Suppose variance of X is finite, how to show X is square-integrable

Question

Suppose $Var(X)<\infty$, how to show that $E(X^2)<\infty$?

$Var(X)=E(X^2)-[E(X)]^2<\infty$, but what about $[E(X)]^2$? Is that also finite?

Thanks for your help!

Answer 1

By definition $\operatorname{Var}X:=\mathbb{E}\left(X-\mu\right)^{2}$ where $\mu$ denotes $\mathbb{E}X$.

So expression $\text{Var}X<\infty$ only makes sense if $\mu$ is properly defined, i.e. if $\mathbb{E}\left|X\right|<\infty$.

In that context it is somehow absurd to prove things like $\mathbb{E}\left|X\right|<\infty$ on base of $\operatorname{Var}X<\infty$. It must be seen as a preassumption.

The rule $\operatorname{Var}X=\mathbb{E}X^{2}-\mu^{2}$ tells us that statement $\text{Var}X<\infty\wedge\mathbb{E}X^{2}=\infty$ cannot be true.