Why Normalize the Population?
In a supercritical Galton–Watson process, the expected population grows geometrically. Dividing the population in generation \(n\) by the corresponding geometric growth factor removes this average growth and produces a process whose conditional expectation does not change from one generation to the next. This martingale property is useful in its own right, and it provides the starting point for later questions about the long-run behavior of a branching population.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge0}\) be a one-type Galton–Watson process with one ancestor, so \(Z_0=1\). Each individual independently produces a number \(\xi\) of children, where \(\xi\) takes values in the nonnegative integers. Assume its offspring mean \(m=\mathbb{E}[\xi]\) satisfies \(1<m<\infty\). The finite mean assumption is essential for the integrability statements below. Let \(\mathcal{F}_n\) record the family tree through generation \(n\), including the identities and number of individuals in that generation, but not their reproduction yet.
Each generation is finite almost surely: every individual has a finite integer number of children almost surely, and induction from the finite value \(Z_0=1\) gives finite \(Z_n\) for every fixed \(n\). Given \(\mathcal{F}_n\), the individuals in generation \(n\) reproduce independently, each with mean offspring number \(m\). This conditional description is the basis of the result.
Proof. Given \(\mathcal{F}_n\), write \(Z_{n+1}\) as the sum of the offspring counts of the \(Z_n\) individuals in generation \(n\). Conditional expectation of this finite sum is \(mZ_n\), so \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=mZ_n\) almost surely. First, \(Z_0=1\) is integrable. If \(Z_n\) is integrable, taking expectations in the conditional identity gives \(\mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]<\infty\). Induction therefore gives \(\mathbb{E}[Z_n]=m^n\) for each \(n\). In particular, \(W_n\) is integrable, since it is nonnegative and \(\mathbb{E}[W_n]=m^{-n}\mathbb{E}[Z_n]=1\); it is also \(\mathcal{F}_n\)-measurable. Finally, \[ \mathbb{E}[W_{n+1}\mid\mathcal{F}_n] =\frac{\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]}{m^{n+1}} =\frac{mZ_n}{m^{n+1}} =W_n \] almost surely. These are precisely the integrability, adaptedness, and conditional-expectation requirements for a martingale, and the expectation identities follow as stated. \(\square\)
The finite offspring mean is enough for this martingale result. In particular, a finite offspring variance is not needed to make \(W_n\) integrable. A finite variance does, however, allow us to calculate the second moments of the normalized population.
Second Moments When They Exist
Suppose in addition that the offspring variance \(\sigma^2=\operatorname{Var}(\xi)\) is finite. The martingale increments are \(D_{n+1}=W_{n+1}-W_n\). Their conditional mean is zero, because the martingale identity gives \(\mathbb{E}[D_{n+1}\mid\mathcal{F}_n]=0\). Their conditional variance can also be computed from the independent offspring counts.
Proof. Conditional on \(\mathcal{F}_n\), \(Z_{n+1}\) is a sum of \(Z_n\) independent offspring counts, so its conditional variance is \(\sigma^2 Z_n\). Since \(W_{n+1}-W_n=(Z_{n+1}-mZ_n)/m^{n+1}\), it follows that \[ \mathbb{E}[D_{n+1}^2\mid\mathcal{F}_n]=\frac{\sigma^2 Z_n}{m^{2n+2}}. \] Taking expectations and using \(\mathbb{E}[Z_n]=m^n\) gives \(\mathbb{E}[D_{n+1}^2]=\sigma^2/m^{n+2}\). For \(0\le r<s\), \(D_{r+1}\) is \(\mathcal{F}_s\)-measurable. Thus \[ \mathbb{E}[D_{r+1}D_{s+1}] =\mathbb{E}\!\left[D_{r+1}\mathbb{E}[D_{s+1}\mid\mathcal{F}_s]\right]=0. \] The increments are therefore pairwise orthogonal in second moment. Since \(W_0=1\), summing their second moments yields \[ \operatorname{Var}(W_n)=\sum_{k=0}^{n-1}\mathbb{E}[D_{k+1}^2] =\sigma^2\sum_{k=0}^{n-1}m^{-(k+2)} =\frac{\sigma^2}{m(m-1)}(1-m^{-n}). \] For \(n=0\), the sum is empty and both sides equal zero. Finally, \(\mathbb{E}[W_n^2]=\operatorname{Var}(W_n)+(\mathbb{E}[W_n])^2\), and \(\mathbb{E}[W_n]=1\). Since \(0\le1-m^{-n}\le1\), the stated uniform bound follows. \(\square\)
This formula is a second-moment statement, not a general condition for the martingale property. When \(\sigma^2\) is infinite, the formula does not apply, even though the finite-mean theorem still guarantees that every \(W_n\) is integrable.
Worked Example: A Zero-or-Two Offspring Law
Suppose each individual has \(0\) children with probability \(\tfrac14\) and \(2\) children with probability \(\tfrac34\). The probabilities sum to \(\tfrac14+\tfrac34=1\), and the offspring mean is \[ m=0\cdot\tfrac14+2\cdot\tfrac34=\tfrac32>1. \] The offspring second moment is \(\mathbb{E}[\xi^2]=0^2\cdot\tfrac14+2^2\cdot\tfrac34=3\), so \(\sigma^2=3-(\tfrac32)^2=\tfrac34\).
In generation \(1\), \(Z_1\) equals \(0\) with probability \(\tfrac14\) and \(2\) with probability \(\tfrac34\). Consequently, \(W_1=Z_1/(\tfrac32)\) equals \(0\) with probability \(\tfrac14\) and \(\tfrac43\) with probability \(\tfrac34\). Direct calculation gives \(\mathbb{E}[W_1]=0\cdot\tfrac14+\tfrac43\cdot\tfrac34=1=W_0\), checking the first martingale step.
The variance formula gives, at generation \(1\), \[ \operatorname{Var}(W_1) =\frac{\tfrac34}{(\tfrac32)(\tfrac12)}\left(1-\tfrac23\right) =1\cdot\tfrac13=\tfrac13. \] Directly, \(\mathbb{E}[W_1^2]=0^2\cdot\tfrac14+(\tfrac43)^2\cdot\tfrac34=\tfrac43\), so \(\operatorname{Var}(W_1)=\tfrac43-1^2=\tfrac13\), as required.
Worked Example: Deterministic Two-Child Reproduction
Suppose every individual has exactly \(2\) children. The offspring law has total probability \(1\) at \(\xi=2\), so \(m=2\) and \(\sigma^2=0\). The population is \(Z_n=2^n\): this holds at \(n=0\), and if \(Z_n=2^n\), then twice as many children give \(Z_{n+1}=2^{n+1}\). Hence \(W_n=2^n/2^n=1\) for every \(n\).
There is no randomness in \(W_n\), so its expectation is \(1\) and its variance is \(0\) at every generation. The second-moment formula also gives \(\operatorname{Var}(W_n)=\frac{0}{2(2-1)}(1-2^{-n})=0\). This example checks that the normalized martingale need not fluctuate: its martingale property allows, but does not require, random increments.
Worked Example: Finite Mean but Infinite Offspring Variance
Let \(\zeta(3)=\sum_{k=1}^{\infty}k^{-3}\), and let \(Y\) have probabilities \(\mathbb{P}(Y=k)=1/(\zeta(3)k^3)\) for integers \(k\ge1\). These probabilities sum to \(1\) by the definition of \(\zeta(3)\). Define \(\xi\) by choosing \(\xi=2\) with probability \(\tfrac12\), and otherwise drawing an independent \(Y\) with probability \(\tfrac12\). This specifies a probability law with total mass \(\tfrac12+\tfrac12=1\).
Because \(\mathbb{E}[Y]=\sum_{k\ge1}k/(\zeta(3)k^3)=\zeta(2)/\zeta(3)<\infty\), where \(\zeta(2)=\sum_{k\ge1}k^{-2}\), the offspring mean is finite and equals \[ m=\tfrac12\cdot2+\tfrac12\cdot\frac{\zeta(2)}{\zeta(3)} =1+\frac{\zeta(2)}{2\zeta(3)}>1. \] Its second moment is infinite: the contribution from the \(Y\) component is \(\tfrac{1}{2\zeta(3)}\sum_{k\ge1}k^{-1}=\infty\). Thus \(\operatorname{Var}(\xi)=\infty\).
The finite-mean theorem still applies, so \(W_1=\xi/m\) is integrable and \(\mathbb{E}[W_1]=\mathbb{E}[\xi]/m=1\). But \(\mathbb{E}[W_1^2]=\mathbb{E}[\xi^2]/m^2=\infty\). This verifies the distinction: the normalized population is a martingale, while its first generation is not square-integrable, so the finite-variance theorem cannot be used.
What the Martingale Property Does—and Does Not—Say
The defining equality is conditional: given the entire population history through generation \(n\), the expected normalized population in the next generation equals its current normalized value. It does not say that \(W_{n+1}=W_n\) on each realization, or that successive generations have the same distribution. In the zero-or-two example, for instance, \(W_1\) can be \(0\) or \(\tfrac43\), even though its expectation is \(W_0=1\).
A common mistake is to normalize \(Z_{n+1}\) by \(m^n\) instead of \(m^{n+1}\). For the zero-or-two example, this incorrect choice at the first step gives \(Z_1/m^0=Z_1\), whose expectation is \(\mathbb{E}[Z_1]=\tfrac32\), not \(W_0=1\). The correction is to divide by \(m^{n+1}\): \(\mathbb{E}[Z_{n+1}/m^{n+1}\mid\mathcal{F}_n]=mZ_n/m^{n+1}=Z_n/m^n\).
The martingale result establishes integrability and a conditional-mean identity at every finite generation. It does not by itself establish that \(W_n\) has an almost-sure or other limiting value. The finite-variance calculation gives a uniform second-moment bound when \(\sigma^2<\infty\); without that hypothesis this bound is unavailable, as the heavy-tailed example demonstrates. Questions about convergence and its precise conditions require additional results.
Check Your Understanding
Use the definition of a martingale and check each expectation or condition explicitly.
- If \(Z_0=1\) and \(m=\tfrac43\), what is \(W_2\) in terms of \(Z_2\)?
- Which conditional expectation identity for \(Z_{n+1}\) implies the martingale identity for \(W_n\)?
- For an offspring law with mean \(m=2\) and variance \(\sigma^2=1\), compute \(\operatorname{Var}(W_1)\) from the formula.
- Why does a finite offspring mean suffice for integrability of each \(W_n\), even if the offspring variance is infinite?
- Does \(\mathbb{E}[W_{n+1}\mid\mathcal{F}_n]=W_n\) imply \(W_{n+1}=W_n\) for every outcome? Explain.