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.
$\endgroup$ 64 Answers
$\begingroup$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 $\frac 1a + \frac1b \ne \frac1{a+b}$.)
$\endgroup$ 1 $\begingroup$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 $\mathrm{E}(X+1)=np+1$, $$ \mathrm{E}\left( \frac{1}{X+1}\right)=\sum_{k=0}^n\frac{1}{k+1}\binom{n}{k}p^{k}(1-p)^{n-k}=\frac{1}{n+1}\neq\frac{1}{np+1}=\frac{1}{\mathrm{E}(X+1)}. $$
$\endgroup$ 3 $\begingroup$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 $\mathbb R^+$
$\endgroup$ 3 $\begingroup$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...?
