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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]

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  • $\begingroup$ 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 '15 at 11:15
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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 $\sigma_g$?

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  • $\begingroup$ Thanks! Let me investigate and see if I can validate it. Any references to go with this, btw? $\endgroup$ – user3264325 Apr 8 '15 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! $\endgroup$ – user3264325 Apr 8 '15 at 6:10
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Perhaps this works : http://en.wikipedia.org/wiki/Geometric_standard_deviation

In particular, see under "Derivation"

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  • $\begingroup$ That's v nice - how did I miss it? : ) $\endgroup$ – user3264325 Apr 10 '15 at 7:44
  • $\begingroup$ Just realised, that the mean has to be +ve. But that's besides the point, the derivation is good $\endgroup$ – user3264325 Apr 10 '15 at 8:49

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