Poisson Counts With Unequal Exposure

Why Exposure Changes the Meaning of a Count

A count of 10 events can be ordinary over a long observation period and surprising over a short one. The count alone does not tell us how much variation to expect: the expected count depends on how much opportunity there was for events to occur. In a Poisson model, known exposure—such as person-time, area sampled, or time observed—enters the expected count, and therefore also enters the scale used to judge a residual.

The key distinction is between a count and a rate. Counts from observations with different exposures are not generally expected to have the same means, even when the underlying event rate is the same. A dispersion check must compare each observed count with its own fitted expected count, not with a common count benchmark.

Definition: Let $Y_i$ be a count observed with known positive exposure $E_i$. A Poisson GLM with an exposure offset specifies $$ Y_i\mid x_i,E_i \sim \operatorname{Poisson}(\mu_i), \qquad \log(\mu_i)=\log(E_i)+x_i^\mathsf{T}\beta. $$ Here $x_i^\mathsf{T}\beta$ models the log event rate, while $\log(E_i)$ is a known offset with coefficient fixed at 1. Thus $\mu_i=E_i\exp(x_i^\mathsf{T}\beta)$. The model assumes $E_i>0$; a zero-exposure observation has a deterministically zero count and is not informative about the rate in this formulation.

For example, if $E_i$ is person-years, then $\exp(x_i^\mathsf{T}\beta)$ is the modeled event rate per person-year and $\mu_i$ is the expected number of events over the observed person-years. Changing exposure units changes the numerical rate and intercept consistently, but it does not turn the offset into an estimated covariate effect.

Residuals Use the Fitted Expected Count

For an observation with fitted mean $\hat{\mu}_i$, the raw count difference is $Y_i-\hat{\mu}_i$. Its size is not directly comparable across observations: under a Poisson model, the variance is $\mu_i$, so counts with larger means naturally have larger typical raw deviations. Two common residuals put the discrepancy on scales that account for the fitted mean.

Definition: The Pearson residual and its contribution to the Pearson statistic are $$ r_{P,i}=\frac{Y_i-\hat{\mu}_i}{\sqrt{\hat{\mu}_i}}, \qquad X_P^2=\sum_{i=1}^n r_{P,i}^2 =\sum_{i=1}^n\frac{(Y_i-\hat{\mu}_i)^2}{\hat{\mu}_i}. $$ For a Poisson model, the signed deviance residual is $$ r_{D,i}=\operatorname{sign}(Y_i-\hat{\mu}_i) \sqrt{2\left\{Y_i\log\left(\frac{Y_i}{\hat{\mu}_i}\right)-(Y_i-\hat{\mu}_i)\right\}}, $$ where the term $Y_i\log(Y_i/\hat{\mu}_i)$ is defined as 0 when $Y_i=0$. The residual deviance is the sum of the squared deviance residuals.

Both residuals depend on exposure through $\hat{\mu}_i$. The Pearson residual divides by the fitted Poisson standard deviation, $\sqrt{\hat{\mu}_i}$. The deviance residual measures the count’s likelihood-based discrepancy from that fitted mean. Neither residual should be calculated by comparing every count with the same mean merely because the observations are all counts.

Worked Example: The Same Rate With Different Exposures

Suppose two independent Poisson counts have exposures $E_1=2$ and $E_2=10$, with a known common rate of 2 events per exposure unit. The expected counts are $\mu_1=2(2)=4$ and $\mu_2=2(10)=20$. The observed counts are $Y_1=5$ and $Y_2=18$.

The raw deviations are $5-4=1$ and $18-20=-2$. The second raw deviation is twice as large in absolute value, but its expected count is five times as large. The Pearson residuals are $$ r_{P,1}=\frac{1}{\sqrt{4}}=0.5, \qquad r_{P,2}=\frac{-2}{\sqrt{20}}\approx-0.447. $$

Their squared contributions are $0.25$ and $4/20=0.20$, so $X_P^2=0.45$. These residuals show that the two count discrepancies are similar in size relative to their Poisson variability. Looking only at the raw deviations would obscure that comparison.

Two Consequences of the Poisson Variance

The basic scaling can be stated precisely when the means are known. It explains why unequal expected counts do not, by themselves, indicate unequal dispersion.

Result: If $Y_i$ are independent Poisson random variables with known means $\mu_i>0$, then $R_i=(Y_i-\mu_i)/\sqrt{\mu_i}$ has mean 0 and variance 1. Consequently, the Pearson sum $Q=\sum_{i=1}^n R_i^2$ has expectation $n$, regardless of whether the $\mu_i$ are equal.

Proof. A Poisson random variable with mean $\mu_i$ has $E[Y_i]=\mu_i$ and $\operatorname{Var}(Y_i)=\mu_i$. Therefore $$ E[R_i]=\frac{E[Y_i]-\mu_i}{\sqrt{\mu_i}}=0, \qquad \operatorname{Var}(R_i)=\frac{\operatorname{Var}(Y_i)}{\mu_i}=1. $$ Since $E[R_i^2]=\operatorname{Var}(R_i)+(E[R_i])^2=1$, linearity of expectation gives $E[Q]=\sum_{i=1}^n E[R_i^2]=n$. Independence is not needed for this expectation calculation, although it is part of the usual model for the observations. $\square$

This result concerns known means. In a fitted GLM, the same observations help determine $\hat{\mu}_i$, so the fitted Pearson statistic does not generally have expectation exactly $n$. Dispersion ratios often compare $X_P^2$ with residual degrees of freedom as a reference summary, but that comparison is not an exact identity for every fitted model. The fitted means, parameter estimation, and model structure matter.

Exposure also changes the precision of observed rates. For $Y\sim\operatorname{Poisson}(E\lambda)$, where $E>0$ is fixed and $\lambda$ is the event rate, $$ E\left[\frac{Y}{E}\right]=\lambda, \qquad \operatorname{Var}\left(\frac{Y}{E}\right) =\frac{\operatorname{Var}(Y)}{E^2} =\frac{E\lambda}{E^2} =\frac{\lambda}{E}. $$ Thus, a rate calculated from greater exposure has lower variance under the same Poisson rate. Dividing a count by exposure gives a rate, but it does not make rates from different exposures equally precise.

Worked Example: Equal Raw Deviations, Unequal Residual Evidence

Compare two observations whose fitted Poisson means are 4 and 100. Suppose the observed counts are 8 and 104, respectively. Both counts exceed their fitted means by 4.

For the first observation, the Pearson residual is $(8-4)/\sqrt{4}=2$, with squared contribution $4$. For the second it is $(104-100)/\sqrt{100}=0.4$, with squared contribution $0.16$. An equal raw deviation therefore has very different standardized size.

The corresponding deviance contributions are also different. For $Y=8,\hat{\mu}=4$, the squared deviance residual is $$ 2\left\{8\log(8/4)-(8-4)\right\} =2(8\log 2-4) \approx 3.090. $$ For $Y=104,\hat{\mu}=100$, it is $$ 2\left\{104\log(104/100)-(104-100)\right\} =2\{104\log(1.04)-4\} \approx 0.158. $$ The larger expected count makes a difference of 4 relatively modest. These calculations illustrate why residual contributions must be judged against their own fitted means.

Fitting a Common Rate Across Unequal Exposures

A simple model makes the role of exposure especially clear. Suppose the observations are independent and share one unknown rate $\lambda>0$, so $Y_i\sim\operatorname{Poisson}(E_i\lambda)$, with fixed known $E_i>0$. The likelihood uses each exposure in its expected count. Maximizing it pools events and exposure, rather than averaging the individual observed rates equally.

Result: Under the independent common-rate model $Y_i\sim\operatorname{Poisson}(E_i\lambda)$, if $T=\sum_iY_i>0$, the maximum-likelihood estimate is $$ \hat{\lambda}=\frac{\sum_{i=1}^nY_i}{\sum_{i=1}^nE_i}. $$ The fitted means are $\hat{\mu}_i=E_i\hat{\lambda}$.

Proof. Write $S=\sum_iE_i$ and $T=\sum_iY_i$. Independence gives the likelihood $$ L(\lambda)=\prod_{i=1}^n \frac{\exp(-E_i\lambda)(E_i\lambda)^{Y_i}}{Y_i!}. $$ For $\lambda>0$, its log-likelihood, omitting no terms that depend on $\lambda$, is $$ \ell(\lambda)=\sum_iY_i\log(\lambda)+\sum_iY_i\log(E_i)-\lambda S-\sum_i\log(Y_i!). $$ Its first derivative is $T/\lambda-S$, and its second derivative is $-T/\lambda^2<0$ when $T>0$. The unique stationary point is $\hat{\lambda}=T/S$; strict concavity makes it the maximum. Substituting into $E_i\lambda$ gives $\hat{\mu}_i=E_i\hat{\lambda}$. If $T=0$, the likelihood decreases as $\lambda$ increases, and its maximum over the closure $\lambda\geq0$ is at the boundary $\hat{\lambda}=0$. $\square$

Worked Example: Fitting a Rate Instead of a Common Count Mean

Three independent observations have exposures $E=(1,2,5)$ and counts $Y=(2,3,10)$. Under a common-rate Poisson model, the total count is $T=2+3+10=15$ and total exposure is $S=1+2+5=8$. Hence $$ \hat{\lambda}=\frac{15}{8}=1.875. $$ The fitted expected counts are $1(1.875)=1.875$, $2(1.875)=3.75$, and $5(1.875)=9.375$.

The Pearson contributions are $$ \frac{(2-1.875)^2}{1.875} =\frac{0.015625}{1.875}\approx0.0083, \qquad \frac{(3-3.75)^2}{3.75} =\frac{0.5625}{3.75}=0.15, $$ and $$ \frac{(10-9.375)^2}{9.375} =\frac{0.390625}{9.375}\approx0.0417. $$ Their sum is $X_P^2\approx0.200$. This is a calculation under the fitted rate model, not proof that the data are less variable than Poisson; with fitted parameters, the distribution and reference for a dispersion summary are not given by the known-means result alone.

For contrast, assigning every observation the common fitted count mean $15/3=5$ would ignore exposure. That incorrect benchmark gives $$ \frac{(2-5)^2+(3-5)^2+(10-5)^2}{5} =\frac{9+4+25}{5}=7.6. $$ The large difference between 7.6 and 0.200 is caused here by using different mean structures, not by a change to the observed counts. Exposure must be included before interpreting a dispersion summary.

Interpreting Dispersion Summaries Carefully

A Pearson statistic or residual deviance summarizes discrepancies across observations, but neither removes the need to specify the expected counts correctly. With unequal exposures, even a common event rate implies unequal Poisson means and unequal raw-count variability. If the fitted means omit the offset, systematic differences in opportunity can be counted as discrepancies from an inappropriate benchmark.

For a fitted Poisson model, inspect whether the exposure is measured and coded consistently, whether it belongs as a fixed offset, and whether each residual uses its own $\hat{\mu}_i$. A large Pearson contribution means an observation is far from its fitted count relative to the fitted Poisson scale; it is not, by itself, an explanation for why it is far away. A dispersion ratio is a summary over observations, not a substitute for a sound mean model.

Common mistake: “The counts differ a lot, so the data must be overdispersed.” This compares raw counts without accounting for their expected values. The correction is to model exposure in $\mu_i$, calculate residuals using those fitted means, and then interpret the resulting summary with the fact of parameter fitting in mind. Unequal expected counts are part of the Poisson model, not evidence of extra variation on their own.

Check Your Understanding

Use the fitted-mean scale, rather than raw counts alone, when answering.

In a Poisson GLM with exposure $E_i$, what term is added to the linear predictor, and why is its coefficient fixed?
Two observations have the same raw deviation from their fitted means, but one has a much larger fitted mean. Which is likely to have the larger absolute Pearson residual, and why?
For a Poisson count with mean $\mu$, derive the variance of the observed rate when the exposure is $E$.
Why is the Pearson statistic’s expectation exactly $n$ in the known-means result, but not generally exactly $n$ after fitting a GLM?
For counts $Y=(4,6)$ with exposures $E=(2,4)$ under a common-rate model, calculate the maximum-likelihood rate and the two fitted means.