Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 9F / Solution

Let $G(s)=\mathbb{E}[s^{\xi }]$ be the offspring probability generating function, and let $G_n(s)=\mathbb{E}[s^{Z_n}]$, for $0\le s\le 1$. Given $Z_n=k$, the next generation is a sum of $k$ independent offspring counts. Multiplication of their probability generating functions givesTaking conditional expectation and then averaging provesThus $G_n$ is the $n$-fold composition of $G$ with itself. Induction makes this precise: it is true for $n=0$, and composing once more gives the next generation. In particular $G_n\circ G=G\circ G_n$, since both are the same iterated function. This is the generating-function recursion for a Galton-Watson process.

Write $m=\mathbb{E}{Z}_1=G^{\prime }(1^{-})$, initially assuming it is finite. The law of total expectation givesStarting at one therefore yields $\boxed{\mathbb{E}{Z}_n=m^n}$, with the value at $n=0$ understood to be one even if $m=0$. Equivalently, differentiating the composition at one gives the same recurrence. If the offspring expectation is infinite, every positive generation has infinite expectation: a positive chance of at least one parent remains at each finite generation, and conditional on any positive parent count the next expected count is infinite.

For a Poisson branching process, $G(s)=e^{\lambda (s-1)}$. Extinction is absorbing because an empty generation has no parents, so extinction by generation $n$ is precisely $Z_n=0$. Hence $x_n=G_n(0)$. Using the composition in the order $G\circ G_n$ givesFor example $x_1=e^{-\lambda }$ is the chance of no offspring in the first generation. The recurrence includes the degenerate case $\lambda =0$, where extinction occurs in the first generation with certainty.