Extinction in the Subcritical Case

Extinction and the Offspring Generating Function

A population can have a declining expected size without disappearing immediately. The subcritical extinction result makes a stronger long-run statement: if each individual has, on average, fewer than one child, then a process begun with one individual eventually reaches zero with probability one. The key is to use the offspring generating function not just to compute generation probabilities, but to constrain all possible fixed points in the unit interval.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge0}\) be a Galton–Watson process in discrete time with state space \(\mathbb{N}_0=\{0,1,2,\ldots\}\), started from one ancestor, so \(Z_0=1\). Each individual independently produces a random number \(\xi\in\mathbb{N}_0\) of children with the same offspring law at every generation. Write \(p_k=\mathbb{P}(\xi=k)\), assume its mean \(m=\mathbb{E}\xi=\sum_{k=0}^{\infty}kp_k\) is finite, and define its probability generating function by \(f(s)=\mathbb{E}[s^\xi]=\sum_{k=0}^{\infty}p_ks^k\) for \(0\le s\le1\). The extinction event is that \(Z_n=0\) for some finite \(n\). Since zero is absorbing, this event is \(\bigcup_{n\ge0}\{Z_n=0\}\).

For each \(n\), let \(q_n=\mathbb{P}(Z_n=0)\). Conditional on the number of children in the first generation, the descendants of those children evolve independently and each has the same generation-size law as a process started from one ancestor. This gives the recursion \(q_0=0\) and \(q_{n+1}=f(q_n)\). The extinction probability is denoted by \(q\).

Proof. The events \(\{Z_n=0\}\) increase with \(n\), because zero is absorbing. Therefore \(q_n\) increases to the probability \(q\) of their union. The recursion gives \(q_{n+1}=f(q_n)\). The function \(f\) is continuous on \([0,1]\): if \(s_r\to s\) in this interval, then \(s_r^\xi\to s^\xi\) for every finite value of \(\xi\), and \(0\le s_r^\xi\le1\), so bounded convergence gives \(f(s_r)\to f(s)\). Taking limits in the recursion yields \(q=f(q)\). To show that \(q\) is the smallest fixed point, let \(r\in[0,1]\) satisfy \(f(r)=r\). Since \(q_0=0\le r\) and \(f\) is nondecreasing on \([0,1]\), induction gives \(q_n\le r\) for every \(n\): if \(q_n\le r\), then \(q_{n+1}=f(q_n)\le f(r)=r\). Taking limits gives \(q\le r\). This holds for every fixed point \(r\), proving the claim. \(\square\)

The proposition turns extinction into a question about the graph of \(f\) and the diagonal \(s\mapsto s\). In the subcritical case, the finite mean supplies a useful bound on the graph over the entire interval; a derivative calculation at one alone would not establish the needed global conclusion.

The Subcritical Extinction Theorem

Proof. First establish a bound for every \(s\in[0,1]\). For an integer \(k\ge1\), factor \(1-s^k=(1-s)(1+s+\cdots+s^{k-1})\). Each of the \(k\) terms in the second factor is at most one, so \(1-s^k\le k(1-s)\), or \(s^k\ge1-k(1-s)\). For \(k=0\), the same inequality holds as equality because \(s^0=1\). Multiply this inequality for \(k\) by \(p_k\) and sum over \(k\). The sum is legitimate because \(\sum_k p_k=1\), the mean \(\sum_k kp_k=m\) is finite, and \(s^k\) lies between zero and one. It follows that

For \(0\le s<1\), subtract \(s\) from both sides to obtain \(f(s)-s\ge(1-m)(1-s)>0\), since \(m<1\) and \(1-s>0\). Hence no \(s<1\) is a fixed point of \(f\). Also, \(f(1)=\sum_k p_k=1\), so one is a fixed point. By the fixed-point characterization, the extinction probability is the smallest fixed point in \([0,1]\); as there is no fixed point below one, it must be \(q=1\). \(\square\)

This argument covers the boundary case \(m=0\) without a separate exception. Indeed, a nonnegative random variable with mean zero is zero with probability one; then \(f(s)=1\), which has no fixed point below one, and extinction occurs by generation one. More generally, the proof uses no assumption that the offspring law has a variance. The finite first moment is what the bound requires.

Worked Examples

What the Result Does—and Does Not—Say

Extinction with probability one is a statement about the eventual fate of the population, not a claim that extinction happens quickly. In the zero-or-two example, the probability of extinction by generation two is \(\tfrac{57}{64}\), which is less than one, even though the probability of extinction at some finite generation is one. The fixed-point argument establishes the latter by taking the limit of the increasing events \(\{Z_n=0\}\).

It is also important to distinguish the derivative \(f'(1)\) from the extinction probability. When the mean is finite, \(f'(1)\) from the left equals \(m\), but this derivative describes the local slope of the generating function at one. Extinction is instead the smallest solution of \(f(s)=s\), a global fixed-point question. For Bernoulli offspring above, \(f'(1)=\tfrac13\), while the extinction probability is \(1\).

Common mistake. A wrong calculation is to identify the extinction probability with the mean, writing \(q=m=\tfrac13\) for the Bernoulli example. This value does not satisfy the fixed-point equation: \(f(\tfrac13)=\tfrac23+\tfrac19=\tfrac79\), whereas \(\tfrac13\ne\tfrac79\). The correction is to solve \(f(s)=s\) and select its smallest solution in \([0,1]\); here the only such solution is \(1\).

The subcritical proof rests on three linked facts: the finite-generation extinction probabilities iterate the offspring generating function; their limit is the smallest fixed point; and a mean below one forces \(f(s)>s\) everywhere below one. Together they show why no positive probability of survival can remain in the limit. The behavior at mean exactly one is a separate case: the strict inequality used here disappears, so this proof does not establish a conclusion there.