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

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?

• That is almost true : It might be "Equal" too. Change $\gt$ to $\ge$.
– Prem
May 11 at 15:05
• Try and look up Jensen's Inequality. If your intuition tells you that expectation behaves just like the "integration" operator then that is because it is an integration operator May 11 at 15:08
• Apply expectation to $-|X| \leq X \leq |X|$. May 11 at 15:26
• Expectation inequality involving absolute values is a duplicate question. May 11 at 15:44
• I made the change > to ≥. May 11 at 21:01

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.
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 $$|f(x)|=f(x)$$