Using Jensen's inequality on $\mathbb{E}[1/x]$ when x can be both positive and negative

Question

We know the function $f(x)=\frac{1}{x}$ is convex when $x$ is positive and concave when $x$ is negative.

I want to show if $\mathbb{E}[\frac{1}{x}]$ is bigger than or smaller than $\frac{1}{\mathbb{E}[x]}$ using Jensen's inequality but what happens if $x$ is on the range for example $-1<x<1$ so that it is concave on some portion and convex on some portion? How do I know the direction of the inequality?

Does the range of the function matter? For example when there are more positives than negatives $-1<x<2$ versus when there are more negative than positive $-2<x<1$?

Thank you so much!

Answer 1

Welcome to MSE!

To know the direction of the inequality, you can try doing some experiments with any simple distribution. However, what I fear is your question is not always well defined when considering $x$ both positive and negative. One example is to consider a uniform distribution on $[-1,1]$ excluding $0$. $$ x\sim Unif[-1,1]/0 $$ Then by symmetry $\mathbb E [x]=0$, $1/\mathbb E [x]$ is not defined.

Similarly, if the distribution has probability density not approaching $0$ around $0$, then the expectation $\mathbb E [1/x]$ is not well defined too. consider the uniform distribution on $[a,0)\cup(0,b]$ $$ \mathbb E [1/x]=\frac1{b-a}[\int_a^0\frac 1xdx+\int_0^b \frac 1xdx] $$ both integrals diverge.