Sharpe Ratio and the t-Statistic

Explain the relationship between the Sharpe ratio and the t-statistic.

Answer

Consider returns $R_1,\dots,R_N$ with mean $\mu$ and standard deviation $\sigma$ , and assume a zero risk-free rate.

Sharpe ratio. The (non-annualised) Sharpe ratio is defined as

\text{SR}=\frac{\bar R}{s},

where $\bar R$ is the sample mean return and $$s$$ is the sample standard deviation.

Relationship to the t-statistic. When returns are observed over $$N$$ periods, the standard error of the mean is $s/\sqrt{N}$ . The usual test statistic for testing the null hypothesis $H_0:\mu=0$ is

t=\frac{\bar R}{s/\sqrt{N}}=\sqrt{N}\,\frac{\bar R}{s}=\sqrt{N}\,\text{SR}.

On the other hand, if returns are i.i.d. across time, the annualised Sharpe ratio is obtained by scaling the periodic Sharpe ratio by $\sqrt{N}$ :

\text{SR}_{\text{ann}}=\sqrt{N}\,\text{SR}.

Hence, under i.i.d. returns, $\text{SR}_{\text{ann}}$ and the t-statistic (score) are approximately the same:

\text{SR}_{\text{ann}} \approx t.

Notes and comments

Comment 1: Note that for short periods, taking the risk-free rate to be $$0$$ makes sense.