Track record to show a Sharpe of 1.2 beats 0.5

Question

Your strategy has an observed annualized Sharpe of 1.2. You want to show, with 95% confidence, 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 z = 1.645.)

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 = 1.2, SR^ = 0.5, z = 1.645: T \geq (1 + 0.72)\left(\frac{1.645}{1.2 - 0.5}ight)^2 = 1.72 	imes \left(\frac{1.645}{0.7}ight)^2 = 1.72 	imes (2.35)^2 = 1.72 	imes 5.52 \approx 9.5 	ext{ years}. Interpretation. You need about 9.5 years (round up to 10). Compared with a Sharpe-1.0 strategy trying to beat the same 0.5 benchmark (which took about 16…

Track record to show a Sharpe of 1.2 beats 0.5

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