Conditional Poisson Models For Counts

Why Model Counts Conditionally?

Counts arise when the outcome records how many times an event occurs in a specified unit, such as the number of reported incidents for a hypothetical policy during one quarter. The expected count may differ with characteristics of that unit. A conditional model describes the distribution of the count given those characteristics; it does not say that every unit has the same expected count.

For unit $i$, let $Y_i$ be a count and let $\mathbf{x}_i$ denote its observed predictors. A Poisson model specifies a distribution for $Y_i$ at each given predictor value. Its mean may vary across units according to a mean model. In this tutorial, the mean model is written generally; choosing a particular link function is a separate modeling decision.

Definition: A response $Y_i$ has a conditional Poisson distribution with mean $\mu_i$ given predictors $\mathbf{x}_i$ if $$ P(Y_i=y\mid \mathbf{x}_i) = \frac{e^{-\mu_i}\mu_i^y}{y!}, \qquad y=0,1,2,\ldots, $$ where $\mu_i=\mu(\mathbf{x}_i;\boldsymbol{\beta})>0$ is the conditional mean specified by a mean model. The parameter $\mu_i$ is both the conditional expectation and conditional variance: $$ E[Y_i\mid \mathbf{x}_i]=\mu_i, \qquad \operatorname{Var}(Y_i\mid \mathbf{x}_i)=\mu_i. $$

The predictors can be categorical, numerical, or a combination. The notation $\mu(\mathbf{x}_i;\boldsymbol{\beta})$ says that the expected count is determined by the predictor values and model parameters $\boldsymbol{\beta}$. It does not, by itself, specify the form of that relationship. Because $\mu_i$ is a Poisson mean, it must be nonnegative; ordinary regression specifications use a strictly positive mean. A zero mean is a boundary case that assigns probability one to $Y_i=0$.

What the Poisson Assumption Implies

A count must be a nonnegative integer, and the Poisson probability mass function assigns probabilities only to those possible values. For any fixed predictor values, the probabilities over all counts sum to one. The mean $\mu_i$ need not be an integer: it is a long-run average for units with those predictor values, not a possible count that every unit must attain.

The equality of conditional mean and conditional variance is a substantive feature of the Poisson distribution. It is not a claim that the observed count equals its mean, nor that counts vary by exactly one. For a particular unit, a count can be well above or below $\mu_i$. The variance describes the spread across repeated outcomes under the same conditional distribution.

Model assumption: For each unit, conditional on its predictors, the count follows the stated Poisson distribution. If the analysis includes multiple units and treats their outcomes as independent given all modeled predictors, their joint conditional probability is the product of their individual conditional probabilities. Conditional Poisson distributions for each unit do not alone guarantee this independence.

The distinction between a distributional assumption and an independence assumption matters. Each count can have the correct conditional Poisson marginal distribution while two counts remain associated because they share an unmodeled source of variation. Independence is justified when, after conditioning on the predictors included in the model, no remaining shared randomness links the outcomes in the way relevant to the analysis.

Why the Conditional Mean and Variance Are Equal

The equality can be verified directly from the probability mass function. Fix the predictors and write $\mu$ for the resulting mean. The expectation is

E[Y\mid\mathbf{x}] = \sum_{y=1}^{\infty}y\frac{e^{-\mu}\mu^y}{y!} = \mu e^{-\mu}\sum_{y=1}^{\infty}\frac{\mu^{y-1}}{(y-1)!} = \mu.

For the second factorial moment, the $y=0$ and $y=1$ terms are zero, so

E[Y(Y-1)\mid\mathbf{x}] = \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.

Since $Y^2=Y(Y-1)+Y$, it follows that $E[Y^2\mid\mathbf{x}]=\mu^2+\mu$. Therefore,

\operatorname{Var}(Y\mid\mathbf{x}) = E[Y^2\mid\mathbf{x}]-\{E[Y\mid\mathbf{x}]\}^2 = (\mu^2+\mu)-\mu^2 = \mu.

The calculation uses the exponential series $\sum_{k=0}^{\infty}\mu^k/k!=e^\mu$. It applies to the Poisson distribution conditional on fixed predictor values. It does not establish that the variance equals the mean after averaging over varying predictors or omitted shared factors.

Worked Examples

Worked Example: Probabilities for a Single Conditional Count

For a hypothetical block of similar service contracts, suppose the number $Y$ of urgent calls in a quarter is Poisson with conditional mean $\mu=2.5$, given the recorded contract characteristics. Calculate $P(Y=0)$, $P(Y=1)$, $P(Y=2)$, and $P(Y\geq 3)$. State the implied conditional variance.

Method. Use the Poisson probability mass function for the first three probabilities, then subtract their sum from one for the tail. The model assumption gives $E[Y\mid\mathbf{x}]=\operatorname{Var}(Y\mid\mathbf{x})=2.5$.

For zero calls, $$P(Y=0\mid\mathbf{x})=e^{-2.5}\frac{2.5^0}{0!}=e^{-2.5}=0.082085\ldots$$ For one call, $$P(Y=1\mid\mathbf{x})=e^{-2.5}\frac{2.5}{1}=0.205212\ldots$$ For two calls, $$P(Y=2\mid\mathbf{x})=e^{-2.5}\frac{2.5^2}{2!} =e^{-2.5}\frac{6.25}{2} =e^{-2.5}(3.125) =0.256516\ldots$$

The probability of fewer than three calls is $$0.082085\ldots+0.205212\ldots+0.256516\ldots=0.543813\ldots$$ Thus, $$P(Y\geq3\mid\mathbf{x})=1-0.543813\ldots=0.456187\ldots$$

To four decimal places, the answers are $0.0821$, $0.2052$, $0.2565$, and $0.4562$, respectively. The first three probabilities sum to $0.5438$ after rounding, and the tail is its complement before rounding. The conditional variance is $2.5$. The common mistake avoided here is treating the non-integer mean $2.5$ as a possible count or rounding it before calculating probabilities.

Worked Example: Different Means for Different Predictor Values

A hypothetical monitoring program records one quarterly incident count for each of two locations. The specified mean model gives conditional means $\mu_1=1.2$ and $\mu_2=2.4$ for their respective predictor values. Assume the counts are conditionally independent given those values. Find the expected total count and the probability that both locations record at least one incident.

Method. Add conditional expectations to find the expected total. For the joint event, calculate each location’s probability of a positive count and multiply because conditional independence is explicitly assumed.

Let $T=Y_1+Y_2$. Linearity of expectation gives $$E[T\mid\mathbf{x}_1,\mathbf{x}_2]=1.2+2.4=3.6.$$ The individual zero-count probabilities are $$P(Y_1=0\mid\mathbf{x}_1)=e^{-1.2}=0.3010\ldots,$$ $$P(Y_2=0\mid\mathbf{x}_2)=e^{-2.4}=0.0907\ldots.$$ Therefore, the probabilities of at least one incident are $1-e^{-1.2}=0.6988\ldots$ and $1-e^{-2.4}=0.9093\ldots$. Conditional independence gives $$ P(Y_1\geq1,Y_2\geq1\mid\mathbf{x}_1,\mathbf{x}_2) =(1-e^{-1.2})(1-e^{-2.4}) =0.6354\ldots. $$

The expected total is $3.6$, and the joint probability is $0.6354$ to four decimal places. The expected total is valid by linearity even without independence; multiplying the two event probabilities is the step that requires independence. Assuming the locations are independent merely because each has a Poisson model would be an unjustified shortcut.

Worked Example: Conditional Independence Does Not Imply Marginal Independence

Two hypothetical facility counts $Y_1$ and $Y_2$ are independent conditional on a common risk category $Z$. The category is low or high, each with probability $0.5$. Conditional on low risk, both counts are Poisson with mean $1$; conditional on high risk, both are Poisson with mean $3$. Find each marginal mean and their marginal covariance. Check whether the counts are marginally independent.

Method. Average the conditional means to obtain each marginal mean. Use the law of total covariance, which adds the expected conditional covariance to the covariance of the conditional means.

For either count, $$E[Y_j]=0.5(1)+0.5(3)=2,\qquad j=1,2.$$ Conditional independence means $\operatorname{Cov}(Y_1,Y_2\mid Z)=0$. The conditional means are both $1$ when $Z$ is low and both $3$ when $Z$ is high. Hence $$ \operatorname{Cov}(Y_1,Y_2) = E[\operatorname{Cov}(Y_1,Y_2\mid Z)] + \operatorname{Cov}(E[Y_1\mid Z],E[Y_2\mid Z]) = 0+\operatorname{Var}(E[Y_1\mid Z]). $$ The conditional mean takes values $1$ and $3$ equally often, so $$ \operatorname{Var}(E[Y_1\mid Z]) = 0.5(1-2)^2+0.5(3-2)^2 = 0.5+0.5 = 1. $$ Thus the marginal covariance is $1$, not zero, and the counts are not marginally independent. The shared risk category explains their association. If $Z$ is included among the conditioned-on predictors, the stipulated conditional independence applies; if it is omitted, it does not.

Common Traps and Model Checks

A conditional model describes a distribution at specified predictor values. If predictor values vary across a population, the unconditional distribution is an average over those values and need not itself be Poisson. Likewise, a common unobserved factor can make observations dependent even when their conditional distributions are Poisson.

Confusing a mean with a count. A mean such as $1.7$ is an expected count, not a claim that $1.7$ events occur for an individual unit.
Assuming independence from marginal distributions. Specifying a Poisson distribution for each response does not specify their joint distribution. Conditional independence must be an additional assumption.
Multiplying probabilities without checking the conditioning. The product rule in the second example is justified by conditional independence given the stated predictors, not by the Poisson marginal distributions alone.
Forgetting the variance implication. Under the conditional Poisson model, the conditional variance equals the conditional mean. If this assumption is unsuitable, the stated model may not adequately represent the count variation.

Key takeaway: A conditional Poisson model specifies $Y_i\mid\mathbf{x}_i\sim\operatorname{Poisson}(\mu_i)$, with $\mu_i$ determined by a mean model and both the conditional mean and variance equal to $\mu_i$. For multiple observations, assess conditional independence separately; it is not automatic from the individual Poisson distributions.

Check Your Understanding

For each problem, use the stated conditional distributions and assumptions. Answers are rounded to four decimal places where probabilities are requested.

A hypothetical inspection count has a Poisson distribution with conditional mean $0.8$. Find the probability of zero or one inspection. (Answer: $0.8088$)
Three conditionally independent count responses have conditional means $0.5$, $1.5$, and $2.0$. Find the conditional expected total and the conditional variance of the total. (Answer: expected total $4.0$; variance $4.0$)
Two conditionally independent counts have means $0.7$ and $1.3$ for their predictor values. Find the probability that both are zero. (Answer: $e^{-2}=0.1353$)
A shared category $Z$ is equally likely to be either of two values. Conditional on $Z$, two counts are independent Poisson responses; both means are $0.5$ in the first category and $2.5$ in the second. Find the marginal mean of either count and the marginal covariance between them. (Answer: marginal mean $1.5$; marginal covariance $1.0$)
Two conditionally independent counts have means $0.6$ and $1.4$. Find the probability that their total is exactly one. (Answer: $2e^{-2}=0.2707$)

What You'll Learn