Reconcile Exposure and Claim-Count Bases

Why the Bases Must Match

An aggregate loss projection combines two quantities: how often claims occur and how much each claim costs under the coverage being modeled. Both quantities must refer to the same exposure unit and the same claim definition. A frequency measured per policy-year cannot be multiplied by a severity average from a different covered population without adjustment. Likewise, a ground-up claim count cannot be paired directly with an average payment calculated only from claims that exceed a deductible.

The practical check is to ask, for every frequency and severity figure: What is one unit of exposure, and which claims are included? A claim count might include every reported ground-up claim, or only claims producing a positive payment. A severity measure might be the original loss $X$, the payment after a deductible, or a payment after both a deductible and a limit. Those are different bases, even when they arise from the same losses.

Definition: The exposure basis is the unit used to measure the population at risk, such as policy-years, vehicle-years, or insured locations for a specified period. The claim-count basis specifies which claims are counted, such as all ground-up claims or only claims with positive payments. The severity basis specifies the amount recorded for each counted claim, such as ground-up loss or payment after coverage modifications.

Frequency and Severity on a Consistent Basis

Let $E$ denote exposure in stated exposure units, and let $N$ be the number of claims on a specified count basis. The observed frequency on that basis is $N/E$. If $Y_j$ is the amount paid for the $j$th claim on the same basis, the average severity is the average of those $Y_j$ values. A basic projection multiplies the expected number of claims by the expected payment per counted claim.

\text{Expected aggregate payment} = \text{expected claim count on the payment basis} \times \text{mean payment per claim on that basis}.

Under a homogeneous frequency-severity model, suppose ground-up claim counts have expected value $\lambda E$ for exposure $E$, and the ground-up severities $X$ are independent and identically distributed, independent of the claim count. Let $Y=g(X)$ be the payment under the coverage terms. Then the expected aggregate payment is $\lambda E[E[Y]]$, where the first $E$ is exposure and $E[Y]$ is expected payment per ground-up claim. This expression counts every ground-up claim, including those with zero payment.

Alternatively, count only claims with positive payment. If $q=P(Y>0)$, the expected positive-payment claim count is $\lambda E q$, and the mean payment conditional on a positive payment is $E[Y\mid Y>0]$. The two approaches agree because $E[Y]=qE[Y\mid Y>0]$. This identity is the central reconciliation: either include zero-payment claims in the count and use payment per ground-up claim, or exclude them and use payment conditional on a positive payment.

\lambda E\,E[Y] = (\lambda E q)\,E[Y\mid Y>0].

This relationship assumes that the ground-up frequency and severity model apply to the exposure being projected and that the covered payment is determined consistently from each ground-up loss. If risks have materially different frequencies or severity distributions, pooling them may produce a misleading average; project by appropriate exposure groups instead.

Why the Reconciliation Identity Holds

Start with ground-up claims.
For each ground-up claim, record its covered payment $Y$, which may be zero. The expected payment per ground-up claim is $E[Y]$.

Separate zero and positive payments.
On the event $Y=0$, the payment contributes nothing. On $Y>0$, its conditional mean is $E[Y\mid Y>0]$. Therefore $E[Y]=P(Y>0)E[Y\mid Y>0]=qE[Y\mid Y>0]$.

Multiply by expected ground-up count.
If expected ground-up count is $\lambda E$, expected aggregate payment is $\lambda E E[Y]$. Substituting the identity from Step 2 gives $(\lambda E q)E[Y\mid Y>0]$, the positive-payment count times its conditional mean.

The argument is an expectation identity; it does not require the claim count to be Poisson. For a compound model with independent claim severities and a random claim count $N$, conditioning on $N$ gives $E[\sum_{j=1}^{N}Y_j\mid N]=N E[Y]$, and hence $E[\sum_{j=1}^{N}Y_j]=E[N]E[Y]$. Independence and a common severity distribution are important for this simple form.

Worked Examples

Worked Example: Projecting a Deductible Payment from Ground-Up Experience

A hypothetical portfolio has 6,000 earned vehicle-years of exposure and 120 ground-up claims. Of those claims, 36 exceed a per-claim deductible of $d=\$500$. Their payments after the deductible total $\$45{,}000$. Project expected payments for 8,000 vehicle-years under unchanged conditions. Assume annual, undiscounted losses; no mortality or interest assumptions are relevant. Use the observed experience directly; no parametric severity distribution is assumed.

Method: reconcile the ground-up and positive-payment bases. Ground-up frequency is $120/6{,}000=0.02$ claims per vehicle-year. The proportion of ground-up claims producing a payment is $36/120=0.30$. Thus positive-payment frequency is $0.02(0.30)=0.006$ claims per vehicle-year.

The mean payment among positive-payment claims is $\$45{,}000/36=\$1{,}250$. For 8,000 vehicle-years, expected positive-payment count is $8{,}000(0.006)=48$. The projected expected aggregate payment is $48(\$1{,}250)=\$60{,}000$.

Check using the ground-up count basis: expected ground-up claims are $8{,}000(0.02)=160$. Expected payment per ground-up claim is $0.30(\$1{,}250)=\$375$. Then $160(\$375)=\$60{,}000$, agreeing with the positive-payment calculation. Common mistake avoided: multiplying ground-up frequency by the conditional mean payment and omitting the 0.30 probability of a positive payment would give $160(\$1{,}250)=\$200{,}000$, an overstatement.

Worked Example: Projecting Directly from Paid-Claim Records

A hypothetical insurer has 4,000 earned policy-years and records 80 claims with positive payments after its existing coverage terms. Those payments total $\$96{,}000$. Estimate expected payments for 2,500 policy-years under the same terms and mix of risks. Assume annual, undiscounted losses, with no mortality or interest assumptions. Use the observed positive-payment data; no parametric distribution is specified.

Method: use the observed paid-claim frequency with paid-claim severity. The positive-payment frequency is $80/4{,}000=0.02$ claims per policy-year. The mean payment per positive-payment claim is $\$96{,}000/80=\$1{,}200$.

For 2,500 policy-years, expected positive-payment count is $2{,}500(0.02)=50$. Therefore expected aggregate payment is $50(\$1{,}200)=\$60{,}000$.

The check is that the projected amount per policy-year is $0.02(\$1{,}200)=\$24$; multiplying by 2,500 policy-years gives $\$60{,}000$. Common mistake avoided: these records already count positive-payment claims. Applying an additional probability of exceeding the coverage threshold would reduce the count a second time. The ground-up claim count and the threshold-exceedance probability are not provided, so they are not needed for this projection.

Worked Example: Changing the Deductible in a Discrete Severity Model

A hypothetical book has ground-up frequency $0.04$ claims per exposure-year. Ground-up severity $X$ is independent of claim count and has this distribution: $P(X=\$400)=0.50$, $P(X=\$1{,}000)=0.30$, and $P(X=\$2{,}500)=0.20$. Compare expected payments for 3,000 exposure-years under deductibles of $\$500$ and $\$1{,}000$, with payment $Y=\max(X-d,0)$. Assume annual, undiscounted losses; no mortality or interest assumptions apply. The specified discrete distribution is the severity model.

Method: calculate positive-payment frequency and conditional payment separately for each deductible. Under $d=\$500$, losses of $\$1{,}000$ and $\$2{,}500$ produce payments of $\$500$ and $\$2{,}000$, respectively. The positive-payment probability is $0.30+0.20=0.50$. The conditional mean payment is $(0.30(\$500)+0.20(\$2{,}000))/0.50=(\$150+\$400)/0.50=\$1{,}100$.

Positive-payment frequency is $0.04(0.50)=0.02$ per exposure-year. In 3,000 exposure-years, expected positive-payment count is $3{,}000(0.02)=60$, giving expected aggregate payment $60(\$1{,}100)=\$66{,}000$.

Under $d=\$1{,}000$, a loss of exactly $\$1{,}000$ produces zero payment. Only the $\$2{,}500$ loss produces a positive payment, of $\$1{,}500$. Thus positive-payment probability is $0.20$, conditional mean payment is $\$1{,}500$, and positive-payment frequency is $0.04(0.20)=0.008$ per exposure-year. The expected count is $3{,}000(0.008)=24$, so expected aggregate payment is $24(\$1{,}500)=\$36{,}000$.

Check on the ground-up basis: under the $\$500$ deductible, expected payment per ground-up claim is $0.30(\$500)+0.20(\$2{,}000)=\$550$; $3{,}000(0.04)(\$550)=\$66{,}000$. Under the $\$1{,}000$ deductible, it is $0.20(\$1{,}500)=\$300$; $3{,}000(0.04)(\$300)=\$36{,}000$. Both agree with the positive-payment calculations. Common mistake avoided: a claim exactly equal to the deductible is not a positive payment when the payment is $\max(X-d,0)$.

Worked Example: Converting Policy Counts to Policy-Year Exposure

A hypothetical portfolio has 1,200 policies, each earning one-half of a policy-year during the observation period. It records 30 positive-payment claims, totaling $\$27{,}000$. Project expected payments for 900 policies, each expected to earn 0.8 policy-year. Assume unchanged risk mix and coverage, annual undiscounted amounts, no mortality or interest assumptions, and empirical paid-claim severities rather than a parametric distribution.

Method: convert both periods to the same exposure unit before calculating frequency. Observed exposure is $1{,}200(0.5)=600$ policy-years, so positive-payment frequency is $30/600=0.05$ per policy-year. Mean payment is $\$27{,}000/30=\$900$.

Projected exposure is $900(0.8)=720$ policy-years. Expected positive-payment count is $720(0.05)=36$, and expected aggregate payment is $36(\$900)=\$32{,}400$. Check: projected expected payment per policy-year is $0.05(\$900)=\$45$, and $720(\$45)=\$32{,}400$. Common mistake avoided: treating 1,200 policies as 1,200 policy-years would halve the observed frequency denominator error in this example and produce an inconsistent projection.

Practical Checks and Exam Traps

Before multiplying a frequency by a severity, write the basis next to each value. For example, label a rate “positive payments per earned policy-year” and a mean “payment dollars per positive-payment claim.” Their product is expected payment dollars per earned policy-year. This unit check often reveals a mismatch before arithmetic begins.

Ground-up count with conditional paid severity: include the positive-payment probability $q$, or convert the frequency to a positive-payment frequency first.
Paid-claim count with paid severity: do not apply $q$ again. The count already excludes zero payments.
Different exposure units: convert counts of policies into earned policy-years, or into another stated unit, before comparing or projecting frequencies.
Changed coverage: do not assume a historical paid-claim frequency remains valid after changing the deductible or limit. Recalculate the positive-payment probability and payment severity on the new terms when ground-up information or a suitable model is available.
Inconsistent populations: check that the count, exposure, and severity observations cover comparable risk groups and periods. A frequency for one group paired with severity from another need not estimate either group’s expected loss.

These checks matter because aggregate projections are often sensitive to frequency errors: a probability applied twice can materially understate cost, while a conditional severity used with an unfiltered ground-up count can overstate it. The essential discipline is to state the exposure unit, identify exactly which claims are counted, and define the payment recorded for each such claim. These methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, keep the exposure unit, claim-count basis, and severity basis explicit.

A hypothetical portfolio has 12,000 exposure-years and 240 ground-up claims. One quarter produce positive payments under a deductible, and their average payment is $\$750$. Project expected payment for 5,000 exposure-years. (Answer: $\$18{,}750$)
A portfolio records 54 positive-payment claims over 1,800 earned policy-years, with total payments of $\$34{,}560$. Under unchanged terms, project expected payment for 2,500 policy-years. (Answer: $\$48{,}000$)
Ground-up frequency is 0.06 claims per exposure-year. Severity is $\$300$ with probability 0.40, $\$900$ with probability 0.40, and $\$1{,}900$ with probability 0.20. For a $\$400$ deductible and 1,000 exposure-years, find expected aggregate payment. (Answer: $\$30{,}000$)
A hypothetical book has 45 positive-payment claims over 900 policy-years and an average payment of $\$1{,}200$. The same basis is expected over 2,400 policy-years. Find projected expected claim count and aggregate payment. (Answer: 120 claims; $\$144{,}000$)
An observed period has 800 policies, each earning 0.75 policy-year, and 24 positive-payment claims averaging $\$500$. For 600 policies each earning 0.8 policy-year, with unchanged terms, find the expected positive-payment count and aggregate payment. (Answer: 19.2 claims; $\$9{,}600$)

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