Stochastic Processes and Financial Mathematics
(part one)

4.3 A branching process

Branching processes are stochastic processes that model objects which divide up into a random number of copies of themselves. They are particularly important in mathematical biology (think of cell division, the tree of life, etc). We won’t study any mathematical biology in this course, but we will look at one example of a branching process: the Bienaymé-Galton-Watson process.

The Bienaymé-Galton-Watson (BGW for short) process is parametrized by a random variable $G$, which is known as the offspring distribution. It is simplest to understand the BGW process by drawing a tree, for example:

Each dot is a ‘parent’, which has a random number of child dots (indicated by arrows). Each parent choses how many children it will have independently of all else, by taking a random sample of $G$. The BGW process is the process $Z_n$, where $Z_n$ is the number of dots in generation $n$.

Formally, we define the BGW process as follows. Let $X_i^n$, where $n,i\ge 1$, be i.i.d. nonnegative integer-valued random variables with common distribution $G$. Define a sequence $(Z_n)$ by $Z_0=1$ and

$$\begin{array}{}\text{(4.5)}& Z_{n+1}=\{\begin{array}{ll}{X}_1^{n+1}+\dots +X_{Z_n}^{n+1},& \text{ if }{Z}_n>0\\ 0,& \text{ if }{Z}_n=0\end{array}\end{array}$$

Then $Z$ is the BGW process. The random variable $X_i^n$ represents the number of children of the $i^{th}$ parent in the $n^{th}$ generation.

Note that if $Z_n=0$ for some $n$, then for all $m>n$ we also have $Z_m=0$.

Let us assume that $G\in L^1$ and write $\mu =\mathbb{E}[G]$. Let ${\mathcal{F}}_n=\sigma (X_m^i;i\in \mathbb{N},m\le n)$. In general, $Z_n$ is not a martingale because

$$\begin{array}{rl}\mathbb{E}[Z_{n+1}]& =\mathbb{E}[X_1^{n+1}+\dots +X_{Z_n}^{n+1}]\\ & =\mathbb{E}[\sum _{k=1}^{\mathrm{\infty }}(X_1^{n+1}+\dots +X_k^{n+1})\text{𝟙}\{Z_n=k\}]\\ & =\sum _{k=1}^{\mathrm{\infty }}\mathbb{E}[(X_1^{n+1}+\dots +X_k^{n+1})\text{𝟙}\{Z_n=k\}]\\ & =\sum _{k=1}^{\mathrm{\infty }}\mathbb{E}\left[(X_1^{n+1}+\dots +X_k^{n+1})\right]\mathbb{E}[\text{𝟙}\{Z_n=k\}]\\ & =\sum _{k=1}^{\mathrm{\infty }}(\mathbb{E}\left[X_1^{n+1}\right]+\dots +\mathbb{E}\left[X_k^{n+1}\right])\mathbb{P}[Z_n=k]\\ & =\sum _{k=1}^{\mathrm{\infty }}k\mu \mathbb{P}[Z_n=k]\\ & =\mu \sum _{k=1}^{\mathrm{\infty }}k\mathbb{P}[Z_n=k]\\ \text{(4.6)}& & =\mu \mathbb{E}[Z_n].\end{array}$$ Here, we use that the $X_i^{n+1}$ are independent of ${\mathcal{F}}_n$, but $Z_n$ (and hence also $\text{𝟙}\{Z_n=k\}$) is ${\mathcal{F}}_n$ measurable. We justify exchanging the infinite $\sum$ and $\mathbb{E}$ using the result of exercise (6.8).

From (4.6), Lemma 3.3.6 tells us that if $(M_n)$ is a martingale that $\mathbb{E}[M_n]=\mathbb{E}[M_{n+1}]$. But, if $\mu <1$ we see that $\mathbb{E}[Z_{n+1}]<\mathbb{E}[Z_n]$ (downwards drift) and if $\mu >1$ then $\mathbb{E}[Z_{n+1}]>\mathbb{E}[Z_n]$ (upwards drift).

However, much like with the asymmetric random walk, we can compensate for the drift and obtain a martingale. More precisely, we will show that

$$M_n=\frac{Z_n}{{\mu }^n}$$

is a martingale.

We have $M_0=1\in m{\mathcal{F}}_0$, and if $M_n\in {\mathcal{F}}_n$ then from (4.5) we have that $M_{n+1}\in m{\mathcal{F}}_{n+1}$. Hence, by induction $M_n\in {\mathcal{F}}_n$ for all $n\in \mathbb{N}$. From (4.6), we have $\mathbb{E}[Z_{n+1}]=\mu \mathbb{E}[Z_n]$ so as $\mathbb{E}[Z_n]={\mu }^n$ for all $n$. Hence $\mathbb{E}[M_n]=1$ and $M_n\in L^1$.

Lastly, we repeat the calculation that led to (4.6), but now with conditional expectation in place of $\mathbb{E}$. The first few steps are essentially the same, and we obtain

$$\begin{array}{rl}\mathbb{E}[Z_{n+1}|{\mathcal{F}}_n]& =\sum _{k=1}^{\mathrm{\infty }}\mathbb{E}[(X_1^{n+1}+\dots +X_k^{n+1})\text{𝟙}\{Z_n=k\}|{\mathcal{F}}_n]\\ & =\sum _{k=1}^{\mathrm{\infty }}\text{𝟙}\{Z_n=k\}\mathbb{E}[X_1^{n+1}+\dots +X_k^{n+1}|{\mathcal{F}}_n]\\ & =\sum _{k=1}^{\mathrm{\infty }}\text{𝟙}\{Z_n=k\}\mathbb{E}[X_1^{n+1}+\dots +X_k^{n+1}]\\ & =\sum _{k=1}^{\mathrm{\infty }}k\mu \text{𝟙}\{Z_n=k\}\\ & =\mu \sum _{k=1}^{\mathrm{\infty }}k\text{𝟙}\{Z_n=k\}\\ & =\mu Z_n.\end{array}$$ Here we use that $Z_n$ is ${\mathcal{F}}_n$ measurable to take out what is known, and then use that $X_i^{n+1}$ is independent of ${\mathcal{F}}_n$. Hence, $\mathbb{E}[M_{n+1}|{\mathcal{F}}_n]=M_n$, as required.