Why Multiplication Needs Conditions
A frequency-severity model represents expected claim cost as a claim rate multiplied by an average payment per claim. This is convenient, but the multiplication is not automatically valid just because a frequency estimate and a severity estimate are both available. The two estimates must refer to compatible risks, exposure periods, coverage terms, and weighting. In particular, severity that varies with a policy’s claim count can make a product of separately averaged quantities incorrect.
Let \(N\) be the number of claims for one risk over a stated period, and let \(Y_j\) be the insurer’s payment for claim \(j\), on a specified coverage basis. The aggregate payment is \(L=\sum_{j=1}^{N}Y_j\), with \(L=0\) when \(N=0\). For one unit of exposure, the pure premium is \(E[L]\). If the period contains \(e\) units of exposure and the model estimates a claim frequency per exposure unit, the corresponding expected aggregate payment per exposure unit is the quantity to be rated.
The term “severity” needs particular care. An average over policies gives each policy equal weight. An average over claim records gives more weight to policies that produce more claims. Those averages need not be equal. Similarly, severity for ground-up losses is not interchangeable with severity for payments after a deductible or limit. The target must match the payment \(Y_j\) whose aggregate cost the pure premium is intended to represent.
Conditional Factorization
Let \(Z\) denote the rating variables used to define a risk class, such as territory or vehicle characteristics. Write \(\lambda(Z)=E[N\mid Z]\). For \(n>0\), let \(m_n(Z)\) be the expected payment for a claim on a risk with \(N=n\), conditional on \(Z\). More precisely, it is the expected average of the \(n\) claim payments for that policy. Conditioning first on the claim count gives the general expression:
This expression does not require a Poisson count distribution. It follows by calculating expected aggregate payment for each possible claim count and then averaging over the count distribution. If, conditional on \(Z\), expected per-claim payment does not vary with \(N\), so \(m_n(Z)=\mu(Z)\) for all \(n>0\) with positive probability, then the expression simplifies:
A standard sufficient assumption is that, conditional on \(Z\), claim count is independent of the claim severities and each claim has the same expected payment \(\mu(Z)\). Independence is sufficient, not necessary: the factorization can also hold if count-related changes in severity happen to cancel in the expectation. The essential condition is that the expected aggregate payment has the stated product.
Worked Example: A Homogeneous Risk Class
Worked Example: A Homogeneous Risk Class
A hypothetical insurer estimates annual cost for one unit of exposure in a narrowly defined class. Annual claim count \(N\) has a Poisson distribution with parameter \(\lambda=0.20\). Conditional on a claim, its payment \(Y_j\) has a Gamma distribution with shape \(k=2\) and scale \(\theta=\$1{,}000\). Claim payments are independent of the count and of one another. Find the annual pure premium. Assume no discounting; mortality is not part of this property-claim model.
Method and assumptions. The mean of a Gamma distribution parameterized by shape \(k\) and scale \(\theta\) is \(k\theta\). Thus the expected payment per claim is \(E[Y_j]=2(\$1{,}000)=\$2{,}000\). The Poisson mean equals its parameter, so \(E[N]=0.20\).
Apply the factorization. Conditional independence and a common severity mean make the severity mean the same at each possible positive claim count. Therefore, \(E[L]=E[N]E[Y_j]=0.20(\$2{,}000)=\$400\) per exposure-year.
Check. The result is an expected aggregate payment per exposure-year, not an average payment per claim: the latter is \(\$2{,}000\). The pure premium is smaller because the expected count is only 0.20. The common mistake this avoids is reporting severity alone as the expected cost per exposure.
Risk Mix: Match the Weighting Before Multiplying
Even if factorization holds within every rating class, multiplying frequency and severity averages over a mixed portfolio can fail. If classes have different frequencies and severities, the portfolio pure premium is the average of class-specific products, not generally the product of the two policy-weighted averages. For classes \(z\) with portfolio proportions \(p_z\), the relevant identity is \(E[L]=\sum_z p_z\lambda_z\mu_z\), provided factorization holds within each class.
A claim-level severity average uses a different weighting. A class producing more claims contributes more claim records. If \(\lambda_z\) is its expected count per policy and \(\mu_z\) its expected payment per claim, the expected payment per claim record across the portfolio is weighted in proportion to \(p_z\lambda_z\). That claim-weighted average can be multiplied by the portfolio frequency to recover the portfolio pure premium, when the targets and exposure bases are aligned.
Worked Example: Two Classes With Different Claim Rates
A hypothetical portfolio has equal numbers of policies in classes A and B. Annual counts are Poisson with means \(\lambda_A=2\) and \(\lambda_B=1\). Each A claim has a deterministic payment of \(\$1{,}000\); each B claim has a deterministic payment of \(\$500\). Find the portfolio pure premium. Assume no discounting; mortality is not applicable.
Step 1: Calculate class pure premiums. In A, \(2(\$1{,}000)=\$2{,}000\) per policy-year. In B, \(1(\$500)=\$500\) per policy-year.
Step 2: Average class costs by policy share. The portfolio pure premium is \(0.5(\$2{,}000)+0.5(\$500)=\$1{,}000+\$250=\$1{,}250\) per policy-year.
Step 3: Check a tempting but wrong calculation. Portfolio frequency is \(0.5(2)+0.5(1)=1.5\) claims per policy-year. The policy-weighted average of the two class severities is \(0.5(\$1{,}000)+0.5(\$500)=\$750\). Multiplying gives \(1.5(\$750)=\$1{,}125\), which is wrong because the severity average gives A and B equal weight even though A produces twice as many claims per policy.
Step 4: Reconcile using claim weighting. Expected claims per policy-year from A are \(0.5(2)=1\); from B they are \(0.5(1)=0.5\). The claim-weighted average severity is \((1(\$1{,}000)+0.5(\$500))/1.5=(\$1{,}250)/1.5=\$833.33\ldots\). Multiplying by 1.5 gives \(\$1{,}250\) per policy-year. The repeating average is retained unrounded until the product, which checks against the class-cost calculation.
When Count and Severity Are Related
Risk mix is one reason for a mismatch; another is that, even within a rating class, risks with different claim counts may have different average severities. For example, a mechanism that produces multiple claims on one policy could also be associated with a different claim-size distribution. In that case, a policy-weighted average of count-specific severities is not generally the severity that should be multiplied by expected frequency.
The general formula makes the issue explicit: each count-specific severity \(m_n(Z)\) is multiplied by \(nP(N=n\mid Z)\). A product using a single severity mean is justified if that mean correctly represents the count-weighted expected payment per claim, or if the conditional mean does not vary with count. It is not justified by multiplying \(E[N\mid Z]\) by an arbitrary average of severities across policies.
Worked Example: Severity Varies With Claim Count
For a hypothetical risk class, annual claim count is 1 with probability 0.50 and 2 with probability 0.50. There are no other possible counts. If the count is 1, the claim payment is deterministically \(\$500\). If the count is 2, both claim payments are deterministically \(\$1{,}000\). Find expected annual aggregate payment and show why a policy-weighted severity can mislead. Assume no discounting; mortality is not applicable.
Step 1: Calculate expected frequency. \(E[N]=0.50(1)+0.50(2)=1.50\) claims per policy-year.
Step 2: Calculate aggregate cost by count. At count 1, aggregate payment is \(\$500\). At count 2, it is \(2(\$1{,}000)=\$2{,}000\). Thus \(E[L]=0.50(\$500)+0.50(\$2{,}000)=\$250+\$1{,}000=\$1{,}250\) per policy-year.
Step 3: Show the incorrect policy-weighted product. Averaging the per-claim severities across policies gives \(0.50(\$500)+0.50(\$1{,}000)=\$750\). The product \(1.50(\$750)=\$1{,}125\) understates the correct expected payment by \(\$125\). The issue is that the higher-severity policies also have more claims.
Step 4: Check with claim weighting. Expected claim records from the count-1 group are \(0.50(1)=0.50\); from the count-2 group they are \(0.50(2)=1.00\). Therefore the expected claim-level average is \((0.50(\$500)+1.00(\$1{,}000))/1.50=\$1{,}250/1.50=\$833.33\ldots\). Multiplying by expected frequency gives \(\$1{,}250\), matching the direct calculation. The common mistake is to use the policy-weighted severity as if it were the claim-weighted severity.
Practical Conditions and Exam Traps
Before multiplying estimates, check that the frequency and severity models refer to the same risks and time period. Frequency should be claims per the exposure unit used for the pure premium; severity should be payment per claim on the same coverage and valuation basis. A severity model for limited payments cannot be combined without adjustment with a frequency model whose target is intended to price a different payment basis.
Also distinguish conditional predictions from portfolio summaries. If a model gives \(\lambda(Z)\) and \(\mu(Z)\) for each risk, calculate the risk-level product and then average those products using the desired policy or exposure weights. Do not generally multiply the separately averaged values:
The two sides agree in special cases, such as constant frequency or constant severity across the mix, or when the weighted association between them happens to cancel. Agreement should not be assumed. A claim-level severity average can be appropriate for a portfolio product, but only when its claim weighting is consistent with the frequency estimate and with the target population.
Define the aggregate payment, exposure unit, period, and coverage basis.
Confirm frequency and severity refer to the same rating class or covariate values.
Determine whether the mean is policy-weighted, claim-weighted, or conditional on claim count.
If severity varies with count, use the count-specific expectation formula or an appropriately claim-weighted mean.
The central exam trap is to treat a frequency average and a severity average as though they automatically describe the same unit. The reliable approach is to begin with expected aggregate payment, condition on the relevant risk information and claim count, and simplify to a product only after its assumptions are checked. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
Check Your Understanding
For each problem, identify the weighting and calculate the requested expected cost.
- A homogeneous class has annual count \(N\sim\operatorname{Poisson}(0.40)\). Claim payment is Gamma with shape 3 and scale \(\$250\), independent of count. Find pure premium per exposure-year. Assume no discounting and no mortality. (Answer: \(\$300.00\) per exposure-year.)
- A portfolio is 30% class A and 70% class B. Their annual Poisson frequency means are 0.5 and 1.5; deterministic claim payments are \(\$1{,}200\) and \(\$400\), respectively. Find pure premium, and compare it with the product of policy-weighted frequency and policy-weighted severity. (Answer: pure premium \(=\$600.00\); the policy-weighted product is \(\$768.00\) and is not valid.)
- Annual count is 0 with probability 0.20, 1 with probability 0.50, and 3 with probability 0.30. Payments are deterministic: \(\$400\) per claim when \(N=1\), and \(\$900\) per claim when \(N=3\). Find expected count and aggregate payment. (Answer: \(E[N]=1.40\); \(E[L]=\$1{,}010.00\) per policy-year.)
- Count is 1 or 2 with equal probabilities. The per-claim payment is \(\$500\) when \(N=1\) and \(\$1{,}000\) when \(N=2\). Calculate the product using the policy-weighted severity and state the correct pure premium. (Answer: policy-weighted product \(=\$1{,}125.00\); correct pure premium \(=\$1{,}250.00\) per policy-year.)
- Two equally sized classes have \((\lambda_A,\mu_A)=(1,\$300)\) and \((\lambda_B,\mu_B)=(3,\$700)\). Assuming factorization within each class, find the portfolio pure premium and the claim-weighted portfolio severity. (Answer: pure premium \(=\$1{,}200.00\) per policy-year; claim-weighted severity \(=\$600.00\) per claim.)