I have just started Poisson processes in my course on stochastic processes. We have just covered random sums and now I got the following tutorial/ exercise class question, but I don't quite see what is expected:
Customers arrive at a cashier according to a Poisson process with intensity $
\lambda$. The service
time of each customer is exponentially distributed with mean $\frac{1}{\mu}$. At time $0$, there is one
customer being served. Let $Z$ be the number of customers that arrive during this customer's
service time. Show that $Z$ has the following generating function:
$$ G_Z(r)= \frac{\mu}{\mu + \lambda (1-r)}$$
This function looks a lot like the exponential generating function $\frac{\mu}{\mu + s}$, which makes sense since the service time is exponentially distributed with parameter $\mu $. However, instead of the variable $s$, we see a $\lambda(1+r)$. Can someone help me a bit through the reasoning?
Best Answer
Suppose the additional service time for the existing customer time is $S \sim \text{Exp}(\mu)$.
Then the number of arrivals in time $S$ is $Z\sim \text{Pois}(\lambda S)$
This implies $$\mathbb P(Z=a) = \int_0^\infty e^{-\lambda s} \frac{\lambda^a s^a}{a!} \mu e^{-\mu s}\, ds = \frac{\lambda^a \mu}{(\lambda+\mu)^{a+1}}= \frac{\frac{\mu}{\lambda}}{\left(1+\frac{\mu}{\lambda}\right)^{a+1}}$$ which is the probability mass function of a geometric random variable on $0,1,2,3,\ldots$ where the probability of "success" (here the service time finishing before another customer joins the queue) is $\frac{\mu}{\lambda}$. You could have jumped straight to this result using a memoryless argument.
The probability generating function is then $$G_Z(r) = \mathbb E(r^Z) =\sum_{a=0}^\infty r^a \frac{\frac{\mu}{\lambda}}{\left(1+\frac{\mu}{\lambda}\right)^{a+1}} = \frac{\frac{\mu}{\lambda}}{\left(1+\frac{\mu}{\lambda}-r\right)}=\frac{\mu}{\mu+\lambda(1-r)}$$
which is almost what you stated in the question, but with a sign change.