Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 10F (Probability)

Conditional on $N=n$, the $n$ type indicators are independent random variables with the Bernoulli distribution of parameter $p$. Thus $F\mid N=n$ has the binomial distribution and its probability generating function is $(1-p+ps{)}^n$. The general rule used here is that for a nonnegative integer-valued random variable $X$, $G_X(s)=\mathbb{E}[s^X]$; for a sum of independent such variables, their probability generating functions multiply, since the expected value of the product $s^{X_1}\cdots s^{X_n}$ factors. The law of total expectation then averages the conditional result:This calculation uses the given independent labeling of objects, including conditional on their number.

A Poisson distribution with parameter $\mu$ has probability generating function $G_N(s)=e^{\mu (s-1)}$. The preceding part gives $G_F(s)=e^{\mu p(s-1)}$, the probability generating function of a Poisson distribution with parameter $\mu p$. To prove independence, retain both counts:Consequently the joint probability generating function factors asComparing the coefficients of $s^f{t}^k$ proves that the joint probabilities are the products of the marginal probabilities. Hence $F$ and $S$ are independent, with distributions $\mathrm{Poisson}(\mu p)$ and $\mathrm{Poisson}(\mu (1-p))$. This is Poisson thinning, including the degenerate counts when $p=0$ or $p=1$.

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.