# How to test signifcance of a sharpe ratio

Let say you have measured a Sharpe Ratio of $$S^*$$. What is the simplest way (ie no fancy distributions) to do a hypothesis that this is different from $$0$$?

So $$H_0: \text{ The sharpe ratio is equal to 0}$$ and $$H_1: \text{ The sharpe ratio is greater than 0}$$.

So given $$S^*$$, $$\mathbb{P}( Y = S^* ) \geq 0.05$$

But what should the $$Y$$ be? I read somewhere online that it could the non centered t distribution, but I am not sure whether this could be centered to the standard t test distribution. Moreover, I would also like to consider the normal distribution and as the sample used to create the statistic should be greater than 30, the t test to normal approxaimtion should apply.

If necessary, here is an introduction to the sharpe ratio, https://www.investopedia.com/terms/s/sharperatio.asp

• The Sharpe ratio is roughly the inverse of the coefficient of variation (CV). Perhaps this connection may give some ideas (just speculating). – Richard Hardy Jun 15 '20 at 10:20
• The Short Sharpe Course on SSRN describes hypothesis testing under normal returns in Chapter 3, and under general returns in Chapter 4. It sounds like you want the latter, which amounts to the standard error approximation given in an answer below. There are higher order approximations due to Mertens' and Bao, which take into account skew of returns. I would also note that for your specific problem of testing whether the Sharpe is zero is equivalent to testing whether the mean is zero, which is a classical statistical problem. – steveo'america Jul 16 '20 at 19:23

Under the simplest assumptions (normally distributed i.i.d. returns), this paper (eq. 9) suggests the following confidence interval at the $$95\%$$ confidence level :
$$\widehat{SR} \pm 1.96 \sqrt{ \frac{\left(1 + \frac{1}{2}\widehat{SR}\right)^2}{T} }$$