Worked Poisson Rate Model With Unequal Exposures

Why Exposure Belongs in a Count Model

Suppose two hypothetical groups are observed for different amounts of time. The group with more exposure may have more recorded events simply because it was observed longer. A comparison of counts alone can therefore be misleading. A Poisson rate model accounts for the amount of exposure while also allowing the expected event rate to differ across predictor groups.

For example, a hypothetical insurer might compare claim counts across policy groups with different total policy-years. The central distinction is between a rate, expected events per unit of exposure, and a count, expected events over a specified exposure amount. The model uses the rate to describe the predictor effect and the exposure to scale the expected count.

Definition: For observation $i$, let $Y_i$ be a count and $e_i>0$ its known exposure amount. A Poisson rate model specifies $Y_i\sim\operatorname{Poisson}(\mu_i)$, with $\mu_i=e_i\lambda_i$, where $\lambda_i$ is the expected event rate per exposure unit. With a log link and predictor vector $x_i$, the model is $$ \log(\mu_i)=\log(e_i)+x_i^\mathsf{T}\beta, $$ or, equivalently, $\log(\lambda_i)=x_i^\mathsf{T}\beta$. The term $\log(e_i)$ is an offset: its coefficient is fixed at one, rather than estimated.

The model assumes that the counts are conditionally independent Poisson random variables, given their predictors and exposures, and that their means follow the stated model. In particular, exposure is assumed to scale the expected count proportionally: doubling exposure while holding predictors fixed doubles the mean. The exposure must be positive, measured on a consistent scale, and relevant to the opportunity for an event to occur. A fitted expected count need not be an integer.

Binary Predictor: Rates, Coefficients, and Predictions

For a binary predictor $x_i$, with $x_i=0$ for the reference group and $x_i=1$ for the comparison group, the model is

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

The reference-group rate is $\exp(\beta_0)$. The comparison-group rate is $\exp(\beta_0+\beta_1)$. Consequently, $\exp(\beta_1)$ is the comparison-to-reference rate ratio. At exposure $e$, the mean count is $e$ times the corresponding rate. This interpretation concerns rates; it does not say that the difference in expected counts is constant across exposure amounts.

Worked Example: Fitting Two Rates From Unequal Policy Exposures

A hypothetical portfolio records claim counts for two policy groups. Group 0 has 2 claims over 100 policy-years and 3 claims over 200 policy-years. Group 1 has 4 claims over 100 policy-years and 5 claims over 150 policy-years. Fit the binary-predictor Poisson rate model, interpret its coefficients, and predict the expected count for 80 policy-years in group 0 and 120 policy-years in group 1.

Method and model. Use $x=0$ for group 0 and $x=1$ for group 1, with $\log(\mu_i)=\log(e_i)+\beta_0+\beta_1x_i$. The fitted rate within each group is its total count divided by its total exposure.

Combine counts and exposures within each group. Group 0 has $2+3=5$ claims and $100+200=300$ policy-years. Its rate estimate is $5/300=1/60=0.0166667$ claims per policy-year. Group 1 has $4+5=9$ claims and $100+150=250$ policy-years. Its rate estimate is $9/250=0.036$ claims per policy-year.

Find the coefficients. The reference-group rate is $\exp(\beta_0)=1/60$, so $\beta_0=\log(1/60)\approx-4.0943$. The rate ratio is $0.036/(1/60)=0.036(60)=2.16$. Thus $\exp(\beta_1)=2.16$ and $\beta_1=\log(2.16)\approx0.7701$.

Predict counts. For 80 policy-years in group 0, the expected count is $80(1/60)=4/3\approx1.3333$. For 120 policy-years in group 1, it is $120(0.036)=4.32$. The predictions are expected counts, so fractional values are appropriate.

Check and common mistake avoided. Multiplying each fitted rate by its stated exposure recovers the two predictions. Comparing only the totals would give $9/5=1.8$, not the rate ratio: that calculation ignores the fact that group 0 has 300 policy-years and group 1 has 250.

Why the Group Rate Estimate Is Count Divided by Exposure

For a binary predictor, observations can be grouped by predictor value when they share the same group-specific rate. The estimate count divided by exposure is not an ad hoc adjustment; it follows from the Poisson likelihood. The argument also explains why unequal exposures are handled by adding the exposures, rather than by taking an unweighted average of observation-specific rates.

Write the likelihood for one group.
Let the independent counts in a group have observations $Y_i=y_i$, exposures $e_i$, and common rate $\lambda$. Since $Y_i\sim\operatorname{Poisson}(e_i\lambda)$, the log likelihood, omitting no parameter-dependent terms, is $\ell(\lambda)=\sum_i[y_i\log(e_i\lambda)-e_i\lambda-\log(y_i!)]$.

Differentiate with respect to the rate.
The score is $d\ell/d\lambda=\sum_i y_i/\lambda-\sum_i e_i=(\sum_i y_i)/\lambda-\sum_i e_i$. Setting the score equal to zero gives $\hat{\lambda}=(\sum_i y_i)/(\sum_i e_i)$, provided the total exposure is positive.

Verify that the stationary point maximizes the likelihood.
The second derivative is $d^2\ell/d\lambda^2=-(\sum_i y_i)/\lambda^2$, which is negative when the total count is positive. Thus the stationary point is a maximum. If the total count is zero, the likelihood is maximized at the boundary $\lambda=0$.

For the two-group model, applying this result separately gives $\hat{\lambda}_0=Y_0/E_0$ and $\hat{\lambda}_1=Y_1/E_1$, where $Y_g$ and $E_g$ are the total count and total exposure in group $g$. The coefficient estimates follow from $\hat{\beta}_0=\log(\hat{\lambda}_0)$ and $\hat{\beta}_1=\log(\hat{\lambda}_1/\hat{\lambda}_0)$, when both estimated rates are positive. If a group has zero total count, its unconstrained group-specific rate estimate is at the boundary zero, and its log-rate coefficient is not finite; the usual finite coefficient calculation does not apply.

Justification: The log link makes the predictor contribution additive on the log-rate scale, so exponentiating a coefficient gives a multiplicative rate effect. The offset contributes $\log(e_i)$, and $\exp(\log(e_i))=e_i$; therefore the mean count is exposure multiplied by the modeled rate.

Exposure Units and the Offset

The numeric value of the intercept depends on the unit used for exposure, but a consistent change of exposure units does not change fitted expected counts or rate ratios. If an exposure value is converted from single units to groups of 100 units, the numerical exposure becomes $e_i/100$, and the numerical rate per new unit becomes 100 times the old rate. The offset and intercept must use the same exposure scale.

Worked Example: Changing Exposure Units Without Changing Predictions

A hypothetical study observes 6 events over 1,200 device-years in group 0 and 6 events over 800 device-years in group 1. Group 1 has $x=1$. Fit the group rates, express the model both per device-year and per 100 device-years, and predict the mean count for 200 device-years in group 1.

Method. Estimate each group rate by total events divided by total exposure. There is one observation in each group, so the totals in this example are the stated counts and exposures.

Rates and rate ratio. Group 0 has rate $6/1{,}200=0.005$ events per device-year. Group 1 has rate $6/800=0.0075$ events per device-year. The rate ratio is $0.0075/0.005=1.5$. Therefore, on the device-year scale, $\beta_0=\log(0.005)\approx-5.2983$ and $\beta_1=\log(1.5)\approx0.4055$.

Convert to 100 device-years. In this scale, 200 device-years is $200/100=2$ exposure units. The group 0 rate is $100(0.005)=0.5$ events per 100 device-years, and the group 1 rate is $100(0.0075)=0.75$. Thus the new intercept is $\log(0.5)\approx-0.6931$, while the group coefficient remains $\log(0.75/0.5)=\log(1.5)\approx0.4055$.

Predict and check. The predicted group 1 count is $2(0.75)=1.5$. Using the original scale gives the same result: $200(0.0075)=1.5$. The exposure conversion changes the intercept but not the fitted count or rate ratio.

Common mistake avoided. Using an exposure of 200 with a rate of 0.75 per 100 device-years would mix units and predict 150 events. The consistent calculation uses either 200 device-years with rate 0.0075 per device-year, or 2 units of 100 device-years with rate 0.75 per 100 device-years.

More Than Two Predictor Groups

A categorical predictor with multiple levels can be represented using a reference group and indicator variables. Each coefficient compares one level with that reference on the log-rate scale. With a separate rate for each category, each rate estimate is its category's total count divided by its total exposure. This is still a rate model with an exposure offset; changing the reference category changes coefficient descriptions, not the fitted rates.

Worked Example: Three Categories and Exposure-Specific Predictions

A hypothetical monitoring program records 5 events over 500 sensor-months in category A, 6 over 400 sensor-months in category B, and 10 over 500 sensor-months in category C. Use category A as the reference. Find the fitted rates and rate ratios, then predict the mean count for 250 sensor-months in category B and 150 sensor-months in category C.

Method. Estimate each category rate by dividing its event total by its exposure total. The log-link model uses category A as the reference and an indicator coefficient for each of B and C.

Estimate the rates. Category A has rate $5/500=0.01$ events per sensor-month. Category B has rate $6/400=0.015$. Category C has rate $10/500=0.02$.

Interpret the coefficients. The intercept is $\log(0.01)\approx-4.6052$. The B-to-A rate ratio is $0.015/0.01=1.5$, so the B coefficient is $\log(1.5)\approx0.4055$. The C-to-A rate ratio is $0.02/0.01=2$, so the C coefficient is $\log(2)\approx0.6931$.

Predict counts. For B at 250 sensor-months, the expected count is $250(0.015)=3.75$. For C at 150 sensor-months, it is $150(0.02)=3$.

Check and common mistake avoided. The B rate is 1.5 times the A rate and the C rate is twice the A rate. The predictions also equal exposure times their respective rates. Treating the coefficients 0.4055 and 0.6931 as rate ratios would be incorrect: exponentiating them gives the rate ratios.

Interpretation and Exam Traps

A reliable calculation keeps three quantities separate: the observed count, the exposure, and the modeled rate. The offset handles exposure; the predictor coefficients describe how the rate changes. For a new observation, first identify the predictor values and exposure units, calculate or select the modeled rate, and then multiply by exposure to obtain the expected count.

Comparing raw counts across unequal exposures. A group with more events may simply have more observation time or more insured exposure. Compare counts per exposure through the fitted rates.
Forgetting the offset in a prediction. The linear predictor includes $\log(e_i)$. Omitting it predicts a rate-scale quantity rather than the expected count for the stated exposure.
Estimating an offset coefficient. In the usual rate model, the coefficient of $\log(e_i)$ is fixed at one to encode proportional scaling. It is not another regression coefficient to interpret or estimate.
Reversing a rate ratio. State the numerator group and denominator group. The coefficient for group 1 relative to group 0 gives $\exp(\beta_1)=\lambda_1/\lambda_0$, not its reciprocal.
Mixing exposure scales. A rate per policy-year must be multiplied by policy-years; a rate per 100 policy-years must be multiplied by the number of 100-policy-year units.

These calculations rely on the stated Poisson rate model: conditionally independent counts, a log-linear rate, and proportional scaling with exposure. If those assumptions are unsuitable, the fitted rates and predictions may not describe the data adequately. The model's interpretation is meaningful only when the event definition and exposure units are consistent across observations.

A practical final check is to write the fitted rate in words, including its exposure unit, and then verify that every predicted mean count is that rate multiplied by the requested exposure. The model is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.

Check Your Understanding

For each problem, use the exposure unit stated in the question and distinguish the rate ratio from the expected count.

A hypothetical group A has 8 events over 400 exposure units, and group B has 9 events over 300 units. With A as the reference, find both rates, the B-to-A rate ratio, and the two log-link coefficients. (Answer: rates $0.0200$ and $0.0300$; rate ratio $1.5$; $\beta_0=\log(0.02)\approx-3.9120$, $\beta_1=\log(1.5)\approx0.4055$.)
A fitted model has $\beta_0=-4$, $\beta_1=\log(2)$, and $x=1$. Find the rate and expected count at exposure 50. (Answer: rate $2e^{-4}\approx0.03663$ per exposure unit; expected count $50(2e^{-4})\approx1.8316$.)
A reference group has 4 events over 200 exposure units, and a comparison group has 9 events over 300 units. Find the comparison-to-reference rate ratio and the expected count for the comparison group at exposure 80. (Answer: rates $0.0200$ and $0.0300$; rate ratio $1.5$; expected count $80(0.03)=2.4$.)
A category has 12 events over 600 exposure units. Express its estimated rate per 100 exposure units, and predict the expected count for 75 exposure units. (Answer: rate $2$ per 100 exposure units; $75$ units are $0.75$ hundreds, so the expected count is $0.75(2)=1.5$.)
Three categories have event totals and exposures A: 4 over 200, B: 9 over 300, and C: 8 over 200. With A as the reference, find the B and C rate ratios and predict the count for 50 exposure units in C. (Answer: rates A $0.0200$, B $0.0300$, C $0.0400$; rate ratios B-to-A $1.5$ and C-to-A $2$; C prediction $50(0.04)=2$.)

What You'll Learn