When variance is finite, is expectation value also finite?

Question

Let expectation value of $X$ be denoted as $E(X)$. Now we define the variance $V(X)$ as below.

$V(X) \equiv E(X^2) - E(X)^2$

Now, if $V(X) < \infty$, is $E(X)$ also finite? My textbook says so but I don't understand the reason.

Jensen's inequality states, for a convex function $h(x)$, $E(h(X)) \geq h(E(X))$ when $E(X)$ and $E(h(X))$ is both finite. For example, $h(x) = x^2$ is a convex function, so we can say $E(X^2) \geq E(X)^2$ but we can use this inequality only when $E(X)$ is finite. So, though Jensen's inequality may be a hint, I don't understand how to use this theorem.

Could anyone give me a hint?

Note: I've already read Does finite variance imply on a finite mean?, but didn't think the answers were correct because they used Jensen's inequality though whether or not the mean was finite was unknown.

Answer 1

Let's work with your definition that $V(X) = E(X^2) - E(X)^2$. First look here for a proof that if $E(X^2) < \infty$ then $E(X)$ is finite which uses Holder rather than Jensen: finiteness of k-th moment implies finiteness of lower moments.

Now suppose that $E(X)$ is infinite, then by the above $E(X^2)$ must also be infinite. But then we can't define the variance as above since $\infty - \infty$ makes no sense.

Even more fundamentally, the form of variance you have given is normally a derivation from the actual definition of variance: $$ V(X) = E[(X - E(X))^2] $$ To make this definition to begin with we actually have to assume $E(X) < \infty$ or this definition makes no sense.

Answer 2

The above famous formula about variance and (square) expectations only hold when $X$ has finite second moments (i.e. variance is finite). Thus from the formula itself you can't reach the desired conclusion.

But the answer is YES. Finite second moments (or finite variance) implies finite first moments (expectations). This is simply because $0 \leq (X - 1)^2 = X^2 - 2X + 1$, then $2X - 1 \leq X^2$. If the integration of $X^2$ (second moments, or of any $(X - c)^2$) is finite, then also the integration of $2X - 1$ (or $X$) is finite.