I was just reading Quantitative Trading: How to Build Your Own Algorithmic Trading Business and it suggests annualizing Sharpe ratio in order to compare performance of strategies:

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

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?

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    $\begingroup$ @noob2 it divides std in denominator, so you can just move sqrt(n) in front. $\endgroup$ Sep 21, 2020 at 11:28
  • $\begingroup$ 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 $T^2$! $\endgroup$
    – shabbychef
    Oct 5, 2021 at 17:18

1 Answer 1


Yes, Sharpe follows a student's t distribution.



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