Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2011 / B / 6 / a / Solution

A Galton-Watson process starts with one ancestor, $X_0=1$. Given generation size $X_r$, definewhere all offspring variables are independent with common distribution $Z$. The empty sum is zero, so zero is absorbing. Its extinction probability isFor the probability generating function $f(s)=\mathbb{E}[s^Z]$, the allowed fixed-point characterization isFor binomial offspring, $f(s)=(1-p+ps{)}^k$, with $f(1)=1$, $f^{\prime }(1)=kp$, and $f$ convex on $[0,1]$.

First assume $0<p<1$. If $k=0$, extinction occurs immediately. For $k=1$, $f(s)-s=(1-p)(1-s)>0$ for $s<1$, so the only fixed point is one. For $k\ge 2$, strict convexity givesIf $kp\le 1$, the right-hand side is at least $s$, so no fixed point below one exists and $\eta =1$. If $kp>1$, $f(s)-s<0$ for $s$ just below one, whereas $f(0)>0$. Continuity supplies a fixed point in $(0,1)$; hence $\eta <1$. Thus in the nondegenerate parameter range,For completeness, if binomial endpoint parameters are allowed, $p=0$ or $k=0$ gives $\eta =1$, while $p=1$ and $k\ge 1$ gives deterministic positive offspring and $\eta =0$. In particular $(k,p)=(1,1)$ survives forever despite mean one. Over $k\in \{0,1,\dots \}$ and $p\in [0,1]$, the full condition is