Second Moments of the Branching Population Martingale

Why Second Moments Matter

A normalized Galton–Watson population is a martingale, but the martingale property alone does not say whether its fluctuations remain controlled as generations pass. The second moment answers that question directly. It also identifies when the martingale is bounded in $L^2$, a condition that permits use of the earlier material on $L^2$-bounded martingale convergence.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $Z_0=1$, and let $(Z_n)_{n\ge0}$ be a Galton–Watson process: every individual independently produces a number $\xi$ of children, where $\xi$ is a nonnegative integer-valued random variable. Write $m=\mathbb{E}[\xi]$ and $\sigma^2=\operatorname{Var}(\xi)$. Throughout, assume $0<m<\infty$ and $\sigma^2<\infty$. Let $\mathcal{F}_n$ record the population and reproduction information through generation $n$, before the individuals in generation $n$ reproduce. The offspring counts in generation $n$ are independent of $\mathcal{F}_n$ and are independent copies of $\xi$.

Definition: The normalized population is $W_n=Z_n/m^n$. A sequence is bounded in $L^2$ if $\sup_{n\ge0}\mathbb{E}[W_n^2]<\infty$. This definition concerns the second moments, not just the means.

Conditional on $\mathcal{F}_n$, the next population is the sum of $Z_n$ independent offspring counts. Thus $\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=mZ_n$, and dividing by $m^{n+1}$ gives $\mathbb{E}[W_{n+1}\mid\mathcal{F}_n]=W_n$. Also, induction gives $\mathbb{E}[Z_n]=m^n$, so $\mathbb{E}[W_n]=1$. These facts establish the martingale property and its constant mean. They do not determine whether $\mathbb{E}[W_n^2]$ stays bounded.

The Second-Moment Formula

Theorem: Under the assumptions above, for every $n\ge0$, $$ \operatorname{Var}(W_n)=\frac{\sigma^2}{m^2}\sum_{r=0}^{n-1}m^{-r}, \qquad \mathbb{E}[W_n^2]=1+\frac{\sigma^2}{m^2}\sum_{r=0}^{n-1}m^{-r}. $$ The sum is empty and equals $0$ when $n=0$. If $m\ne1$, this can also be written as $$ \operatorname{Var}(W_n)=\frac{\sigma^2}{m(m-1)}(1-m^{-n}). $$ If $m=1$, then $\operatorname{Var}(W_n)=n\sigma^2$.

Proof. Put $D_{n+1}=W_{n+1}-W_n=(Z_{n+1}-mZ_n)/m^{n+1}$. Since the conditional mean of $Z_{n+1}$ is $mZ_n$, $\mathbb{E}[D_{n+1}\mid\mathcal{F}_n]=0$. Conditional on $\mathcal{F}_n$, $Z_{n+1}$ is a sum of $Z_n$ independent copies of $\xi$. Therefore its conditional variance is $Z_n\sigma^2$, and $$ \mathbb{E}[D_{n+1}^2\mid\mathcal{F}_n] =\frac{Z_n\sigma^2}{m^{2n+2}}. $$ 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, so the conditional-mean property gives $$ \mathbb{E}[D_{r+1}D_{s+1}] =\mathbb{E}\!\left[D_{r+1}\mathbb{E}[D_{s+1}\mid\mathcal{F}_s]\right]=0. $$ Thus the increments are pairwise orthogonal in $L^2$. Since $W_n=W_0+\sum_{r=0}^{n-1}D_{r+1}$ and $W_0=1$, their means are zero, and their cross-products have expectation zero, expanding the square yields $$ \operatorname{Var}(W_n)=\sum_{r=0}^{n-1}\mathbb{E}[D_{r+1}^2] =\sigma^2\sum_{r=0}^{n-1}m^{-(r+2)} =\frac{\sigma^2}{m^2}\sum_{r=0}^{n-1}m^{-r}. $$ The constant mean $\mathbb{E}[W_n]=1$ implies $\mathbb{E}[W_n^2]=1+\operatorname{Var}(W_n)$. For $m\ne1$, the finite geometric sum is $(1-m^{-n})/(1-m^{-1})$; substituting and simplifying gives the stated expression. For $m=1$, each of the $n$ terms in the sum is $1$, giving $n\sigma^2$. $\square$

The formula gives a sharp distinction between regimes. “Bounded in $L^2$” means the supremum over all $n$ is finite; it is stronger than having a finite second moment at each fixed generation.

Corollary: With $0<m<\infty$ and $\sigma^2<\infty$, $(W_n)$ is bounded in $L^2$ if and only if either $m>1$, or $m=1$ and $\sigma^2=0$.

Proof. If $m>1$, then $0<m^{-1}<1$, and $$ \sum_{r=0}^{n-1}m^{-r}\le \sum_{r=0}^{\infty}m^{-r}=\frac{1}{1-m^{-1}}. $$ The second-moment formula therefore gives a finite bound independent of $n$. If $m=1$, it gives $\mathbb{E}[W_n^2]=1+n\sigma^2$, which is bounded precisely when $\sigma^2=0$. If $0<m<1$ and $\sigma^2>0$, then $m^{-r}\ge1$ for every $r\ge0$, so $\operatorname{Var}(W_n)\ge n\sigma^2/m^2$, which is unbounded. The remaining possibility, $0<m<1$ and $\sigma^2=0$, cannot occur: zero variance forces $\xi=m$ almost surely, while $\xi$ is integer-valued and there is no integer strictly between $0$ and $1$. This proves both directions. $\square$

In the exceptional critical case $m=1,\sigma^2=0$, the offspring count equals $1$ almost surely. Hence $Z_n=1$ and $W_n=1$ at every generation. In the supercritical case $m>1$, the earlier theorem that $L^2$-bounded martingales converge in $L^2$ applies to $W_n$. The second-moment calculation supplies its required boundedness condition; it does not by itself assert that condition in the other regimes.

Worked Calculations

Worked Example: A Supercritical Process with Two Possible Offspring Counts

Suppose each individual has $1$ or $3$ children, each with probability $\tfrac12$. Find the variance and second moment of $W_n$, and verify the first-generation values.

The probabilities sum to $\tfrac12+\tfrac12=1$. The mean is $$ m=1\cdot\tfrac12+3\cdot\tfrac12=2, $$ and the second moment of the offspring count is $1^2\cdot\tfrac12+3^2\cdot\tfrac12=5$. Thus $\sigma^2=5-2^2=1$. Substitution into the formula gives $$ \operatorname{Var}(W_n)=\frac14\sum_{r=0}^{n-1}2^{-r} =\frac12(1-2^{-n}),\qquad \mathbb{E}[W_n^2]=\frac32-\frac{1}{2^{n+1}}. $$ For $n=1$, $W_1=Z_1/2$ is $\tfrac12$ or $\tfrac32$, each with probability $\tfrac12$. Its mean is $\tfrac12\cdot\tfrac12+\tfrac32\cdot\tfrac12=1$, and its variance is $\tfrac12(\tfrac12-1)^2+\tfrac12(\tfrac32-1)^2=\tfrac14$. The formula gives $\tfrac12(1-\tfrac12)=\tfrac14$, and its second-moment expression gives $\tfrac32-\tfrac14=\tfrac54$, matching $\tfrac12(\tfrac12)^2+\tfrac12(\tfrac32)^2=\tfrac54$. Since $m=2>1$, the second moments remain bounded.

Worked Example: A Critical Process with Growing Variance

Suppose an individual has $0$ or $2$ children, each with probability $\tfrac12$. Calculate the second moment at generation $n$, then check it directly at $n=1$.

The probabilities sum to $1$, and $$ m=0\cdot\tfrac12+2\cdot\tfrac12=1,\qquad \sigma^2=(0-1)^2\cdot\tfrac12+(2-1)^2\cdot\tfrac12=1. $$ The critical-case formula gives $\operatorname{Var}(W_n)=n$ and $\mathbb{E}[W_n^2]=1+n$. At $n=1$, $W_1=Z_1$ is $0$ or $2$, each with probability $\tfrac12$. Its mean is $0\cdot\tfrac12+2\cdot\tfrac12=1$; its second moment is $0^2\cdot\tfrac12+2^2\cdot\tfrac12=2$, and its variance is $2-1^2=1$. These agree with the formula at $n=1$. Here the martingale has constant mean $1$, but its second moments grow without bound.

Worked Example: A Subcritical Process with an Unbounded Normalization

Suppose each individual has $0$ or $1$ child, each with probability $\tfrac12$. Find the second moment of $W_n$, and check it at $n=2$.

The probabilities sum to $1$, and $$ m=0\cdot\tfrac12+1\cdot\tfrac12=\tfrac12,\qquad \sigma^2=(0-\tfrac12)^2\cdot\tfrac12+(1-\tfrac12)^2\cdot\tfrac12=\tfrac14. $$ Consequently, $$ \operatorname{Var}(W_n) =\frac{\tfrac14}{(\tfrac12)^2}\sum_{r=0}^{n-1}2^r =2^n-1,\qquad \mathbb{E}[W_n^2]=2^n. $$ For a direct check at $n=2$, survival to generation $2$ requires both individuals in the single ancestral line to have a child, which has probability $\tfrac12\cdot\tfrac12=\tfrac14$. Thus $Z_2=1$ with probability $\tfrac14$ and $Z_2=0$ with probability $\tfrac34$. Since $m^2=\tfrac14$, $W_2=4$ with probability $\tfrac14$ and $W_2=0$ with probability $\tfrac34$. Its mean is $4\cdot\tfrac14=1$, its second moment is $4^2\cdot\tfrac14=4$, and its variance is $4-1^2=3$. The formula gives $\mathbb{E}[W_2^2]=2^2=4$ and $\operatorname{Var}(W_2)=2^2-1=3$, as required.

Interpreting the Formula and Avoiding a Common Error

The three examples show why normalization needs to be interpreted alongside the offspring variance. In the subcritical example, the expected population decreases, but dividing by $m^n$ magnifies the surviving population; the normalized martingale still has mean $1$, while its second moment grows. In the critical example, the mean stays constant and the variance grows linearly. In the supercritical example, the normalized fluctuations have bounded second moments. A martingale property controls conditional means, not second moments.

Common mistake: forgetting the normalization’s extra factor of $m$. In the supercritical example, $m=2$ and $\sigma^2=1$. At the first generation, $W_1=Z_1/2$, so its variance is $\operatorname{Var}(\xi/2)=1/4$. Using $\sigma^2/m^n$ as the variance contribution at step $n$ would give $1$ at $n=0$, which fails this direct check. The correct increment is $D_{n+1}=(Z_{n+1}-mZ_n)/m^{n+1}$; its variance is $\sigma^2/m^{n+2}$, and at $n=0$ this is $1/2^2=1/4$. The second common error is to identify the second moment with the variance: here $\mathbb{E}[W_1^2]=1+\operatorname{Var}(W_1)=\tfrac54$, because the mean is $1$.

When applying the result, check the assumptions before using the formula: there must be one initial ancestor, $m>0$, and finite offspring variance. If $m=0$, the normalization $Z_n/m^n$ is not defined. If the offspring variance is infinite, the conditional variance calculation does not give a finite second moment formula. Under the stated assumptions, the variance sum is the key calculation: it makes explicit whether the accumulated martingale-increment variances stay finite.

Check Your Understanding

Use the offspring mean and variance, then substitute into the finite-generation formula before drawing conclusions.

If $m=3$ and $\sigma^2=2$, write $\operatorname{Var}(W_n)$ as a finite geometric sum and find its limit as $n$ increases.
For $m=1$ and $\sigma^2=0$, what must the offspring count be almost surely, and what is $W_n$?
For $m=1$ and $\sigma^2=\tfrac12$, calculate $\mathbb{E}[W_4^2]$.
For $m=\tfrac12$ and $\sigma^2=\tfrac18$, calculate $\operatorname{Var}(W_2)$ and check whether the second moments are bounded.
Explain why the cross-products of two distinct martingale increments have expectation zero.

Second Moments of the Branching Population Martingale

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