$X$ and $Y$ are independent random variables, $E(X)=0$, $E(Y)$ is finite, how to prove $E(|X+Y|)\geq E(|Y|)$?

Question

I remember the inequality$|a+b|\geq ||a|-|b||$.But this can't be used to this problem. Are there some other inequalities which can work out ? Thanks!

Answer 1

There is a simple proof using conditional Jensen's inequality, but let me give a more elementary proof in case this machinery is not available to you.

Let $F_X$ and $F_Y$ be the c.d.f of $X$ and $Y$, respectively. Then

\begin{align*} E(|X+Y|) &= \int_{\mathbb{R}}\int_{\mathbb{R}} |x+y| \, dF_X(x) dF_Y(y) \\ &\geq \int_{\mathbb{R}} \left| \int_{\mathbb{R}} (x+y) \, dF_X(x) \right| dF_Y(y) \\ &= \int_{\mathbb{R}} |y| dF_Y(y) \\ &= E(|Y|). \end{align*}