Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 23K / i / Solution

In an interval of length $t$, condition on the Poisson process count $X_t=k$. Independently marking its points gives $Y_t\mid X_t=k\sim \mathrm{Bin}(k,p)$. The joint probability generating function of kept and discarded counts isIt factors into the generating functions of independent Poisson variables of means $\lambda pt$ and $\lambda (1-p)t$. The same calculation applies to any interval, and counts and marks in disjoint intervals are independent. Hence $Y$ is a Poisson process of intensity $\boxed{\lambda p}$, and $Y_t$ and $X_t-Y_t$ are independent. This is the Poisson thinning theorem.