Aggregate Loss Data And Credibility

Aggregate Loss Combines Frequency and Severity

For credibility analysis, a portfolio’s aggregate loss per exposure summarizes both how often claims occur and how much they cost. It is therefore different from severity data alone: two groups can have the same average loss per exposure even though one has relatively few costly claims and the other has more frequent, smaller claims. Those differences matter when interpreting how much an observed aggregate average may vary.

Let $N$ be the number of claims during one defined exposure unit, and let $X_j$ be the payment for claim $j$. The aggregate loss $S$ for that exposure is the sum of all claim payments. If there are no claims, the sum is zero. The exposure unit must be specified consistently, such as one policy-year or one vehicle-year.

Definition: Aggregate loss per exposure is total claim payment divided by total exposure. For a single exposure unit, aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Claim frequency is the number of claims per exposure, and claim severity is the payment per claim. For a portfolio with total loss $L$ and total exposure $e$, observed aggregate loss per exposure is $L/e$.

The aggregate measure can also be related to the two component averages. If total claim count is $C$, total loss is $L$, and total exposure is $e$, then, when $C>0$,

\frac{L}{e} = \frac{C}{e}\times\frac{L}{C}.

The first factor is observed claim frequency; the second is observed average severity. If $C=0$, total loss and aggregate loss per exposure are zero, while average severity is undefined because there are no claims to average.

Compound Poisson Model and Aggregate Moments

A standard model for aggregate loss assumes that the claim count $N$ has a Poisson distribution with parameter $\lambda>0$, and that claim payments $X_1,X_2,\ldots$ are independent and identically distributed, independent of $N$, with finite mean and variance. Write $\mu=E[X]$ and $\sigma^2=\operatorname{Var}(X)$. Under these assumptions, $E[N]=\operatorname{Var}(N)=\lambda$. The model is not appropriate without adjustment if claim payments depend on the claim count, if the claims are not identically distributed, or if the Poisson count assumption is not suitable for the situation.

The expected aggregate loss is the expected number of claims times the expected payment per claim. Its variance has two sources: variation in claim payments and variation in claim count.

Result: Under the independent compound Poisson assumptions, $E[S]=\lambda E[X]$ and $\operatorname{Var}(S)=\lambda E[X^2]=\lambda\bigl(\operatorname{Var}(X)+(E[X])^2\bigr)$. The aggregate mean depends on the severity mean; the aggregate variance depends on the severity second moment, not just the severity variance.

A useful justification starts by conditioning on the claim count. Given $N=n$, the aggregate is a sum of $n$ independent claim payments, so its conditional mean is $n\mu$ and its conditional variance is $n\sigma^2$. Applying the laws of total expectation and total variance gives the result.

E[S]=E[E[S\mid N]]=E[N\mu]=\lambda\mu.

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N\sigma^2]+\operatorname{Var}(N\mu)\\ &=\lambda\sigma^2+\lambda\mu^2\\ &=\lambda E[X^2]. \end{aligned}

The last equality uses $E[X^2]=\sigma^2+\mu^2$. The first variance term, $\lambda\sigma^2$, is the contribution from variation in claim severity given a count. The second, $\lambda\mu^2$, is the contribution from variation in the number of claims. This split makes clear why looking at severity variation alone does not determine aggregate variation.

For a non-Poisson claim count independent of identically distributed severities, with finite moments, the corresponding formulas are $E[S]=E[N]\mu$ and $\operatorname{Var}(S)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$. The compound Poisson formula is the special case $E[N]=\operatorname{Var}(N)=\lambda$. For example, if the claim count is fixed rather than random, the count-variation term is zero; one should not apply the compound Poisson variance formula automatically.

Worked Examples

Worked Example: Calculating Aggregate Mean and Variance

Method: Use the compound Poisson moments. A hypothetical equipment portfolio has a Poisson claim count per equipment-year with parameter $\lambda=0.4$. Each claim payment $X$ is independently \$1,000 with probability 0.75 or \$3,000 with probability 0.25, independent of the count. Calculate expected aggregate loss and aggregate variance per equipment-year.

First calculate the severity mean:

E[X]=0.75(1{,}000)+0.25(3{,}000)=750+750=\$1{,}500.

Next calculate its second moment:

E[X^2]=0.75(1{,}000^2)+0.25(3{,}000^2) =750{,}000+2{,}250{,}000 =\$3{,}000{,}000.

The aggregate mean is $0.4(\$1{,}500)=\$600$ per equipment-year. The aggregate variance is $0.4(\$3{,}000{,}000)=\$1{,}200{,}000$ dollars squared. As a check on the variance calculation, the severity variance is $\$3{,}000{,}000-(\$1{,}500)^2=\$750{,}000$. The two aggregate variance contributions are $0.4(\$750{,}000)=\$300{,}000$ from severity and $0.4(\$1{,}500)^2=\$900{,}000$ from frequency; together they equal $\$1{,}200{,}000$.

Common mistake avoided: Using only $\lambda\operatorname{Var}(X)$ would omit the variability in claim count and give \$300,000, not the aggregate variance. The calculation assumes a Poisson count and independent, identically distributed claim payments as stated.

Worked Example: Same Aggregate Mean, Different Variance

Method: Compare each portfolio’s frequency and severity moments separately. Two hypothetical warranty portfolios have independent Poisson claim counts per contract-year and independent, identically distributed severities. Portfolio A has $\lambda=0.5$, with each claim equally likely to cost \$1,000 or \$3,000. Portfolio B has $\lambda=2$, with each claim equally likely to cost \$250 or \$750. Find and compare their expected aggregate losses and aggregate variances.

For Portfolio A, the severity mean is $(\$1{,}000+\$3{,}000)/2=\$2{,}000$. Its second moment is $(\$1{,}000^2+\$3{,}000^2)/2=(\$1{,}000{,}000+\$9{,}000{,}000)/2=\$5{,}000{,}000$. Thus the expected aggregate loss is $0.5(\$2{,}000)=\$1{,}000$, and the aggregate variance is $0.5(\$5{,}000{,}000)=\$2{,}500{,}000$ dollars squared.

For Portfolio B, the severity mean is $(\$250+\$750)/2=\$500$. Its second moment is $(\$250^2+\$750^2)/2=(\$62{,}500+\$562{,}500)/2=\$312{,}500$. Thus the expected aggregate loss is $2(\$500)=\$1{,}000$, and the aggregate variance is $2(\$312{,}500)=\$625{,}000$ dollars squared.

Both portfolios have the same expected aggregate loss, but Portfolio A has the larger variance. The component calculations verify the source of the difference. For A, severity variance is $\$5{,}000{,}000-(\$2{,}000)^2=\$1{,}000{,}000$: the severity contribution is $0.5(\$1{,}000{,}000)=\$500{,}000$, and the frequency contribution is $0.5(\$2{,}000)^2=\$2{,}000{,}000$. For B, severity variance is $\$312{,}500-(\$500)^2=\$62{,}500$: the severity contribution is $2(\$62{,}500)=\$125{,}000$, and the frequency contribution is $2(\$500)^2=\$500{,}000$. Each pair sums to its aggregate variance.

Common mistake avoided: Equal expected aggregate losses do not imply equal variances. Portfolio B has more claims on average, but its much smaller severity second moment makes its aggregate variance lower. The stated frequency and severity distributions are hypothetical and satisfy the model’s independence assumptions.

Worked Example: Reconciling Observed Aggregate Loss and Severity

Method: Calculate aggregate loss per exposure directly, then verify the frequency-times-severity identity. A hypothetical repair plan has four contract-years of exposure. The observed claim counts and claim payments are: contract-year 1, zero claims; contract-year 2, two claims of \$600 and \$1,400; contract-year 3, one claim of \$1,000; contract-year 4, three claims of \$500, \$500, and \$1,000. Calculate observed frequency, average severity, and aggregate loss per exposure.

Total claim count is $0+2+1+3=6$. Total loss is $\$600+\$1{,}400+\$1{,}000+\$500+\$500+\$1{,}000=\$5{,}000$. Therefore observed frequency is $6/4=1.5$ claims per contract-year, and observed average severity is $\$5{,}000/6=\$833.333\ldots$ per claim.

Directly dividing total loss by total exposure gives $\$5{,}000/4=\$1{,}250$ per contract-year. The product check gives $(6/4)(\$5{,}000/6)=\$1{,}250$ exactly. Reporting average severity to cents gives \$833.33, but the product identity is exact when the unrounded value is used.

Common mistake avoided: The zero-claim contract-year remains part of the exposure denominator. Dividing the \$5,000 loss by only the three contract-years with claims would overstate loss per contract-year. Here the empirical calculation is an identity, not a claim that these four observations follow a Poisson model.

Interpreting Aggregate Data for Credibility

Aggregate loss per exposure is a useful experience measure when the question concerns the combined cost of claim frequency and severity. Its value should be interpreted alongside the exposure definition and the underlying claim measure. For example, paid loss and incurred loss are not interchangeable measures; comparisons require a consistent basis. Likewise, a rate per policy-year should not be compared directly with a rate per vehicle-year unless the exposure units have been reconciled.

The compound Poisson model provides a way to reason about the two sources of process variation, but it does not say that every observed portfolio must follow that model. The model’s calculated variance is a consequence of its assumptions. If counts are more variable than Poisson counts, if claim sizes vary systematically with counts, or if the mix of risks differs, the compound Poisson variance may not describe the data. State the model before using its formula, and distinguish a modeled variance from an empirical aggregate average.

Fix the unit and claim measure.
Specify the exposure denominator and whether the loss measure is paid, incurred, or another defined payment basis.

Separate frequency and severity.
Identify the count per exposure and the payment distribution per claim. Check whether the assumed relationship between them is reasonable.

Calculate the aggregate measure.
For observed data use total loss divided by total exposure; for the compound Poisson model use $\lambda E[X]$ for the expected aggregate loss.

Explain variability under the stated model.
Use $\lambda E[X^2]$ for compound Poisson aggregate variance, or show its frequency and severity components.

Common Exam Traps

Using severity variance as aggregate variance. Under the compound Poisson model, aggregate variance is $\lambda E[X^2]$, not simply $\lambda\operatorname{Var}(X)$. The latter omits random-count variation.
Using the wrong severity moment. The expected aggregate loss uses $E[X]$; the aggregate variance uses $E[X^2]$. They are different calculations with different units.
Dropping zero-claim exposures. An exposure with no claims contributes zero loss but still belongs in total exposure when calculating loss per exposure.
Assuming equal means imply equal variability. Frequency and severity mixes can produce the same expected aggregate loss but different aggregate variances.
Applying the formula without its assumptions. The compound Poisson result requires a Poisson count and independent, identically distributed severities independent of that count.

Key takeaway: Aggregate loss per exposure combines claim frequency and claim severity. Under an independent compound Poisson model, its expected value is $\lambda E[X]$, while its variance is $\lambda E[X^2]$. Similar aggregate means can conceal materially different frequency and severity patterns, and therefore different modeled variability.

Check Your Understanding

For each problem, state the relevant calculation and keep the exposure unit clear.

A hypothetical appliance plan has a Poisson claim count with parameter $\lambda=1.2$ per plan-year. Each claim is independently \$200 or \$800 with equal probability, independent of the count. Find expected aggregate loss and aggregate variance per plan-year. (Answer: expected loss \$600; variance \$408,000 dollars squared.)
A hypothetical crop policy has a Poisson count with parameter $0.6$ per policy-year. Severity is \$500 with probability 0.8 and \$2,000 with probability 0.2, independently of count. Find the severity mean, severity second moment, and aggregate variance. (Answer: severity mean \$800; second moment \$1,000,000; aggregate variance \$600,000 dollars squared.)
Two hypothetical portfolios have Poisson counts. Portfolio C has $\lambda=1$ and constant severity \$1,000. Portfolio D has $\lambda=2$ and constant severity \$500. Calculate each expected aggregate loss and variance, then compare. (Answer: both expected losses are \$1,000; C variance is \$1,000,000 dollars squared; D variance is \$500,000 dollars squared.)
A hypothetical service portfolio has five exposure-years, four claims, and total claim payments of \$3,600. Calculate observed frequency, average severity, and aggregate loss per exposure; verify the product identity. (Answer: frequency $0.8$ claims per exposure-year; average severity \$900; aggregate loss \$720 per exposure-year; $0.8\times\$900=\$720$.)
A hypothetical portfolio has $E[N]=3$, $\operatorname{Var}(N)=5$, and independent, identically distributed severities with mean \$400 and variance \$10,000, independent of $N$. Using the general independent-count formula, calculate expected aggregate loss and aggregate variance. (Answer: expected loss \$1,200; variance $3(\$10,000)+5(\$400)^2=\$830,000$ dollars squared.)

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