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

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.