Claim-Type Mixtures Within Aggregate Loss

Why Claim Type Matters

A claim portfolio may contain several kinds of claims, such as damage to different components or claims handled under different coverage categories. The severity distribution can vary by type. If the type of an individual claim is not known in advance, its unconditional severity distribution is a mixture of the type-specific distributions. This lets us use one severity model in a compound aggregate-loss calculation while retaining the effect of different claim types.

The key is to average the right quantities. The overall mean is the probability-weighted average of the type-specific means. The overall second moment is the probability-weighted average of the type-specific second moments—not the square of the overall mean. That second moment is especially useful because aggregate variance depends on it as well as on the claim-count distribution.

Definition: Let $J$ be the claim type, taking values $1,\ldots,k$, with $P(J=j)=\pi_j$, where $\pi_j\geq 0$ and $\sum_{j=1}^k\pi_j=1$. Let $X$ be the payment for one claim, and suppose the conditional distribution of $X$ given $J=j$ has mean $\mu_j=E[X\mid J=j]$ and second moment $m_{2,j}=E[X^2\mid J=j]$. The unconditional claim severity $X$ has the claim-type mixture distribution.

Moments of a Claim-Type Mixture

The law of total expectation gives the mixture mean. Applying the same idea to $X^2$ gives the mixture second moment. Once those are known, the variance follows from $ \operatorname{Var}(X)=E[X^2]-(E[X])^2$.

E[X]=\sum_{j=1}^k \pi_j\mu_j,\qquad E[X^2]=\sum_{j=1}^k \pi_jm_{2,j},\qquad \operatorname{Var}(X)=\sum_{j=1}^k\pi_jm_{2,j} -\left(\sum_{j=1}^k\pi_j\mu_j\right)^2.

If a type-specific variance is given instead of a second moment, use $m_{2,j}=\operatorname{Var}(X\mid J=j)+\mu_j^2$. Thus the overall second moment includes both the conditional variance within each type and the squared conditional mean. The unconditional variance can also be separated into two sources of variation:

\operatorname{Var}(X) =\sum_{j=1}^k \pi_j\operatorname{Var}(X\mid J=j) +\sum_{j=1}^k\pi_j(\mu_j-E[X])^2.

The first term is the probability-weighted average of within-type variances. The second is the variance of the type-specific means around the overall mean. Even if claims within every type have no severity variation, different type means can still make the unconditional severity variable.

Why the second-moment formula holds: Partition the possible claims according to their type. Since the events $J=j$ are mutually exclusive and exhaustive, $E[X^2]=\sum_j E[X^2\mathbf{1}_{\{J=j\}}]$. For each term, $E[X^2\mathbf{1}_{\{J=j\}}]=P(J=j)E[X^2\mid J=j]=\pi_jm_{2,j}$. Adding the terms gives $E[X^2]=\sum_j\pi_jm_{2,j}$. Subtracting $(E[X])^2$ then gives the variance.

Worked Example: Two Claim Types and a Poisson Aggregate

A hypothetical insurer models each claim as type A with probability $0.70$ or type B with probability $0.30$. Given type A, the payment is equally likely to be $100 or $300. Given type B, the payment is always $400. Claim types and payments are generated independently from claim to claim. The claim count $N$ is Poisson with mean $2$, independent of the severities. Find the unconditional severity mean and second moment, then find the aggregate-loss mean and variance.

Method. First calculate the conditional moments for each type and weight them by the type probabilities. Then use the compound Poisson formulas $E[S]=\lambda E[X]$ and $\operatorname{Var}(S)=\lambda E[X^2]$, where $S=\sum_{i=1}^N X_i$ and $\lambda=2$.

For type A, the conditional mean is $(100+300)/2=200$, and the conditional second moment is $(100^2+300^2)/2=(10{,}000+90{,}000)/2=50{,}000$. For type B, the conditional mean is $400$, and the conditional second moment is $400^2=160{,}000$.

Weight the means by their type probabilities: $E[X]=0.70(200)+0.30(400)=140+120=260$ dollars. Weight the second moments in the same way: $E[X^2]=0.70(50{,}000)+0.30(160{,}000)=35{,}000+48{,}000=83{,}000$ dollars squared. As a check, the severity variance is $83{,}000-260^2=83{,}000-67{,}600=15{,}400$ dollars squared.

For a Poisson count with mean $\lambda=2$, the aggregate mean is $E[S]=2(260)=520$ dollars. The aggregate variance is $\operatorname{Var}(S)=2(83{,}000)=166{,}000$ dollars squared.

Check. The compound Poisson variance also equals $\lambda\operatorname{Var}(X)+\lambda(E[X])^2=2(15{,}400)+2(67{,}600)=30{,}800+135{,}200=166{,}000$, agreeing with the second-moment calculation. The mistake to avoid is using only the severity variance in the Poisson formula; that would omit variation from the random claim count.

Using Mixture Moments in a Compound Model

Let $N$ be a nonnegative integer-valued claim count, independent of the severities. Suppose each claim independently receives a type from the same probabilities $\pi_1,\ldots,\pi_k$, and, conditional on its type, has the corresponding type-specific severity distribution. Then the severities $X_1,X_2,\ldots$ are identically distributed according to the unconditional mixture and are independent of $N$. Define aggregate loss by $S=\sum_{i=1}^N X_i$, with $S=0$ when $N=0$.

E[S]=E[N]E[X], \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(X) +\operatorname{Var}(N)(E[X])^2.

These formulas require a finite claim-count mean and variance and finite severity moments as used in the formulas. They do not require $N$ to be Poisson. For example, if $N$ is binomial, its binomial mean and variance go into the same expression. An equivalent variance formula, useful when the mixture second moment is already available, is

\operatorname{Var}(S) =E[N]E[X^2]+\bigl(\operatorname{Var}(N)-E[N]\bigr)(E[X])^2.

To see why the first variance formula holds, condition on $N=n$. Given $N=n$, the aggregate is a sum of $n$ independent severities, so its conditional mean is $nE[X]$ and its conditional variance is $n\operatorname{Var}(X)$. The law of total expectation gives $E[S]=E[N]E[X]$. The law of total variance gives $\operatorname{Var}(S)=E[N\operatorname{Var}(X)]+\operatorname{Var}(NE[X])$, which equals $E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$.

Worked Example: A Binomial Count With Exponential and Fixed Severities

A hypothetical portfolio has $N\sim\operatorname{Binomial}(5,0.20)$, independent of claim severities. Each claim is type A with probability $0.80$, with an exponential payment distribution of mean $100 given type A. It is type B with probability $0.20$, with a fixed payment of $300 given type B. Types are selected independently for all claims. Find the aggregate-loss mean and variance.

Method. Find the unconditional severity moments first. For an exponential payment with mean $100, $E[X\mid A]=100$ and $E[X^2\mid A]=2(100^2)=20{,}000$. For the fixed type B payment, $E[X\mid B]=300$ and $E[X^2\mid B]=300^2=90{,}000$.

The mixture mean is $E[X]=0.80(100)+0.20(300)=80+60=140$ dollars. The mixture second moment is $E[X^2]=0.80(20{,}000)+0.20(90{,}000)=16{,}000+18{,}000=34{,}000$ dollars squared. Therefore $\operatorname{Var}(X)=34{,}000-140^2=34{,}000-19{,}600=14{,}400$ dollars squared.

For $N\sim\operatorname{Binomial}(5,0.20)$, $E[N]=5(0.20)=1$, and $\operatorname{Var}(N)=5(0.20)(0.80)=0.80$. Hence $E[S]=1(140)=140$ dollars, and $\operatorname{Var}(S)=1(14{,}400)+0.80(140^2)=14{,}400+0.80(19{,}600)=14{,}400+15{,}680=30{,}080$ dollars squared.

Check. Using the equivalent formula gives $1(34{,}000)+(0.80-1)(19{,}600)=34{,}000-3{,}920=30{,}080$. A common error is to use $E[N]=1$ as though the count were fixed at one. That would yield only $14{,}400$, ignoring the additional variance from the random binomial count.

Mixture Selection Versus Fixed Type Counts

A claim-type mixture represents uncertainty about the type of each claim: before observing a claim, its type is selected according to the probabilities $\pi_j$. It is appropriate when each claim has the same type probabilities and the type selections are independent across claims. If the portfolio instead has a fixed number of claims of each type, the severities are generally not identically distributed across the whole portfolio. In that setting, calculate the aggregate moments by type or by claim rather than treating every claim as a random draw from one mixture.

For instance, suppose there is exactly one claim of type A and exactly one claim of type B. Their aggregate mean is the sum of their two conditional means, not twice the mean of a randomly selected type. If the type counts themselves are random, they may add another source of variation; the mixture model applies only when its claim-by-claim type-selection assumptions match the situation.

Worked Example: Negative Binomial Frequency and a Discrete Mixture

A hypothetical portfolio has a claim count $N$ equal to the number of failures before the second success in independent Bernoulli trials, each with success probability $0.40$. Each claim is independently type A with probability $0.60$, paying $200, or type B with probability $0.40$, paying $500. The claim count is independent of all types and payments. Find the aggregate-loss mean and variance.

Method. Use the stated negative binomial convention: failures before $r$ successes. Here $r=2$, $p=0.40$, and $1-p=0.60$, so $E[N]=r(1-p)/p=2(0.60)/0.40=3$ and $\operatorname{Var}(N)=r(1-p)/p^2=2(0.60)/(0.40^2)=1.20/0.16=7.5$.

For one claim, $E[X]=0.60(200)+0.40(500)=120+200=320$ dollars. Its second moment is $E[X^2]=0.60(200^2)+0.40(500^2)=0.60(40{,}000)+0.40(250{,}000)=24{,}000+100{,}000=124{,}000$ dollars squared. Thus $\operatorname{Var}(X)=124{,}000-320^2=124{,}000-102{,}400=21{,}600$ dollars squared.

The aggregate mean is $E[S]=3(320)=960$ dollars. The aggregate variance is $\operatorname{Var}(S)=3(21{,}600)+7.5(320^2)=64{,}800+7.5(102{,}400)=64{,}800+768{,}000=832{,}800$ dollars squared.

Check. The equivalent calculation gives $3(124{,}000)+(7.5-3)(102{,}400)=372{,}000+4.5(102{,}400)=372{,}000+460{,}800=832{,}800$. The frequent trap here is confusing the negative binomial parameterization or substituting a success-count variance formula for the failures-before-successes model specified in the problem.

Common Mistakes and Practical Checks

Claim-type mixtures are straightforward once the conditioning is explicit, but several errors can substantially change an aggregate result. Before calculating, identify whether type is random for each claim or fixed by the portfolio description, and distinguish a conditional moment from an unconditional one.

Squaring the mean instead of averaging squared severities. In general, $E[X^2]\neq(E[X])^2$. Calculate each type’s second moment, then weight those second moments.
Dropping between-type variation. The mixture variance is not simply the average of the type-specific variances. Different conditional means also contribute to unconditional variance.
Using the wrong aggregate variance input. The general random-sum formula uses both $E[N]$ and $\operatorname{Var}(N)$. The Poisson simplification $\operatorname{Var}(S)=\lambda E[X^2]$ applies when $N$ is Poisson with mean $\lambda$, not to every frequency model.
Assuming random types when counts are fixed. A fixed portfolio composition is not automatically represented by independent type draws with common probabilities.
Mixing units. If severities are in dollars, their means are in dollars and second moments and variances are in dollars squared; aggregate means and variances retain those respective units.

A useful final check is to calculate $\operatorname{Var}(X)$ in two ways: subtract the squared mixture mean from the mixture second moment, and add the within-type and between-type variance components. For aggregate variance, compare the total-variance formula with its equivalent second-moment form. Agreement helps catch arithmetic and modeling errors.

Key takeaway: First combine the type-specific distributions through probability-weighted conditional moments. Then use the resulting unconditional severity mean and second moment—or mean and variance—with the chosen claim-count model. This separation keeps type variation and frequency variation from being confused. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, treat claim types as independently selected for each claim and independent of the stated frequency unless the problem says otherwise.

A claim is type A with probability $0.75$, paying $80, or type B with probability $0.25$, paying $400. Find $E[X]$, $E[X^2]$, and $\operatorname{Var}(X)$. (Answer: $160; $44,800 dollars squared; $19,200 dollars squared.)
A claim is type A with probability $0.60$, with exponential severity of mean $50, or type B with probability $0.40$, with a fixed severity of $200. Find the unconditional severity mean, second moment, and variance. (Answer: $110; $19,000 dollars squared; $6,900 dollars squared.)
Use the severity mixture in question 2 with an independent Poisson claim count of mean $0.50$. Find the aggregate-loss mean and variance. (Answer: $55; $9,500 dollars squared.)
Use the severity mixture in question 1 with an independent $N\sim\operatorname{Binomial}(4,0.25)$ claim count. Find the aggregate-loss mean and variance. (Answer: $160; $38,400 dollars squared.)
Use the severity mixture in question 1 with $N$ equal to the number of failures before two successes, where each trial has success probability $0.50$. Find the aggregate-loss mean and variance. (Answer: $320; $140,800 dollars squared.)

What You'll Learn