Understanding extinction probability in simple GWP

[psie]

TL;DR

I'm studying a Galton-Watson process (GWP), with the usual assumptions. I am stuck on a theorem regarding the extinction probability.

I am paraphrasing from An Intermediate Course in Probability by Gut.

Background:

Let $$X(n)=\text{ number of individuals in generation } n.$$ We assume $X(0)=1$ and that all individuals give birth according to same probability law, independent of each other. Also, the number of offspring produced by an individual is independent of the number of individuals in their generation. Let $Y$ and $\{Y_k: k\geq1\}$ be generic r.v.s. denoting the number of children obtained by individuals, and set $p_k=P(Y=k), \ k=0,1,2,\ldots$. The case $P(Y=1)=1$ is excluded. $X(1)$ equals the number of children of the ancestor and $X(1)\stackrel{d}{=} Y$, i.e. $X(1)$ and $Y$ are equidistributed. Let $Y_1,Y_2,\ldots$ be the number of children obtained by the first, second, ... child. From the assumptions, $Y_1,Y_2,\ldots$ are i.i.d. and independent of $X(1)$. Since $$X(2)=Y_1+\ldots+Y_{X(1)},$$we've got a random sum of random variables. It is known that the probability generating function (pgf) $g_{X(2)}$ of $X(2)$ is simply $g_{X(1)}(g_Y(t))$. And since $X(1)\stackrel{d}{=} Y$, $g(t)=g_{X(1)}(t)=g_Y(t)$. One can easily derive a formula for the pgf of $X(n)$.

The extinction probability is \begin{align*} \eta&=P(X(n)=0\text{ for some } n) \\ &=P\left(\bigcup_{n=1}^\infty \{X(n)=0\}\right)\end{align*}

Problem:

Theorem: $\eta$ satisfies the equation $t=g(t)$.

Proof. Let $A_k=\{\text{the founding member produces } k\text{ children}\}, k\geq 0$. By the law of total probability we have $$\eta=\sum_{k=0}^\infty P(\text{extinction}\mid A_k)\cdot P(A_k).$$ Now, $P(A_k)=p_k$, and by the independence assumptions we have $$P(\text{extinction}\mid A_k)=\eta^k.$$ This yields $$\eta=\sum_{k=0}^\infty \eta^kp_k=g(\eta).$$

Question:

Why does $P(\text{extinction}\mid A_k)=\eta^k$ hold? I don't see how to justify this formally, given everything above.

 

[psie]

I think I've found an answer from various sources. This is a slightly different proof of the theorem. The idea is that each offspring gives rise to a new GWP

Let $q_n=P(X(n)=0)$ and $p_k=P(Y=k)=P(A_k)$. We've got $$q_n=\sum_{k=0}^\infty p_kq_{n-1}^k.\tag1$$ To prove $(1)$, I quote Durrett:

If $X(1)=k$, an event with probability $p_k$, then $X(n)=0$ if and only if all $k$ families die out in the remaining $n-1$ units of time, an independent event with probability $q_{n-1}^k$. Summing over the disjoint possibilities for each $k$ gives the desired result.

So $q_n=g(q_{n-1})$. Now, since $\{X(n)=0\}\subset \{X(n+1)=0\}$ for all $n\geq 1$, by continuity of measure from below, $\eta=\lim_{n\to\infty}q_n$. Also, a pgf is a continuous function on $[0,1]$ and we have that $$0=q_0\leq q_1\leq\ldots\leq q_n\leq 1,$$so $q=g(q)=\lim_{n\to\infty} g(q_n)$ holds.