I'm stuck on one problem from my stochastic class, which writes as follows:
$(N(t),t\geq 0)$ defines a Poisson process with parameter $\lambda$, and $S_n$'s are the $n$th point of arrival, following $\Gamma(n, \lambda)$. Suppose $Z_n, n\geq 1$ are i.i.d. random variables, with p.d.f. $p(x)$.
Now define $$W = \min_{n\geq1} \{S_n + Z_n\}$$
What is the probability $P(W>w)$?
Obviously $W>w$, if and only if $S_n + Z_n > w, \forall n$
But what then? Since $S_n$'s are not independent, I think there should be a smart way of doing this.
Thanks in advance!
[I'm going to assume that $P[Z_n>0]=1$ and that $(Z_n)$ is independent of $\{N(t),t\ge 0\}$.] The key to a "smart" approach to this is the observation that the collection $\{(S_n,Z_n): n\ge 1\}$ is a Poisson point process (call it $\Pi$) in the quadrant $(0,\infty)\times(0,\infty)$ with intensity $k(x,y)=\lambda p(y)$ with respect to Lebesgue measure. The event $\{W>w\}$ can be expressed as $\{\Pi(T_w)=0\}$, where $T_w$ is the solid triangle $\{(x,y): 0<x\le w, y\le w-x\}$. The random variable $\Pi(T_w)$ has the Poisson distribution with parameter $\int_{T_w}k(x,y)\,dx\,dy$. Once you evaluate this integral, the probability $P[\Pi(T_w)=0]$ will be at hand.
A second approach involves conditioning on the value of $N(w)$. Clearly $\{N(w)=0\}\subset\{W>w\}$ For $n\ge 1$, given that $N(w)=n$, the arrival times $S_1,S_2,\ldots,S_n$ have the same distribution as the order statistics of an iid collection $\{U_1,\ldots,U_n\}$ of random variable uniformly distributed on $(0,w)$. Thus, $$ P[W>z|N(w)=n] =P[U_k+Z_k>w, k=1,2,\ldots,n]=\left(P[U+Z>w]\right)^n, $$ where $Z$ has density $p$ and $U$ is independent of $Z$ and uniformly distibuted on $(0,w)$. Therefore, $$ P[W>z|N(w)=n] =\left(\int_0^w G(w-s)\,ds\right)^n, $$ where $G(y)=\int_y^\infty p(z)\,dz =P[Z>y]$. Use the Law of Total Probability and the known distribution of $N(w)$ to conclude.