I'm not sure if the following statement is right or not. But it looks right to me though I don't know how to prove it. Can someone help?
Suppose $X$ and $Y$ is independent and $EX=EY=0$, do we always have $E(|X+Y|)=E(|X-Y|)$ then? If so, how to prove it?
Thanks.
Let both $X$ and $Y$ be distributed as follows. They have value $-1$ with probability $2/3$ and value $2$ with probability $1/3$, so their expectations are $0$ as required. Since they are independent, $|X+Y|$ takes the value $2$ with probability $4/9$ (when both $X$ and $Y$ are $-1$), the value $1$ with probability $4/9$ (when one of $X$ and $Y$ is $2$ and the other is $-1$), and the value $4$ with probability $1/9$ (when both $X$ and $Y$ are $2$), for an expectation of $16/9$. Meanwhile, $|X-Y|$ takes, with the same probabilities, values $0$, $3$, and $0$, respectively, for an expectation of $12/9$.