From One Generation to the Next
In a Galton–Watson process, each individual produces a random number of offspring, independently of the other individuals and of earlier generations. The offspring mean determines how the expected population changes from one generation to the next. Repeatedly applying that one-generation relation gives a simple formula for every generation. Its interpretation depends on whether the mean number of offspring is below, equal to, or above one.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((Z_n)_{n\ge0}\) be the generation sizes, with state space \(S=\{0,1,2,\ldots\}\) and discrete time \(n\), the generation number. Suppose \(Z_0=z\) is a fixed finite nonnegative integer. Each individual has a number of offspring with the same distribution as a nonnegative-integer-valued random variable \(\xi\), and all individuals’ offspring counts are independent. Assume the offspring counts are also independent of the population history up to the generation in which they reproduce. Write \(m=\mathbb{E}[\xi]\), and assume \(m<\infty\). The population evolves by summing the offspring counts of the individuals in the current generation.
Let \(\mathcal{F}_n\) contain the information about the family tree through generation \(n\), including \(Z_n\) but not the offspring counts that will produce generation \(n+1\). Conditional on \(\mathcal{F}_n\), there are \(Z_n\) individuals, each with expected offspring count \(m\). This gives the basic conditional relation.
Proof. Conditional on \(\mathcal{F}_n\), the value \(Z_n\) is known. If \(Z_n=k\), the next generation is the sum of the offspring counts of those \(k\) individuals. By independence from \(\mathcal{F}_n\), each such count has conditional expectation \(m\); by additivity of expectation, the conditional expectation of the sum is \(km\). This includes \(k=0\), when the sum is empty and equals zero. Thus the conditional expectation is \(mZ_n\). Taking expectations and applying the tower property gives \(\mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]\). Since \(\mathbb{E}[Z_0]=z<\infty\) and \(m<\infty\), induction in this relation shows every \(\mathbb{E}[Z_n]\) is finite. \(\square\)
The Expected Population at Generation \(n\)
The one-generation relation can be applied repeatedly. The factor \(m\) appears once for each passage from one generation to the next, so after \(n\) passages it is raised to the \(n\)th power. The result is about the expected size of the population, not a claim that each realized family tree grows by that factor.
Proof. First take \(Z_0=z\) fixed. At \(n=0\), \(\mathbb{E}[Z_0]=z=zm^0\). Suppose \(\mathbb{E}[Z_n]=zm^n\). The one-generation mean relation then gives $$ \mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]=m(zm^n)=zm^{n+1}. $$ This induction step also holds when \(m=0\): for \(n\ge0\), the recurrence gives \(\mathbb{E}[Z_{n+1}]=0\), matching \(zm^{n+1}=0\). Thus the formula holds for all \(n\ge0\). For random \(Z_0\), the same conditional-mean argument applies, and induction starts from \(\mathbb{E}[Z_0]\) rather than \(z\). At each step, \(\mathbb{E}[Z_{n+1}]=m\mathbb{E}[Z_n]\), which yields the stated formula, with the same convention at \(n=0\). \(\square\)
The value \(m\) separates three regimes for the expected sizes. These labels describe the offspring mean: subcritical means \(m<1\), critical means \(m=1\), and supercritical means \(m>1\). Since offspring counts are nonnegative, \(m\ge0\). When the initial size is positive, the formula shows that expected sizes decrease geometrically in the subcritical case, stay constant in the critical case, and increase geometrically in the supercritical case. If \(Z_0=0\), the population remains zero and the formula gives zero in every regime.
Worked Examples Across the Three Regimes
Worked Example: Subcritical Binomial Offspring
Suppose each individual has a \(\operatorname{Binomial}(2,\tfrac13)\) number of offspring, and \(Z_0=3\). Find the expected population through generation \(3\).
The offspring probabilities are \(\mathbb{P}(\xi=0)=\tfrac49\), \(\mathbb{P}(\xi=1)=\tfrac49\), and \(\mathbb{P}(\xi=2)=\tfrac19\). Therefore \(m=0\cdot\tfrac49+1\cdot\tfrac49+2\cdot\tfrac19=\tfrac69=\tfrac23\), which is less than \(1\). Applying the formula generation by generation gives \(\mathbb{E}[Z_0]=3\), \(\mathbb{E}[Z_1]=3\cdot\tfrac23=2\), \(\mathbb{E}[Z_2]=2\cdot\tfrac23=\tfrac43\), and \(\mathbb{E}[Z_3]=\tfrac43\cdot\tfrac23=\tfrac89\).
Check: The direct expression at generation \(3\) is \(3(\tfrac23)^3=3\cdot\tfrac8{27}=\tfrac89\), agreeing with the repeated recursion. The expectation need not be an integer: it is an average over possible integer-valued population sizes.
Worked Example: Critical Offspring with a Nonconstant Population
Suppose each individual has zero offspring with probability \(\tfrac12\) and two offspring with probability \(\tfrac12\), independently, and \(Z_0=2\). Find the expected population at every generation, and examine the first generation’s distribution.
The offspring mean is \(m=0\cdot\tfrac12+2\cdot\tfrac12=1\), so this is the critical case. The expected-size formula gives \(\mathbb{E}[Z_n]=2\cdot1^n=2\) for every \(n\ge0\). In generation \(1\), both initial individuals must be included in the calculation. The probability that both have zero offspring is \(\tfrac12\cdot\tfrac12=\tfrac14\), giving \(Z_1=0\). The probability that exactly one has two offspring is \(2\cdot\tfrac12\cdot\tfrac12=\tfrac12\), giving \(Z_1=2\). The probability that both have two offspring is \(\tfrac12\cdot\tfrac12=\tfrac14\), giving \(Z_1=4\). Hence \(\mathbb{E}[Z_1]=0\cdot\tfrac14+2\cdot\tfrac12+4\cdot\tfrac14=0+1+1=2\), as the formula predicts.
Check: Applying the mean recursion to generation \(1\) gives \(\mathbb{E}[Z_2]=1\cdot\mathbb{E}[Z_1]=1\cdot2=2\). Thus the expectation is constant even though \(Z_1\) is not constant; the mean alone does not describe the full population-size distribution.
Worked Example: Supercritical Poisson Offspring
Suppose each individual has a Poisson number of offspring with parameter \(\tfrac32\), and \(Z_0=2\). Calculate the expected population through generation \(4\).
For a Poisson random variable with parameter \(\tfrac32\), its mean is $$ m=e^{-3/2}\sum_{k=0}^{\infty}k\frac{(3/2)^k}{k!} =\frac32 e^{-3/2}\sum_{k=1}^{\infty}\frac{(3/2)^{k-1}}{(k-1)!} =\frac32 e^{-3/2}e^{3/2} =\frac32. $$ Thus \(m>1\). Starting from \(\mathbb{E}[Z_0]=2\), the recursion gives \(\mathbb{E}[Z_1]=2\cdot\tfrac32=3\), \(\mathbb{E}[Z_2]=3\cdot\tfrac32=\tfrac92\), \(\mathbb{E}[Z_3]=\tfrac92\cdot\tfrac32=\tfrac{27}4\), and \(\mathbb{E}[Z_4]=\tfrac{27}4\cdot\tfrac32=\tfrac{81}8\).
Check: The closed form gives \(\mathbb{E}[Z_4]=2(\tfrac32)^4=2\cdot\tfrac{81}{16}=\tfrac{81}8\), the same value. The calculation establishes growth of the expectation across generations; it does not establish that every realized population grows.
What the Mean Does—and Does Not—Tell You
The three regimes give a useful first summary of a branching model. In the subcritical example, the expected population shrinks by the factor \(\tfrac23\) per generation. In the critical example, the expected size remains at its initial value. In the supercritical example, the expected size is multiplied by \(\tfrac32\) per generation. These statements follow directly from the mean recursion and require only a finite offspring mean.
Expected size is not the same quantity as the probability that the population is nonempty. For example, in the critical example, generation \(1\) has expected size \(2\), but the probability it is nonempty is \(\tfrac12+\tfrac14=\tfrac34\), obtained by adding the probabilities of \(Z_1=2\) and \(Z_1=4\). Knowing one of these quantities does not in general determine the other. Likewise, a growing expected size in a supercritical model is a statement about an average over possible outcomes, not a guarantee of growth along every sample path. To answer questions about extinction or survival probabilities requires information beyond the mean recursion.
Common mistake: adding the mean instead of multiplying the current expectation. In the Poisson example, a mistaken calculation might give \(\mathbb{E}[Z_2]=2+\tfrac32=\tfrac72\). This treats the offspring mean as an increment that is added once, regardless of how many individuals reproduce. The correct calculation uses the conditional relation: \(\mathbb{E}[Z_1]=2\cdot\tfrac32=3\), then \(\mathbb{E}[Z_2]=3\cdot\tfrac32=\tfrac92\). In general, each generation’s expected size is the preceding expected size multiplied by \(m\), producing the power \(m^n\), not a linear accumulation of \(m\).
When using the formula, identify the initial population and calculate the offspring mean before classifying the regime. Check that the offspring mean is finite, since the recursion and the finite expectation formula rely on that assumption. Then apply the factor once per generation, keeping the generation-zero value separate: \(Z_0\) is already given and corresponds to \(n=0\), not to a reproduction step.
Check Your Understanding
For each question, use the offspring mean and the generation-by-generation recursion; distinguish an expected size from a probability.
- If \(Z_0=4\) and the offspring mean is \(\tfrac34\), find \(\mathbb{E}[Z_1]\), \(\mathbb{E}[Z_2]\), and \(\mathbb{E}[Z_3]\). Identify the regime.
- An offspring law assigns probability \(\tfrac13\) to zero offspring and probability \(\tfrac23\) to two offspring. Compute its mean and classify the regime.
- If \(Z_0\) is random with finite mean \(5\), and \(m=1\), what is \(\mathbb{E}[Z_n]\) for \(n\ge0\)? State the independence assumption needed for the formula.
- Explain why a constant expected population size does not require the population size itself to be constant.
- For a process with \(Z_0=0\), explain what the expected-size formula gives, including at generation \(0\).