Why Exposure Enters a Poisson Model
A count is difficult to compare across observations when the amount of time or opportunity to experience an event differs. A count of four claims over two policy-years and a count of four claims over ten policy-years are not evidence of the same underlying event rate. A Poisson regression handles this distinction by modeling the expected count as exposure multiplied by rate. Taking logarithms then places log exposure in the linear predictor as an offset.
This tutorial derives that offset and explains the condition that gives it its fixed coefficient. The result is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
The usual assumptions are that, conditional on the predictors and exposure, the counts are independent Poisson random variables; the exposure values are known and positive; and the expected count is proportional to exposure at a given rate. In the Poisson model, the conditional variance also equals the conditional mean: \(\operatorname{Var}(Y_i\mid x_i,e_i)=\mu_i\). The proportionality assumption means that doubling exposure doubles the mean count when the rate is unchanged. If that assumption is not suitable, the offset model may not describe the data well.
The rate may vary with predictors. For example, \(x_i\) may contain an intercept and indicators for risk categories, and \(\beta\) contains their coefficients. The exposure is not another risk characteristic in that linear predictor: it has a specific role established by the mean model. A zero-exposure observation cannot be used with \(\log(e_i)\), since that logarithm is undefined; in this setup it provides no positive exposure over which to model a rate.
Deriving the Offset and Its Fixed Coefficient
Start with the mean statement \(\mu_i=e_i\lambda_i\), where \(e_i>0\). Specify a log-linear model for the rate, \(\log(\lambda_i)=x_i^\mathsf{T}\beta\), which is equivalent to \(\lambda_i=\exp(x_i^\mathsf{T}\beta)\). Substitution gives \(\mu_i=e_i\exp(x_i^\mathsf{T}\beta)\). Taking the logarithm yields the offset form.
The coefficient of \(\log(e_i)\) is one because the starting mean is exactly \(e_i\lambda_i\). Equivalently, the model requires \(\mu_i\) to scale linearly with exposure while holding the rate fixed. If exposure is multiplied by a factor \(c>0\), then \(\mu_i\) becomes \(ce_i\lambda_i=c\mu_i\). In the log mean, that change is an addition of \(\log(c)\), which is precisely the change in \(\log(e_i)\) with coefficient one.
It is possible to write a different model with an estimated coefficient \(\gamma\) on log exposure, such as \(\log(\mu_i)=\gamma\log(e_i)+x_i^\mathsf{T}\beta\). That model implies \(\mu_i=e_i^\gamma\exp(x_i^\mathsf{T}\beta)\), not the proportional-exposure mean \(e_i\lambda_i\) unless \(\gamma=1\). It is therefore a different assumption, not a way to estimate the standard offset. The offset is a known component of the linear predictor; it is not an ordinary regression coefficient.
Changing the unit of exposure does not change the underlying rate model, but it does require a consistent change in rate units. For example, if exposure is converted from years to months, a rate per year must be divided by 12 to become a rate per month. The numerical intercept changes accordingly, while the fitted expected counts remain the same. The next examples work through these points explicitly.
Worked Example: Estimating a Common Rate With an Offset
A hypothetical portfolio has three independent policy groups. Their exposures are \(0.5\), \(1.5\), and \(2.0\) policy-years, and their observed claim counts are \(1\), \(3\), and \(4\), respectively. Assume \(Y_i\mid e_i\sim\operatorname{Poisson}(e_i\lambda)\) with one common claim rate \(\lambda\) per policy-year. Find the maximum-likelihood rate using the offset model, and verify the fitted means.
Method and model. With a common rate, \(\log(\lambda)=\beta_0\), so \(\log(\mu_i)=\log(e_i)+\beta_0\). The three Poisson means are \(0.5e^{\beta_0}\), \(1.5e^{\beta_0}\), and \(2e^{\beta_0}\).
Find the estimate. Write \(y=(1,3,4)\). The part of the log-likelihood depending on \(\beta_0\) is \(\ell(\beta_0)=\sum_i[y_i\{\log(e_i)+\beta_0\}-e_i e^{\beta_0}-\log(y_i!)]\). Its derivative is \(d\ell/d\beta_0=\sum_i y_i-e^{\beta_0}\sum_i e_i\). Here \(\sum_i y_i=1+3+4=8\), and \(\sum_i e_i=0.5+1.5+2.0=4.0\). Setting the derivative equal to zero gives \(8-4e^{\beta_0}=0\), so \(e^{\beta_0}=2\) and \(\hat{\lambda}=2\) claims per policy-year. Thus \(\hat{\beta}_0=\log(2)\), approximately \(0.6931\).
Check each fitted mean. The means are \(0.5(2)=1\), \(1.5(2)=3\), and \(2(2)=4\). Their sum is \(1+3+4=8\), equal to the total observed count. The likelihood score is therefore \(8-2(4)=0\), verifying the estimate. The common mistake avoided here is dividing by the number of groups instead of total exposure: the rate estimate is \(8/4=2\), not \(8/3\).
Worked Example: Changing Exposure Units Without Changing the Mean
A hypothetical event model has a baseline rate of \(0.20\) events per policy-year. A risk indicator \(x\) equals 1 for one class and 0 for the baseline class, with rate ratio \(1.5\), so \(\log(\lambda)=\log(0.20)+x\log(1.5)\). Consider a class with \(x=1\) and 18 policy-months of exposure, and a baseline-class observation with \(x=0\) and 30 policy-months. Calculate both expected counts using years and then months.
Convert exposures and identify rates. The first exposure is \(18/12=1.5\) policy-years, and the second is \(30/12=2.5\) policy-years. The class with \(x=1\) has rate \(0.20(1.5)=0.30\) events per year; the baseline class has rate \(0.20\) events per year. The conditional count distributions are Poisson with means equal to exposure times the relevant rate.
Calculate with years. The first mean is \(\mu_1=1.5(0.30)=0.45\). Equivalently, \(\log(\mu_1)=\log(1.5)+\log(0.20)+\log(1.5)=0.4055-1.6094+0.4055=-0.7984\), and \(e^{-0.7984}\) is approximately \(0.45\). The second mean is \(\mu_2=2.5(0.20)=0.50\). Its log mean is \(\log(2.5)+\log(0.20)=0.9163-1.6094=-0.6931=\log(0.50)\).
Calculate with months. The rates per month are \(0.30/12=0.025\) for the first class and \(0.20/12=0.0166667\) for the baseline class. Therefore \(18(0.025)=0.45\) and \(30(0.0166667)=0.50\), matching the year-based means. In log form, the month-scale intercept is \(\log(0.20/12)\), while the offsets are \(\log(18)\) and \(\log(30)\); the risk coefficient remains \(\log(1.5)\). The common mistake avoided is combining an offset in months with a rate per year: \(18(0.30)=5.4\) is not the correct first mean because those units do not match.
Offsets With Categorical Rate Effects
An offset does not prevent the rate from differing across groups. It simply ensures that exposure scales the mean count in the stipulated way. For categorical predictors, one can estimate a separate rate for each category while keeping the exposure coefficient fixed at one. The rates are still modeled through regression coefficients; the exposure remains a known multiplier.
Worked Example: Two Rate Groups With Unequal Exposures
A hypothetical benefit program records counts for two risk classes. Class A has two observations: exposure 2 with count 1, and exposure 3 with count 2. Class B has exposure 4 with count 4, and exposure 6 with count 5. Assume independent Poisson counts and a separate constant rate per policy-year for each class. Fit the rates using a log-exposure offset, then express the model coefficients using A as the reference class.
Estimate each class rate. For a class with a common rate, the likelihood score sets the total observed count equal to the rate times total exposure. For A, total count is \(1+2=3\) and total exposure is \(2+3=5\), giving \(\hat{\lambda}_A=3/5=0.6\). For B, total count is \(4+5=9\) and total exposure is \(4+6=10\), giving \(\hat{\lambda}_B=9/10=0.9\).
Write the offset regression. Let \(I_B=1\) for class B and \(I_B=0\) for class A. The model is \(\log(\mu_i)=\log(e_i)+\beta_0+\beta_1 I_B\). The reference-class estimate is \(\hat{\beta}_0=\log(0.6)\), approximately \(-0.5108\). The class effect is \(\hat{\beta}_1=\log(0.9/0.6)=\log(1.5)\), approximately \(0.4055\). Thus class B's fitted rate is \(e^{\hat{\beta}_0+\hat{\beta}_1}=0.6(1.5)=0.9\).
Verify the fitted means and score conditions. For A, the fitted means are \(2(0.6)=1.2\) and \(3(0.6)=1.8\); their sum is \(3.0\), matching A's total count. For B, the means are \(4(0.9)=3.6\) and \(6(0.9)=5.4\); their sum is \(9.0\), matching B's total count. The class-specific score residual totals are \(3-3=0\) and \(9-9=0\). The common mistake avoided is interpreting \(\hat{\beta}_1\) as an additive increase of \(0.4055\) events per policy-year. It is a log rate ratio; exponentiating gives the rate ratio \(1.5\).
Interpretation and Common Traps
The offset is conceptually simple, but several distinctions matter in applications and calculations:
- The offset is known, not estimated. In the standard proportional-exposure model, \(\log(e_i)\) enters with coefficient one. Treating that coefficient as free specifies a different model.
- The rate and the expected count have different units. A rate is per unit of exposure; multiplying it by exposure produces a mean count. Do not compare or multiply quantities expressed in incompatible units.
- Use the log of exposure. The log link applies to the mean count. Since \(\log(\mu_i)=\log(e_i)+\log(\lambda_i)\), the offset is \(\log(e_i)\), not \(e_i\) itself.
- Exposure must be positive for this formulation. A zero-exposure record has no defined log offset and does not provide a positive-exposure rate observation.
- Do not replace the count model casually with a rate response. Modeling a count with its exposure offset retains the count scale and its Poisson likelihood. Dividing each count by exposure and treating the resulting rate as an ordinary Poisson count changes the response model.
A useful diagnostic check of the algebra is to hold predictors fixed and multiply exposure by \(c\). The model must then multiply the fitted mean by \(c\): \(\exp\{\log(ce_i)+x_i^\mathsf{T}\beta\}=c\exp\{\log(e_i)+x_i^\mathsf{T}\beta\}\). If a proposed formula does not have this property, it is not the proportional-exposure offset model. The central principle is that the offset encodes the exposure scale, while the regression coefficients describe how the rate changes with predictors.
Check Your Understanding
For each problem, identify the relevant rate or offset relationship before calculating.
- A hypothetical count has a Poisson mean equal to exposure times a common rate. Two independent groups have exposures 3 and 7 policy-years and observed counts 2 and 5. Find the maximum-likelihood common rate and the two fitted means. (Answer: rate \(7/10=0.7\) per policy-year; fitted means \(2.1\) and \(4.9\).)
- A model has rate \(0.08\) events per year for the reference class and rate ratio \(2\) for a flagged class. A flagged observation has 9 months of exposure. Find its expected count, expressing the exposure in years. (Answer: \(0.75(0.16)=0.12\).)
- A model is written as \(\log(\mu_i)=\log(e_i)+\beta_0+\beta_1x_i\), with \(e_i>0\). If \(\beta_1=\log(3)\), what is the rate ratio for a one-unit increase in \(x_i\), and by what factor does the mean change when exposure doubles while \(x_i\) stays fixed? (Answer: rate ratio \(3\); mean factor \(2\).)
- Exposure is recorded in days, but a fitted baseline rate is \(0.24\) events per year. For 90 days of exposure and no other rate effect, calculate the mean using 365 days per year. (Answer: \((90/365)(0.24)\approx0.0592\) events.)
- Two risk classes have total counts and exposures \(6\) and \(12\) for class A, and \(9\) and \(12\) for class B. With separate constant rates and class A as reference, find the fitted rates, intercept, and class log-rate coefficient. (Answer: rates \(0.5\) and \(0.75\) per exposure unit; \(\hat{\beta}_0=\log(0.5)\); \(\hat{\beta}_1=\log(1.5)\approx0.4055\).)