Aggregate Loss Variance Decomposition

Why an Aggregate Mean Is Not a Risk Measure

A pure premium is an expected loss per unit of exposure. It is useful for estimating the average cost, but it does not show how widely actual aggregate losses may vary around that average. Two hypothetical portfolios can have the same expected loss and substantially different aggregate loss variances because their claim counts, claim amounts, or both vary differently.

For a compound aggregate loss, two sources of variation can be separated: the number of claims can vary, and the amount of each claim can vary. The standard decomposition quantifies these sources separately. It requires a defined compound model: claim severities are independent and identically distributed, and they are independent of the claim count.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed claim severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Write $\mu=E[X]$ and $\sigma^2=\operatorname{Var}(X)$. If $E[N]$, $\operatorname{Var}(N)$, and $E[X^2]$ are finite, then $E[S]=E[N]\mu$ and $\operatorname{Var}(S)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$.

The first variance term, $E[N]\sigma^2$, is the severity variation component: it captures variation in claim amounts, averaged over the expected number of claims. The second, $\operatorname{Var}(N)\mu^2$, is the claim-count variation component: it captures uncertainty in how many claims occur, with each additional claim contributing an expected amount $\mu$.

Deriving the Decomposition

Condition on the claim count. If $N=n$, the aggregate is the sum of $n$ independent severities. Its conditional mean is $n\mu$, and its conditional variance is $n\sigma^2$. For $n=0$, both are zero. The law of total variance states that $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$.

Find the conditional mean and variance.
For fixed $N=n$, independence of severities gives $E[S\mid N=n]=n\mu$ and $\operatorname{Var}(S\mid N=n)=n\sigma^2$.

Average the conditional variance.
$E[\operatorname{Var}(S\mid N)]=E[N\sigma^2]=E[N]\sigma^2$, since $\sigma^2$ is constant.

Take the variance of the conditional mean.
$\operatorname{Var}(E[S\mid N])=\operatorname{Var}(N\mu)=\mu^2\operatorname{Var}(N)$.

Combine the terms.
The law of total variance adds the two results, giving $\operatorname{Var}(S)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$.

This argument also explains why the components are distinct. Conditional on a particular count, only severity variation remains. Variation in the conditional mean $N\mu$ across possible counts produces the count component.

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

For a Poisson count with parameter $\lambda$, $E[N]=\operatorname{Var}(N)=\lambda$. Thus the compound Poisson aggregate has mean $\lambda\mu$ and variance $\lambda(\sigma^2+\mu^2)=\lambda E[X^2]$. This specialization requires the same count-severity independence and finite second moments. A different count distribution generally has a different relationship between its mean and variance, so the Poisson simplification should not be used automatically.

Worked Example: Poisson Frequency With Exponential Severity

A hypothetical policy portfolio has a Poisson claim count with parameter $\lambda=0.20$ for one year. Each claim severity is exponentially distributed with mean $1,000, independently of the count and of other severities. Find the expected annual aggregate loss and decompose its variance. There is no discounting; interest and mortality assumptions are not applicable.

Method and assumptions. Use the compound Poisson decomposition. The exponential distribution has scale parameter $1,000, so $\mu=$1{,}000$ and $\sigma^2=($1{,}000)^2=$1{,}000{,}000$. The count has mean and variance both equal to $0.20$.

The expected aggregate loss is $E[S]=0.20($1{,}000)=$200$.

The severity component is $E[N]\sigma^2=0.20($1{,}000{,}000)=$200{,}000$. The claim-count component is $\operatorname{Var}(N)\mu^2=0.20($1{,}000)^2=$200{,}000$. Therefore, $\operatorname{Var}(S)=$200{,}000+$200{,}000=$400{,}000$.

Check. For a compound Poisson model, $\operatorname{Var}(S)=\lambda E[X^2]$. Here $E[X^2]=\sigma^2+\mu^2=$2{,}000{,}000$, so $0.20($2{,}000{,}000)=$400{,}000$, agreeing with the decomposition. The common mistake this avoids is using only the severity variance term and omitting variation in the number of claims.

Count Distributions Change the Count Component

The severity component depends on the expected claim count and the severity variance; the count component depends on the variance of the count itself. For a fixed expected frequency, a count distribution with larger variance contributes more aggregate loss variance through $\operatorname{Var}(N)\mu^2$. This is one reason a mean frequency alone does not fully characterize aggregate loss risk.

The units provide a useful check. If severity is measured in dollars, then $\sigma^2$ and $\mu^2$ are measured in squared dollars. The count moments are dimensionless, so both variance components and the resulting $\operatorname{Var}(S)$ are in squared dollars.

Worked Example: Binomial Count and Gamma Severity

For a hypothetical group of four independent exposures, suppose each exposure produces a claim with probability $0.10$. Thus $N$ has a binomial distribution with parameters $n=4$ and $p=0.10$. Given a claim, severity has a gamma distribution with shape $2$ and scale $250, independently across claims and independently of $N$. Find the aggregate mean and both variance components. There is no discounting; interest and mortality assumptions are not applicable.

Method and assumptions. The binomial count has $E[N]=np=4(0.10)=0.40$ and $\operatorname{Var}(N)=np(1-p)=4(0.10)(0.90)=0.36$. For a gamma distribution with shape $k$ and scale $\theta$, mean is $k\theta$ and variance is $k\theta^2$. Hence $\mu=2($250)=$500$ and $\sigma^2=2($250)^2=$125{,}000$.

The expected aggregate loss is $E[S]=0.40($500)=$200$. The severity component is $0.40($125{,}000)=$50{,}000$. The count component is $0.36($500)^2=0.36($250{,}000)=$90{,}000$. Thus $\operatorname{Var}(S)=$50{,}000+$90{,}000=$140{,}000$.

Check. The mean is $200, as expected from four exposures each having claim probability 0.10 and mean claim amount $500: $4(0.10)($500)=$200$. The variance is in squared dollars. The common mistake this avoids is treating the binomial count as Poisson: a Poisson count with mean 0.40 would have count variance 0.40, not 0.36, and would produce a different aggregate variance.

What a Pure Premium Mean Leaves Out

If the exposure base is $e$, the pure premium is often the expected aggregate loss divided by exposure, $E[S]/e$. That value summarizes the average loss cost, not the distribution of loss around the average. Aggregate variance can be divided by $e^2$ to obtain the variance of aggregate loss per exposure for the same portfolio, but that does not make the mean itself a variance measure.

For example, the same aggregate mean can arise from a relatively stable claim count with variable claim amounts, or from a more variable count with different severity variability. A rate based only on expected loss does not reveal which component drives uncertainty. The decomposition is therefore useful for understanding the aggregate loss model, interpreting risk differences, and checking whether a proposed model's assumptions fit the situation.

Key result: Under independent claim count and iid severities with finite second moments, aggregate mean depends on $E[N]$ and $E[X]$, while aggregate variance depends on both the severity variance and the claim-count variance. A pure premium mean alone specifies neither variance component.

When the Standard Formula Does Not Apply

Independence between count and severity is essential to the stated decomposition. If claim severity changes systematically with the number of claims, then the conditional mean and variance of aggregate loss need not be $N\mu$ and $N\sigma^2$ for constants $\mu$ and $\sigma^2$. Common causes of claims can also create dependence among severities. In such cases, use conditional moments that match the dependence structure rather than substituting unconditional severity moments into the independent formula.

A further condition is finiteness of the relevant second moments. If $N$ or claim severity has an infinite second moment, the aggregate variance may be infinite or the displayed decomposition may not yield a finite value. A calculated finite variance should not be reported without checking that the assumed distributions have the required moments.

Worked Example: Detecting Frequency-Severity Dependence

Consider a hypothetical risk with $P(N=1)=0.50$ and $P(N=2)=0.50$. If $N=1$, the claim severity is deterministically $1,000. If $N=2$, both severities are deterministically $500. Find the actual aggregate variance and show why applying the independent formula to the marginal claim severity gives a wrong answer. There is no discounting; interest and mortality assumptions are not applicable.

Method and assumptions. The count and severities are deliberately dependent, so first condition on the count. When $N=1$, $S=$1{,}000$. When $N=2$, $S=2($500)=$1{,}000$. Thus $S$ is always $1,000, $E[S]=$1,000, and \(\operatorname{Var}(S)=0$.

To expose the mistake, calculate the marginal severity distribution across all claims. The expected number of $1,000 claims is $0.50(1)=0.50$; the expected number of $500 claims is $0.50(2)=1.00$. The expected total claim count is $0.50(1)+0.50(2)=1.50$. Therefore, a randomly selected claim has probability $0.50/1.50=1/3$ of being $1,000 and probability $1.00/1.50=2/3$ of being $500.

The marginal severity mean is $(1/3)($1{,}000)+(2/3)($500)=$666.67$, rounded to cents. Its second moment is $(1/3)($1{,}000)^2+(2/3)($500)^2=$500{,}000$. Its variance is $$500{,}000-($666.666\ldots)^2=$55{,}555.56$, rounded to cents.

The count has mean $1.50$ and variance $0.25$. Incorrectly treating count and severity as independent would give a severity component of $1.50($55{,}555.56)=$83{,}333.34$ and a count component of $0.25($666.666\ldots)^2=$111{,}111.11$, for a false total variance of about $194,444.44. The actual variance is zero because aggregate loss is constant. The common mistake is applying the independent compound formula merely because marginal claim severities and count moments are available; the independence assumption fails here.

Exam Checks

A reliable calculation states the count distribution, severity distribution, independence assumptions, and units before substituting into the formula. The following checklist helps prevent errors:

Use $\operatorname{Var}(N)$, not automatically $E[N]$, in the count component.
Use the severity variance $\sigma^2$ in the severity component, not the severity second moment.
For a Poisson count only, replace $\operatorname{Var}(N)$ by $E[N]=\lambda$.
Check whether count and severity are independent and whether severities are independent across claims.
Keep aggregate loss variance in squared currency units, distinct from the expected loss or pure premium.

The decomposition and its assumptions are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, identify the count and severity inputs before calculating the aggregate quantities.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.30)$ and independent exponential claim severities with mean $2,000. Find the aggregate mean and the count and severity components of variance. (Answer: mean $600; count component $1,200,000; severity component $1,200,000; total variance $2,400,000.)
A hypothetical group has $N\sim\operatorname{Binomial}(5,0.20)$. Each claim is a fixed $400. Find aggregate mean and variance. (Answer: mean $400; count component and total variance $128,000; severity component $0.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(0.10)$ and independent gamma severities with shape $4$ and scale $250. Find aggregate mean and variance components. (Answer: mean $100; count component $100,000; severity component $25,000; total variance $125,000.)
A hypothetical count distribution has $E[N]=2$ and $\operatorname{Var}(N)=3$. Independent claim severities have mean $300 and variance $40,000. Find aggregate mean and variance. (Answer: mean $600; severity component $80,000; count component $270,000; total variance $350,000.)
Two independent hypothetical risks have aggregate losses with means $500 and $300, and variances $100,000 and $50,000, respectively. Find the mean and variance of their combined loss and of their average loss per risk. (Answer: combined mean $800 and variance $150,000; average loss per risk has mean $400 and variance $37,500.)

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