Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 10F / c / Solution

With equal type probabilities, $F$ and $S$ have the same marginal probability generating function, namely $G_N((1+s)/2)$. Their assumed independence, together with $N=F+S$, givesSet $H(t)=G_N(1-t)$ for $0\le t\le 1$. The identity becomes $H(t)=H(t/2{)}^2$ and, after $k$ iterations,The finite expected value gives $G_N(1-h)=1-\mu h+o(h)$ as $h\downarrow 0$. To justify this without any analyticity assumption at the endpoint, note that $(1-(1-h{)}^N)/h\to N$ and is bounded above by $N$; dominated convergence gives the derivative $G_N^{\prime }(1-)=\mu$. Substituting the expansion into the iterated identity yieldsThe left side is independent of $k$, so it equals the limit. Thus $G_N(s)=e^{\mu (s-1)}$ on $[0,1]$ and uniqueness of the power-series coefficients proves $N\sim \mathrm{Poisson}(\mu )$. Only finite expected value was needed; the given finite variance is stronger. This proves the Poisson characterization by independent binomial splitting rather than assuming the original count distribution.