0
$\begingroup$

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]

$\endgroup$
1
  • $\begingroup$ 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 $\endgroup$
    – Kamster
    Apr 5, 2015 at 11:15

2 Answers 2

1
$\begingroup$

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 $$\prod_{i=1}^n\Bigg(1+ \bigg[x_i-\prod_{j=1}^n(1+x_j)^{1/n}\,\bigg]^2\, \Bigg)^{1/n}.$$

Addendum ----

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 $\sigma_g$?

$\endgroup$
2
  • $\begingroup$ Thanks! Let me investigate and see if I can validate it. Any references to go with this, btw? $\endgroup$ Apr 8, 2015 at 1:35
  • $\begingroup$ I tried a few numerical calculations. Looks good. Plus I like the approach! Wish I had some intuitive & general way to check with figures. In any case, great thanks again! $\endgroup$ Apr 8, 2015 at 6:10
3
$\begingroup$

Perhaps this works : http://en.wikipedia.org/wiki/Geometric_standard_deviation

In particular, see under "Derivation"

$\endgroup$
2
  • $\begingroup$ That's v nice - how did I miss it? : ) $\endgroup$ Apr 10, 2015 at 7:44
  • $\begingroup$ Just realised, that the mean has to be +ve. But that's besides the point, the derivation is good $\endgroup$ Apr 10, 2015 at 8:49

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.