From a Claim Rate to an Expected Count
A claim rate describes frequency relative to an exposure unit; it is not itself a count. A rate might be stated per policy-year, per vehicle-month, or per thousand operating hours. To find the expected number of claims, first express the amount of exposure in the same units as the rate, then multiply. This unit matching is the central step in Poisson rate scaling.
The model applies when claims occur at a constant rate over the exposure being studied. For example, if the rate is per policy-year, each policy-year contributes the same expected number of claims. If there are more policy-years, the expected count increases in direct proportion. This model does not apply unchanged if the rate varies across the exposure period or across groups; those situations require a rate appropriate to each part.
Exposure can combine the number of insured units with the time each is covered or observed. For example, 80 vehicles observed for half a year contribute 40 vehicle-years of exposure. This multiplication is appropriate when the rate is stated per vehicle-year and the same rate is assumed to apply to each vehicle throughout the observed period.
A useful check is dimensional: multiplying “claims per exposure unit” by “exposure units” leaves “claims.” If the units do not cancel in this way, the rate or exposure has not yet been converted correctly.
Why the Mean Scales Linearly
The linear rule follows from the additivity of independent Poisson counts. Consider total exposure \(E\), divided into \(m\) equal pieces. Under a constant-rate model with independent counts in disjoint exposure pieces, each piece has mean \(rE/m\). The total count is the sum of those counts, so its Poisson mean is the sum of their means.
The same reasoning works for unequal pieces: if piece \(j\) has exposure \(E_j\), its mean is \(rE_j\), and the total mean is \(r\sum_j E_j\). Thus, when the rate stays constant, doubling exposure doubles the mean. The result relies on using one constant rate for all included exposure and on the Poisson count model; it is not a claim that every realized count doubles.
Worked Applications
Worked Example: Vehicles Covered for a Partial Year
A hypothetical insurer models claim counts for a group of 250 vehicles using a constant rate of 0.18 claims per vehicle-year. The vehicles are covered for six months. Find the total exposure and the Poisson mean for the group. State the distribution of the count.
Method. Convert six months to one-half year, then multiply the number of vehicles by their covered time to obtain vehicle-years. The rate is given per vehicle-year.
Each vehicle contributes \(6/12=0.5\) vehicle-year. Therefore total exposure is $$ E=250(0.5)=125\text{ vehicle-years}. $$
The rate is \(r=0.18\) claims per vehicle-year, so $$ \lambda=rE=0.18(125)=22.5. $$ Under the stated constant-rate Poisson model, the group claim count satisfies \(N\sim\operatorname{Poisson}(22.5)\). The expected count is 22.5 claims; the actual count must be a whole number.
Check. At 0.18 claims per vehicle-year, 100 vehicle-years correspond to 18 expected claims. The additional 25 vehicle-years correspond to \(0.18(25)=4.5\), giving \(18+4.5=22.5\). A common mistake is to use 250 as the exposure and get \(0.18(250)=45\), which treats six months as a full year. The corrected exposure is 125 vehicle-years.
Worked Example: Rate per Thousand Operating Hours
A hypothetical equipment program uses a constant claim rate of 3.6 claims per 1,000 operating hours. Fourteen machines each operate for 250 hours during the observation period. Calculate the total exposure in thousands of hours and the Poisson mean. What would the mean be if the same machines instead operated for 300 hours each, with the rate unchanged?
Method. First calculate total operating hours, then express that amount in units of 1,000 hours to match the rate.
For the initial period, $$ E_{\text{hours}}=14(250)=3{,}500\text{ operating hours}, \qquad E=\frac{3{,}500}{1{,}000}=3.5 \text{ units of 1,000 hours}. $$ The rate per rate-unit is \(r=3.6\), so $$ \lambda=3.6(3.5)=12.6. $$ The initial count is modeled as \(N\sim\operatorname{Poisson}(12.6)\).
With 300 hours per machine, total exposure becomes $$ E_{\text{hours,new}}=14(300)=4{,}200\text{ hours}, \qquad E_{\text{new}}=4.2\text{ units of 1,000 hours}. $$ Thus the new mean is $$ \lambda_{\text{new}}=3.6(4.2)=15.12. $$ Equivalently, exposure rose by a factor of \(4{,}200/3{,}500=1.2\), and \(12.6(1.2)=15.12\).
Check. The new exposure is 700 hours greater, or 0.7 units of 1,000 hours. That adds \(3.6(0.7)=2.52\) to the mean, and \(12.6+2.52=15.12\). A common mistake is to multiply 3.6 directly by 3,500, forgetting that the rate is per 1,000 hours. Converting the exposure to 3.5 rate-units avoids that factor-of-1,000 error.
Worked Example: Rescaling Across Policy Counts and Durations
A hypothetical portfolio has a constant claim rate of 0.024 claims per policy-month. A first group consists of 1,800 policies covered for nine months. A second group consists of 2,400 policies covered for six months. Find each group’s total exposure and expected count, then compare the means.
Method. Because the rate is per policy-month, multiply policies by months directly; no conversion to policy-years is needed.
For the first group, $$ E_1=1{,}800(9)=16{,}200\text{ policy-months}, \qquad \lambda_1=0.024(16{,}200)=388.8. $$ Thus \(N_1\sim\operatorname{Poisson}(388.8)\) under the model.
For the second group, $$ E_2=2{,}400(6)=14{,}400\text{ policy-months}, \qquad \lambda_2=0.024(14{,}400)=345.6. $$ Thus \(N_2\sim\operatorname{Poisson}(345.6)\). The second expected count is smaller despite the larger policy count, because its total exposure is smaller.
The ratio of means is $$ \frac{\lambda_2}{\lambda_1} =\frac{345.6}{388.8} =\frac{14{,}400}{16{,}200} =\frac{8}{9}. $$ So the second mean is \(8/9\) of the first. The difference is \(388.8-345.6=43.2\) expected claims.
Check. The exposure ratio is \(14{,}400/16{,}200=8/9\), and \(388.8(8/9)=345.6\). A common mistake is to compare only 2,400 policies with 1,800 policies and conclude that the second mean must be larger. Policy count alone is insufficient when duration also changes.
Worked Example: Converting an Annual Rate Before Scaling
A hypothetical coverage has a constant rate of 0.30 claims per policy-year. Find the Poisson mean for 600 policies covered for four months. Then find the mean for 750 policies covered for eight months, assuming the same rate.
Method. Convert policy-months into policy-years because the rate is annual. One year contains 12 months.
For 600 policies over four months, $$ E_1=600\left(\frac{4}{12}\right)=200\text{ policy-years}, \qquad \lambda_1=0.30(200)=60. $$ The first count has model \(N_1\sim\operatorname{Poisson}(60)\).
For 750 policies over eight months, $$ E_2=750\left(\frac{8}{12}\right)=500\text{ policy-years}, \qquad \lambda_2=0.30(500)=150. $$ The second count has model \(N_2\sim\operatorname{Poisson}(150)\).
Check. The exposure changed from 200 to 500 policy-years, a factor of \(500/200=2.5\). The mean changes by the same factor: \(60(2.5)=150\). A tempting but incorrect calculation is \(0.30(600)(4)=720\), which treats the annual rate as if it were per policy-month. Converting months to a fraction of a year gives the correct exposure.
Choosing and Checking the Exposure Unit
Many scaling errors are unit errors rather than arithmetic errors. Before calculating, identify exactly what one unit in the rate means. “0.30 claims per policy-year” is not interchangeable with “0.30 claims per year” if multiple policies are exposed. Similarly, “3.6 claims per 1,000 hours” requires exposure measured in thousands of hours, not raw hours.
| Rate unit | Exposure calculation | Matching total exposure |
|---|---|---|
| Claims per policy-year | Policies multiplied by covered years | Policy-years |
| Claims per vehicle-month | Vehicles multiplied by covered months | Vehicle-months |
| Claims per 1,000 operating hours | Total hours divided by 1,000 | Thousands of operating hours |
If the rate stays constant but exposure changes, use the exposure ratio to rescale the mean. If both the rate and exposure change, calculate the new mean from the new rate and new exposure; scaling only by exposure would ignore the rate change. If different groups have different constant rates, calculate each group’s mean using its own rate and exposure before combining counts, provided the model’s independence assumptions are appropriate.
Common Exam Traps
- Reporting the rate as the mean. A rate of 0.18 claims per vehicle-year is not the expected count for a portfolio until the vehicle-year exposure is known.
- Ignoring the time unit. If the rate is annual, six months contributes one-half year per exposed unit, not one year.
- Ignoring the number of exposed units. A portfolio’s exposure is usually the sum across its covered units and their covered durations, not just the calendar length of the study.
- Mixing scaled and unscaled units. For a rate per 1,000 hours, use thousands of hours as exposure or convert the rate to claims per hour. Do not multiply the per-1,000 rate by raw hours.
- Scaling an observed count as if it were an expected count. The proportional rule applies to the model mean when the rate is constant. It does not assert that a particular realized count will change in the same proportion.
- Assuming the rate remains constant without justification. If the rate varies by period, group, or operating condition, a single-rate calculation is not justified unless a suitable common rate is specified.
Check Your Understanding
For each problem, state the matching exposure unit before calculating. Assume the stated rate is constant over the relevant exposure.
- A hypothetical insurer uses a rate of 0.12 claims per bicycle-year. It covers 90 bicycles for five months. What is the total exposure in bicycle-years and the Poisson mean? (Answer: 37.5 bicycle-years; \(\lambda=4.5\).)
- A service plan has a rate of 2.5 claims per 500 device-hours. Twelve devices each run for 250 hours. Find the total exposure in units of 500 device-hours and the mean. (Answer: 6 units of 500 device-hours; \(\lambda=15\).)
- A hypothetical portfolio’s rate is 0.015 claims per contract-month. Find the mean for 3,200 contracts covered for eight months. If the contract count instead rises to 4,000 for the same eight months, what is the new mean? (Answer: 384; new mean \(=480\).)
- A constant rate is 0.48 claims per policy-year. Compare the means for 500 policies covered for three months and 300 policies covered for ten months. Which group has the larger expected count, and by how much? (Answer: first mean \(=60\), second mean \(=120\); the second is larger by 60.)
- A hypothetical program has a rate of 1.8 claims per 100 machine-days. It initially observes 40 machines for 75 days. It later observes 50 machines for 60 days at the same rate. Find both means and the exposure ratio. (Answer: initial exposure \(=30\) units of 100 machine-days and mean \(=54\); later exposure \(=30\) units and mean \(=54\); exposure ratio \(=1\).)