Branching processes: Mean and variance

Generational growth

Consider the following scenario (see the figure below):

1. A single individual (cell, organism, etc.) produces $j(=0,1,2,\cdots )$ descendants with probability $p_j$, independently of other individuals.
2. The probability of this reproduction, $\{p_j\}$, is known.
3.  That individual produces no further descendants after the first (if any) reproduction.
4. These descendants each produce further descendants at the next subsequent time with the same probabilities. 
5. This process carries on, creating successive generations.

Figure 1. An example of the branching process.

Let $X_n$ be the random variable representing the population size (number of individuals) of generation $n$. In the above figure, we have $X_0=1$, $X_1=4$, $X_2=7$, $X_3=12$, $X_4=9.$ We shall assume $X_0=1$ as the initial condition. Ideally, our goal would be to find how the population size grows through generations, that is, to find the probability $Pr(X_n=k)$ for each $n=1,2,\cdots$. However, that's not easy, if possible. Therefore, we content ourselves by finding average properties such as the mean population size $\mathbb{E}(X_n)$ or the variance $\mathbb{V}(X_n)$, or some characteristic probabilities such as the probability of extinction.

Let's consider the probability generating function for $p_j$:

$$G(s)=\sum _{j=0}^{\mathrm{\infty }}{p}_j{s}^j.$$

Since we are assuming $X_0=1$, $G(s)$ is the generating function for the random variable $X_1$ (this means $Pr(X_1=j)=p_j$). 

Let $G_2(s)$ be the generating function of the random variable $X_2$ (the population size of the second generation). Since there are $X_1$ individuals, assuming each produce $Y_k(k=1,2,\cdots ,X_1)$ offsprings, we have

$$X_2=Y_1+Y_2+\cdots +Y_{X_1}.$$

In the case of the above Figure 1, we have a realized value of $X_2$ as

$$x_2=y_1+y_2+y_3+y_4=4+1+0+2=7.$$

In general,

$$Pr(Y_k=j)=p_j,Pr(X_1=m)=Pr(X_1=m|X_0=1)=p_m.$$

(Recall that $X_0=1$ is the assumed initial condition.)

Let $$h_n=Pr(X_2=n).$$

Using the partition theorem (or the law of total probability), we have

$$\begin{array}{rcl}{h}_n& =& Pr(X_2=n)=\sum _{r=0}^{\mathrm{\infty }}Pr(X_2=n|X_1=r)Pr(X_1=r)\\ & =& \sum _{r=0}^{\mathrm{\infty }}{p}_{r}Pr(X_2=n|X_1=r).\end{array}$$

Multiply both sides by $s^n$ and sum over $n$,

$$G_2(s)=\sum _{n=0}^{\mathrm{\infty }}{h}_n{s}^n=\sum _{r=0}^{\mathrm{\infty }}{p}_r\sum _{n=0}^{\mathrm{\infty }}P(X_2=n|X_1=r)s^n.$$

The last factor (the second sum) can be calculated as follows.

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}P(X_2=n|X_1=r)s^n& =& \mathbb{E}(s^{X_2}|X_1=r)\text{(conditional expectation)}\\ & =& \mathbb{E}(s^{Y_1+Y_2+\cdots +Y_r})\\ & =& \mathbb{E}(s^{Y_1})\mathbb{E}(s^{Y_2})\cdots \mathbb{E}(s^{Y_r})\text{(by independence)}\\ & =& G(s)G(s)\cdots G(s)\text{(identically distributed)}\\ & =& [G(s){]}^r\end{array}$$

since $\mathbb{E}(s^{Y_1})=\mathbb{E}(s^{Y_2})=\cdots =G(s)$. Each $Y_k$ is (the random variable representing) the number of offsprings produced by an individual, and they are i.i.d. Therefore,

$$G_2(s)=\sum _{r=0}^{\mathrm{\infty }}{p}_r[G(s){]}^r=G(G(s)).$$

Since $G(1)=1$, we have $G(G(1))=G(1)=1$. 

The same result holds for any successive generations. Let $G_m(s)$ be the generating function of $X_m$ (the population size of generation $m$), then

$$G_m(s)=G_{m-1}(G(s))=G(G(\cdots (G(s))\cdots ))$$

where $G(s)$ is nested $m$ times. Note that $G_m(1)=1$ follows from $G(1) = 1$.

Mean and variance

Let ${\mu }_n$ and ${\sigma }_n^2$ be the mean and variance of the population size of the $n$-th generation. Using the pgf $G_n(s)$, we have

$$\begin{array}{rcl}{\mu }_n& =& \mathbb{E}(X_n)=G_n^{\prime }(1),\\ {\sigma }_n^2& =& \mathbb{V}(X_n)=G_n^{″}(1)+{\mu }_n-{\mu }_n^2.\end{array}$$

It is understood that $G_1(s)=G(s)$. For $n=2$, we have

$$\begin{array}{rcl}{\mu }_2& =& \frac{d}{ds}{G}_2(s){|}_{s=1}=\frac{d}{ds}[G(G(s)){]}_{s=1}\\ & =& G^{\prime }(1)G^{\prime }(1)={\mu }_1^2.\end{array}$$

In general,

$$\begin{array}{rcl}{\mu }_n& =& {\frac{\mathrm{d}{G}_n(s)}{\mathrm{d}s}|}_{s=1}\\ & =& {\frac{\mathrm{d}G(G_{n-1}(s))}{\mathrm{d}s}|}_{s=1}\\ & =& G^{\prime }(1)G_{n-1}^{\prime }(1)={\mu }_1{\mu }_{n-1}={\mu }_1^2{\mu }_{n-2}=\cdots ={\mu }_1^n.\end{array}$$

Exercise. Prove this by mathematical induction. □

To obtain the variance ${\sigma }_n^2$, we need second derivatives.

$$\begin{array}{rcl}{G}_n^{\prime }(s)& =& G^{\prime }(G_{n-1}(s))G_{n-1}^{\prime }(s),\\ G_n^{″}(s)& =& G^{″}(G_{n-1}(s))[G_{n-1}^{\prime }(s){]}^2+G^{\prime }(G_{n-1}(s))G_{n-1}^{″}(s).\end{array}$$

With $s=1$ and recalling $G_n(1)=1$ for any $n$, we have

$$\begin{array}{rcl}{G}_n^{″}(1)& =& G^{″}(G_{n-1}(1))[G_{n-1}^{\prime }(1){]}^2+G^{\prime }(G_{n-1}(1))G_{n-1}^{″}(1)\\ & =& G^{″}(1)[G_{n-1}^{\prime }(1){]}^2+G^{\prime }(1)G_{n-1}^{″}(1)\\ & =& ({\sigma }_1^2-{\mu }_1+{\mu }_1^2){\mu }_{n-1}^2+{\mu }_1{G}_{n-1}^{″}(1).\end{array}$$

Let $u_n=G_n^{″}(1)$ and we have

$$u_n-{\mu }_1{u}_{n-1}=({\sigma }_1^2-{\mu }_1+{\mu }_1^2){\mu }_1^{2n-2}$$

since ${\mu }_{n-1}={\mu }_1^{n-1}$. This is a first-order linear inhomogeneous difference equation. We need to distinguish between two cases depending on ${\mu }_1=1$ or not.

Case 1: ${\mu }_1\ne 1$

The corresponding homogeneous difference equation is

$$u_n-{\mu }_1{u}_{n-1}=0$$

which is just a geometric sequence. So the general solution is

$$u_n=B{\mu }_1^n$$

with $B$ being a constant. For the particular solution, try $u_n=C{\mu }_1^{2n}$ with some constant $C$. This tentative solution satisfies the difference equation if

$$C=\frac{{\sigma }_1^2-{\mu }_1+{\mu }_1^2}{{\mu }_1({\mu }_1-1)}.$$

Exercise. Verify this result. □

Combining the two solutions (the general and particular), we have

$$u_n=B{\mu }_1^n+\frac{{\sigma }_1^2-{\mu }_1+{\mu }_1^2}{{\mu }_1({\mu }_1-1)}{\mu }_1^{2n}.$$

The constant $B$ can be determined from the case with $n=1$ where we have

$$u_1={\sigma }_1^2-{\mu }_1+{\mu }_1^2$$

so that

$$B=-\frac{{\sigma }_1^2-{\mu }_1+{\mu }_1^2}{{\mu }_1({\mu }_1-1)}.$$

After all, we have

$$u_n=\frac{{\sigma }_1^2-{\mu }_1+{\mu }_1^2}{{\mu }_1({\mu }_1-1)}{\mu }_1^n({\mu }_1^n-1).$$

I hope you still remember that $u_n=G_n^{″}(1)$. Therefore,

$${\sigma }_n^2=u_n+{\mu }_1^n-{\mu }_1^{2n}=\frac{{\sigma }_1^2{\mu }_1^{n-1}({\mu }_1^n-1)}{{\mu }_1-1}.$$

Case 2: ${\mu }_1=1$

In this case, the original difference equation becomes

$$u_n-u_{n-1}={\sigma }_1^2$$

which is just an arithmetic sequence, so the general solution is

$$u_n=n{\sigma }_1^2+A,$$,

where $A$ is a constant. From $u_1={\sigma }_1^2$, we have $A=0$. Hence,

$${\sigma }_n^2=G_n^{″}(1)=n{\sigma }_1^2.$$