Poisson Likelihood For Unbounded Event Counts

Why a Poisson Likelihood Represents Event Counts

An event count records how many times an event occurred for an observational unit, such as the number of calls received by a service during a defined interval. Counts can be zero and can exceed one. A Poisson likelihood models such counts with a probability distribution whose possible values are all nonnegative integers. In contrast to a count of successes from a fixed number of trials, the Poisson distribution has no fixed upper bound on its support.

In a generalized linear mixed model, the Poisson distribution is commonly specified conditional on the random effects. This tutorial focuses on that response distribution: its support, probability mass function, likelihood, and defining mean–variance relationship. The conditional qualifier matters because observations sharing a random effect need not be independent after that effect is integrated out.

Definition: A random variable $Y$ has a Poisson distribution with mean $\mu$, written $Y\mid b\sim\operatorname{Poisson}(\mu)$, if $\mu>0$ and $$ P(Y=y\mid b)=\frac{e^{-\mu}\mu^y}{y!},\qquad y=0,1,2,\ldots. $$ Here $b$ denotes any conditioning random effect, and the support is the set of all nonnegative integers. Conditional on $b$, the distribution has mean $\mu$ and variance $\mu$.

The mean $\mu$ is the expected count for the observational unit, under the specified conditioning. It is not necessarily an event rate: it is a rate only when an exposure duration or other denominator has been defined and incorporated into the model. In a GLMM with a log mean, one possible conditional specification is

Y_{ij}\mid b_j\sim\operatorname{Poisson}(\mu_{ij}),\qquad \log(\mu_{ij})=\mathbf{x}_{ij}^{\mathsf T}\boldsymbol{\beta}+b_j,

where $Y_{ij}$ is the count for observation $i$ in cluster $j$, $\mathbf{x}_{ij}$ is its covariate vector, $\boldsymbol{\beta}$ is the vector of fixed-effect coefficients, and $b_j$ is a cluster random effect. The exponential of the linear predictor gives a positive conditional mean. A Poisson response specification also assumes that observations are independent conditional on the random effects and other modeled predictors. This assumption does not mean that observations in the same cluster are independent marginally.

Probability, Support, and Likelihood

For a specified mean $\mu$, the probability of observing exactly $y$ events is the probability mass function in the definition. The factorial $y!$ is $1\cdot2\cdots y$, with $0!=1$. For instance, the probability of zero events is $e^{-\mu}$; zero is a valid and often important outcome, not a missing count.

The distribution is normalized because the exponential series gives

\sum_{y=0}^{\infty}\frac{e^{-\mu}\mu^y}{y!} =e^{-\mu}\sum_{y=0}^{\infty}\frac{\mu^y}{y!} =e^{-\mu}e^\mu=1.

If $y_1,\ldots,y_n$ are conditionally independent counts with specified means $\mu_1,\ldots,\mu_n$, their conditional likelihood is the product of their probability masses:

L(\boldsymbol{\mu}\mid\mathbf{y}) =\prod_{i=1}^{n}\frac{e^{-\mu_i}\mu_i^{y_i}}{y_i!}.

The likelihood treats the observed counts as fixed and expresses how their probability varies with the model parameters through the means. Taking logarithms turns the product into a sum:

\ell(\boldsymbol{\mu}\mid\mathbf{y}) =\sum_{i=1}^{n}\left(y_i\log\mu_i-\mu_i-\log(y_i!)\right).

In a GLMM, this is a conditional likelihood when the random effects are held fixed. A marginal likelihood instead averages over their distribution. Because observations in one cluster share a random effect, marginal independence generally does not follow from conditional independence.

Worked Example: Probability of an Exact Count

Suppose, hypothetically, that a defined observation unit has a Poisson count with mean $\mu=2$ events. Find the probability of exactly three events.

The support includes $y=3$, and the probability mass function gives

P(Y=3)=\frac{e^{-2}2^3}{3!} =\frac{8}{6}e^{-2} =\frac{4}{3}e^{-2} \approx 0.1804.

Thus, under this specified Poisson model, the probability of exactly three events for one such unit is about 0.1804. This is a probability for an exact count, not the probability of three or fewer events.

Why the Mean and Variance Are Equal

The equality of the Poisson mean and variance is a defining implication of the distribution, not a rule that holds for every count variable. It can be derived directly from the probability mass function. For $Y\sim\operatorname{Poisson}(\mu)$, first calculate the mean:

\begin{aligned} E(Y) &=\sum_{y=0}^{\infty}y\frac{e^{-\mu}\mu^y}{y!}\\ &=\mu e^{-\mu}\sum_{y=1}^{\infty}\frac{\mu^{y-1}}{(y-1)!}\\ &=\mu e^{-\mu}e^\mu=\mu. \end{aligned}

For the variance, calculate the second factorial moment. The terms for $y=0$ and $y=1$ are zero, so

\begin{aligned} E\{Y(Y-1)\} &=\sum_{y=2}^{\infty}y(y-1)\frac{e^{-\mu}\mu^y}{y!}\\ &=\mu^2 e^{-\mu}\sum_{y=2}^{\infty}\frac{\mu^{y-2}}{(y-2)!} =\mu^2. \end{aligned}

Since $Y^2=Y(Y-1)+Y$, it follows that $E(Y^2)=\mu^2+\mu$. Therefore,

\operatorname{Var}(Y)=E(Y^2)-\{E(Y)\}^2 =(\mu^2+\mu)-\mu^2=\mu.

Consequently, the conditional Poisson assumption in a GLMM is

E(Y_{ij}\mid b_j)=\mu_{ij}, \qquad \operatorname{Var}(Y_{ij}\mid b_j)=\mu_{ij}.

This conditional equality does not assert that the marginal variance, after allowing the random effects to vary, equals the marginal mean. The conditional mean itself varies with the random effect, so marginal variation can include both within-cluster count variation and variation between clusters.

Worked Example: Likelihood for Several Counts

Suppose, hypothetically, that three conditionally independent observational units have counts $0,2,3$, and each has the same specified Poisson mean $\mu=2$. Calculate their joint likelihood and log-likelihood at that mean.

Conditional independence means that the joint likelihood is the product of the three probability masses:

\begin{aligned} L &=P(Y_1=0)P(Y_2=2)P(Y_3=3)\\ &=\left(e^{-2}\right) \left(\frac{2^2e^{-2}}{2!}\right) \left(\frac{2^3e^{-2}}{3!}\right)\\ &=\frac{8}{3}e^{-6} \approx 0.006610. \end{aligned}

The same result follows from the product likelihood: the counts sum to $5$, the means sum to $6$, and the product of the factorial terms is $0!\,2!\,3!=12$, giving $2^5e^{-6}/12=(8/3)e^{-6}$. The log-likelihood is

\ell=\log(L)=\log(8/3)-6\approx -5.019.

The likelihood value is not a posterior probability that $\mu=2$ is true. It is the joint probability mass assigned to these observed counts at the specified mean.

Conditional Likelihood in a Random-Intercept Model

A random intercept allows the conditional mean count to differ across clusters. For example, in the log-mean model $\log(\mu_{ij})=\beta_0+b_j$, two observations in a cluster share the same $b_j$. Conditional on that value, each observation has its own Poisson count distribution and the observations are assumed independent. The shared effect changes their means together.

Worked Example: Cluster-Specific Conditional Means

Consider hypothetical counts $0,2$ in cluster A and $1,4$ in cluster B. For illustration, set $\beta_0=\log 2$, $b_A=0$, and $b_B=\log 2$. Treat these random-effect values as fixed for this conditional calculation; this is not an estimate of model parameters.

For cluster A, $\log(\mu_{iA})=\log 2$, so each conditional mean is $2$. For cluster B, $\log(\mu_{iB})=\log 2+\log 2=\log 4$, so each conditional mean is $4$. Under conditional independence, the likelihood of the four counts is

\begin{aligned} L(\mathbf{y}\mid b_A,b_B) &=\left(e^{-2}\right) \left(\frac{2^2e^{-2}}{2!}\right) \left(4e^{-4}\right) \left(\frac{4^4e^{-4}}{4!}\right)\\ &=(2e^{-4})\left(\frac{128}{3}e^{-8}\right) =\frac{256}{3}e^{-12} \approx 0.0005243. \end{aligned}

The first parenthesized factor is the likelihood for cluster A; the second is the likelihood for cluster B. This product is conditional on the two random effects. A marginal likelihood would require averaging over their distribution rather than substituting these illustrative values.

Assumptions, Diagnostics, and Alternatives

The Poisson specification is appropriate only when the response is a count on the stated observational unit and the model's conditional distribution is plausible. Its principal assumptions here are: counts take nonnegative integer values; each conditional mean is positive; the Poisson conditional variance equals the conditional mean; and observations are conditionally independent given the random effects and modeled predictors. These assumptions describe a statistical model, not a claim that every real event process follows a Poisson distribution exactly.

Useful checks compare observed counts with model-based conditional predictions, including the frequency of zeros and the upper tail. Residual patterns, substantially more conditional variation than the fitted means imply, or repeated clusters of unusually large counts can signal that the specified model is not capturing important structure. Such patterns do not by themselves identify the cause: dependence, omitted predictors, an unsuitable random-effects structure, and a response distribution with greater variation are among the possibilities.

A frequent mistake is to reason that a count is Poisson merely because it is an integer and has no upper bound in the data table. The support and the mean–variance assumption are both relevant; integer-valued data alone do not establish the Poisson model. Conversely, a Poisson distribution's theoretical support being unbounded does not mean a finite sample will display arbitrarily large values. If a count is constrained by a known finite number of opportunities, a binomial representation may better reflect that structure; the earlier material on Bernoulli and binomial data gives those likelihoods. Other count distributions can be considered when the Poisson conditional variance assumption is inadequate, but they do not change what the Poisson model assumes.

A further pitfall is to apply the equality of mean and variance to marginal counts in a mixed model. The equality derived above is conditional on the random effects. Variation in the random effects can make marginal counts more variable, and observations in the same cluster can be marginally associated. Diagnostics and interpretation therefore need to distinguish conditional from marginal quantities.

Key takeaway: A Poisson response assigns probabilities to every nonnegative integer count and, conditional on the model predictors and random effects, assumes that the count's variance equals its mean. In a GLMM, state clearly whether a probability, likelihood, mean, or variance is conditional or marginal.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

For $Y\sim\operatorname{Poisson}(3)$, write the support and calculate $P(Y=0)$.
For $Y\sim\operatorname{Poisson}(4)$, state $E(Y)$ and $\operatorname{Var}(Y)$. What does this equality assume?
Two conditionally independent counts have observed values $1$ and $2$, each with specified mean $1$. Write their joint likelihood.
In a random-intercept Poisson GLMM, why can two observations in the same cluster be associated marginally even when they are independent conditional on the random effect?

Solutions. (1) The support is $0,1,2,\ldots$, and $P(Y=0)=e^{-3}\approx0.04979$. (2) Both the mean and variance are $4$; this equality holds under the Poisson distribution conditional on any specified predictors and random effects. (3) The likelihood is $(e^{-1})(e^{-1}/1!)(e^{-1}/2!)=e^{-3}/2\approx0.02489$. (4) Both observations share the cluster random effect, which changes their conditional means together. Averaging over that shared effect can induce marginal association even though the model assumes conditional independence.