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I'm unsure of how to correctly answer the question below, clarification and help would be much appreciated, thank you.

Let $ W_{1},W_{2},... $ be the waiting times in a Poisson process {$ X(t); t \ge 0 $} of rate $ \lambda$. Independent of the process, let $Z_{1},Z_{2},...$ be iid RVs with common pdf f(x) where 0 < x < $\infty$. Determine $Pr[Z > z]$, where Z = min{$ W_{1}+Z_{1},W_{2}+Z_{2},... $}.

So far I have: Pr[Z>z] = $ Pr[W_{1}+Z_{1} >z, W_{2}+Z_{2} >z,... ] = Pr[W_{1}+Z_{1} >z]Pr[W_{2}+Z_{2} >z]... $

For a particular probability $ Pr[W_{k} +Z_{k} > z] $ the wait time can be interpreted as a uniform distribution $U$ distributed over $(0,z]$.
$ Pr[W_{k} +Z_{k} > z] \\ = \int_{0}^{\infty}Pr[U_{k} +Z_{k} > z |U_{k}=u]Pr[U_{k}=u]du \\ = \int_{0}^{z}Pr[Z_{k} > z-u] \frac{du}{z} \\ = \frac{1}{z} \int_{0}^{z} 1 - Pr[Z_{k} \le z-u] du \\ = \frac{1}{z} \int_{0}^{z} 1 - F_{Z}(z-u) du $

Where $ f(z) = \frac{d}{dz} F_{Z}(z)$

This seems to technically be an answer, but I'm getting the feeling I should be able to simply this down to an actual value with parameters in it.

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+1 for showing your work. –  Did Mar 28 '13 at 9:08
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You are doing all right until the line stating that "the wait time can be interpreted as a uniform distribution", which is a mystery (and wrong).

In fact, you are interested in $p_x=\mathbb P(W+Z\gt x)$, where $x\gt0$, $W$ is exponential with parameter $\lambda$, $Z$ has density $f_Z$, and $Z$ is independent of $W$.

Note that, for every fixed $z$, $\mathbb P(W+Z\gt x\mid Z=z)=\mathbb P(W+z\gt x)$ by independence, and that $\mathbb P(W\gt w)=\mathrm e^{-\lambda\max(0,w)}$ for every $w$. Hence $$ p_x=\mathbb E(\mathbb P(W+Z\gt x\mid Z))=\mathbb E(\mathrm e^{-\lambda\max(0,x-Z)})=\mathbb E(\mathrm e^{-\lambda(x-Z)};Z\lt x)+\mathbb P(Z\gt x), $$ that is, $$ p_x=\int_0^x\mathrm e^{-\lambda(x-z)}f_Z(z)\mathrm dz+\int_x^{\infty}f_Z(z)\mathrm dz. $$

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Thanks for the help. Some questions. Why is $ Pr[W>w] = e^{- \lambda max(0,w)} $? I've learned this is probably an order statistic of sorts, but I don't understand by the max of value of W is chosen. Also I don't understand how the expectation of the probability is possible. –  rhl Mar 28 '13 at 14:14
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The intervals in a homogenous Poisson process are exponential, here every $W$ is exponential of parameter $\lambda$, that is, $P(W\gt w)=e^{-\lambda w}$ for every $w\gt0$. And for $w\leqslant0$, $P(W\gt w)=1$ hence the formula. // $P(W+Z\gt x\mid Z)$ is a random variable, the one written just afterwards, that is $\exp(-\lambda\max(0,x-Z))$. In general, $P(A\mid Z)=\alpha(Z)$ where the function $\alpha$ is defined by the fact that $P(A\mid Z=z)=\alpha(z)$. –  Did Mar 28 '13 at 15:54
    
Thanks. So to solidify my own understanding, the probability in the expectation, which is the complement of the exponential CDF of W, you split up into the two cases Z>x and Z<x. // And also the conditional probability $ Pr[W+Z >x|Z] $ is equal to exp($- \lambda max(0,x-z)$) because we want to integrate over all the possible exponential probabilities, before Z>x where the exponential distribution of W is 1. –  rhl Mar 28 '13 at 17:21
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