Deriving the Mean Population Recursion

From Offspring Counts to a Mean Recursion

A Galton–Watson process describes a population in generations: each individual produces a random number of children, and different individuals reproduce independently according to the same offspring law. The population in the next generation is therefore a random sum. Conditioning on the current population fixes how many terms appear in that sum, which is the key to computing its mean.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(Z_n\) be the number of individuals in generation \(n\), with state space \(S=\{0,1,2,\ldots\}\) and discrete time \(n\geq 0\). Let \(\mathcal{F}_n\) contain the information about the process through generation \(n\), before the individuals in generation \(n\) reproduce. In particular, \(Z_n\) is \(\mathcal{F}_n\)-measurable.

For \(r\geq1\), write \(\xi_{n,r}\) for the number of children produced by the \(r\)th individual in generation \(n\), when that individual exists. Assume these offspring variables are independent and identically distributed, are independent of \(\mathcal{F}_n\) at generation \(n\), and have a common finite mean

Here each \(\xi_{n,r}\) is a nonnegative integer-valued random variable. We assume \(Z_n<\infty\) almost surely, as is the case when the initial population is finite almost surely and every individual has a finite number of children. The offspring-sum representation is

where a sum with no terms is defined to be \(0\). Thus, if \(Z_n=0\), then \(Z_{n+1}=0\): there are no individuals available to reproduce.

Proof. For each \(r\geq1\), define \(Y_r=\mathbf{1}_{\{Z_n\geq r\}}\xi_{n,r}\). The event \(\{Z_n\geq r\}\) is in \(\mathcal{F}_n\), and \(\xi_{n,r}\) is independent of \(\mathcal{F}_n\). Therefore \(\mathbb{E}[Y_r\mid\mathcal{F}_n]=\mathbf{1}_{\{Z_n\geq r\}}\mathbb{E}[\xi_{n,r}\mid\mathcal{F}_n]=m\mathbf{1}_{\{Z_n\geq r\}}\) almost surely. The offspring-sum representation gives \(Z_{n+1}=\sum_{r\geq1}Y_r\): for each outcome, the indicators are one exactly for the first \(Z_n\) terms. Since all \(Y_r\) are nonnegative, conditional monotone convergence gives \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=\sum_{r\geq1}\mathbb{E}[Y_r\mid\mathcal{F}_n]=m\sum_{r\geq1}\mathbf{1}_{\{Z_n\geq r\}}=mZ_n\) almost surely. The last sum has exactly \(Z_n\) terms, including zero terms when \(Z_n=0\). \(\square\)

This conditional statement says that after the generation-\(n\) population is known, each of its \(Z_n\) members contributes an expected \(m\) children. It is an identity of conditional expectations, not an assertion that the realized population is \(mZ_n\). The offspring counts remain random even after \(Z_n\) is known.

Proof. By the conditional-mean proposition and the defining property of conditional expectation, \(\mathbb{E}[Z_{n+1}]=\mathbb{E}[\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]]=\mathbb{E}[mZ_n]=m\mathbb{E}[Z_n]\). The assumption \(\mathbb{E}[Z_n]<\infty\), together with \(m<\infty\), makes the right-hand side finite, so the unconditional expectation is finite as well. If \(Z_0\) is a fixed finite number, it has finite expectation; if \(Z_0\) is random with finite mean, the same calculation applies at generation zero. In either case the offspring variables must have the stated law independently of the information about the initial population. \(\square\)

Worked Examples

Assumptions, Initial Conditions, and a Common Error

The finite-mean condition has a precise role. To conclude a finite numerical identity from the theorem, we need both \(m<\infty\) and \(\mathbb{E}[Z_n]<\infty\). A finite population almost surely is not, by itself, enough to guarantee finite expectation: a random initial population can take finite values on every outcome and still have an infinite mean. In that case the ordinary finite-expectation calculation in the theorem is not available. The conditional identity remains a nonnegative conditional-expectation statement, but it should not be presented as an equality between finite numbers unless the integrability assumptions hold.

The initial population need not be deterministic. If \(Z_0=z\) for a fixed nonnegative integer \(z\), the first expected population is \(mz\). If \(Z_0\) is random with finite mean and is independent of the offspring variables, the first expected population is \(m\mathbb{E}[Z_0]\). The case \(z=0\) is included: the offspring sum is empty, so the next population is zero and the formula gives \(m\cdot0=0\). The case \(m=0\) is also included. Since offspring counts are nonnegative and have mean zero, they equal zero almost surely, and the next population is zero whenever the current population is finite.

Common mistake. A frequent incorrect calculation is to set \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=m\), as if the next generation consisted of the offspring of just one individual. For example, with \(Z_n=3\) and offspring mean \(m=\tfrac23\), this gives the wrong answer \(\tfrac23\). The offspring-sum representation has three terms, so the correct calculation is \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=\tfrac23+\tfrac23+\tfrac23=2\). Equivalently, it is \(mZ_n=\tfrac23\cdot3=2\). The factor \(Z_n\) counts how many individuals reproduce.

The conditional form is the most informative version of the result: it states what the next-generation mean is once the current population is known. Taking expectations then produces the unconditional recursion. Keeping those two steps separate helps prevent confusion between the random population \(Z_{n+1}\), its conditional mean, and its unconditional mean.