Why Counts and Claim Incidence Are Different
A portfolio can have fewer policies with claims than it has claims. One policy may generate two or more claims during the observation period, so counting claims and counting affected policies answer different questions. Confusing these quantities can distort a frequency calculation or lead to an incorrect interpretation of portfolio experience.
This tutorial defines claim count, claim incidence, and claim frequency. It focuses on keeping the numerator and denominator clear: the claim count measures claim events, incidence measures policies with at least one claim, and frequency relates claim events to an exposure base. These distinctions are aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.
A claim count is a count of events. The number of claimant policies is a count of policies. They are equal only if every policy with any claims has exactly one claim. If one policy has three claims, it contributes three to the claim count but only one to the number of claimant policies.
Incidence is a policy-level measure: each policy contributes either zero or one to its numerator. Frequency is an event rate: a policy can contribute more than one claim. If the stated exposure base is policy-years, for example, frequency is measured in claims per policy-year. The denominator is not interchangeable with the number of policies unless the problem establishes that those quantities represent the same exposure basis.
Connecting the Measures
The distinction can be expressed directly. Each policy with claims contributes one claim as its first claim, plus any claims beyond the first. A policy with no claims contributes zero to both the claim count and the count of claimant policies.
Summing this identity over policies shows that total claims equal claimant policies plus claims beyond the first claim on each claimant policy. Thus total claims cannot be less than claimant policies. They are equal precisely when every claimant policy has one claim.
There is also a useful relationship when the data cover the same policies and period and every policy represents exactly one unit of exposure. Define the average number of claims per claimant policy as total claims divided by claimant policies. Multiplying incidence by that average gives frequency per policy:
The claimant-policy term cancels, leaving claims per policy. The relationship depends on using the same cohort and period, and one policy per exposure unit. With a different exposure base, calculate frequency using that base directly; do not assume it equals incidence multiplied by the average claims per claimant policy.
Calculating Each Measure From Observed Counts
Worked Example: Multiple Claims in a Policy Cohort
A hypothetical insurer reviews 120 policies for one full policy period. The claim-count distribution in the observed data is: 90 policies with zero claims, 22 with one claim, 6 with two claims, and 2 with three claims. Calculate total claims, claimant policies, incidence, and frequency per policy-period.
Method and assumptions. Sum policy claim counts to obtain claim events, and count policies with positive claim counts to obtain claimant policies. Each policy represents one full policy-period, so the exposure base is 120 policy-periods. This is a calculation from observed counts; no interest rate, mortality assumption, or stochastic claim-count distribution is used. A common mistake is to treat the 30 claimant policies as if there were only 30 claims.
Step 1: Calculate total claims. The four groups contribute \(90\times0=0\), \(22\times1=22\), \(6\times2=12\), and \(2\times3=6\) claims. Therefore, total claims are \(0+22+12+6=40\).
Step 2: Count claimant policies. Policies with at least one claim are the one-, two-, and three-claim groups: \(22+6+2=30\). As a check, \(90+30=120\) policies in total.
Step 3: Calculate incidence. Incidence is \(30/120=0.25\), or 25% of policies.
Step 4: Calculate frequency. Frequency is \(40/120=0.3333\) claims per policy-period. The average number of claims per claimant policy is \(40/30=1.3333\). Checking the relationship, \(0.25\times1.3333\) is approximately \(0.3333\) claims per policy-period.
Check. Total claims, 40, exceed claimant policies, 30, by 10. The two policies with three claims contribute four claims beyond their first claims, and the six policies with two claims contribute six more; \(4+6=10\). The totals reconcile.
Frequency Requires a Stated Exposure Base
A frequency rate is meaningful only when its numerator, denominator, and time period refer to compatible experience. A problem may provide a particular exposure total, such as policy-years, and ask for claims per that unit. Use the specified total rather than silently substituting the number of policies. Selecting which exposure measure is appropriate is a separate judgment; here the denominator is treated as given.
Worked Example: Claims per Policy-Year Versus Incidence
For a hypothetical book observed over a calendar year, records show 240 policies, 210 policy-years of exposure, 33 claims, and 27 policies with at least one claim. Calculate incidence and frequency, and explain why their denominators differ.
Method and assumptions. Use policies as the denominator for incidence and the stated policy-year total as the denominator for frequency. The policy count, exposure, claims, and claimant-policy count all refer to the same observation period. These are observed data, with no interest rate, mortality assumption, or stochastic claim-count distribution. The common mistake is to divide claims by 240 policies when the requested frequency is per policy-year.
Step 1: Calculate incidence. Incidence is \(27/240=0.1125\), or 11.25% of policies having at least one claim.
Step 2: Calculate frequency. Frequency per policy-year is \(33/210=0.157142\ldots\), which is \(0.1571\) claims per policy-year to four decimal places.
Step 3: Check the distinction. The 33 claims came from 27 claimant policies, so there are \(33-27=6\) claims beyond the first claim on each claimant policy. Incidence uses the 240-policy count, whereas frequency uses 210 policy-years. The two rates answer different questions and have different denominators.
Check. The observed claims per policy, \(33/240=0.1375\), are not the requested claims per policy-year. Because the stated exposure is 210 policy-years, \(0.1571\) is the frequency on that basis.
Aggregating Groups Without Losing the Numerators
When combining groups, add claim counts before calculating frequency, and add claimant-policy counts before calculating incidence. Calculate each aggregate rate from its own combined numerator and denominator. In general, taking an unweighted average of group rates does not produce the overall rate; groups may differ in policy counts or exposure.
Worked Example: Combining Two Groups With Different Exposure
A hypothetical portfolio has two groups observed for a stated period. Group A has 80 policies, 80 policy-years, 24 claims, and 20 claimant policies. Group B has 120 policies, 60 policy-years, 18 claims, and 15 claimant policies. Calculate the portfolio incidence and frequency per policy-year. Also compare the group rates with the aggregate results.
Method and assumptions. Add like quantities across groups, then divide total claimant policies by total policies for incidence and total claims by total policy-years for frequency. The supplied exposure amounts are the applicable denominator. These are observed counts; no interest rate, mortality assumption, or stochastic claim-count distribution is needed. A common mistake is to average the two group frequencies without weighting for their different exposures.
Step 1: Calculate group incidence. Group A incidence is \(20/80=0.25\), or 25%. Group B incidence is \(15/120=0.125\), or 12.5%.
Step 2: Calculate group frequency. Group A frequency is \(24/80=0.30\) claims per policy-year. Group B frequency is \(18/60=0.30\) claims per policy-year.
Step 3: Combine the counts and denominators. There are \(80+120=200\) policies and \(80+60=140\) policy-years. Total claims are \(24+18=42\); total claimant policies are \(20+15=35\).
Step 4: Calculate aggregate rates. Portfolio incidence is \(35/200=0.175\), or 17.5%. Portfolio frequency is \(42/140=0.30\) claims per policy-year.
Step 5: Check the rates. Both groups have frequency 0.30, so their exposure-weighted aggregate frequency is also 0.30. The group incidence rates are different, and the aggregate incidence is based on 35 claimant policies out of 200 policies, not on an unweighted average of group rates. Also, \(42-35=7\) claims are beyond the first claim on each claimant policy.
Check. Group frequencies imply \(0.30\times80=24\) expected observed claims in Group A and \(0.30\times60=18\) in Group B; these sum to the observed 42. The aggregate denominator and numerator reconcile.
Common Mistakes and Practical Checks
Before reporting a rate, write down what is being counted and what is being used as the denominator. A brief reconciliation often catches an error before it propagates into a frequency comparison or a later ratemaking calculation.
- Using claimant policies as the claim count. A policy with several claims contributes several events, but only one claimant policy.
- Using claims to calculate incidence. Incidence counts policies with at least one claim, not claim events. Its numerator cannot exceed its policy denominator.
- Replacing the specified exposure with the policy count. Claims per policy and claims per policy-year are different measures when the denominators differ.
- Averaging group rates without regard to size. For an aggregate frequency, divide total claims by total exposure. For aggregate incidence, divide total claimant policies by total policies.
- Mixing periods or populations. The claims, claimant policies, policy count, and exposure must describe compatible experience.
Check Your Understanding
For each problem, keep claim events, claimant policies, and exposure units separate.
- A hypothetical block has 100 policies, of which 64 have no claims, 28 have one claim, 6 have two claims, and 2 have four claims. Each policy represents one full policy-period. Find total claims, claimant policies, incidence, and frequency per policy-period. (Answer: 48 claims; 36 claimant policies; incidence 36%; frequency 0.48 claims per policy-period.)
- A hypothetical portfolio has 240 policies, 210 policy-years of exposure, 33 claims, and 27 claimant policies. Find incidence and frequency per policy-year. (Answer: incidence 11.25%; frequency 0.1571 claims per policy-year.)
- Group A has 50 policies, 45 policy-years, 12 claims, and 10 claimant policies. Group B has 150 policies, 120 policy-years, 36 claims, and 30 claimant policies. Find combined incidence and frequency per policy-year. (Answer: incidence \(40/200=20\%\); frequency \(48/165=0.2909\) claims per policy-year.)
- Among 150 policies, 18% had at least one claim. Those claimant policies generated 33 claims beyond their first claims. Find the number of claimant policies, total claims, and frequency per policy, assuming one full policy-period per policy. (Answer: 27 claimant policies; 60 claims; frequency 0.40 claims per policy-period.)
- Each of 50 hypothetical policies represents one full policy-period. Eight policies have one claim each, five have two claims each, and two have four claims each; all other policies have no claims. Find incidence, frequency, and average claims per claimant policy. (Answer: incidence \(15/50=30\%\); frequency \(26/50=0.52\) claims per policy-period; average \(26/15=1.7333\) claims per claimant policy.)