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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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This inequality can't be true. Take $X$ to be $1$ with probability $1$. And $Y$ to be $2$ with probability $1$. –  Raskolnikov Mar 11 '12 at 11:45
    
@Raskolnikov $X$ and $Y$ are supposed to have the same law. –  Davide Giraudo Mar 11 '12 at 11:50
    
Silly me, I read over the iid of course. –  Raskolnikov Mar 11 '12 at 11:52
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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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Thanks Davide. A couple of questions - One (silly): What is the reason for the first inequality? I can't see any other condition for $|X-Y|=x$. Two: Anyway one could extend this for the continuous case? It would be great if you can add that as well... –  Bravo Mar 11 '12 at 12:26
    
For the first, write the event $\{|X-Y|=k\}$ as the event $(\{|X-Y|=k\}\cap (X\geq Y))\cup (\{|X-Y|=k\}\cap (X\leq Y))\subset \{X-Y=k\}\cup\{Y-X=k\}$. For the continuous case, if $X$ and $Y$ are independent then $P(X=Y)=0$. –  Davide Giraudo Mar 11 '12 at 13:11
    
I think that his first question was: "why is there an inequality instead of an equality?" (because there is equality, since $x$ is positive). –  D. Thomine Mar 11 '12 at 13:19
    
@Thomine: Do you mean to say there ought to be an equality in the first equation? –  Bravo Mar 11 '12 at 17:01
    
@Davide: I was envisaging something like $f_Z(z)\le 2f_Z(0)$, where $Z=X-Y$. Isn't that true? –  Bravo Mar 11 '12 at 17:01
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