Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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 $(Y,Z)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.