Does finite variance imply on a finite mean?

Question

Assume that a random variable has a finite variance. Does it mean it also has a finite mean?

My approach: I think since it has a finite variance, it means it is in $\mathcal{L}_2$ space of the probability measure $\mu$. Therefore, it should be in $\mathcal{L}_1$ space of the probability measure $\mu$ and thus the mean is finite.

IS that correct?

Answer 1

You just restated the problem. To prove it, use Cauchy Schwarz Inequality: $(E|X|)^2\leq EX^2$ for any r.v. $X$.Thus $E|X|\leq (EX^2)^{1/2}<\infty$ hence you have finite mean of $|X|$. Therefore, $EX^{+}<\infty, EX^{-}<\infty$ so $EX=EX^{+}-EX^{-}$is finite.

Answer 2

By definition, we know$\def\Var{{\textrm{Var}}\,} \def\Cov{{\textrm{Cov}}\,} \def\Corr{{\textrm{Corr}}\,} \def\E{{\textrm{E}}\,}$ $\Var(X) = \E ( X - \E(X))^2$. Hence if variance is finite, $\E(X)$ has to be finite.