Is it true in general that $E(1/X) = 1/E(X)$?

Question

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.

$$\sum \frac{1}{x}p{X=x}=p{X=1}+\frac{p{X=2}}{2}+\frac{p{X=3}}{3}+...$$ and $$\frac{1}{\sum xp{X=x}}=\frac{1}{p{X=1}+2p{X=2}+3p{X=3}+4p{X=4}+...}$$ — idm, Aug 27 '14 at 12:20
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. — Jean-Claude Arbaut, Aug 27 '14 at 12:20
I guess I was looking for some expression showing when it holds but fair enough. Badly posed question :) — , Aug 27 '14 at 12:34

MJD · Accepted Answer · 2014-08-27T13:05:14.253

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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 $\frac 1a + \frac1b \ne \frac1{a+b}$.)

edited Aug 27 '14 at 13:05

answered Aug 27 '14 at 12:48

MJD

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Sorry but the "underlying reason" seems more akin to something like $$\frac12\left(\frac1a+\frac1b\right)\ne\frac1{\frac12(a+b)}.$$ – Did Aug 27 '14 at 13:52

score 5 · Answer 2 · edited Aug 29 '20 at 17:47

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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 $\mathbb R^+$

edited Aug 29 '20 at 17:47

MJD

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answered Feb 26 '17 at 21:03

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It would be good to mention that the inequality here is strict since 1/x is strictly convex. – amars Jun 25 '17 at 11:56
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For non-constant X. – D-Slo Jun 13 '18 at 23:10
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$E(1/X) = 1/E(X)$ when $X$ is a constant with probability $1$. – ghjk Aug 27 '20 at 21:58

score 5 · Answer 3 · edited Aug 27 '14 at 12:57

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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 $\mathrm{E}(X+1)=np+1$, $$ \mathrm{E}\left( \frac{1}{X+1}\right)=\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}p^{k}(1-p)^{n-k}=\frac{1}{n+1}\neq\frac{1}{np+1}=\frac{1}{\mathrm{E}(X+1)}. $$

edited Aug 27 '14 at 12:57

beep-boop

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answered Aug 27 '14 at 12:22

Yssub

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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 – BCLC Aug 27 '14 at 12:34
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@BCLC See my edit. – Yssub Aug 27 '14 at 12:42
Except ${\rm E}\left(\frac{1}{X+1}\right) = \frac{1-(1-p)^{n+1}}{p(n+1)}$ – StubbornAtom Jun 28 '18 at 19:48

score 0 · Answer 4 · answered Aug 27 '14 at 12:47

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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...?

answered Aug 27 '14 at 12:47

CiaPan

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Is it true in general that $E(1/X) = 1/E(X)$?

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