The Normalized Galton-Watson Population

Population Growth and Its Natural Scale

A Galton–Watson process models generations in which each individual produces a random number of offspring, independently of every other individual. If the mean number of offspring is \(m\), then the expected population is multiplied by \(m\) from one generation to the next. This suggests comparing the actual population with the scale \(m^n\). The key point is that this comparison is not merely constant in expectation: relative to the information available through generation \(n\), it satisfies the martingale conditional-mean identity.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let the state space be \(S=\{0,1,2,\ldots\}\), and let time be discrete, \(n\geq 0\). Fix a deterministic, finite initial population \(Z_0=z_0\in S\). For each generation \(n\geq0\) and each potential individual \(k\geq1\), let \(\xi_{n,k}\) be that individual's offspring count. Assume the variables \(\xi_{n,k}\) are independent and identically distributed, take values in \(S\), and have mean \(m=\mathbb{E}[\xi_{n,k}]\) satisfying \(0<m<\infty\). The population evolves by

where a sum with no terms is \(0\). Thus a population of size \(0\) remains \(0\). Each \(Z_n\) is finite almost surely: this holds at \(n=0\), and a finite sum of finite integer-valued offspring counts is finite. Define the generation filtration by \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(\xi_{r,k}:0\leq r<n,\ k\geq1)\) for \(n\geq1\). It records offspring counts used to form generations through \(n\), but not the reproduction of generation \(n\). In particular, \(Z_n\) is \(\mathcal{F}_n\)-measurable, and the offspring counts \(\xi_{n,k}\) are independent of \(\mathcal{F}_n\).

The Conditional Mean and the Martingale Result

The central calculation is conditional on the whole current generation. Each individual in that generation contributes an offspring count with conditional mean \(m\); the number of contributing individuals, \(Z_n\), is already known given \(\mathcal{F}_n\). The identity below makes this precise, including the case \(Z_n=0\).

Proof. For each \(k\geq1\), the event \(\{Z_n\geq k\}\) belongs to \(\mathcal{F}_n\), and \(\xi_{n,k}\) is independent of \(\mathcal{F}_n\). Consequently, \(\mathbb{E}[\mathbf{1}_{\{Z_n\geq k\}}\xi_{n,k}\mid\mathcal{F}_n]=\mathbf{1}_{\{Z_n\geq k\}}m\). The population sum can be written as the nonnegative series \(Z_{n+1}=\sum_{k\geq1}\mathbf{1}_{\{Z_n\geq k\}}\xi_{n,k}\). Applying conditional monotone convergence to its partial sums gives \[ \mathbb{E}[Z_{n+1}\mid\mathcal{F}_n] =\sum_{k\geq1}\mathbf{1}_{\{Z_n\geq k\}}m =mZ_n. \] The last equality holds because exactly \(Z_n\) terms in the sum are \(1\) when \(Z_n\) is finite; if \(Z_n=0\), both sides are \(0\). \(\square\)

Proof. First, \(Z_0\) is integrable. If \(Z_n\) is integrable, the lemma and the defining property of conditional expectation give \(\mathbb{E}[Z_{n+1}]=\mathbb{E}[\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]]=m\mathbb{E}[Z_n]<\infty\). Induction proves that every \(Z_n\), and therefore every \(W_n\), is integrable. The process \(W_n\) is \(\mathcal{F}_n\)-measurable because \(Z_n\) is, and \(m^n\) is a positive constant. Finally, using the lemma and the fact that \(m^{n+1}\) is nonzero, \[ \mathbb{E}[W_{n+1}\mid\mathcal{F}_n] =\frac{\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]}{m^{n+1}} =\frac{mZ_n}{m^{n+1}} =\frac{Z_n}{m^n} =W_n. \] These are all the conditions in the martingale definition. \(\square\)

Proof. The induction in the theorem gives \(\mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]\), starting from \(\mathbb{E}[Z_0]=z_0\). Repeated application yields \(\mathbb{E}[Z_n]=z_0m^n\). Dividing by \(m^n>0\) gives \(\mathbb{E}[W_n]=z_0\). \(\square\)

The normalization therefore keeps the mean population fixed at its initial value. It does not say that \(W_n\) stays equal to \(z_0\) on every outcome, or that the population itself is constant. The martingale claim is about conditional expectation given the current generation's history.

Worked Examples

Assumptions and a Common Mistake

The finite positive mean assumption has two distinct roles. Finiteness ensures that the next population has finite conditional mean when the current population is finite, which supports integrability. Positivity makes the normalization \(m^n\) nonzero. No finite-variance assumption is needed for this martingale result. If \(m=0\), nonnegative offspring counts must equal \(0\) almost surely, but \(Z_n/m^n\) is not defined. If the offspring mean is infinite, the finite-mean calculation and integrability argument do not apply.

The generation filtration also matters. The offspring counts used to form \(Z_{n+1}\) are not yet revealed in \(\mathcal{F}_n\); they are independent of the information being conditioned on. Conditioning on information that includes those counts is a different conditional expectation and need not give the same identity.

Common mistake. It is incorrect to say that \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=m\), as if the next generation consisted of one individual's offspring regardless of the current population. For the process in the preceding example, the event \(Z_1=3\) has probability \(\tfrac14>0\). On that event, the conditional expected next population is \(3\cdot\tfrac54=\tfrac{15}4\), not \(\tfrac54\). The correction is to count all \(Z_n\) parents: the right conditional mean is \(mZ_n\). Dividing this quantity by \(m^{n+1}\), rather than omitting the factor \(Z_n\), gives the required identity.

The result identifies a useful scale for comparing populations across generations: the mean growth factor is removed, leaving a martingale relative to the generation-by-generation information. The identity alone does not claim that individual normalized population paths are constant, nor does it establish a limiting value; it establishes exactly the conditional-mean relation proved above.