# 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.

- \(E(X_n)\);

- \(Var(X_n)\);

- \(P(X_n=0)\) 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 \[\begin{equation} X_{n+1}=\sum_{r=1}^{X_n}\xi_r \tag{8.1} \end{equation}\] 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: \[\begin{equation} \varphi(s)=\sum_{k=0}^{\infty}p_ks^k \tag{8.2} \end{equation}\] and \[\begin{equation} \varphi_n(s)=\sum_{k=0}^{\infty}P(X_n=k)s^k \tag{8.3} \end{equation}\] Obviously from the definition, we have \[\begin{equation} \varphi(1)=\sum_{k=0}^{\infty}p_k=1 \tag{8.4} \end{equation}\] and \[\begin{equation} \varphi_n(1)=\sum_{k=0}^{\infty}P(X_n=k)=1 \tag{8.5} \end{equation}\] for \(n=0,1,\cdots\). We have also found from Lecture 7 that \[\begin{equation} \varphi_{n+1}(s)=\varphi(\varphi_n(s)) \tag{8.6} \end{equation}\] 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 \[\begin{equation} E(X_n)=\sum_{k=1}^{\infty}kP(X_n=k)=\sum_{k=1}^{\infty}s^{k-1}kP(X_n=k)|_{s=1}=\varphi_n^{\prime}(1) \tag{8.7} \end{equation}\] In addition, since \(\varphi_{n+1}(s)=\varphi_n(\varphi(s))\), we have \[\begin{equation} \varphi_{n+1}^{\prime}(s)=[\varphi_n(\varphi(s))]^{\prime}=\varphi_n^{\prime}(\varphi(s))\varphi^{\prime}(s) \tag{8.8} \end{equation}\] Put \(s=1\), from (8.8) we have \[\begin{equation} \varphi_{n+1}^{\prime}(1)=\varphi_n^{\prime}(1)\varphi^{\prime}(1) \tag{8.9} \end{equation}\] (8.9) gives us a recursive relationship, from which we can get \[\begin{equation} \varphi_{n+1}^{\prime}(1)=\varphi_n^{\prime}(1)\varphi^{\prime}(1)=\varphi_{n-1}^{\prime}(1)[\varphi^{\prime}(1)]^2 =\cdots=[\varphi^{\prime}(1)]^{n+1} \tag{8.10} \end{equation}\] 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 \[\begin{equation} \boxed{\mathbf{E(X_n)=m^n}} \tag{8.11} \end{equation}\]

Now we tend to compute \(Var(X_n)\). First note that \[\begin{equation} \varphi_n^{\prime\prime}(s)=\sum_{k=2}^{\infty}k(k-1)P(X_n=k)s^{k-2} \tag{8.12} \end{equation}\] Thus, \[\begin{equation} \varphi_n^{\prime\prime}(1)=E(X_n(X_n-1))=E(X_n^2)-E(X_n) \tag{8.13} \end{equation}\] Then use (8.6) we have \[\begin{equation} \varphi_{n+1}^{\prime}(s)=\varphi^{\prime}(\varphi_n(s))\varphi_n^{\prime}(s) \tag{8.14} \end{equation}\] and \[\begin{equation} \varphi_{n+1}^{\prime\prime}(s)=\varphi^{\prime\prime}(\varphi_n(s))[\varphi_n^{\prime}(s)]^2+\varphi^{\prime}(\varphi_n(s))\varphi_n^{\prime\prime}(s) \tag{8.15} \end{equation}\] Putting \(s=1\), we then have \[\begin{equation} \begin{split} \varphi_{n+1}^{\prime\prime}(1)&=\varphi^{\prime\prime}(\varphi_n(1))[\varphi_n^{\prime}(1)]^2+\varphi^{\prime}(\varphi_n(1))\varphi_n^{\prime\prime}(1)\\ &=\varphi^{\prime\prime}(1)[\varphi_n^{\prime}(1)]^2+\varphi^{\prime}(1)\varphi^{\prime\prime}_n(1) \end{split} \tag{8.16} \end{equation}\] 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 \[\begin{equation} \varphi^{\prime\prime}_{n+1}(1)=M(m^n)^2+m\varphi_n^{\prime\prime}(1)=Mm^{2n}+m\varphi_n^{\prime\prime}(1) \tag{8.17} \end{equation}\] which is also a recursive relationship. Extend it, we have \[\begin{equation} \begin{split} \varphi^{\prime\prime}_{n+1}(1)&=Mm^{2n}+m\varphi_n^{\prime\prime}(1)\\ &=Mm^{2n}+m[Mm^{2(n-1)}+m\varphi_{n-1}^{\prime\prime}(1)]\\ &=\cdots=M\{m^{2n}+m^{2n-1}+\cdots+m^n\} \end{split} \tag{8.18} \end{equation}\] Now note that \[\begin{equation} \begin{split} &\varphi_{n+1}^{\prime\prime}(1)=E(X_{n+1}^2)-E(X_{n+1})\\ &Var(X_{n+1})=E(X_{n+1}^2)-E^2(X_{n+1})=E(X_{n+1}^2)-m^{2(n+1)} \end{split} \tag{8.19} \end{equation}\] Thus, we can get \[\begin{equation} \begin{split} Var(X_{n+1})&=\varphi_{n+1}^{\prime\prime}(1)+m^{n+1}-m^{2(n+1)}\\ &=M\{m^{2n}+m^{2n-1}+\cdots+m^n\}+m^{n+1}-m^{2(n+1)}\\ &=(\sigma^2+m^2-m)\{m^{2n}+m^{2n-1}+\cdots+m^n\}+m^{n+1}-m^{2(n+1)}\\ &=\sigma^2\{m^{2n}+m^{2n-1}+\cdots+m^n\}\\ &+(m^2-m)\{m^{2n}+m^{2n-1}+\cdots+m^n\}+m^{n+1}-m^{2(n+1)}\\ &=\sigma^2\{m^{2n}+m^{2n-1}+\cdots+m^n\}\\ \end{split} \tag{8.20} \end{equation}\] 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 \[\begin{equation} \boxed{\mathbf{Var(X_n)=\left\{\begin{aligned} & n\sigma^2 & m=1\\ & \sigma^2m^{n-1}(\frac{m^{n}-1}{m-1}) & m\neq 1\end{aligned}\right.}} \tag{8.21} \end{equation}\]

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 \[\begin{equation} q_{n+1}=P(X_{n+1}=0)=\varphi_{n+1}(0)=\varphi(\varphi_n(0))=\varphi(q_n) \tag{8.22} \end{equation}\] 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 \[\begin{equation} \pi=\varphi(\pi) \tag{8.23} \end{equation}\] \(\pi\) 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, \[\begin{equation} \varphi^{\prime\prime}(s)=\sum_{k=2}^{\infty}k(k-1)s^{k-2}p_ks^{k-2}>0 \tag{8.24} \end{equation}\] 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 \[\begin{equation} \boxed{\mathbf{\left\{\begin{aligned} & \boldsymbol{\pi}=1 & m=E(X_1)\leq 1\\ & 0<\boldsymbol{\pi}<1 & m=E(X_1)>1\end{aligned}\right. }} \tag{8.25} \end{equation}\]