Why the Normalized Limit Needs More Than a Finite Mean
In a supercritical Galton–Watson process, the expected population grows geometrically. Dividing the generation-\(n\) population by that expected growth removes the leading deterministic trend. The resulting normalized population is a nonnegative martingale, so it has an almost-sure limit. But almost-sure convergence alone does not guarantee that the limit retains its mean or is positive when the population survives. The Kesten–Stigum theorem identifies the extra condition that resolves both questions: a finite offspring \(x\log x\) moment.
Let \((Z_n)_{n\ge0}\) be a Galton–Watson process on a probability space \((\Omega,\mathcal F,\mathbb P)\), started with \(Z_0=1\). Each individual independently has \(\xi\) children, where \(\xi\) is a nonnegative integer-valued random variable with probabilities \(p_k=\mathbb P(\xi=k)\) and mean \(m=\mathbb E[\xi]\). Write \(\mathcal F_n\) for the information about the process through generation \(n\). Throughout, assume \(1<m<\infty\). The survival event is \(\mathcal S=\{Z_n>0\text{ for every }n\ge0\}\), and \(W_n=Z_n/m^n\).
The martingale property follows because, conditional on \(\mathcal F_n\), the \(Z_n\) individuals in generation \(n\) have independent offspring counts with mean \(m\). Thus \(\mathbb E[Z_{n+1}\mid\mathcal F_n]=mZ_n\), and \(\mathbb E[W_{n+1}\mid\mathcal F_n]=W_n\). In particular, \(\mathbb E[W_n]=1\) for every \(n\). The subtle question is whether \(\mathbb E[W]=1\): a nonnegative martingale can lose expectation in its limit.
The Kesten–Stigum Theorem
- If \(\mathbb E[\xi\log\xi]<\infty\), the martingale \((W_n)\) is uniformly integrable, \(\mathbb E[W]=1\), and \(W>0\) almost surely on \(\mathcal S\).
- If \(\mathbb E[\xi\log\xi]=\infty\), then \(W=0\) almost surely, including on \(\mathcal S\).
The distinction is not between survival and extinction: survival has positive probability under the assumptions \(m>1\) and \(m<\infty\). Rather, the distinction is whether the population, when normalized by \(m^n\), has a positive limiting size on surviving paths. A finite mean guarantees the martingale, but does not by itself prevent its limit from being zero.
The main technical input in the theorem’s proof is a standard result about the size-biased branching tree, often called the spine criterion. Define the size-biased offspring variable \(\xi^*\) by \(\mathbb P(\xi^*=k)=kp_k/m\). These probabilities sum to one because \(\sum_k kp_k/m=1\), and \(\mathbb E[\log \xi^*]=\mathbb E[\xi\log\xi]/m\), with the zero-offspring term interpreted as zero. The spine criterion states that the probability measure obtained by biasing generation-\(n\) trees by \(W_n\) has an absolutely continuous limit on the full branching-tree information exactly when \(\mathbb E[\log \xi^*]<\infty\); when this expectation is infinite, that limit is singular to the original tree law. The criterion is a standard deep result: its proof uses a spine construction and a summability argument and is beyond this tutorial. We use it explicitly rather than treating that technical step as proved here.
Proof of the theorem, given the spine criterion. The measures on generation-\(n\) information defined by \(\mathbb Q(A)=\mathbb E[\mathbf 1_A W_n]\), for \(A\in\mathcal F_n\), are consistent: if \(A\in\mathcal F_n\), then \(\mathbb E[\mathbf 1_AW_{n+1}]=\mathbb E[\mathbf 1_A\mathbb E[W_{n+1}\mid\mathcal F_n]]=\mathbb E[\mathbf 1_AW_n]\). The spine criterion applies to their limiting measure. For a nonnegative martingale, the density of the absolutely continuous part of this limiting measure with respect to the original tree law is \(W\); in particular, its total absolutely continuous mass is \(\mathbb E[W]\). This is the standard martingale change-of-measure identity used in the spine criterion.
If \(\mathbb E[\xi\log\xi]<\infty\), then \(\mathbb E[\log\xi^*]<\infty\), so the limiting size-biased measure is absolutely continuous. Its total mass is one, and therefore \(\mathbb E[W]=1\). Since \(W_n\) is a nonnegative martingale with \(\mathbb E[W_n]=1\) and its limit has the same expectation, the martingale is uniformly integrable. If instead \(\mathbb E[\xi\log\xi]=\infty\), the limiting size-biased measure is singular, so its absolutely continuous part has mass zero. The density identity gives \(\mathbb E[W]=0\); since \(W\ge0\), this implies \(W=0\) almost surely.
It remains to establish the positivity statement in the finite-moment case. Let \(q\) be the extinction probability. The branching property gives the distributional identity \(W\overset{d}=m^{-1}\sum_{i=1}^{\xi}W_i\), where, conditional on \(\xi\), the \(W_i\) are independent copies of \(W\), independent of \(\xi\); an empty sum is zero. Because the summands are nonnegative, the sum is zero exactly when every child’s limit is zero. Thus \(r=\mathbb P(W=0)\) satisfies \(r=\mathbb E[r^\xi]\). Also \(r<1\), since \(\mathbb E[W]=1\). The extinction probability \(q\) is the smallest solution in \([0,1]\) of \(s=\mathbb E[s^\xi]\), and for a supercritical Galton–Watson process it is the unique solution below \(1\); this is the standard extinction-probability result from the earlier branching-process material. Hence \(r=q\). Extinction implies \(W=0\), as shown below, so \(\{W=0\}\) and extinction have the same probability and one contains the other. They therefore agree up to a probability-zero event, proving \(W>0\) almost surely on survival. \(\square\)
For the last step, if extinction occurs, then \(Z_n=0\) from some generation onward, and consequently \(W_n=0\) from that generation onward. This proves \(W=0\) on extinction without requiring the \(x\log x\) condition.
Examples: What the Moment Condition Predicts
Worked Example: A Bounded Offspring Law
Suppose each individual has either zero or three children, each with probability \(\tfrac12\). Then \(m=0\cdot\tfrac12+3\cdot\tfrac12=\tfrac32>1\), and the offspring moment is \(\mathbb E[\xi\log\xi]=\tfrac32\log3<\infty\). The theorem predicts a mean-one limit that is positive on survival.
The extinction probability solves \(q=\tfrac12+\tfrac12q^3\). Rearranging gives \(q^3-2q+1=0\), and factoring gives \((q-1)(q^2+q-1)=0\). The root below one in \([0,1]\) is \(q=(\sqrt5-1)/2\), since it is positive, less than one, and solves \(q^2+q-1=0\). Thus the survival probability is \(1-q=(3-\sqrt5)/2>0\).
Check: Substituting the quadratic relation \(q^2=1-q\) gives \(q^3=q(1-q)=q-q^2=2q-1\), so \(\tfrac12+\tfrac12q^3=\tfrac12+\tfrac12(2q-1)=q\). The finite \(x\log x\) moment therefore gives \(W>0\) almost surely on an event whose probability is \((3-\sqrt5)/2\), while extinction has probability \(q\).
Worked Example: Finite Mean but Infinite \(x\log x\) Moment
For integers \(k\ge2\), let \(C=\sum_{k=2}^{\infty}1/[k^2(\log k)^2]\) and set \(p_k=1/[Ck^2(\log k)^2]\), with \(p_0=p_1=0\). The constant \(C\) is positive and finite, so these probabilities sum to one. The mean is \(m=C^{-1}\sum_{k=2}^{\infty}1/[k(\log k)^2]\), which is finite by the integral test. Since every possible offspring count is at least two, \(m\ge2>1\).
For the \(x\log x\) moment, substitution of the probabilities gives \(\mathbb E[\xi\log\xi]=C^{-1}\sum_{k=2}^{\infty}1/[k\log k]=\infty\), again by the integral test. The theorem therefore gives \(W=0\) almost surely. Yet extinction is impossible: each generation has at least twice as many individuals as the previous one, so \(Z_n\ge2^n>0\) for every \(n\). Survival has probability one, but the normalized limit is still zero.
Check: The probabilities sum to \(C^{-1}C=1\); the mean series has terms \(1/[k(\log k)^2]\), while the \(x\log x\) series has terms \(1/[k\log k]\). The first converges and the second diverges. This verifies that the example has finite supercritical mean and infinite \(x\log x\) moment, rather than an infinite mean or a positive extinction probability.
Worked Example: Deterministic Growth
Suppose every individual has exactly two children. Then \(p_2=1\), \(m=2\), and \(Z_n=2^n\) for every \(n\), starting from \(Z_0=1\). Hence \(W_n=Z_n/m^n=2^n/2^n=1\) at every generation, so \(W=1\). The moment is \(\mathbb E[\xi\log\xi]=2\log2<\infty\), in agreement with the theorem.
Check: The offspring probabilities sum to one, the mean is \(2\cdot1=2>1\), and direct substitution gives \(W_n=1\) for each \(n\). Here survival occurs with probability one and the normalized limit is positive on every path.
Interpretation and a Common Pitfall
The \(x\log x\) condition measures how much weight is carried by unusually large families after size-biasing. Under the size-biased law, an individual is more likely to be sampled when it has many children; its expected log offspring count is \(\mathbb E[\xi\log\xi]/m\). If this expectation is finite, the size-biased change of measure remains absolutely continuous, and the normalized martingale retains its mean. If it is infinite, the biased measure becomes singular, and the martingale limit loses all its mean. The infinite-moment example shows why merely checking \(m<\infty\) cannot establish a positive normalized limit.
Common mistake: assuming almost-sure convergence preserves expectation. In the infinite-moment example, \(\mathbb E[W_n]=1\) for every \(n\), because \(\mathbb E[Z_n]=m^n\). A mistaken calculation would then claim \(\mathbb E[W]=\lim_n\mathbb E[W_n]=1\), and conclude that \(W\) is positive with positive probability. The correction is that exchanging expectation and an almost-sure limit requires an additional condition, such as uniform integrability. Here the Kesten–Stigum theorem gives \(W=0\) almost surely, so \(\mathbb E[W]=0\), even though every \(\mathbb E[W_n]=1\). The expectations do not converge to the expectation of the limit.
Check Your Understanding
For each question, distinguish the mean offspring condition from the \(x\log x\) condition, and distinguish survival from positivity of the normalized limit.
- State the Kesten–Stigum conclusion when \(1<m<\infty\) and \(\mathbb E[\xi\log\xi]<\infty\).
- Why does extinction force \(W=0\), regardless of whether the \(x\log x\) moment is finite?
- For the offspring law with \(p_0=p_3=\tfrac12\), calculate \(m\) and \(\mathbb E[\xi\log\xi]\). What does the theorem say about \(W\) on survival?
- In the distribution supported on integers \(k\ge2\) with probabilities proportional to \(1/[k^2(\log k)^2]\), which series establishes finite mean, and which establishes an infinite \(x\log x\) moment?
- Why is it invalid to infer \(\mathbb E[W]=1\) solely from \(\mathbb E[W_n]=1\) for every \(n\) and almost-sure convergence?