Many times when expressing returns, the arithmetic returns are used instead of geometric returns. This is actually quite problematic. But there are ways of actually relating the two returns which I'll share today.

The arithmetic mean, for a random variable $X$, is defined as:

\[\begin{equation*}

A =\frac{1}{n} \sum_{i=1}^n x_i \tag{Arithmetic Mean}

\end{equation*}\]

The geometric mean, for a random variable $X$, is defined as:

\[\begin{equation*}

G = \left( \prod_{i=1}^n x_i\right)^{1/n} \tag{Geometric Mean}

\end{equation*}\]

When talking about returns, however, we actually modify this slightly:

\[\begin{equation*}

g = \left( \prod_{i=1}^n (1+x_i)\right)^{1/n} -1 \tag{Geometric Mean Returns}

\end{equation*}\]

For completion, I'll mention one more mean, the harmonic mean:

\[\begin{equation*}

H = \left( \frac{1}{n}\sum_{i=1}^n x_i^{-1}\right)^{-1} \tag{Harmonic Mean}

\end{equation*}\]

This latter is useful for looking at price multiples.

The problem with the arithmetic mean is that it doesn't reflect what your returns will be. For example, if you gain 50% and lose 50% you're actually down 25%, in spite of the fact that the arithmetic mean is 0%.

But there are ways for relating the two means. I'll give you one. If you're curious about this and 3 other approximations and want to see where they come from, see here: On the Relationship between Arithmetic and Geometric Returns.

The simplest relationship is as follows:

\[\begin{align*}

g &\approx A - 0.5 V(X) \\

&\approx A - 0.5 \sigma^2_x

\end{align*}\]

where $V(X)$ is the variance of the random variable and $\sigma_x$ is the standard deviation. In most cases, you're given a standard deviation so you can find the variance by squaring it.

To see how it works consider the above example of gaining 50% and losing 50%. The arithmetic mean is $A=0\%$. The variance is $V(X)=0.5\%$. As a result, our estimate for the geometric mean returns is:

\[\begin{align*}

g &\approx A - 0.5 V(X) \\

&\approx 0\% - 0.5 (50\%) \\

&\approx -25 \%

\end{align*}\]

So at least in this case it was exact. It won't always be of course.

Running a simulation of $n=10$ annual returns 1000 times assuming a normal distribution with $N(\mu=10\%, \sigma =20\%)$ gives the following relationship between variance and the difference between arithmetic and geometric returns:

As you can see, at least in this case, the multiple is closer to 0.42 than it is to 0.5. But it's really going depend on the situation. The 0.5 factor should be a decent approximation in most cases and it's easier to remember.

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