I want to calculate the variance sigma of a Gaussian (normal) distribution given its mean mu and average deviation d (i.e. average of absolute value of difference from the mean). How can I do this?

If it makes any difference, for the particular case I'm looking at, the mean is 0, and I do know the number of observations n, though I suspect neither of these pieces of info is needed.


For iid normal data, the expected mean deviation is $\sqrt{\frac{2}{\pi}}\sigma$.

For example, see this section, the paragraph beginning "For the normal distribution, the ratio of mean absolute deviation to standard deviation..."

Also see here, which lists the result for the normal.

You can also get it from the mean of a chi(1) distribution. (Note: chi, not chi-squared.)

The derivation isn't particularly hard to do from first principles, though.

Given that result, the square of the population mean-deviation is $\frac{2}{\pi}\sigma^2$. If the mean deviation is $d$, then $\sigma^2 = \frac{\pi}{2}d^2$.

[Of course, if you're doing it from sample statistics, I'm pretty sure this won't give an unbiased estimator.]

  • $\begingroup$ @baixiwei please note the correction above - I had left out the $^2$ off the $d$ in the last part. $\endgroup$
    – Glen_b
    Sep 17 '14 at 8:32

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