Derive Compound Moments After Policy-Level Coverage Changes

Why Coverage Changes Affect Aggregate Moments

An insurer's aggregate payment is shaped by two components: how many claims occur and how much the insurer pays on each claim. A deductible can turn some positive ground-up losses into zero insurer payments, while a policy limit caps payments on large losses. Consequently, the frequency model alone does not determine aggregate moments; the payment distribution after applying the contract matters too.

The key modeling decision is to be consistent about what the claim count represents. One approach counts all ground-up claims and assigns a zero payment to claims that do not pass the deductible. Another counts only claims that produce positive payments and uses the payment distribution conditional on a positive payment. These approaches can agree when implemented consistently, but mixing their frequency and severity assumptions gives incorrect results.

Definition: Let $N$ be the number of ground-up claims in a fixed period. For claim $j$, let $X_j$ be its ground-up loss and $Y_j^L$ the insurer's payment per loss after applying the contract. The aggregate insurer payment is $S=\sum_{j=1}^{N}Y_j^L$, with $S=0$ when $N=0$. In the standard compound model, the payments $Y_j^L$ are independent and identically distributed, and are independent of $N$. Write $m_1=E[Y^L]$ and $m_2=E[(Y^L)^2]$.

Compound Mean and Variance

Under the independence and identical-distribution assumptions above, conditioning on the number of claims gives the aggregate moments. The frequency distribution affects both the expected number of payments and the variability in that number; the modified severity distribution supplies the first and second payment moments.

Theorem: If $E[N]$, $\operatorname{Var}(N)$, and $m_2$ are finite, and the claim payments are independent of $N$, then $$ E[S]=E[N]m_1 $$ and $$ \operatorname{Var}(S)=E[N]\operatorname{Var}(Y^L)+\operatorname{Var}(N)m_1^2 =E[N]m_2+\bigl(\operatorname{Var}(N)-E[N]\bigr)m_1^2. $$

The first variance term reflects variability in payment amounts across the expected number of claims. The second reflects variability in how many claims occur. The equivalent expression using $m_2$ follows from $\operatorname{Var}(Y^L)=m_2-m_1^2$.

Why the Compound Variance Formula Holds

Conditional on $N=n$, the aggregate is a sum of $n$ independent payments. Its conditional mean is $nm_1$, and its conditional variance is $n\operatorname{Var}(Y^L)$. Apply the law of total expectation and the law of total variance:

E[S]=E[E[S\mid N]]=E[Nm_1]=E[N]m_1,

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])\\ &=E[N\operatorname{Var}(Y^L)]+\operatorname{Var}(Nm_1)\\ &=E[N]\operatorname{Var}(Y^L)+\operatorname{Var}(N)m_1^2. \end{aligned}

This derivation also shows why the count variance cannot generally replace the expected count in the first term. Conditional payment variability accumulates across the claims, so its unconditional contribution is $E[N]\operatorname{Var}(Y^L)$, not $\operatorname{Var}(N)\operatorname{Var}(Y^L)$.

For a Poisson count with parameter $\lambda$, $E[N]=\operatorname{Var}(N)=\lambda$, so the formula simplifies to $\operatorname{Var}(S)=\lambda m_2$. This special simplification should not be applied to a non-Poisson count model.

Deductible-and-Limit Payment Moments

For the examples below, $X$ is a nonnegative ground-up loss, $d$ is an ordinary deductible, and $u$ is a limit on the ground-up loss amount, with $0\le d<u$. Thus the insurer pays nothing up to the deductible, then pays the covered amount above it, capped at $u-d$. This convention defines the payment as $Y^L=\min((X-d)_+,u-d)$, where $(z)_+=\max(z,0)$. If a contract instead states a limit $u$ on the insurer's payment, the payment is $\min((X-d)_+,u)$; the two limit conventions must not be confused.

Result: Under the ground-up-loss limit convention just stated, the payment moments can be written using the ground-up survival function $S_X(x)=P(X>x)$: $$ E[Y^L]=\int_d^u S_X(x)\,dx, \qquad E[(Y^L)^2]=2\int_d^u (x-d)S_X(x)\,dx. $$ These tail-integral formulas apply to nonnegative losses with finite moments. They include discrete or continuous severity distributions, including distributions with atoms.

To see why the first integral holds, write the capped payment as the length of the interval from $d$ up to the smaller of $X$ and $u$, when $X>d$. Equivalently, $Y^L=\int_d^u \mathbf{1}_{\{X>x\}}\,dx$. Taking expectations gives the first formula. For any nonnegative random variable $Y$, $E[Y^2]=2\int_0^\infty tP(Y>t)\,dt$. Here $P(Y^L>t)=S_X(d+t)$ for $0\le t<u-d$, and it is zero for larger $t$. Changing variables to $x=d+t$ gives the second formula.

Worked Example: A Poisson Portfolio with a Ground-Up Limit

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$ ground-up claims in a year. For each claim, the ground-up loss $X$ has the following distribution: $P(X=50)=0.2$, $P(X=150)=0.3$, $P(X=300)=0.3$, and $P(X=700)=0.2$, with amounts in dollars. The policy has deductible $d=\$100$ and a ground-up loss limit $u=\$400$. Find the annual aggregate insurer-payment mean and variance.

Method and assumptions. Count all ground-up claims in $N$, including those that produce zero insurer payment. Assume their losses are independent and identically distributed, and independent of $N$. The payment is $Y^L=\min((X-100)_+,300)$.

Step 1: Map ground-up losses to payments. A loss of $\$50$ or $\$150$ produces payment $\$0$ or $\$50$, respectively; a loss of $\$300$ produces $\$200$; and a loss of $\$700$ is capped at $\$300$. Therefore, the payment distribution has probabilities $0.2, 0.3, 0.3, 0.2$ at $\$0, \$50, \$200, \$300$, respectively.

Step 2: Calculate the payment moments. $$ m_1=E[Y^L]=(0.2)(0)+(0.3)(50)+(0.3)(200)+(0.2)(300)=15+60+60=\$135. $$ $$ m_2=E[(Y^L)^2]=(0.2)(0^2)+(0.3)(50^2)+(0.3)(200^2)+(0.2)(300^2) =750+12{,}000+18{,}000=30{,}750\text{ dollars}^2. $$

Step 3: Apply the Poisson compound moments. For this count, $E[N]=\operatorname{Var}(N)=4$. Thus $$ E[S]=4(135)=\$540,\qquad \operatorname{Var}(S)=4(30{,}750)=123{,}000\text{ dollars}^2. $$

Check and common mistake. The payment variance is $30{,}750-135^2=12{,}525$, so the general formula gives $4(12{,}525)+4(135^2)=50{,}100+72{,}900=123{,}000$. The two methods agree. A common mistake is to leave the $\$50$ payment unchanged for the $\$700$ loss; the ground-up limit caps that payment at $\$300$.

Worked Example: Binomial Claim Frequency with Zero Payments

A hypothetical group of five independent exposures each has probability $0.2$ of producing one ground-up claim in the period. Thus $N\sim\operatorname{Binomial}(5,0.2)$. For each ground-up claim, the insurer's payment is $\$0$ with probability $0.5$, $\$100$ with probability $0.3$, and $\$300$ with probability $0.2$, independently of the count and of other payments. Find the aggregate mean and variance.

Method. Use the compound moments with $N$ counting all ground-up claims. In particular, the probability of zero payment is part of the per-claim payment distribution, not a reason to alter the supplied binomial count.

Step 1: Find severity moments. $$ m_1=(0.5)(0)+(0.3)(100)+(0.2)(300)=30+60=\$90, $$ $$ m_2=(0.5)(0^2)+(0.3)(100^2)+(0.2)(300^2)=3{,}000+18{,}000=21{,}000\text{ dollars}^2. $$ Therefore, $\operatorname{Var}(Y^L)=21{,}000-90^2=21{,}000-8{,}100=12{,}900\text{ dollars}^2$.

Step 2: Find count moments. For a binomial count, $E[N]=np=5(0.2)=1$, and $\operatorname{Var}(N)=np(1-p)=5(0.2)(0.8)=0.8$.

Step 3: Calculate aggregate moments. $$ E[S]=1(90)=\$90, $$ $$ \operatorname{Var}(S)=1(12{,}900)+0.8(90^2) =12{,}900+0.8(8{,}100) =12{,}900+6{,}480 =19{,}380\text{ dollars}^2. $$

Check and common mistake. The equivalent second-moment calculation is $1(21{,}000)+(0.8-1)(8{,}100)=21{,}000-1{,}620=19{,}380$. Using the count variance $0.8$ in place of $E[N]=1$ in the first variance term would incorrectly give $0.8(12{,}900)+0.8(8{,}100)=16{,}800$. The first term uses the expected claim count.

Worked Example: Exponential Severity with a Deductible and Limit

A hypothetical insurer has $N\sim\operatorname{Poisson}(2.5)$ ground-up claims in a period. Each ground-up loss is exponential with mean $\$100$, so its survival function is $S_X(x)=e^{-x/100}$ for $x\ge0$. A policy has deductible $d=\$50$ and ground-up limit $u=\$200$. Find the aggregate mean and variance, rounding final dollar values to the nearest cent.

Method and assumptions. Use the tail-integral formulas for the modified severity moments, then use the Poisson compound formulas. All losses are independent and identically distributed and independent of $N$.

Step 1: Calculate the mean payment per loss. $$ m_1=\int_{50}^{200}e^{-x/100}\,dx =100(e^{-0.5}-e^{-2}) \approx 100(0.6065307-0.1353353) \approx \$47.1195. $$

Step 2: Calculate the second payment moment. With $t=x-50$, the upper integration endpoint becomes $150$: $$ \begin{aligned} m_2 &=2\int_{50}^{200}(x-50)e^{-x/100}\,dx\\ &=2e^{-0.5}\int_0^{150}t e^{-t/100}\,dt\\ &=2e^{-0.5}\left[10{,}000\left(1-e^{-1.5}\right)-15{,}000e^{-1.5}\right]\\ &\approx 5{,}363.85\text{ dollars}^2. \end{aligned} $$ The bracketed expression follows by integration by parts, or by evaluating the antiderivative $-100(t+100)e^{-t/100}$ at $0$ and $150$.

Step 3: Compound the severity moments. Since $N$ is Poisson with mean $2.5$, $$ E[S]=2.5(47.1195377)\approx \$117.80, \qquad \operatorname{Var}(S)=2.5(5{,}363.85)\approx 13{,}409.62\text{ dollars}^2. $$

Check and common mistake. The payment cannot exceed $\$150$, the difference $u-d$. The calculated mean payment of about $\$47.12$ is below that cap. Treating $u=\$200$ as the maximum insurer payment would use a different contract and yield different moments.

Choosing a Consistent Frequency Representation

When the ground-up count is Poisson and each ground-up claim independently produces a positive payment with probability $q$, the positive-payment count is Poisson with mean $\lambda q$. If this alternative representation is used, the severity must be the conditional payment distribution given $Y^L>0$. Its mean is $E[Y^L\mid Y^L>0]=m_1/q$, and its second moment is $m_2/q$. Applying the Poisson compound variance formula with these conditional moments gives $\lambda q(m_2/q)=\lambda m_2$, matching the ground-up-count approach. This equivalence depends on the Poisson count and consistent conditioning; it is not a license to combine an unconditional frequency with conditional positive-payment severity.

Exam takeaway: First identify whether $N$ counts ground-up claims or positive insurer payments. Then calculate $m_1$ and $m_2$ for the payment variable that matches that count, and use the correct $E[N]$ and $\operatorname{Var}(N)$. This sequence is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the frequency representation before applying the compound moment formulas.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$ ground-up claims. Each claim payment is $\$0$ or $\$100$, each with probability $0.5$. Find the aggregate mean and variance. (Answer: mean $\$100$; variance $10{,}000$ dollars squared.)
A hypothetical group has $N\sim\operatorname{Binomial}(4,0.25)$. Each claim payment is $\$0$ with probability $0.6$ and $\$200$ with probability $0.4$. Find the aggregate mean and variance. (Answer: mean $\$80$; variance $14{,}400$ dollars squared.)
Ground-up losses have probabilities $0.2, 0.5, 0.3$ at $\$50, \$200, \$500$, respectively. A contract has deductible $d=\$100$ and ground-up limit $u=\$300$, and the ground-up claim count is Poisson with mean $3$. Find the aggregate mean and variance. (Answer: mean $\$330$; variance $51{,}000$ dollars squared.)
A count $N$ has $E[N]=2$ and $\operatorname{Var}(N)=3$. Independent claim payments have $E[Y^L]=\$40$ and $E[(Y^L)^2]=2{,}200$ dollars squared. Find the aggregate mean and variance. (Answer: mean $\$80$; variance $6{,}000$ dollars squared.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(1.5)$ ground-up claims, each with exponential ground-up loss of mean $\$100$. The policy has deductible $d=0$ and ground-up limit $u=\$100$. Find the aggregate mean and variance to the nearest cent. (Answer: mean $\$94.82$; variance $7{,}927.23$ dollars squared.)

What You'll Learn