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This concerns a discrete random variable $X$. I assume the relation doesn't hold in general, but I would like to prove this.

I have tried to use the property that $$ E(g(X)) = \sum_x g(x)f(x) $$ and then simply write $$ \sum_x \frac{1}{x}P(X=x) = \frac{1}{\sum_x x P(X=x)} $$ and then play around with this algebraically without any success.

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  • $\begingroup$ @Jean-ClaudeArbaut what is wrong? $\endgroup$ – Slug Pue Aug 27 '14 at 12:17
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    $\begingroup$ $$\sum \frac{1}{x}p\{X=x\}=p\{X=1\}+\frac{p\{X=2\}}{2}+\frac{p\{X=3\}}{3}+...$$ and $$\frac{1}{\sum xp\{X=x\}}=\frac{1}{p\{X=1\}+2p\{X=2\}+3p\{X=3\}+4p\{X=4\}+...}$$ $\endgroup$ – idm Aug 27 '14 at 12:20
  • $\begingroup$ Sorry, it was rather short. You argument is correct, and you obviously don't have, in general, $\frac{1}{\sum_k x_k p_k}=\sum_k \frac{1}{x_k}p_k$. You can easily find counterexamples. $\endgroup$ – Jean-Claude Arbaut Aug 27 '14 at 12:20
  • $\begingroup$ I guess I was looking for some expression showing when it holds but fair enough. Badly posed question :) $\endgroup$ – Slug Pue Aug 27 '14 at 12:34
  • $\begingroup$ This thread on CV is also relevant. $\endgroup$ – StubbornAtom Jun 28 '18 at 19:44
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Let X be the discrete distribution which takes values 1 and 2 with equal probability. Then $E (X)=\frac32 $ but $ E (\frac1x) = \frac34 $.

(Almost any distribution you choose, discrete or continuous, will confirm that $E(\frac1X)\ne\frac1{E(X)}$. The underlying reason is that $\frac 1a + \frac1b \ne \frac1{a+b}$.)

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    $\begingroup$ Sorry but the "underlying reason" seems more akin to something like $$\frac12\left(\frac1a+\frac1b\right)\ne\frac1{\frac12(a+b)}.$$ $\endgroup$ – Did Aug 27 '14 at 13:52
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Counter-example 1: If $X \sim \text{Bin}(n,p)$, then $\mathrm{E}(X)=np$, but $\mathrm{E}(1/X)$ is not even well defined: $k=0, \, 1/k=?$.

Counter-example 2: If $X \sim \text{Bin(n,p)}$, then $\mathrm{E}(X+1)=np+1$, $$ \mathrm{E}\left( \frac{1}{X+1}\right)=\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}p^{k}(1-p)^{n-k}=\frac{1}{n+1}\neq\frac{1}{np+1}=\frac{1}{\mathrm{E}(X+1)}. $$

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    $\begingroup$ how about a well defined example when it is not true? o/w I can just add at the end of the statement: whenever both expressions are defined $\endgroup$ – BCLC Aug 27 '14 at 12:34
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    $\begingroup$ @BCLC See my edit. $\endgroup$ – Yssub Aug 27 '14 at 12:42
  • $\begingroup$ Except ${\rm E}\left(\frac{1}{X+1}\right) = \frac{1-(1-p)^{n+1}}{p(n+1)}$ $\endgroup$ – StubbornAtom Jun 28 '18 at 19:48
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Another way to see that it is not true, is that for any positive random variable $X$ with $\mathbb E [X] \neq 0$, $$ \frac{1}{\mathbb E[X]} < \mathbb E\left[\frac{1}{X} \right] $$ by Jensen's inequality and the fact that $f(x) = 1/x$ is strictly convex in $\mathbb R^+$

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    $\begingroup$ It would be good to mention that the inequality here is strict since 1/x is strictly convex. $\endgroup$ – amars Jun 25 '17 at 11:56
  • $\begingroup$ For non-constant X. $\endgroup$ – D-Slo Jun 13 '18 at 23:10
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Consider a discrete $X$ which takes value $-2$ or $2$, both with probability $\tfrac 12$.
What is $1/X$ then?
And what are $E$ of the two...?

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