I was asked this question in an interview.

Let's say we have two Sharpe ratio's-
1) Sharpe of 3 computed over two years of data.
2) Sharpe of 2 computed over twelve years of data.

Is there a way to say quantitatively which one is better than the other?

I thought of using confidence intervals, using the fact that if the samples are from a normal distribution, then $\sqrt{n}$(sample mean - population mean)/Sample deviation is T student (n-1).

But then all we know is the Sharpe for the two years and twelve years. We don't have samples for yearly Sharpe to get a mean and variance, neither do we have the returns.

Can you please provide a general framework for solving these kinds of problems or any hint for the particular question?
I know i might be missing a few details, please feel free to make any assumptions to make the case simpler.

  • 1
    $\begingroup$ You can do this with the upsilon distribution, which for two Sharpes is the same as LeCoutre's lambda-prime. See also section 3.5 of my Short Sharpe Course. $\endgroup$
    – shabbychef
    Jan 25, 2019 at 6:30
  • $\begingroup$ I swear there should be a simpler answer here $\endgroup$
    – Trajan
    Jun 15, 2020 at 9:53

1 Answer 1


I can't see how two Sharpe ratios from different times could even be comparable? The logic is flawed here. If the periods are same lenght.->To compare two Sharpe ratios there are a couple of test developed for it. Check Opdyke (2007) or Jobson&Korkies test that Memmel (2003) corrected. The drawback on J&K's test is that it assumes the returns are normally distributed. Opdyke (2007) has fixed that problem.

  • $\begingroup$ Different times as in let's say the Sharpe of 2 is calculated based on past 12 years of returns, and the Sharpe of 3 is calculated based on past 2 years of returns. Which one would you prefer? $\endgroup$
    – novice
    Jan 14, 2017 at 23:02
  • $\begingroup$ J & K do not require normally distributed returns. $\endgroup$
    – shabbychef
    Jan 25, 2019 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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