Total Progeny and Its Generating Function
Generation sizes describe how a branching population evolves one generation at a time. A different question is how many individuals are ever born, counting the initial ancestor. That total may be finite, or it may be infinite. A generating function packages the probabilities of all finite totals into one equation, built directly from the offspring law. Interpreting the value of that function at one requires care: it is the probability of finite total progeny, not automatically one.
Consider a Galton–Watson process \((Z_n)_{n\ge0}\) in discrete time on the state space \(S=\{0,1,2,\ldots\}\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with \(Z_0=1\). Each individual independently has \(\xi\) children, where \(\mathbb{P}(\xi=k)=p_k\) for \(k\ge0\) and \(\sum_{k=0}^\infty p_k=1\). Its offspring generating function is \(G(u)=\mathbb{E}[u^\xi]=\sum_{k=0}^\infty p_k u^k\) for \(0\le u\le1\). Let \(\mathcal{F}_n\) record the process and its reproduction through generation \(n\). Each \(Z_n\) is finite almost surely, though the total over all generations need not be.
For a Galton–Watson process, total progeny is finite exactly when the population becomes extinct in some finite generation. If extinction occurs, only finitely many generations have nonzero size, each with finite size. If extinction never occurs, every generation has at least one individual, so the sum is infinite. Therefore \(H(1)\) is the extinction probability \(q\). The standard extinction classification from the earlier material on Galton–Watson processes says that, when the offspring mean is finite, \(q=1\) in the subcritical case, and also in the nondegenerate critical case; in the supercritical case \(q<1\). There is an essential exception: if \(\mathbb{P}(\xi=1)=1\), the process is critical but never becomes extinct, so \(q=0\).
The Functional Equation
Proof. Let \(\xi\) be the number of children of the initial ancestor. Conditional on \(\xi=k\), the \(k\) child lineages evolve independently, each as a Galton–Watson process started from one individual. Write \(T_1,\ldots,T_k\) for their total progenies, allowing any of them to be infinite. The root contributes one individual, so \(T=1+\sum_{\ell=1}^k T_\ell\) when all the child totals are finite; if any is infinite, \(T=\infty\). For \(0\le s\le1\), the contribution to \(H(s)\) conditional on \(\xi=k\) is therefore \(sH(s)^k\), by independence. For \(k=0\), the empty product is \(1\), giving contribution \(s\), as required for a root with no children. Averaging over \(k\) gives \(H(s)=\sum_{k=0}^\infty p_k sH(s)^k=sG(H(s))\). At \(s=1\), \(H(1)=\mathbb{P}(T<\infty)=q\); the standard extinction result identifies \(q\) as the smallest fixed point of \(G\) in \([0,1]\). \(\square\)
The equation is useful both for solving an entire generating function and for extracting individual probabilities. If \(a_n=\mathbb{P}(T=n)\), then \(H(s)=\sum_{n=1}^\infty a_n s^n\) for \(0\le s<1\). Substitution into \(H(s)=sG(H(s))\) yields the following recursion.
Proof. A total of \(n\) individuals consists of the root and the total progenies of its \(k\) children. If the root has \(k\) children, their finite total sizes \(n_1,\ldots,n_k\) must each be at least \(1\), and they must sum to \(n-1\). The probability of this offspring count and these child totals is \(p_k a_{n_1}\cdots a_{n_k}\), since the child lineages are independent. Summing over all \(k\) and all such positive integer tuples gives the displayed expression. Since \(n_1+\cdots+n_k=n-1\), each \(n_\ell\le n-1\) when \(k\ge1\); hence only already determined coefficients occur. When \(k=0\), there are no children, so total progeny is exactly \(1\), which gives the stated empty-tuple convention. \(\square\)
This recursion and the functional equation describe finite totals even when there is positive probability of infinite progeny. In particular, \(H(s)\) for \(s<1\) does not count infinite outcomes as finite ones; their contribution is zero under the definition of \(H\).
Worked Example: A Subcritical One-Child-or-No-Child Law
Suppose each individual has no children with probability \(\tfrac23\), and one child with probability \(\tfrac13\). Then \(G(u)=\tfrac23+\tfrac13u\). The row of offspring probabilities sums to \(\tfrac23+\tfrac13=1\), and the offspring mean is \(0\cdot\tfrac23+1\cdot\tfrac13=\tfrac13\), so this law is subcritical.
The functional equation is $$ H(s)=s\left(\tfrac23+\tfrac13H(s)\right). $$ Moving the term involving \(H(s)\) to the left gives \(H(s)(1-\tfrac{s}{3})=\tfrac{2s}{3}\), and therefore $$ H(s)=\frac{2s}{3-s}=\frac23s\frac{1}{1-s/3} =\sum_{n=1}^\infty \frac23\left(\frac13\right)^{n-1}s^n,\qquad 0\le s<1. $$ Consequently \(\mathbb{P}(T=n)=\tfrac23(\tfrac13)^{n-1}\) for \(n\ge1\). For example, \(a_1=\tfrac23\), since the root must have no children; \(a_2=\tfrac13\cdot\tfrac23=\tfrac29\), since the root has one child and that child has none; and \(a_3=\tfrac13\cdot\tfrac13\cdot\tfrac23=\tfrac2{27}\).
Check: Substituting the formula into the equation gives \(s(\tfrac23+\tfrac13\cdot\tfrac{2s}{3-s}) =s(\tfrac{2(3-s)+2s}{3(3-s)})=\tfrac{2s}{3-s}=H(s)\). Also \(H(1)=\tfrac{2}{3-1}=1\), in agreement with certain extinction. The probabilities sum to one because \(\sum_{n=1}^\infty \tfrac23(\tfrac13)^{n-1}=\tfrac23/(1-\tfrac13)=1\).
Worked Example: A Critical Binary-Offspring Law
Suppose each individual has either zero or two children, each with probability \(\tfrac12\). The offspring probabilities sum to \(\tfrac12+\tfrac12=1\), and the mean is \(0\cdot\tfrac12+2\cdot\tfrac12=1\). This is critical and nondegenerate. Its generating function is \(G(u)=\tfrac12+\tfrac12u^2\), so $$ H(s)=\frac{s}{2}\bigl(1+H(s)^2\bigr). $$ For \(s>0\), solving the quadratic gives \(H(s)=\frac{1\pm\sqrt{1-s^2}}{s}\). The plus sign gives a value greater than \(1\) for \(0<s<1\), which cannot be a generating function value. Thus $$ H(s)=\frac{1-\sqrt{1-s^2}}{s},\qquad 0<s<1, $$ and \(H(0)=0\) by its definition.
The recursion gives the first finite-total probabilities directly. A total of \(1\) requires zero children, so \(a_1=\tfrac12\). A total of \(2\) is impossible: with two children the root and its children already account for three individuals, so \(a_2=0\). A total of \(3\) requires two children, each with total progeny \(1\), giving \(a_3=\tfrac12(\tfrac12)^2=\tfrac18\). For total \(5\), the two child totals must be \(1\) and \(3\), in either order. Thus \(a_5=\tfrac12(2a_1a_3)=\tfrac12(2\cdot\tfrac12\cdot\tfrac18)=\tfrac1{16}\).
Check: For \(a_3\), the coefficient recursion gives \(\tfrac12a_1^2=\tfrac12(\tfrac12)^2=\tfrac18\). For \(a_5\), it gives \(\tfrac12(2a_1a_3)=\tfrac1{16}\), as above. The explicit solution has \(H(1)=\lim_{s\uparrow1}(1-\sqrt{1-s^2})/s=1\); this agrees with extinction almost surely in the nondegenerate critical case.
Worked Example: Positive Probability of Infinite Total Progeny
Suppose each individual has zero children with probability \(\tfrac14\), and two children with probability \(\tfrac34\). These probabilities sum to \(\tfrac14+\tfrac34=1\); the mean is \(0\cdot\tfrac14+2\cdot\tfrac34=\tfrac32\), so the process is supercritical. Here \(G(u)=\tfrac14+\tfrac34u^2\), and the equation is \(H(s)=s(\tfrac14+\tfrac34H(s)^2)\).
A total of \(1\) requires zero children, giving \(a_1=\tfrac14\). A total of \(3\) requires two children, each with total \(1\), giving \(a_3=\tfrac34(\tfrac14)^2=\tfrac3{64}\). These are finite-total probabilities, not conditional probabilities given eventual extinction.
At \(s=1\), solve the fixed-point equation \(u=\tfrac14+\tfrac34u^2\). Rearranging gives \(3u^2-4u+1=0\), or \((3u-1)(u-1)=0\). Its roots are \(\tfrac13\) and \(1\), so the extinction probability, the smallest root, is \(q=\tfrac13\). Hence the probability of infinite total progeny is \(1-q=\tfrac23\).
Check: The equation at one holds for the generating-function value: \(\tfrac14+\tfrac34(\tfrac13)^2=\tfrac14+\tfrac1{12}=\tfrac13=H(1)\). The explicit solution with \(H(0)=0\) is \(H(s)=\frac{4-\sqrt{16-12s^2}}{6s}\) for \(0<s\le1\); substituting \(s=1\) gives \((4-\sqrt4)/6=1/3\). The remaining probability \(2/3\) belongs to \(T=\infty\), not to any coefficient \(a_n\).
Why the Value at One Needs Care
The equation \(H(s)=sG(H(s))\) is a computational tool, but the solution must retain its probabilistic meaning. At \(s<1\), \(H(s)\) is the generating function of the finite-total probabilities. At \(s=1\), it is their sum, \(\mathbb{P}(T<\infty)\). It need not equal one. In a supercritical process the missing mass is the probability of infinite total progeny. A fixed-point equation alone can have more than one solution at \(s=1\); the correct generating-function value is selected by the extinction probability, the smallest fixed point of \(G\).
Common mistake. It is incorrect to infer that every critical process has \(H(1)=1\). For the deterministic one-child law, \(\mathbb{P}(\xi=1)=1\), so \(G(u)=u\). The equation becomes \(H(s)=sH(s)\), which for each \(0\le s<1\) implies \((1-s)H(s)=0\), hence \(H(s)=0\). Directly, \(Z_n=1\) for every \(n\), so \(T=\infty\) and \(H(1)=\mathbb{P}(T<\infty)=0\). The correction is to apply the critical extinction conclusion only to the nondegenerate critical case; the deterministic one-child case has no extinction.
When calculating coefficients, another useful check is that \(a_n\ge0\), and \(\sum_{n\ge1}a_n=H(1)\), not necessarily \(1\). The coefficient recursion counts only finite child totals. It does not assign a finite size to a lineage that continues forever. These distinctions let the same equation handle subcritical, critical, supercritical, and degenerate models without treating infinite progeny as an ordinary integer-valued outcome.
Check Your Understanding
Use the equation and the probability interpretation together when answering these questions.
- For a law with \(\mathbb{P}(\xi=0)=\tfrac12\) and \(\mathbb{P}(\xi=1)=\tfrac12\), write \(G\) and the equation for \(H(s)\). What is \(H(1)\)?
- Explain why \(T<\infty\) is equivalent to extinction in a finite generation when every generation is finite almost surely.
- For the offspring law in the critical binary example, calculate \(\mathbb{P}(T=7)\) using the coefficient recursion.
- For the supercritical example, what event accounts for the difference between \(1\) and \(\sum_{n=1}^\infty\mathbb{P}(T=n)\)?
- For the deterministic one-child law, solve \(H(s)=sH(s)\) for \(0\le s<1\), and explain why the value at one must be considered separately.