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The geometric mean is like the arithmetic mean on a log scale. with the arithmetic mean it is often useful to find the standard deviation. Can the same sort of thing be done to create a geometric version.

With the arithmetic mean you can find the standard deviation by:

  1. Taking the "arithmetic distance" (difference) from the mean for each data point,
  2. Make them all positive by some function ($n^2$),
  3. taking the arithmetic mean of the results to get the variance and
  4. do the inverse of the step 2 operation ($\sqrt n$) to get a standard deviation.

$$\frac 1n{\sum_{i=0}^n (\overline x-x_i)^2}$$

Can you do a similar processes to get a "geometric standard deviation"? I think it would look something like this:

  1. Taking the "geometric distance" (ratio) from the mean for each data point,
  2. Make them all positive by some function ($|\ln(n)|$?),
  3. taking the geometric mean of the results to get the "geometric variance" and
  4. do the inverse of the step 2 operation ($e^n$?) to get a standard deviation.

$$e^\sqrt[n] {\prod_{i=0}^n\ln(\overline x/x_i)}$$

Is their a reasonable mathematical way to define a geometric standard deviation? And has it already bean done?

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    $\begingroup$ This might answer your question. $\endgroup$ – Prasun Biswas Apr 8 '15 at 13:38
  • $\begingroup$ @PrasunBiswas now I feel sheepish. $\endgroup$ – MegaTom Apr 8 '15 at 15:34

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