# Comparing 2 independent non-central t statistics

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).

• 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? Aug 13, 2010 at 5:44