Harmonic mean estimator is notorious for the possibility of having infinite variance. Now I want to show that it has finite variance when samples are bounded. I am wondering whether my following argument is anyhow flawed.

If we have $n$ i.i.d. random variables $Y_1,\cdots, Y_n>0$, then harmonic mean estimator is defined as

$h_n = \frac{n}{\sum_{k=1}^n \frac{1}{Y_k}}$.

I am wondering whether we can claim that the variance of harmonic mean estimator is finite if $Y_k$ is bounded, that is, $Y_k<C$, where $C$ is some positive constant value.

Here goes my logic.

Let us define a new random variable $Z_k=\frac{1}{Y_k}>0$. Then the harmonic mean can be rewritten as

$h_n = \frac{\sum_{k=1}^n 1}{\sum_{k=1}^n Z_k} = \frac{\sum_{i=1}^n Z_i Y_i}{\sum_{k=1}^n Z_k} = \sum_{i=1}^n Z_i^* Y_i$,


$Z_i^* = \frac{Z_i}{\sum_{k=1}^n Z_k}$.

Now, it is easy to argue that $h_n$ is bounded, because

$h_n \le \sum_{i=1}^n Z_i^* Y_i \le C\sum_{i=1}^n Z_i^* = C$.

As the harmonic mean estimator itself is bounded, it must have finite variance.

  • $\begingroup$ I see how $h_n$ can have infinite variance but I see no way it can have infinite expectation. In what sense then do you mean it can "have infinite mean"? At any rate, since the harmonic mean (being a bona fide mean) is less than the maximum of any of its arguments, obviously when the arguments are bounded the HM will be bounded, too. $\endgroup$ – whuber Oct 16 '14 at 23:16
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    $\begingroup$ My bad. Edited to "infinite variance". $\endgroup$ – absoluteliberty Oct 16 '14 at 23:25
  • $\begingroup$ Are you familiar with the basic inequalities of means? They provide a rigorous basis for your argument. $\endgroup$ – whuber Oct 16 '14 at 23:29
  • $\begingroup$ I think that answers my question. The answer is the above argument is correct and in some sense trivial. Thanks! $\endgroup$ – absoluteliberty Oct 16 '14 at 23:36

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