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.

  • $\begingroup$ 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 $$\mathbb E|X-Y|=\frac{0+1+4}3=\frac53<\mathbb E|X|.$$ $\endgroup$ – Maximilian Janisch 2 days ago

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


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