I'm looking at this video: https://www.youtube.com/watch?v=fZmuJ2A9TC8 @4:43 but the issue is more general.

Here the speaker is taking monthly active returns and averaging them

(8% + 3.6% + ... + 3.7%) / (# months)

This seems near meaningless to me. It's the equivalent of the average return if we start each month with a constant amount of invested capital.

In practice I don't see why that metric makes sense. If any of these months had a -100% then the returns of all the other months are irrelevant to an investor that held the portfolio across the time period.

Can anyone help me get my head around the reasons why this is accepted practice?

  • $\begingroup$ Updated my answer to include your specific questions directly. $\endgroup$ May 18 at 23:05

3 Answers 3


Why are monthly active returns averaged? Should they not be multiplied?

His returns are log returns and assumes that they are normally distributed, hence they are additive. Assuming log returns are normally distributed implies that simple returns are lognormal and, hence not additive but multiplicative.

If any of these months had a -100% then the returns of all the other months are irrelevant to an investor that held the portfolio across the time period.

If any one out of several months has a simple return of $-100\%$, the simple return for the full period is $-100\%$ (ruin). Equivalently, the log-return will be $-\infty$, and arithmetic mean will be the same, and not influenced by the other returns, hence it is intuitively consistent.


If we assume that the price process $S_t$ follows a geometric brownian motion with constant drift and volatility (Lognormal prices), then log $\tau$-period return is given by the model $r_{t+\tau}:=\ln(S_{t+\tau}/S_t) = m\tau + \sigma \sqrt{\tau}Z_{t+1} \sim \mathcal{N}(m\tau, \tau\sigma^2)$.

Now he assumes that the non-observable monthly drift of the portfolio is $m:=\mathbb{E}[r]=10\%$ and volatility $\sigma:=20\%$. That is, $m$ and $\sigma$ are monthly drift and volatility, and since we are looking at monthly returns, $\tau=1$.

Hence with monthly parameters and monthly return, we have that $r_{t+1}:=\ln(S_{t+1}/S_t) = m + \sigma Z_{t+1} \sim \mathcal{N}(m, \sigma^2)$.

He then simulates monthly returns from this model, and then estimate the parameters (which we in reality cannot observe):

$$ \hat{m} =\frac{1}{n}\sum_{i=1}^n r_{i} = \frac{r_{1} + r_{2} + ... + r_{n}}{n} = m + \sigma\sum_{i=1}^n Z_i. $$

Also notice that due to sum of log returns $\hat{m} = r_{0:n}/n$, i.e. the full period return divided by total number of months.

Our estimate of $\hat m$ is en unbiased estimate of the drift (expected monthly return) since $\mathbb{E}[\hat{m}]=m$.

Note that if we did the same thing with simple returns, our estimator would be biased.

Now if you want to know the actual (lognormal) simple return on an investment over $n$ months, you will have to transform the log return back to simple returns. For small returns this is not needed since $R \approx r$, but for longer horizons we ought to do it properly

$$ R_{0:n} = \prod_{i=1}^n (1+R_i)-1 = \exp \left(\sum_{i=1}^n r_i\right)-1 = \exp(r_{0:n})-1 = \exp(n\hat{m})-1. $$

If any one of the simple returns $R_i$ is $-100\%$, then $R_{0:n}=-100\%$. This is the same as if any log-return is $-\infty$. Note that observing a month with minus 100 pct is an event that that won't happen given the model assumptions.


You are correct that with simple returns the time average does not account for compounding effects but this is usually well understood. One reason why this can be preferred to taking geometric averages is that the sample geometric average is a biased estimator of the corresponding population counterpart.

The difference between a geometric and arithmetic average is often small, approximately $0.5\sigma^2$ (2nd order approximation), where $\sigma^2$ is return variance. Using log-returns instead of simple returns would solve the issue but simple returns are additive over securities and can hence be preferred in some cases.


The only explanation for this can be that log-returns are used. K-period log-return is the sum of previous k-1 period returns. Otherwise, your reasons look logical to me.


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