A Worked Poisson Frequency Model With Unequal Exposures

Modeling Frequency When Exposures Differ

Claim counts cannot be compared fairly by looking at totals alone when the groups have different amounts of exposure. A group with more policy-years may have more claims simply because it had more opportunity to produce claims. A Poisson frequency model accounts for this by modeling each expected count as exposure multiplied by a claim rate.

For this tutorial, an observation \(i\) has claim count \(Y_i\), positive exposure \(e_i\), and a class label \(j\). Exposure is measured in the same unit throughout the analysis, such as policy-years. The model assumes that, conditional on the class and exposure, counts are independent and

Y_i \sim \operatorname{Poisson}(e_i\lambda_j), \qquad E[Y_i]=e_i\lambda_j,

where \(\lambda_j\) is the expected claim frequency per exposure unit for class \(j\). The log-link specification writes the same model as

\log E[Y_i]=\log e_i+\beta_0+\beta_j,

where \(\log e_i\) is an offset: its coefficient is fixed at 1, not estimated. One class is chosen as the reference, with its class effect set to zero. Its rate is \(e^{\beta_0}\); another class with effect \(\beta_j\) has rate \(e^{\beta_0+\beta_j}\). This specification assumes a constant rate within each modeled class and a Poisson variance equal to the mean. It does not allow the rate to change with exposure duration or other risk characteristics unless those are represented in the model.

Definition: A fitted class frequency is the fitted expected claim count per unit of exposure. For a class with total claims \(Y_{\cdot j}\) and total exposure \(E_{\cdot j}\), its exposure-adjusted rate estimate is \(Y_{\cdot j}/E_{\cdot j}\). A class relativity is its rate divided by the rate of a stated reference class.

Why Claims Divided by Exposure Gives the Rate Estimate

Suppose a class contains independent observations \(Y_i\sim\operatorname{Poisson}(e_i\lambda)\), with known positive exposures \(e_i\) and one common unknown rate \(\lambda\). Write \(Y=\sum_iY_i\) and \(E=\sum_i e_i\). Ignoring factors that do not depend on \(\lambda\), the likelihood is

L(\lambda)\ \propto\ \lambda^Y e^{-E\lambda}.

For \(Y>0\), the log-likelihood is \(Y\log\lambda-E\lambda\) plus a constant. Its derivative is \(Y/\lambda-E\), and its second derivative is \(-Y/\lambda^2<0\). Thus the unique maximizing rate is \(Y/E\). This is a claim-count total divided by the matching exposure total, not an unweighted average of individual rates.

The condition \(Y>0\) matters. If \(Y=0\), the log-likelihood, apart from a constant, is \(-E\lambda\), which decreases as \(\lambda\) increases over \(\lambda>0\). The likelihood is maximized at \(\lambda=0\) in the closure of the parameter space. Since \(\log 0\) is not finite, there is no finite log-rate coefficient for that class. The ratio \(Y/E=0\) still gives the boundary rate estimate, but setting the score equal to zero does not derive it in this case.

Worked Examples

Worked Example: Two Classes With Unequal Exposure

A hypothetical insurer groups a portfolio into standard and enhanced security classes. The standard class has 36 claims over 1,200 policy-years; the enhanced class has 16 claims over 800 policy-years. Fit the separate-class Poisson rates, take standard as the reference, and predict the expected claims for 250 enhanced-class policy-years.

Method and assumptions. Treat the counts as independent Poisson observations with mean equal to exposure times the class rate. The class rates are constant over the modeled period. Estimate each class rate from its total claims divided by its total exposure.

Standard rate. The exposure is 1,200 policy-years and the count is 36, so

\widehat{\lambda}_{\text{standard}} =\frac{36}{1{,}200} =0.0300

claims per policy-year.

Enhanced rate. Its 16 claims arose over 800 policy-years:

\widehat{\lambda}_{\text{enhanced}} =\frac{16}{800} =0.0200

claims per policy-year. With standard as the reference, the enhanced relativity is

\widehat{r}_{\text{enhanced}} =\frac{0.0200}{0.0300} =\frac{2}{3} \approx 0.6667.

Equivalently, the fitted standard rate is \(e^{\widehat{\beta}_0}=0.0300\), and the enhanced class coefficient is \(\widehat{\beta}_{\text{enhanced}}=\log(2/3)\approx-0.4055\). For 250 enhanced policy-years, the predicted count is

\widehat{\mu} =250(0.0200) =5.00.

The fit reproduces the observed class totals: \(1{,}200(0.0300)=36\) and \(800(0.0200)=16\). The prediction is an expected count, not a guarantee that exactly five claims will occur. The common mistake avoided here is comparing 36 claims with 16 claims without adjusting for the different exposures.

Worked Example: Three-Level Territory Relativities

A hypothetical property program has three territory classes. Territory A records 10 claims over 500 policy-years, B records 30 over 1,000, and C records 60 over 1,500. Use A as the reference. Find each fitted rate, its relativity to A, and the expected claims for 125 policy-years in B and 75 policy-years in C.

Method. Fit a separate rate for each territory, as in a Poisson model with a categorical territory factor and a log-exposure offset.

For A,

\widehat{\lambda}_A=\frac{10}{500}=0.0200.

For B and C,

\widehat{\lambda}_B=\frac{30}{1{,}000}=0.0300, \qquad \widehat{\lambda}_C=\frac{60}{1{,}500}=0.0400.

The relativities to A are therefore

\widehat{r}_B=\frac{0.0300}{0.0200}=1.50, \qquad \widehat{r}_C=\frac{0.0400}{0.0200}=2.00.

Thus B's fitted frequency is 1.50 times A's, and C's is twice A's. In the log-link parameterization, \(\widehat{\beta}_0=\log(0.0200)\), \(\widehat{\beta}_B=\log(1.50)\), and \(\widehat{\beta}_C=\log(2.00)\). The baseline coefficient is approximately \(-3.9120\), while the two territory effects are approximately \(0.4055\) and \(0.6931\).

For 125 policy-years in B, the predicted count is \(125(0.0300)=3.75\). For 75 policy-years in C, it is \(75(0.0400)=3.00\). Across those two exposures, the total predicted count is \(3.75+3.00=6.75\). The fractional prediction is appropriate because it is an expected value. As a check on the class-rate calculations, the fitted counts at the original exposures are \(500(0.0200)=10\), \(1{,}000(0.0300)=30\), and \(1{,}500(0.0400)=60\), exactly the observed totals.

The method assumes that territory is the only modeled rating distinction in this small example. These estimates do not establish that the territory differences are statistically reliable; assessing fit and uncertainty is a separate task.

Worked Example: A Class With No Claims

A hypothetical specialty class has zero claims in 400 policy-years. A comparison class has 24 claims in 800 policy-years. Fit the class rates under separate class-specific Poisson rates, calculate the comparison class's rate relative to the specialty class if possible, and state the prediction for 50 specialty policy-years.

Comparison-class rate. The positive-count class has

\widehat{\lambda}_{\text{comparison}} =\frac{24}{800} =0.0300

claims per policy-year.

Specialty-class rate. Its estimate from claims divided by exposure is \(0/400=0\). This is a boundary maximum, not an interior maximum obtained by solving the score equation. For a positive rate \(\lambda\), the zero-count class likelihood is proportional to \(e^{-400\lambda}\), which decreases as \(\lambda\) increases. Its supremum over positive rates is approached as \(\lambda\) tends to zero.

The log-rate coefficient for the specialty class cannot be finite at this maximum: its rate is \(e^\beta\), which is positive for every finite \(\beta\), and tends to zero only as \(\beta\) tends to negative infinity. Consequently, a ratio using the specialty class as the denominator is undefined, and a finite ordinary log-relativity cannot be reported. The comparison-to-specialty rate ratio would divide by zero.

Under the fitted separate-rate model, the expected count for 50 specialty policy-years is \(50(0)=0\). This is the fitted mean, not a claim that future claims are impossible. The common mistake avoided is to write a finite log coefficient by taking \(\log(0)\), or to claim that setting \(Y/\lambda-E=0\) proves the zero estimate. That score equation has no solution when \(Y=0\).

Reading the Fit and Avoiding Traps

For a categorical frequency model with a separate parameter for each class, the fitted class rate is the total claim count divided by total exposure. With a positive count, the rate is an interior maximum-likelihood estimate. With zero claims and positive exposure, it is a boundary estimate. The fitted count for a new risk is its exposure multiplied by the applicable fitted rate.

The offset makes the exposure adjustment explicit. If exposure is doubled while class characteristics and rate remain unchanged, the expected count doubles. The rate itself does not double. Similarly, a class relativity compares rates, not claim totals or fitted counts from groups with different exposures.

Check the exposure unit.
Use one consistent unit, such as policy-years, for every group and prediction.

Calculate each rate.
Divide a class's total claims by that same class's total exposure.

Choose and state a reference class.
Divide each other rate by the reference rate to obtain its relativity.

Predict counts using exposure.
Multiply the relevant rate by the new exposure; report the result as an expected count.

A tempting but incorrect approach is to compare raw counts or average rates without exposure weights. For example, in the first worked example, an unadjusted comparison of 16 enhanced claims with 36 standard claims might suggest that enhanced frequency is \(16/36\approx0.4444\) of standard frequency. The correct comparison is

\frac{16/800}{36/1{,}200} =\frac{0.0200}{0.0300} \approx0.6667.

The raw-count ratio is wrong because the groups do not have equal exposures. Also avoid treating a predicted expected count as an integer requirement: a model can correctly predict 3.75 claims.

Key takeaway: With a Poisson model and a log-exposure offset, expected count equals exposure times rate. For separate class rates, estimate each rate using claims divided by exposure, then form relativities from the rates. A zero-claim class has a zero boundary rate and no finite log-rate coefficient.

Check Your Understanding

For each problem, assume independent Poisson counts with constant class-specific rates and the stated exposure unit. Give rates to four decimal places where needed, and distinguish expected counts from observed integer counts.

A hypothetical fleet class has 18 claims over 900 vehicle-years; its reference class has 30 claims over 1,500 vehicle-years. Calculate both rates and the first class's relativity to the reference. (Answer: rates \(0.0200\) and \(0.0200\) claims per vehicle-year; relativity \(1.00\).)
A hypothetical building group records 45 claims in 1,500 policy-years, while its reference group has 24 claims in 1,200 policy-years. Find the group's relativity and predicted claims for 80 of its policy-years. (Answer: rate \(0.0300\), reference rate \(0.0200\), relativity \(1.50\), predicted count \(2.40\).)
Three hypothetical service territories have claims and exposures A: 12 and 600, B: 30 and 1,000, C: 0 and 400. Use A as reference. Find the three rates and B's relativity; explain whether C has a finite log-rate coefficient. (Answer: rates A \(0.0200\), B \(0.0300\), C \(0\); B relativity \(1.50\); C is a boundary estimate with no finite log-rate coefficient.)
Two observations in the same hypothetical class have 4 claims over 200 policy-years and 8 claims over 400 policy-years. Estimate the common rate and predict the expected count for 150 policy-years. (Answer: rate \(12/600=0.0200\) claims per policy-year; predicted count \(3.00\).)
A hypothetical class has 20 claims over 500 policy-years. A second class has 42 claims over 1,400 policy-years. Take the first as reference. Calculate the second-class rate, relativity, and predicted count for 350 policy-years in the second class. (Answer: second-class rate \(0.0300\), reference rate \(0.0400\), relativity \(0.75\), predicted count \(10.50\).)

What You'll Learn