Moments of Total Progeny

Total Progeny and Its Moments

Generation sizes describe how a branching population changes from one generation to the next. Total progeny asks a different question: how many individuals are ever born, counting the initial ancestor? In a subcritical Galton–Watson process the expected population shrinks geometrically by generation, but the total includes contributions from every generation. Its mean and variance therefore depend on both the mean and the variability of the offspring count.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(Z_n)_{n\ge0}$ be a discrete-time Galton–Watson process with state space $S=\{0,1,2,\ldots\}$ and $Z_0=1$. Each individual independently produces a number $\xi$ of offspring, where $\xi$ is nonnegative integer-valued, has mean $m=\mathbb{E}\xi$, and has finite variance $\sigma^2=\operatorname{Var}(\xi)$. The reproduction counts for different individuals are independent and have the same distribution. Let $\mathcal{F}_n$ be the information generated by reproduction through generation $n-1$; then $Z_n$ is $\mathcal{F}_n$-measurable, and the offspring counts of individuals in generation $n$ are independent of $\mathcal{F}_n$.

Definition: The total progeny, including the initial ancestor, is $T=\sum_{n=0}^{\infty}Z_n$, with the sum allowed to equal infinity. The process is subcritical when $m<1$. In this tutorial we assume $m<1$ and $\mathbb{E}[\xi^2]<\infty$.

We first work with a process observed only through a fixed generation. For an integer $h\ge0$, define $T^{(h)}=\sum_{n=0}^{h}Z_n$, and write $a_h=\mathbb{E}[T^{(h)}]$ and $v_h=\operatorname{Var}(T^{(h)})$. Truncation makes the calculations finite and lets us pass to total progeny only after establishing suitable bounds.

Proposition: The truncated moments start at $a_0=1$, $v_0=0$, and for $h\ge1$ satisfy $$ a_h=1+ma_{h-1},\qquad v_h=mv_{h-1}+\sigma^2a_{h-1}^2. $$

Proof. At height zero, the only individual counted is the ancestor, so $T^{(0)}=1$, giving $a_0=1$ and $v_0=0$. At height $h\ge1$, condition on the ancestor's offspring count $\xi$. Each of its $\xi$ children starts an independent process truncated at height $h-1$. If $T^{(h-1)}_1,T^{(h-1)}_2,\ldots$ are independent copies of that truncated total, then $$ T^{(h)}=1+\sum_{r=1}^{\xi}T^{(h-1)}_r. $$ Given $\xi=k$, the sum has mean $ka_{h-1}$ and variance $kv_{h-1}$, by independence. Taking expectations gives $a_h=1+\mathbb{E}[\xi]a_{h-1}=1+ma_{h-1}$. For the variance, decompose into the expected conditional variance plus the variance of the conditional mean: $$ v_h=\mathbb{E}[\xi v_{h-1}]+\operatorname{Var}(1+\xi a_{h-1}) =mv_{h-1}+\sigma^2a_{h-1}^2. $$ The conditional means and variances used here are finite: this follows inductively from the finite second moment of $\xi$ and the finite moments at height $h-1$. This proves both recurrences. $\square$

The Subcritical Moment Formulas

Theorem: If $m<1$ and $\mathbb{E}[\xi^2]<\infty$, then total progeny from one ancestor has finite mean and variance, given by $$ \mathbb{E}[T]=\frac{1}{1-m},\qquad \operatorname{Var}(T)=\frac{\sigma^2}{(1-m)^3}. $$

Proof. Iterating the mean recurrence gives $a_h=1+m+\cdots+m^h$. Since $0\le m<1$, this yields $a_h=(1-m^{h+1})/(1-m)$, so $a_h$ increases to $a=1/(1-m)$ and is bounded above by $a$. The variance recurrence and $v_0=0$ imply $$ 0\le v_h\le mv_{h-1}+\sigma^2a^2. $$ Repeated application gives $v_h\le \sigma^2a^2(1+m+\cdots+m^{h-1})\le \sigma^2a^2/(1-m)$. Thus $v_h$ is bounded.

The truncated totals increase pointwise to $T$. By the monotone convergence theorem, $\mathbb{E}[T]=\lim_{h\to\infty}a_h=a$. In particular $T<\infty$ almost surely: if $\mathbb{P}(T=\infty)>0$, its expectation would be infinite. Also, $\mathbb{E}[(T^{(h)})^2]=v_h+a_h^2$ is bounded uniformly in $h$. Applying the monotone convergence theorem to $(T^{(h)})^2$ shows that $\mathbb{E}[T^2]<\infty$. Consequently $v_h=\mathbb{E}[(T^{(h)})^2]-a_h^2$ converges to $\operatorname{Var}(T)$. Taking limits in the variance recurrence is valid because $a_{h-1}\to a$ and $v_h$ converges. If $v=\operatorname{Var}(T)$, then $$ v=mv+\sigma^2a^2. $$ Since $1-m>0$, this gives $v=\sigma^2a^2/(1-m)=\sigma^2/(1-m)^3$, as claimed. The argument also covers $m=0$: then $\xi=0$ almost surely, $\sigma^2=0$, and $T=1$. $\square$

The factor $(1-m)^{-3}$ shows why total-progeny variance can grow quickly near criticality. A small gap between $m$ and $1$ affects the variability of the entire family tree, not just the size of a typical generation. The theorem requires a finite offspring variance; a finite mean alone does not supply the variance calculation or guarantee a finite second moment for $T$.

Worked Example: A Two-Point Offspring Law

Suppose each individual has either zero or two offspring, with probabilities $2/3$ and $1/3$, respectively. Find the mean and variance of total progeny from one ancestor.

The offspring mean is $$ m=0\cdot\frac23+2\cdot\frac13=\frac23. $$ The second moment is $\mathbb{E}[\xi^2]=0^2\cdot\frac23+2^2\cdot\frac13=\frac43$. Therefore $$ \sigma^2=\mathbb{E}[\xi^2]-m^2=\frac43-\left(\frac23\right)^2=\frac89. $$ Since $m=2/3<1$, the theorem applies. Its mean formula gives $$ \mathbb{E}[T]=\frac{1}{1-\frac23}=3. $$ Its variance formula gives $$ \operatorname{Var}(T)=\frac{\frac89}{(1-\frac23)^3} =\frac{\frac89}{\frac1{27}}=24. $$

Check: The limiting mean equation is $3=1+\frac23\cdot3=3$. The limiting variance equation is $24=\frac23\cdot24+\frac89\cdot3^2=16+8=24$. Both moment equations are satisfied.

Bernoulli and Poisson Offspring

These two common offspring laws make the role of $\sigma^2$ especially transparent. With Bernoulli offspring, each individual has at most one child, so the family is a single line until it ends. With Poisson offspring, an individual can have any nonnegative number of children, and the offspring variance equals the offspring mean.

Worked Example: Bernoulli Offspring

Let $\xi$ have a Bernoulli distribution with parameter $p=1/3$: an individual has one child with probability $1/3$, and no children with probability $2/3$. Find the total-progeny moments and verify them from the distribution of $T$.

Here $m=p=1/3$, and $\sigma^2=p(1-p)=\frac13\cdot\frac23=\frac29$. The formulas give $$ \mathbb{E}[T]=\frac{1}{1-\frac13}=\frac32,\qquad \operatorname{Var}(T)=\frac{\frac29}{(1-\frac13)^3} =\frac{\frac29}{\frac8{27}}=\frac34. $$ For a direct check, $T=k$ exactly when the first $k-1$ individuals each have one child and the $k$-th has none. Independence gives $\mathbb{P}(T=k)=\frac23(\frac13)^{k-1}$, for $k\ge1$.

Using the geometric-series identities $\sum_{k\ge1}kx^{k-1}=1/(1-x)^2$ and $\sum_{k\ge1}k(k-1)x^{k-1}=2x/(1-x)^3$, we obtain $$ \mathbb{E}[T]=\frac23\cdot\frac{1}{(1-\frac13)^2}=\frac32, \qquad \mathbb{E}[T(T-1)]=\frac23\cdot\frac{2\cdot\frac13}{(1-\frac13)^3}=\frac32. $$ Thus $\mathbb{E}[T^2]=\mathbb{E}[T(T-1)]+\mathbb{E}[T]=3$, and $\operatorname{Var}(T)=3-(\frac32)^2=\frac34$, agreeing with the theorem.

Worked Example: Poisson Offspring

Suppose offspring counts are Poisson with parameter $\lambda=1/2$. Find the total-progeny moments and check the limiting moment equations.

For a Poisson variable, $\mathbb{E}[\xi]=\lambda$ and $\operatorname{Var}(\xi)=\lambda$; here both equal $1/2$. Since $m=1/2<1$, the formulas give $$ \mathbb{E}[T]=\frac{1}{1-\frac12}=2,\qquad \operatorname{Var}(T)=\frac{\frac12}{(1-\frac12)^3} =\frac{\frac12}{\frac18}=4. $$ Check: Substitution into the limiting mean equation gives $1+\frac12\cdot2=2$. Substitution into the limiting variance equation gives $\frac12\cdot4+\frac12\cdot2^2=2+2=4$. Both calculations reproduce the stated moments.

A Common Variance Error

A frequent mistake is to use $\mathbb{E}[\xi^2]$ where the variance recurrence requires $\operatorname{Var}(\xi)$. The random number of child families contributes variability through the variance of its conditional mean, which is $\operatorname{Var}(\xi)a_{h-1}^2$, not $\mathbb{E}[\xi^2]a_{h-1}^2$.

For the Poisson example with $\lambda=1/2$, $\mathbb{E}[\xi^2]=\lambda+\lambda^2=\frac12+\frac14=\frac34$, whereas $\operatorname{Var}(\xi)=\frac12$. Incorrectly replacing the variance by the second moment in the limiting recurrence would give $$ v=\frac12v+\frac34\cdot2^2, $$ so $v=6$, not $4$. The correct recurrence gives $$ v=\frac12v+\frac12\cdot2^2, $$ so $v=4$. The difference is essential: $\mathbb{E}[\xi^2]$ includes the square of the offspring mean, while the variance term measures only the variation around that mean.

The stated $T$ includes the initial ancestor. If a convention instead counts only descendants, the resulting count is $T-1$: its mean is $1/(1-m)-1=m/(1-m)$, while its variance remains $\sigma^2/(1-m)^3$, since subtracting a constant does not change variance.

Check Your Understanding

Use the truncation recurrences and the total-progeny formulas to answer these questions.

If $m=1/4$ and $\sigma^2=3/16$, calculate $\mathbb{E}[T]$ and $\operatorname{Var}(T)$.
For Bernoulli offspring with parameter $p$, explain why $T$ has a geometric distribution and state its mean and variance when $0\le p<1$.
In the variance recurrence, which term accounts for variability in the number of children, and why is it not based on $\mathbb{E}[\xi^2]$?
What happens to total progeny when $m=0$? Check that the theorem's formulas give the corresponding moments.
If $\mathbb{E}[\xi^2]$ is infinite, which step in the finite-generation argument no longer provides a finite variance bound?

What You'll Learn