Is $\mathbb{E}[|X|] \geq |\mathbb{E}[X]|$?

Question

Given a random variable $X$, from the definition of the variance, $\mathbb{E}[X^2] \geq \mathbb{E}[X]^2$. It seems intuitive that $\mathbb{E}[|X|] \geq |\mathbb{E}[X]|$ should be true. Is this the case?

Answer 1

They could be equal, but other than that you're basically right. The intuition is that $E[X]$ is a sum with both positive and negative terms, some of which cancel each other out, but $E[|X|]$ has the same terms but all made positive. (This assumes a discrete random variable; if your random variables are continuous, replace sums with integrals.)

More generally: These are both examples of Jensen's inequality, which says that $E(f(X)) \ge f(E(X))$ for a random variable X and a convex function $f$. When we have $f(x) = x^2$ this becomes $E(X^2) \ge E(X)^2$ (the variance is positive); when we have $f(x) = |x|$ this becomes $E(|X|) \ge |E(X)|$, as you observed.

Answer 2

If $f(x)$ is the pdf of $X$ you have

$$\mathbb{E}[|X|]=\int_Df(x)|x|\mathrm{d}x$$ and $$|\mathbb{E}[X]|=\left|\int_D f(x)x\mathrm{d}x\right|$$ Appying the absolute value properties: $$|\mathbb{E}[X]|=\left|\int_D f(x)x\mathrm{d}x\right|\leq \int_D |f(x)\cdot x|\mathrm{d}x=\int_D |f(x)|\cdot|x|\mathrm{d}x=\int_D f(x)|x|\mathrm{d}x=\mathbb{E}[|X|]$$ So $$|\mathbb{E}[X]|\leq\mathbb{E}[|X|]$$

Note that a $f(x)$ is always non negative, so $|f(x)|=f(x)$