Extinction by a Specified Generation
In a branching process, extinction is absorbing: once a generation contains no individuals, every later generation is also empty. This makes the probability of being extinct by a fixed generation a natural quantity to compute. For a Galton–Watson process, the calculation reduces to repeatedly evaluating the offspring probability generating function. It gives a finite-generation answer without requiring a calculation of the probability of eventual extinction.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let the state space be \(S=\{0,1,2,\ldots\}\), and let \((Z_n)_{n\ge0}\) be a Galton–Watson process in discrete generation time. Each individual has a number of children with distribution \(\mathbb{P}(K=k)=p_k\), for \(k\in S\); reproduction is independent across individuals and generations. The population evolves according to \(Z_{n+1}=\sum_{\ell=1}^{Z_n}K_{n,\ell}\), where the variables \(K_{n,\ell}\) are independent copies of \(K\), and an empty sum is zero. Let \(\mathcal{F}_n\) record the population and reproduction history through generation \(n\).
The offspring probability generating function is \(f(s)=\mathbb{E}[s^K]=\sum_{k=0}^{\infty}p_ks^k\), for \(0\le s\le1\), with \(s^0=1\), including at \(s=0\). For one initial ancestor, \(Z_0=1\), define \(q_n=\mathbb{P}(Z_n=0\mid Z_0=1)\). Since the process is already alive at generation zero, \(q_0=0\). The event \(Z_n=0\) is extinction by generation \(n\); because zero is absorbing, it includes paths that became extinct at any earlier generation.
Proof. First take one ancestor. Suppose the ancestor has \(k\) children in generation one. The \(k\) family lines evolve independently, and each is extinct after a further \(n\) generations with probability \(q_n\). Thus the probability that all \(k\) lines are extinct at that time is \(q_n^k\). When \(k=0\), this probability is \(1\), as required for the empty generation. Averaging over the ancestor’s offspring count gives \(q_{n+1}=\sum_{k=0}^{\infty}p_kq_n^k=f(q_n)\). This sum is well-defined because \(0\le q_n\le1\) and \(\sum_kp_k=1\). Starting from \(q_0=0\), induction yields \(q_n=f^{\circ n}(0)\): it holds at \(n=0\), and if it holds at \(n\), then \(q_{n+1}=f(q_n)=f(f^{\circ n}(0))=f^{\circ(n+1)}(0)\). Now start with \(r\) ancestors. Their family lines are independent, and the total population at generation \(n\) is zero exactly when every one of the \(r\) lines is extinct then. The probability is therefore \(q_n^r\) for \(r\ge1\). For \(r=0\), the population is zero at every generation, so the probability is \(1\), matching the stated convention. \(\square\)
The recursion \(q_{n+1}=f(q_n)\) is also a practical calculation rule. Begin at \(q_0=0\), then evaluate \(f\) at the previous value once for each additional generation. For several initial ancestors, first compute the one-ancestor probability \(q_n\), then take its \(r\)th power. This order matters: the generations correspond to function composition, while the initial ancestors contribute an ordinary power.
Worked Finite-Generation Calculations
Worked Example: Offspring Are Either Zero or Two
Suppose each individual has zero children with probability \(\tfrac12\) and two children with probability \(\tfrac12\). Find the extinction probabilities by generations one, two, and three for one ancestor, then find the probability of extinction by generation two when there are three initial ancestors.
The offspring generating function is \(f(s)=\tfrac12+\tfrac12s^2\). Starting with \(q_0=0\), the recursion gives \(q_1=f(0)=\tfrac12+\tfrac12(0)^2=\tfrac12\). Next, \(q_2=f(\tfrac12)=\tfrac12+\tfrac12(\tfrac12)^2=\tfrac12+\tfrac18=\tfrac58\). For generation three, \(q_3=f(\tfrac58)=\tfrac12+\tfrac12(\tfrac58)^2=\tfrac12+\tfrac{25}{128}=\tfrac{64}{128}+\tfrac{25}{128}=\tfrac{89}{128}\).
With three initial ancestors, independence of their family lines gives \(\mathbb{P}(Z_2=0\mid Z_0=3)=q_2^3=(\tfrac58)^3=\tfrac{125}{512}\). As a check of the recursion at generation three, \(f(\tfrac58)=\tfrac{64}{128}+\tfrac{25}{128}=\tfrac{89}{128}\), the value just obtained. For the three-ancestor calculation, the probability of extinction by generation two for each line is \(\tfrac58\), so the probability all three lines are extinct is \(\tfrac58\cdot\tfrac58\cdot\tfrac58=\tfrac{125}{512}\).
Worked Example: At Most One Child Per Individual
Suppose each individual has one child with probability \(\tfrac23\) and no children with probability \(\tfrac13\). Find the probability of extinction by generation four for one ancestor and for two initial ancestors.
Here \(f(s)=\tfrac13+\tfrac23s\). The recursion suggests the exact expression \(q_n=1-(\tfrac23)^n\). To verify it, at \(n=0\) the expression gives \(1-1=0=q_0\). If \(q_n=1-(\tfrac23)^n\), then \(q_{n+1}=f(q_n)=\tfrac13+\tfrac23[1-(\tfrac23)^n]=\tfrac13+\tfrac23-(\tfrac23)^{n+1}=1-(\tfrac23)^{n+1}\). Thus the expression holds for every \(n\) by induction.
For one ancestor at generation four, \(q_4=1-(\tfrac23)^4=1-\tfrac{16}{81}=\tfrac{65}{81}\). For two ancestors, the probability that both family lines are extinct by generation four is \(q_4^2=(\tfrac{65}{81})^2=\tfrac{4225}{6561}\).
Check: The generation-three value is \(q_3=1-(\tfrac23)^3=1-\tfrac8{27}=\tfrac{19}{27}\). Applying the recursion once gives \(f(q_3)=\tfrac13+\tfrac23\cdot\tfrac{19}{27}=\tfrac{27}{81}+\tfrac{38}{81}=\tfrac{65}{81}=q_4\). This verifies the generation-four result directly.
Worked Example: Zero, One, or Two Children
Suppose the offspring probabilities are \(p_0=\tfrac12\), \(p_1=\tfrac14\), and \(p_2=\tfrac14\). Find the probability of extinction by generation three for one ancestor and for two initial ancestors.
The generating function is \(f(s)=\tfrac12+\tfrac14s+\tfrac14s^2\). The first value is \(q_1=f(0)=\tfrac12+\tfrac14(0)+\tfrac14(0)^2=\tfrac12\). The second is \(q_2=f(\tfrac12)=\tfrac12+\tfrac14\cdot\tfrac12+\tfrac14(\tfrac12)^2=\tfrac12+\tfrac18+\tfrac1{16}=\tfrac8{16}+\tfrac2{16}+\tfrac1{16}=\tfrac{11}{16}\).
Now substitute \(q_2=\tfrac{11}{16}\) into \(f\): \(q_3=\tfrac12+\tfrac14\cdot\tfrac{11}{16}+\tfrac14(\tfrac{11}{16})^2=\tfrac12+\tfrac{11}{64}+\tfrac{121}{1024}=\tfrac{512}{1024}+\tfrac{176}{1024}+\tfrac{121}{1024}=\tfrac{809}{1024}\). For two initial ancestors, the probability is \(q_3^2=(\tfrac{809}{1024})^2=\tfrac{654481}{1048576}\), since \(809^2=(800+9)^2=640000+14400+81=654481\) and \(1024^2=1048576\).
Check: Substituting the generation-two value into the recursion gives \(f(\tfrac{11}{16})=\tfrac{512}{1024}+\tfrac{176}{1024}+\tfrac{121}{1024}=\tfrac{809}{1024}\), confirming the one-ancestor probability. The two-ancestor probability is the product of the two independent line probabilities, \(\tfrac{809}{1024}\cdot\tfrac{809}{1024}=\tfrac{654481}{1048576}\).
How the Probabilities Change with the Deadline
Proof. If \(Z_n=0\), then the defining population recursion gives \(Z_{n+1}=\sum_{\ell=1}^{0}K_{n,\ell}=0\). Therefore \(\{Z_n=0\}\subseteq\{Z_{n+1}=0\}\), and taking probabilities gives \(q_n\le q_{n+1}\). Each \(q_n\) is a probability, so \(q_n\le1\), and \(q_0=0\) for one initial ancestor. These statements together give the asserted bounds and monotonicity. \(\square\)
The probability of becoming extinct for the first time at generation \(n\ge1\) is consequently \(q_n-q_{n-1}\). Indeed, the event of extinction by generation \(n\) contains the event of extinction by generation \(n-1\), and their difference consists exactly of paths that first reach zero at generation \(n\). This is different from \(q_n\), which includes all paths already extinct by earlier generations.
For a finite offspring law, the recursion can be evaluated using exact arithmetic at each step. For an offspring law with infinitely many possible values, the same rule applies: evaluate \(f(q_n)=\sum_{k=0}^{\infty}p_kq_n^k\). The defining series converges for \(q_n\in[0,1]\), because its nonnegative terms are bounded in sum by \(\sum_kp_k=1\). No assumption about a finite offspring mean is needed for this fixed-generation formula.
A Common Mistake
A frequent error is to treat the generation number as an ordinary exponent on the first-generation extinction probability. The generations do not consist of independent repeated trials with a fixed extinction probability: the number of individuals exposed to reproduction can change from one generation to the next. The generating function must be composed with itself.
For the zero-or-two offspring law above, the one-generation probability is \(q_1=\tfrac12\). Incorrectly squaring this number to estimate extinction by generation two gives \((\tfrac12)^2=\tfrac14\). The correct calculation evaluates the generating function at \(q_1\): \(q_2=f(q_1)=\tfrac12+\tfrac12(\tfrac12)^2=\tfrac12+\tfrac18=\tfrac58\). The value \(\tfrac14\) is not the probability that the process is extinct at generation two; the correct value is \(\tfrac58\). For \(r\) initial ancestors, the ordinary power appears only after the one-ancestor probability has been computed: the formula is \([f^{\circ n}(0)]^r\).
Check Your Understanding
For each question, specify whether you are considering one initial ancestor or several, and use the generation recursion before taking any power for multiple ancestors.
- For an offspring law with \(p_0=\tfrac13\) and \(p_2=\tfrac23\), write \(f\) and calculate \(q_1\) and \(q_2\).
- Explain why \(q_0=0\) for one initial ancestor but the probability of \(Z_0=0\) is \(1\) when there are no initial ancestors.
- If one-ancestor extinction probabilities by generation \(n\) and \(n+1\) are \(q_n\) and \(q_{n+1}\), express the probability of first becoming extinct at generation \(n+1\).
- For two initial ancestors, explain why the probability of extinction by generation \(n\) is \(q_n^2\), rather than \(f^{\circ n}(0^2)\).
- Give a reason the finite-generation recursion is defined even when the offspring distribution has infinite mean.