Offset Versus Estimated Predictor

Why the Exposure Coefficient Matters

A policy observed for more time generally has more opportunity to generate claims. A frequency model represents that opportunity by relating expected claim count to exposure. The key modeling choice is whether exposure’s effect is specified in advance or estimated from the data.

With an offset, the model fixes the coefficient of log exposure at one. This encodes proportionality: doubling exposure doubles expected count, all else equal. If the coefficient is estimated instead, the data determine how expected count changes with exposure. That flexibility can be useful, but it changes the model’s meaning: the implied claim rate per unit of exposure need not be constant.

Definition: Let $N_i$ be the claim count for observation $i$, $e_i>0$ its exposure, and $x_i$ its rating variables. A Poisson frequency model with a log link and an exposure offset is $$ N_i\sim\operatorname{Poisson}(\mu_i),\qquad \log(\mu_i)=\log(e_i)+\beta_0+x_i^\mathsf{T}\beta. $$ The term $\log(e_i)$ is an offset: its coefficient is fixed at one, rather than estimated. An alternative with an estimated log-exposure coefficient is $$ N_i\sim\operatorname{Poisson}(\mu_i),\qquad \log(\mu_i)=\alpha+\gamma\log(e_i)+x_i^\mathsf{T}\beta. $$

In these models, $\mu_i=E[N_i]$. The Poisson assumption specifies that, conditional on the modeled predictors, the count has mean and variance $\mu_i$; independent observations are assumed for the likelihood calculations below. The offset model implies $\mu_i/e_i=\exp(\beta_0+x_i^\mathsf{T}\beta)$, a rate that does not vary with exposure when rating variables are held fixed. The estimated-predictor model implies $\mu_i/e_i=\exp(\alpha+x_i^\mathsf{T}\beta)e_i^{\gamma-1}$. Thus $\gamma=1$ recovers proportional exposure scaling, while another value implies a different relationship.

What the Two Specifications Assume

The offset is appropriate when the intended quantity is a rate per unit of exposure and the exposure measure represents proportional time at risk or another suitable denominator. For example, if all other rating characteristics are held fixed, a half-year policy should have half the expected count of a full-year policy under the offset model. The model does not estimate whether that proportionality holds; it imposes it.

In the estimated-predictor model, $\gamma$ is the elasticity of expected count with respect to exposure: multiplying exposure by $c$ multiplies expected count by $c^\gamma$, all else equal. A coefficient below one means count grows less than proportionally with exposure; a coefficient above one means it grows more than proportionally. Neither outcome automatically identifies the reason. Differences in the mix of risks, exposure measurement, or other model misspecification may contribute.

When $\gamma=1$, the two expressions have the same structure after identifying $\alpha$ with $\beta_0$. When $\gamma\ne1$, the estimated-predictor model no longer represents a constant rate per exposure. It should therefore be chosen because that alternative relationship is relevant and supportable, not merely because estimating another coefficient is possible. Exposure values must be positive to use $\log(e_i)$.

Deriving the Constant-Rate Offset Estimate

A simple derivation shows what the offset does in a model with no rating variables and a common claim rate $r$. Assume independent counts $N_i\sim\operatorname{Poisson}(e_i r)$, with positive exposures. The likelihood is the product of the Poisson probabilities. Ignoring terms that do not depend on $r$, its log is

\ell(r)=\sum_i \left[N_i\log(e_i r)-e_i r\right]+\text{constant}.

Differentiate with respect to $r$, set the result equal to zero, and solve. The second derivative is negative for $r>0$, so the solution is the maximum-likelihood estimate when there is at least one claim in total.

\frac{d\ell}{dr} =\frac{\sum_i N_i}{r}-\sum_i e_i=0 \quad\Longrightarrow\quad \widehat r=\frac{\sum_i N_i}{\sum_i e_i}.

This result is claims divided by total exposure, not the unweighted average of the individual ratios $N_i/e_i$. With rating variables, the coefficients generally require likelihood-based estimation, but the offset still fixes the exposure coefficient at one.

Worked Example: Fitting Both Specifications to Two Exposure Groups

A hypothetical insurer observes 20 otherwise comparable policies at $e=0.5$ years, with 10 claims in total, and 20 policies at $e=2$ years, with 20 claims in total. Compare (1) a constant-rate Poisson model with an exposure offset and (2) a Poisson model with an estimated coefficient on log exposure. Assume independent Poisson policy counts and the same model parameters within each exposure group. No interest or mortality assumption is relevant.

Method and offset fit. The total exposure in the first group is $20(0.5)=10$ policy-years; in the second it is $20(2)=40$ policy-years. Thus total exposure is $10+40=50$, and total claims are $10+20=30$. The constant-rate estimate is

\widehat r=\frac{30}{50}=0.6\text{ claims per policy-year}.

The fitted expected count per policy is $0.5(0.6)=0.3$ for the first group and $2(0.6)=1.2$ for the second. The fitted group totals are $20(0.3)=6$ and $20(1.2)=24$, adding to 30 as required by the fitted overall rate.

Method and estimated-predictor fit. Use $\mu(e)=\exp(\alpha)e^\gamma$. With two exposure levels and two parameters, the group means can be matched exactly. The observed mean counts per policy are $10/20=0.5$ and $20/20=1$. Taking the ratio cancels $\exp(\alpha)$:

\frac{1}{0.5} =\left(\frac{2}{0.5}\right)^\gamma \quad\Longrightarrow\quad 2=4^\gamma \quad\Longrightarrow\quad \widehat\gamma=\frac{\log 2}{\log 4}=0.5.

Using the first group to find the intercept gives $\exp(\widehat\alpha)(0.5)^{0.5}=0.5$, so $\exp(\widehat\alpha)=0.5/\sqrt{0.5}=0.7071$ and $\widehat\alpha=\log(0.7071)\approx-0.3466$. Checking the predictions, $0.7071\sqrt{0.5}=0.5$ and $0.7071\sqrt{2}=1$, matching both observed group means. The fitted rates per policy-year are $0.5/0.5=1$ and $1/2=0.5$; they are not constant.

Check and common mistake avoided. The estimated coefficient is $0.5$, not $1$, because the observed mean count only doubles when exposure quadruples. Treating this fitted model as though it were a constant-rate offset model would miss the different implied rates. The two specifications answer different modeling questions; exact fit to these two group means does not by itself establish which relationship will generalize.

Using an Estimated Coefficient in a Rating Model

An estimated coefficient can describe a fitted relationship, but its practical effect is easiest to see by calculating predictions and rates separately. The following example treats fitted coefficients as given; it does not claim that they have been estimated from the small prediction scenario itself.

Worked Example: Interpreting a Fitted Exposure Elasticity

For a hypothetical portfolio, suppose the fitted count model is $\log(\mu)=\log(0.06)+0.8\log(e)+\log(1.5)$ for a specified territory. Assume $N\sim\operatorname{Poisson}(\mu)$. Find the expected count and expected count per exposure unit for a policy with $e=0.25$. Compare with the prediction if the exposure coefficient were fixed at one. No interest or mortality assumption is relevant.

Estimated-predictor calculation. The territory multiplier is $1.5$, and $0.25^{0.8}=\exp(0.8\log(0.25))\approx0.3299$. Therefore

\mu=0.06(0.3299)(1.5)\approx0.02969, \qquad \frac{\mu}{e}=\frac{0.02969}{0.25}\approx0.1188.

So the expected count is approximately $0.02969$, and the implied rate is approximately $0.1188$ per exposure unit. These are model expectations, not a claim that a fraction of a claim will occur.

Offset comparison. If the coefficient were fixed at one while retaining the same reference rate and territory factor, the expected count would be $0.06(0.25)(1.5)=0.0225$. Its rate is $0.0225/0.25=0.09$ per exposure unit. The difference arises from the exposure assumption: the estimated coefficient $0.8$ does not enforce a constant rate. In particular, doubling exposure multiplies expected count by $2^{0.8}\approx1.741$, rather than by 2.

Common mistake avoided. Do not interpret $0.8$ as a multiplicative rating factor of $0.8$. It is an exponent on exposure. Also do not add a fixed $\log(e)$ offset to this same $0.8\log(e)$ predictor unless the intended model truly has exposure exponent $1.8$.

Comparing the Fits and Avoiding Specification Errors

The offset model is a constrained version of the model with an estimated log-exposure coefficient: it imposes $\gamma=1$. A likelihood-ratio comparison can quantify how much fit improves when that constraint is relaxed, provided the models use the same observations and other predictors. The statistic is twice the difference in maximized log likelihoods; under the usual regularity conditions, its large-sample reference distribution for one added coefficient is chi-square with one degree of freedom. This is evidence about fit, not proof that the more flexible relationship is appropriate for every use.

For the two-group example above, the offset fit has fitted group totals 6 and 24, while the estimated-predictor fit has fitted totals 10 and 20. The latter is saturated at the group level. The likelihood-ratio statistic can be calculated as the Poisson deviance of the offset fit against those observed totals:

D=2\left[ 10\log\left(\frac{10}{6}\right) +20\log\left(\frac{20}{24}\right) \right] \approx2.9237.

The terms involving observed and fitted totals cancel here because both pairs sum to 30. Against a chi-square distribution with one degree of freedom, $D$ has an approximate $p$-value of $0.0873$. This comparison does not establish that the offset is true; it indicates that these data alone do not provide strong evidence against the restriction at a 5% level under this approximation. In practice, assess the model in the context of the intended rating target and the data available.

Keep these distinctions clear when specifying or interpreting a model:

Offset: $\log(e)$ enters with coefficient fixed at one, so expected count is proportional to exposure.
Estimated predictor: $\gamma\log(e)$ enters with $\gamma$ estimated, so expected count scales as $e^\gamma$.
Rate interpretation: under the estimated-predictor form, divide expected count by exposure before calling the result a rate; it varies with exposure unless $\gamma=1$.
Model meaning: adding both $\log(e)$ as an offset and $\gamma\log(e)$ as a predictor gives total exponent $1+\gamma$, not $\gamma$.
Data support: estimating $\gamma$ requires informative variation in exposure. A flexible fit cannot compensate for a poorly chosen exposure measure or omitted risk differences.

Key takeaway: Use a frequency offset when the model should enforce expected count proportional to exposure. Estimate a coefficient on log exposure only when the intended model allows a nonproportional relationship, and interpret that coefficient as a count elasticity rather than as a rate multiplier.

Check Your Understanding

For each hypothetical situation, identify the relevant exposure model and calculate the requested quantity.

A Poisson count model has a common rate of $0.08$ claims per exposure unit and uses an offset. What is the expected count at $e=0.75$? (Answer: $0.08(0.75)=0.06$ claims.)
A fitted model is $\mu=0.10e^{0.7}$. By what factor does expected count change when exposure doubles, and does this model impose a constant rate? (Answer: $2^{0.7}\approx1.6245$; no, its implied rate varies with exposure.)
In two otherwise comparable exposure groups, the observed mean counts per policy are $0.4$ at $e=1$ and $0.8$ at $e=4$. In a model $\mu=\exp(\alpha)e^\gamma$, find $\gamma$. (Answer: $\gamma=\log(0.8/0.4)/\log(4/1)=0.5$.)
A constant-rate offset model has 18 claims over 30 exposure units. Find its fitted rate and expected count for a policy with $e=0.5$. (Answer: rate $=18/30=0.6$; expected count $=0.6(0.5)=0.30$.)
A fitted model has $\mu=0.05e^{1.2}$. At $e=2$, find the expected count and implied rate per exposure unit. Does the model enforce proportionality? (Answer: $\mu=0.05(2^{1.2})\approx0.1149$; rate $\approx0.0574$; no, because the exposure exponent is $1.2$, not 1.)

What You'll Learn