Critical Total Progeny and Power-Law Tails

Critical Total Progeny Has a Heavy Tail

In a critical branching process, each individual has one child on average. That balance does not mean that population sizes remain near one: the process eventually dies out, but occasionally it produces a large family tree. When the offspring variance is finite and positive, the probability that the entire tree contains exactly $n$ individuals decays at a power-law rate proportional to $n^{-3/2}$. The lattice span of the offspring distribution determines the leading constant and which tree sizes are possible.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(Z_n)_{n\ge0}$ be a Galton–Watson branching process, in discrete generation time $n$, with countable state space $S=\{0,1,2,\ldots\}$, initial population $Z_0=1$, and offspring variable $\xi$. Each individual independently produces a nonnegative integer number of children distributed as $\xi$. Assume throughout that $\mathbb{E}\xi=1$ and $0<\operatorname{Var}(\xi)=\sigma^2<\infty$. These assumptions exclude the deterministic one-child process.

Definition: The total progeny is $T=\sum_{r=0}^{\infty}Z_r$, allowing $T=\infty$. The lattice span $d$ of the offspring distribution is the greatest positive integer dividing every difference $a-b$ between support points $a,b$ of $\xi$. Thus, for any fixed support point $a$, every possible offspring value lies in $a+d\mathbb{Z}$, and $d$ is the largest such span.

Write $\xi_1,\ldots,\xi_n$ for independent copies of $\xi$, and $S_n=\xi_1+\cdots+\xi_n$. The earlier material on Dwass’s formula gives, for every integer $n\ge1$,

\mathbb{P}(T=n)=\frac{1}{n}\mathbb{P}(S_n=n-1).

This identity reduces a branching-tree probability to a sum of independent offspring counts. Here the sum has mean $n$, while the required value $n-1$ is one below its mean. The local central limit theorem evaluates the probability of this single lattice value.

The Lattice Local Limit Step

Standard result (lattice local central limit theorem): If independent identically distributed integer-valued variables have mean $\mu$, finite positive variance $\sigma^2$, and lattice span $d$, then, uniformly over lattice points $x\in na+d\mathbb{Z}$, $$ \frac{\sigma\sqrt{n}}{d}\mathbb{P}(S_n=x) = \frac{1}{\sqrt{2\pi}} \exp\!\left(-\frac{(x-n\mu)^2}{2n\sigma^2}\right)+o(1), $$ where $a$ is any support point. For points outside $na+d\mathbb{Z}$, the probability is zero. This local central limit theorem is a standard result; its proof is beyond this tutorial.

For the branching process, take $\mu=1$ and $x=n-1$. This value is in the required lattice precisely when $n-1\equiv na\pmod d$. Call such an $n$ lattice-admissible. At an admissible $n$, the centered, scaled target is $-1/(\sigma\sqrt{n})$, which tends to zero. The local central limit theorem therefore gives

\mathbb{P}(S_n=n-1) = \frac{d}{\sigma\sqrt{2\pi n}}\bigl(1+o(1)\bigr).

The theorem also implies that this probability is positive for all sufficiently large lattice-admissible $n$. Congruence is necessary for attainability at every $n$, but it need not be sufficient at small $n$.

Theorem (Critical total-progeny tail): Under the assumptions above, along lattice-admissible integers $n$ tending to infinity, $$ \mathbb{P}(T=n)\sim \frac{d}{\sigma\sqrt{2\pi}}\,n^{-3/2}. $$ For every lattice-inadmissible $n$, $\mathbb{P}(T=n)=0$.

Proof. Dwass’s formula applies to the total progeny of a Galton–Watson process started from one individual, so $\mathbb{P}(T=n)=n^{-1}\mathbb{P}(S_n=n-1)$. At an admissible $n$, the lattice local central limit theorem applies because the offspring variables are identically distributed, have mean $1$, finite positive variance $\sigma^2$, and span $d$. The target $n-1$ differs from the mean $n$ by $-1$, so its standardized displacement tends to zero. Hence $\mathbb{P}(S_n=n-1)=d(\sigma\sqrt{2\pi n})^{-1}(1+o(1))$. Dividing by $n$ proves the stated asymptotic. If $n$ is inadmissible, $S_n$ lies in $na+d\mathbb{Z}$ and $n-1$ does not, so the sum probability and, by Dwass’s formula, the total-progeny probability are both zero. $\square$

Worked Examples

Worked Example: Binary Offspring and Odd Total Sizes

Suppose $\mathbb{P}(\xi=0)=\mathbb{P}(\xi=2)=\tfrac12$. The mean is $0\cdot\tfrac12+2\cdot\tfrac12=1$. Also, $\mathbb{E}[\xi^2]=0^2\cdot\tfrac12+2^2\cdot\tfrac12=2$, so $\sigma^2=2-1^2=1$. The support differences are $2$ and $-2$, giving span $d=2$.

For a support point $a=0$, the sum $S_n$ is always even. The target $n-1$ is even exactly when $n$ is odd. Thus even total sizes have probability zero. At $n=3$, the target sum is $2$. Exactly one of the three offspring values must equal $2$, so

\mathbb{P}(S_3=2)=\binom{3}{1}\left(\frac12\right)^3=\frac{3}{8}, \qquad \mathbb{P}(T=3)=\frac{1}{3}\cdot\frac{3}{8}=\frac18.

Check: A tree of total size $3$ must have a root with two children, each of whom has no children. Its probability is $\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18$, agreeing with Dwass’s formula. Along odd $n$, the asymptotic constant is $d/(\sigma\sqrt{2\pi})=2/\sqrt{2\pi}$. The factor $2$ matters: the asymptotic is along the odd integers, while the probabilities at even integers are exactly zero.

Worked Example: Poisson Offspring

Let $\xi$ have a Poisson distribution with parameter $1$: $\mathbb{P}(\xi=k)=e^{-1}/k!$ for $k=0,1,\ldots$. Its mean and variance are both $1$, and its support has span $d=1$. The sum $S_n$ is Poisson with parameter $n$. Dwass’s formula gives, for every $n\ge1$,

\mathbb{P}(T=n) =\frac{1}{n}\left(e^{-n}\frac{n^{n-1}}{(n-1)!}\right) =e^{-n}\frac{n^{n-1}}{n!}.

For example, at $n=2$, this is $e^{-2}\cdot 2^1/2!=e^{-2}$. Directly, a tree of size $2$ requires the root to have one child and that child to have none, with probability $(e^{-1})(e^{-1})=e^{-2}$.

Stirling’s formula, $n!\sim\sqrt{2\pi n}(n/e)^n$, yields

e^{-n}\frac{n^{n-1}}{n!} \sim e^{-n}\frac{n^{n-1}}{\sqrt{2\pi n}\,n^ne^{-n}} = \frac{1}{\sqrt{2\pi}}n^{-3/2}.

Check: The general constant is $d/(\sigma\sqrt{2\pi})=1/\sqrt{2\pi}$, exactly the constant obtained from the explicit formula.

Worked Example: Congruence Need Not Ensure Small-Size Attainability

Consider the offspring law $\mathbb{P}(\xi=0)=0.56=\tfrac{14}{25}$, $\mathbb{P}(\xi=2)=0.40=\tfrac{10}{25}$, and $\mathbb{P}(\xi=5)=0.04=\tfrac{1}{25}$. These probabilities sum to $\tfrac{14+10+1}{25}=1$, and the mean is $2\cdot\tfrac{10}{25}+5\cdot\tfrac{1}{25}=1$. The second moment is $4\cdot\tfrac{10}{25}+25\cdot\tfrac{1}{25}=\tfrac{65}{25}=\tfrac{13}{5}$, so the variance is $\tfrac{13}{5}-1=\tfrac85>0$. The support differences have greatest common divisor $1$, so $d=1$.

Every integer $n$ satisfies the lattice congruence when $d=1$. But for $n=2$, the Dwass target is $S_2=1$, and a sum of two values from $\{0,2,5\}$ cannot equal $1$. Therefore $\mathbb{P}(T=2)=0$. Equivalently, a tree of size $2$ would require the root to have exactly one child, which this offspring law does not allow.

Check: Dwass’s formula gives $\mathbb{P}(T=2)=\tfrac12\mathbb{P}(S_2=1)=0$. The local central limit theorem’s positive approximation concerns sufficiently large admissible $n$; it does not assert positivity at every small $n$.

Extinction Is Certain, but the Mean Total Progeny Is Infinite

The tail asymptotic describes the size distribution, but almost-sure extinction is a separate fact. Let $G(s)=\mathbb{E}[s^\xi]$, $0\le s\le1$, be the offspring probability generating function. Because the offspring mean is $1$ and the variance is positive, the offspring law is not concentrated at $1$. It must have positive probability of zero offspring and positive probability of at least two offspring: if $\xi\ge1$ almost surely, mean $1$ would force $\xi=1$ almost surely; and if $\xi\le1$ almost surely, mean $1$ would do the same.

Proposition: For a Galton–Watson process started from one individual with mean-one, nondegenerate offspring, extinction occurs almost surely. If, in addition, the offspring mean is finite, then $\mathbb{E}T=\infty$.

Proof. Let $q_n=\mathbb{P}(Z_n=0)$. Since zero is absorbing, this is also the probability of extinction by generation $n$. Independence of offspring gives $q_0=0$ and $q_{n+1}=G(q_n)$. The sequence $q_n$ is nondecreasing and bounded above by $1$, so it has a limit $q$. Continuity of $G$ on $[0,1]$ gives $q=G(q)$. Also $G(1)=1$ and $G'(1)=\mathbb{E}\xi=1$. Since $\mathbb{P}(\xi\ge2)>0$, $G$ is strictly convex on $(0,1)$; its tangent at $1$ is the line $s\mapsto s$. Thus $G(s)>s$ for $0\le s<1$, so $1$ is its only fixed point in $[0,1]$. Consequently $q=1$, proving extinction almost surely. For the expectation, let $\mathcal{F}_n$ contain the population and reproduction information through generation $n$. Conditional on $\mathcal{F}_n$, the expected next generation is $\mathbb{E}[Z_{n+1}\mid\mathcal{F}_n]=Z_n\mathbb{E}\xi=Z_n$. Induction from $Z_0=1$ gives $\mathbb{E}Z_n=1$ for every $n$. By monotone convergence applied to the partial sums of $T$, $\mathbb{E}T=\sum_{n\ge0}\mathbb{E}Z_n=\sum_{n\ge0}1=\infty$. $\square$

There is no contradiction between these conclusions. Almost-sure extinction says that the realized tree has finitely many individuals with probability one; it does not bound the average size of that tree. Rare, very large trees make the expectation infinite. The tail gives the same warning: along the admissible sizes, $n\mathbb{P}(T=n)$ is asymptotic to a positive constant times $n^{-1/2}$, whose sum diverges. The generation-by-generation expectation argument proves infinite mean without requiring a separate summation over the lattice-admissible sizes.

Common pitfalls. First, omitting the lattice-span factor gives the wrong constant whenever $d>1$. In the binary example, a span-one calculation would miss the factor $2$, and it would fail to record that even sizes have probability zero. Second, lattice congruence is not a guarantee of attainability at every finite $n$; it is the local limit theorem that gives positivity for sufficiently large admissible $n$. Third, almost-sure extinction does not imply finite expected total progeny. In the critical finite-variance case, extinction is certain while $\mathbb{E}T$ is infinite.

Check Your Understanding

Use the Dwass identity and check the lattice condition before applying the asymptotic.

For offspring values $0$ and $2$, which total-progeny sizes are lattice-admissible, and what is the asymptotic constant along those sizes?
For a critical offspring law with span $3$ and variance $4$, write the leading constant in terms of $\pi$ along admissible sizes.
In the Poisson-offspring example, compute $\mathbb{P}(T=1)$ directly from Dwass’s formula.
Why does the lattice local central limit theorem not imply that every congruence-admissible small $n$ has positive total-progeny probability?
Give the generation-by-generation calculation that proves $\mathbb{E}T=\infty$, and explain why it does not contradict almost-sure extinction.

Critical Total Progeny and Power-Law Tails

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