# For random variables in the space $L^{2}(\mathbb{P)}$, do the variances always exist?

For random variables in the space $L^{2}(\mathbb{P)}$, do the variances always exist?

Does this imply that $L^{1}(\mathbb{P)\supset }L^{2}(\mathbb{P)}$?

Many thanks!

Yes.

For finite measure spaces:

1st question

$$\operatorname{Var}(f) = E\left(f - E(f)\right)^2 = \left\|f - E(f)\right\|^2_2.$$

Since constant functions belong to $L^2$, the above expression is just a norm of some function from $L^2$, hence finite.

2nd question

$L^2(\mathbb{P}) \subseteq L^1(\mathbb{P})$ since

($1$ denotes function that is constantly 1)

$$\|f\|_2 = \|f\|_2 \| \mathbf{1} \|_2 \geq \left| \langle |f|, \mathbf{1} \rangle \right| = \|f\|_1$$

From Cauchy-Schwarz inequality.

In fact, something more general holds.

Yes. Since $$0 \leq |x| < 1+x^2 ~\forall x,$$ it follows that $$0 \leq E[|X|] < E[1+X^2] = 1+E[X^2].$$ Thus, if $E[X^2]$ is finite, then so is $E[|X|]$ finite and thus $E[X]$ is properly defined (not of the indeterminate form $\infty - \infty$ as happens with Cauchy random variables) and finite too. It follows that $$\operatorname{var}(X) = E[(X-E[X])^2] = E[X^2] - (E[X])^2$$ is finite.