Show that the E(|X|) is finite.

Question

Show that if $E(X^2)<\infty$ then $E(|X|)<\infty$.

My try:

In other word, if $$\int x^2f(x)dx<\infty\Rightarrow\int xf(x)dx<\infty$$ for continuous case which $\int f(x)dx=1$

or $$\sum x^2f(x)dx<\infty\Rightarrow\sum xf(x)dx<\infty$$ for discrete case which $\sum f(x)dx=1$

then I get stuck already because I can't continue.

And I found similar question on this link but I not really understand their answer, is there any simple way to solve this problem? I appreciate your help.

Answer 1

$0 \leq |X| < 1+X^2~ \Rightarrow ~ 0 \leq E[|X|] < 1 + E[X^2] < \infty$.

Answer 2

$$\int x f(x)\,dx=\int x \sqrt{f(x)}\sqrt{f(x)}\,dx\leqslant\left(\int x^2f(x)\,dx\right)^\frac12\left(\int f(x)\,dx\right)^\frac12$$ by Cauchy inequality.

Answer 3

Since $h(z)=z^2$ is a convex function, by Jensen's inequality:

$$h(E(|X|)) \le E(h(|X|))\\ \Rightarrow (E(|X|))^2 \le E(|X|^2) <\infty \\ \Rightarrow E(|X|) < \infty $$

Another way to intuitively think of this is this: Consider the random variable $|X|$. Then: $$ 0 \le V(|X|)\\ \Rightarrow 0 \le E|X^2|-(E(|X|))^2 \\ \Rightarrow (E(|X|))^2 \le E|X^2| = E(X^2) < \infty \\ \Rightarrow E(|X|) < \infty $$