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

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 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 $$-1 versus when there are more negative than positive $$-2?

Thank you so much!

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.