Solving Extinction Probabilities for Quadratic Generating Functions

From a Fixed-Point Equation to an Explicit Answer

For a Galton–Watson process, the extinction probability is characterized by a fixed point of the offspring generating function. When offspring counts are limited to zero, one, or two, that generating function is quadratic, so the fixed-point equation can be solved exactly. The algebra is short; choosing the correct root requires care. In particular, the root \(1\) is always present, but it need not be the extinction probability.

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 \(S=\{0,1,2,\ldots\}\), and let \(\mathcal{F}_n\) record the process and reproduction information through generation \(n\). Start with one ancestor, \(Z_0=1\). Each individual independently has \(\xi\) offspring, where \(\mathbb{P}(\xi=k)=p_k\). In this tutorial, \(p_k=0\) for \(k\ge3\), so \(p_0+p_1+p_2=1\), with each \(p_k\ge0\). The offspring generating function is \(f(s)=\mathbb{E}[s^\xi]=p_0+p_1s+p_2s^2\) for \(0\le s\le1\).

The earlier fixed-point characterization says that \(q\) is the smallest solution in \([0,1]\) of \(f(s)=s\). We first give the short argument that explains why the smallest solution, rather than an arbitrary one, is required.

Proof. Set \(q_n=\mathbb{P}(Z_n=0)\) for \(n\ge0\). Because \(Z_0=1\), \(q_0=0\). The branching property gives \(q_{n+1}=f(q_n)\): conditional on the ancestor having \(k\) children, extinction by generation \(n+1\) requires all \(k\) descendant processes to be extinct by their generation \(n\), an event of probability \(q_n^k\); averaging over \(k\) gives \(f(q_n)\). The events \(\{Z_n=0\}\) increase with \(n\), because zero is absorbing, so \(q_n\) increases and is bounded above by \(1\). Its limit \(q_\infty\) is the probability of their union, which is the extinction event. Since \(f\) is a polynomial and hence continuous, taking limits in \(q_{n+1}=f(q_n)\) gives \(q_\infty=f(q_\infty)\). Now let \(r\in[0,1]\) be any fixed point. The function \(f\) is nondecreasing on \([0,1]\), since its coefficients are nonnegative. Starting from \(q_0=0\le r\), induction gives \(q_n\le r\): if \(q_n\le r\), then \(q_{n+1}=f(q_n)\le f(r)=r\). Passing to the limit yields \(q_\infty\le r\). Thus \(q=q_\infty\) is a fixed point no larger than any other fixed point in \([0,1]\). \(\square\)

Solving the Quadratic, Including Its Boundary Cases

Since \(p_0+p_1+p_2=1\), the equation \(f(s)=s\) can be rewritten as

This form exposes the root \(s=1\), which follows from the normalization of the offspring probabilities. When \(p_2>0\), the quadratic factors exactly. When \(p_2=0\), one must instead solve the resulting linear equation, and the case \(p_0=0\) is exceptional because the equation then vanishes identically.

Proof. Suppose first that \(p_2>0\). The fixed-point polynomial factors as \(p_2s^2-(p_0+p_2)s+p_0=(s-1)(p_2s-p_0)\), because multiplying the right-hand side gives \(p_2s^2-(p_0+p_2)s+p_0\). Its roots are \(1\) and \(p_0/p_2\), counting a repeated root twice if \(p_0=p_2\). The second root is nonnegative. If \(p_0<p_2\), it lies in \([0,1)\), so the smaller fixed point is \(p_0/p_2\). If \(p_0=p_2\), both roots equal \(1\); if \(p_0>p_2\), the second root exceeds \(1\), leaving \(1\) as the only fixed point in \([0,1]\). The smallest-fixed-point proposition therefore gives the stated answers for \(p_2>0\). Now suppose \(p_2=0\). Normalization gives \(p_1=1-p_0\), and the equation becomes \(p_0+(1-p_0)s=s\), or \(p_0(1-s)=0\). If \(p_0>0\), its only solution is \(s=1\), so \(q=1\). If \(p_0=0\), then \(p_1=1\) and \(f(s)=s\) for every \(s\in[0,1]\). The smallest fixed point is \(0\), so \(q=0\). In this last case every individual has exactly one child, and the population never reaches zero. This proves every case. \(\square\)

The ratio \(p_0/p_2\) is used only when \(p_2>0\); it must not be used as a formula when \(p_2=0\). The theorem also includes the boundary \(p_0=0<p_2\), for which the ratio gives \(q=0\). If there are no offspring of count zero, the process cannot become extinct, even though \(1\) remains a fixed point.

Worked Calculations

Boundary Cases and a Common Error

When \(p_2=0\) and \(p_0>0\), the offspring count is either zero or one. For example, with \(p_0=\tfrac25\), \(p_1=\tfrac35\), \(p_2=0\), the probabilities sum to \(\tfrac25+\tfrac35=1\), and \(f(s)=\tfrac25+\tfrac35s\). Solving \(f(s)=s\) gives \(\tfrac25(1-s)=0\), hence \(s=1\) and \(q=1\). By contrast, if \(p_0=p_2=0\), normalization forces \(p_1=1\); then \(Z_n=1\) for every \(n\), so \(q=0\), even though every \(s\in[0,1]\) solves \(f(s)=s\). This is the deterministic one-child exception: the smallest fixed point rule still applies, but the equation no longer has a unique root.

Common pitfall: choosing \(1\) just because it is always a root. In the supercritical example, one might stop at \(f(1)=1\) and report extinction probability \(1\). That is wrong because the equation also has the smaller root \(\tfrac13\). As a check using iteration, \(q_0=0\), \(q_1=f(0)=\tfrac16\), and

Here \(\tfrac{17}{72}<\tfrac13=\tfrac{24}{72}\), consistent with the increasing extinction-by-generation probabilities approaching the smaller fixed point. The correction is not to discard the root \(1\), but to compare all roots in \([0,1]\) and select the smallest.

A reliable calculation therefore has three steps: check that the offspring probabilities are nonnegative and sum to one; solve the equation without dividing by a coefficient that might be zero; then retain the smallest root in the unit interval. The repeated-root case \(p_0=p_2>0\), the zero-offspring boundary \(p_0=0\), and the deterministic case \(p_0=p_2=0\) each deserve an explicit check. These steps turn a compact algebraic equation into the correct probabilistic answer.