Is $E(X^{-1})=\frac{1}{E(X)}$?

Question

If $E$ is the expected value of a random variable, is $E(X^{-1})=\frac{1}{E(X)}$? From the properties of the expected value it looks like this shouldn't be true. However, I have used it in a couple of execises and it seemed to work out okey.

EDIT: For example I would like to use this property to calculate $E\left(\frac{X^2}{X^2+Y^2} \right)$ where $X, Y$ are independent stardard normally distributed random variables

"However, I have used it in a couple of execises and it seemed to work out okey." I don't think so. — leonbloy, Apr 14 '20 at 12:05
What examples did you try? Suppose $X={\frac 1{10^9},1}$ with equal probability. — lulu, Apr 14 '20 at 12:05
It may be harder to find a nontrivial example where it does hold. — anomaly, Apr 14 '20 at 12:07
Also relevant: https://math.stackexchange.com/q/2312138/321264. — StubbornAtom, Apr 14 '20 at 14:10

leonbloy · Accepted Answer · 2020-04-14T13:00:12.833

4

Now, it's not. In particular, if the variable is strictly positive (and not -almost surely- constant), then, because $g(x)=1/x$ is convex, Jensen inequality tells us that

$$E(1/X) > \frac{1}{E(X)}$$

In some cases (eg asymptotics) the equation can be assumed as approximately true (namely if the variance tends to zero) - see here.

edited Apr 14 '20 at 13:00

answered Apr 14 '20 at 12:07

leonbloy

63,430

2

Assuming $X$ is not almost surely constant, of course. – Dark Malthorp Apr 14 '20 at 12:21
@DarkMalthorp Right, fixed- thanks – leonbloy Apr 14 '20 at 13:00

score 3 · Answer 2 · answered Apr 14 '20 at 12:05

3

If a random variable has an equal chance of being 1 or 2, the expected value is 1.5, so the reciprocal of the expected value is 2/3, but the expected value of the reciprocal is 3/4 (equal chance of being 1/2 or 1).

answered Apr 14 '20 at 12:05

Paul

8,153

score 1 · Answer 3 · answered Apr 14 '20 at 13:53

One way to tell that this should not hold is that it's quite easy to "break" one side while leaving the other intact. For instance:

If $X = \pm 1$ with equal probability, then $\mathbb E[X^{-1}]$ is $0$, but $1/\mathbb E[X]$ is undefined.
If $X \in \{0, 1\}$ with equal probability, then $\mathbb E[X^{-1}]$ is undefined, while $1 / \mathbb E[X]$ is $2$.

Is $E(X^{-1})=\frac{1}{E(X)}$?

3 Answers3