Extinction at Criticality

Criticality and the Extinction Question

A critical branching process has mean offspring count one: on average, each individual replaces itself with one child. This balance does not mean that the population is likely to stay constant. Random variation can produce generations with too few descendants, and once the population reaches zero it stays there. The key question is whether that variation eventually causes extinction with probability one.

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 \(S=\{0,1,2,\ldots\}\), started from one ancestor, \(Z_0=1\). Each individual independently has a number \(\xi\) of children with probabilities \(p_k=\mathbb{P}(\xi=k)\), where \(k\in S\). Assume the offspring mean is finite and critical:

The offspring probability generating function is \(f(s)=\mathbb{E}[s^\xi]=\sum_{k=0}^{\infty}p_ks^k\), for \(0\le s\le1\). Define the extinction probability by \(q=\mathbb{P}(Z_n=0\text{ for some }n\ge0)\). The zero state is absorbing, so this is also the probability that the population is eventually zero.

The exception is necessary: when every individual has exactly one child, the population remains exactly one in every generation. To prove the other case, we connect extinction to the fixed points of \(f\), then show that criticality leaves no fixed point below \(1\).

Extinction as a Fixed Point

Proof. For \(n\ge0\), let \(q_n=\mathbb{P}(Z_n=0)\). Since \(Z_0=1\), \(q_0=0\). Conditional on the ancestor having \(k\) children, the \(k\) descendant processes are independent, each with probability \(q_n\) of having no individuals in generation \(n\). Thus the probability that all \(k\) are extinct by that generation is \(q_n^k\), including \(q_n^0=1\) when \(k=0\). Averaging over \(k\) gives \(q_{n+1}=\sum_{k=0}^{\infty}p_kq_n^k=f(q_n)\). The events \(\{Z_n=0\}\) increase with \(n\), because zero is absorbing, and their union is the event of eventual extinction. Therefore \(q_n\) increases to \(q\). Since \(0\le q_n\le1\), dominated convergence applied to the series gives \(f(q_n)\to f(q)\), so \(q=f(q)\). Now let \(r\in[0,1]\) be any fixed point of \(f\). The map \(f\) is nondecreasing on \([0,1]\), since each function \(s^k\) is nondecreasing there. Starting from \(q_0=0\le r\), induction using \(q_{n+1}=f(q_n)\) gives \(q_n\le f(r)=r\) for every \(n\). Taking limits yields \(q\le r\). Thus \(q\) is the smallest fixed point. \(\square\)

This proposition reduces the extinction problem to a geometric one. In the critical case, \(f(1)=1\) and the mean condition gives \(f'(1)=1\) when the derivative is understood from the left. The following argument establishes the needed geometry directly, including at \(s=0\), without relying on a picture or on a derivative at the endpoint.

Why the Critical Fixed Point Is Unique

Proof. For \(0\le s\le1\), the mean-one identity gives

This equality follows by expanding the right-hand side: the constant terms sum to \(-1\), and the terms involving \(s-1\) sum to \(-(s-1)\mathbb{E}\xi=-(s-1)\). Together with \(\sum_k p_ks^k=f(s)\), the result is \(f(s)-1-(s-1)=f(s)-s\). The sums are finite or convergent because \(0\le s^k\le1\) and \(\sum_k kp_k=1\).

For \(k=0\) and \(k=1\), the quantity \(s^k-1-k(s-1)\) is zero. For \(k\ge2\), define \(g_k(s)=s^k-1-k(s-1)\). It has \(g_k(1)=0\), and for \(0\le s<1\), its derivative on \((0,1)\) is \(g_k'(s)=k(s^{k-1}-1)<0\). Hence \(g_k(s)>g_k(1)=0\) for \(0<s<1\); at \(s=0\), \(g_k(0)=k-1>0\). Each term in the sum is therefore nonnegative, and terms with \(k\ge2\) are strictly positive for \(s<1\).

It remains to check there is positive probability of at least two children. If \(p_k=0\) for every \(k\ge2\), then \(\mathbb{E}\xi=p_1=1\), which forces \(p_1=1\) and contradicts \(\mathbb{P}(\xi\ne1)>0\). Thus some \(p_k>0\) for \(k\ge2\), and its summand makes \(f(s)-s>0\) for every \(s<1\). At \(s=1\), \(f(1)=1\). So \(1\) is the only fixed point in \([0,1]\). \(\square\)

The theorem now follows from the proposition and lemma: in the non-deterministic critical case, the smallest fixed point is \(1\), so \(q=1\). In the remaining case, \(\mathbb{E}\xi=1\) and no probability is assigned to \(k\ne1\), so \(p_1=1\). Then \(f(s)=s\), and the recursion \(q_{n+1}=f(q_n)\), \(q_0=0\), gives \(q_n=0\) for every \(n\), hence \(q=0\).

Worked Examples

A Common Pitfall and Why the Result Matters

A tempting but incorrect calculation is to identify extinction probability with the chance that the ancestor has no children. For the zero-or-two law above, that gives \(q=\mathbb{P}(\xi=0)=\tfrac12\). This is only the probability of extinction by generation \(1\), namely \(q_1=f(0)=\tfrac12\); it ignores the possibility that descendants die out in later generations. The next iterate already differs: \(q_2=f(q_1)=f(\tfrac12)=\tfrac58\). Extinction probability is the limit of the iterates, or equivalently the smallest fixed point, not generally \(f(0)\).

The critical result also clarifies why the mean alone does not describe a branching process. Mean one does not imply that the population stays near its initial size, or that extinction is impossible. Unless reproduction is exactly one child per individual, the offspring law has positive mass on outcomes with at least two children; the fixed-point geometry then forces eventual extinction with probability one. This is an almost-sure statement, not a claim that extinction occurs by any particular finite generation.