# Chapter 8 Branching Process (Lecture on 01/28/2021)

Branching process is first introduced to understand the population size of the British royal family. It is assumed that each individual can give birth to $$k$$ sons with probability $$p_k$$, and define $$X_n$$ as the number of individuals in the family at the $$n$$th generation. We are interested in the following things.

1. $$E(X_n)$$;
1. $$Var(X_n)$$;
1. $$P(X_n=0)$$ as $$n\to\infty$$.

Since $$X_{n+1}$$ denotes the number of individuals in the family at generation $$n+1$$, it can therefore be written as $$$X_{n+1}=\sum_{r=1}^{X_n}\xi_r \tag{8.1}$$$ where $$\xi_r$$ is the number of sons born out of the $$r$$th individual in the $$n$$th generation. $$P(\xi_r=k)=p_k$$ for $$k=0,1,\cdots$$.

We define a few generating functions: $$$\varphi(s)=\sum_{k=0}^{\infty}p_ks^k \tag{8.2}$$$ and $$$\varphi_n(s)=\sum_{k=0}^{\infty}P(X_n=k)s^k \tag{8.3}$$$ Obviously from the definition, we have $$$\varphi(1)=\sum_{k=0}^{\infty}p_k=1 \tag{8.4}$$$ and $$$\varphi_n(1)=\sum_{k=0}^{\infty}P(X_n=k)=1 \tag{8.5}$$$ for $$n=0,1,\cdots$$. We have also found from Lecture 7 that $$$\varphi_{n+1}(s)=\varphi(\varphi_n(s)) \tag{8.6}$$$ Let $$m=E(X_1)=\sum_{k=1}^{\infty}kp_k$$ and $$\sigma^2=Var(X_1)$$, assume they both exist and are finite. Then by definition of $$\varphi_n(s)$$, we have $$$E(X_n)=\sum_{k=1}^{\infty}kP(X_n=k)=\sum_{k=1}^{\infty}s^{k-1}kP(X_n=k)|_{s=1}=\varphi_n^{\prime}(1) \tag{8.7}$$$ In addition, since $$\varphi_{n+1}(s)=\varphi_n(\varphi(s))$$, we have $$$\varphi_{n+1}^{\prime}(s)=[\varphi_n(\varphi(s))]^{\prime}=\varphi_n^{\prime}(\varphi(s))\varphi^{\prime}(s) \tag{8.8}$$$ Put $$s=1$$, from (8.8) we have $$$\varphi_{n+1}^{\prime}(1)=\varphi_n^{\prime}(1)\varphi^{\prime}(1) \tag{8.9}$$$ (8.9) gives us a recursive relationship, from which we can get $$$\varphi_{n+1}^{\prime}(1)=\varphi_n^{\prime}(1)\varphi^{\prime}(1)=\varphi_{n-1}^{\prime}(1)[\varphi^{\prime}(1)]^2 =\cdots=[\varphi^{\prime}(1)]^{n+1} \tag{8.10}$$$ If we assume that $$\varphi^{\prime}(1)=m=E(X_1)$$, then $$\varphi_{n+1}^{\prime}(1)=m^{n+1}$$. Now note that by (8.7), $$\varphi_{n+1}^{\prime}(1)=E(X_{n+1})$$, we finally obtain that $$$\boxed{\mathbf{E(X_n)=m^n}} \tag{8.11}$$$

Now we tend to compute $$Var(X_n)$$. First note that $$$\varphi_n^{\prime\prime}(s)=\sum_{k=2}^{\infty}k(k-1)P(X_n=k)s^{k-2} \tag{8.12}$$$ Thus, $$$\varphi_n^{\prime\prime}(1)=E(X_n(X_n-1))=E(X_n^2)-E(X_n) \tag{8.13}$$$ Then use (8.6) we have $$$\varphi_{n+1}^{\prime}(s)=\varphi^{\prime}(\varphi_n(s))\varphi_n^{\prime}(s) \tag{8.14}$$$ and $$$\varphi_{n+1}^{\prime\prime}(s)=\varphi^{\prime\prime}(\varphi_n(s))[\varphi_n^{\prime}(s)]^2+\varphi^{\prime}(\varphi_n(s))\varphi_n^{\prime\prime}(s) \tag{8.15}$$$ Putting $$s=1$$, we then have $$$\begin{split} \varphi_{n+1}^{\prime\prime}(1)&=\varphi^{\prime\prime}(\varphi_n(1))[\varphi_n^{\prime}(1)]^2+\varphi^{\prime}(\varphi_n(1))\varphi_n^{\prime\prime}(1)\\ &=\varphi^{\prime\prime}(1)[\varphi_n^{\prime}(1)]^2+\varphi^{\prime}(1)\varphi^{\prime\prime}_n(1) \end{split} \tag{8.16}$$$ The second equation uses the result of (8.5). Denote $$\varphi^{\prime\prime}(1)=E(X_1^2)-E(X_1)=\sigma^2+m^2-m=M$$, then from (8.16) we get $$$\varphi^{\prime\prime}_{n+1}(1)=M(m^n)^2+m\varphi_n^{\prime\prime}(1)=Mm^{2n}+m\varphi_n^{\prime\prime}(1) \tag{8.17}$$$ which is also a recursive relationship. Extend it, we have $$$\begin{split} \varphi^{\prime\prime}_{n+1}(1)&=Mm^{2n}+m\varphi_n^{\prime\prime}(1)\\ &=Mm^{2n}+m[Mm^{2(n-1)}+m\varphi_{n-1}^{\prime\prime}(1)]\\ &=\cdots=M\{m^{2n}+m^{2n-1}+\cdots+m^n\} \end{split} \tag{8.18}$$$ Now note that $$$\begin{split} &\varphi_{n+1}^{\prime\prime}(1)=E(X_{n+1}^2)-E(X_{n+1})\\ &Var(X_{n+1})=E(X_{n+1}^2)-E^2(X_{n+1})=E(X_{n+1}^2)-m^{2(n+1)} \end{split} \tag{8.19}$$$ Thus, we can get $$$\begin{split} Var(X_{n+1})&=\varphi_{n+1}^{\prime\prime}(1)+m^{n+1}-m^{2(n+1)}\\ &=M\{m^{2n}+m^{2n-1}+\cdots+m^n\}+m^{n+1}-m^{2(n+1)}\\ &=(\sigma^2+m^2-m)\{m^{2n}+m^{2n-1}+\cdots+m^n\}+m^{n+1}-m^{2(n+1)}\\ &=\sigma^2\{m^{2n}+m^{2n-1}+\cdots+m^n\}\\ &+(m^2-m)\{m^{2n}+m^{2n-1}+\cdots+m^n\}+m^{n+1}-m^{2(n+1)}\\ &=\sigma^2\{m^{2n}+m^{2n-1}+\cdots+m^n\}\\ \end{split} \tag{8.20}$$$ The last equation is becasue $$(m^2-m)\{m^{2n}+m^{2n-1}+\cdots+m^n\}$$ is a telescoping sequence, and it cancels with the last two terms. Therefore, we finally have the formula for $$Var(X_n)$$ as \boxed{\mathbf{Var(X_n)=\left\{\begin{aligned} & n\sigma^2 & m=1\\ & \sigma^2m^{n-1}(\frac{m^{n}-1}{m-1}) & m\neq 1\end{aligned}\right.}} \tag{8.21}

Now we tend to calculate the extinction probability, which is defined as the probability that the population of the royal families will eventually die out, denote at $$P(X_n=0.\exists n)$$. Note that $$X_n=0$$ implies that $$X_k=0,\forall k\geq n$$. Note also that extinction nevers occurs if $$p_0=0$$. This is because $$p_0$$ is the probability of having no son. Therefore, $$p_0=0$$ means each individual have probability one to have at least one son and the extinction never happens. We assume $$p_0>0$$ in the following calculation. We also assume $$p_0+p_1<1$$.

Note that $$\varphi(s)=\sum_{k=0}^{\infty}p_ks^k$$, which is a power series in $$s$$ with non-negative coefficients and $$p_0>0$$ and $$p_0<1$$. Hence, $$\varphi(s)$$ is a strictly increasing function. Let $$q_n=P(X_n=0)=\varphi_n(0)$$ (by (8.3)), then $$$q_{n+1}=P(X_{n+1}=0)=\varphi_{n+1}(0)=\varphi(\varphi_n(0))=\varphi(q_n) \tag{8.22}$$$ which is another recursive relationship. Now since $$q_1=\varphi_1(0)=P(X_1=0)=P(\xi_1=0)=p_0>0$$ (we assume $$X_0=1$$), and from (8.2), $$q_2=\varphi(q_1)>\varphi(0)=q_1$$ since $$\varphi(s)$$ is a strictly increasing function. In particular, if we assume $$q_n>q_{n-1}$$, we have $$\varphi(q_n)>\varphi(q_{n-1})$$, that is, $$q_{n+1}>q_n$$. Therefore, by induction, $$\{q_n\}_{n\geq 1}$$ is an increasing sequence of real numbers and also $$q_n\leq 1$$. Thus, $$\{q_n\}_{n\geq 1}$$ is a monotonically increasing sequence and bounded above by 1, $$\lim_{n\to\infty}q_n$$ exists. Let $$\pi:=\lim_{n\to\infty}q_n$$, obviously $$0<\pi\leq 1$$. Now, from (8.22), taking limit on both side, we have $$$\pi=\varphi(\pi) \tag{8.23}$$$ $$\pi$$ is the probability of eventual extinction and $$\pi$$ is the root of the equation $$\varphi(s)=s$$.

Claim: $$\pi$$ is the smallest root of $$\varphi(s)=s$$.

Proof: Let $$s_0$$ be any other root of the equation, it implies $$q_1=\varphi(0)<\varphi(s_0)=s_0$$. Very similarly, we can show any $$q_n<s_0$$, which implies that $$\pi=\lim_{n\to\infty}q_n<s_0$$.

Notice also $$\varphi(1)=\sum_{k=0}^{\infty}p_k=1$$, so 1 is a root of the equation $$\varphi(s)=s$$. In addition, $$$\varphi^{\prime\prime}(s)=\sum_{k=2}^{\infty}k(k-1)s^{k-2}p_ks^{k-2}>0 \tag{8.24}$$$ since by our assumption, $$p_0+p_1<1$$ and there exist $$p_i,i\geq 2$$ such that $$p_i>0$$. Since any function with the second derivatives greater than 0 is a convex function, we have $$\varphi$$ is a convex function. Therefore, there can be only two cases about $$\pi$$ (Figure 8.1 illustrate this):

• Case 1: If $$\varphi^{\prime}(1)>1$$, then $$f_1(s)=\varphi(s)$$ and $$f_2(s)=s$$ intersect twice in the region of $$s\in[0,1]$$, we have $$\pi<1$$.

• Case 2: If $$\varphi^{\prime}(1)\leq1$$, then $$f_1(s)=\varphi(s)$$ and $$f_2(s)=s$$ intersect only once in the region of $$s\in[0,1]$$, which is just 1. Therefore, $$\pi=1$$.

Notice that $$\varphi^{\prime}(1)=E(X_1)=m$$, summing up the above result, we have \boxed{\mathbf{\left\{\begin{aligned} & \boldsymbol{\pi}=1 & m=E(X_1)\leq 1\\ & 0<\boldsymbol{\pi}<1 & m=E(X_1)>1\end{aligned}\right. }} \tag{8.25}

This result makes sense intuitively. $$E(X_1)$$ is the expected number of people in the first generation. The result basically says that, if the expected number of people in the first generation is less than or equal to 1, the population will be distinct soon or later for sure. If the expected number of people in the first generation is greater 1, the population will have a positive chance of not being extinct.