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?
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asked Nov 19, 2009 at 15:30
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2$\begingroup$ 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 … $\endgroup$– Harald Hanche-OlsenCommented Nov 19, 2009 at 17:45
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4 Answers
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It just occurred to me that since Cauchy random variables are closed under reciprocals and closed under sums, they're closed under harmonic means.
answered Nov 19, 2009 at 17:02
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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.
answered Nov 20, 2009 at 13:02
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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 $1/2$.
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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.
answered Nov 19, 2009 at 15:44
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$\begingroup$ No perhaps: it's a trivial exercise to show this works. $\endgroup$ Commented Nov 19, 2009 at 15:46
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$\begingroup$ 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). $\endgroup$ Commented Nov 19, 2009 at 15:54
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3$\begingroup$ The harmonic mean is the arithmetic mean conjugated by x \mapsto 1/x. $\endgroup$ Commented Nov 19, 2009 at 19:25