A Complete Analysis of a Three-Point Offspring Law

One Offspring Law, Four Calculations

A Galton–Watson process with offspring counts restricted to zero, one, and three is simple enough to analyze exactly, yet it displays the main distinctions between generation-by-generation behavior and eventual extinction. Its generating function is cubic, finite-generation extinction probabilities come from repeated substitution, and the ultimate extinction probability can be found by factoring a cubic fixed-point equation.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $Z_n$ be the number of individuals in generation $n$, with $Z_0=1$. Each individual independently has $\xi$ children, where $\mathbb{P}(\xi=0)=a$, $\mathbb{P}(\xi=1)=b$, and $\mathbb{P}(\xi=3)=c$. Throughout, $a,b,c\ge0$ and $a+b+c=1$; allowing probabilities to be zero makes the boundary cases explicit. Given the population through generation $n$, the offspring counts of the $Z_n$ individuals are independent copies of $\xi$, and $Z_{n+1}$ is their sum.

Definition: The offspring probability generating function is $f(s)=\mathbb{E}[s^\xi]$, for $0\le s\le1$. For this law, $f(s)=a+bs+cs^3$. Define $q_n=\mathbb{P}(Z_n=0)$, the probability the population has become extinct by generation $n$. The ultimate extinction probability is $q=\mathbb{P}(Z_n=0\text{ for some }n\ge0)$.

Because zero population remains zero, the events $\{Z_n=0\}$ increase with $n$. The calculation of $q_n$ and the calculation of $q$ are related, but they are not interchangeable: $q_n$ is obtained after finitely many iterations, while $q$ is a limit.

Generation Means and Variances

The offspring mean and variance are denoted by $m=\mathbb{E}\xi$ and $\sigma^2=\operatorname{Var}(\xi)$. Directly from the three possible offspring counts,

m=b+3c,\qquad \sigma^2=b+9c-(b+3c)^2.

The generation moments follow from conditioning on the current population. The next result gives the formulas, including the case $m=1$.

Theorem: For every integer $n\ge0$, $\mathbb{E}Z_n=m^n$. For $n\ge1$, $$ \operatorname{Var}(Z_n)= \begin{cases} \sigma^2 n,&m=1,\\[2mm] \sigma^2 m^{n-1}\dfrac{m^n-1}{m-1},&m\ne1. \end{cases} $$

Proof. Given $Z_n$, the next generation is a sum of $Z_n$ independent offspring counts. Thus $\mathbb{E}[Z_{n+1}\mid Z_n]=mZ_n$ and $\operatorname{Var}(Z_{n+1}\mid Z_n)=\sigma^2Z_n$. Taking expectations gives $\mathbb{E}Z_{n+1}=m\mathbb{E}Z_n$, and $\mathbb{E}Z_0=1$, so induction yields $\mathbb{E}Z_n=m^n$. The law of total variance gives $\operatorname{Var}(Z_{n+1})=m^2\operatorname{Var}(Z_n)+\sigma^2\mathbb{E}Z_n$. Writing $V_n=\operatorname{Var}(Z_n)$, this is $V_{n+1}=m^2V_n+\sigma^2m^n$, with $V_0=0$. Repeated substitution gives $V_n=\sigma^2\sum_{j=0}^{n-1}m^{2n-2-j}$. If $m=1$, the sum has $n$ terms equal to $1$, so $V_n=n\sigma^2$. If $m\ne1$, the sum is a finite geometric sum: $\sum_{j=0}^{n-1}m^{2n-2-j}=m^{n-1}(m^n-1)/(m-1)$. This also covers $m=0$, using $0^0=1$ in the $n=1$ term. Substitution proves the stated formulas. $\square$

Worked Example: A Supercritical Law

Take $a=\tfrac14$, $b=\tfrac12$, and $c=\tfrac14$. The probabilities sum to $\tfrac14+\tfrac12+\tfrac14=1$, so this is a valid offspring law. Its generating function is $f(s)=\tfrac14+\tfrac12s+\tfrac14s^3$.

The mean is $m=\tfrac12+3(\tfrac14)=\tfrac54$. The second moment is $\mathbb{E}[\xi^2]=\tfrac12+9(\tfrac14)=\tfrac{11}4$, so $\sigma^2=\tfrac{11}4-(\tfrac54)^2=\tfrac{44}{16}-\tfrac{25}{16}=\tfrac{19}{16}$. Consequently, $\mathbb{E}Z_1=\tfrac54$ and $\mathbb{E}Z_2=(\tfrac54)^2=\tfrac{25}{16}$. The variances are $V_1=\tfrac{19}{16}$ and \[ V_2=(\tfrac54)^2(\tfrac{19}{16})+\tfrac{19}{16}(\tfrac54) =\tfrac{475}{256}+\tfrac{380}{256} =\tfrac{855}{256}. \]

For extinction, $q_0=0$, $q_1=f(0)=\tfrac14$, and \[ q_2=f(\tfrac14)=\tfrac14+\tfrac12(\tfrac14)+\tfrac14(\tfrac14)^3 =\tfrac{64}{256}+\tfrac{32}{256}+\tfrac{1}{256} =\tfrac{97}{256}. \] As a check, $q_2$ is greater than $q_1$, as required because extinction by generation $1$ implies extinction by generation $2$. The ultimate probability, derived below, is $(\sqrt5-1)/2$, the smaller fixed point in $[0,1]$. Substitution into $s^2+s-1=0$ verifies that value is a fixed point; $q_2=\tfrac{97}{256}$ is a finite-generation probability, not the ultimate probability.

Finite-Generation and Ultimate Extinction

Conditioning on the first generation gives the recursion for finite-generation extinction. If there are $k$ children, extinction by generation $n+1$ requires each of those $k$ descendant processes to be extinct by its generation $n$. The descendant processes are independent and each has extinction probability $q_n$. Therefore, for $n\ge0$,

q_0=0,\qquad q_{n+1}=f(q_n)=a+bq_n+cq_n^3.

The ultimate probability is the limit of these iterates and is determined by the fixed points of $f$.

Theorem: The ultimate extinction probability $q$ is the smallest solution in $[0,1]$ of $f(s)=s$. For $c>0$, it is $$ q= \begin{cases} \dfrac{-1+\sqrt{1+4a/c}}{2},&a\le2c,\\[2mm] 1,&a>2c. \end{cases} $$ For $c=0$, $q=1$ if $a>0$, and $q=0$ if $a=0$.

Proof. Since extinction is absorbing, the events $\{Z_n=0\}$ increase to the event of eventual extinction. Continuity of probability for increasing events gives $q_n\to q$. The recursion $q_{n+1}=f(q_n)$, together with continuity of the polynomial $f$, implies $q=f(q)$. To see that $q$ is the smallest such fixed point, let $r\in[0,1]$ satisfy $f(r)=r$. We have $q_0=0\le r$. The function $f$ is nondecreasing on $[0,1]$, because its coefficients are nonnegative; hence $q_n\le r$ implies $q_{n+1}=f(q_n)\le f(r)=r$. Induction gives $q_n\le r$ for every $n$, and taking limits gives $q\le r$.

It remains to identify the smallest fixed point. Since $a+b+c=1$, the fixed-point equation factors as \[ f(s)-s=a-(a+c)s+cs^3=(s-1)(cs^2+cs-a)=0. \] If $c>0$, the roots of the quadratic factor are $\bigl(-1\pm\sqrt{1+4a/c}\bigr)/2$. The root with the minus sign is negative. The root with the plus sign is nonnegative and is at most $1$ exactly when $a\le2c$: this follows by rearranging $\sqrt{1+4a/c}\le3$, which is equivalent to $a\le2c$. When $a\le2c$, that root is no greater than the other fixed point $1$, so it is the smallest fixed point in $[0,1]$. When $a>2c$, the quadratic root with the plus sign exceeds $1$, leaving $1$ as the only fixed point in $[0,1]$. If $c=0$, then $b=1-a$ and $f(s)=a+(1-a)s$. For $a>0$, $f(s)=s$ has only the solution $s=1$. For $a=0$, $f(s)=s$ for every $s\in[0,1]$, and the iteration starts at $q_0=0$ and stays at zero. These cases prove the formula and the smallest-fixed-point claim. $\square$

The mean can be used to state the same classification when $c>0$: $m-1=2c-a$. Thus $a>2c$ is subcritical, $a=2c$ is critical, and $a<2c$ is supercritical. The explicit fixed-point calculation shows extinction probability $1$ in the first two cases and less than $1$ in the third. When $c=0$, the exceptional law $a=0,b=1$ produces exactly one child forever, so its extinction probability is zero despite $m=1$; this is why the boundary case must be stated separately.

Worked Example: The Critical Case

Let $a=\tfrac12$, $b=\tfrac14$, and $c=\tfrac14$. Their sum is $\tfrac12+\tfrac14+\tfrac14=1$. The generating function is $f(s)=\tfrac12+\tfrac14s+\tfrac14s^3$, and $m=\tfrac14+3(\tfrac14)=1$. The second moment is $\tfrac14+9(\tfrac14)=\tfrac52$; hence $\sigma^2=\tfrac52-1=\tfrac32$.

The generation formulas give $\mathbb{E}Z_n=1$ and $\operatorname{Var}(Z_n)=\tfrac32n$. For finite extinction probabilities, $q_1=f(0)=\tfrac12$, while \[ q_2=f(\tfrac12)=\tfrac12+\tfrac14(\tfrac12)+\tfrac14(\tfrac12)^3 =\tfrac{16}{32}+\tfrac4{32}+\tfrac1{32} =\tfrac{21}{32}. \] The fixed-point factorization is $(s-1)(\tfrac14s^2+\tfrac14s-\tfrac12)=0$. The quadratic factor is zero when $s^2+s-2=0$, whose roots are $1$ and $-2$. Thus the only fixed point in $[0,1]$ is $1$, and the ultimate extinction probability is $q=1$. This agrees with the critical classification, while the finite value $q_2=\tfrac{21}{32}$ shows why ultimate extinction does not mean extinction within a specified finite number of generations.

Check: The variance formula at $n=2$ gives $\operatorname{Var}(Z_2)=2(\tfrac32)=3$. The recursion gives the same value: $V_2=m^2V_1+\sigma^2\mathbb{E}Z_1=1^2(\tfrac32)+\tfrac32(1)=3$.

Worked Example: A Subcritical Law

Let $a=\tfrac35$, $b=\tfrac15$, and $c=\tfrac15$. The probabilities sum to $\tfrac35+\tfrac15+\tfrac15=1$. Here $f(s)=\tfrac35+\tfrac15s+\tfrac15s^3$, and $m=\tfrac15+3(\tfrac15)=\tfrac45$. The second moment is $\tfrac15+9(\tfrac15)=2$, giving $\sigma^2=2-(\tfrac45)^2=2-\tfrac{16}{25}=\tfrac{34}{25}$.

Thus $\mathbb{E}Z_2=(\tfrac45)^2=\tfrac{16}{25}$, and \[ \operatorname{Var}(Z_2) =\tfrac{34}{25}\left((\tfrac45)^2+\tfrac45\right) =\tfrac{34}{25}\left(\tfrac{16}{25}+\tfrac{20}{25}\right) =\tfrac{1224}{625}. \] For extinction, $q_1=\tfrac35$ and \[ q_2=f(\tfrac35) =\tfrac35+\tfrac15(\tfrac35)+\tfrac15(\tfrac35)^3 =\tfrac{375}{625}+\tfrac{75}{625}+\tfrac{27}{625} =\tfrac{477}{625}. \] Since $a=\tfrac35>\tfrac25=2c$, the fixed-point formula gives $q=1$. To check the moment calculation independently, the variance recursion yields $V_2=(\tfrac45)^2(\tfrac{34}{25})+(\tfrac{34}{25})(\tfrac45) =\tfrac{544}{625}+\tfrac{680}{625} =\tfrac{1224}{625}$, as above.

How to Read the Results Correctly

Three quantities are especially easy to conflate. The mean $m$ is the expected number of children of one individual. The mean $\mathbb{E}Z_n=m^n$ describes the expected population in a particular generation. The extinction probability $q$ concerns whether any individuals remain eventually. In particular, $q_n$ is not obtained by solving $f(s)=s$ at each finite $n$; it is obtained by iterating $q_{n+1}=f(q_n)$, starting from $q_0=0$.

Common mistake. In the subcritical example, a mistaken calculation might replace the finite-generation recursion with the fixed-point equation and claim $q_1=1$, because the ultimate extinction probability is $1$. That is incorrect: $q_1=\mathbb{P}(Z_1=0)=a=\tfrac35$, directly from the offspring law. The fixed point gives the limit of the sequence, not its first iterate. Indeed, the correct next value is $q_2=f(\tfrac35)=\tfrac{477}{625}$, which is greater than $\tfrac35$ but still less than $1$.

The explicit cubic factorization is particularly useful because it exposes all the cases without numerical root-finding. It also makes the role of the three-offspring probability visible: when $c>0$, comparing $a$ with $2c$ determines whether the nontrivial quadratic root lies below $1$. The recursion remains the practical way to calculate a small number of finite-generation probabilities; the factored equation is the tool for the ultimate probability.

Check Your Understanding

Use the formulas and verify each numerical answer by substitution.

For $a=\tfrac14$, $b=\tfrac14$, $c=\tfrac12$, compute $f(s)$, $m$, and $\sigma^2$. Is the process subcritical, critical, or supercritical?
For any valid offspring probabilities $a,b,c$, express $q_1$ and $q_2$ in terms of $a,b,c$.
In the critical worked example, calculate $q_3$ from $q_2=\tfrac{21}{32}$ using the displayed generating function.
Explain why the smallest-fixed-point argument needs $f$ to be nondecreasing on $[0,1]$.
What happens to the ultimate extinction probability when $c=0$ and $a=0$? Describe the population path as well as the fixed-point equation.

A Complete Analysis of a Three-Point Offspring Law

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