The error bar on a Sharpe of 1.0 over 4 years

Question

A strategy reports an annualized Sharpe ratio of 1.0 estimated from 4 years of returns. Treat the returns as roughly independent and normal.

Estimate the standard error of the Sharpe ratio, build an approximate 95% confidence interval, and say whether the Sharpe convincingly beats zero.

Accepted Answer

The formula. For roughly independent, normal returns, the standard error of an annualized Sharpe over T years is approximately \operatorname{SE}(\widehat{SR}) \approx \sqrt{\frac{1 + 	frac{1}{2}SR^2}{T}}. Plug in the numbers. With SR = 1.0 and T = 4: \operatorname{SE} \approx \sqrt{\frac{1 + 	frac{1}{2}(1.0)^2}{4}} = \sqrt{\frac{1.5}{4}} = \sqrt{0.375} \approx 0.61. Confidence interval. Roughly two standard errors on each side: 1.0 \pm 1.96 	imes 0.61 \approx 1.0 \pm 1.20 = (-0.20,\ 2.20). The interval dips below zero, so a Sharpe of 1.0 over four years is not convincingly different from z…

The error bar on a Sharpe of 1.0 over 4 years

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