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:
- Taking the "arithmetic distance" (difference) from the mean for each data point,
- Make them all positive by some function ($n^2$),
- taking the arithmetic mean of the results to get the variance and
- 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:
- Taking the "geometric distance" (ratio) from the mean for each data point,
- Make them all positive by some function ($|\ln(n)|$?),
- taking the geometric mean of the results to get the "geometric variance" and
- 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?