From a Fitted Rate to a Predicted Count
A frequency model may report a predicted claim rate per policy-year, while the policy being priced covers only part of a year. A portfolio may also contain policies with different terms or different fitted rates. To predict counts correctly, match each fitted rate to the exposure period over which it applies. A rate per policy-year must be multiplied by exposure measured in policy-years; a rate per policy-month must be multiplied by policy-months.
In the Poisson frequency model, the exposure-adjusted mean is the predicted number of claims, not the annualized rate itself. The mean may be fractional even though an observed claim count is an integer. For instance, an expected count of 0.12 does not mean that a policy will have 0.12 of a claim; it is the model’s average over repeated comparable risks.
Why Counts Scale with Exposure
Suppose the modeled claim process for a risk has a constant rate \(\lambda\) claims per policy-year while its rating characteristics remain unchanged. In an exposure interval of length \(e\) policy-years, the Poisson model specifies a count \(N\) with mean \(e\lambda\). Thus, if the exposure is halved while the rate remains the same, the expected count is also halved.
The same result follows from the exposure-offset form of a Poisson generalized linear model. If \(x_i\) is the vector of rating variables and \(\beta\) is the fitted coefficient vector, then the model with a log link and exposure offset is:
Exponentiating both sides gives \(E[N_i\mid x_i,e_i]=e_i\exp(x_i^\mathsf{T}\beta)\). The fitted rate per exposure unit is \(\lambda_i=\exp(x_i^\mathsf{T}\beta)\); multiplying by \(e_i\) gives the fitted mean count. This is why the exposure enters multiplicatively on the mean scale and additively as \(\log(e_i)\) on the log scale.
For separate policy groups with rates \(\lambda_g\) and exposures \(e_g\), the group mean is \(e_g\lambda_g\). If the group counts are independent Poisson variables, their total is Poisson with mean equal to the sum of the group means. This addition is essential: a portfolio’s predicted total is found by summing exposure-adjusted counts, not by averaging rates without regard to exposure.
Worked Examples
Worked Example: A Short-Term Policy Block
A hypothetical insurer’s fitted frequency for a rating class is 0.24 claims per policy-year. It covers 45 otherwise comparable policies for five months each. Find the predicted count for the block and the probability that the block has no claims. Assume the fitted rate is constant during each term, and the policy counts are independent Poisson variables. No interest or mortality assumption is involved.
Step 1: Convert the term to the rate’s exposure unit. Five months is \(5/12\) of a year per policy. The total exposure is therefore \(45(5/12)=18.75\) policy-years.
Step 2: Calculate the predicted count. The fitted rate is \(\lambda=0.24\) claims per policy-year, so the block mean is \(18.75(0.24)=4.50\) claims. Equivalently, each policy has mean \((5/12)(0.24)=0.10\), and \(45(0.10)=4.50\). The aggregate count is Poisson with parameter 4.50 under the stated independence and rate assumptions.
Step 3: Calculate the probability of zero claims. For a Poisson count with mean 4.50, \(P(N=0)=e^{-4.50}=0.0111\), to four decimal places.
Check. The per-policy and aggregate exposure calculations both give a mean of 4.50. The zero-claim probability is small because the expected total count is 4.50. The method avoids treating the five-month rate as though it were a full-year expected count.
Worked Example: Combining Groups with Different Terms
A hypothetical portfolio has three groups. Group A has 120 policies with a fitted rate of 0.10 claims per policy-year and a three-month term. Group B has 50 policies at 0.22 per policy-year with a nine-month term. Group C has 20 policies at 0.35 per policy-year with a six-month term. Find each group’s predicted count, the portfolio total, and the exposure-weighted portfolio rate. Assume independent Poisson counts within and across groups, with each group’s rate constant during its term. No interest or mortality assumption is involved.
Group A. Its exposure is \(120(3/12)=30\) policy-years. Its predicted count is \(30(0.10)=3.00\).
Group B. Its exposure is \(50(9/12)=37.5\) policy-years. Its predicted count is \(37.5(0.22)=8.25\).
Group C. Its exposure is \(20(6/12)=10\) policy-years. Its predicted count is \(10(0.35)=3.50\).
The portfolio’s total exposure is \(30+37.5+10=77.5\) policy-years, and its predicted total count is \(3.00+8.25+3.50=14.75\). Under the assumptions, the total count is Poisson with parameter 14.75. The exposure-weighted portfolio rate is \(14.75/77.5=0.1903226\ldots\), or 0.1903 claims per policy-year to four significant figures.
Check and common mistake. Multiplying 77.5 policy-years by 0.1903226 gives approximately 14.75, as required. A wrong calculation would take the unweighted average rate, \((0.10+0.22+0.35)/3=0.223333\ldots\), then multiply it by 77.5 to get approximately 17.31 claims. That average gives each group’s rate equal weight even though their exposures differ. The correction is to multiply each group’s rate by its own exposure, add the predicted counts, and divide by total exposure only if an overall rate is also needed.
Worked Example: A Rating Change During the Term
A hypothetical block contains 25 policies, each exposed for four months in one rating class and then eight months in another. The fitted rate is 0.18 claims per policy-year in the first class and 0.30 in the second. Find the expected total count. Assume the indicated rates apply throughout their respective segments and that disjoint segment counts are independent Poisson variables. No interest or mortality assumption is involved.
Step 1: Calculate exposure and mean for the first segment. The block has \(25(4/12)=25/3\) policy-years in the first class. Its expected count is \((25/3)(0.18)=1.50\).
Step 2: Calculate exposure and mean for the second segment. The block has \(25(8/12)=50/3\) policy-years in the second class. Its expected count is \((50/3)(0.30)=5.00\).
Step 3: Add the segment means. The expected total is \(1.50+5.00=6.50\) claims. With independent Poisson segment counts, the aggregate count is Poisson with parameter 6.50.
Check and common mistake. Each policy’s expected count is \((4/12)(0.18)+(8/12)(0.30)=0.06+0.20=0.26\); multiplying by 25 gives 6.50. Applying only the second, higher rate to all 12 months would instead give \(25(0.30)=7.50\), which ignores the first four months in the original class. Segmenting exposure at the rating change avoids that error.
Practical Rules for Exposure Scaling
The arithmetic is simple, but reliable predictions depend on the model inputs and exposure definition being aligned. Apply these checks before combining results:
Confirm whether the fitted rate is per policy-year, vehicle-year, or another exposure unit. Do not assume that a rate labeled “annual” uses the same denominator as the exposure field.
For a term of \(t\) months measured on a twelve-month policy-year basis, use \(t/12\) policy-years per policy. Apply the same conversion consistently across the portfolio.
If rating variables or the applicable fitted rate change during a term, divide the term into segments and use the appropriate rate for each segment.
Calculate \(e_g\lambda_g\) for each group or segment and sum those means. Use total exposure as the denominator only when calculating an exposure-weighted portfolio rate.
A simple proportional scaling is appropriate when the fitted rate is intended to remain constant per unit of exposure over the period being projected. If claim frequency varies materially by season, calendar period, or duration within a term, one constant annual rate may not represent each subperiod. Use rates appropriate to the periods being projected and sum their exposure-adjusted means. Likewise, this calculation predicts the model’s expected count; it does not by itself establish that the frequency model is well calibrated or that its assumptions hold.
Check Your Understanding
For each problem, keep the rate’s time unit and the exposure unit consistent.
- A hypothetical class has a fitted rate of 0.16 claims per policy-year. What count is predicted for 75 policies insured for nine months each? (Answer: 9.00 claims.)
- A fitted rate is 0.27 claims per policy-year. Find the expected count for 40 policies with four-month terms. (Answer: 3.60 claims.)
- A portfolio has 60 policies at 0.12 claims per policy-year for six months and 30 policies at 0.28 per policy-year for three months. What is the predicted total count? (Answer: 5.70 claims.)
- Eighteen policies each spend five months at a fitted rate of 0.24 claims per policy-year and seven months at 0.12. What is their expected total count? (Answer: 3.06 claims.)
- A portfolio has 40 policies at 0.20 claims per policy-year for three months, 25 at 0.32 for nine months, and 15 at 0.08 for twelve months. Find the total predicted count and the exposure-weighted portfolio rate. (Answer: 9.20 claims and 0.2103 claims per policy-year.)