How to obtain $\lim_{n\to \infty}\frac{\mathbb{P}(T>n)}{p^n}=?$ for Galton-Walson tree

Question

Consider a Galton-Walson tree with $Z_0\geq 1$ and denote by $Z_n$ the number of particles in nth generation. For some $p\in (0,1$, assume the offspring distribution $p_0=1-p, p_1=p$. Define the extinction time $$T:=\min\{n\geq 0: Z_n=0\}$$. How to derive the value of the $$\lim_{n\to \infty}\frac{\mathbb{P}(T>n)}{p^n}=?$$

Answer 1

Let $(\xi_n)$ be a sequence of i.i.d random variables with distribution $\mathcal{B}(p )$ (that is $\mathbb{P}(\xi_1=1) = 1- \mathbb{P}(\xi_1 = 0) = p$). It should be clear that $Z_{n+1} = \xi_{n+1} \mathbf{1}_{Z_n = 1}$. In words, conditionally on the number $Z_n$ of individuals in generation $n$, the number $Z_{n+1}$ of individuals in generation $n+1$ is either $0$ if $Z_n = 0$ or Bernoulli distributed if $Z_n =1$. Now we have $$T := \min\left\lbrace n \geq 0 :\, Z_n = 0 \right\rbrace = \min\lbrace n \geq 1: \, \xi_n = 0\rbrace.$$ This means that $T$ is the number of Bernoulli trials needed to get one failure so $T$ is geometrically distributed with parameter $1-p$. Finally, $$\mathbb{P}(T>n) = \mathbb{P}(\xi_1 = 1, \ldots ,\xi_n = n) = p^n.$$

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Another (simpler, IMO) way to view this is in terms of trees. Notice that since every individual has either $0$ or $1$ offspring, the resulting tree consists of one branch with height $T$. Furthermore, to each node corresponds a $\mathcal{B}( p )$-distributed random variable giving the number of offspring. So you see directly that the extinction time $T$ is the number of Bernoulli trials needed to get one failure.