Chapter 0: A Branching-Process Example

The purpose of this chapter is threefold: to take something which is probably well known to you from books such as the immortal Feller (1957) or Ross (1976), so that you start on familiar ground; to make you start to think about some of the problems involved in making the elementary treatment into rigorous mathematics; and to indicate what new results appear if one applies the somewhat more advanced theory developed in this book. We stick to one example: a branching process. This is rich enough to show that the theory has some substance.

Number of children

• $\mu=f'(1)$ .

Size of generations

• $f_{n}=\underbrace{f\circ\dotsb\circ f}_{n}$ . Indeed, $\mathbb{E}(\theta^{Z_{n}})=$ $\mathbb{E}(\mathbb{E}(\theta^{Z_{n}}|Z_{n-1}))=$ $\mathbb{E}(\mathbb{E}(\prod_{k=1}^{Z_{n-1}}\theta^{X_{k}^{(n-1)}}|Z_{n-1}))=$ $\mathbb{E}(f^{Z_{n-1}}(\theta))=$ $f_{n-1}(f(\theta))$ .

Probability of extinction

• Since $f$ is continuous, $f(0)=\mathbb{P}(X=0)>0$ , and $f$ is convex, $f(\pi)=\pi$ determines the unique solution for $\pi$ .
• The cases are divided into subcritical $\mu<1$ , $\pi=1$ , critical $\mu=1$ , $\pi=1$ , and supercritical $\mu>1$ , $\pi\in(0,1)$ .

Martingale

• $\mathbb{E}(Z_{n}|Z_{0},\dotsc,Z_{n-1})=\mathbb{E}(Z_{n}|Z_{n-1}).$
• $\mathbb{E}(Z_{n}|Z_{n-1})=$ $\sum_{k\ge0}k\mathbb{P}(Z_{n}=k|Z_{n-1})=$ $[\sum_{k\ge0}k\theta^{k-1}\mathbb{P}(Z_{n}=k|Z_{n-1})]|_{\theta=1}=$ $\tfrac{d}{d\theta}[\sum_{k\ge0}\theta^{k}\mathbb{P}(Z_{n}=k|Z_{n-1})]|_{\theta=1}=$ $\tfrac{d}{d\theta}\mathbb{E}(\theta^{Z_{n}}|Z_{n-1})|_{\theta=1}=$ $\tfrac{d}{d\theta}f^{Z_{n-1}}(\theta)|_{\theta=1}=$ $Z_{n-1}f^{Z_{n-1}-1}(1)f'(1)=$ $\mu Z_{n-1}$ .
• Let $M_{n}=Z_{n}/\mu^{n}$ for $n\ge0$ . Then, $$\mathbb{E}(M_{n}|Z_{o},Z_{1},\dotsc,Z_{n-1})=M_{n-1}.$$ Hence, $M_{n}$ is a martingale relative to the process $Z$ .

$M_{\infty}$

• The Martingale Convergence Theorem (to be studied later) implies that $M_{\infty}=\lim M_{n}$ exists with probability $1$ .
• However, if $\mu\le1$ , then $\mathbb{E}(M_{\infty})=0\neq1=\lim\mathbb{E}(M_{n})$ !
• In fact, $\mathbb{E}(M_{\infty})=1=\lim\mathbb{E}(M_{n})$ iff $\mu>1$ and $\mathbb{E}(X\log X)<\infty$ . Otherwise, $M_{\infty}=0$ a.s.
• Let $L(\lambda)=\mathbb{E}\exp(-\lambda M_{\infty})$ . Since for $\lambda>0$ , $(\exp(-\lambda M_{n}))_{n\ge0}$ is bounded uniformly and converges to $\exp(-\lambda M_{\infty})$ , by the Bounded Convergence Theorem, $L(\lambda)=$ $\lim\mathbb{E}\exp(-\lambda M_{n})=$ $\lim f_{n}(\exp(-\lambda/\mu^{n}))=$ $f(\lim f_{n-1}(\exp(-\lambda/\mu/\mu^{n-1})))=$ $f(L(\lambda/\mu))$ . $L(\lambda)$ helps finding the distribution of $M_{\infty}$ .

Geometric distribution

• $f(\theta)=$ $\sum_{k\ge0}pq^{k}\theta^{k}=$ $\tfrac{p}{1-q\theta}$ , $\mu=$ $f'(\theta)|_{\theta=1}=$ $\tfrac{q}{p}$ , $\pi=$ $\tfrac{p}{1-q\pi}=$ $\tfrac{\min\{p,q\}}{q}$ .
• $f_{n}$ : consider the map $h$ from $2\times2$ matrices to fractional linear transformations such that $G=$ $\left(\begin{smallmatrix}a & b\\ c & d \end{smallmatrix}\right)\mapsto$ $h_{G}(\theta)=$ $\tfrac{a\theta+b}{c\theta+d}$ , then $(h_{G}\circ h_{H})(\theta)=h_{GH}(\theta)$ , hence, $f_{n}(\theta)=h_{G^{n}}(\theta)$ where $G^{n}=$ $\left(\begin{smallmatrix}0 & p\\ -q & 1 \end{smallmatrix}\right)^{n}=$ $\left(\begin{smallmatrix}\mu-\mu^{n} & \mu^{n}-1\\ \mu-\mu^{n+1} & \mu^{n+1}-1 \end{smallmatrix}\right)$ .
• If $\mu\le1$ , then $f_{n}(\theta)\to1$ , and the process dies out.
  • If $\mu<1$ , then $\lim_{n\to\infty}\mathbb{P}(Z_{n}=k|Z_{n}\neq0)=(1-\mu)\mu^{k-1}$ , $k\in\mathbb{N}$ .
    • Note that in this case as $n\to\infty$ , $\mathbb{E}(M_{n})\approx$ $\underbrace{\tfrac{1}{\mu^{n}}\tfrac{1}{1-\mu}}_{\approx\mathbb{E}(Z_{n}|Z_{n}\neq0)}\underbrace{\mu^{n}(1-\mu)}_{\approx\mathbb{P}(Z_{n}\neq0)}=1$ .
  • If $\mu=1$ , then $\lim_{n\to\infty}\mathbb{P}(Z_{n}/n>x|Z_{n}\neq0)=e^{-x}$ , $x\ge0$ .
    • Note that in this case $f_{n}(\theta)=\tfrac{n-(n-1)\theta}{(n+1)-n\theta}$ , and as $n\to\infty$ , $\mathbb{E}(M_{n})\approx$ $\underbrace{n+1}_{\approx\mathbb{E}(Z_{n}|Z_{n}\neq0)}\underbrace{\tfrac{1}{n+1}}_{\approx\mathbb{P}(Z_{n}\neq0)}=1$ .
• If $\mu>1$ , then $\mathbb{P}(M_{\infty}=0)=\pi=\tfrac{1}{\mu}=\tfrac{p}{q}$ and $\mathbb{P}(M_{\infty}>x)=(1-\pi)\exp(-(1-\pi)x)$ .