What a Population-Martingale Bound Measures
A branching process can survive for a long time while having a small population in a particular generation, or produce a very large population with small probability. The normalized Galton–Watson population martingale gives a direct way to bound the second kind of event. It also makes an important distinction visible: a bound on unusually large population sizes is not, on its own, a bound on the probability of eventual survival.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. Consider a Galton–Watson process \((Z_n)_{n\ge0}\) with state space \(S=\{0,1,2,\ldots\}\) and discrete time \(n\). The initial population is a fixed integer \(Z_0=z_0\ge1\). Each individual independently produces a random number \(\xi\) of children, with \(\mathbb{E}\xi=m\), where \(0<m<\infty\). Write \(\mathcal{F}_n\) for the information about the process and reproduction through generation \(n\). Extinction is absorbing: if \(Z_n=0\), then \(Z_{n+1}=0\).
Given \(\mathcal{F}_n\), the \(Z_n\) individuals reproduce independently, so \(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=mZ_n\). Taking expectations successively gives \(\mathbb{E}Z_n=z_0m^n\). Define the normalized population by \(W_n=Z_n/m^n\). Then \(\mathbb{E}[W_{n+1}\mid\mathcal{F}_n]=W_n\), so \((W_n)_{n\ge0}\) is a nonnegative martingale with \(\mathbb{E}W_n=z_0\) for every \(n\).
Proof. Since \(W_n\ge0\), the pointwise inequality \(W_n\ge a\mathbf{1}_{\{W_n\ge a\}}\) holds. Taking expectations gives \(z_0=\mathbb{E}W_n\ge a\mathbb{P}(W_n\ge a)\), proving the first bound. Since \(W_n=Z_n/m^n\) and \(m^n>0\), the events \(\{W_n\ge r/m^n\}\) and \(\{Z_n\ge r\}\) are equal. Substituting \(a=r/m^n\) in the first bound gives \(\mathbb{P}(Z_n\ge r)\le z_0m^n/r\). \(\square\)
The estimate applies to either real or integer thresholds, but the event must be interpreted as written: it uses \(\ge\). If \(r\) is not an integer, \(\{Z_n\ge r\}\) is the same event as \(\{Z_n\ge\lceil r\rceil\}\), because \(Z_n\) is integer-valued. The bound may be informative when \(r\) is much larger than the expected population, and weak when \(r\) is small.
Worked Example: A Large Generation in a Supercritical Process
Suppose \(Z_0=1\), and each individual has either \(0\) or \(4\) children, each with probability \(\tfrac12\). These probabilities sum to \(\tfrac12+\tfrac12=1\), and \(m=0\cdot\tfrac12+4\cdot\tfrac12=2\). Find an upper bound for the probability that generation \(3\) has at least \(32\) individuals.
The expected population is \(\mathbb{E}Z_3=1\cdot2^3=8\). Applying the generation-size bound with \(r=32\) gives $$ \mathbb{P}(Z_3\ge32)\le\frac{8}{32}=\frac14. $$ In normalized form, \(W_3=Z_3/8\), so the event is \(\{W_3\ge4\}\), and Markov’s inequality gives \(\mathbb{P}(W_3\ge4)\le\mathbb{E}W_3/4=1/4\). The two calculations agree.
Check: Multiplying the normalized threshold \(4\) by \(m^3=8\) gives the original threshold \(32\), and \(z_0/a=1/4\). This is an upper bound; it does not assert that the probability equals \(\tfrac14\).
A Bound for the Entire Path
A generation-by-generation estimate does not immediately control whether the martingale ever becomes large. For a finite time horizon, a stopping-time argument gives a corresponding path bound.
Proof. Define \(\tau=\inf\{k\ge0:W_k\ge a\}\), with \(\inf\varnothing=\infty\), and let \(A=\{\tau\le N\}\). The event \(A\) is exactly \(\{\max_{0\le k\le N}W_k\ge a\}\). For each \(k\le N\), \(\{\tau=k\}\in\mathcal{F}_k\). The martingale property implies \(\mathbb{E}[W_N\mid\mathcal{F}_k]=W_k\); hence $$ \mathbb{E}[W_N\mathbf{1}_{\{\tau=k\}}] =\mathbb{E}[W_k\mathbf{1}_{\{\tau=k\}}] \ge a\,\mathbb{P}(\tau=k). $$ Summing over \(k=0,\ldots,N\) and using \(W_N\ge0\) gives $$ z_0=\mathbb{E}W_N \ge \mathbb{E}[W_N\mathbf{1}_A] \ge a\sum_{k=0}^N\mathbb{P}(\tau=k) =a\,\mathbb{P}(A). $$ Dividing by \(a>0\) proves the result. \(\square\)
The same bound gives a correct infinite-horizon statement, provided the threshold is handled carefully. For \(b>0\), the events \(\{\max_{0\le k\le N}W_k>b\}\) increase with \(N\), and their union is \(\{\sup_{k\ge0}W_k>b\}\). The finite-horizon proof, with a strict crossing of \(b\), bounds each event by \(z_0/b\). Continuity of probability for increasing events therefore gives \(\mathbb{P}(\sup_{k\ge0}W_k>b)\le z_0/b\). In particular, for \(a>0\), the event \(\{\sup_{k\ge0}W_k\ge a\}\) is contained in \(\{\sup_{k\ge0}W_k>b\}\) whenever \(0<b<a\). Letting \(b\) increase to \(a\) yields $$ \mathbb{P}\!\left(\sup_{k\ge0}W_k\ge a\right)\le\frac{z_0}{a}. $$ This threshold argument matters: a sequence can have supremum \(a\) without ever attaining \(a\), so the event \(\{\sup_k W_k\ge a\}\) need not be the union of the finite-horizon events \(\{\max_{k\le N}W_k\ge a\}\).
Worked Example: Bounding a Crossing at Any Generation
Suppose the offspring number is always \(1\), with \(Z_0=2\). The offspring distribution has total probability \(1\), mean \(m=1\), and \(Z_n=2\) for every \(n\). Thus \(W_n=2\) for every \(n\). For the threshold \(a=5\), the infinite-horizon bound gives $$ \mathbb{P}\!\left(\sup_{k\ge0}W_k\ge5\right)\le\frac{2}{5}. $$ In this example the event is impossible, so its actual probability is \(0\), which satisfies the bound.
Check: The normalized value is \(Z_n/1^n=2\), which is strictly below \(5\) at every time. The bound is valid but not sharp. It controls the probability of crossing the chosen level; it does not claim a crossing will occur.
Why Large-Population Bounds Are Not Survival Bounds
Let \(\mathcal{S}=\{Z_n>0\text{ for every }n\ge0\}\) be the event of survival. Since extinction is absorbing, \(\mathcal{S}=\bigcap_{n\ge0}\{Z_n>0\}\). By contrast, the generation-size inequality concerns \(\{Z_n\ge r\}\) for a specified threshold \(r\). Taking \(r=1\) only gives \(\mathbb{P}(Z_n>0)\le z_0m^n\), which may be no better than the trivial bound \(1\). The estimate does not give a useful lower bound on \(\mathbb{P}(Z_n>0)\), nor does a small upper bound on \(\mathbb{P}(Z_n\ge r)\) show that extinction is likely: a population can be positive but below \(r\).
There is also a distinction between a martingale being zero at some time and the process surviving. If \(Z_n=0\), then \(W_k=0\) for all \(k\ge n\). But a bound on large values of \(W_k\) does not decide whether \(W_k\) is zero or merely small. Any inference from the limit of \(W_n\) to survival requires additional results and hypotheses; the inequalities proved here make no such identification.
Worked Example: Critical Branching Can Become Extinct
Start with one individual. Each individual has \(0\) or \(2\) children, each with probability \(\tfrac12\); these probabilities sum to \(1\), and the mean is \(m=1\). Let \(s_n=\mathbb{P}(Z_n>0)\). Then \(s_0=1\). For the population to be positive at generation \(n+1\), the ancestor must have two children and at least one of their descendant processes must be positive at generation \(n\). Independence gives $$ s_{n+1}=\frac12\bigl[1-(1-s_n)^2\bigr] =s_n-\frac{s_n^2}{2}. $$
Thus \(0\le s_{n+1}\le s_n\), so \(s_n\) has a limit \(s\ge0\). Passing to the limit in the displayed polynomial recursion gives \(s=s-s^2/2\), hence \(s=0\). The events \(\{Z_n>0\}\) decrease with \(n\), and their intersection is survival. Continuity of probability for decreasing events gives \(\mathbb{P}(\mathcal{S})=\lim_{n\to\infty}s_n=0\). This example shows how survival requires looking at positivity across all generations, not only at whether the population is unusually large.
Check: At generation \(1\), \(Z_1\) is \(0\) or \(2\), each with probability \(\tfrac12\); therefore \(s_1=\tfrac12\), also given by \(1-\tfrac12(1)^2=\tfrac12\). The generation-size bound at threshold \(2\) gives \(\mathbb{P}(Z_1\ge2)\le\mathbb{E}Z_1/2=1/2\), which is exact here, but it does not by itself establish the eventual-extinction conclusion.
Common Pitfalls
Wrong calculation: “Since \(\mathbb{E}W_n=z_0\), the probability of survival is at most \(z_0/a\).” This applies Markov’s inequality to the wrong event. Markov bounds \(\mathbb{P}(W_n\ge a)\), or a maximal event such as \(\mathbb{P}(\sup_k W_k\ge a)\); survival is the event that every population remains positive, with no fixed large threshold.
Correction: First specify the event. For a single generation, use \(\mathbb{P}(Z_n\ge r)\le z_0m^n/r\). For a crossing at any time, use the maximal bound, with the strict-threshold argument when passing to an infinite horizon. To determine survival or extinction, use an argument that addresses positivity across generations, as in the critical example.
These estimates are useful when a question concerns an exceptionally large generation or a path that crosses a high normalized level. They also provide a quick diagnostic: if the resulting upper bound exceeds \(1\), it remains mathematically valid but gives no information beyond the basic probability bound. Always check the normalization, the event’s threshold, and whether the question is about large values or about being nonzero.
Check Your Understanding
For each question, identify the event being bounded before applying an inequality.
- If \(z_0=2\), \(m=3\), and \(n=2\), what upper bound does Markov’s inequality give for \(\mathbb{P}(Z_2\ge36)\)?
- State the infinite-horizon bound for \(\mathbb{P}(\sup_{k\ge0}W_k\ge a)\), and explain why a threshold \(b<a\) is used in its proof.
- For offspring number always equal to \(1\), initial population \(4\), and \(a=5\), compare the actual probability of \(\sup_kW_k\ge a\) with the maximal bound.
- Why does an upper bound on \(\mathbb{P}(Z_n\ge r)\) not generally give an upper bound on \(\mathbb{P}(Z_n>0)\)?
- In the critical \(0\)-or-\(2\) example, verify the recursion for \(s_{n+1}\) by conditioning on the ancestor’s number of children.