Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 26J (Applied Probability)

For an interval of length $t$, the joint probability generating function of the marked counts isThis factors into two Poisson generating functions. Independent parent increments on disjoint intervals and independent marks give independent increment vectors, so the factorization holds jointly over all intervals. Thus the two processes are independent Poisson processes of rates $\lambda p$ and $\lambda (1-p)$. For $m$ types the same calculation gives ${\prod }_j\exp [\lambda p_{j}t(s_j-1)]$, proving independent Poisson processes of rates $\lambda p_j$. A zero probability gives an identically zero process.

By Poisson thinning, the first arrival times of the coupon types are independent exponential distribution $T_j$ of rates $\lambda p_j$. Completion requires all those arrivals, so $T={\max }_j{T}_j$ andTo relate time to count rigorously, let $E_k$ be the successive independent exponential waiting times of mean $1/\lambda$. The completion count $L$ depends only on the coupon labels, which are independent of all $E_k$. ThereforeNonnegative integration justifies the sum, avoiding any unproved optional-stopping assertion. Thus the Poissonized coupon completion time and count identity is $\lambda \mathbb{E}T=\mathbb{E}L$. Changing variables $s=\lambda t$ in the tail integral giveswhich depends only on the type probabilities, not $\lambda$. These expectations are finite when all $p_j>0$; a missing type makes both infinite and completion impossible.