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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.

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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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