Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 26I / b

An arrival at time $u\in [0,t]$ is still present at $t$ exactly when its independent service time exceeds $t-u$. Thinning the marked Poisson process with this retention probability gives a Poisson count with parameterThusThis reasoning allows atoms in the service-time distribution; the strict survival inequality correctly treats a customer whose service has ended.