500 trials over 10 years: how high must the winner be?

Question

A quant runs a massive search of 500 configurations over 10 years and reports the best, which has an annualized Sharpe of 1.2. Suppose none of the 500 has any genuine edge.

What Sharpe would the best of 500 zero-edge configurations show by luck alone, and what does that imply about the reported 1.2?

Accepted Answer

Each junk backtest is noise. A zero-edge strategy over T years yields a sample Sharpe with standard error about \operatorname{SE}0 \approx \sqrt{1/T}. Best of many. The expected maximum of N standard-normal draws is about \sqrt{2\ln N}, so the luck bar for the best-of-N Sharpe is 	ext{luck bar} \approx \sqrt{\frac{2\ln N}{T}}. Plug in the numbers. With N = 500 and T = 10: \sqrt{2\ln 500} = \sqrt{2 	imes 6.215} = \sqrt{12.43} \approx 3.53,\qquad 	ext{luck bar} \approx \frac{3.53}{\sqrt{10}} = \frac{3.53}{3.162} \approx 1.12. Interpretation. Two forces pull in opposite directions here. The h…

500 trials over 10 years: how high must the winner be?

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