Separate Poisson Rates Across Risk Classes

Why Estimate Separate Rates?

A portfolio may contain risk classes with different expected claim frequencies. A rate estimated for one class need not describe another: for example, a hypothetical insurer might observe different claim frequencies for two types of insured equipment. When the model assigns each class its own rate, the likelihood must preserve that distinction rather than force all observations to share one parameter.

This tutorial considers complete claim counts observed over known exposures. It focuses on constructing their joint likelihood and estimating the separate class-specific Poisson rates. No claim severities are needed for this calculation.

Definition: Suppose there are $J$ risk classes. For class $j$, let $N_j$ be the number of claims observed over known exposure $e_j>0$, measured in a stated exposure unit, and let $\lambda_j\geq 0$ be the claim rate per exposure unit. The model is $$ N_j\sim\operatorname{Poisson}(e_j\lambda_j),\qquad j=1,\ldots,J, $$ with the counts independent across classes. The parameters $\lambda_1,\ldots,\lambda_J$ are estimated separately.

Here, “complete” means that all claims in each observation window are counted; the observations are not truncated or censored. The exposure amounts are known, the rate is constant within each class over its recorded exposure, and the class counts are independent under the model. Exposure might be policy-years, vehicle-years, or another consistently defined quantity. The rate’s units are the reciprocal of the exposure unit.

The Poisson parameter is the expected number of claims in the observed exposure, $e_j\lambda_j$, not the rate by itself. This distinction matters whenever classes have different exposure amounts: a class with more exposure can have more claims even if its underlying rate is lower.

Joint Likelihood and Separate Estimates

The probability of observing $n_j$ claims in class $j$ is the Poisson probability with mean $e_j\lambda_j$. Independence means that the joint likelihood is the product of these class-specific probabilities.

L(\lambda_1,\ldots,\lambda_J) =\prod_{j=1}^{J} \frac{\exp(-e_j\lambda_j)(e_j\lambda_j)^{n_j}}{n_j!}.

Taking logarithms turns this product into a sum. Terms involving the observed counts and exposures but not the unknown rates may be retained or omitted when maximizing over the rates.

\ell(\lambda_1,\ldots,\lambda_J) =\sum_{j=1}^{J} \left[-e_j\lambda_j+n_j\log(e_j\lambda_j)-\log(n_j!)\right].

Because each $\lambda_j$ appears only in its own summand, maximizing the joint likelihood amounts to maximizing each class’s likelihood separately. For a class with $n_j>0$, differentiating its log-likelihood gives the score equation

\frac{\partial\ell}{\partial\lambda_j} =-e_j+\frac{n_j}{\lambda_j}=0, \qquad \widehat{\lambda}_j=\frac{n_j}{e_j}.

Result: For independent complete Poisson counts with known positive exposures and separate class-specific rates, the maximum likelihood estimate for class $j$ is the observed claim count divided by that class’s exposure: $$ \widehat{\lambda}_j=\frac{n_j}{e_j}. $$ If $n_j=0$ and the parameter space includes zero, the maximum occurs at $\widehat{\lambda}_j=0$.

Why the Estimate Is a Claim Count per Exposure

The score equation gives the estimate directly, but its interpretation can also be seen from the likelihood’s shape. For $n_j>0$, the second derivative of the class-specific log-likelihood is

\frac{\partial^2\ell}{\partial\lambda_j^2} =-\frac{n_j}{\lambda_j^2}<0\qquad(\lambda_j>0).

Thus the log-likelihood is strictly concave for positive rates. It increases where its score is positive and decreases where the score is negative, so its unique stationary point, $n_j/e_j$, is the maximum. At that estimate, the fitted expected count is

e_j\widehat{\lambda}_j =e_j\frac{n_j}{e_j} =n_j.

In other words, the fitted mean count equals the observed count for each class. The rate is the amount of observed count per unit of exposure. This is a likelihood result under the stated Poisson model; it does not mean that a rate estimated from a small count has little uncertainty.

If $n_j=0$, the class log-likelihood, apart from terms not involving the rate, is $-e_j\lambda_j$. It decreases as $\lambda_j$ increases, so its maximum over $\lambda_j\geq0$ occurs at zero. If a model instead requires $\lambda_j>0$, there is no positive rate that attains the maximum: the likelihood approaches its supremum as $\lambda_j$ approaches zero. If $e_j=0$, there is no information about that class’s rate from the observation, so the estimate cannot be obtained by dividing by exposure.

Identify the exposure unit.
For each class, record the total exposure in the same units used to define its rate.

Specify the Poisson mean.
Write the expected count for class $j$ as $e_j\lambda_j$, not just $\lambda_j$.

Form the joint likelihood.
Multiply the class-specific Poisson probabilities, using a distinct rate parameter for each class.

Estimate class by class.
For positive exposure, divide each class’s observed count by its own exposure; check the zero-count boundary separately.

Worked Applications

Worked Example: Two Classes With Different Exposures

A hypothetical equipment insurer records 18 claims over 1,200 equipment-years in class A and 14 claims over 700 equipment-years in class B. Estimate each class’s Poisson claim rate. All counts are complete and the exposures are known. No interest rate or mortality assumption is involved. Assume independent Poisson counts with constant rates $\lambda_A$ and $\lambda_B$ per equipment-year.

Method. Use the separate-rate maximum likelihood estimate, count divided by exposure. The class-specific likelihood is $$ L(\lambda_A,\lambda_B) =\frac{\exp(-1200\lambda_A)(1200\lambda_A)^{18}}{18!} \frac{\exp(-700\lambda_B)(700\lambda_B)^{14}}{14!}. $$

For class A, $$ \widehat{\lambda}_A=\frac{18}{1200}=0.015 $$ claims per equipment-year.

For class B, $$ \widehat{\lambda}_B=\frac{14}{700}=0.020 $$ claims per equipment-year. Equivalently, these are 1.5 and 2.0 claims per 100 equipment-years.

Check. The fitted expected counts are $1200(0.015)=18$ and $700(0.020)=14$, matching the observed counts class by class. A common mistake is to pool the data and calculate $32/1900\approx0.0168$ for both classes. That estimates a common-rate model, not the specified model with separate rates; it would erase the difference between the two class estimates.

Worked Example: Combining Records Within Each Class

A hypothetical fleet portfolio reports two separate observation periods for each of two vehicle classes. Class C has 4 claims over 160 vehicle-years in the first period and 3 claims over 120 vehicle-years in the second. Class D has 7 claims over 200 vehicle-years and 5 claims over 150 vehicle-years. Estimate the rate for each class. Treat all counts as complete and independent, with constant class-specific Poisson rates per vehicle-year. No interest rate or mortality assumption applies.

Method. Multiply the likelihood contributions for records within each class. Since a sum of independent Poisson counts is Poisson with mean equal to the sum of their means, the class totals have the following counts and exposures.

For class C, total claims are $4+3=7$, and total exposure is $160+120=280$ vehicle-years. Thus $$ \widehat{\lambda}_C=\frac{7}{280}=0.025 $$ claims per vehicle-year.

For class D, total claims are $7+5=12$, and total exposure is $200+150=350$ vehicle-years. Thus $$ \widehat{\lambda}_D=\frac{12}{350} =\frac{6}{175} \approx0.03429 $$ claims per vehicle-year.

Check. The fitted total expected counts are $280(0.025)=7$ for C and $350(6/175)=12$ for D. For D, the separate period means are $200(6/175)=48/7$ and $150(6/175)=36/7$, which sum to 12. A common mistake is to average the two period-specific rates equally. The correct estimate weights each period through its exposure: the combined rate is total claims divided by total exposure.

Worked Example: A Class With No Observed Claims

In a hypothetical two-class portfolio, class M has zero claims over 80 policy-years, while class N has 11 claims over 220 policy-years. Estimate both rates, allowing a rate of zero. Assume complete counts, independent Poisson models, known exposures, and constant rates per policy-year. No interest rate or mortality assumption is involved.

Method. Maximize the two class likelihood contributions separately, using the boundary solution when a count is zero. For M, the likelihood contribution is $$ L_M(\lambda_M)=\exp(-80\lambda_M), $$ because the probability of observing zero claims is the Poisson probability at zero.

For $\lambda_M\geq0$, this likelihood is at most 1 and decreases as $\lambda_M$ increases. Its maximum is therefore at $$ \widehat{\lambda}_M=0 $$ claims per policy-year.

For N, the positive-count estimate is $$ \widehat{\lambda}_N=\frac{11}{220}=0.05 $$ claims per policy-year.

Check. At the estimates, the fitted means are $80(0)=0$ for M and $220(0.05)=11$ for N. A zero estimate does not imply that the true rate is known to be zero; it is the maximum likelihood estimate for the observed zero count under this model. A common mistake is to use a positive-count score formula $n/e$ without checking the boundary. Here it gives zero, which is valid only because zero is included in the parameter space.

Assumptions, Interpretation, and Exam Traps

The estimate $n_j/e_j$ depends on the model and data structure. It is appropriate when the count is complete, exposure is known and positive, and a single Poisson rate applies throughout the class’s observed exposure. If the exposure is recorded in policy-months, for instance, the estimated rate is per policy-month unless the exposure is converted to policy-years. Changing the exposure unit changes the numerical rate by the corresponding scale factor, not the fitted expected count.

The separate-rate model does not assert that rates across classes are equal. It assigns one unknown parameter to each class and estimates each using that class’s own count and exposure. Pooling data across classes changes the model unless a common rate is explicitly assumed. Similarly, multiplying the rate by exposure is necessary to obtain the Poisson mean; treating the rate itself as the mean is generally wrong unless exposure equals one unit.

A useful likelihood check is that the joint log-likelihood separates into a sum with one rate per class. If a proposed estimate for one class depends on another class’s count, that is a warning sign for this separate-rate model. This direct factorization relies on the stated independent Poisson structure; it should not be carried over automatically to models with dependent counts or exposure-dependent rates.

Key takeaway: With complete independent Poisson counts and known exposure, estimate each class rate using that class’s count divided by its own exposure. Aggregate counts and exposure within a class, keep different classes separate, and treat a zero count as a boundary case when zero rates are allowed.

Check Your Understanding

For each problem, use the stated exposure unit and assume complete counts with independent Poisson distributions and constant rates within risk class.

A hypothetical repair plan records 15 claims over 600 machine-years in class A and 9 claims over 300 machine-years in class B. Find each separate rate per machine-year. (Answer: $\widehat{\lambda}_A=0.025$, $\widehat{\lambda}_B=0.030$ claims per machine-year.)
A class has 2 claims over 40 policy-years in one period and 5 claims over 160 policy-years in another. Find its combined maximum likelihood rate per policy-year. (Answer: $7/200=0.035$ claims per policy-year.)
A portfolio observes 8 claims over 160 unit-years in class R and 8 claims over 320 unit-years in class S. Find the separate rates and explain why equal claim counts do not imply equal rates. (Answer: $\widehat{\lambda}_R=0.05$ and $\widehat{\lambda}_S=0.025$ claims per unit-year; the exposures differ.)
A hypothetical class has zero claims over 125 policy-years. If the rate parameter may be zero, find its maximum likelihood estimate and fitted expected count. (Answer: $\widehat{\lambda}=0$ claims per policy-year; fitted expected count $=0$.)
Class U has 3 claims over 120 exposure units, while class V has 9 claims over 180 exposure units. Calculate the separate rates and the rate that would result from incorrectly imposing one common rate; identify which model the pooled value estimates. (Answer: separate rates are $0.025$ and $0.05$; the pooled rate is $12/300=0.04$ per exposure unit and estimates a common-rate model.)

What You'll Learn