MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The arithmetic mean of normal random variables is normal. The geometric mean of log-normal random variables is log-normal. But is there a common distribution family closed under taking harmonic means?

share|cite|improve this question
It seems to me that, since makes little sense to take the harmonic mean of variables that aren't both of the same sign, you should restrict your question to variables that are positive (almost surely). At least, I am almost sure … – Harald Hanche-Olsen Nov 19 '09 at 17:45

It just occurred to me that since Cauchy random variables are closed under reciprocals and closed under sums, they're closed under harmonic means.

share|cite|improve this answer

Any class of distributions which is closed under independent sums and almost surely nonzero will work here (and of course will also give an example for geometric means corresponding to the log-normal) the same way as in Michael's example. So besides normal (p=2) and Cauchy (p=1) random variables, the reciprocals of any p-stable random variables work. Of course, only Cauchy have the property of being distributed the same as their reciprocals, so John's answer doesn't generalize this far.

share|cite|improve this answer

Perhaps variables of the form 1/X, where X is normal? (We can ignore the problem of division by zero because X is zero only with probability zero.) Of course this isn't exactly common.

share|cite|improve this answer
No perhaps: it's a trivial exercise to show this works. – Mark Meckes Nov 19 '09 at 15:46
It was so trivial that I didn't want to actually sit down and do it, and so I was afraid there was some reason it wouldn't work (say, that x -> 1/x isn't monotone). – Michael Lugo Nov 19 '09 at 15:54
The harmonic mean is the arithmetic mean conjugated by x \mapsto 1/x. – Theo Johnson-Freyd Nov 19 '09 at 19:25

In general you want to take $X=1/Y$, where $Y$ is a stable random variable. A surprising example is $X=N(0,1)^2$, since $N(0,1)^{-2}$ is stable of index $1/2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.