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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?2012-09-13
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    Basically, when $P(Z\geqslant0)=1$ or when $P(Z\leqslant0)=1$.2012-09-13
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    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.2012-09-13
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    @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 02012-09-13
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    Like @MichaelChernick said. In general the comparison is impossible. If you have a specific situation in mind, explain it.2012-09-13
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    I will try to explain my problem to you: $px_u^c = O_u + (q_u^1 + \cdots + q_u^{c-1}) T_u + X_u^c \\ l_u^c = F_u - 1 + q_1^c (W_1 - 1) + \cdots + q_{u-1}^c (W_{u-1} - 1) \\ tx_u^c = (c - 1) f + s + l_u^c \delta \\ (q_u^c = 1) \mbox{ iff } (px_u^c \leq tx_u^c) \and (l_u^c + W_u \leq K)$ $X_u^c$ are some random variables and $q_u^c$ are some binary variables which will be 0 or 1 depending on how both the inequalities will be true. All the other letters represent constants. I am interested in evaluating the prob of the following cond.: Pr($tx_u^c - px_u^c + X_u^c + W_u \delta \geq q_u^c D_u)$2012-09-14
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    Whoops... :-) $ $2012-09-14