How long to show a Sharpe of 1.5 beats a benchmark of 1.0?

Question

Your strategy has an observed annualized Sharpe of 1.5. You want to show, with 95% confidence, that its true Sharpe exceeds a benchmark Sharpe of 1.0. Treat returns as roughly independent and normal.

Roughly how many years of track record do you need? (Use z = 1.645.)

Accepted Answer

Minimum track record length. You need the gap over the benchmark to be large relative to the Sharpe's standard error, \operatorname{SE}(SR) \approx \sqrt{(1 + 	frac{1}{2}SR^2)/T}. Solving (SR - SR^)/\operatorname{SE} \geq z for the number of years: T \geq \left(1 + 	frac{1}{2}SR^2ight)\left(\frac{z}{SR - SR^}ight)^2. Plug in the numbers. With SR = 1.5, SR^ = 1.0, z = 1.645: T \geq (1 + 1.125)\left(\frac{1.645}{1.5 - 1.0}ight)^2 = 2.125 	imes (3.29)^2 = 2.125 	imes 10.82 \approx 23.0 	ext{ years}. Interpretation. You need about 23 years. Two effects push this higher than a lower-Sha…

How long to show a Sharpe of 1.5 beats a benchmark of 1.0?

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