1
$\begingroup$

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.

Can someone please help me with the details here?

$\endgroup$

2 Answers 2

6
$\begingroup$

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:

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

Edit: Here is a reference

$\endgroup$
11
  • $\begingroup$ Take your point about sample vs population sigma, @phdstudent. But still respectfully disagree ;-) Will edit mine below, to elaborate. $\endgroup$
    – demully
    Jun 15, 2020 at 14:51
  • 1
    $\begingroup$ Adding a reference for clarity: alo.mit.edu/wp-content/uploads/2017/06/… $\endgroup$
    – phdstudent
    Jun 15, 2020 at 19:58
  • $\begingroup$ thanks - interesting paper, need to Monte Carlo to work on how the inconsistency happens. $\endgroup$
    – demully
    Jun 15, 2020 at 22:32
  • $\begingroup$ 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! $\endgroup$
    – demully
    Jun 15, 2020 at 23:49
  • $\begingroup$ The questions is whether (Mu-Rf) is statistically different from zero to start with! $\endgroup$
    – phdstudent
    Jun 16, 2020 at 10:24
1
$\begingroup$

This one divides people ;-)

There is a very simple answer to your question; usually proposed by people with decades of markets experience, for whom (a) a Monte Carlo proof of statistical consistency is "enough". And (b) who tend to think that market uncertainty will always trump model uncertainty many times over. Which can upset a different group, who get upset by the lack of greek letters, associated lengthy calculus. and formal proofs. The difference is more philosophical than substantive; because the two approaches don't tend to suggest very different outcomes when applied to real-world data.

It comes down to whether you are happy making the following intuitive statement, or not. "The significance of Sharpe>0 is the same as that of Returns>0 given Time and Volatility". Assuming a large sample (and thus Student's T ~ Normal Z, as you you say), then:

Returns = Sharpe * Time * Vol

Timed Vol = Vol * root(Time)

The one-tailed p-value is Inv-Normal(Returns/Timed Vol), equals N'(SR * root(Time)). Simples...

Many (often more scholarly) commentators are not happy with the initial intuitive assumption above. IE P(SR>0) = P(Returns>0 | Vol). They do not think of the Sharpe as a convenient ratio for comparing different securities; but as a phenomenon in its own right; with its own distribution. In which case, they would argue that it has its own distribution and its own standard error, in its own right.

As opposed to volatility already being the standard error for returns; and Sharpe being returns/SE, equals already the Z-score (or T-stat) for the simplest hypothesis test as per Stats 101.

Whichever logic is "correct" depends on my intellectual and practical priors here, I suppose ;-)

$\endgroup$
1
  • $\begingroup$ "As opposed to volatility already being the standard error for returns; and Sharpe being returns/SE, equals already the Z-score (or T-stat) for the simplest hypothesis test as per Stats 101." But for your first solution you used returns/timed vol not returns/vol? $\endgroup$
    – Trajan
    Aug 18, 2021 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.