Worked Poisson Model With Rate Ratios And Offsets

Why Exposure Belongs in a Count Model

A count of claims, incidents, or service calls depends both on the underlying event rate and on how much opportunity there was for an event to occur. A portfolio observed for 500 policy-years would generally be expected to generate more claims than an otherwise similar portfolio observed for 100 policy-years. A Poisson regression model handles this difference by including exposure as an offset.

The offset is especially useful when a problem asks for fitted rates as well as expected counts. The rate describes the modeled frequency per unit of exposure; the expected count also incorporates the amount of exposure. These are related quantities, but they are not interchangeable.

Definition: Suppose $Y_i$ is a count observed over positive exposure $E_i$. A Poisson log-link model with predictors $x_{i1},\ldots,x_{ip}$ assumes that, conditional on the predictors and exposure, $Y_i$ has a Poisson distribution with mean $\mu_i$, and $$ \log(\mu_i)=\log(E_i)+\beta_0+\beta_1x_{i1}+\cdots+\beta_px_{ip}. $$ The term $\log(E_i)$ is an offset: its coefficient is fixed at 1. The fitted rate per unit of exposure is $$ \lambda_i=\frac{\mu_i}{E_i} =\exp(\beta_0+\beta_1x_{i1}+\cdots+\beta_px_{ip}), $$ and the fitted expected count is $\mu_i=E_i\lambda_i$.

The standard Poisson model also assumes that observations are conditionally independent and that, for each observation, the conditional mean and variance are both $\mu_i$. The model is appropriate when the response is a count and the expected count scales proportionally with exposure, given the included predictors. If exposure does not represent comparable opportunities for events, or if the count-generating process does not meet the model assumptions, the offset formula may not be appropriate as stated.

From Coefficients to Fitted Rates

For a categorical predictor, one category is chosen as the reference category and coded as 0; another category can be coded as 1. If its estimated coefficient is $\hat\beta$, holding other predictors fixed, changing that predictor from 0 to 1 multiplies the fitted rate by $\exp(\hat\beta)$. For a numeric predictor, increasing it by one unit multiplies the fitted rate by $\exp(\hat\beta)$, provided the model includes it linearly on the log-rate scale.

Rate-ratio result: For two predictor profiles $A$ and $B$, the fitted rate ratio is $$ \frac{\hat\lambda_A}{\hat\lambda_B} =\exp\left(\sum_{j=1}^{p}\hat\beta_j(x_{Aj}-x_{Bj})\right). $$ The intercept cancels. Exposure does not appear in this rate ratio because the comparison is between rates, not expected counts. If comparing fitted counts instead, the exposure ratio must also be included: $$ \frac{\hat\mu_A}{\hat\mu_B} =\frac{E_A}{E_B}\frac{\hat\lambda_A}{\hat\lambda_B}. $$

The coefficient interpretation follows directly from the model. For two profiles that differ only in predictor $x_j$ by one unit, their log fitted rates differ by $\hat\beta_j$. Exponentiating that difference gives the ratio of the rates. Exposure cancels when each mean is divided by its own exposure; it remains in the expected-count ratio when the exposures differ.

\log\left(\frac{\hat\lambda_A}{\hat\lambda_B}\right) =\log(\hat\lambda_A)-\log(\hat\lambda_B) =\sum_{j=1}^{p}\hat\beta_j(x_{Aj}-x_{Bj}).

Worked Calculations

The following examples use hypothetical event-count data. In each case, the parameter values are treated as given fitted model estimates; the task is to apply the model, not to estimate its coefficients.

Worked Example: Two Predictors and Different Exposures

A hypothetical insurer models claim counts using annual policy exposure. Let $x=1$ indicate a coastal territory and $z=1$ indicate a specified vehicle class. The fitted model is $$ \log(\mu_i)=\log(E_i)+\log(0.02)+x_i\log(1.5)+z_i\log(2). $$ Calculate the fitted rate and expected count for a coastal policy group in the specified vehicle class with $E=50$ policy-years. Also calculate them for a coastal group in the reference vehicle class with $E=80$ policy-years. Find the rate ratio for the first group relative to the second. Assume the conditional counts are independent Poisson variables with the model means shown.

Method. First remove the offset to obtain each fitted rate. Then multiply each rate by its own exposure to obtain the expected count.

For the first group, $x=1$ and $z=1$. Its fitted rate is $$ \hat\lambda_1=\exp\{\log(0.02)+\log(1.5)+\log(2)\} =0.02(1.5)(2)=0.06 $$ claims per policy-year. Its expected count is $$ \hat\mu_1=50(0.06)=3. $$ Thus, $Y_1$ is modeled as Poisson with parameter 3.

For the second group, $x=1$ and $z=0$. Its fitted rate is $$ \hat\lambda_2=\exp\{\log(0.02)+\log(1.5)\} =0.02(1.5)=0.03 $$ claims per policy-year. Its expected count is $$ \hat\mu_2=80(0.03)=2.4. $$ Thus, $Y_2$ is modeled as Poisson with parameter 2.4.

The groups have the same territory, so the only predictor difference is vehicle class. The fitted rate ratio is $0.06/0.03=2$. As a check, the count ratio is $3/2.4=1.25$, not 2, because the first group has only $50/80=0.625$ as much exposure: $0.625(2)=1.25$. The common mistake is to treat the expected-count ratio as the rate ratio.

Worked Example: An Exposure-Adjusted Group Comparison

A hypothetical service network models incident counts per operating hour. Let $u=1$ indicate an urban site and $s=1$ indicate that a site uses a specified safety system. Its fitted model is $$ \log(\mu_i)=\log(E_i)+\log(0.02)+u_i\log(1.4)+s_i\log(0.75). $$ Site A is urban, uses the safety system, and has 300 operating hours. Site B is rural, does not use the system, and has 100 operating hours. Calculate both fitted rates and expected counts, and compare their rates. Assume independent Poisson counts with these fitted means.

Method. Apply each predictor multiplier to the baseline rate, then apply the site’s exposure. For the adjusted rate comparison, compare the predictor profiles without including exposure.

For Site A, $u=1$ and $s=1$, so $$ \hat\lambda_A=0.02(1.4)(0.75)=0.021 $$ incidents per operating hour. Its expected count is $$ \hat\mu_A=300(0.021)=6.3. $$ The modeled count has a Poisson distribution with parameter 6.3.

For Site B, $u=0$ and $s=0$, so its fitted rate is the baseline: $$ \hat\lambda_B=0.02 $$ incidents per operating hour. Its expected count is $$ \hat\mu_B=100(0.02)=2. $$ The modeled count has a Poisson distribution with parameter 2.

The exposure-adjusted rate ratio for Site A relative to Site B is $$ \frac{\hat\lambda_A}{\hat\lambda_B} =\frac{0.021}{0.02}=1.05. $$ Equivalently, the predictor multipliers give $1.4(0.75)=1.05$. Site A’s fitted rate is 5% higher, holding the model’s specified effects together. Its fitted expected count is $6.3/2=3.15$ times Site B’s expected count because it also has three times the exposure; indeed, $3(1.05)=3.15$. The common mistake is to call 3.15 the adjusted rate ratio.

Worked Example: A Numeric Predictor and an Exposure-Unit Check

A hypothetical equipment operator models event counts using exposure measured in thousands of machine-hours. The fitted model is $$ \log(\mu_i)=\log(E_i)+\log(0.04) +\frac{\text{age}_i}{5}\log(1.2) +r_i\log(1.5)+q_i\log(0.8), $$ where age is in years, $r=1$ indicates a designated region, and $q=1$ indicates use of an inspection program. Calculate the fitted rate and expected count for equipment aged 10 years, in the designated region, with the inspection program, observed for 25,000 machine-hours. Compare its fitted rate with that of 5-year-old equipment in the reference region without the program. Assume independent Poisson counts conditional on the model variables.

Method. Convert the exposure to the model’s stated unit, evaluate the age effect in five-year increments, and then multiply the rate by exposure to obtain the expected count.

The exposure is $E=25{,}000/1{,}000=25$ thousand machine-hours. For the first equipment profile, age is 10 years, so age contributes two five-year increments. Also, $r=1$ and $q=1$. Its fitted rate is $$ \hat\lambda_A=0.04(1.2)^2(1.5)(0.8) =0.04(1.44)(1.2)=0.06912 $$ events per thousand machine-hours. Its expected count is $$ \hat\mu_A=25(0.06912)=1.728. $$ The count is modeled as Poisson with parameter 1.728.

The comparison profile is 5 years old, in the reference region, with no program. Its age effect is one five-year increment, so $$ \hat\lambda_B=0.04(1.2)=0.048 $$ events per thousand machine-hours. The fitted rate ratio is $$ \frac{\hat\lambda_A}{\hat\lambda_B} =\frac{0.06912}{0.048}=1.44. $$ This also follows by comparing multipliers: the age difference contributes $1.2$, region contributes $1.5$, and the inspection program contributes $0.8$, giving $1.2(1.5)(0.8)=1.44$. The comparison does not depend on exposure because it is a rate ratio. The common mistake is to use $E=25{,}000$ in the model even though its exposure unit is one thousand machine-hours; the correct offset is $\log(25)$, not $\log(25{,}000)$.

Practical Checks and Common Traps

An offset model is often easiest to calculate by separating the work into three pieces: exposure, baseline rate, and predictor multipliers. This keeps the rate calculation distinct from the count calculation and makes the units visible.

Confirm the exposure unit.
Express the exposure in the same units used to define the rate, such as policy-years or thousands of machine-hours.

Calculate the fitted rate.
Exponentiate the predictor portion of the linear predictor, or multiply the baseline rate by the relevant rate-ratio factors.

Calculate the expected count.
Multiply the fitted rate by the stated exposure. This value is the Poisson mean; it need not be an integer.

Make the requested comparison.
Use a rate ratio for frequency per exposure unit. For a count ratio, include the ratio of exposures as well.

A useful final check is dimensional: rate multiplied by exposure must have units of a count. If the exposure is doubled with predictors held fixed, the expected count doubles while the fitted rate remains unchanged. If a predictor changes, the rate changes according to its multiplicative effect. These checks can reveal an omitted offset, a reversed comparison, or an exposure-unit error.

The offset formulation also encodes a substantive assumption: within the modeled conditions, twice as much exposure corresponds to twice the expected count. It is not merely a correction applied after fitting. If observations are dependent, or if their variability is not compatible with the Poisson assumptions, the fitted values may still be computable but the stated probability model may not describe the data adequately.

Key takeaway: In a Poisson log-link model with an exposure offset, the fitted rate is the exponential of the predictor part of the model. Multiply that rate by exposure to obtain the expected count. Predictor coefficients produce adjusted rate ratios; unequal exposures affect count ratios, not rate ratios.

Check Your Understanding

For each hypothetical setting, use the stated model and exposure units. Check whether the requested quantity is a rate or an expected count.

A model has fitted rate $0.03$ per policy-year for a specified risk profile. What is the expected count for 40 policy-years? (Answer: $1.2$ claims.)
A baseline fitted rate is $0.01$ per operating hour. A site is in a category with rate ratio 1.4 and uses a device with rate ratio 0.5. For exposure of 100 hours, calculate its fitted rate and expected count. (Answer: Rate $0.007$ incidents per hour; expected count $0.7$.)
A model has a baseline rate of $0.02$ per exposure unit, a vehicle-class rate ratio of 1.6, and a driver-category rate ratio of 0.75. Find the fitted rate and expected count for 80 exposure units in both nonreference categories. (Answer: Rate $0.024$ per unit; expected count $1.92$.)
Two profiles have fitted rates of $0.025$ and $0.02$ per unit, with exposures 120 and 60 units, respectively. Find the first-to-second rate ratio and expected-count ratio. (Answer: Rate ratio $1.25$; expected-count ratio $2.5$.)
A numeric predictor has a rate ratio of 1.1 per five-unit increase. What is the fitted rate ratio for a 10-unit increase, holding all other predictors fixed? (Answer: $1.1^2=1.21$.)

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