The Conditional Poisson Response Model

Why Specify a Count Distribution?

A regression model needs more than a rule connecting predictors to a response. It also needs a stochastic component: a probability distribution describing the response once the predictors are held fixed. For count outcomes—such as the number of visits, events, or defects recorded for each observational unit—the Poisson distribution is a basic model for nonnegative integer responses.

The conditional Poisson response model assigns a Poisson distribution to each response after conditioning on its covariates. This specification says what possible counts are allowed and how their probabilities are organized. It does not, by itself, specify how the distribution’s parameter changes with the covariates. That additional regression structure is a separate part of the model.

The Poisson Distribution for One Response

A count can be zero, but it cannot be negative or fractional. The Poisson distribution places probability on exactly these nonnegative integer values. Its parameter is a positive number that controls the distribution.

Definition: A random variable \(Y\) has a Poisson distribution with parameter \(\lambda>0\), written \(Y\sim\operatorname{Poisson}(\lambda)\), if its support is \(\{0,1,2,\ldots\}\) and its probability mass function is

P(Y=y)=\frac{e^{-\lambda}\lambda^y}{y!}, \qquad y=0,1,2,\ldots.

For values outside this support, the probability is zero. Here \(0!=1\).

The formula assigns a probability to every possible count. In particular, \(P(Y=0)=e^{-\lambda}\), and the probability of a particular positive count is found by substituting that count for \(y\). The factorial in the denominator is part of the probability mass function; omitting it generally gives an incorrect probability.

Worked Example: Calculating Probabilities for One Count

Suppose the number of calls received by a service desk during a specified hour is modeled as \(Y\sim\operatorname{Poisson}(2)\). Calculate the probabilities of zero, one, and three calls.

For zero calls, substitute \(y=0\) and \(\lambda=2\):

P(Y=0)=\frac{e^{-2}2^0}{0!}=e^{-2}\approx 0.1353.

For one call:

P(Y=1)=\frac{e^{-2}2^1}{1!}=2e^{-2}\approx 0.2707.

For three calls, the factorial is \(3!=6\):

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

These are probabilities of three separate outcomes, not a complete distribution: other counts, such as two or four calls, also have positive probability.

Conditional Poisson Responses

In regression, observations come with covariates. Write \(Y_i\) for the count response on observational unit \(i\), and \(x_i\) for that unit’s covariate information, for \(i=1,\ldots,n\). A conditional distribution describes the response while holding that information fixed. The Poisson parameter may differ from one covariate profile to another.

Definition: A conditional Poisson response model specifies, for each \(i=1,\ldots,n\), that

Y_i\mid x_i\sim\operatorname{Poisson}(\lambda_i), \qquad \lambda_i>0.

Equivalently, for \(y_i\in\{0,1,2,\ldots\}\),

P(Y_i=y_i\mid x_i) =\frac{e^{-\lambda_i}\lambda_i^{y_i}}{y_i!},

and this conditional probability is zero for \(y_i\notin\{0,1,2,\ldots\}\). The parameter \(\lambda_i\) can depend on the covariates \(x_i\). A regression model must specify that dependence separately.

This is the stochastic component of a regression model: it selects the distributional family and response support, and it gives the conditional mass function. To complete a regression model, one must also specify how the parameters \(\lambda_i\) are related to the covariates. The conditional Poisson specification alone does not choose that relationship, and it does not require every observation to have the same parameter.

The notation \(Y_i\mid x_i\) emphasizes that the distribution is conditional. If covariates are random variables, the same statement can be written as a conditional distribution given their observed values. The parameter \(\lambda_i\) is then understood as the value associated with unit \(i\)’s covariate information. Throughout, the model requires \(\lambda_i>0\) for every covariate profile to which the specification applies.

Worked Example: Different Covariate Profiles, Different Count Distributions

Suppose two patients have covariate profiles \(x_1\) and \(x_2\), and a model specification assigns them Poisson parameters \(\lambda_1=0.8\) and \(\lambda_2=2.4\), respectively. Find the probability of no events for the first patient and exactly two events for the second, conditional on both profiles.

For the first patient:

P(Y_1=0\mid x_1) =\frac{e^{-0.8}(0.8)^0}{0!} =e^{-0.8} \approx 0.4493.

For the second patient:

P(Y_2=2\mid x_2) =\frac{e^{-2.4}(2.4)^2}{2!} =\frac{5.76}{2}e^{-2.4} =2.88e^{-2.4} \approx 0.2613.

The two conditional distributions have the same count support but different parameters, so their probabilities for particular outcomes differ. These parameter values specify the distributions at the two profiles; by themselves they do not tell us what functional rule relates a parameter to a covariate.

Conditional Independence and the Joint Model

A model for each response’s conditional distribution does not, on its own, state how responses are related to each other. A common regression specification adds conditional independence: once the covariates are given, the responses are independent. This assumption yields a convenient joint conditional probability.

Definition: The responses \(Y_1,\ldots,Y_n\) are conditionally independent given their covariates if, for every set of count values \(y_1,\ldots,y_n\),

P(Y_1=y_1,\ldots,Y_n=y_n\mid x_1,\ldots,x_n) =\prod_{i=1}^{n}P(Y_i=y_i\mid x_1,\ldots,x_n),

whenever the conditional probabilities are defined. In the usual unit-specific model, each response’s conditional distribution depends on its own covariate information \(x_i\).

Theorem: Suppose that, conditional on the covariates \(x_1,\ldots,x_n\), the responses \(Y_1,\ldots,Y_n\) are independent and each has a Poisson distribution with parameter \(\lambda_i>0\). Then, for nonnegative integer values \(y_1,\ldots,y_n\),

P(Y_1=y_1,\ldots,Y_n=y_n\mid x_1,\ldots,x_n) =\prod_{i=1}^{n}\frac{e^{-\lambda_i}\lambda_i^{y_i}}{y_i!}.

If any \(y_i\) is outside the nonnegative integers, the joint probability is zero.

Proof. By conditional independence, the joint conditional probability is the product of the individual conditional probabilities. By the conditional Poisson specification, the \(i\)th factor is \(e^{-\lambda_i}\lambda_i^{y_i}/y_i!\) when \(y_i\) is a nonnegative integer. Substituting these \(n\) factors into the product gives the stated expression. If any value is outside the support, its individual conditional probability is zero, so the product is zero. \(\square\)

Worked Example: A Joint Probability Under Conditional Independence

Suppose two responses are conditionally independent given their covariates, with \(Y_1\mid x_1\sim\operatorname{Poisson}(1.5)\) and \(Y_2\mid x_2\sim\operatorname{Poisson}(0.5)\). Calculate the conditional probability that the first count is two and the second count is zero.

Apply the joint probability formula and substitute the two observed counts:

\begin{aligned} &P(Y_1=2,Y_2=0\mid x_1,x_2)\\ &\quad=P(Y_1=2\mid x_1)P(Y_2=0\mid x_2)\\ &\quad=\left(\frac{e^{-1.5}(1.5)^2}{2!}\right) \left(\frac{e^{-0.5}(0.5)^0}{0!}\right)\\ &\quad=\left(1.125e^{-1.5}\right)e^{-0.5}\\ &\quad=1.125e^{-2}\approx 0.1523. \end{aligned}

The factorization is justified by the stated conditional independence assumption. Without that assumption, the individual Poisson distributions alone would not determine this joint probability.

Why the Conditioning Matters

Conditional independence is not the same as unconditional independence. If covariates are random, two responses can be related marginally because their distributions both depend on related or shared covariate information, even when they are independent after conditioning. Thus, the independence assumption belongs to the conditional model, not automatically to the data before covariates are accounted for.

The same distinction matters when describing a study. Saying that observations are independent may be a reasonable modeling assumption for separate units after their covariates are fixed, but it is not a consequence of assigning each response a Poisson distribution. The stochastic specification and the dependence assumptions are distinct components that should be stated clearly.

Checking the Probability Mass Function

A probability mass function must assign nonnegative probabilities whose total over the support is one. The Poisson formula meets this requirement for every \(\lambda>0\). Its normalization follows directly from the exponential series.

Result: For \(\lambda>0\), the Poisson probabilities sum to one:

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

Proof. Since \(e^{-\lambda}\) is a positive constant, factor it out of the sum. The power-series expansion \(e^\lambda=\sum_{y=0}^{\infty}\lambda^y/y!\) converges for every real \(\lambda\). Therefore,

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

All terms are nonnegative, and \(\lambda>0\) makes the displayed probabilities well-defined. \(\square\)

Common Mistakes

Allowing impossible count values. A Poisson response takes values \(0,1,2,\ldots\), not negative values or fractions. A probability assigned to a negative count is zero.
Assuming all units have the same distribution. The response distributions may have different parameters \(\lambda_i\), depending on the covariates. A common Poisson family does not require identical probabilities across units.
Inferring independence from the marginal distributions. Individual conditional Poisson specifications do not imply conditional independence. The joint assumption must also be made before multiplying the individual probabilities.
Treating the distributional choice as a complete regression model. The Poisson specification gives the response distribution conditional on covariates, but it does not specify how \(\lambda_i\) varies with those covariates.
Forgetting the factorial. For a count \(y\), the denominator is \(y!\). For instance, the probability of two events includes \(2!=2\), not \(2\) omitted from the formula.

Key takeaway: The conditional Poisson response model specifies a nonnegative integer response and its probability mass function given covariates. A full regression specification additionally describes how its positive Poisson parameter depends on those covariates and, when multiple responses are modeled, states any conditional dependence assumptions.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

For a Poisson response, what values belong to the support, and what is the probability mass function?
If \(Y\mid x\sim\operatorname{Poisson}(1)\), calculate \(P(Y=0\mid x)\) and \(P(Y=2\mid x)\).
Two responses each have a conditional Poisson distribution. What additional assumption justifies multiplying their individual conditional probabilities to obtain a joint probability?
Why does specifying a conditional Poisson distribution not, by itself, specify a complete regression model?
Explain why conditional independence given covariates does not necessarily imply unconditional independence.