At 99% confidence, how long to beat a benchmark of 0.5?

Question

Your strategy has an observed annualized Sharpe of 1.0. You want to show, with 99% confidence (a stricter bar), that its true Sharpe exceeds a benchmark Sharpe of 0.5. Treat returns as roughly independent and normal.

Roughly how many years of track record do you need? (Use the one-sided 99% cutoff, z = 2.326.)

Accepted Answer

Minimum track record length. Solving (SR - SR^)/\operatorname{SE} \geq z with \operatorname{SE}(SR) \approx \sqrt{(1 + 	frac{1}{2}SR^2)/T}: T \geq \left(1 + 	frac{1}{2}SR^2ight)\left(\frac{z}{SR - SR^}ight)^2. Plug in the numbers. With SR = 1.0, SR^ = 0.5, and the stricter z = 2.326: T \geq (1 + 0.5)\left(\frac{2.326}{1.0 - 0.5}ight)^2 = 1.5 	imes (4.652)^2 = 1.5 	imes 21.64 \approx 32.5 	ext{ years}. Interpretation. You need about 33 years, double the roughly 16 years the same claim required at 95% confidence. The only thing that changed is the confidence cutoff, from z = 1.645 to…

At 99% confidence, how long to beat a benchmark of 0.5?

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