TITLE: Prove: $\operatorname{E}[X^2]<\infty\Longrightarrow \operatorname{E}[X]$ exists
QUESTION [5 upvotes]: I don't know how to prove this. Lets assume that $X$ is a discrete random variable. I've just come this far: If we do a direct proof of the implication, then we start with the assumption:
$$
\operatorname{E}[X^2]=\sum_{x\in Im(X)} x^2\cdot p_X(x) <\infty.
$$
However, I don't know how I can follow from this that $\operatorname{E}[X]$ exists.

REPLY [4 votes]: I'm a big fan of Jenson's inequality for probability measures, for $f(x)=x^r$ with $r\geq 1$, implies:
$$f(E[X]])\leq E[f(X)],$$
which gives, after rearranging:
$$E[X]\leq E[X^r]^{1/r}$$

REPLY [2 votes]: Here's another; Use the Cauchy-Schwarz inequality.
$$E|X| = E|X|\cdot 1 \le \|1\|_2 \|X\|_2 = \left(E|X|^2\right)^{1/2}.$$