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I need to prove this seemingly simple inequality. If $X$ and $Y$ are iid discrete random variables, how does one prove that

$2P(|X-Y|=0)\ge P(|X-Y|=x)$ where $x$ is any other positive integer.

Is there any analogous result in the continuous case?

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    Silly me, I read over the iid of course.2012-03-11

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Let $a_k:=P(X=k)=P(Y=k)$. We have \begin{align*} P(|X-Y|=x)&= P(X-Y=x)+P(Y-X=x)\\ &=\sum_{k\geq 0}P(Y=k)P(X=x+k)+\sum_{k\geq 0}P(X=k)P(Y=x+k)\\ &=2\sum_{k\geq 0}a_ka_{x+k}\\ &\leq 2\left(\sum_{k\geq 0}a_k^2\right)^{1/2}\left(\sum_{k\geq 0}a_{h+k}^2\right)^{1/2}\\ &\leq 2\sum_{k\geq 0}a_k^2\\ &=2P(X=Y). \end{align*}

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    @D.Thomine: I see.2012-03-13