Past exam of the options II for mathematics and philosophy course of the University of Oxford / 2013 / N2

The probability generating function iswhere the series is defined, with the usual power-series convention $s^0=1$ even at $s=0$. In particular $G_X(0)=p_0$ and $G_X(1)=1$.

If $|s|\le 1$, then ${\sum }_k|p_k{s}^k|\le {\sum }_k{p}_k=1$. Thus the power series converges absolutely at every such real $s$, including both endpoints. ThereforeThe radius of convergence is at least one, though it can be larger for particular distributions.

Inside its radius of convergence a power series can be differentiated term by term. HenceEvery mass is therefore recovered uniquely from the function, proving that the generating function determines the entire distribution. No derivatives at the potentially delicate endpoint one are needed.

For the Poisson distribution, summing the exponential series givesThe series is absolutely convergent for every real or complex $s$; its derivative at one is $\lambda$, the expected offspring count.

Count the original infected person first. Conditional on their producing $X=k$ new infections, each of the $k$ new people initiates an independent descendant family with the same law as the original total. If $T_i$ denotes that family size, the branching property givesThe family variables are mutually independent and independent of $X$. At $X=0$ the sum is empty and equals zero, so the total is one. This is the total progeny decomposition of a Galton-Watson process.

For $|s|\le 1$, all the generating-function expectations are absolutely bounded. Conditional on $X=k$, independence givesAt $k=0$ the product is one, consistent with the empty-family case. The law of total expectation now yieldsThis is the total progeny generating function of a branching process. The given almost-sure finiteness of $T$ ensures $G_T(1)=1$. The conditional-expectation proof avoids assuming a finite expected total before it has been established.

Write $g(s)=G_T(s)$. For $0<s<1$, differentiate the power-series identity from part (ii):Because $g(s)\le 1$ and $\lambda <1$, its denominator is positive. Using $g(s)=s{e}^{\lambda (g(s)-1)}$ givesAs $s\uparrow 1$, almost-sure finiteness and bounded convergence give $g(s)\to 1$. On the other hand, monotone convergence in $g^{\prime }(s)={\sum }_{k\ge 1}k\mathbb{P}(T=k)s^{k-1}$ gives $g^{\prime }(s)\to \mathbb{E}T$, initially allowing infinity. The displayed formula makes this limit finite and provesThis establishes finiteness rather than assuming it when rearranging an expectation equation. It is the subcritical branching-process total-progeny mean.