The limit theorems for the Galton-Watson processes are important for many applications. Also, they suggest what to expect in more complicated processes. The limit laws are different in the supercritical, subcritical, and critical cases. In the supercritical case, the principal result is that the growth is asymptotically exponential, and that with probability 1 the random variable \(M_n=X_n/\mu^n\) tends to a limit \(M_\infty\). As a consequence, \(X_n\) is approximated by \(M_\infty\mu^n\) for large n. This is an extension of the exponential or Malthusian law of growth in the realm of stochasticity.
However, here the analogy ends. In the subcritical and critical cases, the probability of extinction is equal to 1 and the limit of \(\{X_n>0\}\) is exponential.
Recall from the end of Lecture 8, that the population size \(X_n\) of a branching process, normalized by its own mean \[ M_n =\frac{X_n}{\mu^n}, \] is a martingale, i.e., \[ E(M_{n+1}\ |\ X_0, X_1, \ldots, X_n)=M_n. \]
In other words, given the history of \(M_n\) up to stage \(n\), the next value \(M_{n+1}\) of \(M\) is on average what is is now: \(M\) is constant on average in this very sophisticated sense of conditional expectation given past and present. The true statement \[ EM_n=1 \qquad \mbox{for any} \ n \] is of course infinitely cruder.
Martingale Convergence Theorem If \(\{ M_n, \ n\ge 0\}\) is a non-negative martingale, then there exists a proper random variable \(M_\infty\) with finite expectation such that \[ \lim_{n\to\infty}M_n=M_\infty \qquad \mbox{a.s.}, \] where a.s. stands for almost surely (or with probability 1).
Note A statement \(S\) is said to be true almost surely (a.s.) or with probability 1 if 9surprise, surprise!) \(P(S \ \mbox{is true}) = 1\).
Because our martingale is non-negative \((M_n\ge 0)\), the above theorem implies that it is almost surely true that \[ M_\infty = \lim_{n\to \infty} M_n \qquad \mbox{exists}. \]
Note that if \(M_\infty>0\) for some outcome (which can happen with positive probability only when \(\mu>1\)), then teh statement \[ \frac{X_n}{\mu^n} \to M_\infty \qquad \mbox{a.s.} \] is a precise formulation of exponential growth.
We know that \(M_\infty=\lim M_n\) exists with probability 1, and that \(E(M_n)=1\) for any \(n\). We might be tempted to believe taht \(E(M_\infty)=1\). However, we already know that if \(\mu\le 1\), then, almost surely, the process dies out and \(M_n\) is eventually 0. Hence if \(\mu\le 1\), then \(M_\infty =0\) (a.s.) and \[ 0=E(M_\infty)\ne lim E(M_n)=1. \]
What is going wrong is that (when \(\mu\le 1\)) for large \(n\), the chances are that \(M_n\) will be large if \(M_n\) is not 0 and, very roughly speaking, this large values multiplying its small probability will keep \(E(M_n)\) at 1.
In the critical and subcritical case, \(M_\infty\equiv 0\) since \(p_e=1\). Therefore, \(M_\infty\) might be non-degenerate only if \(\mu>1\). This is indeed true if an additional condition of finite offspring (reproduction) variance \(Var(Y)\) is imposed. We will formulate (without proof) the following limit theorem in the supercritical case.
Limit Theorem (supercritical Processes) If \(\mu>1\), \(Var(Y)<\infty\), and \(X_0=1\), then \[ (i)\qquad \lim_{n\to \infty} E(M_n-M_\infty)^2=0; \] \[ (ii) \qquad E( M_\infty)=1, \qquad Var (M_\infty)=\frac{Var(Y)}{\mu^2-\mu}; \] \[ (iii) \qquad P(M_\infty=0)=p_e. \]
Since \(M_n\to M_\infty\) (a.s.), it is obvious that for \(\lambda>0\), \[ \exp (-M_n)\to \exp (-\lambda M_\infty)\qquad \mbox{(a.s.)}. \]
Furthermore, it can be show (using the Bounded Convergence Theorem) that
\[ (a) \qquad E \exp(-\lambda M_\infty)=\lim E \exp(-\lambda M_n). \]
Since \(M_n=X_n/\mu^n\) and the population PGF \(E\left(s^{X_n}\right)\) equals the \(n\)-th composition \(\Pi_{X_0}(\ldots (\Pi_{X_0}(s)))\), we have \[ (b) \qquad E \exp(-\lambda M_n)= \Pi_{X_0}(\ldots (\Pi_{X_0}(s_n))),\qquad s_n=\exp\left(-\frac{\lambda}{\mu^n}\right). \]
Using (a) and (b), in principle (if very rarely in practice), we can calculate teh left-hand side of (a) and determine the distribution of \(M_n\).
This concrete example is just about the only one in which one can calculate everything explicitly, but, in the way oof mathematics, it is useful in many contexts.
We take the offspring distribution (for the typical number of children \(Y\)) to be a geometric distribution: \[ P(Y=k)=pq^k, \qquad (k=0,1,\ldots), \] where \(0<p<1\) and \(q=1-p\). Then, as we already know, \[ \Pi_{X_0}(s)=\frac{p}{1-qs}=\frac{1}{1+\mu(1-s)}, \qquad \mu=\frac{q}{p}, \] and \[ p_e= \left\{ \begin{array}{ll} \frac{p}{q} & \mbox{if }\ q>p,\\ 1 & \mbox{if}\ q\le p. \end{array} \right. \] Suppose now that \(\mu>1\). The \(n\)-th composition of \(\Pi_{X_0}\) can be calculated explicitly to be \[ \Pi_{X_0}(\ldots (\Pi_{X_0}(s)))= 1-\frac{\mu^n(\mu-1)(1-s)}{\mu(\mu^n-1)(1-s)+\mu-1}. \] Then you can check (using the L’Hospital rule) that, for \(\lambda>0\), \[ E \exp(-\lambda M_\infty)= 1-\frac{\mu^n(\mu-1)(1-\exp(-\lambda\mu^{-1}))}{\mu(\mu^n-1)(1-\exp(-\lambda\mu^{-1})+\mu-1} \] \[ \to 1-\frac{\lambda(\mu-1)}{\lambda\mu+\mu-1}=\frac{1}{\mu}+\left(1-\frac{1}{\mu}\right)\frac{(1-\mu^{-1})}{\lambda+1-\mu^{-1}} \] From this we can deduce that \[ P(M_\infty\le x)=p_e+(1-p_e)e^{-(1-p_e)x}, \] i.e., the distribution of the limit \(M_\infty\) is exponential with atom \(p_e\) at 0.
Consider a supercritical population given that its dies sooner or later. Contrary to what one might guess, the population size does not first grow exponentially and later drops drastically to 0.
Let, as usual, \(X_0=1\) and denote by \(T_n\) the accumulated population size up to generation \(n\), i.e., \[ T_n=1+X_1+\ldots+X_n. \]
We know from Lecture 7 that if \(\mu<1\), then the process dies our rapidlyand the total number of individuals ever born is \[ T=1+X_1+\ldots+X_n+\ldots \] is finite and \[ E(T)=\sum_{k=0}^\infty \mu^k=\frac{1}{1-\mu}. \] On the other hand, if \(\mu\ge 1\), then \(E(T_n)\to \infty\) as \(n\to \infty\). However, if we condition on the event that a supercritical process dies sooner of later and denote the extinction time (generattion index) by \(\tau\) we have the following result.
Theorem If \(\mu>1\) then \[\begin{equation} \tag{9.1} E(T\ |\ \tau<\infty)=\frac{1}{1-\Pi'_{X_0}(p_e)}, \end{equation}\] where \(p_e=P(\tau<\infty)\) is the extinction probability of the process (note that \(\Pi'_{X_0}(p_e)<1\)).
Proof By the definition of conditional expectation, we write \[\begin{equation}\tag{9.2} E(T\ |\ \tau<\infty)=\frac{E(T\ ;\ \tau<\infty)}{P(\tau<\infty)}. \end{equation}\] Further, \[\begin{equation} \tag{9.3} E(T\ ;\ \tau<\infty)=\sum_{n=0}^\infty E(X_n\ ;\ n<\tau<\infty). \end{equation}\] Now observe that \[ E(X_n\ ;\ n<\tau<\infty)=\sum_{k=1}^\infty kP(X_n=k\ ; \ n<\tau<\infty)=\sum_{k=1}^\infty kP(X_n=k)p_e^k, \]
Substituting this first in (9.3) and then in (9.2), gives (9.1). The proof of the theorem is complete.
In particular, in the supercritical process with geometric offspring distribution, we have \(p_e=q/p=\mu^{-1}<1\) and \[ \Pi'_{X_0}(p_e)=\frac{qp}{(1-pp_e)^2}=\frac{1}{\mu}. \] Hence, \[ E(T\ |\ \tau<\infty)=\frac{1}{1-\mu^{-1}}=\frac{\mu}{\mu-1}=1+\frac{1}{\mu-1}. \] Note that this expectation is monotone decreasing to 1 when \(\mu\) increases.
We obtained an (at the first sight) unexpected result: if a supercritical process dies out rather rapidly (and, therefore, the total amount of ever born individuals in the population is small) it may be an indicator to the fact that the expected value of the o§spring number in this process is large. However, after some thoughts one can see that since the process is supercritical it either dies out almost of the beginning of the evolution (and therefore, the accumulated size indeed, is small) or if this is not the case it hardly be able to extinct since the size of the population becomes large after several (surviving) generations.