The Conditional Recursion for Generation-Size Distributions

From One Generation to the Next

In a Galton–Watson process, each individual reproduces independently according to the same offspring law. The population in the next generation is therefore a sum whose number of terms is itself random: there are \(Z_n\) terms, one for each individual in generation \(n\). Conditioning on \(Z_n\) fixes that number, making the generating function of the sum a product. Averaging over \(Z_n\) then gives a recursion for the unconditional generation-size distribution.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge 0}\) be a discrete-time process with state space \(\{0,1,2,\ldots\}\), where \(Z_n\) is the number of individuals in generation \(n\). Let \(\xi_{n,r}\) denote the number of offspring of individual \(r\) in generation \(n\). Assume that all the variables \(\xi_{n,r}\), for \(n\ge 0\) and \(r\ge 1\), are independent and have the same distribution \[ \mathbb{P}(\xi_{n,r}=j)=p_j,\qquad j\ge 0,\qquad \sum_{j=0}^{\infty}p_j=1. \] Assume also that \(Z_0\) is independent of these offspring variables. The population evolves by \[ Z_{n+1}=\sum_{r=1}^{Z_n}\xi_{n,r}, \] where an empty sum, occurring when \(Z_n=0\), is zero.

For conditioning, let \(\mathcal{F}_n=\sigma(\xi_{m,r}:0\le m<n,\ r\ge1;\,Z_0)\). This filtration contains the information used to construct the population through generation \(n\), so \(Z_n\) is \(\mathcal{F}_n\)-measurable. The offspring variables \(\xi_{n,r}\) for generation \(n\) are independent of \(\mathcal{F}_n\). These assumptions are the reason the conditional calculation below works.

The Conditional Generating Function

Proof. First fix \(s\in[0,1]\). On the event \(Z_n=k\), the population update is \(Z_{n+1}=\sum_{r=1}^{k}\xi_{n,r}\). Therefore \(s^{Z_{n+1}}=\prod_{r=1}^{k}s^{\xi_{n,r}}\). Given \(\mathcal{F}_n\), the value of \(k=Z_n\) is known and the variables in this product are independent of \(\mathcal{F}_n\) and mutually independent. Each factor has conditional expectation \(G(s)\). Thus the conditional expectation of the product is \(G(s)^k\). When \(k=0\), the product is empty and equals \(1=G(s)^0\), so the same identity holds. This proves the assertion conditioned on \(\mathcal{F}_n\). Taking conditional expectation again with respect to the smaller information \(\sigma(Z_n)\), which is contained in \(\mathcal{F}_n\), gives \(\mathbb{E}[s^{Z_{n+1}}\mid Z_n]=G(s)^{Z_n}\). Finally, on an event \(Z_n=k\) of positive probability, the conditional PGF is the conditional expectation given that event, namely \(G(s)^k\). \(\square\)

The exponent is the current population size: each current individual contributes one independent offspring count. This identity also specifies the whole conditional distribution, not just a single probability. For fixed \(k\), the next population is a sum of \(k\) independent copies of the offspring variable. Hence its conditional distribution is the \(k\)-fold convolution of the offspring distribution; when \(k=0\), it is concentrated at zero.

Unconditional Composition

Proof. By the definition of \(H_{n+1}\) and the rule for conditional expectation, \[ H_{n+1}(s)=\mathbb{E}[s^{Z_{n+1}}] =\mathbb{E}\!\left[\mathbb{E}[s^{Z_{n+1}}\mid\mathcal{F}_n]\right]. \] The conditional theorem makes the inner expectation \(G(s)^{Z_n}\), so \(H_{n+1}(s)=\mathbb{E}[G(s)^{Z_n}]\). Since \(0\le G(s)\le1\), it is a valid argument of \(H_n\); by the definition of \(H_n\), this last expectation is \(H_n(G(s))\). This proves the rule. \(\square\)

The order of composition records which randomness is averaged first. Starting with the offspring PGF, \(G(s)\) describes one individual’s contribution. Given \(Z_n\), the contributions multiply to \(G(s)^{Z_n}\). Averaging that power over the distribution of \(Z_n\) applies \(H_n\) to \(G(s)\). If there is one initial individual, then \(H_0(s)=s\), and this one-step rule gives \(H_1(s)=G(s)\). For a general initial population, its distribution remains visible through \(H_0\).

Common Composition Errors

A common error is to use \(G(s)\) as the conditional PGF regardless of the number of individuals in the current generation. That expression describes the offspring of one individual; with \(k\) individuals the correct conditional PGF is \(G(s)^k\). For example, in the Poisson example with \(Z_n=3\), the incorrect expression \(\exp(2(s-1))\) gives probability \(e^{-2}\) of zero offspring in total. The correct expression is \(\exp(6(s-1))\), which gives probability \(e^{-6}\). Three individuals all having zero offspring is less likely than one individual having zero offspring.

Another error is reversing the order in the unconditional rule. In the random-initial-population example, the correct PGF is \(H_0(G(s))\), not \(G(H_0(s))\). At \(s=0\), the correct formula gives \(H_0(G(0))=\tfrac12+\tfrac12(\tfrac12)^2=\tfrac58\), the probability of zero individuals in generation one. The reversed expression gives \(G(H_0(0))=\tfrac12+\tfrac12\cdot\tfrac12=\tfrac34\), which disagrees with the direct conditioning calculation. The order follows from the two stages: first form the offspring contribution for a fixed current population, then average over the current population’s distribution.

This recursion is useful whenever a question concerns the distribution of the next generation. It handles a fixed or random initial population and accommodates any offspring law with a well-defined probability distribution on the nonnegative integers. The essential conditions are independence across individuals and generations, a common offspring law, and independence of the initial population from the offspring variables. Without these assumptions, the product \(G(s)^{Z_n}\) need not describe the conditional PGF.