I have a branching process where each individual has a
$$\sum_{k=0}^{\infty} \lambda^k \frac{e^{-\lambda}}{k!}$$
(Poisson) probability of producing $k$ descendants. I want to show that the probability of ultimate extinction is certain if $\lambda \le 1$.
Well the expected value $E[Z_n]$ of the $n$-th generation will be $\lambda^n$. Therefore, as we go towards infinity if, $\lambda < 1$ then $E[Z_n] = 0$. So we expect that eventually there will be $0$ individuals, i.e. extinction will have occurred.
But if $\lambda = 1$, then $E[Z_n] = 1 \ne 0$. So extinction doesn't occur here?
Where am I going wrong? Am I going about this in the right way by checking the expected value or should I be using some other method to do determine that the probability of ultimate extinction is certain if $\lambda \le 1$?
Best Answer
Why not apply the general argument, valid for every subcritical or critical branching process?
Conditionally on $Z_1=k$, the probability to go extinct is $\theta^k$, where $\theta$ is the probability that the whole process, starting from $Z_0=1$, goes extinct. Thus, $$\theta=\sum_kP(Z_1=k)\theta^k=G(\theta),\qquad G(s)=E(s^{Z_1}),$$ where $G$ denotes the generating function of the number of offspring of each individual. If $E(Z_1)\leqslant1$ (and $P(Z_1=1)\ne1$), the function $G$ is strictly convex with $G'(1)\leqslant1$ hence $G(t)\gt t$ for every $t$ in $[0,1)$, in particular, $\theta=1$.
$\hspace{25ex}$
Source: Any textbook on branching processes (which one are you following?).
Note: The argument in your question, based on the expectations, works when $\lambda\lt1$ but it needs to be made more precise. Indeed, $E(Z_n)\to0$ and, because $Z_n$ is nonnegative and integer valued, $Z_n\geqslant\mathbf 1_{Z_n\ne0}$ almost surely hence $P(Z_n\ne0)\leqslant E(Z_n)$ and one knows that $P(Z_n\ne0)\to0$. Note that the argument would fail in general for real valued random variables.