Writing Exposure Offsets In Frequency Models

Why Frequency Models Need an Exposure Offset

A policy observed for a full year and an otherwise identical policy observed for one quarter do not have the same expected claim count. A frequency model must account for that difference while separating the amount of time at risk from the underlying claim rate. The log-exposure offset does this directly: it scales the expected count by known exposure without asking the model to estimate an exposure effect.

Let $N_i$ be the claim count for observation $i$, and let $e_i$ be its positive exposure, measured in a specified unit such as insured-unit years. A rate $\lambda_i$ is the expected number of claims per exposure unit. Under a Poisson frequency model, the conditional mean count is $\mu_i=e_i\lambda_i$. The predictors in the model describe how the rate varies with risk characteristics; exposure scales that rate into a count for the observation period.

Definition: A Poisson frequency model with a log link and log-exposure offset specifies $$ N_i\mid \mathbf{x}_i,e_i \sim \operatorname{Poisson}(\mu_i),\qquad \log(\mu_i)=\log(e_i)+\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}. $$ Here $e_i>0$ is known exposure, $\mathbf{x}_i$ is the vector of modeled rating characteristics, and $\boldsymbol{\beta}$ is estimated from the data. The term $\log(e_i)$ is an offset: its coefficient is fixed at 1, not estimated. The fitted rate per exposure unit is $\lambda_i=\mu_i/e_i=\exp(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta})$.

The Poisson assumption means that, conditional on the modeled characteristics and exposure, the count has mean and variance both equal to $\mu_i$. The offset construction also assumes that expected claims scale proportionally with exposure over the observation period, with the modeled rate applying throughout that exposure. It does not, by itself, guarantee that the Poisson variance assumption is appropriate.

Exposure must be positive because $\log(0)$ is undefined. A record with zero exposure cannot be given this offset as written; exposure and claim-count data should be checked for inconsistencies before fitting. If $e_i=1$ in the chosen units, then $\log(e_i)=0$, so the offset does not change that record’s linear predictor.

Why the Offset Has Coefficient One

The coefficient of the log exposure is fixed by the relationship between a claim rate and an expected count. Suppose the conditional claim rate is $\lambda_i$ claims per exposure unit. Over $e_i$ exposure units, the expected count is $e_i\lambda_i$. If the log-rate predictor is $\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}$, then $\lambda_i=\exp(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta})$. Taking the logarithm of the expected count gives:

\begin{aligned} \mu_i &= e_i\lambda_i\\ &=e_i\exp(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}),\\ \log(\mu_i)&=\log(e_i)+\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}. \end{aligned}

Thus the offset is not a fitted relationship between exposure and risk. It encodes the known multiplication needed to convert a rate into an expected count. Estimating a coefficient for $\log(e_i)$ would allow the data to change that scaling, so the model would no longer impose the stated proportional relationship.

This specification is consistent with the Poisson-process interpretation: when the rate is constant over time, the expected count over an interval is its rate multiplied by the interval’s exposure. For example, a rate of $0.04$ claims per insured-unit year implies an expected count of $0.01$ over $0.25$ insured-unit years. The rate is not itself the expected count unless exposure equals one unit.

Writing the Model and Reading Its Predictions

A practical specification starts by identifying the count being modeled, the exposure unit, and the rating predictors. For a model with an intercept and one binary rating characteristic $x_i$, write:

\log(\mu_i)=\log(e_i)+\beta_0+\beta_1x_i.

The implied rate is $\exp(\beta_0+\beta_1x_i)$, while the expected count is $e_i\exp(\beta_0+\beta_1x_i)$. The intercept and rating coefficients therefore determine the rate structure; the offset determines how much exposure is represented by the record. The Poisson distribution describes the count around its conditional mean.

A useful specification check is to divide the fitted expected count by the record’s exposure. The result should recover the fitted rate:

\frac{\widehat{\mu}_i}{e_i} = \frac{e_i\exp(\mathbf{x}_i^\mathsf{T}\widehat{\boldsymbol{\beta}})}{e_i} = \exp(\mathbf{x}_i^\mathsf{T}\widehat{\boldsymbol{\beta}}).

Worked Example: Predicting a Half-Year Claim Count

A hypothetical insurer models annual claim frequency for a class of equipment. For a specified risk characteristic $x=1$, the fitted rate is represented by $\beta_0=\log(0.04)$ and $\beta_1=\log(1.5)$. A policy has $e=0.5$ insured-unit years. Find its fitted rate, expected count, and probability of at least one claim.

Method and assumptions. Use a Poisson model with log link and fixed log-exposure offset. The fitted count distribution is $N\mid x,e\sim\operatorname{Poisson}(\mu)$. Its mean and variance are both $\mu$. The rate is constant over the half-year of exposure. Interest and mortality assumptions are not applicable because no discounting or life-contingency calculation is involved.

Step 1: Find the rate. The predictor contribution for the risk characteristic is $\beta_0+\beta_1x=\log(0.04)+\log(1.5)=\log(0.06)$. Therefore, $\lambda=\exp(\log(0.06))=0.06$ claims per insured-unit year.

Step 2: Apply exposure. The offset is $\log(0.5)$, so $\log(\mu)=\log(0.5)+\log(0.06)=\log(0.03)$. Hence $\mu=0.5(0.06)=0.03$ claims.

Step 3: Calculate the probability. For a Poisson variable, $P(N=0)=\exp(-\mu)=\exp(-0.03)=0.9704$, to four decimal places. Thus $P(N\geq 1)=1-P(N=0)=1-0.9704=0.0296$, to four decimal places.

Check. Dividing the expected count by exposure gives $0.03/0.5=0.06$, the rate found in Step 1. This confirms the distinction between the half-year expected count and the annual rate. The common mistake is to report $0.06$ as the half-year expected count; the correction is to multiply the rate by $0.5$.

Worked Predictions for Unequal Exposures

Unequal exposure is the reason the offset matters even when two observations have the same modeled rate. The rate may be the same, but a longer observation period has a larger expected count. Conversely, the model’s rating coefficients should not change solely because one record represents more exposure.

Worked Example: Separating a Rate from a Count

A hypothetical insurer’s fitted model for a risk class has $\beta_0=\log(0.05)$ and a binary characteristic coefficient $\beta_1=\log(1.2)$. A policy has $x=1$ and $e=0.4$ insured-unit years. Calculate the fitted expected count and rate. Also identify the result of omitting the offset and explain the error.

Method and assumptions. Use $\log(\mu)=\log(e)+\beta_0+\beta_1x$, with $N\mid x,e\sim\operatorname{Poisson}(\mu)$. The exposure unit is one insured-unit year, the rate is constant over the observed term, and the model’s Poisson mean and variance are both $\mu$. Interest and mortality assumptions are not applicable.

Step 1: Calculate the fitted rate. The rating predictor is $\beta_0+\beta_1x=\log(0.05)+\log(1.2)=\log(0.06)$. Therefore, $\lambda=\exp(\log(0.06))=0.06$ claims per insured-unit year.

Step 2: Calculate the expected count using the offset. Since $\log(e)=\log(0.4)$, $\log(\mu)=\log(0.4)+\log(0.06)=\log(0.024)$. Thus $\mu=0.4(0.06)=0.024$, and $N\mid x,e\sim\operatorname{Poisson}(0.024)$.

Step 3: Diagnose the omission error. If the offset is omitted, the model calculation is $\exp(\beta_0+\beta_1x)=0.06$. That value is the annual rate, not the expected count for $0.4$ exposure units. Treating it as the count overstates the expected count by $0.06-0.024=0.036$. Correcting the omission means multiplying by exposure: $0.4(0.06)=0.024$.

Check. The fitted expected count divided by exposure is $0.024/0.4=0.06$, which recovers the fitted rate. No rounding was needed for the expected count or rate.

Worked Example: Combining Predictions Across Policies

A hypothetical portfolio has three independent policies. A Poisson model has base rate $0.04$ claims per insured-unit year. The policies have class relativities $1.5$, $1.0$, and $0.5$, and exposures $0.5$, $1.0$, and $2.0$ insured-unit years, respectively. Find each expected count, the portfolio expected count, and the probability of at least one claim in the portfolio.

Method and assumptions. For policy $i$, use $\lambda_i=0.04r_i$ and $\mu_i=e_i\lambda_i$, where $r_i$ is its stated class relativity. Each policy count is conditionally Poisson with mean and variance $\mu_i$, and the three counts are independent. Interest and mortality assumptions are not applicable.

Policy 1. The rate is $0.04(1.5)=0.06$. The expected count is $0.5(0.06)=0.03$.

Policy 2. The rate is $0.04(1.0)=0.04$. The expected count is $1.0(0.04)=0.04$.

Policy 3. The rate is $0.04(0.5)=0.02$. The expected count is $2.0(0.02)=0.04$.

Portfolio expected count. Add the individual means: $0.03+0.04+0.04=0.11$ claims. Because independent Poisson counts sum to a Poisson count, the portfolio total has distribution $\operatorname{Poisson}(0.11)$.

Probability of at least one claim. The probability of no claims is $\exp(-0.11)=0.8958$, to four decimal places. Therefore, the probability of at least one claim is $1-0.8958=0.1042$, to four decimal places.

Check. Total exposure is $0.5+1.0+2.0=3.5$ insured-unit years, and the exposure-weighted mean rate is $0.11/3.5=0.03143$ claims per insured-unit year, to five decimal places. Multiplying that rate by total exposure gives $3.5(0.03143)\approx0.11$, consistent with the sum of the three expected counts. The common mistake is to add the three rates, $0.06+0.04+0.02$, as though they were expected counts; each rate must first be multiplied by its own exposure.

Specification Checks and Common Traps

Before relying on a fitted frequency model, verify that the count, exposure, and offset refer to the same observational unit and period. If exposure is measured in insured-unit years, the resulting rate is per insured-unit year. An exposure entered in a different unit changes the numerical interpretation of the rate, so the unit should be documented and used consistently.

Common errors include:

Omitting the offset. The predictor then gives a modeled rate under this construction, but the model treats it as the count mean for every observation, regardless of exposure.
Estimating an exposure coefficient. In the offset specification, the coefficient on $\log(e_i)$ is fixed at one. It is not an ordinary rating coefficient.
Confusing the rate and expected count. The count mean is $\mu_i=e_i\lambda_i$; the rate is $\lambda_i=\mu_i/e_i$.
Taking the logarithm of zero exposure. The offset is undefined at zero exposure. Investigate such records rather than inserting an arbitrary small positive value.
Forgetting the distributional assumption. The offset determines the mean structure; it does not establish that counts are Poisson or that their conditional variance equals their mean.

In a Poisson likelihood, the offset enters through the specified mean $\mu_i$. The log-exposure term is known for each record, while the model estimates the coefficients describing the rate. This preserves the intended distinction: exposure measures how much risk-time is observed, and the rating structure describes the expected frequency per exposure unit. Correct specification is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Key takeaway: Write $\log(\mu_i)=\log(e_i)+\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}$. The offset coefficient is fixed at one, the fitted rate is $\exp(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta})$, and the expected count is exposure multiplied by that rate.

Check Your Understanding

For each problem, distinguish the fitted rate from the expected count and apply exposure in the stated unit.

A hypothetical class has fitted rate $0.03$ claims per insured-unit year. A policy has relativity $2.0$ and exposure $0.2$ insured-unit years. What is its expected count? (Answer: $0.03(2.0)(0.2)=0.012$ claims.)
A fitted model has $\beta_0=\log(0.05)$, $\beta_1=\log(0.8)$, $x=1$, and exposure $1.5$ insured-unit years. Find the fitted rate and expected count. (Answer: Rate $=0.05(0.8)=0.04$; expected count $=1.5(0.04)=0.06$.)
Two independent policies have the same fitted rate of $0.08$ claims per insured-unit year, with exposures $0.25$ and $0.75$ insured-unit years. What is the expected total count, and what is its distribution? (Answer: Expected total $=0.08(0.25)+0.08(0.75)=0.08$; total count is $\operatorname{Poisson}(0.08)$.)
A model gives $\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}=\log(0.07)$, and a policy has exposure $0.4$ insured-unit years. Write the offset model’s expected count and recover its rate. (Answer: $\mu=0.4(0.07)=0.028$; rate $=0.028/0.4=0.07$ claims per insured-unit year.)
A record shows one claim but has zero exposure. Can the ordinary log-exposure offset be applied to this record as entered? (Answer: No. $\log(0)$ is undefined, and a positive claim count with zero exposure is inconsistent with the stated exposure-count setup; the data must be investigated.)

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