Expected Extinction Time in Branching Process

A branching process is a sequence ${(X_i)}_{i=0}^\infty$ of $\mathbb{N}$ valued random variables on a probability space $(\Omega,\mathcal{F},\mu)$ such that there is some distribution $p$ over $\mathbb{N}$ such that the distribution of $X_{i+1}$ conditional on $X_i$ is the sum of $X_i$ random variables distributed according to $p$. We may think of this as individuals in a species that reproduces asexually having offspring. Extinction is then the event $E$ that for some $i\in\mathbb{N}$, $X_i(\omega)=0$. This has probability $1$ if $p$ has mean $\mu$ less than or equal to $1$, and probability less than $1$ otherwise. Let extinction time $T(\omega)$ be defined as follows: for $\omega\in E$, $\min\{i\ |\ X_i(\omega)=0\}$, or else $\infty$. It is easy to see that if $\mu<1$ then $\textrm{E}[T(\omega)]<\infty$, and if $\mu>1$, then $\textrm{E}[T(\omega)]=\infty$. But I was wondering if when $\mu=1$, we always have $\textrm{E}[T(\omega)]=\infty$?

Solution 1:

The extinction time $T:=\inf\{n\ge1:X_n=0\}$ will have either finite or infinite mean depending on the offspring distribution $p$, equivalently on its probability generating function $$F(s):=\sum_{k=0}^\infty p_ks^k,\quad s\in[0,1].$$ Suppose that the mean offspring $\mu:=F'(1-)$ equals $1$. Then we have $\mathbb E[T]<\infty$ if and only if $$I:=\int_0^1\frac{1-u}{F(u)-u}\,\mathrm du<\infty,$$ see the Lemma on page 494 of Seneta, “The Galton-Watson Process with Mean One”, Journal of Applied Probability , 1967, Vol. 4, No. 3, pp. 489-495.

If the offspring distribution has finite variance, $\sigma^2:=F''(1-)<\infty$, then we always have $\mathbb E[T]=\infty$ (because $F(u)-u\sim\frac12(1-u)^2\sigma^2$ as $u\to1^-$ makes $I$ infinite). See Proposition 5 of Prof. Lalley's notes for an easy proof.

In the infinite variance case, $F(s):=s+\frac23(1-s)^\frac32$ is an example of p.g.f. for which $\mathbb E[T]<\infty$.