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I have learnt that the Sharpe ratio is a measure of the annualized return rate mean over the annualised standard deviation of return rate distribution. I also learnt that when compounding, the mean of the return rate distribution does not correspond to the overall return rate at the end of the test period (the classic example is : I have 100 USD, then I loose 50%, then I gain 50% I end up with 75 USD which is an overall return of -25%, while return mean is 0%).

Since the return mean does not correspond to reality in most of the case (i.e., when the return is compounded), why the Sharpe ratio does not take the cumulative return (i.e, exp(sum of log returns)) as a numerator rather than the mean of return rates ?

Please note that I've made a lot of research on Google and StackExchange and there seem not to be a definitive standard response to this question.

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    $\begingroup$ Are you using Sharpe as a one-off calculation or as part of a portfolio optimization? $\endgroup$
    – amdopt
    Commented May 4, 2023 at 19:52
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    $\begingroup$ Your denominator doesn't use compounded returns so you'd bias the SR (i.e. overstate it) $\endgroup$
    – oronimbus
    Commented May 4, 2023 at 20:33
  • $\begingroup$ @amdopt yes exactly, I run optimization of my parameter on a test period using BFGS, my loss function is the negative of the SR. But the "mean" that I see in many website seem to use the return rate mean annualized instead of exp(sum(log return)) annualized (cumulative return annualized) $\endgroup$
    – Jerem Lachkar
    Commented May 4, 2023 at 22:00
  • $\begingroup$ @oronimbus Denominator is the standard deviation, you mean that I should "compound" the standard deviation ? Could you please precise your comment ? $\endgroup$
    – Jerem Lachkar
    Commented May 4, 2023 at 22:01
  • $\begingroup$ @JeremLachkar Searching this site for other questions will yield a lot of results. I would recommend doing that. Something like "sharpe ratio returns" in the search bar. Good luck! $\endgroup$
    – amdopt
    Commented May 4, 2023 at 23:07

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Sharpe uses log returns, not simple.

The log return of 50/100 = -0.6931

The log return of 75/50 = 0.4054

The average is = -0.1438. This is what Sharpe uses.

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    $\begingroup$ Can you pls cite some sources which show that the log returns are typically used in calculating the Sharpe ratio rather than the simple returns? Morningstar, for example, appears to use simple returns as described here. $\endgroup$
    – Alper
    Commented May 4, 2023 at 19:34
  • $\begingroup$ Exactly, this looks very different from one website to another, that's why I was asking.. $\endgroup$
    – Jerem Lachkar
    Commented May 4, 2023 at 21:58
  • $\begingroup$ Note that you could use the cumulative returns ( assuming that you use log returns so that the cumulative return is the cumulative sum of the returns ) but then you would have to the version of the CLT that deals with sums. So, the answer you get if you use the correct formula when resorting to the CLT ( you need to calculate the standard deviation of the sum of the returns rather than the standard deviation of the mean return ) will be identical to the answer you get when using the mean return. $\endgroup$
    – mark leeds
    Commented May 5, 2023 at 4:50
  • $\begingroup$ One more thing: If you went the cumulative route, then the risk free rate that gets subtracted will also need to be either A) a sum of the various risk free rates over the appropriate period or B) an estimate of the risk free rate per period, multiplied by the number of periods used. $\endgroup$
    – mark leeds
    Commented May 5, 2023 at 4:51
  • $\begingroup$ So if I may sum up, the SR is usually more calculated on mean return and not cumulative return ? And if I go for the cumulative return way, I didn't understand what should I put as a denominator ? "you need to calculate the standard deviation of the sum of the returns rather than the standard deviation of the mean return" The sum of the return is a single number, so I can't compute a standard deviation on that. Maybe you mean the standard deviation of the log-return as a denominator ? $\endgroup$
    – Jerem Lachkar
    Commented May 10, 2023 at 15:37

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