From Offspring Probabilities to Population Moments
A Galton–Watson process turns one offspring distribution into a sequence of random generation sizes. The computations are systematic: first find the offspring mean and variance; then condition on the current population to obtain the next generation’s moments. Normalizing by the expected population produces a martingale, whose variance gives bounds on finite-generation deviations and extinction.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge0}\) be a Galton–Watson process started from one ancestor, so \(Z_0=1\). Each individual independently has a number \(\xi\) of children, with \(\xi\) taking values in the nonnegative integers. Write \(m=\mathbb{E}\xi\) and \(\sigma^2=\operatorname{Var}(\xi)\). Assume \(m<\infty\) and \(\sigma^2<\infty\). Let \(\mathcal{F}_n\) record the population and reproduction information through generation \(n\). Given \(\mathcal{F}_n\), the \(Z_n\) individuals reproduce independently according to the same offspring distribution.
The conditional calculation behind the moment formulas is simple but important. Given \(\mathcal{F}_n\), the next generation is a sum of \(Z_n\) independent copies of \(\xi\). Therefore its conditional mean is \(mZ_n\) and its conditional variance is \(\sigma^2 Z_n\). These identities apply also when \(Z_n=0\), since an empty sum is zero.
Proof. Conditional expectation gives \(\mathbb{E}Z_{n+1}=m\mathbb{E}Z_n\); starting from \(Z_0=1\), induction yields \(\mathbb{E}Z_n=m^n\). Write \(v_n=\operatorname{Var}(Z_n)\). The law of total variance gives $$ v_{n+1}=\mathbb{E}[\operatorname{Var}(Z_{n+1}\mid\mathcal{F}_n)] +\operatorname{Var}(\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]) =\sigma^2m^n+m^2v_n, $$ with \(v_0=0\). Iterating this recurrence gives \(v_n=\sigma^2\sum_{k=0}^{n-1}m^{n-1-k}m^k\) only if the terms are indexed incorrectly: instead, expanding the recurrence yields $$ v_n=\sigma^2\sum_{k=0}^{n-1}m^{2(n-1-k)}m^k =\sigma^2m^{n-1}\sum_{r=0}^{n-1}m^r. $$ For \(m\ne1\), the geometric sum is \((m^n-1)/(m-1)\); for \(m=1\), it equals \(n\). Since \(W_n=Z_n/m^n\), its variance is \(v_n/m^{2n}\). For \(m\ne1\), this simplifies to \(\sigma^2[m(m-1)]^{-1}(1-m^{-n})\); for \(m=1\), division changes nothing. \(\square\)
The same conditional calculation verifies the martingale property: \(\mathbb{E}[W_{n+1}\mid\mathcal{F}_n]=mZ_n/m^{n+1}=W_n\). In particular, \(\mathbb{E}W_n=1\). A finite-generation probability bound follows by applying Chebyshev’s inequality to this mean-one random variable. For \(\varepsilon>0\), $$ \mathbb{P}(|W_n-1|\ge\varepsilon)\le \frac{\operatorname{Var}(W_n)}{\varepsilon^2}. $$ This is a bound at a specified generation, not a claim about an infinite-time event.
Worked Calculations
Worked Example: A Supercritical Binomial Offspring Law
Suppose each individual has a binomial \(\operatorname{Binomial}(2,\tfrac34)\) number of children. Compute the offspring moments, the first two generation-size moments, the normalized variance at generation two, and a bound on extinction by that generation.
The offspring probabilities are \(\mathbb{P}(\xi=0)=\tfrac1{16}\), \(\mathbb{P}(\xi=1)=\tfrac6{16}=\tfrac38\), and \(\mathbb{P}(\xi=2)=\tfrac9{16}\). Thus $$ m=0\cdot\tfrac1{16}+1\cdot\tfrac38+2\cdot\tfrac9{16} =\tfrac38+\tfrac{18}{16}=\tfrac32, $$ and $$ \mathbb{E}[\xi^2]=0+\tfrac38+4\cdot\tfrac9{16} =\tfrac38+\tfrac94=\tfrac{21}{8}. $$ Therefore \(\sigma^2=\tfrac{21}{8}-(\tfrac32)^2=\tfrac{21}{8}-\tfrac{18}{8}=\tfrac38\).
The expectation formula gives \(\mathbb{E}Z_1=\tfrac32\) and \(\mathbb{E}Z_2=(\tfrac32)^2=\tfrac94\). At generation one, \(Z_1=\xi\), so \(\operatorname{Var}(Z_1)=\tfrac38\). At generation two, the variance recurrence gives $$ \operatorname{Var}(Z_2)=\sigma^2\mathbb{E}Z_1+m^2\operatorname{Var}(Z_1) =\tfrac38\cdot\tfrac32+\tfrac94\cdot\tfrac38 =\tfrac9{16}+\tfrac{27}{32}=\tfrac{45}{32}. $$ Consequently \(\operatorname{Var}(W_2)=\tfrac{45}{32}/(\tfrac94)^2=\tfrac{45}{32}\cdot\tfrac{16}{81}=\tfrac5{18}\). Substitution in the proposition gives the same value: $$ \frac{\tfrac38}{\tfrac32(\tfrac32-1)}(1-(\tfrac32)^{-2}) =\tfrac12(1-\tfrac49)=\tfrac5{18}. $$
Extinction by generation two is the event \(Z_2=0\). It occurs if \(Z_1=0\), or if \(Z_1=1\) and that individual has no children, or if \(Z_1=2\) and both individuals have no children. Hence $$ \mathbb{P}(Z_2=0)=\tfrac1{16}+\tfrac38\cdot\tfrac1{16}+\tfrac9{16}\cdot(\tfrac1{16})^2 =\tfrac{256+96+9}{4096}=\tfrac{361}{4096}. $$ Since \(Z_2=0\) implies \(W_2=0\), it implies \(|W_2-1|\ge1\). Chebyshev’s inequality therefore gives \(\mathbb{P}(Z_2=0)\le\operatorname{Var}(W_2)=\tfrac5{18}\). The exact probability satisfies this bound: \(361\cdot18=6498<20480=5\cdot4096\).
Worked Example: A Critical Process and a Loose Bound
Let \(\mathbb{P}(\xi=0)=\mathbb{P}(\xi=2)=\tfrac12\). Find the first two generation expectations and variances, then assess the Chebyshev bound at generation two.
Here \(m=0\cdot\tfrac12+2\cdot\tfrac12=1\), while \(\mathbb{E}[\xi^2]=0+4\cdot\tfrac12=2\), so \(\sigma^2=2-1^2=1\). The critical-case formulas give \(\mathbb{E}Z_1=\mathbb{E}Z_2=1\), \(\operatorname{Var}(Z_1)=1\), and \(\operatorname{Var}(Z_2)=2\). Because \(m=1\), \(W_2=Z_2\). Thus for any \(\varepsilon>0\), $$ \mathbb{P}(|W_2-1|\ge\varepsilon)\le\frac{2}{\varepsilon^2}. $$ For example, at \(\varepsilon=2\) the bound is \(\tfrac12\).
For comparison, the exact probability of extinction by generation two can be computed recursively. Extinction by generation one has probability \(\tfrac12\). If there are two individuals in generation one, both lines must be extinct by the next generation; their independent probabilities are each \(\tfrac12\). Thus \(\mathbb{P}(Z_2=0)=\tfrac12+\tfrac12(\tfrac12)^2=\tfrac58\). The Chebyshev bound at threshold \(1\) is \(2\), which is valid but gives no information because a probability is at most \(1\). This example shows that a correct finite-generation bound need not be numerically useful.
Worked Example: A Subcritical Process with at Most One Child
Suppose \(\mathbb{P}(\xi=0)=\mathbb{P}(\xi=1)=\tfrac12\). Compute the probability the population survives through generation \(n\), and compare it with a general expectation bound.
The offspring mean is \(m=0+\tfrac12=\tfrac12\); the second moment is \(\mathbb{E}[\xi^2]=\tfrac12\), so \(\sigma^2=\tfrac12-(\tfrac12)^2=\tfrac14\). Since no individual can have more than one child, \(Z_n\) is either zero or one. It equals one exactly when each of the first \(n\) individuals in the lineage has one child. Independence gives \(\mathbb{P}(Z_n>0)=(\tfrac12)^n\). This agrees with \(\mathbb{E}Z_n=m^n=(\tfrac12)^n\).
For any nonnegative random variable \(Y\) and \(r>0\), Markov’s inequality says \(\mathbb{P}(Y\ge r)\le\mathbb{E}Y/r\). Taking \(Y=Z_n\) and \(r=1\), and using that \(Z_n\) is integer-valued, gives $$ \mathbb{P}(Z_n>0)=\mathbb{P}(Z_n\ge1)\le\mathbb{E}Z_n=2^{-n}. $$ In this example the bound is exact. In other subcritical processes it may be strict, because a surviving population can contain more than one individual.
Using the Bounds Carefully
For a one-ancestor process, the expectation formula gives a second useful finite-generation estimate. Since \(Z_n\ge0\), for every \(r>0\), $$ \mathbb{P}(Z_n\ge r)\le\frac{m^n}{r}. $$ In particular, because \(Z_n\) is a nonnegative integer, \(\mathbb{P}(Z_n>0)=\mathbb{P}(Z_n\ge1)\le m^n\). When \(m<1\), this bound tends to zero with \(n\); when \(m>1\), it may exceed one and then is uninformative. The inequality remains correct in both cases.
When \(m>1\), the normalized variance formula also gives the explicit extinction bound $$ \mathbb{P}(Z_n=0)\le \operatorname{Var}(W_n) =\frac{\sigma^2}{m(m-1)}(1-m^{-n}). $$ Indeed, \(W_n=0\) on extinction by generation \(n\), and \(|W_n-1|=1\) there, so the event is contained in \(\{|W_n-1|\ge1\}\); Chebyshev then applies. The bound requires a finite offspring variance. It is an upper bound for extinction by a particular generation, not an equality or an exact extinction probability.
Common mistake. The variance is not the second moment. In the binomial example, using \(\operatorname{Var}(\xi)=\mathbb{E}[\xi^2]=\tfrac{21}{8}\) would incorrectly inflate every population-variance calculation. The correction is to subtract the squared mean: \(\tfrac{21}{8}-(\tfrac32)^2=\tfrac38\). Another frequent error is to use the supercritical variance expression at \(m=1\), where its denominator vanishes. At criticality, the geometric sum has \(n\) terms each equal to one, giving \(\operatorname{Var}(Z_n)=n\sigma^2\).
These calculations separate three ingredients that are easy to conflate: the offspring law determines \(m\) and \(\sigma^2\); conditioning determines the generation-size recurrences; and probability inequalities turn the resulting moments into bounds. Checking each ingredient before combining them makes the bounds easier to interpret and less prone to algebraic errors.
Check Your Understanding
For each question, state any offspring moments you use and keep strict and non-strict event thresholds distinct.
- If \(\xi\) is Bernoulli with success probability \(\tfrac14\), compute \(m\), \(\sigma^2\), and \(\mathbb{E}Z_3\).
- For a process with \(m=1\) and offspring variance \(\sigma^2=2\), find \(\operatorname{Var}(W_4)\).
- Explain why the event \(Z_n=0\) is contained in \(\{|W_n-1|\ge1\}\) when \(m>0\) and \(Z_0=1\).
- If \(m=\tfrac12\), what upper bound does Markov’s inequality give for \(\mathbb{P}(Z_4>0)\)?
- For the offspring law taking values \(0\) and \(2\) with probability \(\tfrac12\) each, compute \(\mathbb{E}[\xi^2]\) and \(\operatorname{Var}(\xi)\). Why are these quantities different?