Deriving the Variance of Compound Aggregate Loss

Why Aggregate-Loss Variance Has Two Components

An insurer’s aggregate loss can vary because individual claims have different sizes and because the number of claims changes from period to period. A variance calculation that accounts for only one of these sources of uncertainty can materially understate or misstate the risk. The compound-loss variance formula separates the two contributions and applies to a broad class of frequency models.

The focus here is the variance of aggregate loss, not its full probability distribution. The formula is useful when the claim count and severity distribution are known well enough to calculate their moments, even when constructing the entire distribution would be impractical. The treatment is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be claim severities. The aggregate loss is $S_N=\sum_{j=1}^{N}X_j$, with $S_N=0$ when $N=0$. Write $\mu=E[X_j]$ and $\sigma^2=\operatorname{Var}(X_j)$ for the common severity mean and variance.

The standard compound model assumes the severities are identically distributed and mutually independent, and that the entire severity sequence is independent of $N$. Assume also that $E[N]$ and $\operatorname{Var}(N)$ are finite and that severities have finite mean and variance. Under these conditions, the variance of aggregate loss is

\operatorname{Var}(S_N)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2.

The first term reflects variation in claim sizes, accumulated over the expected number of claims. The second reflects variation in the number of claims, scaled by the square of the average claim size. Even if severities are constant, random claim counts can make aggregate loss variable; even if the count is fixed, random severities can do so.

Deriving the Compound Variance Formula

Condition on a particular claim count $N=n$. The aggregate is then the sum of exactly $n$ independent severities. Independence makes the conditional variance the sum of their variances, while linearity of expectation makes the conditional mean the sum of their means:

E[S_N\mid N=n]=n\mu,\qquad \operatorname{Var}(S_N\mid N=n)=n\sigma^2.

To account for the fact that $N$ itself is random, apply the law of total variance: the unconditional variance equals the expected conditional variance plus the variance of the conditional mean. Substituting the two conditional expressions gives the result.

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

Variance of the conditional mean.
$\operatorname{Var}(E[S_N\mid N])=\operatorname{Var}(N\mu)=\mu^2\operatorname{Var}(N)$, since $\mu$ is a constant.

Add the two components.
$\operatorname{Var}(S_N)=E[N]\sigma^2+\operatorname{Var}(N)\mu^2$. The first component is severity variation; the second is frequency variation.

This derivation also shows exactly where the assumptions enter. Identically distributed severities give a common $\mu$ and $\sigma^2$; independence among severities lets the conditional variances add without covariance terms; independence between $N$ and the severity sequence keeps those conditional moments valid for every count. If those assumptions fail, the formula need not apply as written.

Useful special case: For Poisson frequency with parameter $\lambda$, $E[N]=\operatorname{Var}(N)=\lambda$. Therefore $\operatorname{Var}(S_N)=\lambda(\sigma^2+\mu^2)$. Since $E[X^2]=\sigma^2+\mu^2$, this is also $\lambda E[X^2]$. For other frequency models, do not replace $\operatorname{Var}(N)$ by $E[N]$ unless the model actually has equal mean and variance.

Applying the Formula

A reliable calculation has three parts: find the claim-count mean and variance, find the severity mean and variance, and substitute each into its corresponding term. Keep units visible. If severity is measured in thousands of dollars, the aggregate variance is in squared thousands of dollars.

Worked Example: Poisson Frequency and Two-Point Severity

A hypothetical portfolio has a Poisson claim count with parameter $\lambda=4$ per year. Claim severities are independent of the count and of one another. Each claim is either $ \$5$ or $ \$15$, each with probability $0.5$. Find the annual aggregate-loss mean and variance.

Method and assumptions. Use the compound model. For a Poisson count, $E[N]=\operatorname{Var}(N)=4$. The severity distribution has two outcomes, so calculate its moments directly.

Severity mean. $\mu=0.5(5)+0.5(15)=2.5+7.5=\$10$.

Severity variance. The deviations from the mean are $-\$5$ and $\$5$, so $\sigma^2=0.5(25)+0.5(25)=25$ dollars squared.

Aggregate mean. $E[S_N]=E[N]\mu=4(10)=\$40$.

Aggregate variance. The severity contribution is $E[N]\sigma^2=4(25)=100$. The frequency contribution is $\operatorname{Var}(N)\mu^2=4(10^2)=400$. Thus $\operatorname{Var}(S_N)=100+400=500$ dollars squared.

Check. For Poisson frequency, the variance is $\lambda E[X^2]$. Here $E[X^2]=0.5(5^2)+0.5(15^2)=12.5+112.5=125$, and $4(125)=500$, agreeing with the two-component calculation. The common mistake this check guards against is omitting the frequency contribution and reporting only $100$.

Worked Example: Binomial Frequency and Exponential Severity

A hypothetical service contract has at most five claims in a year. Each of the five independent exposure units produces one claim with probability $0.2$, so $N\sim\operatorname{Binomial}(5,0.2)$. Claim severities are independent of the count and each other, and have an exponential distribution with mean $\$3$. Find the aggregate variance.

Method and assumptions. For $N\sim\operatorname{Binomial}(n,p)$, $E[N]=np$ and $\operatorname{Var}(N)=np(1-p)$. For an exponential severity with mean $\theta$, the variance is $\theta^2$. Here $n=5$, $p=0.2$, and $\theta=\$3$.

Frequency moments. $E[N]=5(0.2)=1$, and $\operatorname{Var}(N)=5(0.2)(0.8)=0.8$.

Severity moments. $\mu=\$3$, and $\sigma^2=3^2=9$ dollars squared.

Substitute. The severity contribution is $1(9)=9$. The frequency contribution is $0.8(3^2)=0.8(9)=7.2$. Therefore, $\operatorname{Var}(S_N)=9+7.2=16.2$ dollars squared.

Check. The count can vary even though it is capped at five, so a frequency contribution is required. Treating the count as fixed at its mean of one would give only $9$, not $16.2$. A noninteger expected count is a moment, not an actual fixed number of claims.

Worked Example: Binomial Frequency With Severity in Thousands

A hypothetical book of three policies has independent claim indicators, each with probability $0.4$, so $N\sim\operatorname{Binomial}(3,0.4)$. Each claim is independently $ \$2{,}000$ or $ \$8{,}000$, each with probability $0.5$. Find the aggregate variance in thousands of dollars squared and the aggregate standard deviation in dollars.

Method and assumptions. Apply the compound formula, expressing severity in thousands of dollars to keep the arithmetic compact. Thus each severity is $2$ or $8$ thousand dollars.

Frequency moments. $E[N]=3(0.4)=1.2$, and $\operatorname{Var}(N)=3(0.4)(0.6)=0.72$.

Severity mean and variance. The mean is $\mu=0.5(2)+0.5(8)=1+4=5$ thousand dollars. The squared deviations from the mean are both $9$, so $\sigma^2=0.5(9)+0.5(9)=9$ squared thousands of dollars.

Aggregate variance. The severity term is $1.2(9)=10.8$. The frequency term is $0.72(5^2)=0.72(25)=18$. The aggregate variance is $10.8+18=28.8$ squared thousands of dollars.

Standard deviation and check. The standard deviation is $\sqrt{28.8}=5.366563\ldots$ thousand dollars, or $\$5{,}366.56$ to the nearest cent. The units check: variance is in squared thousands, while its square root is in thousands. The common unit trap is to report $28.8$ as a dollar amount rather than a variance in squared units.

Common Mistakes and Model Limits

The formula is compact, but each term has a distinct role. Before calculating, verify which variance belongs to which random quantity: $\sigma^2$ is claim-severity variance, while $\operatorname{Var}(N)$ is claim-count variance. The squared mean severity multiplies the latter because a change in the count changes expected aggregate loss by $\mu$ per claim.

Leaving out frequency variation. Using only $E[N]\sigma^2$ assumes away the uncertainty in $N$. It is correct only when $\operatorname{Var}(N)=0$, such as a fixed claim count.
Using the wrong frequency moment. The second term uses $\operatorname{Var}(N)$, not $E[N^2]$. These are related by $\operatorname{Var}(N)=E[N^2]-(E[N])^2$, but are not interchangeable.
Squaring the wrong quantity. The frequency term is $\operatorname{Var}(N)\mu^2$, not $\operatorname{Var}(N)\sigma^2$. Count variation changes the conditional mean of aggregate loss, which is $N\mu$.
Ignoring dependence. If claim severities are dependent, conditional variances may include covariance terms. If frequency is related to severity, the stated compound formula may also fail. The independence assumptions must be assessed before applying it.

A useful check is to consider limiting cases. If the claim count is fixed at $n$, then $\operatorname{Var}(N)=0$ and aggregate variance reduces to $n\sigma^2$, the variance of a sum of $n$ independent severities. If severity is constant at $\mu$, then $\sigma^2=0$ and aggregate variance reduces to $\mu^2\operatorname{Var}(N)$, because only the count varies.

Key takeaway: Under independent, identically distributed severities that are independent of the claim count, aggregate variance is the expected number of claims times severity variance, plus claim-count variance times squared mean severity. Calculate and label both terms before adding them.

Check Your Understanding

For each problem, state the frequency and severity moments before applying the compound variance formula.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2.5)$. Independent severities have mean $\$4$ and variance $16$ dollars squared. Find the aggregate variance. (Answer: $80$ dollars squared)
A hypothetical group has $N\sim\operatorname{Binomial}(4,0.25)$. Each claim has a fixed severity of $\$6$. Find the aggregate variance and identify which component creates it. (Answer: $27$ dollars squared, entirely from frequency variation)
A hypothetical program has $N\sim\operatorname{Binomial}(6,0.3)$. Independent severity is equally likely to be $1,3,5,$ or $7$ thousand dollars. Find the aggregate variance in squared thousands of dollars. (Answer: $29.16$ squared thousands of dollars)
A hypothetical portfolio has Poisson frequency with parameter $3$, and independent exponential severities with mean $\$2$. Find the aggregate variance. (Answer: $24$ dollars squared)
A hypothetical claim count has probabilities $P(N=0)=0.5$, $P(N=1)=0.3$, and $P(N=2)=0.2$. Independent severities have mean $\$10$ and variance $4$ dollars squared. Find the aggregate variance. (Answer: $63.8$ dollars squared)

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