Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 12F / b / Solution

Given $X_n=k$, the next size is a sum of $k$ independent reproduction counts. Multiplication of their probability generating functions gives $\mathbb{E}[s^{X_{n+1}}\mid X_n=k]=G(s{)}^k$. Averaging this conditional expectation provesWith one initial individual, $G_0(s)=s$ and $G_1(s)=G(s)$, so $G_n$ is the $n$-fold composition of $G$ with itself.