# 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.

The answer above is not correct.

Let's go by parts:

Denote the mean of returns $$\mu$$. Denote the standard deviation of returns: $$\sigma$$.

Therefore the sharpe ratio is:

$$SR = \frac{\mu-r_f}{\sigma}$$

The corresponding standard errors are:

$$se(\hat{mu}) = \frac{\sigma}{\sqrt{t}}$$ $$se(\hat{\sigma}) = \frac{\sqrt{2} \sigma^2}{\sqrt{T}}$$ $$se(\hat{SR}) = \frac{\sqrt{1+SR^2/2}}{\sqrt{T}}$$

So the t-stat for the sharpe ratio is:

$$t-stat(\hat{SR})= \frac{\hat{SR}}{se(\hat{SR})}$$

• Take your point about sample vs population sigma, @phdstudent. But still respectfully disagree ;-) Will edit mine below, to elaborate. – demully Jun 15 at 14:51
• Adding a reference for clarity: alo.mit.edu/wp-content/uploads/2017/06/… – phdstudent Jun 15 at 19:58
• thanks - interesting paper, need to Monte Carlo to work on how the inconsistency happens. – demully Jun 15 at 22:32
• OK, it's clear gaming this that variation in realised sigmas do have an effect on SR. Assuming of course the concept of a true population volatility/variance is a meaningful concept (one for another day). This can be material testing for non-zero Sharpes. This said, it remains a tautology that if (Mu-Rf)>0 then Sharpe>0. So there's a paradox in the special case of a test for zero, if one's confidence in a positive numerator is any different to one's confidence in a positive ratio (given a positive denominator, by definition). That makes no sense at all. Tx for interesting food for thought! – demully Jun 15 at 23:49
• The questions is whether (Mu-Rf) is statistically different from zero to start with! – phdstudent Jun 16 at 10:24