# Geometric Variance

If the arithmetic mean is:

$\frac { \Sigma (x_i) }{n}$

and the geometric mean is

$(\prod (1+x_i) ) ^{1/n}$

The arithmetic variance is

$\frac { \Sigma(x_i-\mu)^2 } {n}$

then what is the geometric variance?

[I actually have an answer, while it gets a decent result I have to think about a way to check it, and it looks funny]

• Geometric variance is the interest rate per period over a n period time frame you need to compound to get some growth. It's good for when talking about rates over a period of time since arithmetic means will almost never gets this correct, but arithmetic are usually used as a single period estimate – Kamster Apr 5 '15 at 11:15

For a random variable $\xi$, the variance is defined by $$mean\Big(\big(\xi -mean (\xi)\big)^2\Big).$$ Then the geometric variance should be defined by $$\prod_{i=1}^n\Bigg(1+ \bigg[x_i-\prod_{j=1}^n(1+x_j)^{1/n}\,\bigg]^2\, \Bigg)^{1/n}.$$

The definition given in the link below is only a way of thinking. However, it does not provide a consistent definition. For example, for the variance var, it would be defined by something like $$\ln var = \frac{\sum_{i=1}^n (\ln A_i - \ln u_g)^2}{n}.$$ If the standard deviation is defined by $$\ln \sigma_g = \sqrt{\frac{\sum_{i=1}^n (\ln A_i - \ln u_g)^2}{n}},$$ Then what is the relationship between $var$ and $\sigma_g$?