Exposure And Influence In Poisson Regression

Exposure Separates Rates From Counts

A Poisson regression often models event counts observed over unequal amounts of time, population, or opportunity. Ten events over one day and ten events over one year are not comparable as estimates of an underlying rate. The exposure records how much opportunity each observation had to generate events; the model uses it to connect the expected count to the rate.

This distinction also matters for influence. Exposure affects how much information an observation contributes to estimating rates. But a fitted count is a rate multiplied by the target exposure, so the same change in an estimated rate can imply very different changes in predicted counts.

Definition: Let $Y_i$ be an observed count, $t_i>0$ a known exposure, and $x_i$ a $p$-vector of covariates, for $i=1,\ldots,n$. The Poisson log-link rate model is $$ Y_i \mid x_i,t_i \sim \operatorname{Poisson}(\mu_i), \qquad \mu_i=t_i\lambda_i,\qquad \log \lambda_i=x_i^\mathsf{T}\beta. $$ Equivalently, $\log \mu_i=\log t_i+x_i^\mathsf{T}\beta$, where $\log t_i$ is a known offset, not an estimated coefficient. The expected rate per exposure unit is $\lambda_i=\exp(x_i^\mathsf{T}\beta)$; the expected count is $\mu_i=t_i\lambda_i$.

Assume the observations are conditionally independent, the design matrix $X$ with rows $x_i^\mathsf{T}$ has full column rank, and the fitted model has a finite maximum-likelihood estimate $\hat\beta$. The log-likelihood, apart from no omitted terms, is

\ell(\beta)=\sum_{i=1}^n \left\{y_i\bigl(\log t_i+x_i^\mathsf{T}\beta\bigr) -t_i\exp(x_i^\mathsf{T}\beta)-\log(y_i!)\right\}.

The score contribution from case $i$ is $x_i(y_i-\mu_i)$. Thus exposure is not an extra response or a case weight entered by hand: it appears in the fitted mean and consequently in the score and information. The rate observation $Y_i/t_i$ has conditional expectation $\lambda_i$ and variance $\lambda_i/t_i$. Larger exposure therefore makes the observed rate more precise under the Poisson model.

Information Weights and Leverage

For the log link, differentiating the log-likelihood gives the score and observed information

U(\beta)=\sum_{i=1}^n x_i(y_i-\mu_i),\qquad \mathcal I(\beta)=X^\mathsf{T}WX,\qquad W=\operatorname{diag}(\mu_1,\ldots,\mu_n).

At the fitted model, write $\hat\mu_i=t_i\exp(x_i^\mathsf{T}\hat\beta)$ and $\hat W=\operatorname{diag}(\hat\mu_i)$. The GLM hat matrix in weighted least-squares geometry is $H=\hat W^{1/2}X(X^\mathsf{T}\hat W X)^{-1}X^\mathsf{T}\hat W^{1/2}$. Its diagonal $h_i$ is the leverage of case $i$.

Result: Provided $X$ has full column rank and each fitted mean is positive, the Poisson leverage is $$ h_i=\hat\mu_i\,x_i^\mathsf{T}(X^\mathsf{T}\hat W X)^{-1}x_i. $$ Exposure enters both the case’s own information weight $\hat\mu_i$ and the total information matrix. Consequently, increasing $t_i$ does not in general multiply $h_i$ by a fixed factor while leaving other leverages unchanged.

The formula follows by taking the $i$th diagonal element of $H$: row $i$ of $\hat W^{1/2}X$ is $\sqrt{\hat\mu_i}x_i^\mathsf{T}$, so the diagonal element is $\hat\mu_i x_i^\mathsf{T}(X^\mathsf{T}\hat W X)^{-1}x_i$. Since $H$ is a projection matrix, $0\le h_i\le 1$, and its diagonal sums to the model rank $p$. These facts concern the fitted weighted geometry; they do not say that a high-leverage observation must have a large effect on the estimate.

Worked Example: Exposure Weights in an Intercept-Only Rate Model

Suppose two independent counts follow a common rate $\lambda=\exp(\beta)$, with exposures $t_1=1$ and $t_2=9$, and observed counts $y_1=4$ and $y_2=9$. Find the fitted rate, fitted means, and leverages.

The log-likelihood score equation is $y_1+y_2-\lambda(t_1+t_2)=0$. Therefore $$ \hat\lambda=\exp(\hat\beta)=\frac{4+9}{1+9}=1.3,\qquad \hat\mu_1=1(1.3)=1.3,\qquad \hat\mu_2=9(1.3)=11.7. $$

With an intercept-only design, $x_1=x_2=1$, so $X^\mathsf{T}\hat W X=\hat\mu_1+\hat\mu_2=13$. Hence $$ h_1=\frac{1.3}{13}=0.1,\qquad h_2=\frac{11.7}{13}=0.9. $$ In this common-rate model the leverage is the case’s share of total exposure, $h_i=t_i/(t_1+t_2)$. The larger-exposure case carries more information about the common rate, even though both observations have the same covariate pattern.

The exposure-share identity in this example is special to the intercept-only common-rate model. With covariates, the direction $x_i$ also matters: an observation at a distinctive covariate pattern can have high leverage even when its exposure is not the largest. It is therefore useful to distinguish exposure weight, which controls the scale of an observation’s contribution, from design leverage, which also reflects where that contribution lies in the design.

Worked Example: Leverage in a Two-Rate Group Model

Consider a model with separate rates for groups A and B, represented by two indicator coefficients. Group A has exposures $1$ and $3$, and counts $2$ and $5$. Group B has exposures $2$ and $6$, and counts $1$ and $5$. Calculate the fitted means and leverages.

For group A, the fitted rate is total count divided by total exposure: $$ \hat\lambda_A=\frac{2+5}{1+3}=\frac{7}{4}=1.75. $$ The fitted means are $1.75$ and $3(1.75)=5.25$. Since the two group-A observations share one indicator column, their leverage is their fitted mean divided by the sum of fitted means in that group. Thus $h_1=1.75/7=0.25$ and $h_2=5.25/7=0.75$.

For group B, $$ \hat\lambda_B=\frac{1+5}{2+6}=\frac{6}{8}=0.75. $$ The fitted means are $2(0.75)=1.5$ and $6(0.75)=4.5$. Therefore $h_3=1.5/6=0.25$ and $h_4=4.5/6=0.75$. Each group’s leverages sum to one, matching its one estimated rate parameter. Within each group, the case with three times the exposure has three times the leverage.

Case Deletion and Rate Influence

Leverage describes the fitted design and weights; influence also depends on how the observed response differs from its fitted mean. Let $r_i=y_i-\hat\mu_i$ be the raw Poisson residual and $A=X^\mathsf{T}\hat W X$. A first-order approximation to the coefficient change after deleting case $i$ can be derived directly from the score equations.

Result: If $A$ is nonsingular and $1-h_i\ne0$, the one-step case-deletion approximation is $$ \hat\beta_{(-i)}-\hat\beta\ \approx\ -\frac{A^{-1}x_i r_i}{1-h_i}. $$ This is a local approximation, not an identity for the exact deleted-data maximum-likelihood estimate. It is most informative when the deletion does not move the estimate far from $\hat\beta$.

To derive it, let $\delta=\hat\beta_{(-i)}-\hat\beta$. The deleted score is $U_{(-i)}(\beta)=U(\beta)-x_i(y_i-\mu_i(\beta))$. At the full estimate, $U(\hat\beta)=0$, so $U_{(-i)}(\hat\beta)=-x_i r_i$. The derivative of the deleted score at $\hat\beta$ is $-A+\hat\mu_i x_i x_i^\mathsf{T}$. A first-order Taylor expansion of the deleted score equation therefore gives $$ 0\ \approx\ -x_i r_i-\left(A-\hat\mu_i x_i x_i^\mathsf{T}\right)\delta. $$

Solving this linear equation uses the rank-one inverse identity $$ \left(A-\hat\mu_i x_i x_i^\mathsf{T}\right)^{-1}x_i =\frac{A^{-1}x_i}{1-\hat\mu_i x_i^\mathsf{T}A^{-1}x_i} =\frac{A^{-1}x_i}{1-h_i}. $$ Substitution yields the stated approximation. The denominator shows why a high-leverage case can amplify the effect of its residual. If deletion makes the design unidentified, or $h_i=1$, this approximation is not available; the deleted fit must be assessed as a separate fitting problem. $\square$

Worked Example: Exact Rate Changes and Approximate Coefficient Changes

Return to the intercept-only data with exposures $1,9$, counts $4,9$, fitted rate $1.3$, and fitted means $1.3,11.7$. Compare exact deletion estimates with the one-step approximation.

For case 1, $r_1=4-1.3=2.7$ and $h_1=0.1$. The exact rate after deleting it is $9/9=1$, so the exact coefficient change is $\log(1)-\log(1.3)\approx-0.262$. Here $A=13$, and the one-step formula gives $$ \delta_1\approx-\frac{(1/13)(2.7)}{0.9}\approx-0.231. $$

For case 2, $r_2=9-11.7=-2.7$ and $h_2=0.9$. The exact rate after deleting it is $4/1=4$, giving coefficient change $\log(4)-\log(1.3)\approx1.124$. The one-step approximation is $$ \delta_2\approx-\frac{(1/13)(-2.7)}{0.1}\approx2.077. $$ The second approximation is much less accurate because deletion changes the fitted rate substantially. The exact calculations verify that the high-exposure case has the larger effect here, but the approximation’s size should not be mistaken for an exact deletion result.

From Estimated Rates to Predicted Counts

For a prediction at covariate vector $x_*$, the fitted rate is $\hat\lambda_*=\exp(x_*^\mathsf{T}\hat\beta)$. At target exposure $t_*$, the fitted count is $\hat\mu_*=t_*\hat\lambda_*$. For a small coefficient change $\delta$, the rate changes approximately by $\hat\lambda_*x_*^\mathsf{T}\delta$, while the count changes approximately by $$ \Delta\hat\mu_*\approx\hat\mu_*x_*^\mathsf{T}\delta. $$ Thus a coefficient change has the same proportional effect on rate and count at a fixed covariate pattern, but its absolute effect on the count is multiplied by target exposure. If two target settings have different exposures, their absolute count changes can rank very differently from their rate changes.

Worked Example: One Rate Change, Two Count Changes

In the preceding intercept-only example, deleting case 1 changes the exact fitted rate from $1.3$ to $1$. The estimated rate therefore falls by $0.3$, regardless of the exposure chosen for a future prediction. At target exposure $t_*=1$, the fitted count falls from $1(1.3)=1.3$ to $1(1)=1$, a change of $-0.3$. At target exposure $t_*=100$, it falls from $100(1.3)=130$ to $100(1)=100$, a change of $-30$. The rate change is identical; the absolute count change is one hundred times larger for the second target.

This separates two questions that should not be conflated: how much a case changes the estimated rate function, and how much that rate change affects a count prediction at a specified exposure. The latter depends on the prediction target as well as on the deleted case.

Common Pitfall: Treating Exposure as an Ordinary Predictor

A common mistake is to add $\log t_i$ as an ordinary covariate and estimate its coefficient freely, even when $t_i$ is known exposure intended to represent opportunity. That replaces the model $\log\mu_i=\log t_i+x_i^\mathsf{T}\beta$, whose exposure coefficient is fixed at one, with a different model that estimates an exposure effect. Another mistake is to assume that high leverage alone means a case is influential. In the deletion approximation, influence depends on both $r_i$ and the geometry $A^{-1}x_i$, amplified by $1/(1-h_i)$. A case with a very small residual can have high leverage but little effect on the fitted coefficients.

Key takeaway: In Poisson regression with known exposure, the offset sets the expected-count scale and contributes to information through $\hat\mu_i=t_i\hat\lambda_i$. Leverage is exposure-weighted and design-dependent. Influence on rates concerns changes to $\hat\beta$ and $\hat\lambda$; influence on counts also depends on the exposure at which the prediction is made.

Check Your Understanding

Work through these questions to check how exposure, leverage, and influence fit together.

In a common-rate intercept-only Poisson model, explain why $h_i=t_i/\sum_j t_j$.
For an observation with exposure $t_i$, what are the conditional expectation and variance of the observed rate $Y_i/t_i$?
Why can a high-leverage Poisson case have little influence on the fitted coefficients?
State the one-step deletion approximation and identify the two features of case $i$ that affect it.
A deleted case changes a fitted rate by the same amount at two target settings. Why can the resulting absolute changes in predicted counts differ?