Here's the statement of the problem (from Ross's Introduction to Probability Models):
For those unfamiliar with "infinite server queues," they are described here. In this case, however, the service times are not exponentially distributed; rather, they are distributed according to some common distribution $G$. It follows that $X(t)$, the number of customers that have completed service by time $t$ and that arrived at time $s, s\le t,$ is Poisson distributed with mean
$$E[X(t)]=\lambda \int_{0}^{t}G(t-s)ds=\lambda \int_{0}^{t}G(y)dy.$$
Similarly, the distribution of $Y(t)$, the number of customers being served at time $t$ and that arrived at time $s, s \le t,$ is Poisson distributed with mean
$$E[Y(t)]=\lambda \int_{0}^{t}\bar G(t-s)ds=\lambda \int_{0}^{t}\bar G(y)dy$$
where $\bar G(t-s) = 1 – G(t-s)$.
Now, for part $(a)$, let $A =\{\text{the first customer to arrive is also the first to depart} \}$, i.e., our desired event; and suppose the first customer arrives at time $0$ and departs at time $t$. Then, we consider the event $A$ conditioned on the event in which $0$ customers have completed service by time $t$, i.e.,
$$ \mathbb P[A | X(t) = 0] = \exp\left\{ -\lambda \int_{0}^{t}G(y)dy\right\}. $$
Ok, I get that. But then, for some reason, the following is the answer:
$$ \mathbb P[A] = \int_{0}^{\infty} \left( \exp\left\{ -\lambda \int_{0}^{t}G(y)dy\right\} \right) dG(t). $$
And I don't really understand where this comes from. If anybody could shed some light on this, I'd really appreciate it. Thanks.
Best Answer
Given an empty system at time $0$ the number of departures by time $t$ is a non homogeneous Poisson process with rate $$\lambda(t)=\lambda\int^{t}_{0}G(y)dy$$ and so the probability of no departures by time $t$ is given by: $$exp(-\lambda\int^{t}_{0}G(y)dy)$$ now if we set time $0$ to be time of the first arrival and let $G_1$ be the service time of the first arrival then we have: $$P(A|G_1=t)=exp(-\lambda\int^{t}_{0}G(y)dy)$$ now we have $$P(A)=\int^{\infty}_{0}P(A|G_1=t)dG_1(t)=$$ $$\int^{\infty}_{0}P(A|G_1=t)dG(t)=$$ $$\int^{\infty}_{0}exp\left(-\lambda\int^{t}_{0}G(y)dy\right)dG(t)$$