One estimate of the 'quality' of a portfolio of stocks is the Sharpe ratio, which is defined as the mean of the returns divided by the standard deviation of the returns (modulo adjustments for risk free rate, etc). The sample Sharpe ratio is the sample mean divided by the sample standard deviation. Up to a constant factor ($\sqrt{n}$, where $n$ is the number of observations), this is distributed as a (possibly non-central) $t$-statistic.

Are there known techniques for comparing the mean of independent variables distributed as non-central $t$-statistics? Of course, there are non-parametric tests of mean, but is there something specific to the case of noncentral $t$? (I'm not sure what I meant by that.)

edit: the original question is somewhat ambiguous (well, it's actually not what I want). Is there a way to test the null hypothesis: population Sharpe ratio of $X$ equals population Sharpe ratio of $Y$, given independent collections of observations drawn from $X$ and $Y$? Here Sharpe ratio is mean divided by standard deviation.

edit: given $n_x, n_y$ observations of $X, Y$, construct sample means, standard deviations, to get sample Sharpe ratios: $\hat{S}_x = \frac{\hat{\mu}_x}{\hat{\sigma}_x}, \hat{S}_y = \frac{\hat{\mu}_y}{\hat{\sigma}_y}$. Then $t_x = \sqrt{n_x}\hat{S}_x$, and $t_y = \sqrt{n_y}\hat{S}_y$ are distributed as non-central t-statistics with noncentrality parameters $\sqrt{n_x}S_x$ and $\sqrt{n_y}S_y$, where $S_x, S_y$ are the population Sharpe ratios of $X, Y$. Given these independent observations, I wish to test the null hypothesis $H_0: S_x = S_y$. In one form of the problem, one only has the summary statistics $n_x, n_y, \hat{\mu}_x, \hat{\mu}_y, \hat{\sigma}_x, \hat{\sigma}_y$.

For large sample sizes, $t_x, t_y$ are approximately normal, I believe but the small sample size case is also of interest (funds often quote performance based on monthly returns).

  • $\begingroup$ When you say that the Sharpe ratio is the mean divided by the sample standard deviation... "Up to a constant factor (sqrt(n), where n is the number of observations)"... what does this last part mean? Is it really the mean divided by the standard error of the mean? $\endgroup$ Aug 13, 2010 at 5:44

1 Answer 1


I'm probably missing something important, but why does the fact that your observed variable is a Sharpe ratio change the statistic you would use to test the difference in Sharpe ratios? I understand that they are already distributed like 2 independent non-central t statistics, but what forces you to treat them that way?

Presumably the central limit theorem would hold even for Sharpe ratios and as such you should be able to apply a parametric test of mean differences, e.g. independent-samples Z.

More importantly, if your data is financial data wouldn't it be better to treat these as paired samples paired by the times at which they were observed?

  • $\begingroup$ yes, my question was poorly worded and ambiguous. to your last point: there are well known techniques for comparing Sharpe given paired observations. My Q is for independent observations, applicable for: 1. non-overlapping times, or 2. the case where summary statistics are provided by a third party (e.g. in a fund prospectus). $\endgroup$
    – shabbychef
    Aug 13, 2010 at 16:46
  • $\begingroup$ @shabbychef I realize this is pretty late to the party but could you point me in the direction of some the the techniques for comparing Sharpe given paired observations? $\endgroup$
    – mgilbert
    Sep 22, 2015 at 21:14
  • 1
    $\begingroup$ Leung & Wong test is probably still the best approach: papers.ssrn.com/sol3/papers.cfm?abstract_id=907270 $\endgroup$
    – shabbychef
    Sep 22, 2015 at 22:00

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