Is there an analytical method of looking at the geometric mean that will allow one to break it down to its various components?

The focus of the question is more for financially related returns, but I am open to consider other fields as well to see if it is applicable.

So the aim of the question is to get some sort of results along the lines of


where $X,Y,Z$ are all factors/variables that help determine the value of the geometric mean.

I know there is a difference in the way its calculated etc, but I guess what I am looking for is what specific factor drives the difference between arithmetic and geometric mean. Most of the time, in financially related contexts arithmetic mean tends to be larger in magnitude than geometric mean due to the variability of return values, i.e. where $x_i$ does not equal $x_j$ where the arithmetic mean is the $\sum \limits_{i=1}^n\frac{x_i}{n}$, and the geometric mean is $\prod \limits_{i}^n {(x_i+1)}^{\frac{1}{n}} - 1$, but just saying that all the returns are not the same is what causes the difference to not quite feel exact enough.

Is there something that is better that will show me exactly what causes the gap between geometric and arithmetic means?

Approximations using Taylor series would be ok too...

  • $\begingroup$ 1) Geometric mean is the exponent of the arithmetic mean of the logarithm'ized values. 2) The formula you display for geometric mean is not genuine one. It adds 1 to values, which suggests being a trick to cope with zero values; and this affects the result. $\endgroup$
    – ttnphns
    Commented Nov 2, 2012 at 8:46
  • $\begingroup$ @ttnphs It's not a 'trick' to deal with zeros - h.l.m is dealing with returns data; the $x_i$ are like interest rates, -- the way it's presented is how returns are normally presented $\endgroup$
    – Glen_b
    Commented Nov 2, 2012 at 11:05

1 Answer 1


What you hope to do is not possible in general, except approximately.

Edit: Approximate Taylor series result:

$g(X) = g(\mu_X + X-\mu_X) = g(\mu_X) + (X-\mu_X)\cdot g'(\mu_X) + \frac{(X-\mu_X)^2}{2!}\cdot g''(\mu_X) + ...$

$E(g(X)) = g(\mu_X) + E(X-\mu_X)\cdot g'(\mu_X) + E(X-\mu_X)^2/2!\cdot g''(\mu_X) + ...$

i.e. $E(g(X)) \approx g(\mu_X) + \sigma^2_X/2\cdot g''(\mu_X)$

The following works in terms of a population, but the result can be seen to apply to a sample by treating the ECDF as a CDF.

Let $R_i$ be the $i^\rm{th}$ member of a population of returns.

Let $Y = 1+R$ and $Z = \log(Y)$.

Let $g(Y) = \log(Y)$

$E(\log(Y)) \approx \log(\mu_Y) - \sigma^2_Y/(2 \mu^2_Y)$

$\exp(E(\rm{log(Y)})) \approx \mu_Y \exp(-\frac{\sigma^2_Y}{2 \mu^2_Y})$

i.e. $\rm{GM}(Y) \approx \rm{AM}(Y) \exp(-\frac{\rm{Var}(Y)}{2 \rm{AM}(Y)})$

Approximate result based on lognormal $(1+R)$:

If, as is often assumed to approximately hold, $Z \sim \rm{N}(\mu_Z, \sigma^2_Z)$, then

$\rm{GM}(Y) \approx \rm{AM}(Y) (\frac{1}{\sqrt{\rm{Var}(Y)/\rm{AM}(Y)^2 + 1}})$

(If I haven't made an error there somewhere(!), then for $\rm{Var}(Y)$ small the two should give similar answers.)

As you can see from either formula, the arithmetic and geometric means get further apart when the coefficient of variation of the $Y$'s is larger (the variance of the $Z$'s is larger).

Because these are based on population approximations, the variance formula would normally use the $n$ denominator, but since we're already approximating, that's of no matter.

  • $\begingroup$ Why is it only possible as an approximate? $\endgroup$
    – h.l.m
    Commented Nov 4, 2012 at 20:22
  • $\begingroup$ Essentially you can't compute the quantity exactly from a fixed subset of 'factors' unless one or more of the factors effectively contains the information in the geometric mean. You can approximate it in various ways, such as via a Taylor expansion. $\endgroup$
    – Glen_b
    Commented Nov 4, 2012 at 21:39
  • $\begingroup$ Would it please be possible if you could edit your answer to view the results of this Taylor expansion to be able to view how one might be able to approximately analytically view the geometric mean? $\endgroup$
    – h.l.m
    Commented Nov 4, 2012 at 22:10
  • $\begingroup$ Could you rephrase your request? I couldn't guess what you wanted to say. Are you asking for how to use Taylor expansion to approximate geometric means in terms of arithmetic means (and some other quantities)? $\endgroup$
    – Glen_b
    Commented Nov 4, 2012 at 23:32
  • $\begingroup$ Yes I am asking for how to use Taylor expansion to approximate geometric means in terms of arithmetic means. $\endgroup$
    – h.l.m
    Commented Nov 5, 2012 at 1:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.