Why An Offset Is Not An Exposure Weight

Why Exposure Belongs in the Mean

In a Poisson frequency model, an exposure amount determines how long, or how much, a risk is observed. If a policy is observed for three times as long as another otherwise comparable policy, the expected claim count should be three times as large. That scaling belongs in the mean of the count distribution. It is not the same operation as giving an observation more weight in a likelihood.

The distinction matters in fitting as well as interpretation. Replacing the exposure offset with a likelihood weight can change fitted counts and the apparent amount of information in the data. Applying both when only an offset is called for can also change the estimate and make model-based uncertainty look too small. The distinction is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Let $Y_i$ be the observed claim count for observation $i$, and let $e_i>0$ be its known exposure in a consistent unit, such as policy-years. In a Poisson frequency model with predictors $x_i$, the mean is $$ Y_i\mid x_i,e_i \sim \operatorname{Poisson}(\mu_i), \qquad \log(\mu_i)=\log(e_i)+x_i^\mathsf{T}\beta. $$ The term $\log(e_i)$ is an offset: a known term in the linear predictor whose coefficient is fixed at 1. The likelihood contribution is not multiplied by $e_i$.

With a log link, the offset makes the expected count proportional to exposure while the fitted frequency rate is $\exp(x_i^\mathsf{T}\beta)$. The unit for $e_i$ must match the unit in which the rate is interpreted. If exposure is measured in policy-years, the modelled rate is claims per policy-year.

For independent Poisson counts, the log-likelihood, apart from terms not involving $\beta$, is

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

A likelihood weight $w_i$, by contrast, multiplies a likelihood contribution. For example, a weighted log-likelihood has contributions

\ell_w(\beta)=\sum_i w_i\left[y_i\log(\mu_i)-\mu_i\right],

where constants independent of the parameters have been omitted. This changes the contribution observation $i$ makes to the score and curvature. A weight can be appropriate for a particular sampling or analysis design, but it does not, by itself, make the modelled mean proportional to exposure.

Deriving the Common-Rate Estimate

Consider the simplest case: all observations have the same underlying frequency rate $\lambda>0$, so $Y_i$ is Poisson with mean $e_i\lambda$. The offset model has $\log(\mu_i)=\log(e_i)+\beta$, where $\lambda=\exp(\beta)$. Its log-likelihood, including the terms that depend on $\lambda$, is

\ell(\lambda)=\sum_i \left[y_i\log(e_i\lambda)-e_i\lambda-\log(y_i!)\right].

Differentiating with respect to $\lambda$ gives the score. When the total observed count is positive, setting this score to zero produces the maximum-likelihood estimate:

\frac{d\ell}{d\lambda} =\frac{\sum_i y_i}{\lambda}-\sum_i e_i=0 \quad\Longrightarrow\quad \widehat{\lambda}=\frac{\sum_i y_i}{\sum_i e_i}.

Thus the rate estimate is total claims divided by total exposure, and the fitted count for observation $i$ is $e_i\widehat{\lambda}$. The result follows because exposure enters the count mean as $e_i\lambda$, while the log-likelihood score adds observed counts in the numerator and exposure in the denominator.

There is an important boundary case. If $\sum_i y_i=0$, the score is $-\sum_i e_i$, which is negative for every $\lambda>0$. The likelihood increases as $\lambda$ approaches zero, so its maximum is at $\lambda=0$ if that boundary is allowed. In that case there is no finite estimate of $\beta=\log(\lambda)$. If the parameter space requires $\lambda>0$, there is no attained maximum-likelihood estimate.

Under this common-rate model, the Fisher information for $\beta=\log(\lambda)$ is $\lambda\sum_i e_i$. The usual model-based standard error of $\widehat{\beta}$ is therefore $1/\sqrt{\widehat{\lambda}\sum_i e_i}$, when the estimate is positive. This uncertainty calculation assumes independent Poisson counts and the specified mean model.

Worked Example: Replacing an Offset With a Likelihood Weight

Two hypothetical policy groups have exposures $e_1=1$ and $e_2=3$ policy-years, with observed counts $y_1=2$ and $y_2=9$. Assume independent Poisson counts and a common frequency rate. Find the correct fitted counts and model-based uncertainty for the log rate. Then calculate what happens if exposure is used as a likelihood weight in place of the offset.

Correct method: use an offset. Total claims are $2+9=11$, and total exposure is $1+3=4$ policy-years. Therefore

\widehat{\lambda}=\frac{11}{4}=2.75 \quad\text{claims per policy-year}.

The fitted counts are $\widehat{\mu}_1=1(2.75)=2.75$ and $\widehat{\mu}_2=3(2.75)=8.25$. They sum to $2.75+8.25=11$, matching the total observed count.

The estimated information for $\beta=\log(\lambda)$ is the sum of fitted counts, $2.75+8.25=11$. Thus

\operatorname{SE}(\widehat{\beta})=\frac{1}{\sqrt{11}}\approx 0.3015.

By the delta method, the standard error of the rate estimate is approximately $\widehat{\lambda}\operatorname{SE}(\widehat{\beta})=2.75/\sqrt{11}\approx 0.8292$ claims per policy-year.

Incorrect substitution: use exposure as a likelihood weight, with no offset. Fit a common Poisson mean $\mu$ and multiply each log-likelihood contribution by $e_i$. The score with respect to $\log(\mu)$ is $\sum_i e_i(y_i-\mu)$, so

\widehat{\mu} =\frac{e_1y_1+e_2y_2}{e_1+e_2} =\frac{1(2)+3(9)}{1+3} =\frac{29}{4} =7.25.

This specification predicts 7.25 claims for each group, rather than predictions proportional to exposure. Its weighted likelihood curvature for $\log(\mu)$ is $\sum_i e_i\widehat{\mu}=4(7.25)=29$, giving a naive model-based standard error $1/\sqrt{29}\approx0.1857$ for $\log(\mu)$. The smaller value does not establish that the data contain more information about a frequency rate: this is a different, exposure-inappropriate mean model with a weighted likelihood.

Method and trap avoided. The offset model fits a rate and scales each expected count by exposure. The weighted model above fits a common count mean and gives the longer-exposure row more influence. Treating those two operations as interchangeable changes both the fitted counts and the uncertainty calculation.

Why Applying Both Can Also Be Wrong

Keeping the offset but also multiplying each likelihood contribution by exposure is not a harmless alternative. It changes the score even though the mean still scales with exposure. In the common-rate model, using weights $w_i=e_i$ as well as the offset gives the estimating equation

\sum_i e_i(y_i-e_i\lambda)=0 \quad\Longrightarrow\quad \widehat{\lambda}_{\text{both}} =\frac{\sum_i e_i y_i}{\sum_i e_i^2}.

That is generally not the total-count-over-total-exposure estimate. It also produces a different likelihood curvature. The next example calculates both differences.

Worked Example: The Effect of Using an Offset and Exposure Weights Together

A hypothetical portfolio has two independent Poisson count observations with exposures $e_1=1$ and $e_2=2$ policy-years and counts $y_1=1$ and $y_2=7$. Assume a common frequency rate. Compare the correct offset fit with a fit that uses both the offset and likelihood weights $w_i=e_i$.

Correct offset fit. The total count is $1+7=8$, and total exposure is $1+2=3$. Thus $\widehat{\lambda}=8/3$. The fitted counts are

\widehat{\mu}_1=1\left(\frac{8}{3}\right)=\frac{8}{3}, \qquad \widehat{\mu}_2=2\left(\frac{8}{3}\right)=\frac{16}{3}.

They sum to $8/3+16/3=8$. The estimated information for the log rate is 8, so its model-based standard error is $1/\sqrt{8}\approx0.3536$.

Offset plus exposure weights. The weighted score equation gives

\widehat{\lambda}_{\text{both}} =\frac{1(1)+2(7)}{1^2+2^2} =\frac{15}{5} =3.

The fitted counts become $1(3)=3$ and $2(3)=6$. Although these fitted counts still scale in proportion to exposure, the estimate is different from $8/3$ and the fitted total is $3+6=9$, not the observed total of 8.

For this weighted likelihood, the curvature for the log rate at the estimate is $\sum_i e_i\widehat{\mu}_i=1(3)+2(6)=15$. Its naive model-based standard error is $1/\sqrt{15}\approx0.2582$, rather than $0.3536$. The likelihood weights have changed both the estimating equation and the information calculation.

Method and trap avoided. Use the offset alone when exposure represents time at risk and the intended assumption is a constant frequency rate conditional on the predictors. Do not add exposure weights merely because exposure already appears in the model.

When a Likelihood Weight Has a Separate Purpose

A likelihood weight is not inherently incorrect. Its meaning depends on how it is defined and why it is used. For example, an integer frequency weight may represent repeated identical records under a specified model; other weights may arise from a sampling design or a chosen weighted estimating procedure. Those interpretations require their own assumptions. They do not turn a count mean that lacks an exposure term into a frequency model.

When rows are grouped counts, the count itself and its exposure already determine the Poisson likelihood contribution under the independent-count model. If a row has $y_i$ claims over $e_i$ policy-years, the model mean is $e_i\lambda_i$, not merely $\lambda_i$. A weight should therefore not be introduced as a substitute for the known exposure in that mean.

For grouped data, it is also important to distinguish a row containing one count over a long exposure from multiple independent rows. A single Poisson count over $e_i$ exposure has mean $e_i\lambda_i$. Duplicating the row as though it were several independent observations is a different data and likelihood construction. The appropriate model follows from how the observations were collected, not from a desire to make an exposure value act as a multiplier somewhere in the calculation.

Common Exam Traps

Putting exposure in the wrong place. In a log-link frequency model, $\log(e_i)$ is added to the linear predictor with coefficient fixed at 1. Multiplying the likelihood by $e_i$ does not impose that relationship on the mean.
Reporting a count as though it were a rate. With an offset, $\exp(x_i^\mathsf{T}\beta)$ is the fitted rate and $e_i\exp(x_i^\mathsf{T}\beta)$ is the fitted count. State the exposure unit when reporting the rate.
Assuming weights only affect standard errors. A likelihood weight changes the score as well as the curvature, and can therefore change parameter estimates, fitted values, and uncertainty.
Using the positive-rate score formula for all-zero data. If all observed counts are zero, the common-rate likelihood has its maximum at the boundary $\lambda=0$, if allowed. There is no finite log-rate estimate.

Key takeaway: An exposure offset specifies how exposure scales the expected count: $\mu_i=e_i\exp(x_i^\mathsf{T}\beta)$. A likelihood weight changes how strongly an observation contributes to the fitting criterion. Use each only when its distinct modelling interpretation is justified.

Check Your Understanding

For each problem, state whether exposure is being used as an offset, a likelihood weight, or both, and check how that choice affects the result.

A common-rate Poisson model has exposures 1 and 3 policy-years and counts 3 and 9. Find the offset estimate and fitted counts. What common mean results if exposure is instead used only as a likelihood weight? (Answer: Offset rate $=3$ per policy-year and fitted counts $3,9$; weight-only common mean $=7.5$ per row.)
One count of 5 is observed over 2 policy-years. Under a common-rate Poisson model, find the offset rate estimate and the fitted count. What is the fitted common mean if the model has no offset and the sole row is given likelihood weight 2? (Answer: Offset rate $=2.5$ per policy-year and fitted count $=5$; weight-only mean $=5$.)
A risk class has two rows: exposures 1 and 3 policy-years, counts 2 and 8. Estimate its common rate with an offset and calculate its fitted counts. Then replace the offset with likelihood weights equal to exposure and fit a common mean. (Answer: Offset rate $=2.5$, fitted counts $2.5,7.5$; weight-only mean $=6.5$ per row.)
For exposures 1 and 2 with counts 1 and 7, calculate the common-rate estimate if both the offset and exposure likelihood weights are used. Compare it with the correctly offset estimate. (Answer: Using both gives $3$ per policy-year; the offset-only estimate is $8/3$ per policy-year.)
A common-rate Poisson model has positive total exposure but zero counts in every row. What is the maximum-likelihood estimate if rate zero is allowed, and is there a finite estimate of the log rate? (Answer: The rate MLE is the boundary value $0$; there is no finite log-rate estimate.)

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