Most papers in the literature measure expected returns using the simple average of past returns. Why is this? When is it more correct to use geometric returns instead? Any good references?

I know that using log returns solves many of the issues with arithmetic/geometric returns, but not ideal to construct portfolios.

In any case most empirical asset pricing papers tend to use arithmetic returns. Why is that?

  • 2
    $\begingroup$ Contemporaneous relative returns are additive at the portfolio level, while log returns are additive over time. But the more likely answer is that relative returns are higher than log returns so the results look more impressive. (These are academic papers, right?) $\endgroup$ Mar 2, 2021 at 16:29
  • $\begingroup$ I am reluctant to comment because you know more about this than I do. My understanding is average returns correctly represent the expected return for the next short time period, while geometric is a better measure of long run returns. $\endgroup$
    – nbbo2
    Mar 2, 2021 at 16:55
  • 1
    $\begingroup$ Perhaps a mixture of the following reasons? (1) The differences aren't too large: 0.57% vs 0.47% for MKTRF or 0.25% vs 0.21% for HML (from Ken French's website). (2) Arithmetic averages are simpler to compute. (3) The ``natural'' way to estimate $\mathbb{E}[r]$ is $\frac{1}{T}\sum\limits_{t=1}^T r_t$. $\endgroup$
    – Kevin
    Mar 2, 2021 at 17:23
  • $\begingroup$ The sentence may say average return but it's a shortcut to annualized average return? $\endgroup$ Mar 2, 2021 at 17:47
  • $\begingroup$ This seems like a second-order effect. Uncertainty around the expected return would dwarf the arithmetic/geometric difference. $\endgroup$
    – user42108
    Mar 2, 2021 at 18:34

1 Answer 1


When using arithmetic returns the right way to calculate an average is via the geometric average. The reason is that there is a multiplicative relationship between the returns. Example: Let $P_t$ denote the stock price at time $t$, then the simple (arithmetic) net return is defined as: \begin{equation} r_t=\frac{P_t-P_{t-1}}{P_{t-1}}=\frac{P_t}{P_{t-1}}-1 \end{equation} Now look at: \begin{align} r_t[k]&=\frac{P_t}{P_{t-k}}-1=\frac{P_t}{P_{t-1}}\cdot \frac{P_{t-1}}{P_{t-2}} \dots \frac{P_{t-k+1}}{P_{t-k}}-1=(1+r_t)\cdot(1+r_{t-1})\dots(1+r_{t-k+1})-1\\&=\prod_{j=0}^{k-1}(1+r_{t-j})-1 \end{align} The annualized average return (geometric average of $k-1$ period returns) is given by \begin{align} \overline{r_t[k]}&=\left(\prod_{j=0}^{k-1}(1+r_{t-j})\right)^{\frac{1}{k}}-1\\&=\exp\left(\ln\left(\prod_{j=0}^{k-1}(1+r_{t-j})\right)^\frac{1}{k} \right)-1\\&=\exp\left(\frac{1}{k} \sum_{j=0}^{k-1}\ln(1+r_{t-j}) \right)-1 \end{align} However, notice that this expression is quite difficult to compute. So it is often approximated by the arithmetic average of the $k-1$-period returns \begin{equation} \overline{r_t[k]} \approx \frac{1}{k}\sum_{j=0}^{k-1}r_{t-j} \end{equation} This linear approximation obtains from a 1-order Taylor expansion around zero. So if the returns $r_{t-j}$ are small you get almost the same result with less effort. Especially when looking at daily returns this approximation is very good in most cases, because $r_{j-i} \approx 0$.

When you are using log-returns it is even simpler, because there is no multiplicative relationship and the "right" average is the simple average.


  • $\begingroup$ Going to upvote this. Will wait a couple more weeks to see if other answers come around. Else will accept your answer. Thanks. $\endgroup$
    – phdstudent
    Mar 4, 2021 at 11:44
  • $\begingroup$ I forgot. A good reference for that is: Tsay (2010): Analysis of Financial Time Series, 2nd edition, chapter 1. $\endgroup$
    – Lars
    Mar 4, 2021 at 14:49

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