Estimating Rate Models With Exposure Offsets

Why Exposure Belongs in the Mean

A policy observed for a full year has more opportunity to generate claims than an otherwise identical policy observed for only a fraction of a year. A frequency model must account for that difference before comparing risks. In a Poisson rate model, known exposure does this through a fixed log-exposure offset. The offset adjusts the expected claim count while leaving the rating coefficients to describe differences in claim rate.

Suppose observation $i$ has claim count $N_i$, exposure $e_i>0$, and predictor vector $x_i$. Exposure might be measured in policy-years, vehicle-years, or another consistent unit. The model assumes independent Poisson counts conditional on the predictors and exposures. With a log link, the expected count is exposure multiplied by the modeled rate.

Definition: An exposure-offset Poisson rate model is $$ N_i\mid x_i,e_i\sim \operatorname{Poisson}(\mu_i),\qquad \mu_i=e_i\lambda_i,\qquad \log(\mu_i)=\log(e_i)+x_i^\mathsf{T}\beta. $$ Here $e_i$ is known and positive, $x_i$ is the specified predictor vector, and $\beta$ is the vector of unknown coefficients. The term $\log(e_i)$ is a fixed offset: its coefficient is fixed at 1, not estimated.

The modeled rate is $\lambda_i=\exp(x_i^\mathsf{T}\beta)$, measured per unit of exposure. The expected count is $\mu_i=e_i\exp(x_i^\mathsf{T}\beta)$. Thus, doubling exposure doubles the expected count when the rating predictors are held fixed; it does not, by itself, double the underlying rate.

Likelihood and the Role of the Offset

Under the stated independent Poisson model, the probability of observing $N_i=n_i$ is $e^{-\mu_i}\mu_i^{n_i}/n_i!$. Substituting the exposure-offset mean gives the likelihood and log-likelihood:

L(\beta)=\prod_i \frac{\exp\{-e_i\exp(x_i^\mathsf{T}\beta)\} \left[e_i\exp(x_i^\mathsf{T}\beta)\right]^{n_i}}{n_i!},

\ell(\beta)=\sum_i\left[ n_i\{\log(e_i)+x_i^\mathsf{T}\beta\} -e_i\exp(x_i^\mathsf{T}\beta)-\log(n_i!) \right].

The exposure appears in two places in the log-likelihood: as $n_i\log(e_i)$, and in the mean term $-e_i\exp(x_i^\mathsf{T}\beta)$. It is not a separate coefficient to estimate. For fixed data, $n_i\log(e_i)$ does not vary with $\beta$, but the exposure in the mean term affects the fitted coefficients and their likelihood.

Differentiating with respect to $\beta$ gives the score equations. With $X$ denoting the design matrix, $n$ the vector of observed counts, and $\mu$ the vector of modeled means, the score is $X^\mathsf{T}(n-\mu)$. The maximum likelihood estimates solve $X^\mathsf{T}(n-\mu)=0$, when a finite solution exists.

U(\beta)=\frac{\partial\ell}{\partial\beta} =\sum_i x_i(n_i-\mu_i) =X^\mathsf{T}(n-\mu).

Worked Example: Estimating a Constant Claim Rate

A hypothetical portfolio has three independent policy groups with claim counts $n=(2,3,5)$ and exposures $e=(1,2,2)$ policy-years. Assume a common claim rate $\lambda=\exp(\beta)$ per policy-year and no other predictors. Find the maximum likelihood rate and the fitted expected counts. Then check the score.

Method. With an intercept-only rate, $\mu_i=e_i\exp(\beta)=e_i\lambda$. The log-likelihood, omitting terms that do not depend on $\lambda$, is $(\sum_i n_i)\log(\lambda)-\lambda\sum_i e_i$. Its derivative with respect to $\lambda$ is $(\sum_i n_i)/\lambda-\sum_i e_i$.

Set the derivative equal to zero. The observed total is $2+3+5=10$ claims, and total exposure is $1+2+2=5$ policy-years. Therefore $\hat\lambda=10/5=2$ claims per policy-year, or $\hat\beta=\log(2)$.

The fitted means are $\hat\mu_1=1(2)=2$, $\hat\mu_2=2(2)=4$, and $\hat\mu_3=2(2)=4$. Their total is $2+4+4=10$, equal to the observed total.

Check. The intercept score is $\sum_i(n_i-\hat\mu_i)=(2-2)+(3-4)+(5-4)=0$, as required at the maximum likelihood estimate. This total-count-over-total-exposure formula applies to a common rate with an intercept only; it does not generally estimate a model with varying rating effects.

Common mistake avoided: Averaging the three group-specific observed rates without regard to exposure would give $(2/1+3/2+5/2)/3=2$ in this particular data set, but that agreement is accidental. The likelihood weights the groups through their exposure and counts; the general intercept-only estimate is the ratio of total counts to total exposure.

Why the Offset Enters the IRLS Equations

For a log-link Poisson model, the linear predictor for the mean is $\eta_i=\log(\mu_i)=\log(e_i)+x_i^\mathsf{T}\beta$. In Fisher scoring, the Poisson variance is $\operatorname{Var}(N_i\mid x_i,e_i)=\mu_i$. Consequently, the IRLS weight for observation $i$ is $\mu_i$, and the working response for the full linear predictor is $z_i=\eta_i+(n_i-\mu_i)/\mu_i$.

The coefficient update must fit the part of that working response that can change with $\beta$. Since $\log(e_i)$ is fixed, subtract it from the working response before fitting the predictor terms. The update is therefore weighted least squares of $z_i-\log(e_i)$ on the design matrix $X$, with weights $\mu_i$.

z_i=\log(e_i)+x_i^\mathsf{T}\beta+\frac{n_i-\mu_i}{\mu_i}, \qquad \beta_{\text{new}} =\arg\min_b\sum_i\mu_i \left[z_i-\log(e_i)-x_i^\mathsf{T}b\right]^2.

This is a local quadratic approximation to the log-likelihood, not a replacement of the Poisson likelihood by an ordinary least-squares model. The offset is removed from the working response because it is fixed, not because exposure is irrelevant to the update: exposure affects $\mu_i$, and therefore affects both the working response and the weights.

Calculate the current means.
For the current coefficient vector, calculate $\mu_i=e_i\exp(x_i^\mathsf{T}\beta)$ for every observation.

Construct the working quantities.
Set $z_i=\log(\mu_i)+(n_i-\mu_i)/\mu_i$ and assign weight $w_i=\mu_i$.

Remove the fixed offset for the coefficient fit.
Fit $z_i-\log(e_i)$ to the predictor design $x_i$ by weighted least squares using $w_i=\mu_i$.

Update and repeat.
Use the resulting coefficients to recalculate the means and repeat until the coefficient and likelihood changes satisfy the chosen convergence criteria.

Worked Example: One IRLS Update With Unequal Exposures

Consider two hypothetical rating groups. Group 1 has $e_1=1$ policy-year, $n_1=2$ claims, and predictor row $x_1=(1,0)$. Group 2 has $e_2=2$ policy-years, $n_2=4$ claims, and predictor row $x_2=(1,1)$. Start from $\beta=(0,0)$. Calculate one IRLS coefficient update, showing the offset adjustment and weighted least-squares solution.

Method. The model is $\log(\mu_i)=\log(e_i)+\beta_0+\beta_1x_{i2}$. At the starting coefficients, both modeled rates are $\exp(0)=1$. Thus $\mu_1=1(1)=1$, and $\mu_2=2(1)=2$. The offsets are $\log(1)=0$ and $\log(2)$.

For group 1, $\eta_1=\log(1)=0$, so $z_1=0+(2-1)/1=1$. After subtracting the offset, the adjusted working response is $z_1-\log(e_1)=1$; its weight is $w_1=\mu_1=1$.

For group 2, $\eta_2=\log(2)$, so $z_2=\log(2)+(4-2)/2=\log(2)+1$. Subtracting $\log(e_2)=\log(2)$ gives adjusted working response $1$; its weight is $w_2=\mu_2=2$.

The weighted least-squares rows are $(1,0)$ and $(1,1)$, with adjusted responses $1,1$ and weights $1,2$. The first row requires $\beta_{0,\text{new}}=1$; the second then requires $\beta_{0,\text{new}}+\beta_{1,\text{new}}=1$. Hence the update is $(1,0)$.

Check. The updated means are $\mu_1=\exp(1)=2.7183$ and $\mu_2=2\exp(1)=5.4366$. The updated intercept score is $(2-2.7183)+(4-5.4366)=-2.1549$, which is not zero. One IRLS update is not necessarily the maximum likelihood estimate; the iteration must continue.

Common mistake avoided: Treating $z_i$ itself as the response for fitting the coefficients would leave the known $\log(e_i)$ inside the coefficient fit. The correct response is $z_i-\log(e_i)$. The offset still influences the update through both the original $z_i$ and the weights.

Exposure Units and Interpretation

The chosen exposure unit determines the numerical unit of the estimated rate and the numerical value of the intercept. It must be consistent across observations. Changing all exposures from years to months changes the offset by a constant, so the intercept shifts to compensate; fitted expected counts remain unchanged when the model is transformed correctly.

Worked Example: Converting the Exposure Unit

A hypothetical book has 9 claims over 6 policy-years in one rating class and 6 claims over 3 policy-years in another. Fit an intercept-only common-rate model using years, then express the same model using policy-months. Verify that the fitted expected counts are unchanged.

Method in years. Total claims are $9+6=15$, and total exposure is $6+3=9$ policy-years. The fitted annual rate is $\hat\lambda_{\text{year}}=15/9=1.6667$ claims per policy-year. The fitted means are $6(1.6667)=10$ and $3(1.6667)=5$, with the unrounded rate giving exactly 10 and 5.

Convert to months. The exposures are $6(12)=72$ and $3(12)=36$ policy-months, totaling $108$. The fitted monthly rate is $15/108=0.1388889$ claims per policy-month. Annualizing gives $12(0.1388889)=1.6667$ claims per policy-year.

The fitted means in months are $72(15/108)=10$ and $36(15/108)=5$, matching the year-based results. Rounding rates to four decimal places only for display, the annual rate is 1.6667 and the monthly rate is 0.1389; the mean checks use the unrounded values.

Check. The change of unit changes the rate by a factor of 12 and changes the exposure by the reciprocal factor in the mean calculation. The predicted claim counts therefore remain the same.

Common mistake avoided: Entering the numerical exposures in months while interpreting the fitted rate as claims per year creates a factor-of-12 error. The offset’s units and the reported rate’s units must match.

Common Errors and Practical Checks

An exposure offset is appropriate when exposure is known, positive, and represents proportional time or volume at risk under the model. The Poisson specification also assumes the conditional mean equals the conditional variance. If observations are dependent, exposure is measured with error, or claim-count variation is materially greater than the Poisson variance, the stated likelihood and its standard errors may not be appropriate without a different modeling treatment.

Omitting the offset: Modeling $\log(\mu_i)=x_i^\mathsf{T}\beta$ treats expected counts as if they did not change with exposure. This can confound duration with risk.
Estimating an exposure coefficient by default: The standard rate model fixes the coefficient of $\log(e_i)$ at 1. Estimating it instead changes the model assumption and is not the fixed-offset formulation.
Using exposure only as an IRLS weight: This does not generally produce the same model. Exposure determines the Poisson mean through $\mu_i=e_i\exp(x_i^\mathsf{T}\beta)$; the resulting IRLS weight is $\mu_i$, not automatically $e_i$.
Mixing exposure units: A value recorded as 6 months is $0.5$ policy-years if the model uses years, not 6. Check units before taking the logarithm.
Assuming one IRLS step solves the model: The weighted least-squares problem is a local update. Recompute means and working quantities until convergence, then check the score or another suitable convergence criterion.

Key takeaway: In a Poisson rate model, the known exposure enters the log mean as $\log(e_i)$ with its coefficient fixed at 1. In IRLS, subtract that offset from the working response before fitting the estimated coefficients, while retaining exposure’s effect on the means and weights.

Check Your Understanding

For each problem, identify the exposure unit and distinguish the fitted rate from the fitted count.

A hypothetical class has 12 claims over 8 policy-years. Under an intercept-only Poisson rate model, what is the maximum likelihood rate per policy-year? (Answer: 1.5 claims per policy-year.)
Two hypothetical groups have counts 3 and 5 and exposures 1 and 4 policy-years. Fit a common rate and give each group’s fitted mean. (Answer: Rate $=8/5=1.6$ claims per policy-year; fitted means $1.6$ and $6.4$.)
For an intercept-only model, one observation has $n_1=3,e_1=2$, and another has $n_2=1,e_2=1$. Starting at $\beta=0$, calculate one IRLS update. (Answer: $\beta_{\text{new}}=1/3$, so the updated rate is $\exp(1/3)\approx1.3956$ per exposure unit.)
A hypothetical portfolio has 7 claims over 2 policy-years. Re-express the exposure in months, calculate the monthly rate, and annualize it. (Answer: 24 policy-months; $7/24=0.2917$ claims per policy-month; $12(7/24)=3.5$ claims per policy-year.)
At a current intercept-only rate of 1 claim per policy-year, an observation has $e=3$ policy-years and $n=6$ claims. Find its current mean, IRLS weight, and adjusted working response after subtracting the offset. (Answer: $\mu=3$, weight $=3$, and $z-\log(e)=0+(6-3)/3=1$.)

What You'll Learn