Deriving Aggregate-Loss Variance

Why Aggregate-Loss Variance Has Two Sources

An insurer’s total loss over a period can vary because the number of claims varies, because the cost of each claim varies, or because both vary. A variance calculation that captures only severity variation misses the uncertainty in claim count. A calculation that uses only the average severity also misses variation among individual losses.

The random-sum model expresses aggregate loss as a random number of individual claim amounts added together. The key result is that aggregate variance depends on both the expected variation in claim amounts and the variation in the number of claims. This distinction is useful when assessing how volatile an aggregate loss total may be; it is also a foundation for later credibility calculations.

Definition: Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be individual claim amounts with common mean $\mu_X=E[X]$ and variance $\sigma_X^2=\operatorname{Var}(X)$. The aggregate loss is $$ S_N=\sum_{j=1}^{N}X_j, $$ with $S_N=0$ when $N=0$. For the formulas in this tutorial, assume the claim amounts are identically distributed, mutually independent, independent of $N$, and have finite second moments. Also assume $E[N]$ and $\operatorname{Var}(N)$ are finite.

The model is an aggregate of claim amounts for a specified period, such as one hypothetical policy year. Unless stated otherwise, amounts are nominal and are not discounted: interest and mortality assumptions do not enter these calculations. If claim amounts are dependent on one another or on the count, the formulas below need not apply.

The Mean and Variance Formulas

Write $m_N=E[N]$ and $v_N=\operatorname{Var}(N)$. Under the assumptions above, the aggregate mean and variance are

E[S_N]=m_N\mu_X

and

\operatorname{Var}(S_N)=m_N\sigma_X^2+v_N\mu_X^2.

The first variance term, $m_N\sigma_X^2$, is the contribution from random claim amounts, averaged over the expected number of claims. The second term, $v_N\mu_X^2$, is the contribution from random claim count: when the count fluctuates, the aggregate changes roughly by the average claim amount for each claim added or removed.

Equivalent form: Since $E[X^2]=\sigma_X^2+\mu_X^2$, the variance can also be written as $$ \operatorname{Var}(S_N)=E[N]E[X^2]+\bigl(\operatorname{Var}(N)-E[N]\bigr)\mu_X^2. $$ Use this form only with the count moments shown; it is not generally correct to add $E[N]E[X^2]$ and $\operatorname{Var}(N)\mu_X^2$.

A particularly useful special case is a Poisson claim count. If $N\sim\operatorname{Poisson}(\lambda)$, then $E[N]=\operatorname{Var}(N)=\lambda$, so

E[S_N]=\lambda E[X], \qquad \operatorname{Var}(S_N)=\lambda E[X^2].

The second moment $E[X^2]$ is not the same as the severity variance $\operatorname{Var}(X)$. It includes both severity variance and the squared severity mean. This difference matters whenever claim count is random.

Why the Variance Formula Holds

Condition on the claim count. If $N=n$, the aggregate is the sum of $n$ independent claim amounts. Its conditional mean is $n\mu_X$, and its conditional variance is $n\sigma_X^2$. The law of total variance separates the average conditional variance from the variance of the conditional mean:

Find the conditional mean and variance.
For a fixed count $N=n$, independence gives $E[S_N\mid N=n]=n\mu_X$ and $\operatorname{Var}(S_N\mid N=n)=n\sigma_X^2$.

Average the conditional mean.
By iterated expectation, $E[S_N]=E[E[S_N\mid N]]=E[N\mu_X]=E[N]\mu_X$.

Apply total variance.
$\operatorname{Var}(S_N)=E[\operatorname{Var}(S_N\mid N)]+\operatorname{Var}(E[S_N\mid N])$.

Substitute and simplify.
The first term is $E[N\sigma_X^2]=E[N]\sigma_X^2$; the second is $\operatorname{Var}(N\mu_X)=\operatorname{Var}(N)\mu_X^2$. Adding them gives the stated formula.

This derivation shows exactly where the independence assumptions enter. Conditional on $N=n$, the variance of a sum of $n$ independent claim amounts is $n\sigma_X^2$. If the claim amounts are dependent, covariance terms may also contribute to the conditional variance.

Applying the Formula to Different Count Models

The severity distribution and claim-count distribution play different roles. First obtain the count mean and variance from the count model. Then obtain the severity mean and variance—or its second moment—from the severity model. Finally, substitute into the aggregate formulas, keeping units consistent. For example, if severity is measured in dollars, aggregate variance is measured in dollars squared.

Worked Example: Poisson Count and Two-Point Severity

A hypothetical one-year coverage has $N\sim\operatorname{Poisson}(4)$. Each claim amount is independently $X=500$ dollars with probability $0.5$ and $X=1{,}500$ dollars with probability $0.5$, independently of $N$. Find the aggregate mean and variance. Amounts are nominal with no discounting; interest and mortality assumptions are not applicable.

Method. Use the random-sum formulas, first calculating the severity moments. The Poisson parameter is $\lambda=4$, so $E[N]=4$ and $\operatorname{Var}(N)=4$.

The severity mean is $$ E[X]=0.5(500)+0.5(1{,}500)=250+750=1{,}000\text{ dollars}. $$ The severity second moment is $$ E[X^2]=0.5(500^2)+0.5(1{,}500^2) =0.5(250{,}000)+0.5(2{,}250{,}000) =1{,}250{,}000\text{ dollars}^2. $$ Thus $\operatorname{Var}(X)=1{,}250{,}000-1{,}000^2=250{,}000\text{ dollars}^2$.

The aggregate mean is $$ E[S_N]=4(1{,}000)=4{,}000\text{ dollars}. $$ The variance using the two-component formula is $$ \operatorname{Var}(S_N) =4(250{,}000)+4(1{,}000^2) =1{,}000{,}000+4{,}000{,}000 =5{,}000{,}000\text{ dollars}^2. $$ As a check, the Poisson special case gives $4E[X^2]=4(1{,}250{,}000)=5{,}000{,}000\text{ dollars}^2$, the same result.

Common mistake avoided: Using $4\operatorname{Var}(X)=1{,}000{,}000$ as the aggregate variance omits the variability in the Poisson count. The correct variance includes the additional $4E[X]^2=4{,}000{,}000$ dollars squared.

Worked Example: Binomial Count and Exponential Severity

For a hypothetical one-year book, let $N\sim\operatorname{Binomial}(10,0.2)$. Given the count, individual claim amounts are independent exponential random variables with mean $3{,}000$ dollars, independent of $N$. Find the aggregate mean and variance. Use nominal losses without discounting; interest and mortality assumptions are not applicable.

Method. For a binomial count, $E[N]=np$ and $\operatorname{Var}(N)=np(1-p)$. Therefore, $$ E[N]=10(0.2)=2, \qquad \operatorname{Var}(N)=10(0.2)(0.8)=1.6. $$ For an exponential severity with mean $3{,}000$, the rate parameter is $1/3{,}000$ per dollar. Its variance is the square of its mean: $$ E[X]=3{,}000,\qquad \operatorname{Var}(X)=3{,}000^2=9{,}000{,}000\text{ dollars}^2. $$

The expected aggregate is $$ E[S_N]=2(3{,}000)=6{,}000\text{ dollars}. $$ For the variance, calculate the frequency and severity contributions separately: $$ E[N]\operatorname{Var}(X)=2(9{,}000{,}000)=18{,}000{,}000\text{ dollars}^2, $$ $$ \operatorname{Var}(N)E[X]^2 =1.6(3{,}000^2) =1.6(9{,}000{,}000) =14{,}400{,}000\text{ dollars}^2. $$ Thus $$ \operatorname{Var}(S_N)=18{,}000{,}000+14{,}400{,}000 =32{,}400{,}000\text{ dollars}^2. $$

Check. The exponential second moment is $E[X^2]=9{,}000{,}000+3{,}000^2=18{,}000{,}000$. The equivalent form gives $2(18{,}000{,}000)+(1.6-2)(9{,}000{,}000)=36{,}000{,}000-3{,}600{,}000=32{,}400{,}000$ dollars squared.

Common mistake avoided: Do not substitute $E[X^2]$ for $\operatorname{Var}(X)$ in the first term while retaining the count-variance term. The two-component formula uses severity variance in its first term; the equivalent second-moment form adjusts the coefficient of $E[X]^2$.

Worked Example: A Count Distribution Given by Probabilities

A hypothetical portfolio’s one-year claim count has probabilities $P(N=0)=0.2$, $P(N=1)=0.5$, and $P(N=3)=0.3$. Independent claim amounts are $2{,}000$ dollars with probability $0.75$ and $6{,}000$ dollars with probability $0.25$, independently of $N$. Find the aggregate mean and variance. No discounting is applied; interest and mortality assumptions are not applicable.

Method. Calculate the count moments directly from its probability distribution: $$ E[N]=0(0.2)+1(0.5)+3(0.3)=0+0.5+0.9=1.4, $$ $$ E[N^2]=0^2(0.2)+1^2(0.5)+3^2(0.3)=0+0.5+2.7=3.2. $$ Therefore, $$ \operatorname{Var}(N)=E[N^2]-E[N]^2=3.2-(1.4)^2=3.2-1.96=1.24. $$

The severity mean is $$ E[X]=2{,}000(0.75)+6{,}000(0.25)=1{,}500+1{,}500=3{,}000\text{ dollars}. $$ Its second moment and variance are $$ E[X^2]=4{,}000{,}000(0.75)+36{,}000{,}000(0.25) =3{,}000{,}000+9{,}000{,}000 =12{,}000{,}000\text{ dollars}^2, $$ $$ \operatorname{Var}(X)=12{,}000{,}000-3{,}000^2 =12{,}000{,}000-9{,}000{,}000 =3{,}000{,}000\text{ dollars}^2. $$

The aggregate mean is $1.4(3{,}000)=4{,}200$ dollars. The variance components are $$ 1.4(3{,}000{,}000)=4{,}200{,}000\text{ dollars}^2 $$ and $$ 1.24(3{,}000^2)=1.24(9{,}000{,}000)=11{,}160{,}000\text{ dollars}^2. $$ Therefore, $$ \operatorname{Var}(S_N)=4{,}200{,}000+11{,}160{,}000 =15{,}360{,}000\text{ dollars}^2. $$

Check. The equivalent form gives $1.4(12{,}000{,}000)+(1.24-1.4)(9{,}000{,}000)=16{,}800{,}000-1{,}440{,}000=15{,}360{,}000$ dollars squared. The negative adjustment is valid here because the count variance, $1.24$, is less than its mean, $1.4$.

Common mistake avoided: A count distribution given by probabilities is not necessarily Poisson. Its mean and variance must be calculated from the probabilities rather than assumed equal.

Common Traps and Practical Checks

A reliable calculation keeps the two sources of variation visible until the final addition. In particular, do not treat a random claim count as if it were fixed at its expected value. If the count were fixed at $n$, the aggregate variance would be $n\operatorname{Var}(X)$. For a random count, there is an additional term $\operatorname{Var}(N)E[X]^2$.

Confusing a second moment with a variance. $E[X^2]=\operatorname{Var}(X)+E[X]^2$. They are generally different.
Assuming the Poisson identity for every count model. Only for Poisson counts do $E[N]$ and $\operatorname{Var}(N)$ coincide.
Dropping the count-variation term. This understates aggregate variance whenever $\operatorname{Var}(N)>0$ and $E[X]\ne 0$.
Mixing units. If claim amounts are in thousands of dollars, the resulting variance is in thousands of dollars squared; convert consistently before reporting dollars squared.
Ignoring dependence. If severity depends on the count, or claim amounts within a period are dependent, the independent random-sum formula may not hold.

A useful final check is to calculate the variance in two ways when the required moments are available: first as $E[N]\operatorname{Var}(X)+\operatorname{Var}(N)E[X]^2$, then using the equivalent second-moment form. Agreement is a strong check against confusing the severity variance with its second moment. The result is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each hypothetical one-year loss model, find the requested aggregate quantity. Assume independent, identically distributed claim amounts independent of the count, with no discounting; interest and mortality assumptions are not applicable.

$N\sim\operatorname{Poisson}(6)$. Each claim amount has mean $800$ dollars and variance $160{,}000$ dollars squared. Find the aggregate mean and variance. (Answer: mean $4{,}800$ dollars; variance $4{,}800{,}000$ dollars squared.)
$N\sim\operatorname{Binomial}(8,0.25)$. Each claim amount has mean $2{,}000$ dollars and variance $4{,}000{,}000$ dollars squared. Find the aggregate mean and variance. (Answer: mean $4{,}000$ dollars; variance $14{,}000{,}000$ dollars squared.)
The claim count has $E[N]=3$ and $\operatorname{Var}(N)=4$. Severity has mean $500$ dollars and variance $90{,}000$ dollars squared. Find the aggregate mean and variance. (Answer: mean $1{,}500$ dollars; variance $1{,}270{,}000$ dollars squared.)
$N\sim\operatorname{Poisson}(5)$. Each claim is $1{,}000$ dollars with probability $0.8$ and $4{,}000$ dollars with probability $0.2$. Find the aggregate mean and variance. (Answer: mean $8{,}000$ dollars; variance $20{,}000{,}000$ dollars squared.)
A candidate uses $E[N]=4$, $\operatorname{Var}(N)=2$, $E[X]=1{,}000$ dollars, and $\operatorname{Var}(X)=250{,}000$ dollars squared, but reports aggregate variance as $4(250{,}000)=1{,}000{,}000$ dollars squared. Identify the omitted component and give the corrected aggregate variance and mean. (Answer: omitted count-variation component $2(1{,}000^2)=2{,}000{,}000$ dollars squared; corrected variance $3{,}000{,}000$ dollars squared; mean $4{,}000$ dollars.)

What You'll Learn