Consider iid RVs $\{X_n\}$, and $E\left[\frac{1}{X_n}\right]=+\infty$. I'm looking for an upper bound for the tail of the harmonic mean of $\{X_n\}$, i.e. I want to upper bound the following

\begin{equation*} P\left(\sum_{n=1}^N \frac{1}{X_n} > t\right) = P\left(\frac{N}{\sum_{n=1}^N \frac{1}{X_n}} < \frac{N}{t}\right). \end{equation*}

But I'm out of luck because $E\left[\sum_{n=1}^N\frac{1}{X_n}\right]=+\infty$, and I cannot apply Markov's inequality; I don't think I can invoke a Chernoff bound either.

Any ideas?

  • 1
    $\begingroup$ You may use the fact that the hyperbolic mean of the $X_n$ is lower than their arithmetic mean, and find an upper bound for the arithmetic mean... Just my 2 cents. $\endgroup$ – Nicolas FRANCOIS Jul 19 '17 at 15:11
  • $\begingroup$ @NicolasFRANCOIS you're absolutely correct, but I just realized I need to upper bound the CDF of the harmonic mean, and NOT the tail. $\endgroup$ – ToniAz Jul 19 '17 at 16:09
  • $\begingroup$ I suspect that for given $t$ and $N$ there will be particular distributions meeting your requirements for $X_n$ giving values for the probability anywhere in $(0,1]$ $\endgroup$ – Henry Jul 19 '17 at 17:40

If $\sum_{n=1}^N 1/X_n > t>0$ then at least one $X_n < N/t$, so $$ P\left(\sum_{n=1}^N \frac{1}{X_n} > t\right) \le P\left(\bigcup_{n=1}^N (X_n < N/t)\right) = 1 - P(X_n \ge N/t)^N $$

  • $\begingroup$ Fantastic! I established a lower bound in a similar way, using the fact that if $\forall n,\,1/X_n > t/N$, then $\sum_{n=1}^N \,1/X_n > t$. Thank you for the upper bound! $\endgroup$ – ToniAz Jul 19 '17 at 17:54

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