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I have the following question: Assuming that $X$ and $Y$ are two independent discrete random variables and $\mathrm{Pr}(X \leq Y)$ is known, how easily one can compute the following probability: $\mathrm{Pr}(X \leq Y + Z)$, where $Z$ is a another discrete random variable. It is also known that $Y$ and $Z$ are dependent. I think this makes things a bit complicated .

Do you have any ideas? Bogdan.

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If $Y$ is independent of $X$, then $\mathrm P(X\leqslant Y)=\mathrm E(F(Y))$, where $F:x\mapsto\mathrm P(X\leqslant x)$ is the CDF of $X$. If $Y+Z$ is independent of $X$, then $\mathrm P(X\leqslant Y+Z)=\mathrm E(F(Y+Z))$.

The comparison of $F(Y)$ and $F(Y+Z)$, or of $\mathrm E(F(Y))$ and $\mathrm E(F(Y+Z))$, depends on the specific hypotheses made on $(Y,Z)$.

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Thanks! Can you give me some particular cases where F(Y) and F(Y+Z) can be easily compared? –  Bogdan Sep 13 '12 at 13:09
    
Basically, when $P(Z\geqslant0)=1$ or when $P(Z\leqslant0)=1$. –  Did Sep 13 '12 at 13:13
    
This cases are too simple:). I was thinking more it there is a way to approximate E(F(Y+Z)) based on E(F(Y)) or anything similar. –  Bogdan Sep 13 '12 at 13:19
    
@Bogdan I think you are missing an important point. There are infinitely many ways that dependence relationships can be constructed between Y and Z as well as infinitely many chocies for random variables X, Y and Z such that say P(X<=Y)=p where p is a known value 0<p<1. –  Michael Chernick Sep 13 '12 at 17:10
    
Like @MichaelChernick said. In general the comparison is impossible. If you have a specific situation in mind, explain it. –  Did Sep 13 '12 at 18:48
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