How long to prove your Sharpe beats a benchmark of 0.5?

Question

Your strategy has an observed annualized Sharpe of 1.0. You want to show, with 95% confidence, that its true Sharpe exceeds a benchmark Sharpe of 0.5 (not just zero). Treat returns as roughly independent and normal.

Roughly how many years of track record do you need? (Use the one-sided 95% cutoff, z = 1.645.)

Accepted Answer

The idea: minimum track record length. It is not enough to beat zero; often you must show you beat a benchmark Sharpe SR^. You need the gap between your Sharpe and the benchmark to be large relative to the Sharpe's standard error. Recall \operatorname{SE}(SR) \approx \sqrt{(1 + 	frac{1}{2}SR^2)/T}. Requiring \frac{SR - SR^}{\operatorname{SE}(SR)} \geq z and solving for the number of years T gives the minimum track record length: T \geq \left(1 + 	frac{1}{2}SR^2ight)\left(\frac{z}{SR - SR^}ight)^2. In words: the smaller your edge over the benchmark, the more years you need, and it grows …

How long to prove your Sharpe beats a benchmark of 0.5?

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