Only 20 trials over 5 years: is a Sharpe of 1.0 impressive?

Question

A researcher tries just 20 configurations over 5 years and reports the best, which has an annualized Sharpe of 1.0. Suppose none of the 20 has any genuine edge.

What Sharpe would the best of 20 zero-edge configurations show by luck alone, and what does that say about the reported 1.0?

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}, giving a luck bar for the best-of-N Sharpe of 	ext{luck bar} \approx \sqrt{\frac{2\ln N}{T}}. Plug in the numbers. With N = 20 and T = 5: \sqrt{2\ln 20} = \sqrt{2 	imes 2.996} = \sqrt{5.99} \approx 2.45,\qquad 	ext{luck bar} \approx \frac{2.45}{\sqrt{5}} = \frac{2.45}{2.236} \approx 1.09. Interpretation. Even with a modest search of only 20 trials, luck alo…

Only 20 trials over 5 years: is a Sharpe of 1.0 impressive?

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