Exact Extinction Calculations for Finite Offspring Laws

From an Offspring Law to an Exact Extinction Probability

For a Galton–Watson process, extinction is determined by repeated reproduction across generations. When the number of children has finite support, its probability generating function is a polynomial. That makes the extinction equation an algebraic equation, but finding its roots is only part of the calculation: one must identify which root corresponds to extinction and account for exceptional cases such as an individual who always has exactly one child.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge0}\) be a discrete-time Galton–Watson process with state space \(S=\{0,1,2,\ldots\}\), started from one individual, so \(Z_0=1\). Each individual has an independent number \(\xi\) of offspring, with \(\mathbb{P}(\xi=k)=p_k\), where \(p_k\ge0\), \(\sum_{k=0}^Kp_k=1\), and \(p_k=0\) for \(k>K\). Once \(Z_n=0\), all later population sizes are zero. Extinction means that \(Z_n=0\) at some finite generation; write \(q\) for its probability.

The fixed-point equation comes from conditioning on the first generation. If the first individual has \(k\) children, extinction occurs exactly when each of the \(k\) independent descendant processes becomes extinct, an event with probability \(q^k\). Averaging over \(k\) gives \(q=G(q)\). But the equation may have more than one root in \([0,1]\), so that relation alone does not identify \(q\).

Proof. Define \(q_n=\mathbb{P}(Z_n=0)\), so \(q_0=0\). Given \(Z_n=r\), the \(r\) individuals in generation \(n\) have independent offspring, and the conditional probability that \(Z_{n+1}=0\) is \(p_0^r\). Equivalently, by the generating-function recursion for generation sizes, \(q_{n+1}=G(q_n)\). The events \(\{Z_n=0\}\) increase with \(n\), because zero is absorbing. Their union is the extinction event, so \(q_n\) increases to \(q\). Continuity of the polynomial \(G\) implies \(q=G(q)\). If \(r\in[0,1]\) is any fixed point, then \(q_0=0\le r\); since \(G\) is nondecreasing on \([0,1]\), induction gives \(q_n\le r\) for every \(n\), and hence \(q\le r\). Thus \(q\) is the smallest fixed point.

For the classification, set \(h(s)=G(s)-s\). Then \(h(1)=0\) and \(h'(1)=m-1\). If \(m<1\), convexity gives \(h(s)\ge h(1)+h'(1)(s-1)>0\) for \(0\le s<1\), so there is no fixed point below one. If \(m=1\) and the law is not concentrated at one, some \(p_k>0\) for \(k\ge2\), so \(h''(s)=\sum_{k=2}^K k(k-1)p_ks^{k-2}>0\) for \(0<s\le1\). Consequently \(h'(s)<h'(1)=0\) for \(s<1\), and \(h(s)>h(1)=0\) for \(s<1\). If \(m=1\) and the law is concentrated at one, then \(G(s)=s\) for every \(s\), while \(Z_n=1\) for every \(n\), so \(q=0\).

Finally, if \(m>1\), then some \(p_k>0\) for \(k\ge2\), so \(h\) is strictly convex on \((0,1]\). Since \(h'(1)>0\), \(h(s)<0\) for \(s<1\) sufficiently close to one. If \(p_0>0\), then \(h(0)=p_0>0\), so continuity gives a root in \((0,1)\). There cannot be two distinct roots below one as well as the root at one: strict convexity would force the value at the middle of three such roots to be strictly below zero, contradicting its being a root. If \(p_0=0\), then \(h(0)=0\); strict convexity gives \(h(s)<0\) for \(0<s<1\), so zero is the only root below one. The smallest-root result identifies these roots as the extinction probabilities. \(\square\)

Constructing and Solving the Polynomial

For a finite offspring law, the calculation has a consistent sequence: put each probability \(p_k\) in front of \(s^k\), check that the probabilities sum to one, compute \(m\), and solve \(G(s)-s=0\). The root at \(s=1\) is always present because \(G(1)=1\). In a supercritical law, that root is not the extinction probability; the relevant root is the smallest one in \([0,1]\). Factoring out \(s-1\) is often the step that reveals a simpler exact root.

Reading the Roots Correctly

The mean classifies whether extinction is certain, but in the supercritical case the polynomial determines its exact probability. A finite-support law need not yield a root expressible by a simple radical; “exact calculation” means forming and analyzing the correct polynomial, then giving an exact root when it can be identified algebraically. Numerical root-finding is a separate method for cases where a convenient exact factorization is unavailable.

Common mistake: choosing the root at one in a supercritical example. For the law with probabilities \(\tfrac12\) at offspring counts zero and three, the equation factors as \((s-1)(s^2+s-1)=0\). Choosing the largest root \(1\) as the extinction probability is wrong: it is a fixed point, but not the smallest one. The root \(\tfrac{\sqrt5-1}{2}\) lies in \([0,1)\), and the smallest-root result shows that it is \(q\). The correction is not to discard \(s=1\), but to apply the smallest-root rule after solving the equation.

Two checks help prevent errors. First, verify \(\sum_{k=0}^Kp_k=1\) before using the polynomial; otherwise the coefficients do not describe an offspring distribution. Second, compute \(m=\sum_k kp_k\) directly from the law, not from an incorrectly selected root. The cases \(m=1\) and \(p_0=0\) deserve particular attention: criticality alone does not describe the deterministic one-child process, and supercriticality alone does not ensure a positive extinction probability.