Extinction Probability as a Fixed Point

From Finite Generations to Ultimate Extinction

For a Galton–Watson process, the offspring generating function encodes the number of children produced by one individual. Iterating that function gives probabilities of extinction by successive generations. The central point of this tutorial is that the limit of those probabilities is not just a solution to a fixed-point equation: it is the smallest solution in the unit interval. The distinction matters because an equation can have more than one solution, and not every solution represents the probability of eventual extinction.

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 \(\{0,1,2,\ldots\}\), started from one individual, so \(Z_0=1\). Each individual independently produces a number of children with probabilities \(p_k=\mathbb{P}(\xi=k)\), \(k\ge0\). Write \((\mathcal{F}_n)_{n\ge0}\) for the filtration describing the family tree through generation \(n\). The offspring generating function is \(f(s)=\mathbb{E}[s^\xi]=\sum_{k=0}^\infty p_k s^k\) for \(0\le s\le1\), using \(s^0=1\), including when \(s=0\). In particular, \(\sum_{k=0}^\infty p_k=1\) and \(f(0)=p_0\).

Define \(q_n=\mathbb{P}(Z_n=0)\), the probability of extinction by generation \(n\), and \(q=\mathbb{P}(Z_n=0\text{ for some }n\ge0)\), the ultimate extinction probability. If \(Z_n=0\), then every later generation is also zero: there are no individuals to produce offspring. Thus the events \(\{Z_n=0\}\) increase with \(n\), and their union is the event of ultimate extinction.

Proof. Since \(Z_0=1\), \(q_0=0\). To find \(q_{n+1}\), condition on the number of children of the initial individual. If that number is \(k\), extinction by generation \(n+1\) occurs exactly when each of the \(k\) resulting descendant processes is extinct within its first \(n\) generations. These descendant processes are independent, each with the same law as the original process started from one individual. The conditional probability is therefore \(q_n^k\), including \(k=0\), for which it is \(1\). Averaging over \(k\) gives \(q_{n+1}=\sum_{k=0}^\infty p_kq_n^k=f(q_n)\).

The events \(\{Z_n=0\}\) increase, so countable additivity of probability gives \(q_n\uparrow q\). It remains to pass to the limit in the recursion. For each fixed \(k\), \(q_n^k\to q^k\); for \(k=0\), both sides are \(1\). To justify the infinite sum, fix \(\varepsilon>0\). Choose \(K\) so that \(\sum_{k>K}p_k<\varepsilon\), which is possible because the probabilities sum to \(1\). The finite sum \(\sum_{k=0}^Kp_kq_n^k\) converges to \(\sum_{k=0}^Kp_kq^k\). The absolute difference between the two omitted tails is at most \(\sum_{k>K}p_k<\varepsilon\), since every power lies in \([0,1]\). As \(\varepsilon\) can be arbitrarily small, \(f(q_n)\to f(q)\). Taking limits in \(q_{n+1}=f(q_n)\) gives \(q=f(q)\).

Finally, let \(r\in[0,1]\) satisfy \(f(r)=r\). For \(0\le s\le t\le1\), each \(s^k\le t^k\), so summing with nonnegative weights \(p_k\) shows \(f(s)\le f(t)\). Since \(q_0=0\le r\), induction gives \(q_n\le r\): if \(q_n\le r\), then \(q_{n+1}=f(q_n)\le f(r)=r\). Taking the limit yields \(q\le r\). As \(q\) itself is a solution in \([0,1]\), it is the smallest such solution. \(\square\)

The proof separates three ideas that are useful in calculations. First, the recursion concerns extinction by a particular generation, not ultimate extinction directly. Second, increasing finite-generation probabilities converge to the probability of their union. Third, the smallest-root conclusion follows from the iteration starting at \(0\) and the monotonicity of \(f\); it does not follow merely from solving \(f(s)=s\).

What the Smallest-Root Statement Does—and Does Not—Say

The fixed-point equation is a necessary equation for ultimate extinction, but it may have multiple solutions. The branching construction selects one of them through the sequence \(q_0=0\), \(q_{n+1}=f(q_n)\). The proof that this sequence lies below every fixed point is what identifies the extinction probability. Merely finding roots of a polynomial or other expression for \(f(s)-s\) is not enough to decide which root is the probability.

Common mistake. In the two-child example, one might solve \(3s^2-4s+1=0\), choose the root \(1\), and report certain extinction. The algebraic root is valid, but the conclusion is wrong: the finite-generation probabilities are generated from \(q_0=0\), and they remain at most \(\tfrac13\). The limit must therefore be \(\tfrac13\), not \(1\). The correction is to use the iteration or the smallest-root theorem after solving the equation, and then check the chosen root by substituting it into \(f\).

These conclusions depend on having a probability distribution for the offspring count and on the independent, identical reproduction defining a Galton–Watson process. They do not require a finite expected number of children: the proof uses the probabilities \(p_k\), bounded powers on \([0,1]\), and finite-tail control of their sum. This fixed-point result identifies \(q\); more detailed classifications of when \(q\) equals \(1\), or when it is strictly below \(1\), require additional analysis.