```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```


## Introduction

A **branching process** with **offspring** distribution $p$ on $\mathbb{N}$ is a  Markov chain $Z = (Z_0, Z_1, \ldots)$ with state space $\mathbb{N}$,  where $Z_0=1$ with transitions given by
$$\mathbb{P}(Z_{n+1} = t | Z_n =s) = \mathbb{P}(X_1 + \cdots + X_s = t),$$
where $X_i$ are iid random variables with the distribution $p$. We will always assume that $p_i \not =1$ and $p_0 + p_1 <1$, which implies that the generating function $\hat{p}$ is strictly convex. We are concerned the **extinction** probability $q \in [0,1]$.   Notice that $\{ Z_n = 0 \} \subseteq \{ Z_{n+1} = 0 \}$, so that we have
$$
\begin{eqnarray*}
\mathbb{P}(\lim_{n \to \infty} Z_n =0 ) &=& \mathbb{P}(\bigcup_{n=1}^{\infty} \{ Z_n = 0 \}) \\
&=&\lim_{n \to \infty} \mathbb{P}(Z_n =0) \\
&=& q.
\end{eqnarray*}
$$

## An analysis of the generating function

Notice that $g_{Z_0}(s) = s$ and $f(s) := g_{Z_1}(s) = \hat{p}(s)$.  Set $f_n(s):= g_{Z_{n}}(s)$.  By the random sums lemma, we have that $f_{n+1}(s) := f_{n}(f(s))$, so that more generally, we have that $f_n(s)$ is given by a $n$-fold composition.

### Collecting some computations

We set $\mu:=\mathbb{E(Z_1)}$ and $\sigma^2 := \mathrm{var}(Z_1) \not =0$.  Induction gives:
$$f_n'(1) = \mu^n = \mathbb{E}(Z_n).$$
This can also be obtained via the conditional expectations and induction:

$$\mathbb{E}(Z_{n+1} | Z_n) = \mu Z_n.$$

We also have, for $\mu \not=1$: 

$$\mathrm{var}(Z_n) = \frac{\sigma^2 \mu^n(\mu^n-1)}{\mu^2 - \mu}$$

and for $\mu =1$:

$$\mathrm{var}(Z_n) = n^2 \sigma^2.$$


## The fixed point of the generating function

**Theorem.**

Let $p$ be the offspring distribution for a branching process $Z$.  Let $\mu = \mathbb{E}(Z_1)$ and let $q \in [0,1]$ be the extinction probability.   

* If $\mu \leq 1$, then $q=1$.
* If $\mu >1$, then $q \in (0,1)$ is the unique solution to $\hat{p}(s) =s$ in the interval $[0,1)$.

*Proof.*

Notice that $\mathbb{P}(Z_n=0) = f_n(0) <1$ for all $n$, and $f_n(0)$ is non-decreasing in $n$ and convergent to $q$.  Since $f$ is continuous, we have
$$q=\lim_{n \to \infty} f_{n+1}(0) = f( \lim_{n \to \infty} f_n(0) ) = f(q).$$

* If $\mu \leq 1$, then $f'(s)<1$ for $s \in (0,1)$ and the  mean-value theorem gives that $f(s) > s$, so that we must have $q=1$.

* If $\mu >1$, then a similar argument gives that $f(s) < s$, when $s \in (1-\epsilon, 1)$, for some small $\epsilon$.  Convexity considerations give that $f(s)=s$ has an unique solution in $[0,1)$; moreover, $q <1$, since $f(s) < s$ near $s=1$--note that $f_n(0) <1$, if the limit is $q=1$, then we must at some point enter the epsilon range where $f(s)<s$, where the sequence would be decrease.

## Convergence when $\mu >1$

We consider $W_n = Z_n/\mu^n$.   Notice that with this normalization, we have
$$\mathbb{E}(W_{n+1} | W_n) = W_n.$$
Some calculations give

$$\mathbb{E}(W_n) = 1+ \frac{\sigma^2}{\mu^2 - \mu} (1- \mu^{-n}).$$

$$\mathbb{E}(W_{n+k} - W_n)^2 = \frac{\sigma^2 \mu^{-n}}{\mu^2 - \mu} (1- \mu^{-k}).$$
Thus we see that the sequence $W_1, W_2, \ldots$ is Cauchy in $L^2$ and by completeness, the sequence converges to a **random variable** $W$ in $L^2$, with mean $1$; furthermore, by taking $k\to \infty$, we see that by the Borel-Cantelli lemma and Markov's inequality we have almost-sure convergence.  





### Endnotes

* Standard references include
  * Grimmett and Stirzaker [(2001)](https://zbmath.org/1015.60002)
  * Harris [(1963)](https://zbmath.org/1037.60001)

* Version: `r format(Sys.time(), '%d %B %Y')`
  * [Source code](https://tsoo-math.github.io/lttc/branch.Rmd)

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