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Galton-Watson Branching Processes · Tutorial 344 of 994

Geometric Offspring Laws and Their Generating Functions

Learn how the two geometric conventions produce different offspring laws, generating functions, moments, and probabilities.

Advanced 12 min read

What You'll Learn

  • Define geometric offspring counts on the nonnegative integers and on the positive integers.
  • Derive each probability generating function from its probability mass function.
  • Calculate the mean, variance, and second factorial moment under both conventions.
  • Recover exact offspring probabilities from the parameters or generating functions.
  • Identify and correct the common error of shifting the geometric law incorrectly.

Two Geometric Laws for Offspring Counts

A geometric distribution can describe the number of failures before the first success, or the number of trials up to and including the first success. These differ by one, so using the word “geometric” without specifying the support can change an offspring model substantially. Here we derive both conventions and compare their generating functions, moments, and probabilities.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). In a Galton–Watson process, an individual’s offspring count is a random variable \(\xi\) taking values in \(\mathbb{N}_0=\{0,1,2,\ldots\}\); the population process \((Z_n)_{n\ge0}\) is indexed by discrete generation \(n\) and has state space \(\mathbb{N}_0\). This tutorial concerns the possible laws for \(\xi\). We use a parameter \(p\) with \(0<p\le1\), and write \(q=1-p\). The endpoint \(p=1\) is included because it gives a valid degenerate law; \(p=0\) is excluded because the proposed probabilities would not sum to one.

Definition: The nonnegative-integer geometric law has \(\mathbb{P}(\xi=k)=p q^k\) for \(k\in\mathbb{N}_0\). The positive-integer geometric law has \(\mathbb{P}(\xi=k)=p q^{k-1}\) for \(k\in\{1,2,\ldots\}\), and probability zero at \(k=0\). In both cases \(q=1-p\). The probability generating function (PGF) is \(G(s)=\mathbb{E}[s^\xi]=\sum_k\mathbb{P}(\xi=k)s^k\), for \(0\le s\le1\).

These are probability distributions: for \(0<p<1\), the geometric-series identity \(\sum_{k=0}^{\infty}q^k=1/(1-q)=1/p\) gives total mass one in the first convention, and \(\sum_{k=1}^{\infty}q^{k-1}=1/p\) gives total mass one in the second. When \(p=1\), \(q=0\): the first law puts probability one at zero, while the second puts probability one at one.

Generating Functions and Moments

For the nonnegative-integer convention, summing the probability mass function gives \(G(s)=p\sum_{k=0}^{\infty}(qs)^k\). For the positive-integer convention the corresponding sum starts at \(k=1\), so one factor of \(s\) is taken out. These series converge for \(0\le s\le1\), since \(q<1\).

$$ G_{\text{nonnegative}}(s)=\frac{p}{1-qs}, \qquad G_{\text{positive}}(s)=\frac{ps}{1-qs}. $$

The second formula is \(s\) times the first. This reflects the shift in support: if \(Y\) has the nonnegative-integer geometric law, then \(1+Y\) has the positive-integer geometric law. Indeed, for every \(k\ge1\), \(\mathbb{P}(1+Y=k)=\mathbb{P}(Y=k-1)=p q^{k-1}\).

Theorem: For the nonnegative-integer geometric law, \(\mathbb{E}[\xi]=q/p\), \(\operatorname{Var}(\xi)=q/p^2\), and \(\mathbb{E}[\xi(\xi-1)]=2q^2/p^2\). For the positive-integer geometric law, \(\mathbb{E}[\xi]=1/p\), \(\operatorname{Var}(\xi)=q/p^2\), and \(\mathbb{E}[\xi(\xi-1)]=2q/p^2\). These formulas also hold at \(p=1\).

Proof. For \(0<p<1\), differentiate the displayed PGFs at \(s<1\), where their geometric series and derivatives converge. The nonnegative-integer PGF has derivatives \(G'(s)=pq/(1-qs)^2\) and \(G''(s)=2pq^2/(1-qs)^3\). The positive-integer PGF has derivatives \(G'(s)=p/(1-qs)^2\) and \(G''(s)=2pq/(1-qs)^3\). Since \(1-q=p\), evaluating at \(s=1\) gives the means and second factorial moments stated in the theorem. For any nonnegative integer-valued random variable with finite second moment, \(\mathbb{E}[\xi^2]=\mathbb{E}[\xi(\xi-1)]+\mathbb{E}[\xi]\). Thus the nonnegative-integer variance is \(2q^2/p^2+q/p-(q/p)^2=q/p^2\). The positive-integer variance is \(2q/p^2+1/p-(1/p)^2=q/p^2\), using \(q=1-p\). When \(p=1\), the two laws are constant, respectively zero and one, so their means, variances, and factorial moments are \(0,0,0\) and \(1,0,0\), as the formulas give. \(\square\)

For general PGFs, the derivative identities require care at \(s=1\). For \(0\le s<1\), termwise differentiation gives \(G'(s)=\sum_{k\ge1}k\mathbb{P}(\xi=k)s^{k-1}\) and \(G''(s)=\sum_{k\ge2}k(k-1)\mathbb{P}(\xi=k)s^{k-2}\). As \(s\) increases to one, the nonnegative terms increase, so their limits are the corresponding moments, possibly infinite. In particular, \(G'(1)=\mathbb{E}[\xi]\) when the mean is finite, and \(G''(1)=\mathbb{E}[\xi(\xi-1)]\) when the second factorial moment is finite. Differentiability once at \(1\) does not by itself imply a finite second factorial moment. For the two geometric PGFs above, the rational formulas have finite derivatives of every order near \(1\), so the moment calculations are valid.

Worked Calculations

Worked Example: A Nonnegative Geometric Offspring Count

Suppose the number of offspring has the nonnegative-integer geometric law with \(p=\tfrac23\). Find its PGF, mean, variance, and probabilities of zero, one, and two offspring. Also check the remaining probability.

Here \(q=1-p=1-\tfrac23=\tfrac13\). Substituting into the PGF and moment formulas gives

$$ G(s)=\frac{2/3}{1-s/3}, \qquad \mathbb{E}[\xi]=\frac{1/3}{2/3}=\frac12, \qquad \operatorname{Var}(\xi)=\frac{1/3}{(2/3)^2}=\frac34. $$

The requested probabilities follow directly from \(\mathbb{P}(\xi=k)=p q^k\): \(\mathbb{P}(\xi=0)=\tfrac23(\tfrac13)^0=\tfrac23\), \(\mathbb{P}(\xi=1)=\tfrac23(\tfrac13)=\tfrac29\), and \(\mathbb{P}(\xi=2)=\tfrac23(\tfrac19)=\tfrac2{27}\). Their sum is \(\tfrac{18}{27}+\tfrac6{27}+\tfrac2{27}=\tfrac{26}{27}\). The remaining probability is \(\mathbb{P}(\xi\ge3)=\sum_{k=3}^{\infty}\tfrac23(\tfrac13)^k=(\tfrac23)(\tfrac1{27})/(1-\tfrac13)=\tfrac1{27}\). Hence the total is \(\tfrac{26}{27}+\tfrac1{27}=1\).

Check: \(G(1)=\frac{2/3}{1-1/3}=1\), as required for a PGF. The second factorial moment is \(2q^2/p^2=2(1/9)/(4/9)=\tfrac12\); adding the mean gives \(\mathbb{E}[\xi^2]=\tfrac12+\tfrac12=1\), and \(1-(\tfrac12)^2=\tfrac34\), which confirms the variance.

Worked Example: A Positive Geometric Offspring Count

Suppose every individual has at least one offspring and the count follows the positive-integer geometric law with \(p=\tfrac34\). Calculate its PGF, mean, variance, and probabilities of one, two, and three offspring, then verify the probability mass not listed.

Now \(q=1-\tfrac34=\tfrac14\). Therefore

$$ G(s)=\frac{(3/4)s}{1-s/4}, \qquad \mathbb{E}[\xi]=\frac{1}{3/4}=\frac43, \qquad \operatorname{Var}(\xi)=\frac{1/4}{(3/4)^2}=\frac49. $$

Using \(\mathbb{P}(\xi=k)=p q^{k-1}\), we obtain \(\mathbb{P}(\xi=1)=\tfrac34\), \(\mathbb{P}(\xi=2)=\tfrac34\cdot\tfrac14=\tfrac3{16}\), and \(\mathbb{P}(\xi=3)=\tfrac34\cdot\tfrac1{16}=\tfrac3{64}\). Their sum is \(\tfrac{48}{64}+\tfrac{12}{64}+\tfrac3{64}=\tfrac{63}{64}\). The tail from four onward is \(\sum_{k=4}^{\infty}\tfrac34(\tfrac14)^{k-1}=(\tfrac34)(\tfrac1{64})/(1-\tfrac14)=\tfrac1{64}\). The four contributions total one.

Check: \(G(1)=\frac{3/4}{1-1/4}=1\). The second factorial moment is \(2q/p^2=2(1/4)/(9/16)=\tfrac89\). Since the mean is \(\tfrac43\), the second moment is \(\tfrac89+\tfrac43=\tfrac{20}{9}\), and the variance is \(\tfrac{20}{9}-(\tfrac43)^2=\tfrac49\).

Worked Example: Equal Means, Different Offspring Probabilities

Choose one law of each type with mean \(2\), and compare their PGFs and probabilities of small offspring counts. This shows why a mean alone does not determine the offspring law.

For the nonnegative-integer law, \(q/p=2\) and \(p+q=1\), so \(q=2p\), \(3p=1\), and \(p=\tfrac13,\ q=\tfrac23\). Its PGF and first three probabilities are

$$ G_{\text{nonnegative}}(s)=\frac{1/3}{1-2s/3}, \quad \mathbb{P}(\xi=0)=\frac13,\quad \mathbb{P}(\xi=1)=\frac29,\quad \mathbb{P}(\xi=2)=\frac4{27}. $$

For the positive-integer law, its mean is \(1/p\), so \(1/p=2\) gives \(p=\tfrac12\) and \(q=\tfrac12\). Its PGF and probabilities of one, two, and three offspring are

$$ G_{\text{positive}}(s)=\frac{s/2}{1-s/2}, \quad \mathbb{P}(\xi=1)=\frac12,\quad \mathbb{P}(\xi=2)=\frac14,\quad \mathbb{P}(\xi=3)=\frac18. $$

Both means check: the first is \((2/3)/(1/3)=2\), and the second is \(1/(1/2)=2\). But the nonnegative law has probability \(\tfrac13\) of no offspring, while the positive law has probability zero of no offspring. The PGFs also differ, not merely by their parameter values: the positive-law PGF has a factor \(s\), encoding its support restriction.

Parameter Identification and a Common Mistake

Result: If the mean \(m\) is specified, then a nonnegative-integer geometric law with that mean must have \(p=1/(m+1)\), while a positive-integer geometric law must have \(p=1/m\). The first requires \(m\ge0\); the second requires \(m\ge1\).

Proof. In the first convention \(m=q/p=(1-p)/p=1/p-1\), so \(m+1=1/p\) and \(p=1/(m+1)\). Since \(0<p\le1\), this gives exactly \(m\ge0\). In the second convention \(m=1/p\), so \(p=1/m\); the condition \(0<p\le1\) is equivalent to \(m\ge1\). Conversely, for every mean in the stated range, the displayed parameter belongs to \((0,1]\) and substitution into the relevant mean formula returns \(m\). Thus the parameter is both necessary and sufficient. \(\square\)

Common mistake. For a positive-integer geometric law with \(p=\tfrac12\), using \(p q^k\) at \(k=0,1,\ldots\) incorrectly assigns probability \(\tfrac12\) to zero offspring. That contradicts the positive support. The correct formula is \(p q^{k-1}\) for \(k\ge1\): it gives \(\mathbb{P}(\xi=1)=\tfrac12\), \(\mathbb{P}(\xi=2)=\tfrac14\), and, in general, zero probability at \(k=0\). Its total mass is \(\sum_{k=1}^{\infty}\tfrac12(\tfrac12)^{k-1}=1\).

When specifying a geometric offspring model, state its support along with its parameter. The nonnegative version permits no offspring and has mean \(q/p\); the positive version forces at least one offspring and has mean \(1/p\). Their PGFs make the shift explicit, and their probability formulas allow individual offspring probabilities to be computed without ambiguity.

Check Your Understanding

For each question, specify which geometric convention is being used and check numerical probabilities by summing the relevant mass or tail.

  1. For the nonnegative-integer law with \(p=\tfrac12\), find \(G(s)\), \(\mathbb{P}(\xi=0)\), \(\mathbb{P}(\xi=2)\), and the mean.
  2. For the positive-integer law with \(p=\tfrac13\), find the mean and variance. What is the probability of zero offspring?
  3. A nonnegative-integer geometric offspring law has mean \(3\). Find \(p\), \(q\), and \(\mathbb{P}(\xi=2)\).
  4. Why can two geometric offspring laws with the same mean have different probabilities of zero offspring?
  5. State the condition needed to identify \(G''(1)\) with a finite second factorial moment, and explain why the geometric laws here satisfy it.