From a Fitted Rate to a New Count
A Poisson regression model can describe how a count rate varies with predictors. To make a prediction for a new case, first use its predictor values to calculate its fitted rate, then use its exposure to calculate its expected count. Keeping those steps separate helps avoid a common error: reporting a rate as if it were already the expected number of events.
For example, a hypothetical insurer might model claim counts using vehicle-years of exposure. The fitted rate is measured in claims per vehicle-year; the expected count for a particular policy depends on both that rate and the number of vehicle-years observed. This topic is aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus.
The exposure offset has coefficient fixed at 1. Thus, when predictor values stay the same, multiplying exposure by a factor multiplies the expected count by that factor, while leaving the fitted rate unchanged. The exposure unit must be stated: a rate per vehicle-year and a rate per 1,000 vehicle-years differ by a factor of 1,000.
Why the Rate and Count Formulas Hold
The model specifies the logarithm of the expected count. Exponentiating both sides gives the expected count directly; dividing by exposure then gives the rate. This also shows why the rate is calculated from the predictor part of the linear predictor, while the count includes exposure.
\(\log(\mu)=\log(e)+\mathbf{x}^{\mathsf T}\boldsymbol{\beta}\).
Since the exponential of a sum is the product of exponentials, \(\mu=e\exp(\mathbf{x}^{\mathsf T}\boldsymbol{\beta})\).
\(\mu/e=\exp(\mathbf{x}^{\mathsf T}\boldsymbol{\beta})\), so the expected count per exposure unit is the fitted rate.
In practice, the coefficients are estimated. Replacing \(\boldsymbol{\beta}\) by \(\widehat{\boldsymbol{\beta}}\) gives a plug-in fitted rate and fitted expected count. The conditional Poisson model also implies \(\operatorname{Var}(Y\mid\mathbf{x},e)=\mu\). Therefore, the fitted expected count is a mean, not a guarantee of the count that will occur. A future count is integer-valued and may be above or below its mean.
Worked Example: A New Policy’s Rate and Expected Count
A hypothetical model for annual claim counts uses vehicle-years as exposure and a binary predictor \(x\), equal to 1 for a specified vehicle class and 0 otherwise. The fitted model is \(\log(\mu)=\log(e)+\widehat{\beta}_0+\widehat{\beta}_1x\), with \(\widehat{\beta}_0=\log(0.02)\) and \(\widehat{\beta}_1=\log(1.5)\). Find the fitted rate and expected count for a policy with \(x=1\) and \(e=4\) vehicle-years. Also find the probability of at least one claim, treating the fitted coefficients as fixed.
Method. Evaluate the predictor portion to get the rate, multiply by exposure to get the mean count, and then use the Poisson distribution for the probability. The coefficient values imply a baseline rate of 0.02 claims per vehicle-year and a 1.5 rate multiplier when \(x=1\).
Fitted rate. The predictor portion is \(\widehat{\beta}_0+\widehat{\beta}_1x=\log(0.02)+\log(1.5)=\log(0.03)\). Exponentiating gives a fitted rate of \(0.03\) claims per vehicle-year.
Fitted expected count. Multiply by the four vehicle-years: \(\widehat{\mu}=4(0.03)=0.12\) claims.
Probability check. Under the model, \(Y\sim\operatorname{Poisson}(0.12)\). Thus \(P(Y\geq 1)=1-P(Y=0)=1-\exp(-0.12)=1-0.886920\ldots=0.1131\), to four decimal places.
The results are consistent: the mean is 0.12 claims over four vehicle-years, equivalent to a rate of \(0.12/4=0.03\) claims per vehicle-year. The fitted mean is not a claim count that must occur; the future count remains an integer. This calculation uses the fitted coefficients as fixed and does not include coefficient-estimation uncertainty.
Predicting a Future Count
When a problem asks for an expected count, the fitted mean is the answer. When it asks for the probability of a specified count or event, use the full conditional Poisson distribution with that mean. If \(Y\sim\operatorname{Poisson}(\mu)\), then \(P(Y=k)=\exp(-\mu)\mu^k/k!\) for \(k=0,1,2,\ldots\). For example, \(P(Y\geq 1)=1-\exp(-\mu)\). A predicted count should not be rounded to an integer merely because observed counts are integers.
The model assumption matters: these probability calculations apply when the conditional count is Poisson with the stated mean. They are not obtained from the mean alone if a different count distribution is assumed. For the Poisson model, the conditional variance equals the conditional mean, so the fitted mean also determines the model’s count variability when coefficients are treated as fixed.
Worked Example: Using the Count Distribution at New Predictor Values
A hypothetical model for service incidents per machine-year has fitted rate \(\exp(\widehat{\eta})=0.01(2^a)(3^b)\), where \(a\) and \(b\) are two recorded predictors. For a new machine, \(a=1\), \(b=1\), and exposure is 25 machine-years. Find the fitted rate, expected count, and probability of exactly two incidents.
Method. Substitute the new predictors into the rate, multiply by the exposure, then apply the Poisson probability formula. The model assumes the count conditional on these predictors and exposure is Poisson.
Fitted rate. \(0.01(2^1)(3^1)=0.01(2)(3)=0.06\) incidents per machine-year.
Expected count. \(\widehat{\mu}=25(0.06)=1.5\) incidents.
Probability of exactly two. With \(Y\sim\operatorname{Poisson}(1.5)\), \(P(Y=2)=\exp(-1.5)(1.5^2/2!)=\exp(-1.5)(2.25/2)=0.223130\ldots(1.125)=0.2510\), to four decimal places.
As a check, the probability is less than 1 and the mean is 1.5; an outcome of exactly two is plausible, but it is not the model’s expected count. The expected count remains 1.5, not 2.
Combining Predictions Across Exposures
For several groups, calculate each fitted rate from that group’s predictors and then multiply by its own exposure. The expected total count is the sum of the group expected counts. If the group counts are conditionally independent Poisson variables, their sum is Poisson with mean equal to the sum of their means.
An overall rate across groups is not generally the simple average of the group rates. It is the total expected count divided by total exposure, which weights each group’s rate by its exposure. This is useful when exposures differ, but it does not mean that every group has that same fitted rate.
Worked Example: Aggregating Two Groups With Different Rates
A hypothetical model for equipment failures gives a fitted rate of 0.01 failures per machine-year for group A and 0.02 failures per machine-year for group B. The groups have exposures of 12.5 and 7.5 machine-years, respectively. Assuming their counts are conditionally independent Poisson variables, find each group’s expected count, the total expected count, the overall exposure-weighted rate, and the probability of no failures in either group.
Method. Multiply each fitted rate by its own exposure; add the means; divide the total mean by total exposure for the overall rate. Independence and the Poisson assumptions allow the total-count probability to be calculated from the summed mean.
Group A. Its expected count is \(12.5(0.01)=0.125\) failures.
Group B. Its expected count is \(7.5(0.02)=0.15\) failures.
Total expected count. \(0.125+0.15=0.275\) failures across \(12.5+7.5=20\) machine-years.
Overall rate. Divide the total mean by the total exposure: \(0.275/20=0.01375\) failures per machine-year. Equivalently, the exposure-weighted calculation is \(\{12.5(0.01)+7.5(0.02)\}/20=0.01375\).
Probability of no failures. The total count is Poisson with mean \(0.275\), so \(P(Y_{\mathrm{total}}=0)=\exp(-0.275)=0.759572\ldots=0.7596\), to four decimal places. The probability of at least one failure is therefore \(1-0.7596=0.2404\), to four decimal places using the unrounded zero probability.
The calculations check: \(20(0.01375)=0.275\), recovering the total expected count. The overall rate lies between the two group rates because both exposures are positive.
Common Mistakes and a Reliable Workflow
A frequent error is to exponentiate only the predictor part and report the result as an expected count. That result is a rate per exposure unit. Another is to average rates without accounting for exposure, even though the groups contribute different amounts of exposure. A third is to round the fitted mean to an integer and describe it as the future count.
For instance, in the first worked example, \(\exp(\widehat{\beta}_0+\widehat{\beta}_1x)=0.03\) is the rate in claims per vehicle-year. Calling it an expected count of 0.03 claims for a four-vehicle-year policy is incorrect. The correction is to multiply by exposure: \(4(0.03)=0.12\) expected claims. The corresponding future count has a Poisson distribution with mean 0.12; it is not a deterministic count of 0.12.
A dependable sequence is to check the units, evaluate the rate at the new predictor values, multiply by the new exposure, and only then determine whether the question asks for a mean or a probability. For aggregated predictions, retain each group’s own exposure and rate until its expected count has been calculated.
- Rate: \(\exp(\mathbf{x}^{\mathsf T}\widehat{\boldsymbol{\beta}})\), in counts per exposure unit.
- Expected count: \(e\exp(\mathbf{x}^{\mathsf T}\widehat{\boldsymbol{\beta}})\), in counts.
- Future count: a random integer-valued outcome, modeled here as Poisson with that fitted mean when the model assumptions apply.
Check Your Understanding
For each problem, distinguish the fitted rate from the fitted expected count. Unless stated otherwise, treat fitted coefficients as fixed and assume the conditional count follows the stated Poisson model.
- A fitted rate model is \(0.04(1.5^x)\) events per unit of exposure. For \(x=1\) and exposure 10 units, calculate the fitted rate and expected count. (Answer: rate \(=0.06\) events per exposure unit; expected count \(=0.6\).)
- A hypothetical count is Poisson with fitted mean 0.32 for a new case. Find the probability of no events, to four decimal places. (Answer: \(\exp(-0.32)=0.7261\).)
- A fitted rate is \(0.005(3^x)\) per exposure unit. For \(x=1\) and exposure 40 units, find the expected count and the probability of exactly one event, to four decimal places. (Answer: expected count \(=0.6\); \(P(Y=1)=0.6\exp(-0.6)=0.3293\).)
- Two independent groups have rates 0.03 and 0.05 events per unit of exposure and exposures 20 and 10 units. Find the total expected count, overall exposure-weighted rate, and probability of no events across both groups, to four decimal places where applicable. (Answer: total expected count \(=1.1\); overall rate \(=1.1/30=0.0367\); probability of no events \(=\exp(-1.1)=0.3329\).)
- A fitted rate is \(0.01(2^a)(1.5^b)\) incidents per unit of exposure. For a case with \(a=1\), \(b=2\), and exposure 20 units, find the fitted rate, expected count, and probability of at least one incident, to four decimal places. (Answer: rate \(=0.045\); expected count \(=0.9\); probability \(=1-\exp(-0.9)=0.5934\).)