Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 12F (Probability)

The probability generating function isTermwise differentiation givesThe coefficients are nonnegative, and $\mathbb{P}(X\ge 2)>0$, so both derivatives are positive and nondecreasing on $(0,1]$.

Let $d_n$ be the probability of extinction by generation $n$. Conditioning on the initial individual's offspring givesThe events increase to eventual extinction, so $d_n\uparrow d$ and continuity gives $d=G_X(d)$. If $r\ge 0$ is any other fixed point, induction from $d_0=0\le r$ and monotonicity of $G_X$ gives $d_n\le r$ for every $n$. Hence $d\le r$, proving the Galton-Watson extinction fixed point characterization.

Since $G_X(1)=1$ and $G_X^{\prime }(1)=\mu >1$, for $t<1$ sufficiently close to one,At zero, $G_X(0)-0\ge 0$. Continuity therefore gives a fixed point in $[0,1)$, and part (b) implies $d<1$.

Taylor's theorem about $1$ gives, for some $\xi \in [t,1]$,Since $G_X^{\prime \prime }$ is nondecreasing andthe stated inequality follows. At the fixed point $d<1$,Therefore the quadratic Galton-Watson extinction bound isThe denominator is at least $2(\mu -1)$ because $X$ is integer-valued and $\mathbb{E}[(X-1)(X-2)]\ge 0$, so $d^{\ast }\ge 0$.