Long live, bacteria!

Let $P(x)$ be the probability generating function of the number $Z$ of descendants of one individual, and let $P_n(x)$ be the PGF of the number $Z_n$ of individuals at the $n$ th generation. Thus $P_0(x) = x$ and conditional probability thinking tells us that $P_{n+1}(x) = P_n(P(x))$ . By induction we deduce that $P_{n+1}(x) \; = \; P(P_n(x))$ If we define $\pi_n = P_n(0) = \mathbb{P}[Z_n = 0]$ , then we have $\pi_0 = 0$ and $\pi_{n+1} = P(\pi_n)$ . Moreover

$\begin{aligned} \mathbb{P}[\mathrm{Extinction}] & = \sum_{n=1}^\infty \mathbb{P}[\mathrm{Extinction\; first\; occurs\; at\; generation \;} n] \\ & = \sum_{n=1}^\infty \mathbb{P}[Z_n = 0 < Z_{n-1}] \; = \; \sum_{n=1}^\infty \big(\pi_n - \pi_{n-1}\big) \\ & = \lim_{n \to \infty} \pi_n \end{aligned}$

Certainly $P(1) =1$ , and so there exists $\alpha > 0$ , the least positive solution of the equation $P(x) = x$ . If we assume that $\mathbb{P}[Z = 0] \,>\, 0 \hspace{1cm} \mathbb{P}[Z > 1] \,>\, 0$ then it is clear that $P(0) = 0$ and that $P'(x) > 0$ for all $0 < x < 1$ . It is a simple induction, using the Mean Value Theorem, to show that $(\pi_n)$ is a strictly increasing sequence with $0 < \pi_n < \alpha$ for all $n \ge 1$ . Hence $0 < \mathbb{P}[\mathrm{Extinction}] = \lim_{n\to\infty}\pi_n \le \alpha$ . On the other hand, since $P$ is continuous and since $\pi_{n+1} = P(\pi_n)$ for all $n$ , it is clear that $\mathbb{P}[\mathrm{Extinction}]$ is a solution of the equation $P(x) = x$ . We are done!