An analytical framework for considering the geometric mean 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 
GEOMETRIC MEAN = f(ARITHMETIC MEAN,X ,Y,Z)

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...
 A: 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.
