# Prove that $\mathbb{E}|X-Y|\geq \mathbb{E}|X|$ if $X$ and $Y$ are independent and $\mathbb{E}{X} = 0$

Given that $$X$$ and $$Y$$ are independent random variables and $$\mathbb{E}{X} = 0$$, how to prove that this inequality $$\mathbb{E}|X-Y|\geq \mathbb{E}|X|$$ holds?

Surely, it's true that $$\mathbb{E}|X-Y|\geq \mathbb{E}|X| - \mathbb{E}|Y|$$, but somehow the independence and $$\mathbb{E}{X} = 0$$ condition should sharpen this lower bound. However, I have no idea, how to use these conditions.

Any help would be appreciated.

• It is wrong. Counter-example: Let $$\mathbb P(X=-1)=\mathbb P(X=-2)=\mathbb P(X=3)=\frac13,$$ and let $Y=-1$ almost surely. Then $\mathbb E(X)=0$ and $\mathbb E|X|=2$ but $$\mathbb E|X-Y|=\frac{0+1+4}3=\frac53<\mathbb E|X|.$$ – Maximilian Janisch 2 days ago

In the original question, $$E|X|$$ should be replaced by $$E|Y|$$. Independence of $$X$$ and $$Y$$ means that $$E|X+Y|=\int E[|y+X|] P_Y(dy)\geq \int |E(y+X)|P_Y(dy)=\int|y|P_Y(dy)=E|Y|$$