Expectation of the function of a random variable

Question

If a random variable $X$ has finite expectation, is the expectation of the function of $X$, e.g. $$f(X)=\exp(X)$$ also finite? How to prove or disprove?

Answer 1

Let $X$ be uniformly distributed on $(0,1)$, in symbols $X \sim \mathcal U(0,1). $ Then it is well known that $E[X]=\frac12$ which is finite. Now consider two functions of $X$:

1. $f(x)=\dfrac1{x^2}$ and
2. $g(x)=\exp(x)$

to obtain that $E[f(x)]=\infty$ but $E[g(X)]=1<\infty$. The conclusion is that the there is no predetermint result (it depends on the distribution of $X$ and on the function $f$ of $X$ that you want to compute).

Answer 2

Let $X$ have density function $\frac{2}{x^3}$ for $x\ge 1$, and $0$ elsewhere.

Then $E(X)$ is finite, for $\int_1^\infty x\cdot \frac{2}{x^3}\,dx$ converges. But $\int_1^\infty e^x\cdot \frac{2}{x^3}\,dx$ diverges, so $E(e^X)$ does not exist.

Answer 3

An easy counter example is, if you choose the function $f$ to be $f(x) = x^2$.

Then you see, that the fact that $X$ has finite expectation does not mean that $X^2$ has finite expectation. Choose for example $\mathbb{P} (X\leq y) = y^{-1.5}$ for $y\geq 1$ or zero otherwise. That has finite mean but not finite second moment. It does also not have a finite exponential moment (i.e. if $f(x) = \exp(x)$).