In a Galton-Watson process with offspring distribution P(X=0) = 1/4, P(X=1) = 1/2, P(X=2) = 1/4, the mean offspring number μ is 1. The extinction probability is:
A0 — the population survives forever with probability 1
B1 — extinction is certain, since μ = 1 (critical case)
C1/2 — the process is a fair coin flip between survival and extinction
D3/4 — determined by P(X=0) + P(X=1)
When μ = 1 (critical case), the extinction probability is 1, regardless of the offspring distribution (assuming the distribution is not degenerate at 1). The process is a non-negative martingale with mean 1 in every generation, but it is absorbed at 0. The expected population stays constant, but the variance accumulates: Var(Z_n) = nσ² grows without bound. The process drifts to extinction through random fluctuations — the expected value is maintained only because increasingly rare large populations compensate for the increasingly likely extinction.
Question 2 True / False
The extinction probability q of a supercritical Galton-Watson process (μ > 1) is the smallest non-negative fixed point of the probability generating function G(s) = E[s^X].
TTrue
FFalse
Answer: True
The extinction probability satisfies q = P(Z_n → 0) = G(q), derived by conditioning on the first generation: q = Σ_k p_k · q^k = G(q). The PGF G is convex on [0,1] with G(1) = 1. When μ = G'(1) > 1, the convexity ensures G has a second fixed point q* < 1 in addition to s = 1. The extinction probability is this smaller fixed point. Geometrically, G(s) dips below the diagonal near s = q* and returns to it at s = 1. When μ ≤ 1, the only fixed point in [0,1] is s = 1, giving certain extinction.
Question 3 Short Answer
The normalized population W_n = Z_n/μ^n in a supercritical Galton-Watson process is a martingale. Under what condition on the offspring distribution does W_n converge to a non-degenerate (not identically zero) limit?
Think about your answer, then reveal below.
Model answer: W_n converges a.s. to a limit W by the martingale convergence theorem (since W_n ≥ 0 and E[W_n] = 1). The limit W is non-degenerate (P(W > 0) = 1 - q > 0 on the survival event) if and only if E[X log X] < ∞, the Kesten-Stigum condition (also called the X log X condition). When E[X log X] = ∞, the martingale converges to 0 a.s. even on the survival event — the population grows at rate μ^n but with such extreme variability that the normalized version degenerates.
The Kesten-Stigum theorem (1966) is the sharpest result on this question. The condition E[X log X] < ∞ controls the fluctuations: it ensures W_n is uniformly integrable, so L¹ convergence holds and E[W] = 1. When it fails, the offspring distribution has such a heavy right tail that rare individuals with enormous numbers of children dominate the population, and the ratio Z_n/μ^n cannot stabilize. This connects directly to the theory of martingale convergence — non-negative L¹-bounded martingales always converge a.s., but L¹ convergence (non-degeneracy) requires uniform integrability.
Question 4 Multiple Choice
In a Galton-Watson process with Poisson(λ) offspring distribution, the PGF is G(s) = e^{λ(s-1)}. For λ = 2, the extinction probability q satisfies q = e^{2(q-1)}. This equation has a solution q ≈ 0.203, meaning:
AAbout 20.3% of individuals in each generation will die
BThe population goes extinct with probability ≈ 0.203, and survives forever with probability ≈ 0.797
CThe population reaches 0 within 0.203 × n generations on average
DEach individual has a 20.3% chance of producing no offspring
The extinction probability q ≈ 0.203 means that starting from a single individual, the entire population line eventually dies out with probability about 20.3%. With probability about 79.7%, the population grows exponentially (roughly as 2^n since μ = 2). Note P(X=0) = e^{-2} ≈ 0.135, which is smaller than q — extinction can occur even when the founder produces offspring, because all descendant lines must also eventually die. The extinction probability is the fixed point of G, not the zero-offspring probability.
Question 5 Short Answer
A critical (μ = 1) Galton-Watson process with σ² = Var(X) ∈ (0,∞) satisfies P(Z_n > 0) ~ 2/(nσ²) as n → ∞. Interpret this result.
Think about your answer, then reveal below.
Model answer: The survival probability decays as 1/n, meaning extinction is certain but slow — it takes on the order of n generations for the process to die with high probability. Conditioned on survival to generation n, the population size Z_n is of order n (specifically, Z_n/n converges in distribution to an Exponential(2/σ²) random variable, the Yaglom limit). So the critical process shows a tension: it is doomed to extinction, but the rare surviving populations are large (order n), which keeps the unconditional expectation E[Z_n] = 1 despite the vanishing survival probability.
This asymptotic result, due to Kolmogorov (1938) and Yaglom (1947), reveals the delicate behavior at criticality. The product P(Z_n > 0) · E[Z_n | Z_n > 0] ≈ (2/nσ²) · (nσ²/2) = 1, consistent with E[Z_n] = 1 for all n. The critical case is a phase boundary between subcritical (exponential extinction) and supercritical (positive survival probability), and this 1/n decay rate characterizes the phase transition.