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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.

user84325
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1 Answers1

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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 $12/9$.

Andreas Blass
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  • A little dispressed to see an "example" instead of a prove. Nevertheless, thanks for the nice solution. – user84325 Jun 29 '13 at 14:11
  • @user84325 I suspect (but haven't checked) that counterexamples are all over the place --- almost any not-too-symmetric distributions for $X$ and $Y$ should give a counterexample. The one in my answer was just the first not-too-symmetric distribution that happened to occur to me. – Andreas Blass Jul 01 '13 at 01:24