# Geometric means, standard deviation, and sharpe ratios

I have 3 related questions:

a) I've seen formulas for GM and GS which eithier do, or do not, involve taking the exponent. Which is right?

i.e. for GM I've seen both $$\text{mean}(\ln(1+r_{t}))$$ and $$\exp(\text{mean}(\ln(1+r_{t})))-1$$

For GS again I've seen both $$\text{std}(\ln(1+r_{t}))$$ and $$\exp(\text{std}(\ln(1+r_{t})))-1$$

I don't know if there's a right answer to this question or it's just a question of preference, but it seems logical to me that we should reverse the log operation by applying an exponent.

b) It's a well known approximation that the geometric mean (GM) is roughly equal to the arithmetic mean (AM) minus half the variance (V).

(this is a nice paper discussing this, and other approximations)

Does anyone know of a simple approximation for GS similar to $$GM = AM - \frac{V}{2}$$

Empirically, based on simulating Gaussian returns and also from real data, the geometric standard deviation (GS) seems to be very close to AS. This does however depend on a number of factors:

• Whether we use $$\text{std}(\ldots)$$ or $$\exp(\text{std}(\ldots)) -1$$ [See below]. Clearly the latter will always be larger.
• The level of AS. At higher levels of AS, GS will generally be higher than AS.
• The sharpe ratio of the returns. For Sharpe ratio of 0.25 GS is a little higher than AS. With a sharpe ratio of 1.0 GS is lower than AS.

c) Given GM less than AM and AS~GS, I don't understand why it's usually quoted that "geometric Sharpe ratios $$(\frac{GM}{GS})$$ are higher than arithmetic $$(\frac{AM}{AS})$$"

Again with experiments this only seems to hold true for unrealistically high Sharpes. For annualised Sharpes below around 0.7 the Geometric sharpe is lower than the arithmetic. In the real world there aren't many assets with Sharpes above 0.7....

It also depends on if what the answer to question (a) is. The exponent version of GS and GM gives a slightly lower sharpe ratio figure than the simple versions do.

The geometric mean of quantities $$\{a_1, \dots, a_n\}$$ is $$\bar{a}_g = \left( \prod_{i=1}^n a_i \right)^{1/n}$$ Taking the logarithm of both sides gives $$\log \bar{a}_g = \frac{1}{n} \sum_{i=1}^n \log a_i$$ so the log of the geometric mean is equal to the arithmetic mean of the logs.

In your case, the relevant quantities $$a_i$$ are the growth rates over each period, $$a_i = 1 + r_i$$ and plugging this into the above equation gives $$\log(1 + \bar{r}_g) = \frac{1}{n}\sum_{i=1}^n \log(1 + r_i)$$ which can be rearranged to $$\bar{r}_g = \exp\left(\frac{1}{n}\sum_{i=1}^n \log(1 + r_i) \right) - 1$$ so you are correct to exponentiate the arithmetic mean of the logs (and subtract 1) in order to calculate the geometric growth rate. The reason that you sometimes see $$\bar{r}_g = \frac{1}{n}\sum_{i=1}^n \log(1 + r_i)$$ is because when the growth rate is small, $$\log(1 + \bar{r}_g) \approx \bar{r}_g$$ so it is often an approximation that you can get away with.

To defined the geometric standard deviation of $$\{a_1,\dots,a_n\}$$ we use the idea above that the log of the geometric mean is the arithmetic mean of the logs. Similarly, define the log of the geometric variance $$\sigma_g^2$$ to be the arithmetic variance of the logs -

$$\log \sigma_g^2 = \frac{1}{n} \sum_{i=1}^n \left( \log a_i - \log \bar{a}_g \right)^2$$

Again, you can substitute $$a_i = 1 + r_i$$ on the right-hand side, and then use the approximation $$\log(1 + x) \approx x - \tfrac{1}{2}x^2$$ to compute an approximation to $$\sigma_g^2$$ in terms of the arithmetic standard deviation $$\sigma_a^2$$, and higher moments of the distribution.

For the last part of your question "I don't understand why it's quoted that geometric sharpes are higher than arithmetic" - is that true? I have never seen anyone make that claim.

• This is great. To answer your question, the claim is in this other stack answer: Apr 12, 2016 at 14:39
• quant.stackexchange.com/questions/3607/… Also in a couple of other places. It does seem unlikely, since usually the always lower GM effect almost always dominates even a slightly lower GV. Apr 12, 2016 at 14:46
• @robcarver The comment at that link is incorrect. Probably the author mis-spoke. In general you should expect the arithmetic average of returns to dominate the geometric average, and the arithmetic sharpe ratio to dominate the geometric sharpe ratio. Apr 12, 2016 at 15:12
• "Again, you can substitute $a_i=log(1+r_i)$" should read simply "$a_i = (1+r_i)$" Feb 6, 2021 at 11:28