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Galton-Watson Branching Processes · Tutorial 342 of 994

Bernoulli and Binomial Offspring Models

Learn how Bernoulli and binomial offspring laws determine generating functions, moments, and the distribution of the first generation.

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What You'll Learn

  • Identify the support and parameter restrictions of Bernoulli and binomial offspring laws.
  • Derive their probability generating functions from their probability mass functions.
  • Calculate offspring means and variances from generating-function derivatives.
  • Find first-generation population probabilities from one or several initial individuals.
  • Interpret zero, unit, and boundary parameter cases without confusing offspring with reproduction opportunities.

Two Basic Offspring Models

In a Galton–Watson process, each individual produces a random number of children, independently of the other individuals and of reproduction in other generations. Two useful models arise when reproduction is described by a fixed number of opportunities: an individual may have at most one child, or may have at most \(r\) children. The first gives a Bernoulli offspring law; the second gives a binomial offspring law. Their simple formulas make them useful for checking calculations and for representing limited reproduction opportunities.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). The generation sizes \((Z_n)_{n\ge0}\) take values in \(S=\{0,1,2,\ldots\}\), with discrete generation index \(n\). Write \(\xi_{n,i}\) for the number of children of the \(i\)th individual in generation \(n\). In the Galton–Watson model these offspring counts are independent and identically distributed. The population update is \(Z_{n+1}=\sum_{i=1}^{Z_n}\xi_{n,i}\), where a sum over no individuals is zero. Here the focus is the offspring law and the distribution of \(Z_1\) for a fixed initial size \(Z_0=z\), not the behavior of later generations.

Definition: A Bernoulli offspring law with parameter \(p\) assigns probability \(1-p\) to \(0\) children and probability \(p\) to \(1\) child, where \(0\le p\le1\). A binomial offspring law with parameters \(r,p\) assigns probability \(\binom{r}{k}p^k(1-p)^{r-k}\) to \(k\) children, for \(k=0,\ldots,r\), where \(r\) is a nonnegative integer and \(0\le p\le1\).

For a nonnegative integer-valued offspring count \(\xi\), its probability generating function is \(G(s)=\mathbb{E}[s^\xi]\), for \(0\le s\le1\). Thus \(G(s)=\sum_k\mathbb{P}(\xi=k)s^k\). The earlier material on offspring distributions and probability generating functions establishes this encoding: the coefficient of \(s^k\) is the probability of \(k\) offspring. For both models here, the sum is finite, so \(G\) is a polynomial and its derivatives at \(1\) are well-defined.

Generating Functions and Moments

Theorem: If \(\xi\) has finite support and generating function \(G\), then \(\mathbb{E}[\xi]=G'(1)\) and \(\operatorname{Var}(\xi)=G''(1)+G'(1)-\bigl(G'(1)\bigr)^2\).

Proof. Write \(a_k=\mathbb{P}(\xi=k)\), so \(G(s)=\sum_k a_k s^k\), with only finitely many nonzero terms. Differentiating the polynomial gives \(G'(1)=\sum_k k a_k=\mathbb{E}[\xi]\), and \(G''(1)=\sum_k k(k-1)a_k=\mathbb{E}[\xi(\xi-1)]\). Since \(\xi^2=\xi(\xi-1)+\xi\), it follows that \(\mathbb{E}[\xi^2]=G''(1)+G'(1)\). Subtracting \((\mathbb{E}[\xi])^2\) from \(\mathbb{E}[\xi^2]\) proves the variance formula. \(\square\)

For Bernoulli reproduction, the only possible values are \(0\) and \(1\). Directly from the probabilities,

$$ G(s)=(1-p)+ps. $$

Its derivatives are \(G'(s)=p\) and \(G''(s)=0\). The theorem therefore gives mean \(p\) and variance \(p-p^2=p(1-p)\). These expressions include the boundary cases: if \(p=0\), the individual has no children with probability one; if \(p=1\), the individual has exactly one child with probability one. In either case the variance is zero.

For binomial reproduction, the binomial theorem gives

$$ G(s)=\sum_{k=0}^{r}\binom{r}{k}p^k(1-p)^{r-k}s^k =\bigl((1-p)+ps\bigr)^r. $$

For \(r\ge1\), differentiation yields \(G'(s)=rp((1-p)+ps)^{r-1}\), so \(G'(1)=rp\), and \(G''(1)=r(r-1)p^2\). The variance formula gives \(r(r-1)p^2+rp-r^2p^2=rp(1-p)\). If \(r=0\), the offspring count is identically zero, and the same mean and variance formulas give zero. At \(p=0\) the count is also identically zero; at \(p=1\) it is identically \(r\). Both boundary cases have variance zero.

Worked Example: Bernoulli Reproduction

Suppose one initial individual has Bernoulli offspring with \(p=\tfrac25\). Find the generating function, mean, variance, and the probabilities for the first-generation size \(Z_1\).

The probability of zero children is \(1-p=1-\tfrac25=\tfrac35\); the probability of one child is \(\tfrac25\). Thus \(G(s)=\tfrac35+\tfrac25s\). Its derivatives are \(G'(s)=\tfrac25\) and \(G''(s)=0\), so \(\mathbb{E}[\xi]=\tfrac25\) and \(\operatorname{Var}(\xi)=0+\tfrac25-(\tfrac25)^2=\tfrac25-\tfrac4{25}=\tfrac6{25}\).

With \(Z_0=1\), the first-generation size is the single individual's offspring count. Hence \(\mathbb{P}(Z_1=0)=\tfrac35\) and \(\mathbb{P}(Z_1=1)=\tfrac25\). Check: the probabilities sum to \(\tfrac35+\tfrac25=1\), and the mean from the probabilities is \(0\cdot\tfrac35+1\cdot\tfrac25=\tfrac25\), as given by \(G'(1)\).

Worked Example: Binomial Reproduction with Three Opportunities

Suppose each individual has \(r=3\) independent opportunities to produce one child, each succeeding with probability \(p=\tfrac13\). Find the offspring generating function, mean, variance, and all possible first-generation probabilities from one initial individual.

The probability of \(k\) children is \(\binom{3}{k}(\tfrac13)^k(\tfrac23)^{3-k}\), for \(k=0,1,2,3\). The generating function is \(G(s)=(\tfrac23+\tfrac13s)^3\). From the binomial formulas, the mean is \(rp=3\cdot\tfrac13=1\), and the variance is \(rp(1-p)=3\cdot\tfrac13\cdot\tfrac23=\tfrac23\).

For the first-generation probabilities, substitute each value of \(k\):

$$ \begin{aligned} \mathbb{P}(Z_1=0)&=\binom30(\tfrac13)^0(\tfrac23)^3=\tfrac8{27},\\ \mathbb{P}(Z_1=1)&=\binom31(\tfrac13)^1(\tfrac23)^2=\tfrac{12}{27}=\tfrac49,\\ \mathbb{P}(Z_1=2)&=\binom32(\tfrac13)^2(\tfrac23)^1=\tfrac6{27}=\tfrac29,\\ \mathbb{P}(Z_1=3)&=\binom33(\tfrac13)^3(\tfrac23)^0=\tfrac1{27}. \end{aligned} $$

Check: their sum is \((8+12+6+1)/27=27/27=1\). The mean computed from these probabilities is \((0\cdot8+1\cdot12+2\cdot6+3\cdot1)/27=27/27=1\), matching \(rp\). Also, the generating function derivatives give \(G'(1)=1\) and \(G''(1)=3\cdot2\cdot(\tfrac13)^2=\tfrac23\); therefore the variance is \(\tfrac23+1-1^2=\tfrac23\), matching the formula above.

Several Initial Individuals

When \(Z_0=z\) is fixed, the first generation pools the offspring of \(z\) independent individuals. For Bernoulli reproduction, each of those individuals has one opportunity, so \(Z_1\) is binomial with \(z\) opportunities and success probability \(p\). For binomial offspring with \(r\) opportunities per individual, there are \(rz\) independent opportunities in total. This aggregation gives a direct way to compute the first-generation distribution.

Theorem: Let \(z\) be a fixed nonnegative integer and suppose every initial individual has binomial offspring with parameters \(r,p\), where \(r\) is a nonnegative integer and \(0<p<1\). Then, if \(rz>0\), \(Z_1\) has the binomial distribution with parameters \(rz,p\): \(\mathbb{P}(Z_1=k)=\binom{rz}{k}p^k(1-p)^{rz-k}\) for \(k=0,\ldots,rz\). If \(rz=0\), then \(Z_1=0\) with probability one. The Bernoulli case is obtained by setting \(r=1\).

Proof. Each of the \(z\) individuals has \(r\) independent opportunities, and every opportunity succeeds with probability \(p\). Independence of offspring across individuals means all \(rz\) opportunities are independent. The total \(Z_1\) is the number of successful opportunities. For \(0\le k\le rz\), any specified set of \(k\) successful opportunities and \(rz-k\) failures has probability \(p^k(1-p)^{rz-k}\), and there are \(\binom{rz}{k}\) such sets. Summing these disjoint possibilities gives the stated probability. If \(rz=0\), there are no opportunities, so the total is zero. When \(p=0\), all opportunities fail, and when \(p=1\), all succeed; these give \(Z_1=0\) and \(Z_1=rz\), respectively. \(\square\)

Equivalently, for \(rz>0\), the generating function of \(Z_1\) is \(((1-p)+ps)^{rz}\). This follows by multiplying the generating function \((1-p)+ps\) for each independent opportunity. The coefficients of the resulting polynomial are exactly the first-generation probabilities in the theorem. This is a statement about \(Z_1\) for fixed \(Z_0\); it does not require a formula for later generation sizes.

Worked Example: Two Individuals with Two Opportunities Each

Let \(Z_0=2\). Each individual has binomial offspring with \(r=2\) and \(p=\tfrac34\). Find the distribution, mean, and variance of \(Z_1\).

There are \(rz=2\cdot2=4\) independent opportunities, so \(Z_1\) is binomial with parameters \(4,\tfrac34\). For \(k=0,1,2,3,4\), the probabilities are

$$ \begin{aligned} \mathbb{P}(Z_1=0)&=\binom40(\tfrac34)^0(\tfrac14)^4=\tfrac1{256},\\ \mathbb{P}(Z_1=1)&=\binom41(\tfrac34)(\tfrac14)^3=\tfrac{12}{256}=\tfrac3{64},\\ \mathbb{P}(Z_1=2)&=\binom42(\tfrac34)^2(\tfrac14)^2=\tfrac{54}{256}=\tfrac{27}{128},\\ \mathbb{P}(Z_1=3)&=\binom43(\tfrac34)^3(\tfrac14)=\tfrac{108}{256}=\tfrac{27}{64},\\ \mathbb{P}(Z_1=4)&=\binom44(\tfrac34)^4(\tfrac14)^0=\tfrac{81}{256}. \end{aligned} $$

The mean is \(rzp=4\cdot\tfrac34=3\), and the variance is \(rzp(1-p)=4\cdot\tfrac34\cdot\tfrac14=\tfrac34\). Check: the probability numerators sum to \(1+12+54+108+81=256\). The mean computed from the list is \((0\cdot1+1\cdot12+2\cdot54+3\cdot108+4\cdot81)/256=768/256=3\). For a second variance check, the second factorial moment is \(4\cdot3\cdot(\tfrac34)^2=\tfrac{27}{4}\); adding the mean gives \(\mathbb{E}[Z_1^2]=\tfrac{27}{4}+3=\tfrac{39}{4}\), and \(\tfrac{39}{4}-3^2=\tfrac34\).

Interpreting Parameters and Avoiding a Common Error

The parameter \(p\) is the success probability of one reproduction opportunity, not the expected number of children by itself in the binomial model. The parameter \(r\) counts the available opportunities and sets the largest possible offspring count. Thus the binomial mean is \(rp\), while the Bernoulli mean is \(p\) because Bernoulli reproduction has just one opportunity. The restriction \(r\in\{0,1,2,\ldots\}\) is essential: a non-integer number of opportunities does not define this binomial offspring model. The restriction \(0\le p\le1\) ensures that the stated probabilities are valid.

A common mistake is to treat the second derivative \(G''(1)\) as the variance. It is instead the second factorial moment \(\mathbb{E}[\xi(\xi-1)]\); the mean and its square must also be used. For example, with \(r=3\) and \(p=\tfrac12\), \(G''(1)=3\cdot2\cdot(\tfrac12)^2=\tfrac32\), but the variance is \(\tfrac32+\tfrac32-(\tfrac32)^2=\tfrac34\), not \(\tfrac32\). The correction is to apply \(\operatorname{Var}(\xi)=G''(1)+G'(1)-(G'(1))^2\) after computing both derivatives.

These models also differ in their possible offspring counts. Bernoulli reproduction permits only \(0\) or \(1\), so one individual cannot produce multiple children. Binomial reproduction permits every count from \(0\) through \(r\) when \(0<p<1\) and \(r>0\). At the boundary \(p=0\) it permits only zero children; at \(p=1\) it permits exactly \(r\). A formula for a generating function or moment remains useful at these boundaries, but the resulting law is deterministic rather than variable.

Check Your Understanding

Use the generating-function derivative formulas and check each probability list by summing its entries.

  1. For Bernoulli offspring with \(p=\tfrac35\), write \(G(s)\), and calculate the mean and variance.
  2. For binomial offspring with \(r=4\) and \(p=\tfrac12\), what are the maximum offspring count, mean, and variance?
  3. If \(Z_0=3\) and each individual has Bernoulli offspring with \(p=\tfrac14\), identify the distribution of \(Z_1\) and calculate \(\mathbb{P}(Z_1=2)\).
  4. Explain why \(G''(1)\) alone is not generally the variance, and give the correct formula.
  5. What are the offspring distributions when \(p=0\) and \(p=1\) in the binomial model with a fixed nonnegative integer \(r\)?