Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 26J / b

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.