How many years of returns are needed before a true Sharpe ratio of 0.5 is two standard errors above zero, and does sampling daily instead of monthly shorten the wait?

How many years of returns are needed before a true Sharpe ratio of 0.5 is two standard errors above zero, and does sampling daily instead of monthly shorten the wait?

Approach: Write the standard error of an annualised Sharpe estimate in years, form the t statistic, and check how the frequency of observation enters both the estimate and its error.

16 years. The standard error of an annualised Sharpe estimate over T years is about sqrt((1 + S^2/2)/T), which is close to 1/sqrt(T) for small S, so the t statistic against a zero Sharpe is t = S*sqrt(T) whatever the sample length in observations. Setting 0.5*sqrt(T) = 2 gives T = 16 years, and the exact standard error sqrt(1.125/T) pushes it to 18 years. Sampling daily does not help, because the per period Sharpe falls by sqrt(252) while the number of observations rises by 252, so the two effects cancel and only the calendar span counts. A two year record showing a Sharpe of 0.5 carries a t statistic of 0.71 and is statistically indistinguishable from luck.

Follow-up: How does the required span change if returns have excess kurtosis of 6 and negative skew?

Key concepts: sharpe ratio standard error, t statistic, sample length, annualisation.