.3108, which, when divided by 12, gives a mean monthly return of 0.0259, or 2.59%.\n\nNext, I need to calculate the standard deviation. To do this, I need to compute the sum of squares of the monthly returns. That means squaring each return and summing those squared values. I get a sum of squares of 0.03056740. Now, I need to calculate the sample variance. I could also directly calculate the sum of squares from the mean but I'll continue as I have. First, I square the mean return and multiply by the number of samples: `n * mean^2`. This is 12 * (0.0259)^2 = 0.00804972. I can subtract this from the sum of squares to get `SS = 0.03056740 - 0.00804972 = 0.02251768`.\n\nNow, the sample variance is calculated by dividing the sum of squares `SS` by `n-1`, which in this case is 11. So the sample variance is 0.02251768 / 11 = 0.00204706. The sample standard deviation is then the square root of the variance: sqrt(0.00204706) = 0.04524447, which is 4.52%.\n\nFinally, I calculate the Sharpe Ratio. This is simply the mean return divided by the standard deviation. So, 0.0259 / 0.04524447 = 0.5724456. Since this is a monthly value, I need to annualize it. I do this by multiplying the monthly Sharpe Ratio by the square root of 12 (the number of months in a year), which is roughly 3.4641016. The annual Sharpe Ratio is then 0.5724456 * 3.4641016 = 1.9830. So, the annualized Sharpe Ratio for this data set is approximately 1.98. This indicates a good risk-adjusted return, but the significance would depend on the benchmark.\n\n\n\n{\"answer\": 1.983, \"confidence\": 99}" }, { "taskId": "sharpe_annualized-5", "type": "sharpe_annualized", "tier": "spec", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 1.3915400475486268, "answer": 1.3915, "confidence": 98, "grade": "exact", "relError": 0.000028779300098047822, "cost": 0.006385, "reply": "**My Analysis of the Returns**\n\nAlright, let's break this down. First, I'm given a set of returns: 0.0603, -0.0347, 0.0392, and so on, twelve in total. My initial thought is to calculate the sum of these returns, which I meticulously calculate step-by-step. The sum of these returns comes out to be 0.1695. Then, I calculate the mean by dividing the sum of the returns (0.1695) by the number of returns (12), which results in a mean of approximately 0.014125.\n\nNext, I shift focus to calculating the standard deviation. Since the mean is 0.014125, which can be thought of as 141.25 when we consider the original values multiplied by 10^4. I compute the deviations of each return from the mean, and then square each of these deviations. These values are meticulously calculated and stored.\n\nAfter squaring each deviation, I sum up all of the squared deviations. This sum of squares is crucial for calculating the variance.\n\nI divide the sum of squares by *n-1*, which is 11 in this case (since there are 12 data points), to get the sample variance, and after calculating this the variance I arrive at approximately 123642.2045 (in 10^-8). I take the square root of the variance to get the standard deviation. Doing so, I obtain a standard deviation of approximately 0.0351628.\n\nFinally, I calculate the Sharpe ratio. The monthly Sharpe ratio is the mean return divided by the standard deviation. Thus, the ratio becomes approximately 0.401703.\n\nTo annualize this, I multiply the monthly Sharpe by the square root of 12. Ultimately, I obtain an annualized Sharpe ratio of roughly 1.3915.\n\n\n\n```json\n{\"answer\": 1.3915, \"confidence\": 98}\n```" }, { "taskId": "kelly_binary-1", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-3.8-flash", "expected": 0.4848484848484847, "answer": 0.4848, "confidence": 100, "grade": "exact", "relError": 0.00009999999999966636, "cost": 0.001895, "reply": "**Kelly Criterion for This Binary Bet**\n\nOkay, let's break this down using the Kelly criterion. I'm looking at a binary bet and need to figure out the optimal fraction of my capital to wager. The standard Kelly formula for a binary bet is:\n\n$f^* = \\frac{p \\cdot b - q}{b}$.\n\nWhere:\n* `p` is the probability of winning. In this case, `p` equals 0.83.\n* `q` is the probability of losing, which is 1 - p. So, `q` = 0.17.\n* The contract price, `c`, is 0.67.\n* If I bet `c`, net profit will be 1 - c = 0.33.\n* Odds `b = (1 - c)/c = 0.33/0.67`\n\nI can rearrange the formula to:\n\n`f* = (p - c) / (1 - c) = (0.83 - 0.67)/(1 - 0.67) = 0.16/0.33`\n\nThus, `f* = 16/33`.\n\nNow, let's double-check this to make sure I'm following the correct logic. I can also express the formula as:\n\n`f* = (p(b+1) - 1)/b = (p/c - 1)/((1-c)/c) = (p - c)/(1 - c)`\n\nSo yes, the optimal fraction is `f* = (p - c)/(1 - c)`.\n\nTherefore, 16/33 is the correct solution. It's approximately 0.484848... which means I should bet roughly 48.5% of my capital on this.\n\n\n\n```json\n{\"answer\": 0.4848, \"confidence\": 100}\n```" }, { "taskId": "kelly_binary-2", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-3.8-flash", "expected": 0.28985507246376807, "answer": 0.2899, "confidence": 100, "grade": "exact", "relError": 0.0001550000000001245, "cost": 0.002581, "reply": "**Kelly Criterion Calculation Verification**\n\nOkay, let's break this down methodically. I'm given a binary contract with a cost of $0.31 and a payout of $1. That means the net odds *b* are (1-0.31)/0.31, or roughly 0.69/0.31. The probability of winning *p* is 0.51, and therefore, the probability of losing *q* is 1-0.51 = 0.49.\n\nNow, applying the Kelly fraction formula, *f*^*, I can calculate it in a couple of ways. The first way, *f*^* = (b*p - q)/b, gives me *f*^* = p - q/b. Substituting the values, I compute 0.51 - (0.49 * 0.31) / 0.69. Multiplying 0.49 by 0.31, I get 0.1519. Dividing that by 0.69, I arrive at roughly 0.220145. Subtracting that from 0.51 gives me approximately 0.289855.\n\nHowever, I also have the shortcut *f*^* = (p - c) / (1 - c). Let me double-check this. Plugging in the same values, that's (0.51 - 0.31) / (1 - 0.31) which works out to 0.20 / 0.69. This results in the same approximate value of 0.289855.\n\nWait a second, is the shortcut really correct? Let's verify by expanding the standard Kelly fraction formula, starting from (b*p - q)/b. That equals p - q/b. Because *b* = (1-c)/c, and *q* = (1-p), I can substitute that giving p - ((1-p)*c)/(1-c) which can then be manipulated. Putting it all over the common denominator, that results in (p*(1-c) - c + p*c)/(1-c). Simplifying, this does, in fact, reduce to (p - c) / (1 - c). Perfect, this confirms the shortcut method is valid!\n\nSo, the Kelly fraction for this contract, as I’ve calculated it both ways, is about 0.289855, or 20/69. I can move forward with confidence, using this result to determine optimal bet sizing.\n\n\n\n```json\n{\"answer\": 0.2899, \"confidence\": 100}\n```" }, { "taskId": "kelly_binary-3", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-3.8-flash", "expected": 0.09859154929577463, "answer": 0.09859, "confidence": 100, "grade": "exact", "relError": 0.000015714285714109126, "cost": 0.002513, "reply": "**Analyzing the Kelly Criterion for a Binary Prediction Market**\n\nOkay, let's break this down. I'm looking at a binary prediction market contract. The setup is straightforward: it costs $0.29, pays out $1 if the event happens, and $0 if it doesn't. My potential net profit is $0.71, and my potential loss is the initial $0.29. The model estimates a 36% probability of winning (p = 0.36), which means a 64% chance of losing (q = 0.64).\n\nNow, to determine the optimal bet size using the Kelly Criterion, I need to calculate the odds. Net odds (b) are calculated as net profit divided by the amount risked, which equals $(1 - 0.29) / 0.29 = 0.71 / 0.29$.\n\nThere are two ways to get to the Kelly fraction (f*). The first is the more general formula: f* = (b*p - q) / b. Plugging in the numbers, I end up with [(0.71 / 0.29) * 0.36 - 0.64] / (0.71 / 0.29). Simplifying this, I get [0.2556 - 0.1856] / 0.29 divided by 0.71/0.29 which gives me 0.07 / 0.71.\n\nAlternatively, for a binary contract, there's a more streamlined version: f* = (p - c) / (1 - c), where 'c' is the contract price. Here, c = 0.29, and p = 0.36. So, f* = (0.36 - 0.29) / (1 - 0.29) = 0.07 / 0.71. This gives me approximately 0.09859.\n\nSo, in both formulations, I arrive at the same answer. The optimal Kelly fraction to use is about 0.07 / 0.71. I can now evaluate how much of my capital I should allocate to this trade.\n\n\n\n```json\n{\"answer\": 0.09859, \"confidence\": 100}\n```" }, { "taskId": "kelly_binary-4", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-3.8-flash", "expected": 0.8333333333333334, "answer": 0.8333, "confidence": 100, "grade": "exact", "relError": 0.00003999999999999559, "cost": 0.00221, "reply": "**Kelly Criterion Calculation - A Quick Check**\n\nOkay, let's see, I'm quickly running through a Kelly criterion calculation for a binary bet to confirm my mental math. The cost is 0.16, and the net payoff per unit invested, *b*, is (1 - 0.16) / 0.16, or 5.25. The probability of a win is given as 0.86, and naturally, that means the loss probability, *q*, is 0.14.\n\nNow, I'm thinking about the standard Kelly formula: *f*^* = (b*p - q) / b. Plugging in the numbers, I get: 0.86 - 0.14/5.25. Converting 5.25 to 21/4, it's 0.86 - 0.56/21. Simplifying further, this gets me to roughly 0.86 - 0.08/3 which is about 0.8333.\n\nWait a second, there's a simpler alternative formula specific to this binary case, right? It's *f*^* = (p - c) / (1 - c). Let's try that one! That should be (0.86 - 0.16) / (1 - 0.16), or 0.70 / 0.84 which becomes 70/84 which simplifies to 5/6. Yep, that's the same, 0.833333...\n\nExcellent, both approaches yield the same result, and it confirms the initial calculation. Confidence is high here. 5/6 looks to be the optimal fraction to wager, as expected.\n\n\n\n```json\n{\"answer\": 0.8333, \"confidence\": 100}\n```" }, { "taskId": "kelly_binary-5", "type": "kelly_binary", "tier": "spec", "domain": "prediction", "model": "gemini:gemini-3.8-flash", "expected": 0.576271186440678, "answer": 0.5763, "confidence": 100, "grade": "exact", "relError": 0.00005000000000002715, "cost": 0.00284, "reply": "**Kelly Criterion Derivation and Calculation**\n\nAlright, let's break this down. I need to figure out the optimal bet sizing using the Kelly criterion for this binary bet. The basic formula is: $f^* = \\frac{p \\cdot b - q}{b}$. I see that $p$ is the probability of winning (0.75 in this case) and $q$ is the probability of losing, which is simply $1-p$ (so 0.25). The variable '$b$' represents the net fractional odds. For this specific case, it's calculated as (payout - cost) / cost, which translates to (1 - 0.41) / 0.41. Simplifying that, it comes out to 0.59 / 0.41.\n\nNow, wait a minute, I remember an alternative representation of the Kelly criterion: $f^* = \\frac{p}{c} - \\frac{q}{1-c}$. Is that right? Ah, hold on. Let's derive this from scratch to make sure. I'll need to define some variables. Let's say my bankroll is $X$ and I'm staking a fraction $f$.\n\nIf I win, my bankroll becomes $X(1 - f + f/c)$, and if I lose, it's $X(1 - f)$. Now, to get the expected log return, I need to compute $E[\\ln(X_1/X_0)]$. This looks like $p \\ln(1 + f(1-c)/c) + (1-p) \\ln(1 - f)$. Okay, that's the setup. To optimize, I need to take the derivative with respect to $f$ and set it to zero.\n\nThe derivative is: $p \\frac{(1-c)/c}{1 + f(1-c)/c} - \\frac{1-p}{1-f} = 0$. Let's simplify this by setting $b = (1-c)/c$. That means: $p b (1-f) = (1-p)(1 + b f)$. Now, let's expand and solve for $f$: $p b - p b f = 1 - p + b f - p b f$, which simplifies to $b f = p b - (1 - p)$. So, $b f = p(b+1) - 1$. Since $b+1 = 1/c$, then $b f = p/c - 1 = (p - c)/c$. Now, since $b = (1-c)/c$, then $f = \\frac{(p - c)/c}{(1-c)/c} = \\frac{p - c}{1 - c}$. Bingo. We've derived it. Now, it makes sense.\n\nLet's plug in the numbers to check: $p = 0.75$ and $c = 0.41$. So $f = \\frac{0.75 - 0.41}{1 - 0.41} = \\frac{0.34}{0.59}$. Doing the math... that's approximately 0.576. So, the optimal stake fraction for this bet is about 57.6%. Now, this seems right.\n\n\n\n```json\n{\"answer\": 0.5763, \"confidence\": 100}\n```" }, { "taskId": "inverse_liquidation-1", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 24920.824742268043, "answer": 24920.82, "confidence": 99, "grade": "exact", "relError": 1.9029338285041182e-7, "cost": 0.006623, "reply": "**Analyzing the Liquidation Price for a Short BTCUSD Inverse Perpetual**\n\nOkay, let's break this down systematically. I'm short a BTCUSD perpetual contract. My entry was at $P_0 = 24173.2$ with 25x leverage. I'm holding a position size of $Q$ USD. My initial thinking was straightforward. At entry, the position value in BTC is $V_0 = Q / P_0$. Initial margin is $IM = 0.04 V_0$ and maintenance margin is $MM = 0.01 V_0$.\n\nNow, the core of the problem: what price triggers liquidation? For an inverse contract, the PnL calculation is a bit tricky, but I know it's PnL is in BTC. If the price moves to $P$, the position value is $V = Q / P$. The PnL, as initially calculated, was $Q / P_0 - Q / P = V_0 - Q / P = V_0 (1 - P_0 / P)$. I jumped to the conclusion that liquidation would occur when $IM + PnL = MM$, and solved the equation $0.04 V_0 + V_0 (1 - P_0 / P) = 0.01 V_0$, which yielded $P = P_0 / 1.03$. But, wait a minute! Is that right? I shorted, so the price going UP means I lose money.\n\nIt means the initial PnL calculation was incorrect. I have to think more carefully about how the inverse contract works. I'm selling USD worth of BTC at $P_0$ which is $Q/P_0$ BTC. To close the short, I have to buy back $Q$ USD worth of BTC at $P$. That costs $Q/P$ BTC. So PnL = $Q / P_0 - Q / P$.\n\nIf $P > P_0$, then $Q/P < Q/P_0$, so $Q/P_0 - Q/P > 0$? That means positive PnL when price goes up. Something is not right.\n\nOkay, let's step back. Contract size is 1 USD. If I'm short $N$ contracts, then PnL = $N \\times (1/P_{\\text{entry}} - 1/P_{\\text{exit}})$?\n\nNow, I understand, in an inverse short, when price goes up I lose. Let's work out PnL. If I short 1 BTC at 20,000, that's 20,000 contracts. If the price doubles to 40,000 USD, I lose 20,000 USD, but that is 0.5 BTC. Thus PnL is $-(Q/P - Q/P_0)$. Is that it?\n\nOkay, I have it. The profit from a short position comes from a fall in price, the formula for inverse short PnL: PnL = $Q \\times (1/P_{\\text{exit}} - 1/P_{\\text{entry}})$ and thus PnL in BTC = $Q \\times (1/P - 1/P_0)$ and it's positive if price goes down and vice versa. Then, for the long, PnL is $Q \\times (1/P_0 - 1/P)$, and that's the base to check against.\n\nLet's check the short position: Price rises from 20,000 to 40,000, then PnL = $Q \\times (1/40000 - 1/20000) = - Q / 40000 < 0$. Losses! Correct!\n\nTherefore, PnL = $Q(1/P - 1/P_0) = V_0 (P_0/P - 1)$. As $P \\to \\infty$, loss approaches $-V_0$.\n\nLet's re-solve the liquidation condition: $IM + PnL = MM$ so $0.04 V_0 + Q(1/P - 1/P_0) = 0.01 V_0$, which simplifies to $0.04 + P_0/P - 1 = 0.01$, and that leads to $P_0/P = 0.97$. This gives $P = P_0 / 0.97$, where $P > P_0$, as expected for a short liquidation. At this point the position PnL in BTC =\n\n\n```json\n{\"answer\": 24920.82, \"confidence\": 99}\n```" }, { "taskId": "inverse_liquidation-2", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 53342.9702970297, "answer": 53342.97, "confidence": 99, "grade": "exact", "relError": 5.5683007055427045e-9, "cost": 0.002855, "reply": "**Inverse Perpetual Contract Liquidation Analysis**\n\nOkay, let's break this down. I'm looking at an inverse perpetual contract, and I need to understand the liquidation mechanics. The entry price is $P_0$, and the leverage is *L*. The position size is *Q* USD, and I need to find the liquidation price.\n\nFirst, the initial position value in BTC, *V0*, is simply the position size in USD divided by the entry price. The initial margin, *M*, is *V0* divided by the leverage, meaning I only need to put up a fraction of the total position value. The maintenance margin, *MM*, is lower than the initial margin.\n\nNow, for the key part: the profit and loss (PnL). This is an inverse contract, so the payoff is in BTC, not USD. When the price of BTC rises, my long position should profit. The contract size is in USD, so the payoff in BTC is related to how the BTC/USD rate changes. Let's see... a long position means I'm effectively buying BTC with USD, so I profit if BTC goes up. Thinking about it, the PnL should be the difference in the inverse price. If *P* goes up, then *1/P* will go down, but I am in a long contract. I should be long *1/P*. Yes! The payoff in BTC is correctly calculated as *Q x (1/P0 - 1/P)*.\n\nNow, the liquidation condition. Liquidation happens when the margin plus the PnL equals the maintenance margin. So:\n\n*M + PnL = MM*\n\nWhich is:\n\n*0.02V0 + Q(1/P0 - 1/P) = 0.01V0*\n\nSince *V0 = Q / P0*, and I can divide all parts of the equation by *Q* (since *Q* cannot be zero).\n\nThen, after some algebra:\n*1 / P = 1.01 / P0*\nSo, the liquidation price, *P*, is *P0 / 1.01*.\n\nLet's plug in the numbers to check:\n\n*P = 53876.4 / 1.01 = 53342.97...*\n\nTherefore, if the price drops to approximately $53,342.97, my position gets liquidated. That's the key.\n\n\n\n```json\n{\"answer\": 53342.97, \"confidence\": 99}\n```" }, { "taskId": "inverse_liquidation-3", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 53194.15525114154, "answer": 53194.16, "confidence": 99, "grade": "exact", "relError": 8.927406475571944e-8, "cost": 0.00349, "reply": "**Inverse Long Position Liquidation Analysis**\n\nOkay, let's break this down systematically. I'm looking at a long inverse position, and need to confirm the liquidation price. The key here is the inverse contract; it's priced in USD, but settled in Bitcoin, which makes the PnL calculation a little counterintuitive if you're not careful.\n\nFirst, I define my variables: Entry price ($P_0$) is 58247.6, leverage ($L$) is 10, position size is $Q$ (in USD), and the position's value in BTC at entry is $Q / P_0$. The initial margin ($M$) is then calculated as $(Q / P_0) / L$, or equivalently $Q / (P_0 * L)$.\n\nThe PnL calculation for a long inverse is where I have to be precise: it's not simply the difference in prices. I have to express it as the change in the inverse of the price. The proposed PnL formula is $\\text{PnL} = Q \\left( \\frac{1}{P_0} - \\frac{1}{P} \\right)$. I have to confirm this is correct and I do this by working it out by hand and checking against an example.\n\nIf the price goes up ($P > P_0$), the position should profit. With an example where $P_0 = 50$, and $P = 100$, the PnL $= Q(1/50 - 1/100) = Q(0.01)$, which is indeed positive. This checks out; the PnL increases as the price goes up.\n\nNow, for the liquidation condition. It's when the margin plus the PnL falls to the maintenance margin level (MM). I'm using an MM of 0.5%.\n\nThe initial margin ($M$) is $Q / (10 * P_0)$. The maintenance margin is $0.005 * (Q / P_0)$. The liquidation condition is therefore:\n\nInitial Margin + PnL = Maintenance Margin, or:\n\n$\\frac{Q}{10 P_0} + Q \\left( \\frac{1}{P_0} - \\frac{1}{P} \\right) = 0.005 \\frac{Q}{P_0}$.\n\nI divide by $Q$ to get:\n$\\frac{1}{10 P_0} + \\frac{1}{P_0} - \\frac{1}{P} = \\frac{0.005}{P_0}$.\n\nThis simplifies to:\n$\\frac{1.1 - 0.005}{P_0} = \\frac{1}{P}$.\n$\\frac{1.095}{P_0} = \\frac{1}{P}$\n\nAnd finally, to solve for P, the liquidation price:\n$P = \\frac{P_0}{1.095} = \\frac{58247.6}{1.095} \\approx 53194.155$\n\nSo, the liquidation price should be around 53194.155. Now I've confirmed the formula and found the liquidation level for this specific position.\n\n\n\n```json\n{\"answer\": 53194.16, \"confidence\": 99}\n```" }, { "taskId": "inverse_liquidation-4", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 170355.2216748768, "answer": 170355, "confidence": 99, "grade": "exact", "relError": 0.0000013012508488557827, "cost": 0.00804075, "reply": "**Liquidation Price Calculation for a Short Inverse Contract**\n\nOkay, so I'm dealing with a short position in an inverse BTCUSD contract, and I need to figure out the liquidation price. The entry price is $P_0 = 115273.7$, leverage $L = 3$, and the maintenance margin rate $MMR = 0.01$. I'm thinking in terms of contract size, let's call it $Q$ USD.\n\nFirst, I need to define my initial variables. The position value at entry in BTC is $V_0 = Q / P_0$. The initial margin $IM$ is then $V_0 / L = Q / (L \\cdot P_0)$. The maintenance margin $MM$ is calculated as $MMR \\cdot V_0 = MMR \\cdot Q / P_0$.\n\nThe crucial part here is the unrealized PnL for a short position. Since it's an inverse contract, the standard PnL formula is flipped. A short contract profits as the price drops. After a bit of mental juggling, and considering the contract specifications, I conclude the PnL should be $PnL = Q \\cdot (1/P - 1/P_0)$.\n\nLiquidation happens when the initial margin plus PnL equals the maintenance margin: $IM + PnL = MM$. Substituting the expressions for $IM$, $PnL$, and $MM$, I get $Q / (L \\cdot P_0) + Q \\cdot (1/P - 1/P_0) = MMR \\cdot Q / P_0$. Now, divide everything by $Q$, and I have $1 / (L \\cdot P_0) + 1/P - 1/P_0 = MMR / P_0$.\n\nMy goal is to solve for $P$, the liquidation price. I rearrange the equation and get $1/P = 1/P_0 + MMR / P_0 - 1 / (L \\cdot P_0) = (1/P_0) \\cdot [1 + MMR - 1/L]$. Now plug in the numbers. I have $1/P = (1/115273.7) \\cdot [1 + 0.01 - 1/3]$. This simplifies to $1/P = (1/115273.7) \\cdot [1.01 - 0.33333333] = (1/115273.7) \\cdot [0.676666667]$. Therefore, $P = P_0 / (1 + MMR - 1/L)$.\n\nPlugging in the known numbers, I see $P = 115273.7 / (1.01 - 1/3) = 115273.7 / (0.676666667)$, so $P = 170355.22$\n\nI am still a bit concerned about the PnL equation, but after reviewing the inverse contract mechanics, it should hold. Then, I double-check my work. I re-calculate the liquidated position with the new price, and confirm. \n\n\n\n```json\n{\"answer\": 170355, \"confidence\": 99}\n```" }, { "taskId": "inverse_liquidation-5", "type": "inverse_liquidation", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 115003.24607329845, "answer": 115003.2, "confidence": 99, "grade": "exact", "relError": 4.006260694662902e-7, "cost": 0.005455, "reply": "**Analyzing the Liquidation Price for a Short Inverse Perpetual Contract**\n\nOkay, so I'm short an inverse perpetual with an entry at $109828.1 and 20x leverage. Let's see... the initial setup: I've got a position size of *Q* USD contracts. The position value at entry in BTC is *Q* divided by the entry price. The initial margin is that value divided by the leverage, and the maintenance margin is a percentage (0.5%) of the position value.\n\nRight, the key is the unrealized P&L. I had a quick moment of confusion but I sorted it out. For a short position, the P&L in BTC is going to be size times the *change* in the inverse of the price: *Q* multiplied by the quantity (1/*P* - 1/*P_e*). If the price *goes up* (*P* > *P_e*), this is bad and negative P&L. The position is good if the price goes down. The critical point is the liquidation condition, and it occurs when Initial Margin + PnL equals Maintenance Margin. This lets me determine what the liquidation price needs to be.\n\nPlugging this into the liquidation equation and dividing by *Q*, I get a relationship for the inverse of the liquidation price (1/*P_liq*). Some algebra leads me to 1/*P_liq* equals 1/*P_e* multiplied by the quantity 1 - (1/*L*) + *MMR*. Doing the arithmetic, 1 - (1/20) + 0.005 is equal to 0.955. So my final liquidation price equation is *P_liq* = *P_e* / 0.955. I just have to plug in the values and calculate the results.\n\nOkay, let's plug in the entry price. I get a liquidation price of about $115003.25.\n\n\n\n```json\n{\"answer\": 115003.2, \"confidence\": 99}\n```" }, { "taskId": "inverse_avg_entry-1", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 2398.0256188582507, "answer": 2397.97, "confidence": 100, "grade": "exact", "relError": 0.000023193604694428027, "cost": 0.0029145, "reply": "**Calculating Average Entry Price for Inverse Contracts**\n\nOkay, let's break this down. I'm looking at inverse contract PnL in ETH. The key is remembering the formula for long positions: `PnL = contracts * (1/entry - 1/exit)`. Which also means that the total ETH value at entry is calculated by summing `C_i / P_i`. My goal here is to calculate the average entry price, represented as `P̄`, which is found by dividing total contracts `∑C_i` by the sum of `C_i / P_i`.\n\nNow, let's put it into practice with the given data. I see the total contracts across several trades: 7000, 34000, 36000, and 9000. Summing these, I get a total of 86000 contracts, which is our numerator.\n\nNext, I need the `C_i / P_i` values. I've got the contract size and the entry price for each trade, so let's compute those:\n\n1. 7000 / 2309.36 ≈ 3.031143\n2. 34000 / 3563.54 ≈ 9.541074\n3. 36000 / 1854.34 ≈ 19.413915\n4. 9000 / 2321.56 ≈ 3.876704\n\nThen, I summed these results:\n\n3.031143 + 9.541074 = 12.572217\n12.572217 + 19.413915 = 31.986132\n31.986132 + 3.876704 = 35.862836\n\nNow, for calculating the Average entry price, I divide the total contracts by the total value of `C_i / P_i`, to get `86000 / 35.862836 ≈ 2397.97`. But I want to be precise, so I’ll re-calculate, making sure to avoid rounding errors:\n\n1. 7000 / 2309.36 = 3.0311432\n2. 34000 / 3563.54 = 9.5410743\n3. 36000 / 1854.34 = 19.4139155\n4. 9000 / 2321.56 = 3.8767036\nSum = 35.8628366\nFinally, `86000 / 35.8628366 = 2397.9698...` confirming my result. The average entry price is approximately 2397.97. This gives us a good base value for making future predictions.\n\n\n\n{\"answer\": 2397.97, \"confidence\": 100}" }, { "taskId": "inverse_avg_entry-2", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 3712.5303509582454, "answer": 3712.53, "confidence": 100, "grade": "exact", "relError": 9.453343462758414e-8, "cost": 0.002512, "reply": "**Calculating the Average Entry Price for an Inverse Perpetual Contract**\n\nAlright, let's break this down. We're dealing with an inverse perpetual contract, and I need to figure out the average entry price, which is key for understanding the PnL. The PnL in ETH is derived from the formula: `Position_USD * (1/Entry_Price - 1/Exit_Price)` for a long position. Therefore, I need to get the average entry price, which I'm calling $\\bar{P}$.\n\nTo find $\\bar{P}$, I'm using the relationship where `Total_USD / $\\bar{P}$ = $\\sum (USD_i / P_i)$`. This implies that $\\bar{P}$ is calculated as the total USD divided by the sum of USD at each entry price divided by the corresponding price. In formula form, that's $\\bar{P} = \\frac{\\sum USD_i}{\\sum (USD_i / P_i)}$.\n\nOkay, let's plug in the given values. We have two positions, with USD\\_1 = 34000 at P\\_1 = 3528.19 and USD\\_2 = 16000 at P\\_2 = 4176.2. The total USD is just the sum of these, so 34000 + 16000 = 50000.\n\nNow for the calculations. Dividing USD\\_1 by P\\_1 gives me 34000 / 3528.19 = 9.6366692. Then, USD\\_2 / P\\_2 is 16000 / 4176.2 = 3.8312341. Adding these together, I have a total ETH base value of 9.6366692 + 3.8312341 = 13.4679033.\n\nFinally, $\\bar{P}$ = 50000 / 13.4679033 ≈ 3712.5303. To ensure this calculation is correct, I'll quickly double-check: multiplying the price by the previously-derived ETH amounts should come very close to the USD amounts, and it does. So, I'm confident my average entry price for the inverse perpetual is 3712.5303.\n\n\n\n```json\n{\"answer\": 3712.53, \"confidence\": 100}\n```" }, { "taskId": "inverse_avg_entry-3", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 2495.29829080704, "answer": 2495.3, "confidence": 99, "grade": "exact", "relError": 6.849653873569959e-7, "cost": 0.00439725, "reply": "**Analyzing the Average Entry Price of Inverse Contracts**\n\nOkay, so I see a trader has entered some inverse contracts. My immediate thought is to calculate the average entry price for these positions. We're dealing with inverse perpetuals, so the PnL is calculated in coin (ETH) using the formula Contract_Qty * (1/Entry - 1/Exit). The combined PnL is the sum of these individual contract PnLs. We want to find the average entry price.\n\nI know that the total PnL can also be expressed as Total_Qty * (1/Avg_Entry - 1/Exit), which, when expanded, equals the sum of each individual contract. Therefore, Total_Qty / Avg_Entry = sum(Qty_i / Entry_i). And so, the average entry price, Avg_Entry, is Total_Qty / sum(Qty_i / Entry_i). This means it's the harmonic mean of the entry prices, weighted by the contract USD size.\n\nNow, let's look at the specific contracts. I see the trader bought:\n\n1. $41,000 at 3517.25\n2. $10,000 at 2976.86\n3. $27,000 at 4229.06\n4. $40,000 at 1545.09\n\nThe total quantity, Total_Qty, is $118,000 (41000 + 10000 + 27000 + 40000).\n\nNow, to calculate the sum of ETH amounts (Qty_i / Entry_i), I need to do the following:\n\n1. 41000 / 3517.25 = 11.656834 ETH\n2. 10000 / 2976.86 = 3.359244 ETH\n3. 27000 / 4229.06 = 6.384397 ETH\n4. 40000 / 1545.09 = 25.888459 ETH\n\nThe sum is approximately 47.288934 ETH (11.656834 + 3.359244 + 6.384397 + 25.888459).\n\nNow, I calculate the average entry price: 118000 / 47.288934 = 2495.298...\n\nJust to be absolutely certain, let me recalculate a few of those divisions to make sure there are no rounding errors, even though my confidence is high at this point. Recalculating the same calculations gets me a total of 47.288936, for an average entry price of $2495.30. There it is.\n\n\n\n{\"answer\": 2495.3, \"confidence\": 99}" }, { "taskId": "inverse_avg_entry-4", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 2610.72093572179, "answer": 2610.72, "confidence": 100, "grade": "exact", "relError": 3.584150941701721e-7, "cost": 0.003437, "reply": "**Calculating Average Entry Price in Inverse Contracts**\n\nOkay, let me break down this PnL calculation for inverse contracts. It's crucial to understand how to determine the average entry price, which is essential for accurate profit and loss assessment. I'm focusing on a long position here.\n\nThe core idea is that the PnL in ETH is derived from the contract value multiplied by the difference between the inverse of the entry price and the inverse of the exit price. Extending this, the total PnL is the sum of these individual gains. Through some manipulation, I can arrive at the following. The sum of the USD amount for each trade, divided by its price, is equal to the sum of the USD amount divided by the average price. From here, it becomes clear that the average price is the sum of all USD amounts divided by the sum of each USD amount divided by its price. This confirms that I'm dealing with a harmonic mean, weighted by the USD amounts involved in each fill.\n\nNow, let's look at the specific fills I have. I have four trades:\n\n1. 10000 USD at 3282.44\n2. 14000 USD at 2147.97\n3. 12000 USD at 1744.23\n4. 23000 USD at 3736.80\n\nI need to calculate the average entry price based on these fills. First, I compute the ETH value for each fill by dividing the USD amount by the price. Then, I sum the total USD traded and the total ETH involved. The total USD is 59000. Summing the ETH amounts, I got around 22.599121 ETH. I then find the average price by dividing total USD by total ETH.\n\nHowever, I want precision, so let me do the calculations again with a bit more accuracy. After recalculating, I have:\n\n1. 10000 / 3282.44 = 3.04651433 ETH\n2. 14000 / 2147.97 = 6.51778191 ETH\n3. 12000 / 1744.23 = 6.87982663 ETH\n4. 23000 / 3736.80 = 6.15499893 ETH\n\nSumming these ETH values gives me 22.59912180 ETH, so my average entry price, P_avg, is 59000 / 22.59912180, which is approximately 2610.7209. This is the precise harmonic mean weighted by USD amounts. I'm satisfied.\n\n\n\n```json\n{\"answer\": 2610.72, \"confidence\": 100}\n```" }, { "taskId": "inverse_avg_entry-5", "type": "inverse_avg_entry", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 3424.594098647905, "answer": 3424.6, "confidence": 100, "grade": "exact", "relError": 0.0000017232267313904394, "cost": 0.003578, "reply": "**Calculating the Average Entry Price for Inverse Contracts**\n\nOkay, let me break this down. For these inverse contracts, which are USD-denominated and settled in the native coin, the profit and loss (PnL) calculation is straightforward. It's essentially the position size in USD multiplied by the difference between the inverse of the entry price and the inverse of the exit price.\n\nNow, if I want to have the same PnL regardless of the exit price, I need a single average entry price. This means that the total USD value divided by this average entry price should be equal to the sum of each individual position (in coin) divided by its corresponding entry price.\n\nLet's look at the numbers I have. I have four positions:\n\n1. 31,000 USD at an entry price of 3315.48. This translates to 31000/3315.48 = 9.3499885 ETH\n2. 36,000 USD at 3410.8. That gives me 36000/3410.8 = 10.5547086 ETH\n3. 40,000 USD at 3481.71, or 40000/3481.71 = 11.4886076 ETH\n4. 5,000 USD at 3813.27, which is 5000/3813.27 = 1.3112106 ETH\n\nAdding them up, the total USD invested is 112,000 USD, and the total coin position is 32.7045153 ETH. This calculation seems to verify, when checked multiple times.\n\nNow, to calculate the average entry price, I divide the total USD (112,000) by the total coin position (32.7045153). That gives me roughly 3424.604 as the average entry price.\n\n\n\n```json\n{\"answer\": 3424.6, \"confidence\": 100}\n```" }, { "taskId": "inverse_pnl_fees-1", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 0.02840777213808475, "answer": 0.02841, "confidence": 98, "grade": "exact", "relError": 0.0000784243799345561, "cost": 0.012323, "reply": "**Inverse Contract PnL Calculation for a Short Position**\n\nOkay, let's break this down meticulously. We have a short position with a size of $44,000 USD. The entry price, P1, is $42,887.5, and the exit price, P2, is $41,686.7. The key here is to determine the correct formula for the Profit and Loss (PnL) in an inverse contract, particularly for a short.\n\nInitially, my instinct was to use a basic formula involving the contract size and the difference in 1/prices. But I immediately needed to verify. Short profits when the price drops, so how does this reflect in the formula? The thought process involves several mental checks.\n\nFirst, consider the mechanics: I *sold* a certain amount of Bitcoin at P1. This represents a certain amount of Bitcoin that I received. When closing the short, I *buy* Bitcoin at P2. The profit is the difference in Bitcoin amounts received at entry versus paid at exit.\n\nI started revisiting the formula: Is PnL = Size * (1/P1 - 1/P2)? No, this doesn't capture the inverse nature. So I tested with concrete example values: If prices are 10k, 20k, a rising price should mean positive Long PnL. This confirmed Long PnL = Size * (1/Pentry - 1/Pexit)\n\nTherefore, Short PnL = -Size * (1/Entry - 1/Exit) = Size * (1/Exit - 1/Entry). Or, if price drops from P1 to P2, then Short PnL = Size * (1/P2 - 1/P1).\n\nThis aligns with my understanding: if the price drops, P2 is smaller than P1, making the calculation result > 0.\n\nSo let's compute the Gross PnL: 44000 * (1/41686.7 - 1/42887.5). The difference in the reciprocals is calculated, and the Gross PnL equals 0.02955255 BTC.\n\nHowever, fees need to be accounted for. The fee rate is 0.055%. The position value at open is Size/P1 = 1.02594 BTC. The fee at open is 0.000564267 BTC. The position value at close is Size/P2 = 1.05549 BTC. The fee at close is 0.000580521 BTC. Total fees are therefore 0.001144788 BTC.\n\nFinally, calculate the Net PnL = Gross PnL - Total Fees = 0.02840776 BTC.\n\n\n\n```json\n{\"answer\": 0.02841, \"confidence\": 98}\n```" }, { "taskId": "inverse_pnl_fees-2", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -0.06525146382766053, "answer": -0.065251, "confidence": 100, "grade": "exact", "relError": 0.00000710831042433914, "cost": 0.005034, "reply": "**Inverse Contract PnL Calculation**\n\nOkay, let's break this down. I've got a long position in an inverse BTCUSD contract. The position size is $45,000. Entry was at $49,446, and I closed at $46,182.6. Initially, I had to double-check the PnL formula because it's an inverse contract. For a long position, the PnL in BTC should be calculated as `Size * (1/Entry - 1/Exit)`. I need to convert the USD position size to BTC at both entry and exit to use this formula effectively.\n\nI was a bit worried because the price dropped – a loss in the traditional sense – but with inverse contracts, a long position benefits from a price increase. A drop in price is a loss. I verified this by working through a hypothetical example where the price rises. That confirms the formula is correct for long positions: the PnL in BTC is indeed `Size * (1/Entry - 1/Exit)`.\n\nSo, plugging in the numbers: `45000 * (1/49446 - 1/46182.6)`. That equates to `45000 / 49446 - 45000 / 46182.6`. Calculating those fractions gives me approximately 0.9100837 BTC at entry and 0.9743929 BTC at exit. Subtracting, my gross PnL is roughly -0.0643092 BTC – a loss.\n\nNow, fees. I calculated the open and close fees at 0.05% of the BTC value at each trade, which is `0.0005 * (45000 / Entry)` and `0.0005 * (45000 / Exit)`, respectively. Adding these gives me a total fee of approximately 0.00094224 BTC.\n\nTo determine my net PnL, I deduct the total fees from the gross PnL. This comes out to -0.06525144 BTC.\n\nLet me just recalculate for precision:\nEntry price in BTC: `45000 / 49446 = 0.9100837277 BTC`\nExit price in BTC: `45000 / 46182.6 = 0.9743929532 BTC`\n\nGross PnL = 0.9100837277 - 0.9743929532 = -0.0643092255 BTC\nTotal fee = 0.0005 * (0.9100837277 + 0.9743929532) = 0.0009422383 BTC\nNet PnL = -0.0643092255 - 0.0009422383 = -0.0652514638 BTC\n\nSo, the precise net PnL is a loss of approximately -0.0652514638 BTC. Time to evaluate this trade and adjust the strategy.\n\n\n\n{\"answer\": -0.065251, \"confidence\": 100}" }, { "taskId": "inverse_pnl_fees-3", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 0.22922018359225318, "answer": 0.2292, "confidence": 99, "grade": "exact", "relError": 0.00008805329415973827, "cost": 0.003923, "reply": "**Analyzing the Net Profit/Loss of an Inverse Contract**\n\nOkay, let's break this down systematically. I see a long position with a size of $189,000 USD. The entry price, $P_1$, was $91,422.1, and the exit price, $P_2$, was $102,932.1. Now, with an inverse contract, the PnL is calculated using the formula: Size * (1/$P_1$ - 1/$P_2$). But we need to factor in the fees.\n\nThe fee rate is 0.05%, or 0.0005. The open fee in BTC is then Size / $P_1$ * 0.0005, and the close fee is Size / $P_2$ * 0.0005. Combining the formulas, we get Net PnL = Size * (1/$P_1$ - 1/$P_2$) - 0.0005 * Size * (1/$P_1$ + 1/$P_2$).\n\nThat can be simplified to: Net PnL = Size * [ (1 - 0.0005)/$P_1$ - (1 + 0.0005)/$P_2$ ].\n\nNow, the math. Let's substitute the values: 189000 * [ 0.9995 / 91422.1 - 1.0005 / 102932.1 ]. Alright, let's calculate each term.\n\nTo be precise, I can calculate the values separately to compute the gross profit and the fees, which is likely better, to avoid the potential for minor rounding errors that might be amplified.\n\nI calculate 189000 / 91422.1 = 2.06733383, and 189000 / 102932.1 = 1.83616190. Then, PnL_gross = 2.06733383 - 1.83616190 = 0.23117193 BTC.\n\nNext, Fee_open = 2.06733383 * 0.0005 = 0.00103367 BTC and Fee_close = 1.83616190 * 0.0005 = 0.00091808 BTC. The total fees are 0.00195175 BTC.\n\nTherefore, the Net PnL is the gross profit minus the fees. Net PnL = 0.23117193 - 0.00195175 = 0.22922018 BTC. So, the net profit is approximately 0.22922018 BTC.\n\n\n\n```json\n{\"answer\": 0.2292, \"confidence\": 99}\n```" }, { "taskId": "inverse_pnl_fees-4", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -0.026151408867684533, "answer": -0.02615, "confidence": 98, "grade": "exact", "relError": 0.00005387349078061194, "cost": 0.005663, "reply": "**Inverse Contract PnL Calculation Deep Dive**\n\nOkay, let's break this down meticulously. We're looking at an inverse contract, or coin-margined contract, which I know inside and out. The position size is $14,000 USD. The initial price, $P_1$, is $89,747.2, and the closing price, $P_2$, is $76,940.3. A key is that this is a *long* position, so I need to make sure my logic is correct about profits and losses relative to price movements.\n\nInitially, I quickly calculate PnL using the standard formula: `Position size * (1/P1 - 1/P2)`. Wait, something feels off. A long position *should* profit when the price *rises*, not falls. I need to make sure the formula aligns with the inherent economics. I need to flip the prices in the correct places in the formula, because an inverse contract measures PnL in the base currency (BTC in this case).\n\nLet's test this with a simplified example to nail down the formula. If I enter at 10,000 and exit at 20,000, then I would buy 14000 USD of BTC, which is 1.4 BTC at entry. If the price doubles, that now equals 0.7 BTC. So, in terms of BTC: I should have a *loss* of 0.7 BTC... or a profit of USD: 10,000. If I start at 10,000 and the price goes to 20,000, that is profit.\n\nOk, so with a long BTC position I profit when the BTC price *rises*. The USD profit is then position size * (P2 - P1) / P1. The formula in BTC *must* align with this.\n\nThe correct formula for a long inverse contract, after some mental checks, is `PnL = size * (1/P_entry - 1/P_exit)`. Yes, that formula holds! If P2 is greater than P1, then the PnL is positive.\n\nNow, let's plug in the actual values. With P1 being 89,747.2 and P2 being 76,940.3, the calculation is: `14000 * (1/89747.2 - 1/76940.3)`. The difference inside the parentheses is negative, leading to a loss in BTC as expected. The gross PnL is approximately -0.02596553 BTC.\n\nNow, I need to account for fees. The fee rate is 0.055%. The opening fee is the fee rate times the position size divided by the entry price, or `0.00055 * (14000 / 89747.2)`, which is about 0.0000857965 BTC. The closing fee is the fee rate times position size divided by the exit price, `0.00055 * (14000 / 76940.3)`, which yields around 0.0001000776 BTC. The total fees come to approximately 0.000185874 BTC.\n\nFinally, the *net* PnL is the gross PnL minus the total fees: `-0.02596553 - 0.00018587`, which gives us a final net PnL of approximately -0.0261514 BTC. This is our final result, taking into account price movement and all the fees. A loss of about 0.0261514 BTC!\n\n\n\n```json\n{\"answer\": -0.02615, \"confidence\": 98}\n```" }, { "taskId": "inverse_pnl_fees-5", "type": "inverse_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -0.011586282591352507, "answer": -0.011586, "confidence": 98, "grade": "exact", "relError": 0.00002439016572207775, "cost": 0.00419325, "reply": "**Inverse Perpetual Contract Short PnL Calculation**\n\nOkay, let's break this down. We have an inverse perpetual contract, and I'm short with a position size of $44,000 USD. The open price was $102,777.6, and the close price is $105,593.7. I know that because it's a short, the price going up means I'm losing, but let's make sure our formula is correct.\n\nSince this is an inverse contract, the PnL is going to be in BTC. I'm going to start with the basic formula: PnL = Position Size in USD * (1 / P_open - 1 / P_close). But that doesn't seem right for a short, if price goes up, I should have a loss! Let's think about this... if I short at 100 and buy back at 200, I clearly lose money. Okay, so P_open < P_close, which means 1/P_open > 1/P_close, which means 1/P_open - 1/P_close is positive. That's a loss! That does't make sense.\n\nI need the standard inverse contract short PnL formula, which I recall as: Short PnL = Quantity * (1 / Exit Price - 1 / Entry Price). I'll test this out... Entry at 100, exit at 200: 1/200 - 1/100, so 0.005 - 0.01 = -0.005. Negative! Perfect.\n\nSo, the correct formula is: Short PnL = Quantity * (1/P_close - 1/P_open). Now I see it.\n\nSo, in this case:\nSize = 44000 USD\nOpen BTC value = 44000 / 102777.6 = 0.428108848 BTC\nClose BTC value = 44000 / 105593.7 = 0.416691522 BTC.\n\nWait, here's what's happening... I sold 44000 USD of BTC at 102777.6, which gave me 0.4281 BTC. Then, to close the short, I bought 44000 USD of BTC at 105593.7, costing me 0.41669 BTC. Okay, this is right. Inverse contracts, priced in USD, settled in BTC. When I go short, I profit when the price goes down. And here, since it goes up from 102k to 105k, I must lose BTC (and USD).\n\nThe loss in BTC is: PnL = 44000 * (1 / 105593.7 - 1 / 102777.6) = 0.416691522 - 0.428108848 = -0.011417326 BTC.\n\nNext, I have to include fees. My fee rate is 0.02% or 0.0002.\nFee at opening = 0.0002 * (44000 / 102777.6) = 0.000085622 BTC\nFee at closing = 0.0002 * (44000 / 105593.7) = 0.000083338 BTC\nSo, total fees = 0.000085622 + 0.000083338 = 0.000168960 BTC.\n\nFinally, the net PnL is -0.011417326 - 0.000168960 = -0.011586286 BTC.\n\n\n\n```json\n{\"answer\": -0.011586, \"confidence\": 98}\n```" }, { "taskId": "linear_pnl_fees-1", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 6057.334255752001, "answer": 6057.334, "confidence": 100, "grade": "exact", "relError": 4.222187356162598e-8, "cost": 0.002262, "reply": "**My Short BTC Trade Analysis**\n\nOkay, let's break this down. I've got a short position open, 1.703 BTC, and looking at the details. My entry was at 26380.89, and I closed it out at 22814.19. Excellent price action, it seems! Now, to calculate the profit...\n\nFirst, the gross profit. That's a straightforward linear short calculation: size times the difference between entry and exit. So, 1.703 multiplied by (26380.89 - 22814.19), which gives me 1.703 * 3566.7, or 6074.0901. Solid start.\n\nNow, I can't forget about fees. I'm operating at a fee rate of 0.02%, or 0.0002. Gotta factor those in. The opening notional value was the size times the entry: 1.703 * 26380.89, giving 44926.65567. The opening fee is that notional times the fee rate, or 44926.65567 * 0.0002, which works out to 8.985331134.\n\nFor the closing fee, it's the size times the exit: 1.703 * 22814.19, so 38852.56557. Then, multiply that by the fee rate to get the closing fee: 38852.56557 * 0.0002, or 7.770513114.\n\nTotal fees are simply the sum of the opening and closing fees. Adding those two: 8.985331134 + 7.770513114, resulting in 16.755844248.\n\nFinally, the net PnL. The gross profit minus the total fees, simple. 6074.0901 - 16.755844248. The result is 6057.334255752. Not bad. Not bad at all. I have to say, it looks like a good trade!\n\n\n\n```json\n{\"answer\": 6057.334, \"confidence\": 100}\n```" }, { "taskId": "linear_pnl_fees-2", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 2844.8890875599996, "answer": 2844.889, "confidence": 100, "grade": "exact", "relError": 3.0778001086810425e-8, "cost": 0.002049, "reply": "**Analyzing the Trade's Profitability**\n\nOkay, let's break this down. My position size was 1.164 BTC, entered at 36718.19 and exited at 39204.01. I need to calculate the profitability, considering those entry and exit prices. The fee rate is 0.055%, which is critical to consider.\n\nFirst, the gross profit and loss. It's simply the size multiplied by the difference between the exit and entry price. The price difference (exit - entry) is 2485.82. Multiplying that by the size (1.164 BTC) gives me a gross profit of 2893.49448. That's a good start.\n\nNow, to calculate the fees accurately, I have to consider the notional value. I calculated that by multiplying the size by both the entry and exit prices. The entry notional was 42739.97316, and the exit was 45633.46764. The sum of these two provides a total notional value of 88373.4408. This is what I'll base the fees on.\n\nThe total fees are the total notional multiplied by the fee rate (0.00055), which comes out to 48.60539244.\n\nFinally, the net PnL is the gross PnL minus the total fees. So, 2893.49448 minus 48.60539244, giving me a final net profit of 2844.88908756. This is the figure that truly matters to me. It's a healthy return.\n\n\n\n{\"answer\": 2844.889, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-3", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -3783.556934676003, "answer": -3783.56, "confidence": 100, "grade": "exact", "relError": 8.101699141391267e-7, "cost": 0.002207, "reply": "**Analyzing the Trade Outcome**\n\nAlright, let's break down this Bitcoin trade. The position size was 0.612 BTC, and I entered long at 57972.37. Unfortunately, the market moved against me, and the position was closed at 51850.49.\n\nFirst, I calculate the gross P&L. It's simply the size multiplied by the difference between the close and open prices: 0.612 * (51850.49 - 57972.37). That's a negative 6121.88, which becomes -3746.59056 after multiplication. Not great.\n\nNow, let's factor in the fees. The open notional was 0.612 BTC * 57972.37, equating to 35479.09044. With a fee of 0.00055, that's 19.513499742 in fees just to open the position.\n\nThe close notional calculation is 0.612 BTC * 51850.49, which equals 31732.49988. Applying the same 0.00055 fee, I get 17.452874934 for closing.\n\nSumming the open and close fees (19.513499742 + 17.452874934), I get a total of 36.966374676 in transaction costs.\n\nFinally, the net P&L is the gross P&L minus the total fees. So, -3746.59056 - 36.966374676, landing me at a net loss of -3783.556934676. That's a pretty clear indicator of a losing trade. Time to learn from this.\n\n\n\n{\"answer\": -3783.56, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-4", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -1108.5578196799993, "answer": -1108.5578, "confidence": 100, "grade": "exact", "relError": 1.775279456924296e-8, "cost": 0.0021555, "reply": "**Analyzing the Trade's P&L**\n\nOkay, let's break this down meticulously. The trade size is 2.648 BTC. I opened at 4857.54 and closed at 4440.76. First, I need the notional values. The notional open is simply the size multiplied by the open price, which comes out to 12862.76592. The notional close is the size multiplied by the close price, equaling 11759.13248.\n\nNow, for the gross profit and loss. It's the difference between the notional close and notional open, which is the size multiplied by the difference between the close and open prices. That works out to 2.648 * (-416.78) which equals -1103.63344. So, we're already seeing a loss there.\n\nNext, I need to account for fees. The open fee is 0.0002 times the notional open, which is 2.572553. The close fee is 0.0002 times the notional close, yielding 2.351826. Thus, the total fees are the sum of those, or alternatively, 0.0002 times the sum of the notional open and close, resulting in 4.92437968.\n\nFinally, the net P&L is the gross P&L minus the total fees. That calculates to -1103.63344 minus 4.92438, which equals -1108.55782.\n\n\n\n{\"answer\": -1108.5578, \"confidence\": 100}" }, { "taskId": "linear_pnl_fees-5", "type": "linear_pnl_fees", "tier": "spec", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 3351.620043262996, "answer": 3351.62, "confidence": 100, "grade": "exact", "relError": 1.2908084972596765e-8, "cost": 0.002139, "reply": "**My Analysis of the Short Position**\n\nOkay, let's break this down. I had a short position open, a decent-sized one at 1.949 BTC. The entry price was 28983.12, and it closed out at 27232.54. That's a solid move in my favor, as expected.\n\nMy gross profit and loss, that's straightforward. Size times the price difference. So, 1.949 multiplied by (28983.12 - 27232.54) which comes out to 1.949 multiplied by 1750.58, equals a nice gross profit of 3411.88042 USDT.\n\nNow to factor in the fees. I need the open and close notional values: 1.949 multiplied by the open price (28983.12) which is 56488.10088 USDT and 1.949 multiplied by the close price (27232.54) which is 53076.22046 USDT.\n\nThe total notional is then the sum of the open and close notional, which is 109564.32134 USDT. My fee rate is 0.055%, which translates to 0.00055. This will be an important factor. I can get my total fees by multiplying the total notional by the fee rate: 109564.32134 multiplied by 0.00055, results in 60.260376737 USDT.\n\nFinally, to calculate my net profit, I subtract the total fees from the gross profit. 3411.88042 minus 60.260376737, which gives me a net profit of 3351.620043... USDT. Not bad, a worthwhile trade after accounting for those transaction fees.\n\n\n\n```json\n{\"answer\": 3351.62, \"confidence\": 100}\n```" }, { "taskId": "r_fx_lots_usd_account-1", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 0.11472881355932601, "answer": 0.1147, "confidence": 98, "grade": "exact", "relError": 0.0002511449254299066, "cost": 0.001924, "reply": "**My Position Sizing Calculation**\n\nOkay, let's break this down. My account sits at $5000, and I'm comfortable risking 1%, which means I'm looking at a $50 risk for this trade. I'm trading USD/CAD, and I've identified an entry at 1.3538 with a stop loss at 1.3479.\n\nThe stop loss is 59 pips away (1.3538 - 1.3479 = 0.0059). Now, I need to figure out the pip value in USD. Remember, with USD/CAD, USD is the base currency, and CAD is the quote. A standard lot in this case is 100,000 USD. Each pip is 0.0001 CAD per unit traded, or 10 CAD per standard lot. That pip value translates to 10 CAD / 1.3538 (the current exchange rate) which gets me 7.386615 USD per pip per standard lot.\n\nWith a risk of 59 pips, my total risk per standard lot is 59 * (10 / 1.3538) = 435.80998 USD. Now I can calculate how many standard lots I can trade to stay within my $50 risk tolerance: 50 USD / 435.80998 USD per lot = 0.1147288. So, I need to trade approximately 0.1147 standard lots to stay within my risk parameters.\n\n\n\n```json\n{\"answer\": 0.1147, \"confidence\": 98}\n```" }, { "taskId": "r_fx_lots_usd_account-2", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 0.5939090909090843, "answer": 0.5939, "confidence": 99, "grade": "exact", "relError": 0.000015306903402306825, "cost": 0.002131, "reply": "**Calculating My Trade Size**\n\nAlright, let's break this down. My account size is $25,000, and I'm sticking to a 1% risk, so that's $250 I'm comfortable risking on this USD/CAD trade. My entry point is 1.3066, and I've set my stop loss at 1.3011. That's a stop loss of 55 pips, which is critical information.\n\nNow, let's figure out the lot size. A standard lot is 100,000 of the base currency, which is USD in this case. With USD/CAD, a single pip for a standard lot is worth 10 CAD (0.0001 * 100,000 CAD). Since I'm working in USD terms, I need to convert that pip value. At the entry price of 1.3066, each pip is worth approximately $7.65 (10 CAD / 1.3066 CAD/USD).\n\nMy target risk is $250. The formula is: Risk = Lots * (Pip Value per Lot) * (Number of Pips). So, I'm trying to solve for Lots: $250 = Lots * ($7.65/pip) * (55 pips). Rearranging, Lots = $250 / ($7.65/pip * 55 pips). This simplifies to Lots = $250 / (550 / 1.3066). Calculating the required lot size gives us approximately 0.59 lots. This is the optimal trade size to risk $250 on this trade.\n\n\n\n{\"answer\": 0.5939, \"confidence\": 99}" }, { "taskId": "r_fx_lots_usd_account-3", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 1.1982352941177008, "answer": 1.198, "confidence": 99, "grade": "exact", "relError": 0.00019636720672136493, "cost": 0.002626, "reply": "**My Trading Strategy for USD/CHF**\n\nOkay, here's the breakdown of my planned USD/CHF trade. My account balance is $25,000, and I'm sticking to my 1% risk rule, which means I can risk $250 on this trade.\n\nI'm entering the market at 0.8148, and my stop-loss is set at 0.8131. That gives me a stop distance of 17 pips (0.0017). Now, let's figure out the lot size.\n\nSince the base currency is USD and the quote currency is CHF, one pip is 0.0001 CHF per USD. For a standard lot (100,000 USD), one pip is 10 CHF. To convert that CHF value to USD at my entry price of 0.8148, the pip value in USD is 10 CHF / 0.8148 CHF/USD, which equals approximately $12.27 per pip.\n\nNow, calculating the total risk per lot, it's 17 pips multiplied by $12.27, which gives me roughly $208.64 per lot. To determine how many lots I can trade, I divide my total risk ($250) by the risk per lot ($208.64) which results in 1.198 lots.\n\nLet me double-check this:\nIf I put the stop loss at 17 pips, it comes to a stop loss of 0.0017 CHF on L * 100,000 USD position. This equals 170 * L CHF. Converting to USD at the entry price, this is (170 * L) / 0.8148 USD. I need to make sure this is 250 USD, so L equals 250 * 0.8148 / 170 which is exactly the same number: 1.198235... lots.\n\nSo, I'm confident that 1.198 lots is the correct position size for this USD/CHF trade, given my risk parameters. I'll need to round this to the nearest allowable increment given by my broker.\n\n\n\n```json\n{\"answer\": 1.198, \"confidence\": 99}\n```" }, { "taskId": "r_fx_lots_usd_account-4", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 1.0178846153846508, "answer": 1.0179, "confidence": 99, "grade": "exact", "relError": 0.000015114301873409756, "cost": 0.002602, "reply": "**My Trade Calculation Thoughts**\n\nOkay, here's the breakdown of this USD/JPY trade, meticulously calculated to ensure I'm adhering to my risk parameters. My account holds $25,000, and I'm sticking to a 1% risk per trade, so that's $250 I'm willing to put on the line.\n\nThe plan is to buy USD/JPY at 158.79 and set my stop loss at 158.40. That gives me a risk of 39 pips, calculated as the difference between the entry and stop levels (158.79 - 158.40 = 0.39, and since it's USD/JPY, that 0.39 translates to 39 pips).\n\nNow, let's figure out the appropriate lot size. Knowing a standard lot is 100,000 USD, and that for USD/JPY, one pip is 0.01 JPY, means that with a standard lot, one pip is worth 1,000 JPY (0.01 JPY * 100,000). To get the pip value in USD, I have to convert that 1,000 JPY using the entry price, which is 158.79 JPY/USD. This means that each pip is worth approximately $6.30 (1,000 JPY / 158.79 JPY/USD). With 39 pips of risk, that's roughly $245.61 risk on a standard lot, which is very close to my $250 risk target.\n\nAlternatively, I could have calculated it this way: Let X represent the number of standard lots. My lot size is X * 100,000. The total loss in JPY is then X * 100,000 * 0.39 JPY, or 39,000 * X JPY. Converting to USD, the loss would be 39,000 * X / 158.79 USD. I want that loss to equal $250, so I set up the equation: 39,000 * X / 158.79 = 250. Solving for X, I get X = 250 * 158.79 / 39,000 = 1.0178846.\n\nTherefore, for this trade, the optimal lot size is around 1.0178846 standard lots (or very close to it). This guarantees that I am following my 1% risk assessment of $250.\n\n\n\n```json\n{\"answer\": 1.0179, \"confidence\": 99}\n```" }, { "taskId": "r_fx_lots_usd_account-5", "type": "r_fx_lots_usd_account", "tier": "realistic", "domain": "forex", "model": "gemini:gemini-3.8-flash", "expected": 7.182727272727309, "answer": 7.1827, "confidence": 98, "grade": "exact", "relError": 0.000003796987728178971, "cost": 0.00233, "reply": "**Calculating Optimal Lot Size for a USD/JPY Trade**\n\nOkay, let's break this down systematically. My account size is $100,000, and I'm comfortable with a 1% risk per trade, so that's $1,000 I'm willing to lose on this one. The entry point is 158.02, and my stop-loss is set at 157.80. That gives me a stop-loss distance of 0.22 JPY, or 22 pips.\n\nNow, I need to figure out the appropriate lot size. Since this is USD/JPY, I'm buying USD, meaning USD is the base currency and JPY is the quote currency. One standard lot is 100,000 of the base currency, so that's 100,000 USD. With JPY pairs, a pip is 0.01 JPY. Therefore, one pip per standard lot translates to 100,000 * 0.01 = 1,000 JPY.\n\nHowever, I need to express this in USD terms. At the entry price of 158.02, the pip value in USD is 1,000 JPY / 158.02 USD/JPY = approximately $6.3283 per pip. My stop-loss is 22 pips, meaning the risk per standard lot is 22 pips * ($1,000 JPY / 158.02) = ($22,000 / 158.02) = $139.22288.\n\nFinally, to calculate the optimal lot size, I divide my total risk allowed ($1,000) by the risk per standard lot ($139.22288), so 1,000 / 139.22288, which is 7.182727 standard lots. I'll need to round this number down, as I cannot trade a partial lot. The maximum amount of lots I can trade for this trade is 7.\n\n\n\n```json\n{\"answer\": 7.1827, \"confidence\": 98}\n```" }, { "taskId": "r_inverse_pnl_plain-1", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -0.23573231643213824, "answer": -0.2357, "confidence": 99, "grade": "exact", "relError": 0.00013708952861178644, "cost": 0.007115, "reply": "**Analyzing an Inverse Perpetual Contract Short Position**\n\nOkay, let's break this down. I've got an inverse perpetual contract, short position, with a quantity of 103,000 USD. The entry price was 49553.7 USD, and the exit price 55827.2 USD. The initial thought was to use the formula PnL = Quantity * (1 / Entry - 1 / Exit). Wait a second, this is a *short* position, so the profit should occur when the price *drops*. Since the exit price is *higher* than the entry, this is a *loss*. So, the formula must be slightly different.\n\nLet's rethink: with a short, you want the price to go down so the formula is: Short PnL = Quantity * (1 / Exit - 1 / Entry). If the price *Exit > Entry*, then *1/Exit < 1/Entry* and thus PnL is negative, which is a loss, as expected.\n\nI had to clarify this: Since I am short BTCUSD, I am bearish on BTC. If BTC price rises, I lose. So, you are selling BTC, and buying it back at a higher price.\n\nLet's trace: I sell BTC at P_entry and hold USD, then I buy BTC back at P_exit to close. The quantity is measured in USD, but this is an *inverse* contract so it represents the amount of the underlying in BTC.\n\nThe Bybit formula is: Short PnL = Contract Quantity * (1 / Exit Price - 1 / Entry Price). I check this is logical: If Exit > Entry, 1/Exit < 1/Entry, (1/Exit - 1/Entry) < 0 and there is a loss.\n\nNow, let's plug in the numbers.\n\nPnL = 103,000 * (1 / 55827.2 - 1 / 49553.7) = 103,000 * (-0.0000022677) = -0.233575 BTC.\n\nSo, trading PnL = -0.233574434 BTC.\n\nNow the fees. The taker fee is 0.055%.\n\nAt entry, position value in BTC = Quantity / Entry = 103000 / 49553.7 = 2.078553 BTC. Entry fee = 2.078553 * 0.00055 = 0.0011432 BTC.\n\nAt exit, position value in BTC = Quantity / Exit = 103000 / 55827.2 = 1.844978 BTC. Exit fee = 1.844978 * 0.00055 = 0.0010147 BTC.\n\nTotal fees = 0.0011432 + 0.0010147 = 0.0021579 BTC.\n\nSo, including fees: Realized PnL = -0.233575 - 0.002158 = -0.235733 BTC.\n\nRecalculating everything with higher precision:\nRealized PnL = -0.235732377 BTC\n```python\ndef calculate_pnl(quantity, entry_price, exit_price):\n \"\"\"\n Calculates PnL for a short inverse perpetual contract.\n\n Args:\n quantity (float): Quantity in USD.\n entry_price (float): Entry price in USD.\n exit_price (float): Exit price in USD.\n\n Returns:\n float: PnL in BTC.\n \"\"\"\n pnl = quantity * (1 / exit_price - 1 / entry_price)\n return pnl\n\ndef calculate_fees(position_value, fee_rate):\n \"\"\"\n Calculates trading fees.\n\n Args:\n position_value (float): Position value in BTC.\n fee_rate (float): Fee rate (e.g., 0.00055 for 0.055%).\n\n Returns:\n float: Fees in BTC.\n \"\"\"\n return position_value * fee_rate\n\n# Inputs\nquantity = 103000 # USD\nentry_price = 49553.7 # USD\nexit_price =\n\n\n```json\n{\"answer\": -0.2357, \"confidence\": 99}\n```" }, { "taskId": "r_inverse_pnl_plain-2", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -0.0669895883004742, "answer": -0.06699, "confidence": 99, "grade": "exact", "relError": 0.0000061457240779507516, "cost": 0.005622, "reply": "**Inverse Perpetual Contract PnL Calculation**\n\nOkay, let's break this down. I'm looking at an inverse perpetual contract with a quantity of 78,000 USD, where 1 contract equals 1 USD. My long entry price was $95,195.8, and I exited at $88,094.2. So, the price actually *dropped* during my position.\n\nFirst, I need to confirm the PnL formula. For an inverse long, is it Q * (1/Entry - 1/Exit) or Q * (1/Exit - 1/Entry)? Let me think. If the price goes up, I want the PnL to be positive. If Entry = 10,000 and Exit = 20,000, 1/10000 - 1/20000 = +0.00005, which results in a profit. So it *is* Q * (1/Entry - 1/Exit). Got it.\n\nNow, let's apply it to this case. Entry = 95195.8 and Exit = 88094.2. Because Entry > Exit, meaning the price fell, 1/Entry - 1/Exit will be negative, resulting in a loss, as expected.\n\nI need to calculate:\n\n1 / 95195.8 = 0.000010504665...\n1 / 88094.2 = 0.000011351485...\nDifference = -0.00000084682 BTC/contract.\n\nSo, the gross PnL for my 78,000 contracts is 78000 * -0.00000084682 = -0.06605196 BTC.\n\nNow, fees. The taker fee is 0.055% (0.00055). The fee is based on the order value in BTC.\n\nAt entry, the order value was 78000 / 95195.8 = 0.81936388 BTC. The entry fee is 0.81936388 * 0.00055 = 0.00045065 BTC.\n\nAt exit, the order value was 78000 / 88094.2 = 0.88541584 BTC. The exit fee is 0.88541584 * 0.00055 = 0.00048698 BTC.\n\nTotal fees are 0.00045065 + 0.00048698 = 0.00093763 BTC.\n\nFinally, the Net PnL is Gross PnL minus Total Fees: -0.06605196 - 0.00093763 = -0.06698959 BTC.\n\nLet me just double-check those calculations to make sure I haven't messed anything up. Okay, looks like everything checks out. The net PnL is confirmed at -0.066989588 BTC, a slight loss. All good.\n\n\n\n```json\n{\"answer\": -0.06699, \"confidence\": 99}\n```" }, { "taskId": "r_inverse_pnl_plain-3", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 0.0024428835303150374, "answer": 0.002443, "confidence": 99, "grade": "exact", "relError": 0.00004767713381220255, "cost": 0.002725, "reply": "**Analyzing the Inverse Perpetual Contract P&L**\n\nOkay, let's break this down. We're looking at an inverse perpetual contract, specifically a short position with a quantity of 42,000 USD. The entry price was 117914.1 USD, and we exited at 116982.6 USD. Now, the P&L formula for this inverse short is pretty straightforward: `Quantity * (1/Exit - 1/Entry)`.\n\nSo, the math is: 42000 * (1/116982.6 - 1/117914.1). Let's calculate the reciprocals. 1/116982.6 gives us approximately 8.54828025 * 10^-6, and 1/117914.1 is roughly 8.48074996 * 10^-6. The difference is 6.753029 * 10^-8. Multiplying that difference by the quantity (42000) gives us a P&L of 0.002836272 BTC. Excellent.\n\nNow, we need to factor in the fees. The taker fee is 0.055%, or 0.00055. The fee on opening the position is `Quantity / Entry * 0.00055`, which calculates to approximately 0.000195905 BTC. The fee on closing the position is `Quantity / Exit * 0.00055`, which comes out to about 0.000197465 BTC. Thus, the total fees are the sum of these two figures, which is about 0.000393370 BTC.\n\nFinally, we calculate the realized P&L after fees. This is the initial P&L (0.002836272 BTC) minus the total fees (0.000393370 BTC), leaving us with a net profit of 0.002442902 BTC. Pretty standard procedure.\n\n\n\n```json\n{\"answer\": 0.002443, \"confidence\": 99}\n```" }, { "taskId": "r_inverse_pnl_plain-4", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 0.001886499688812526, "answer": 0.0018865, "confidence": 98, "grade": "exact", "relError": 1.6495495644927526e-7, "cost": 0.003814, "reply": "**Analyzing the Bitcoin Inverse Perpetual Trade**\n\nOkay, let's break this down systematically. I'm looking at a Bitcoin inverse perpetual contract here, where one contract represents 1 USD. The quantity is 192,000 contracts, and I entered at 20434 USD and exited at 20460.6 USD. This is a long position.\n\nFor an inverse contract like this, the profit and loss (PnL) calculation is key. I'm using the formula PnL (in BTC) = Qty * (1/Entry - 1/Exit). First, I should calculate the entry and exit values in BTC, just to be sure. So the entry value is 192000 / 20434 = 9.3961045 BTC and the exit value is 192000 / 20460.6 = 9.3838890 BTC. The gross PnL formula is correct for a long: Qty * (1/Entry - 1/Exit). I can calculate that directly: 1/20434 - 1/20460.6 = 26.6 / (20434 * 20460.6), which then multiplied by 192000 contracts results in a gross PnL of approximately 0.0122155 BTC.\n\nNow, fees. The taker fee is 0.055%, or 0.00055. The fee on entry is (192000 / 20434) * 0.00055 = 0.00516786 BTC. The fee on exit is (192000 / 20460.6) * 0.00055 = 0.00516114 BTC. This gives us total fees of 0.0103290 BTC.\n\nFinally, the net PnL is the gross PnL minus the total fees: 0.0122155 BTC - 0.0103290 BTC = 0.0018865 BTC.\n\nLet's double-check all of this. The gross PnL is, indeed, approximately 0.01221550 BTC, as calculated earlier. The fees are 0.00055 * (9.39610453 + 9.38388903) = 0.010328996 BTC, which is consistent. The net PnL then comes out to 0.001886504 BTC.\n\nLooks like everything is consistent and checks out.\n\n\n\n```json\n{\"answer\": 0.0018865, \"confidence\": 98}\n```" }, { "taskId": "r_inverse_pnl_plain-5", "type": "r_inverse_pnl_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": -0.0962309957129481, "answer": -0.09623, "confidence": 98, "grade": "exact", "relError": 0.000010347112598492922, "cost": 0.010659, "reply": "**Inverse Contract PnL Calculation Deep Dive**\n\nOkay, so I'm looking at this short position with 139,000 contracts. My entry was at $59,513.2 and I bought back at $62,000.9. This is an inverse perpetual contract, and the profit and loss (PnL) calculation is *not* a simple linear one. It's inverse, so the formula is a bit counterintuitive. I need to be very precise here to avoid any costly mistakes.\n\nMy initial thought was the standard formula –Contracts * (1/Entry - 1/Exit) – but that's for a long position. For a short, I have to flip that. So, it's Contracts * (1/Exit - 1/Entry), right? Let's break this down.\n\nThe core of this is the behavior of short positions. I'm essentially *selling* high and *buying* low, but in the inverse contract world, the *settlement* is in Bitcoin (BTC). So, I need to think about my BTC exposure here.\n\nAt entry, I *received* 139,000 / $59,513.2 BTC. At exit, I *paid* 139,000 / $62,000.9 BTC. And since the price went up, my intuition is that I've lost money on the short. The *key* is to recognize that in an inverse contract, a short position profits when the price goes *down* and loses when the price goes *up*.\n\nLet's look at a simpler example to make sure this is right. If I short at $10,000 and close at $5,000, then the PnL is positive.\n\nThen: 1/5000 - 1/10000 = 0.0002 - 0.0001 = +0.0001 BTC per USD.\n\nTherefore, for this short, I have to use, *Contracts * (1/Exit - 1/Entry)*\n\nSo, back to the numbers: 1/62000.9 is approximately 0.0000161288, and 1/59513.2 is approximately 0.0000168030. Subtracting them, I get a *negative* result: -0.0000006742. Multiplying by 139,000 gives me -0.093714 BTC.\n\nNow, let's verify this, and this is where it's crucial I don't get the settlement mixed up. If I short 1 BTC at $10,000, I receive 10,000 USD and owe 1 BTC. Then, if the price drops to 5,000 USD, I need only 5,000 to buy back 1 BTC. This means I have 5,000 in USD left over, which is equivalent to 1 BTC.\n\nTherefore, the Bybit formula is correct.\n\n*Short PnL = Position Size * (1/Exit Price - 1/Entry Price)*.\n\nNow, I'm confident in the formula and direction of the outcome (negative). It's time to refine the calculation to be exact. So, PnL = 139,000 * (1/62000.9 - 1/59513.2) = -0.093716452 BTC. Now, fees need to be taken into account.\n\nTaker fee is 0.055%.\n\nFees, entry: (139000 / 59513.2) * 0.00055 = 0.00128459 BTC.\n\nFees, exit: (139000 / 62000.9) * 0.00055 = 0.00123304 BTC.\n\nTotal fees are 0.00251763 BTC.\n\nRealized PnL is PnL minus Total fees, so I'm looking at -0.096234086 BTC. This confirms the loss due to the price increase. I lost a little less than 0.1 BTC on this trade after fees.\n\n\n\n```json\n{\"answer\": -0.09623, \"confidence\": 98}\n```" }, { "taskId": "r_inverse_avg_plain-1", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 70867.19713259465, "answer": 70867.2, "confidence": 100, "grade": "exact", "relError": 4.046167285427324e-8, "cost": 0.002472, "reply": "**Calculating Average Entry Price for an Inverse Contract**\n\nOkay, so I'm looking at these two positions in an inverse contract, where one contract equals one USD. The key here is to get the average entry price across both trades. First, I need to figure out the BTC value of each position. For Position 1, I have 76,000 contracts entered at a price of 62771.7. That's simple, 76000 divided by 62771.7 gives me roughly 1.2107366 BTC.\n\nThen for Position 2, I see 55,000 contracts with an entry price of 86235.1. Dividing 55000 by 86235.1 gets me about 0.6377914 BTC. Now, to get the average, I need the totals. Total contracts are just 76,000 plus 55,000, that's 131,000 contracts in total. Adding up the BTC values, I get about 1.848528.\n\nNow, to calculate the average entry price, I divide the total contracts by the total BTC value. So, 131,000 divided by 1.848528, resulting in roughly 70867.197. Let me just double-check that with a bit more precision. Okay, recalculating with more decimal places to confirm - the BTC totals are indeed 1.21073667187 and 0.63779134018, and the sum of that gives me the 1.84852801205. The precise calculation then is 131,000 divided by 1.84852801205 which is 70867.196. Since we're dealing with prices and tick sizes, let's round this appropriately. Considering standard decimal places on exchanges, I'll keep at least four significant figures, therefore the final average entry price is around 70867.2.\n\n\n\n```json\n{\"answer\": 70867.2, \"confidence\": 100}\n```" }, { "taskId": "r_inverse_avg_plain-2", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 34209.14079495059, "answer": 34209.1, "confidence": 99, "grade": "exact", "relError": 0.0000011925160831806673, "cost": 0.001883, "reply": "**Calculating My Average Entry Price for a Bybit BTCUSD Inverse Perpetual Position**\n\nOkay, let me break this down. I'm looking at a Bybit BTCUSD inverse perpetual contract, and the goal is to determine my average entry price for a blended position. Remember, one contract here is equivalent to one USD. I've got three distinct entry points, and I need to figure out the weighted average.\n\nFirst, I need to know the total contract value. I have 76,000, 73,000, and 42,000 contracts, summing to a total of 191,000 USD. Then, I need to translate these contracts into their equivalent BTC values. Since this is an inverse contract, the position size in BTC is the contract value divided by the price.\n\nSo, I'll calculate the BTC holdings for each entry:\n\n* For the first position: 76,000 USD / 33,536.1 USD/BTC = 2.2662146 BTC.\n* Second position: 73,000 USD / 24,929.9 USD/BTC = 2.9282107 BTC.\n* And finally, the third entry: 42,000 USD / 108,003.1 USD/BTC = 0.3888777 BTC.\n\nNow, I'll add up all the BTC holdings: 2.2662146 + 2.9282107 + 0.3888777 = 5.583303 BTC.\n\nWith a total of 191,000 USD in contract value and 5.583303 BTC, the average entry price is simply the total USD value divided by the total BTC. Therefore, 191,000 USD / 5.583303 BTC = 34,209.14 USD/BTC. So, my average entry price for this blended position is 34,209.14 USD.\n\n\n\n```json\n{\"answer\": 34209.1, \"confidence\": 99}\n```" }, { "taskId": "r_inverse_avg_plain-3", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 88678.94192491287, "answer": 88678.94, "confidence": 100, "grade": "exact", "relError": 2.1706538502484576e-8, "cost": 0.002725, "reply": "**Analyzing the Inverse Perpetual Contract**\n\nOkay, so I'm looking at this inverse perpetual contract, and immediately I see the structure: 1 contract equals 1 USD. That's straightforward enough. The total USD value is simply the sum of all the contracts, which in this case totals to 155,000 USD. No surprises there.\n\nNow, with inverse perpetuals, the position size in Bitcoin is calculated using a specific formula: the sum of each contract amount divided by its corresponding price. Let me break that down to make sure I understand this.\n\nI've got three positions here:\n\n* **Position 1:** 59,000 contracts at a price of 82124.2. That gives me roughly 0.7184 BTC.\n* **Position 2:** 34,000 contracts at a price of 81534. That's approximately 0.4170 BTC.\n* **Position 3:** 62,000 contracts at a price of 101232.7. That's about 0.6125 BTC.\n\nSumming the BTC, I get a total position size of roughly 1.7479 BTC.\n\nTo determine the average entry price, I'll divide the total USD value (155,000) by the total Bitcoin value (1.74787837 BTC). I calculate that this gives me an average entry price of approximately 88678.94.\n\nI like to double-check my calculations, especially given the various decimal places involved. Doing so, I'm satisfied that my total calculations of the average entry price is correct.\n\n\n\n```json\n{\"answer\": 88678.94, \"confidence\": 100}\n```" }, { "taskId": "r_inverse_avg_plain-4", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 46924.5904555392, "answer": 46924.59, "confidence": 100, "grade": "exact", "relError": 9.707899338219995e-9, "cost": 0.001647, "reply": "Here's my analysis of this Bybit inverse perpetual average entry price calculation:\n\n**Calculating the Average Entry Price on a Bybit Inverse Perpetual Contract**\n\nOkay, so I'm looking at how to calculate the average entry price for a position on a Bybit inverse perpetual contract. Let's break this down. First, I need to figure out the total contract value in BTC. The formula is, of course: Value = sum(Contracts / Price)\n\nI see two entries here: 42,000 contracts at a price of 30128.4, and 67,000 contracts at a price of 72132.8. Easy enough, let's calculate the total contracts and total BTC value.\n\nThe total contracts are simply 42,000 + 67,000 = 109,000. That part's straightforward. Now for the BTC value. I have to divide each contract amount by its respective price and sum them. So, for the first entry, it's 42,000 / 30128.4 = 1.3940335 BTC. For the second entry, 67,000 / 72132.8 = 0.9288423 BTC.\n\nSumming those, the total BTC is 1.3940335 + 0.9288423 = 2.3228758 BTC.\n\nFinally, to get the average entry price, I divide the total contracts by the total BTC value. That's 109,000 / 2.3228758 = 46924.59. So, the average entry price for this position is 46924.59.\n\n\n\n{\"answer\": 46924.59, \"confidence\": 100}" }, { "taskId": "r_inverse_avg_plain-5", "type": "r_inverse_avg_plain", "tier": "realistic", "domain": "perps", "model": "gemini:gemini-3.8-flash", "expected": 37529.651947786224, "answer": 37529.65, "confidence": 100, "grade": "exact", "relError": 5.189992769330071e-8, "cost": 0.002843, "reply": "**Calculating the Average Entry Price for a Bybit Inverse Perpetual Contract**\n\nOkay, let's break this down. I'm dealing with a Bybit inverse perpetual contract, and I need to calculate the average entry price given two positions. The key here is understanding that in inverse contracts, the contract value is in USD, but the position value is in BTC. So, first, I need to convert both contract values into BTC.\n\nI've got the total contracts: 36,000 and 61,000, summing to 97,000 total contracts. To find the equivalent BTC value for each position, I divide the contract amount by the entry price (in USD). So, it's a simple: Contracts / Price.\n\nFor the first position, 36,000 contracts at an entry price of $23,930.3, that's 36000 / 23930.3, which equals roughly 1.504368934 BTC.\n\nFor the second position, 61,000 contracts at $56,468.2, that's 61000 / 56468.2, which comes out to approximately 1.080254019 BTC.\n\nNow, I sum those BTC values: 1.504368934 + 1.080254019, giving me a total of approximately 2.584622953 BTC.\n\nFinally, to get the average entry price, I divide the total number of contracts (97,000) by the total BTC value (2.584622953). That calculation yields roughly 37,529.65.\n\nI've double-checked my calculations, just to be sure. I can confirm the individual BTC calculations, and my final average entry price calculation. I'm noting to round to the nearest tick size but the question is for 4 significant figures, and the tick size is 0.5 for Bybit BTCUSD inverse, so the correct answer is 37529.65.\n\n\n\n{\"answer\": 37529.65, \"confidence\": 100}" }, { "taskId": "r_var_from_returns-1", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 4.917273424034056, "answer": 4.917, "confidence": 98, "grade": "exact", "relError": 0.00005560480585021505, "cost": 0.007274, "reply": "**Calculating the 1-Day 95% Parametric VaR**\n\nAlright, let's get down to brass tacks. I've got a set of 40 returns, and I need to calculate the 1-day 95% parametric Value at Risk (VaR). The goal is to express this as a percentage. First, I'll calculate the mean. I've already summed the returns: the sum of the first row is 9.20, the second is -12.73, the third is -3.35 and the fourth is -3.31. The total of these sums comes out to -10.19. Dividing that by the number of observations, 40, gives me a mean (μ) of -0.25475%.\n\nNext, I need the standard deviation. This requires calculating the sum of squares, which I've done painstakingly for each return. I've summed the squares for each row, and then summed the sums: Row 1 = 68.4440, Row 2 = 81.9993, Row 3 = 112.1389, and Row 4 = 53.3799, for a total of 315.9621. Now the sum of squares formula is SS = sum of squared values - (sum of values)^2 / N. SS = 315.9621 - (-10.19)^2 / 40. SS=315.9621 - 2.5959 = 313.3662. Now I can calculate the sample variance (s^2) using SS/ (N-1) = 313.3662/39 = 8.03503. The square root of the variance is the sample standard deviation (s), which comes out to be 2.8346%.\n\nNow for the VaR calculation. Using the standard formula VaR = -(μ - z * s) or z * s - μ, where z is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score (z_0.95) is approximately 1.64485. I can quickly calculate z * s = 1.644853 * 2.83461 = 4.6625%. Then I subtract the mean, which is already a negative, so VaR = 4.6625 - (-0.25475) = 4.9173%. So, the 1-day 95% parametric VaR for this dataset is approximately 4.917%.\n\n\n\n{\"answer\": 4.917, \"confidence\": 98}" }, { "taskId": "r_var_from_returns-2", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 4.2934299989968965, "answer": 4.411, "confidence": 95, "grade": "wrong", "relError": 0.027383700451753444, "cost": 0.008576, "reply": "**Analysis of Return Distribution and VaR Calculation**\n\nOkay, here's what I've got. I'm starting with a set of 40 returns. First, I summed the positive and negative returns separately to get a total sum. The sum of the positive returns is 58.61, and the negative returns sum to -40.17. The overall sum is 18.44. The total count checks out – 22 positive and 18 negative, totaling 40 returns. Therefore the arithmetic mean comes to 0.461.\n\nNow, let's look at the volatility. To get a handle on the distribution, I've calculated the sum of squares for each return. I've broken it down into positive and negative returns for clarity. Sum of squares positive values gives 218.89 and negative gives 131.82. The total sum of squares is therefore 350.71.\n\nI can make a decent approximation for the standard deviation. A crude estimate of the mean absolute return puts it around 2.5 and standard deviation between 2.9 and 3. I've crunched the numbers, using the formula `Variance = (sum(x_i^2) - N*mean^2) / (N-1)`. Subtracting `sum(x)^2/N` from sum(x^2) gives a value of 342.21, and dividing by 39 gives the sample variance of 8.7746 and a sample standard deviation of 2.9622. This is close to the earlier rough estimate.\n\nFinally, I'm calculating the 95% Parametric Value at Risk (VaR). This is a standard measure of potential loss. The formula for VaR is `VaR = z_0.95 * s - mean`, where z_0.95 is the z-score for the 95th percentile (approximately 1.64485), 's' is the sample standard deviation, and 'mean' is the average return. Plugging in the numbers, VaR_0.95 = 1.64485 * 2.9622 - 0.461 = 4.4114. Therefore, at a 95% confidence level, the portfolio could lose up to 4.4114.\n\n\n\n```json\n{\"answer\": 4.411, \"confidence\": 95}\n```" }, { "taskId": "r_var_from_returns-3", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 5.2422646771168795, "answer": 5.21, "confidence": 98, "grade": "correct", "relError": 0.006154721118473615, "cost": 0.009239, "reply": "**Analysis of Returns and Value at Risk (VaR)**\n\nOkay, let's break this down systematically. I'm given 40 returns, and the initial calculation of the sums of positive and negative returns seems like a good starting point. I quickly see a small discrepancy in the sums, probably a computational error. I need to be precise, so let's re-calculate both the positive and negative sums to ensure accuracy. Okay, so re-calculating the sum of the negative returns gives me -57.47, a slightly different value, and the sum of the positives is 60.47, for a total sum of 3.00, confirming that there were some errors in the original calculation. From that I derive the mean, 0.075%.\n\nNow, moving on, I need to compute the sample standard deviation, so the sum of squares is needed. I'm calculating the sum of squares for both positive and negative returns, separating them for clarity. It looks like the original calculation was performed correctly, so no need to recalculate that. I'm calculating the sample variance by correcting for the mean. I'll need to remember that this gives a correction factor, as this is a sample variance not the population variance. Now the sample standard deviation, which is the square root of that. Now I have everything I need to calculate the VaR.\n\nI'm aiming for a 95% confidence level here, so I need the corresponding z-score, which I've got stored away (or would quickly look up). I remember the formula for VaR: it's the z-score multiplied by the standard deviation, minus the mean return. Applying this, I find that the 95% VaR is approximately 5.21%. That's a useful number, and I'd need to consider what it means in the context of the underlying asset, trading strategy, or risk tolerance. I'd also need to think about how robust this VaR calculation is, and how it might change under different market conditions or using different methodologies. For example, did I account for skew, or did I assume normal distribution? The small sample size also makes me a little cautious. I would expect further analysis to be required.\n\n\n\n```json\n{\"answer\": 5.210, \"confidence\": 98}\n```" }, { "taskId": "r_var_from_returns-4", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 7.037483251314367, "answer": 7.037, "confidence": 95, "grade": "exact", "relError": 0.00006866820099023418, "cost": 0.00590925, "reply": "**My Analysis of the Return Data**\n\nOkay, let's break this down. I've been given a series of 40 return values, and my task is to perform some basic statistical analysis. First, I'll calculate the sum and the mean. I can see that the individual returns are a mix of positive and negative values. I've summed them, row by row, and then got the grand total which I've used to compute the overall mean, which turns out to be -0.4705. This gives me a quick sense that the distribution is, on average, slightly negative.\n\nNext, I need to look at the data's variability. I've calculated the sum of squares for each value. This allows me to calculate the sum of squares, using a row-by-row breakdown to help me keep track of the computations. From the sum of squares, I can compute the SS_dev using the formula: Sum(x^2) - N*mean^2 = 319.6167 - 40 * (-0.4705)^2. This yields a SS_dev of approximately 310.7619. This gives me a good start on the variance.\n\nKnowing this, I can proceed with the sample variance, calculated as SS_dev / (N-1), or 310.7619 / 39, which is approximately 7.96825. Taking the square root of the variance, I get the sample standard deviation, which is about 2.8228. This standard deviation gives me a solid quantitative measure of the data's dispersion around the mean, and, based on the magnitude of the returns, it seems reasonably high, as it is several times the magnitude of the mean.\n\nFinally, I'm asked to calculate a 99% Value at Risk (VaR). This is a crucial risk management metric. The formula I'm using is VaR_alpha = z_alpha * sigma - mu. Where z_alpha is the standard normal quantile for 99% confidence, which is 2.3263 (derived by using norm.ppf(0.99)). Substituting into the formula, I've got VaR = 2.3263 * 2.8228 - (-0.4705), which gives me a VaR of 7.0373. So, under this model, I'm 99% confident that my losses will not exceed 7.0373.\n\n\n\n{\"answer\": 7.037, \"confidence\": 95}" }, { "taskId": "r_var_from_returns-5", "type": "r_var_from_returns", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 6.7877378350475945, "answer": 6.727, "confidence": 95, "grade": "correct", "relError": 0.008948170439639311, "cost": 0.008388, "reply": "**Analyzing the Return Series**\n\nOkay, here's what I'm seeing. I've been given a series of 40 returns, and my immediate goal is to get a sense of their distribution and calculate a Value at Risk (VaR) at the 99% confidence level. First, I'm summing the positive and negative returns separately to get a handle on the overall directional bias, which results in 50.51 and 48.72 respectively. This gives a total sum of 1.79. Then, calculating the mean of this return series, which is 0.04475.\n\nNext, to estimate the sample standard deviation, I square each return value and sum the squares. I'm noting that many of the returns cluster around the +/- 3 to 4 range. After calculating both the sum of positive return squares (167.54) and negative return squares (162.96), I get a total sum of squares of 330.50. From that, after using the correction for sample variance, I calculate the sample variance (s^2) to be 8.4723, which gives a standard deviation (s) of 2.9107.\n\nFinally, with a mean of 0.04475 and a standard deviation of 2.9107, I use the z-score associated with a 99% confidence level (which is 2.32635) to calculate a 1-day parametric VaR. Given my calculations, the VaR value is 6.7266.\n\n\n\n{\"answer\": 6.727, \"confidence\": 95}" }, { "taskId": "r_sharpe_daily-1", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.40527021190035356, "answer": 0.4053, "confidence": 98, "grade": "exact", "relError": 0.00007350182365181334, "cost": 0.007124, "reply": "**My Analysis of the Return Data**\n\nOkay, let's break this down. I'm looking at a set of percentage returns – thirty values in all. My initial step is to calculate the sum of these returns; I'm getting a total of 1.43%. Then I calculate the mean, which comes out to be 0.0476667%. This is the basic average return, but we need more to fully understand the data's characteristics.\n\nNow, it's time to compute the sum of squares, since I will want to derive the standard deviation. After summing the squared values, and then calculating the sum of squares I end up with a value of 146.4985, from which I adjust for the mean. This results in the sum of squares value of 146.4303. Using this, I can calculate the sample variance (5.04932) and ultimately the sample standard deviation, which turns out to be 2.24707%. The standard deviation is the measure of the data's volatility.\n\nFinally, to get an understanding of the risk-adjusted return, I need to compute the Sharpe ratio. The daily Sharpe ratio is simply the mean return divided by the standard deviation. That gives me 0.0212128. I can annualize this by multiplying by the square root of 365, which is 19.10497. This gives me an annualized Sharpe ratio of 0.40527. This suggests the investment generates a very small amount of excess return relative to the risk.\n\n\n\n```json\n{\"answer\": 0.4053, \"confidence\": 98}\n```" }, { "taskId": "r_sharpe_daily-2", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.37258355219979333, "answer": 0.3726, "confidence": 98, "grade": "exact", "relError": 0.00004414526650342787, "cost": 0.008106, "reply": "**Analysis of Portfolio Returns**\n\nOkay, let's break down this data. I've got a series of percentage returns here: 0.49, 3.50, 1.12... right, and a whole list of them. First, I need to get a handle on the basic descriptive statistics. I see a mix of gains and losses, so I'll start by summing them up. I'll split the positives and negatives to keep it clean. Okay, Positive sum is 30.31, negative sum is -28.96. The total sum is 1.35.\n\nThat gives me a simple mean of 0.045%. Not very exciting, but it's a starting point. Now, to get a sense of volatility, I need to calculate the sum of squares, and here it is: 154.4683. Since variance gives you a relative level of volatility, I subtract the squared mean times the sample size. Then I calculate the sample variance to be 5.324398.\n\nFrom the variance, I can find the sample standard deviation, which is the square root of that. That gives me a standard deviation of 2.307466. Now, I have all the key pieces: mean and standard deviation.\n\nTo get a sense of the risk-adjusted return, I'll calculate the daily Sharpe ratio. This is simple, I just divide the mean by the standard deviation: 0.0195019. Now, let's annualize it to make it useful. Assuming 365 days a year, I multiply the daily Sharpe by the square root of 365, which comes out to roughly 19.104973, giving me an annualized Sharpe Ratio of approximately 0.372583. That is the final answer!\n\n\n\n```json\n{\"answer\": 0.3726, \"confidence\": 98}\n```" }, { "taskId": "r_sharpe_daily-3", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": -0.6855095501276544, "answer": -0.6855, "confidence": 98, "grade": "exact", "relError": 0.000013931429040744886, "cost": 0.008076, "reply": "**Analysis of Returns: A Deep Dive**\n\nOkay, here's the breakdown. I've been given a dataset of returns, and my initial goal is to extract some key statistical insights. I'm seeing 30 data points. First, I'm summing the positive and negative returns separately, just to get a feel for the distribution and see if I need to calculate an additional metric. The positive values come to about 28.29%, and the negatives total -30.79%. This gives me a total sum of -2.50%. The mean, therefore, is -0.083333%, quite close to zero, which could be an indicator of stationarity.\n\nNext, I calculated the sum of squares of each return, because I am keen on standard deviation. I calculated each squared value, accumulating it. The final sum of squares is 156.6316. Using the sum of squares, I calculated the sample variance: 156.4233 / 29 = 5.3939. This gives me a standard deviation (s) of 2.3225%.\n\nWith the mean and standard deviation in hand, I can now compute the Sharpe ratio. The daily Sharpe ratio is -0.083333 / 2.322478, which equals -0.035881. Annualizing this requires multiplying by the square root of the number of trading days in a year (assuming 365 days): -0.035881 * sqrt(365) = -0.6855. I also know that there may be adjustments required, but the result is a negative annualized Sharpe ratio of approximately -0.6855, this suggests poor risk-adjusted returns for this period.\n\n\n\n```json\n{\"answer\": -0.6855, \"confidence\": 98}\n```" }, { "taskId": "r_sharpe_daily-4", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 0.5941064255855656, "answer": 0.5941, "confidence": 98, "grade": "exact", "relError": 0.000010815546321174527, "cost": 0.00712, "reply": "**Analyzing Daily Returns and Calculating Sharpe Ratio**\n\nOkay, let's break this down. I'm given a list of 30 daily returns, expressed as percentages. My primary goal here is to calculate the annualized Sharpe ratio, which will give me a measure of risk-adjusted return. First, I need to get some descriptive statistics from the data.\n\nI'll start by summing the daily returns. I carefully separate the positive and negative values to maintain clarity. After tallying, the sum of all returns is 2.18%. From this, the mean return can be easily found to be 0.0726667%.\n\nNext, to calculate the standard deviation, I have to compute the sum of squares and then apply the standard formula. I calculate the square of each daily return, ensuring I maintain precision and accuracy in each step. Then, I sum these squared values, and then make further calculations for the variance and standard deviation of the sample.\n\nWith the sum of squares in hand, I can calculate the sample variance. I compute the total sum of squares and subtract the mean squared. This yields a value of 158.3554. I divide this by 29, representing the degrees of freedom (n-1), resulting in a sample variance of 5.46053. The standard deviation, the square root of the variance, is 2.33678%.\n\nNow, with both the mean return and the standard deviation calculated, I have all the pieces I need to compute the Sharpe ratio. The daily Sharpe ratio is simply the mean divided by the standard deviation: 0.0310969.\n\nFinally, to annualize this, I'll assume 365 trading days in a year (even though I know, practically, the number varies). I multiply the daily Sharpe ratio by the square root of 365, giving me an annualized Sharpe ratio of 0.5941. That's a good start. From here, I would look to see how to compare this result to other assets.\n\n\n\n```json\n{\"answer\": 0.5941, \"confidence\": 98}\n```" }, { "taskId": "r_sharpe_daily-5", "type": "r_sharpe_daily", "tier": "realistic", "domain": "risk", "model": "gemini:gemini-3.8-flash", "expected": 6.721949210975506, "answer": 6.722, "confidence": 95, "grade": "exact", "relError": 0.000007555698935003174, "cost": 0.006557, "reply": "**Analyzing Daily Returns and Calculating Sharpe Ratio**\n\nOkay, let's break this down. I'm given a set of 30 daily returns, expressed as percentages. My immediate thought is to sum them to get the total return, and then find the mean return. I'm going to sum the returns sequentially, grouping the sum in sets of 10 for ease of checking, as follows:\n\n* I summed the first 10 values, obtaining a sum of 7.71.\n* The second set of 10 values sum up to 13.40 (21.11 total - 7.71 from first set).\n* The final set of 10 returns yielded a sum of 4.63 (25.74 total - 21.11 from the prior two sets).\n\nThus, the overall sum is 25.74. With 30 data points, the mean daily return is simply this sum divided by 30, which gives me 0.858%.\n\nNext, I need to calculate the standard deviation to get a volatility measure. This requires summing the squares of each daily return. Again, I'll calculate this in three groups of 10 for easier verification. I've taken the square of each return and summed:\n\n* For the first 10, the sum of squares is 65.4175.\n* For the second 10, the sum of squares is 56.0964.\n* For the final 10, the sum of squares is 73.0257.\n\nThis gives a total sum of squares of 194.5396. To obtain the sample variance, I need to adjust for the mean. I'll use the formula SS = SumSq - (Sum^2)/30. Using the values calculated: 194.5396 - (25.74^2)/30, the result is approximately 172.45468. Dividing this sum of squares by (N-1), or 29, gives a sample variance of 5.94671. Then, taking the square root of that value, I get a sample standard deviation of 2.43859.\n\nNow for the Sharpe ratio. The daily Sharpe ratio is simply the mean return divided by the standard deviation. So, 0.858 / 2.43859 gives me a daily Sharpe ratio of approximately 0.35183.\n\nFinally, I need to annualize the Sharpe ratio. I'm assuming 365 trading days in a year, so I multiply the daily Sharpe by the square root of 365, which is roughly 19.10497. This gives me an annualized Sharpe ratio of 6.7218. This ratio is quite high, indicating a strong risk-adjusted return based on the data provided, but I'd need more context to evaluate this fully.\n\n\n\n```json\n{\"answer\": 6.722, \"confidence\": 95}\n```" }, { "taskId": "r_iv_then_delta-1", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.44697422776459944, "answer": 0.447, "confidence": 98, "grade": "exact", "relError": 0.000057659332014418574, "cost": 0.006146, "reply": "**Black-Scholes Delta Calculation**\n\nAlright, let's break this down. We're given a set of option parameters: $S = 447.74$, $K = 463$, $T = 23/365$, $r=0$, $q=0$, and $C = 21.6661$. My goal is to determine the Black-Scholes delta, which is $N(d_1)$. I know the Black-Scholes formula for a call option is $C = S N(d_1) - K N(d_2)$, where $d_1 = \\frac{\\ln(S/K) + \\frac{1}{2}\\sigma^2 T}{\\sigma \\sqrt{T}}$ and $d_2 = d_1 - \\sigma \\sqrt{T}$.\n\nFirst, I need to figure out the implied volatility, $\\sigma$. The intrinsic value is zero since the option is out-of-the-money ($S < K$), but the call price of $21.6661$ is surprisingly high for such a short-term option. I'll need to estimate sigma iteratively. I define $x = \\sigma \\sqrt{T}$. Let's calculate $m = \\ln(S/K) = \\ln(447.74/463) \\approx -0.033514$. This means $d_1 = m/x + x/2$ and $d_2 = m/x - x/2$.\n\nLet's start with a guess: if $\\sigma = 0.5$, then $x \\approx 0.1255$. That seems a bit low. A quick approximation suggests the price would be around 17 in this case. Let's try $\\sigma = 0.6$. This leads to $x \\approx 0.1506$. Plugging those values, $d_1 \\approx -0.1472$ and $d_2 \\approx -0.2978$. That gives me $N(d_1) \\approx 0.4415$ and $N(d_2) \\approx 0.3829$, leading to $C \\approx 20.40$. A little low, it's still about $1.26 too low.\n\nNow, let's try $\\sigma = 0.64$, implying $x = 0.160656$. This gives me $d_1 \\approx -0.1283$ and $d_2 \\approx -0.28896$. Then $N(d_1) \\approx 0.44896$ and $N(d_2) \\approx 0.3863$, which leads to a price of $C \\approx 22.16$. This suggests our volatility $x$ value is between 0.1506 and 0.1607.\n\nTime for interpolation. Using the two most recent price guesses: $(21.6661 - 20.40) / (22.16 - 20.40) \\approx 0.72$. So I calculate $x \\approx 0.1506 + 0.72 * 0.0101 \\approx 0.1579$.\n\nLet's recalculate the Black Scholes values. With this in mind, let's try $x = 0.1578$, which gives $d_1 \\approx -0.13348$ and $d_2 \\approx -0.29128$, or $N(-0.1335) \\approx 0.44689$ and $N(-0.2913) \\approx 0.38541$. That yields $C \\approx 21.646$. This is incredibly close!\n\nTherefore, the Black-Scholes delta $N(d_1)$ is approximately $0.4470$.\n\n\n\n{\"answer\": 0.4470, \"confidence\": 98}" }, { "taskId": "r_iv_then_delta-2", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.25544432164825037, "answer": 0.2554, "confidence": 98, "grade": "exact", "relError": 0.00017350805828985117, "cost": 0.005471, "reply": "**My Implied Volatility Calculation**\n\nOkay, so I've got a call option here with these parameters: S = 357.63, K = 383, T = 78/365, r = 0, q = 0, and a market price C = 5.0781. The goal is to back out the implied volatility, sigma, using the Black-Scholes model. First, let's get the standard inputs calculated: T is approximately 0.2137 years, which means the square root of T is about 0.4623, and the ratio S/K is 0.93376; I'll take the natural log of that to get -0.068535.\n\nI need to calculate d1 and d2, and they both depend on this \"x\" which is sigma * sqrt(T). I know the Black-Scholes formula, so I'll try to iterate, essentially guessing and checking for a value of x, and corresponding implied sigma, that gets me close to the market price. The delta I'll need is N(d1).\n\nLet's begin by approximating. Since the call price is small, and in relation to the stock price (C/S = 0.0142), a small x, in relation to the implied sigma should work.\n\nFirst guess x = 0.15; I plug that into the formulas for d1 and d2. From there I calculate the cumulative normal distribution, N(d1) and N(d2). The result is C = 11.72, which is way too high.\n\nLet's adjust. I'll drop x to 0.10. Now, with a new d1 and d2 calculated, my N(d1) and N(d2) are updated, and now I arrive at C = 5.41; that's closer to the market value.\n\nI can zero in better. I'll test x = 0.096. Now C = 4.955. This is slightly lower than the market value.\n\nI'll interpolate between x=0.096 and x=0.10. Using the change in the call price (dC) for a change in volatility (dx), I can approximate dC/dx with the slope between my two closest points. I also use a quick vega approximation here - it's vega and it doesn't change much with the change in x. Using that, I'm approximating that the slope is about 113.75, which means that the change in x to get the perfect call price is small.\n\nI solve to get my optimal value of x, which is approximately 0.09707.\n\nThe only thing left is to calculate d1 and then the cumulative normal distribution. d1 is -0.6575. After looking up the standard normal distribution tables, I found N(-0.6575) to be about 0.2554. Therefore, delta is 0.2554\n\n\n\n{\"answer\": 0.2554, \"confidence\": 98}" }, { "taskId": "r_iv_then_delta-3", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.4262987878297406, "answer": 0.4263, "confidence": 98, "grade": "exact", "relError": 0.0000028434757357895263, "cost": 0.006138, "reply": "**Black-Scholes Delta Calculation for an Out-of-the-Money Call**\n\nOkay, let's break this down. I've got a call option with the following parameters: underlying asset price ($S$) at 152.85, strike price ($K$) at 162, a time to maturity ($T$) of 32/365, which is roughly 0.08767, or do I need to think about 252? No, calendar days makes sense here. Interest rates ($r$) and dividends ($q$) are zero. The call price ($C$) is given as 8.6404. It's obviously out-of-the-money since the spot is less than the strike. The intrinsic value is zero.\n\nThe goal is to figure out the delta, which is $N(d_1)$, using the Black-Scholes formula. I can start with $C = S N(d_1) - K N(d_2)$. Let's introduce some shorthand: $d_1 = x$ and $d_2 = y = x - \\sigma \\sqrt{T}$ where $\\sigma$ is the volatility, and $v = \\sigma \\sqrt{T}$.\n\nRewriting the formula: $C = S N(x) - K N(x - v)$. Now, I know that $d_1 = \\frac{\\ln(S/K) + \\frac{1}{2}v^2}{v} = \\frac{\\ln(S/K)}{v} + \\frac{v}{2}$ and consequently $d_2 = d_1 - v = \\frac{\\ln(S/K)}{v} - \\frac{v}{2}$. Let's calculate $L = \\ln(S/K) = \\ln(152.85/162) = \\ln(0.9435185) \\approx -0.058145$. Therefore, $d_1 = \\frac{L}{v} + \\frac{v}{2}$.\n\nNow, I'm going to iterate to find the value of v (volatility times the square root of time) that gets me close to $C = 8.6404$.\n\nFirst, I'll test $v = 0.3$. This gives me $d_1 = -0.0438$ and $d_2 = -0.3438$. $N(d_1) \\approx 0.4825$ and $N(d_2) \\approx 0.3655$, leading to a call price of approximately 14.54, which is too high.\n\nThen I'll try $v = 0.2$. With this input, I get $d_1 = -0.1907$ and $d_2 = -0.3907$. This makes $N(d_1) \\approx 0.4244$ and $N(d_2) \\approx 0.3480$, so the calculated $C \\approx 8.49$, which is close.\n\nNow I test $v = 0.203$. This leads to $C \\approx 8.67$, so let's interpolate between $v = 0.2$ and $v = 0.203$. Since $C = 8.6404$ is our target, it seems $v$ is approximately 0.2025. Then, $d_1 = -0.18588$.\n\nLet's calculate $N(-0.18588)$ precisely. This requires the cumulative distribution function for a standard normal variable. Approximating the integral from 0 to 0.18588 of $\\frac{1}{\\sqrt{2\\pi}} e^{-t^2/2}$ is $0.07372$. Then $\\Phi(-0.18588) = 0.5 - 0.07372 = 0.42628 \\approx 0.4263$. Therefore, Delta $\\approx 0.4263$.\n\n\n\n```json\n{\"answer\": 0.4263, \"confidence\": 98}\n```" }, { "taskId": "r_iv_then_delta-4", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.6063214899790841, "answer": 0.6063, "confidence": 98, "grade": "exact", "relError": 0.00003544320865960742, "cost": 0.005131, "reply": "**My Thoughts on Implied Volatility and Delta Calculation**\n\nAlright, let's break this down. I've got a call option here with a price of 28.2965. The underlying asset is trading at 198.88, the strike price is 192, and we're looking at a time to maturity of approximately 0.29589 years (108 days). The interest rate, *r*, and the dividend yield, *q*, are both zero in this scenario.\n\nFirst, I calculate the intrinsic value, which is simply the difference between the current underlying price and the strike price: 198.88 - 192 = 6.88. The difference between the option's price and its intrinsic value gives me the time value: 28.2965 - 6.88 = 21.4165. This is the premium above its \"in-the-money\" value.\n\nNow, my goal is to find the implied volatility, which is the value of volatility that, when plugged into the Black-Scholes formula, gives me the observed market price of the option. The formula itself is:\n\nC = S * N(d1) - K * N(d2)\n\nwhere:\n* C is the call price\n* S is the underlying price\n* K is the strike price\n* N() is the cumulative standard normal distribution function\n* d1 = (ln(S/K) + (1/2) * sigma^2 * T) / (sigma * sqrt(T))\n* d2 = d1 - sigma * sqrt(T)\n\n* sigma is the volatility I need to find\n* T is the time to maturity\n\nThe key here is that Delta, or the sensitivity of the option price to changes in the underlying asset, is simply N(d1). So, the ultimate goal is to find *sigma*!\n\nI calculate the initial ln(S/K) to be ln(198.88/192), which is approximately 0.035207.\n\nI'm going to iterate with different values for *x*, defined as *sigma* * sqrt(T), to refine this number.\n\nLet's try *x* = 0.6. Plugging that in gives d1 = 0.3587, and d2 = -0.2413. This gives me a call price of 49.60, which is too high.\n\nThen let's try *x* = 0.3. This gives a price of 26.90. This is too low.\n\nThen I tried *x* = 0.32. This gives me a price of 28.433. A little high, but closer.\n\nThen, *x* = 0.318. This gives me a price of 28.279, which is close to our target of 28.2965.\n\nI will need to be very precise to find the \"true\" implied volatility. We need a value between 0.318 and 0.32, which appears to be about 0.3182.\n\nFinally, I estimate the Delta, N(d1), is approximately 0.60634.\n\nSo, implied volatility is roughly 0.3182 / sqrt(0.29589) = 0.5843. Therefore, the value of delta for this call is near 0.60634. I'm satisfied.\n\n\n\n{\"answer\": 0.6063, \"confidence\": 98}" }, { "taskId": "r_iv_then_delta-5", "type": "r_iv_then_delta", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 0.6017546889052645, "answer": 0.6018, "confidence": 98, "grade": "exact", "relError": 0.00007529828279015368, "cost": 0.008373, "reply": "**Option Pricing Implied Volatility Calculation**\n\nOkay, let's break this down. I'm given the Black-Scholes formula, simplified with risk-free rate and dividend yield set to zero, and the observed call option price ($C = 4.3243$). The goal is to back out the implied volatility, specifically estimating $x = \\sigma \\sqrt{T}$. I know $S = 108.83$, $K = 107$, and $T = 14/365$. The time value ($C - (S-K)$) is around 2.49.\n\nFirst, I express $\\ln(S/K)$ and approximate it. Then, I define $d_1$ and $d_2$ in terms of $x$. The Brenner-Subrahmanyam approximation for ATM options gives me a starting point. Since the time value is around 2.5 and the spot is ~108, I can use that to estimate $x$. This initial approximation, $x \\approx 0.058$, is a crucial starting point.\n\nI then test this initial estimate. Plugging it back into the Black-Scholes formula, I compute $d_1$, $d_2$, and the option price. This gives me a value that's too low, 3.51. The error lies in assuming the time value, so I need to iterate and adjust. Since the option price is too low, the implied volatility is too low.\n\nThen I test a higher value $x = 0.086$, leading to $C = 4.68$, a value that is too high. This gives a bracket. Given $x = 0.058$, $C=3.51$. Given $x=0.086$, $C=4.68$. Now I can use interpolation to narrow the value down further, calculating the slope and iterating.\n\nI do a linear interpolation to estimate $x$ and reach $x=0.0775$. Calculating $C$, I get 4.3323, which is very close to the target of 4.3243. So now I need to refine.\n\nI estimate the Vega. I have to make a tiny correction based on the delta to get a more accurate volatility. The implied volatility, $x \\approx 0.0773$, is near enough, I can check the final answer using this. From that value, I have calculated the delta $N(0.2580)$ and the value of it.\n\nThis process highlights the iterative nature of implied volatility calculation when an analytical solution is unavailable. Each iteration brings me closer to the solution by refining the implied volatility based on the difference between the calculated price and the observed market price.\n\n\n\n```json\n{\"answer\": 0.6018, \"confidence\": 98}\n```" }, { "taskId": "r_deribit_straddle_breakeven-1", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 113713.12767394731, "answer": 113713.1, "confidence": 99, "grade": "exact", "relError": 2.433663365945778e-7, "cost": 0.001939, "reply": "**Analyzing the Break-Even Point of a BTC Straddle**\n\nOkay, so I'm looking at a straddle position on BTC, with a strike price (K) of 101,000. The total premium I paid is 0.1118 BTC, which is the sum of the premiums for the call and put. This is an inverse option, settled in BTC. My main focus is to determine the BTC/USD price above the strike at which this straddle becomes profitable, or, more precisely, breaks even.\n\nSince we are above the strike, the put expires worthless, and the payoff comes entirely from the call option. The Deribit call payoff, as I recall, is calculated as `max(0, S - K) / S`, where S is the spot price. Conveniently, that simplifies to `1 - K / S`.\n\nMy objective here is to figure out the point where the payoff from this call covers the initial investment - the premium I paid. So, my break-even condition is essentially: the payoff in BTC equals the total premium I paid.\n\nThis translates to the equation: `1 - K / S = 0.1118`.\nFrom here, I'm just doing the algebra:\n`K / S = 1 - 0.1118 = 0.8882`\nand solving for S, which is `S = K / 0.8882`.\n\nNow it's a matter of plugging in the numbers. I'll compute `101000 / 0.8882`. That gives me approximately 113,713.12767. Therefore, the straddle breaks even when the BTC/USD price is approximately 113,713.13.\n\n\n\n```json\n{\"answer\": 113713.1, \"confidence\": 99}\n```" }, { "taskId": "r_deribit_straddle_breakeven-2", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 93608.77985797288, "answer": 93608.8, "confidence": 100, "grade": "exact", "relError": 2.1517241384055273e-7, "cost": 0.002058, "reply": "**Analyzing the Straddle Break-Even Point**\n\nOkay, so I'm looking at this straddle scenario. The strike price, K, is 87000. We've got a total premium paid of 0.0706 BTC (0.015 + 0.0556). Since this is settled in BTC, specifically inverse or crypto-margined options, the payoff calculation is in BTC terms. I'm focusing on the call side here, specifically figuring out the upper break-even point as the question implies.\n\nThe payoff structure for the Deribit call is max(S - K, 0) in USD. But since we are settling in BTC, we'll need to express it that way. The payoff in BTC is, therefore, max(S - K, 0) / S, or max(1 - K/S, 0).\n\nMy goal is to find the Bitcoin price at expiry where the straddle breaks even, focusing on the call side of the straddle. Profit is Payoff in BTC - Premium in BTC. When S > K (we are considering the call side), the payoff is simply 1 - K/S.\n\nThe break-even condition is when profit is zero. So, to solve for the upper break-even point, I set the payoff equal to the premium paid, i.e., 1 - K/S = P.\n\nPlugging in the numbers, 1 - K/S = 0.0706. So, K/S = 1 - 0.0706 = 0.9294.\n\nTherefore, S = K / 0.9294, and S = 87000 / 0.9294.\n\nThat gives me a value of approximately 93608.78. So, the straddle breaks even when the BTC/USD price is approximately $93608.78.\n\n\n\n```json\n{\"answer\": 93608.8, \"confidence\": 100}\n```" }, { "taskId": "r_deribit_straddle_breakeven-3", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 64193.23251006419, "answer": 64193.23, "confidence": 99, "grade": "exact", "relError": 3.910169481716397e-8, "cost": 0.001939, "reply": "**Analyzing a BTC-Settled Call Option Straddle on Deribit**\n\nAlright, let's break this down. I'm looking at a BTC-settled call option on Deribit. The payout, expressed in BTC, is essentially the maximum of zero and the difference between the spot price (S) and the strike price (K), all divided by S. That simplifies to max(0, 1 - K/S).\n\nI see the total cost of this straddle is 0.0809 BTC. Now, I need to figure out the break-even price, but I'm focusing on the scenario where the price is *above* the strike, meaning the call option is in the money and the put is out of the money. So, in this scenario, the payout equation simplifies to just 1 - K/S.\n\nThe key is that I need the payout to equal the total cost. So, I set 1 - K/S equal to 0.0809. Solving for K/S, I get 0.9191. I know that the strike price, K, is 59000. Now, I just need to find the spot price (S) that satisfies this. So, S = K / 0.9191. That gives me S = 64193.23.\n\nOkay, let's double-check this. Plugging that spot price (S=64193.23) back into the payout formula, I get (64193.23 - 59000) / 64193.23 = 0.0809 BTC, which is the total cost. Perfect!\n\nTherefore, the break-even price is approximately 64193.23, or 64193.\n\n\n\n```json\n{\"answer\": 64193.23, \"confidence\": 99}\n```" }, { "taskId": "r_deribit_straddle_breakeven-4", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 102706.75082914304, "answer": 102706.75, "confidence": 99, "grade": "exact", "relError": 8.072916682855671e-9, "cost": 0.002058, "reply": "**Analyzing the BTC Straddle Strike's Break-Even Point**\n\nOkay, so I'm looking at this straddle strike where K equals 96000. These are inverse options settled in BTC, which means the payoff is a bit different. I paid a total premium of 0.0653 BTC. My goal is to determine the BTC/USD price, S, at expiry where my profit in BTC is zero, meaning I break even. Since we're looking for a break-even point with profit of 0, I'm only concerned with the scenario where S is greater than K.\n\nBecause S is greater than K, the call option is in the money, and the put option expires worthless, which simplifies things. The payoff of the Deribit BTC call option, expressed in BTC terms, needs a bit of manipulation. The initial payoff in USD is straightforward: max(S - K, 0). But since it's settled in BTC, we divide that USD payoff by the spot price S. This gives me the payoff in BTC: max(S - K, 0) / S, which simplifies to 1 - K/S. Yes, let's verify that. That is correct.\n\nNow, the total cost in BTC is 0.0653 BTC. To find my break-even point, I need to set the payoff in BTC equal to the total cost. This gives me the equation: 1 - K/S = 0.0653. Now the math starts: 1 - 0.0653 = K/S, which turns to 0.9347 = 96000/S. Solving for S, I get S = 96000 / 0.9347. That's approximately 102706.75. So, at a price of roughly $102,706.75, I'll break even.\n\n\n\n```json\n{\"answer\": 102706.75, \"confidence\": 99}\n```" }, { "taskId": "r_deribit_straddle_breakeven-5", "type": "r_deribit_straddle_breakeven", "tier": "realistic", "domain": "options", "model": "gemini:gemini-3.8-flash", "expected": 73994.63806970509, "answer": 73994.6, "confidence": 100, "grade": "exact", "relError": 5.144927535165799e-7, "cost": 0.002138, "reply": "**Analyzing a BTC Straddle on Deribit**\n\nOkay, so I'm looking at this straddle position on Deribit, with the strike price at $69,000. It's an inverse BTC-margined contract, which is important to remember. The options are settled in BTC. I've shelled out a premium of 0.0412 BTC for the call and 0.0263 BTC for the put. That brings the total cost to 0.0675 BTC. Now, let's break down the potential profit.\n\nThe call option payoff at expiry, when the spot BTC/USD price ($S$) is above the strike ($K$), is simply $\\max(S - K, 0) / S$, or equivalently, $\\max(1 - K/S, 0)$.\n\nThus, for $S > K$, the profit in BTC from the call is $(1 - K/S) - \\text{Total cost}$. I need to figure out the break-even point when the price goes up.\n\nI know I have to set the profit equal to zero and solve for the price to break even, so:\n$1 - K/S = \\text{Total cost}$\n$1 - 69000/S = 0.0675$\n\nSo, the math is:\n$69000/S = 1 - 0.0675 = 0.9325$\n$S = 69000 / 0.9325 = 73994.638...$\n\nOkay, so the break-even is roughly $73,995, above the $69,000 strike. That checks out, that's what I was expecting. I've got to make sure I get more than the total cost of the straddle. And hey, the put has no value when $S > K$, that is important.\n\n\n\n```json\n{\"answer\": 73994.6, \"confidence\": 100}\n```" } ] }