How precise is a Sharpe of 0.8 measured over 10 years?

Question

A strategy reports an annualized Sharpe ratio of 0.8 estimated from 10 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 is reliably above zero.

Accepted Answer

The formula. For roughly independent, normal returns, the standard error of an annualized Sharpe measured over T years is approximately \operatorname{SE}(\widehat{SR}) \approx \sqrt{\frac{1 + 	frac{1}{2}SR^2}{T}}. The uncertainty falls with the square root of the number of years. Plug in the numbers. With SR = 0.8 and T = 10: \operatorname{SE} \approx \sqrt{\frac{1 + 	frac{1}{2}(0.8)^2}{10}} = \sqrt{\frac{1 + 0.32}{10}} = \sqrt{0.132} \approx 0.36. Confidence interval. Roughly two standard errors on each side: 0.8 \pm 1.96 	imes 0.36 \approx 0.8 \pm 0.71 = (0.09,\ 1.51). The interval is ab…

How precise is a Sharpe of 0.8 measured over 10 years?

Your answer

More Statistics questions