# Inequalities between random variables

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 $(Y,Z)$.
Basically, when $P(Z\geqslant0)=1$ or when $P(Z\leqslant0)=1$. – Did Sep 13 '12 at 13:13