I have daily log returns of my asset that run over several years and I would like to calculate a time series of the Rolling Sharpe Ratio.

This Sharpe Ratio asks specifically for:

  • Annualized simple returns;
  • And annualized standard deviation of simple returns.

This is not standard procedure, and I'm confused. My questions are about how to calculate the annualized simple returns and annualized standard deviation of simple returns.

Is it correct to calculate the annualized returns given historical daily data as follows: Given $r_{n} = \ln (P_{n} / P_{n-1})$ so that my daily returns are $ R_{n} = e^{r_{n}} - 1$; will my annualized returns be the geometric mean:

$$ 1 + R_{\text{annualized}} = \left( \prod_{j = 0}^{252 - 1} ( 1 + R_{n - j} ) \right)^{1/252}?$$

About the annualized standard deviation. I calculate the daily standard deviation of simple returns $R_{n}$ and multiply by $\sqrt{252}$? I know that this works with log returns because these are normally distributed, but do simple returns work in the same way?

I apologize if this question is too basic. Any reference recommendation is highly appreciated.


1 Answer 1


to answer the question about the annual return: First your product can only start with $i=1$ because otherwise you have negative indices. If this is corrected: with your definitions and so on, doesn't this simply reduce to $\log P_252 - \log P_1$ which is the log return of the last year? What you might try to express is the average annual return which makes sense if you have more than one year. E.g. if $R_2$ i the return over two years then the $R^a$ that satisfies $$ (1+R^a)^2 = 1 + R_2 $$ is the average annual return. With your formula this could be an average daily return though.

For the standard deviation you usually use log-returns (depending whether you agreed on something else or there is a law background that prescribes to use something else). The difference will be negligible. More important: with the square-root you get an annualized volatility and should compare it to an annual return.

EDIT: to answer the comment: By your definition: $$ \prod_{j = 0}^{252 - 1} ( 1 + R_{n - j} ) = \prod_{j = 0}^{252 - 1} e^{r_{n-j}} $$ and $$ e^{r_{n-j}} = \exp \left( \log P_{n-j} - \log P_{n-j-1} \right). $$ Thus the above product of telescope form. For $j=1$ you get $$ \exp \left( \log P_{n-1} - \log P_{n-2} \right) = P_{n-1} / P_{n-2} $$ and for $j=2$ it is $$ P_{n-2}/ P_{n-3} $$ thus $P_{n-2}$ cancels. For $j = 252$ you get $P_{n-252}/P_{n-253}$. Thus if all indices are correct then all terms except the first and the last cancel.

  • $\begingroup$ About the returns. I've been specifically asked to calculate the annualized simple returns given the historical daily data from several years. I really thought this would be just the geometric average over 252. I do not see why this is equal to $\log (P_{252}/P_{1} )$. Could you explain me why or point me to some reference. About the standard deviation. I've been specifically asked to deal with the annualized standard deviation of simple returns given historical daily data. $\endgroup$
    – OkThen
    Commented Dec 23, 2018 at 20:23
  • $\begingroup$ @OkThen I edited the answer and tried to show the telescope property of the product. Go to the section "in general" in this link en.wikipedia.org/wiki/Telescoping_series to see what I mean. $\endgroup$
    – Richi Wa
    Commented Dec 29, 2018 at 21:08

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