Equidispersion And Support In Poisson Regression

What Poisson Regression Assumes About Counts

Poisson regression describes a count response through its distribution at each covariate value. Two features of that distribution are central: the response takes nonnegative integer values, and its conditional variance equals its conditional mean. These are precise distributional assumptions, not statements that every observed data set must have a sample variance exactly equal to its sample mean.

The distinction between conditional and marginal behavior matters. A model can be equidispersed at every fixed covariate profile while counts pooled across profiles have variance greater than their mean. To interpret the model correctly, first identify the response’s state space and then keep the conditioning on predictors explicit.

Definition: In Poisson regression, for a response $Y$ and covariates $X=x$, the conditional model is $Y\mid X=x\sim\operatorname{Poisson}(\lambda(x))$, where $\lambda(x)\geq 0$. Its conditional probability mass function is $$ P(Y=y\mid X=x)=e^{-\lambda(x)}\frac{\lambda(x)^y}{y!},\qquad y=0,1,2,\ldots, $$ with the probability at $y=0$ defined as $1$ when $\lambda(x)=0$. The conditional state space is the set of nonnegative integers. If $\lambda(x)>0$, every value in that state space has positive probability, so the support is the entire set. If $\lambda(x)=0$, the support is $\{0\}$.

In the usual log-link specification, $\lambda(x)=\exp(\eta(x))$, where $\eta(x)$ is a finite linear predictor. Then $\lambda(x)>0$ for every finite covariate profile, so the conditional support is all nonnegative integers. The zero-mean case is still useful to state because it is a valid boundary Poisson distribution, even though a finite linear predictor under the log link cannot produce it.

Conditional Mean and Variance

The equality of the conditional mean and variance is called equidispersion. In a regression model, it is a statement made separately at each covariate value: the mean and variance are both determined by $\lambda(x)$. The following derivation includes the boundary case $\lambda(x)=0$.

Theorem: If $Y\mid X=x\sim\operatorname{Poisson}(\lambda(x))$, then $$ E[Y\mid X=x]=\lambda(x),\qquad \operatorname{Var}(Y\mid X=x)=\lambda(x). $$

Handle a zero mean.
If $\lambda(x)=0$, then $Y=0$ with probability one. Its conditional mean and variance are both zero, as claimed.

Compute the mean for a positive mean.
Write $\lambda=\lambda(x)>0$. From the probability mass function, $$ E[Y\mid X=x]=\sum_{y=1}^{\infty}y e^{-\lambda}\frac{\lambda^y}{y!} =\lambda e^{-\lambda}\sum_{y=1}^{\infty}\frac{\lambda^{y-1}}{(y-1)!} =\lambda. $$ The final series is $e^\lambda$, by the exponential series.

Compute the second factorial moment.
Similarly, $$ E[Y(Y-1)\mid X=x] =\sum_{y=2}^{\infty}y(y-1)e^{-\lambda}\frac{\lambda^y}{y!} =\lambda^2 e^{-\lambda}\sum_{y=2}^{\infty}\frac{\lambda^{y-2}}{(y-2)!} =\lambda^2. $$

Convert to the variance.
Since $Y^2=Y(Y-1)+Y$, the conditional second moment is $\lambda^2+\lambda$. Therefore $$ \operatorname{Var}(Y\mid X=x) =E[Y^2\mid X=x]-E[Y\mid X=x]^2 =(\lambda^2+\lambda)-\lambda^2 =\lambda. $$ This proves both claims. $\square$

For a response with conditional mean $\lambda(x)=4$, the model therefore specifies a conditional variance of $4$, not $4^2$. The corresponding conditional standard deviation is $2$. A larger mean permits a larger spread in the count scale, but the variance is fixed by the mean rather than estimated as an independent conditional parameter in the ordinary Poisson model.

Worked Example: Support and Probabilities at a Fitted Mean

Suppose a fitted Poisson regression gives $\lambda(x)=4$ for a particular covariate profile. Identify the support and calculate the conditional mean, variance, and two count probabilities.

Because $4>0$, every nonnegative integer has positive conditional probability, and no negative or fractional value does. By the mean–variance result, $$ E[Y\mid X=x]=4,\qquad \operatorname{Var}(Y\mid X=x)=4. $$ For zero counts, $$ P(Y=0\mid X=x)=e^{-4}\frac{4^0}{0!}=e^{-4}\approx 0.0183. $$ For two counts, $$ P(Y=2\mid X=x)=e^{-4}\frac{4^2}{2!}=8e^{-4}\approx 0.1465. $$ These probabilities concern individual possible outcomes; they do not imply that a finite sample must contain exactly the number of zeroes or twos obtained by multiplying these probabilities by its size.

Worked Example: The Boundary Case of a Zero Mean

Consider the Poisson distribution with $\lambda(x)=0$. What values can $Y$ take with positive probability?

At $y=0$, the boundary convention gives $P(Y=0\mid X=x)=1$. For every integer $y\geq 1$, the numerator contains $0^y=0$, so $P(Y=y\mid X=x)=0$. Thus the state space remains the nonnegative integers, but the support—the values with positive probability—is only $\{0\}$. In this case $E[Y\mid X=x]=0$ and $\operatorname{Var}(Y\mid X=x)=0$.

This is distinct from a very small positive mean. If $\lambda(x)>0$, however small, every nonnegative integer still has positive probability. Under a log link, a finite linear predictor yields a positive mean; an exact zero mean is not obtained by setting a finite linear predictor to a value.

What Equidispersion Does—and Does Not—Say

Equidispersion is conditional on the predictors in the model. It does not assert that all observations have the same mean, since $\lambda(x)$ can vary across covariate profiles. Nor does it assert that the empirical variance in a particular finite sample must equal the empirical mean. A sample is random: even if the model is correct, its observed variance may lie above or below its observed mean.

The assumption does impose a relationship on the conditional distribution. At a fixed $x$, the conditional variance is not free to be much smaller or larger than the conditional mean. Consequently, a count such as $2.7$ cannot be an individual response under this model, and a negative count is also outside its state space. Data recorded in those forms require a different response definition or a different probability model.

A sample variance substantially larger than its sample mean can alert an analyst to possible lack of fit, but it is not by itself a proof that conditional equidispersion fails. The comparison may pool observations with different fitted means, and random sampling also creates variation in the comparison. The conditional structure must be considered before interpreting a pooled summary.

Theorem: Suppose $X$ is random and, conditional on $X$, $Y\mid X\sim\operatorname{Poisson}(\lambda(X))$. If $E[\lambda(X)]<\infty$ and $\operatorname{Var}(\lambda(X))<\infty$, then $$ E[Y]=E[\lambda(X)],\qquad \operatorname{Var}(Y)=E[\lambda(X)]+\operatorname{Var}(\lambda(X)). $$ Thus the marginal variance equals the marginal mean plus the variance of the conditional mean.

Proof. The conditional mean and variance are both $\lambda(X)$, by the result above. The law of total expectation gives $$ E[Y]=E\{E[Y\mid X]\}=E[\lambda(X)]. $$ The law of total variance gives $$ \operatorname{Var}(Y) =E\{\operatorname{Var}(Y\mid X)\} +\operatorname{Var}\{E[Y\mid X]\} =E[\lambda(X)]+\operatorname{Var}(\lambda(X)). $$ The stated moment conditions ensure that these quantities are finite. Since $\operatorname{Var}(\lambda(X))\geq 0$, the marginal variance is at least the marginal mean, and it is strictly greater when $\lambda(X)$ is not almost surely constant. $\square$

This result explains an important distinction: mixing Poisson counts with different conditional means can produce marginal overdispersion, even when each conditional distribution satisfies equidispersion exactly. “The variance equals the mean” is therefore not generally true after ignoring predictors.

Worked Example: Conditional Equidispersion but Marginal Overdispersion

Let $X$ take values $a$ and $b$, each with probability $1/2$. Suppose $Y\mid X=a\sim\operatorname{Poisson}(1)$ and $Y\mid X=b\sim\operatorname{Poisson}(3)$. Find the marginal mean and variance and compare them.

The conditional means are $1$ and $3$, so $$ E[\lambda(X)]=\frac12(1)+\frac12(3)=2. $$ The variance of the conditional mean is $$ \operatorname{Var}(\lambda(X)) =\frac12(1-2)^2+\frac12(3-2)^2=1. $$ Each conditional distribution is equidispersed: its variance is respectively $1$ or $3$. But the marginal variance is $$ \operatorname{Var}(Y)=E[\lambda(X)]+\operatorname{Var}(\lambda(X))=2+1=3. $$ Therefore the marginal variance, $3$, exceeds the marginal mean, $2$. The marginal probability of zero also illustrates that pooling does not simply yield a Poisson distribution with mean $2$: $$ P(Y=0)=\frac12e^{-1}+\frac12e^{-3}\approx 0.2088, \qquad e^{-2}\approx 0.1353. $$ The first value is the probability under the mixture; the second is the zero probability for a single $\operatorname{Poisson}(2)$ distribution. Variation in the conditional means changes the pooled distribution.

Reading Count Data Through the Model

The support and equidispersion assumptions help clarify what a Poisson regression fit means. At each covariate profile, it supplies a distribution over possible integer counts, with probabilities determined by a single mean parameter. The fitted mean is not merely a point prediction: under the model it also determines the conditional variance and the probability of every possible count.

For applied work, keep these checks separate:

Response scale: confirm that each response is a count of events, recorded as a nonnegative integer. A rate, proportion, or continuous measurement is not itself a Poisson count response.
Conditional interpretation: compare variability with the mean after accounting for the predictors, rather than treating a pooled mean–variance comparison as decisive.
Boundary interpretation: distinguish a zero observed count, which is an ordinary possible outcome at positive means, from a zero conditional mean, which gives probability one to a zero response.
Sampling variation: do not expect a finite sample’s empirical mean and variance to match exactly just because a Poisson model is assumed.

A practical pitfall is to conclude, “The sample variance is larger than the sample mean, so the Poisson model is impossible.” That reasoning treats a random sample summary as if it were the exact conditional population variance and may also ignore variation in fitted means across covariates. The correction is to distinguish the model’s conditional moment statement from pooled sample summaries, then assess whether the chosen conditional model adequately represents the data.

Key takeaway: The Poisson regression response has nonnegative-integer state space; its support is all nonnegative integers at a positive conditional mean and only $\{0\}$ at a zero mean. Equidispersion applies conditionally, while pooling different conditional means can create a marginal variance larger than the marginal mean.

Check Your Understanding

Answer each question using the distinction between conditional and marginal behavior.

For a Poisson response with positive conditional mean, what is its support, and why is a fractional observed response incompatible with that model?
If $\lambda(x)=0$, what is the support? How does it differ from the state space?
At a covariate profile with conditional mean $5$, what are the conditional variance and standard deviation?
Why can counts pooled across two covariate profiles have marginal variance greater than marginal mean even when each conditional distribution is Poisson?
Does an observed sample variance larger than its sample mean by itself establish that conditional equidispersion fails? Explain.

Equidispersion And Support In Poisson Regression

What You'll Learn