Deriving Aggregate Mean and Variance From Component Moments

Why Component Moments Determine Aggregate Risk

An aggregate loss combines two sources of uncertainty: how many claims occur and how large each claim is. A model for claim count alone does not describe total loss, and a severity model alone does not describe how many losses are added together. The mean and variance of the total depend on both components, but they can be calculated from a small set of moments without first finding the full aggregate distribution.

The key is to keep the two sources of variability separate. Conditional on a known claim count, only the severities vary. Across different possible claim counts, the conditional mean also varies. This distinction explains why aggregate variance is generally not just the expected number of claims multiplied by severity variance.

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

The independence assumptions matter. The count must be independent of the severity sequence, and the severities must have the same distribution and be mutually independent. If, for example, a high claim count is associated with larger severities, the formulas below need not apply as stated.

Deriving the Aggregate Mean and Variance

Condition first on the claim count. If $N=n$, the aggregate is a sum of exactly $n$ independent severities. Therefore its conditional mean is $n\mu$, and its conditional variance is $n\sigma^2$. For $n=0$, both are zero, consistent with an empty sum.

Find the conditional mean.
For $N=n$, linearity of expectation gives $E[S\mid N=n]=n\mu$. Thus $E[S\mid N]=N\mu$.

Average over possible counts.
The law of total expectation gives $E[S]=E[E[S\mid N]]=E[N\mu]=E[N]\mu$.

Separate the two variance sources.
The law of total variance gives $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$.

Substituting the conditional variance $N\sigma^2$ and conditional mean $N\mu$ into the last step gives the central result.

Theorem: Under the stated independence and identical-distribution assumptions, $$ E[S]=E[N]E[X_1] $$ and $$ \operatorname{Var}(S)=E[N]\operatorname{Var}(X_1)+\operatorname{Var}(N)\big(E[X_1]\big)^2. $$ The first variance term is the expected within-count severity variation. The second is variation in the conditional aggregate mean as the claim count changes.

A useful equivalent expression uses the severity second raw moment $E[X_1^2]$. Since $\operatorname{Var}(X_1)=E[X_1^2]-\mu^2$,

\operatorname{Var}(S) =E[N]E[X_1^2]+\big(\operatorname{Var}(N)-E[N]\big)\mu^2.

This form is especially helpful when a problem gives $E[X_1^2]$ directly. It also makes clear that the shortcut $\operatorname{Var}(S)=E[N]E[X_1^2]$ is not universal: it holds for a Poisson count, for which $\operatorname{Var}(N)=E[N]$, but not generally.

Worked Applications

Worked Example: Compound Poisson Loss With Two Severity Amounts

A hypothetical portfolio has $N\sim\operatorname{Poisson}(4)$ claims in a period. Claim severities are independent of $N$ and of one another. Each severity is \$2 with probability $0.75$ and \$6 with probability $0.25$. Find the mean and variance of aggregate loss $S$, in dollars and squared dollars.

Method and assumptions. Use the random-sum moment formulas. For a Poisson random variable with parameter $\lambda=4$, $E[N]=4$ and $\operatorname{Var}(N)=4$. The specified severity distribution has finite moments and satisfies the independence assumptions.

First calculate the severity mean:

$$ E[X_1]=(0.75)(2)+(0.25)(6)=1.50+1.50=3. $$

Next calculate the second raw moment and variance:

E[X_1^2]=(0.75)(2^2)+(0.25)(6^2) =(0.75)(4)+(0.25)(36)=3+9=12, $$ $$ \operatorname{Var}(X_1)=12-3^2=12-9=3.

The aggregate mean is

E[S]=4(3)=12\text{ dollars}.

The aggregate variance is

\operatorname{Var}(S)=4(3)+4(3^2)=12+36=48\text{ dollars}^2.

As a check, the Poisson form is $\operatorname{Var}(S)=\lambda E[X_1^2]$, which gives $4(12)=48\text{ dollars}^2$, agreeing with the calculation. The common mistake avoided here is using the Poisson shortcut for every count model; it works in this example because the count is Poisson.

Worked Example: Compound Binomial Loss

A hypothetical device fleet has five units, each independently producing a claim during the period with probability $0.2$. Thus $N\sim\operatorname{Binomial}(5,0.2)$. Conditional on a claim, its severity is \$1 or \$5, each with probability $0.5$, independently of the count and of other severities. Find the aggregate mean and variance.

Method and assumptions. Apply the random-sum formulas using the binomial count moments $E[N]=np$ and $\operatorname{Var}(N)=np(1-p)$. Here,

E[N]=5(0.2)=1,\qquad \operatorname{Var}(N)=5(0.2)(0.8)=0.8.

The severity moments are

E[X_1]=(0.5)(1)+(0.5)(5)=3, $$ $$ E[X_1^2]=(0.5)(1^2)+(0.5)(5^2)=0.5+12.5=13, $$ $$ \operatorname{Var}(X_1)=13-3^2=4.

Therefore,

E[S]=1(3)=3\text{ dollars}, $$ $$ \operatorname{Var}(S)=1(4)+0.8(3^2)=4+7.2=11.2\text{ dollars}^2.

Check with the raw-moment form:

E[N]E[X_1^2]+\big(\operatorname{Var}(N)-E[N]\big)\big(E[X_1]\big)^2 =1(13)+(0.8-1)(9)=13-1.8=11.2.

The answers agree. The common mistake avoided is treating this binomial count as Poisson and reporting $E[N]E[X_1^2]=13$ for the variance. The actual variance is $11.2\text{ dollars}^2$, because the binomial count has variance $0.8$, not $1$.

Worked Example: Using Count and Severity Moments Directly

In a hypothetical compound model, $N$ is negative binomial with parameters $r=8$ and $p=2/3$, using the convention that $N$ counts failures before the $r$th success. The severities are independent and identically distributed, independent of $N$, with $E[X_1]=3$ and $E[X_1^2]=13$. Find the aggregate mean and variance.

Method and assumptions. For this negative binomial convention,

E[N]=\frac{r(1-p)}{p},\qquad \operatorname{Var}(N)=\frac{r(1-p)}{p^2}.

Since $1-p=1/3$,

E[N]=\frac{8(1/3)}{2/3}=4,\qquad \operatorname{Var}(N)=\frac{8(1/3)}{(2/3)^2} =\frac{8/3}{4/9}=6.

The severity variance is obtained from its second raw moment:

\operatorname{Var}(X_1)=E[X_1^2]-\big(E[X_1]\big)^2 =13-3^2=13-9=4.

Thus the aggregate mean is

$$ E[S]=4(3)=12. $$

The aggregate variance is

\operatorname{Var}(S)=4(4)+6(3^2) =16+54=70.

A second check using raw moments gives

$$ 4(13)+(6-4)(9)=52+18=70. $$

The mean is 12 amount units and the variance is 70 squared-amount units. The common mistake avoided is to multiply the claim-count variance by the severity second moment. The formula requires $\operatorname{Var}(N)(E[X_1])^2$, not $\operatorname{Var}(N)E[X_1^2]$.

What Changes With the Claim-Count Model?

Once the severity moments are known, the count distribution enters through just $E[N]$ and $\operatorname{Var}(N)$. The same severity model can therefore have different aggregate variance under different frequency assumptions, even when the expected claim count is identical.

Claim-count model	$E[N]$	$\operatorname{Var}(N)$	Aggregate variance
Poisson with parameter $\lambda$	$\lambda$	$\lambda$	$\lambda E[X_1^2]$
Binomial with parameters $n,p$	$np$	$np(1-p)$	$np\operatorname{Var}(X_1)+np(1-p)(E[X_1])^2$
Negative binomial, failures before $r$ successes	$r(1-p)/p$	$r(1-p)/p^2$	$E[N]\operatorname{Var}(X_1)+\operatorname{Var}(N)(E[X_1])^2$

The table is a substitution aid, not a replacement for checking the model’s parameterization. In particular, negative binomial conventions differ: some count failures before a fixed number of successes, while others use a different parameter definition. State the convention before using its moments.

Common Mistakes and a Reliable Check

The most frequent variance error is confusing severity variance with severity second moment. These are related but not interchangeable:

E[X_1^2]=\operatorname{Var}(X_1)+\big(E[X_1]\big)^2.

For a general claim count, the correct formula weights severity variance by $E[N]$ and the squared severity mean by $\operatorname{Var}(N)$. Substituting $E[X_1^2]$ as though it were $\operatorname{Var}(X_1)$, or using the Poisson shortcut when the count is not Poisson, changes the result.

A useful exam check is to calculate both forms of the variance when the inputs allow it:

Use $E[N]\operatorname{Var}(X_1)+\operatorname{Var}(N)(E[X_1])^2$.
Use $E[N]E[X_1^2]+(\operatorname{Var}(N)-E[N])(E[X_1])^2$.
For a Poisson count only, verify the simplification to $E[N]E[X_1^2]$.

If both general forms do not agree, check the arithmetic and the conversion from the severity second moment to severity variance. If the Poisson shortcut fails to match the general formula for a Poisson count, check the count moments and the severity moments.

Key takeaway: For an independent and identically distributed severity sequence independent of $N$, aggregate mean is $E[N]E[X_1]$, while aggregate variance has two parts: $E[N]\operatorname{Var}(X_1)$ from severity fluctuation and $\operatorname{Var}(N)(E[X_1])^2$ from count fluctuation.

These moment identities are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. They are useful both for direct calculations and for checking aggregate-distribution results obtained by conditioning or recursion, but they do not by themselves specify the full aggregate distribution.

Check Your Understanding

For each problem, identify the count and severity moments before calculating the aggregate moments. Assume all severity variables are independent and identically distributed and independent of the claim count.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1.2)$. Each claim is \$10 or \$30 with probability $0.5$ each. Find $E[S]$ and $\operatorname{Var}(S)$. (Answer: mean \$24; variance \$600 dollars$^2$)
A hypothetical group has $N\sim\operatorname{Binomial}(6,0.3)$. Each claim severity is 1 or 3 amount units with probability $0.5$ each. Find the aggregate mean and variance. (Answer: mean 3.6 amount units; variance 6.84 squared-amount units)
Let $N$ count failures before the fifth success, with success probability $p=0.5$. Each severity is 2 or 4 amount units, with probabilities $0.25$ and $0.75$, respectively. Find the aggregate mean and variance. (Answer: mean 17.5 amount units; variance 126.25 squared-amount units)
A compound model has $E[N]=7$, $\operatorname{Var}(N)=14$, $E[X_1]=4$, and $E[X_1^2]=22$. Find the aggregate mean and variance. (Answer: mean 28 amount units; variance 266 squared-amount units)
A hypothetical model has a fixed claim count $N=3$. Each severity has mean 5 amount units and variance 2 squared-amount units. Find the aggregate mean and variance, and identify which variance component is zero. (Answer: mean 15 amount units; variance 6 squared-amount units; the claim-count component is zero)

Claim-count model	\(E[N]\)	\(\operatorname{Var}(N)\)	Aggregate variance
Poisson with parameter \(\lambda\)	\(\lambda\)	\(\lambda\)	\(\lambda E[X_1^2]\)
Binomial with parameters \(n,p\)	\(np\)	\(np(1-p)\)	\(np\operatorname{Var}(X_1)+np(1-p)(E[X_1])^2\)
Negative binomial, failures before \(r\) successes	\(r(1-p)/p\)	\(r(1-p)/p^2\)	\(E[N]\operatorname{Var}(X_1)+\operatorname{Var}(N)(E[X_1])^2\)

What You'll Learn