Let $Z_0,Z_1,\ldots$ be a Galton-Watson process with $Z_0=1$. Let $\mu = \mathbb E(Z_1)$.
Let $\phi(s) = \mathbb E (s^{Z_1})$ be the generating function. When $\mu > 1$, there is an unique solution $\rho$ for $\phi(\rho) = \rho$ on $[0,1)$. And we have $\mathbb P\{Z_n = 0 \text{ for some } n\} = \rho$.
It is also known that $Z_n/\mu^n$ is a non-negative martingale, which converges almost surely to a random variable $W$ with $\mathbb E(W)<\infty$.
The above can be find in many probability textbooks.
What I'm trying to solve is the following exercise
Show that if $\mathbb P (\lim Z_n/\mu^n = 0) < 1$ then it is $= \rho$ and hence
$\{\lim Z_n /\mu^n > 0\} = \{Z_n > 0 \text { for all $n$} \}$ a.s.
Let $A = \{ \lim Z_n/\mu^n = 0\}$. Let $B = \{Z_n = 0 \text { for some $n$}\}$. We are asked to show then given $\mathbb P(A) <1$, we have $\mathbb P(A) = \mathbb P(B)$.
My idea is that since $B \subseteq A$, it suffices to show that $\mathbb P(A \setminus B)=0$, i.e.,
$$
\mathbb P(\lim Z_n/\mu^n = 0, Z_n > 0 \text { for all $n$}) = 0.
$$
I also know that with probability $1$, $Z_n \to 0$ or $Z_n \to \infty$. So the above equation is the same as
$$
\mathbb P(\lim Z_n/\mu^n = 0, Z_n \to \infty) = 0.
$$
I'm stuck here, because although in this set, $Z_n \to \infty$, it still could be $o(\mu^n)$. How can I show this can not happen?
Best Answer
I'm going to go another route and use the fact that we know that $\rho$ is a solution of $\sum_{k}s^k p_k=:\phi(\rho)=\rho$, $p_k$ the offspring distribution. Define $$ W_n=\frac{Z_n}{\mu^n}, W=\lim_{n\to \infty} W_n. $$ We seek to show that, in the case $1<\mu<\infty, \mathbb{P}(W=0)=\rho$. Note that $W$ exists and is a.s. finite because $W_n$ is a martingale.
We use the following decomposition for $Z_{n+1}$:
\begin{equation} Z_{n+1}=\sum_{i=1}^{Z_1} Z_n^{i}, \end{equation} where $Z_n^{i}$ are some copies of $Z_n$ independent of each other. In words, we are looking at the process $Z_{n+1}$ as a sum of some $Z_n$ processes, sum over the first descendant of the initial particle.Note $Z_{n}^{i}$ is the $i^{th}$ child of ancestor $\varnothing$.
Now divide the above by $\mu^n$ and let $n\to \infty$. You get on the left hand side $mW$ and on the right hand side $\sum_{i=1}^{Z_1} W^{i}$, where $W^{i}$ are, as above, independant copies of $W$. This tells you that $mW$ is distributed as $\sum_{i=1}^{Z_1} W^{i}$, and so $$ \begin{align} \mathbb{P}(W=0)&=\mathbb{P}(mW=0)\\ &=\mathbb{P}(\sum_{i=1}^{Z_1} W^{i}=0)\\ &=\sum_{k=1}^\infty p_k \mathbb{P}(\sum_{i=1}^{k} W^{i}=0~|~Z_1=k)\\ &=\sum_{k=1}^{\infty} p_k\mathbb{P}(W=0)^k\\ &= \phi(\mathbb{P}(W=0)), \end{align} $$ It then follows that $\mathbb{P}(W=0)$ must be equal to $\rho$.