The Deterministic One-Child Exception

The One-Child Law

In a Galton–Watson process, individuals reproduce independently according to a common offspring distribution. Most extinction conclusions depend not only on the mean number of children, but also on whether that number varies. The deterministic law in which every individual has exactly one child is the sharpest illustration: the mean offspring count is one, but a population started by one individual never becomes extinct and never grows.

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 \(S=\{0,1,2,\ldots\}\). Here \(Z_n\) is the number of individuals in generation \(n\). For each \(n\ge0\) and individual label \(k\ge1\), let \(X_{n,k}\) be the number of children of individual \(k\) in generation \(n\). These offspring counts are independent and have the same distribution as a nonnegative integer-valued random variable \(X\). The population recursion is \(Z_{n+1}=\sum_{k=1}^{Z_n}X_{n,k}\), with an empty sum equal to zero. Let \(\mathcal{F}_n\) record the population and reproduction history through generation \(n\).

Now assume \(X=1\) almost surely. The offspring generating function is \(G(s)=s\), and the offspring mean is \(\mathbb{E}[X]=1\). This is a critical mean, but the reproduction is not random: each present individual is replaced by exactly one individual in the next generation.

Proof. Since \(X_{n,k}\) has the same distribution as \(X\), \(\mathbb{P}(X_{n,k}=1)=1\) for each \(n,k\). There are only countably many pairs \((n,k)\), so with probability one all of these equalities hold simultaneously. On that event, suppose \(Z_n=r\). If \(r=0\), the defining empty sum gives \(Z_{n+1}=0=r\). If \(r\ge1\), the recursion gives \(Z_{n+1}=\sum_{k=1}^{r}1=r\). Starting from \(Z_0=r\), induction yields \(Z_n=r\) for every \(n\). For \(r=1\), none of the generations is zero, so \(E\) is empty up to a probability-zero event and \(\mathbb{P}(E)=0\). \(\square\)

Generation Sizes and Extinction

The proposition gives more than the eventual extinction probability. It specifies the entire population path: from one ancestor, every generation has size one. Thus \(\mathbb{E}[Z_n]=1\) and \(\operatorname{Var}(Z_n)=0\) for every \(n\). There is no contradiction between the mean remaining one and the population not going extinct: the mean here comes from a population size that is exactly one, not from averaging over growth and extinction outcomes.

To describe extinction by a fixed generation, define \(q_n=\mathbb{P}_1(Z_n=0)\), where the subscript indicates \(Z_0=1\). Since generation zero contains one individual, \(q_0=0\). The extinction-by-generation recursion is \(q_{n+1}=G(q_n)\): the initial ancestor has \(X\) children, and, conditional on that number, all their descendant lines must be extinct by the corresponding later generation. In this case the recursion simply keeps zero at zero.

Proof. At generation zero, \(q_0=\mathbb{P}_1(Z_0=0)=0\). If \(q_n=0\), the recursion and \(G(s)=s\) give \(q_{n+1}=G(q_n)=G(0)=0\). Induction proves \(q_n=0\) for all \(n\). The events \(\{Z_n=0\}\) are increasing in \(n\), because zero individuals have no children and the population stays zero thereafter. Their union is \(E\), so countable additivity (continuity of probability for increasing events) gives \(\mathbb{P}(E)=\lim_{n\to\infty}q_n=0\). Finally, substituting \(G(q)=q\) into the fixed-point equation gives \(q=q\), which is satisfied by every \(q\in[0,1]\). Thus the equation alone cannot select the extinction probability. \(\square\)

In general, iteration from \(q_0=0\) is informative: its successive values are the probabilities of extinction by successive generations, and their limit is eventual extinction probability. Here iteration produces only zero, even though \(q=1\) is also a fixed point. The smallest fixed point is the relevant one in this example; merely finding a fixed point is not enough.

Why Criticality Does Not Settle This Case

The critical extinction theorem states that a Galton–Watson process with finite offspring mean one becomes extinct almost surely when its offspring law is not concentrated at one. The qualification matters. For \(X=1\) almost surely, the mean is one, but the law is concentrated at one, so the theorem's hypothesis fails. The deterministic one-child process is precisely the excluded case: starting from one ancestor, it has zero extinction probability.

The generating-function geometry makes the exception visible without applying a general critical-case argument. Here \(G(s)=s\) throughout \([0,1]\), so the fixed-point equation has an entire interval of solutions, rather than identifying a unique probability. The generation-by-generation calculation resolves the ambiguity: the probabilities of extinction by generation \(n\) are all zero, and their limit is zero. The process survives, but that statement does not mean its population grows; its size remains exactly one.

Common mistake. A tempting but incorrect calculation is: “The mean is one, so the process is critical; therefore the extinction probability is \(q=1\).” The conclusion applies the critical extinction theorem without checking its nondegeneracy condition. In this example, \(G(s)=s\), so \(q=G(q)\) reduces to \(q=q\), and substituting \(q=1\) only verifies that one is a fixed point—not that it is the extinction probability. Iterating from the correct initial value gives \(q_0=0\), \(q_1=G(0)=0\), and \(q_2=G(0)=0\), with every later iterate also zero. The correct extinction probability is therefore \(0\).

Keep three quantities separate: the offspring mean, the population path, and the extinction probability. In this case, the offspring mean is one; the population path from one ancestor is constantly one; and the extinction probability is zero. None of these statements implies that the population becomes arbitrarily large. The exception is a useful reminder that conclusions stated in terms of the mean often require an explicit nondegeneracy assumption.