# Is scaled Sharpe ratio a t-statistic?

$$\text{Annualized Sharpe Ratio} = \sqrt{N_T} \frac{\bar{R_s} - R_{b}}{\sigma_{R_s}}$$

where $$\bar{R_s}$$ are strategy returns for a certain period, $$R_b$$ -- benchmark returns and $$N_T$$ is number of periods in a year (e.g. 12 if $$R_s$$ is computed monthly). This seems like a t-statistic?

$$t_{\bar{x}} = \sqrt{n}\frac{\bar{x} - \mu}{\text{s.e.}({\bar{x}})}$$

So Sharpe ratio can be interpreted as a number of standard deviations from a benchmark returns?

• @noob2 it divides std in denominator, so you can just move sqrt(n) in front. Sep 21 '20 at 11:28
• yes, it is basically a $t$ statistic. More interesting, though, is when you move to the multivariate case: the (squared) Sharpe of the Markowitz portfolio is Hotelling's $T^2$! Oct 5 at 17:18