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.