Why Consider a Geometric Count Response?
Some event counts vary much more than a Poisson model allows. The geometric distribution is one specific way to represent that extra variability: its variance is determined by its mean and grows quadratically as the mean increases. It can be a useful candidate when that particular mean-variance relationship is plausible, rather than simply because a sample contains some large counts.
The convention matters. Here, a geometric response counts the number of failures before the first success, so its possible values are \(0,1,2,\ldots\). This is an unbounded count with zero in its support. Other conventions place the support at \(1,2,\ldots\); those describe a different count and are not used here.
Under this parameterization, the mean determines the entire response distribution: once \(\mu\) is specified, the success probability is \(q=1/(1+\mu)\). There is no separate dispersion parameter to estimate. In the usual negative binomial parameterization by size and mean, this is the special case with size equal to one. It is therefore more restrictive than a negative binomial model whose size can vary.
Deriving the Mean and Variance
A short derivation makes the model's fixed variance relationship explicit. The probability-generating function of \(Y\) is \(G(s)=E(s^Y)\). Summing the geometric series gives \(G(s)=q\sum_{y=0}^{\infty}[(1-q)s]^y=q/[1-(1-q)s]\), for values of \(s\) where the series converges. The generating-function identities are \(E(Y)=G'(1)\) and \(E[Y(Y-1)]=G''(1)\).
Evaluating at \(s=1\) gives \(E(Y)=(1-q)/q=\mu\) and \(E[Y(Y-1)]=2(1-q)^2/q^2=2\mu^2\). Since \(Y^2=Y(Y-1)+Y\), it follows that \(E(Y^2)=2\mu^2+\mu\). Therefore \(\operatorname{Var}(Y)=E(Y^2)-[E(Y)]^2=\mu+2\mu^2-\mu^2=\mu+\mu^2\).
For every positive \(\mu\), the geometric variance exceeds the mean. At small means the additional \(\mu^2\) term can be modest; as the mean grows, it becomes increasingly important. This is the defining feature of the candidate family, not a guarantee that it will describe every highly variable count.
Worked Example: Probabilities From a Geometric Mean
Suppose a hypothetical response is modelled as geometric with mean \(\mu=4\) events per observation. Find the corresponding success probability, variance, probability of zero events, probability of exactly three events, and probability of at least three events.
The success probability is \(q=1/(1+\mu)=1/5=0.2\). The variance is \(\mu(1+\mu)=4(5)=20\) events squared. Thus the standard deviation is \(\sqrt{20}\approx4.472\) events.
Using \(P(Y=y)=q(1-q)^y\), the probability of zero events is \(P(Y=0)=0.2(0.8)^0=0.2\). The probability of exactly three events is \(P(Y=3)=0.2(0.8)^3=0.1024\).
For at least three events, sum the probabilities from \(y=3\) onward, or use the geometric-series tail: \(P(Y\geq3)=(1-q)^3=(0.8)^3=0.512\). This exceeds the probability of exactly three events because it also includes counts of four or more.
Using the Geometric Family in a GLMM
In a generalized linear mixed model, the response family describes the distribution conditional on the random effects. For observation \(i\) in cluster \(j\), let \(Y_{ij}\) be the count and \(b_j\) the cluster-level random effect or vector of random effects. A geometric-response GLMM can specify \(\operatorname{E}(Y_{ij}\mid b_j)=\mu_{ij}\), with a log link \(\log(\mu_{ij})=x_{ij}^{T}\beta+z_{ij}^{T}b_j\). Conditional on the random effects, the geometric model imposes \(\operatorname{Var}(Y_{ij}\mid b_j)=\mu_{ij}+\mu_{ij}^{2}\).
This formulation assumes that each response has support \(0,1,2,\ldots\), that its conditional distribution is geometric at the stated conditional mean, and that the link and linear predictor describe how covariates and random effects determine that mean. As in a conventional GLMM, responses are typically assumed conditionally independent given the random effects and covariates. The random-effects distribution is a separate part of the model; a common specification is a multivariate normal distribution with mean zero and an estimated covariance matrix.
The quadratic variance relationship applies conditional on the random effects. It does not generally describe the marginal relationship between the response and its marginal mean after the random effects are integrated out. This distinction is important when assessing variation or interpreting predictions.
Worked Example: Conditional and Marginal Variation With a Random Intercept
Consider a hypothetical count model with a random intercept \(b\sim N(0,0.25)\) and conditional mean \(\mu(b)=0.5\exp(b)\). Conditional on \(b\), suppose the response is geometric. Find the conditional mean and variance at \(b=0\), then find the marginal mean and variance across clusters.
At \(b=0\), \(\mu(0)=0.5\exp(0)=0.5\) events. The conditional geometric variance is \(0.5(1+0.5)=0.75\) events squared.
For a normal random variable with variance \(0.25\), \(E[\exp(b)]=\exp(0.25/2)=\exp(0.125)\). Therefore the marginal mean is \(E[\mu(b)]=0.5\exp(0.125)\approx0.5666\) events. Also, \(E[\mu(b)^2]=0.25E[\exp(2b)]=0.25\exp(2^2(0.25)/2)=0.25\exp(0.5)\approx0.4122\) events squared.
By the law of total variance, \(\operatorname{Var}(Y)=E[\operatorname{Var}(Y\mid b)]+\operatorname{Var}(E[Y\mid b])\). The first term is \(E[\mu(b)+\mu(b)^2]\approx0.5666+0.4122=0.9788\) events squared. The second is \(E[\mu(b)^2]-[E\{\mu(b)\}]^2\approx0.4122-(0.5666)^2\approx0.0912\) events squared. Thus the marginal variance is approximately \(0.9788+0.0912=1.0699\) events squared.
Evaluating the geometric variance formula at the marginal mean instead gives \(0.5666(1+0.5666)\approx0.8876\) events squared. That is not the marginal variance: the random-effect mixture adds between-cluster variation, and the marginal response is not generally geometric.
Checking Whether the Geometric Model Is Plausible
The geometric family is a focused candidate, not a general remedy for overdispersion. Its conditional variance is fixed once its mean is known. A useful assessment therefore asks whether that particular conditional distribution is credible after accounting for covariates and random effects, not merely whether the raw sample variance exceeds the raw sample mean.
For a simple model with no random effects, an observed sample mean gives a direct moment-based parameter estimate \(\hat{q}=1/(1+\bar{x})\). In a GLMM, the means vary with covariates and random effects, so fitting the model requires estimating the fixed effects, random-effects distribution, and any other model components together. The elementary calculation below is useful for understanding the parameterization, but does not replace that joint model fitting.
Worked Example: A Moment-Based Check on Hypothetical Counts
Suppose eight hypothetical observations have counts \(0,1,1,2,2,3,4,7\). Use their mean to obtain the geometric moment estimate, calculate the variance predicted at that mean, and compare it with the sample variance.
The counts sum to \(20\), so \(\bar{x}=20/8=2.5\) events. The moment estimate of the success probability is \(\hat{q}=1/(1+\bar{x})=1/3.5=2/7\approx0.2857\). The corresponding geometric variance is \(\bar{x}(1+\bar{x})=2.5(3.5)=8.75\) events squared.
The squared counts sum to \(84\), and the sample variance is \(s^2=[84-20^2/8]/(8-1)=(84-50)/7=34/7\approx4.857\) events squared. This small hypothetical sample has a sample variance below the model variance evaluated at its mean. That difference alone does not establish that the geometric model is unsuitable: sample variances fluctuate, and model assessment must account for sample size, covariates, and any random effects.
In practice, examine whether the fitted model reproduces relevant features of the counts, including their spread and tail behavior, and inspect residual or simulation-based diagnostics appropriate to the fitted GLMM. A systematic mismatch in large counts, zero frequencies, or patterns across fitted means can indicate that the specified family or other model components are inadequate. Diagnostics should be interpreted conditionally on the design and fitted structure rather than by comparing one unadjusted variance with one mean.
Comparisons, Limitations, and a Common Pitfall
| Candidate | Variance pattern | Useful distinction |
|---|---|---|
| Poisson | Conditional variance equals the conditional mean | Does not include a separate quadratic excess-variance term |
| Geometric | Conditional variance is \(\mu+\mu^2\) | Fixes the negative binomial size at one |
| More general negative binomial | Allows a different mean-variance relationship through its dispersion or size parameterization | Less restrictive when the geometric relationship does not fit |
| NB1 variance form | Uses a variance increasing linearly with the mean, with a dispersion multiplier | Represents a different variance pattern from the geometric quadratic form |
The geometric model can be appealing for highly variable counts because its variance can greatly exceed its mean. Its limitation is the same feature: it enforces a particular quadratic relationship rather than estimating the strength of quadratic overdispersion freely. If the observed conditional variability is not compatible with that relationship, a different count family or a revised model structure may be more appropriate. A count with a known upper bound also does not become a geometric response merely because its variance is large; the response support and event-generating process still matter.
Common mistake: “The marginal mean is \(0.5666\), so the geometric variance must be \(0.8876\).” This applies a conditional family formula to a marginal mean in a mixed model. The correction is to use the law of total variance: average the conditional variances and add the variance of the conditional means. In the random-intercept example, that gives a marginal variance of about \(1.0699\) events squared, not \(0.8876\).
The central selection question is therefore not simply whether counts are overdispersed relative to Poisson. It is whether the geometric response distribution, with its zero-based support and fixed conditional variance \(\mu+\mu^2\), is a defensible description of the data after the GLMM's covariates and random effects are taken into account.
Check Your Understanding
Try each question, then check the solutions below.
- Under the zero-based convention, what are the support and mean-parameterized probability mass function of a geometric response?
- If the conditional mean is \(3\) events, what success probability and conditional variance does the geometric model imply?
- Why is the geometric family a restricted negative binomial model?
- In a GLMM, why can the marginal variance differ from the geometric variance evaluated at the marginal mean?
Solutions. (1) The support is \(y=0,1,2,\ldots\), and \(P(Y=y)=\mu^y/(1+\mu)^{y+1}\). (2) \(q=1/(1+3)=0.25\), and \(\operatorname{Var}(Y\mid b)=3(1+3)=12\) events squared. (3) It fixes the negative binomial size at one, so it does not estimate that parameter freely. (4) The marginal variance equals the expected conditional variance plus the variance of the conditional mean; the latter term reflects variation across random effects, and the resulting marginal distribution is not generally geometric.