# Poisson Processes: waiting times

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

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, $$p_x=\int_0^x\mathrm e^{-\lambda(x-z)}f_Z(z)\mathrm dz+\int_x^{\infty}f_Z(z)\mathrm dz.$$
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
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