Second Moments Of Aggregate Loss

Why the Second Moment Matters

For an aggregate loss, the expected value describes its average, but it does not describe how widely outcomes vary around that average. The second raw moment, \(E[S^2]\), is a direct route to the variance because \(\operatorname{Var}(S)=E[S^2]-(E[S])^2\). In a compound model, this moment depends on both the claim count and the severity distribution. Keeping those contributions straight is especially important when checking a variance calculation.

Throughout this tutorial, let \(N\) be a nonnegative integer-valued claim count. Let \(X_1,X_2,\ldots\) be independent and identically distributed claim severities, independent of \(N\), with finite second moment. The aggregate loss is \(S=\sum_{j=1}^{N}X_j\), with \(S=0\) when \(N=0\). These are the standard compound aggregate assumptions; the formula below need not apply unchanged if count and severities are dependent, or if severities are not identically distributed.

Definition: The second raw moment of a random variable \(Z\) is \(E[Z^2]\). It is not the variance. The two are related by \(\operatorname{Var}(Z)=E[Z^2]-(E[Z])^2\), provided the second moment is finite.

The Aggregate Second-Moment Formula

The formula uses the frequency mean \(E[N]\), the frequency factorial moment \(E[N(N-1)]\), and the first two raw moments of an individual severity. The factorial moment counts ordered pairs of distinct claim positions when there are \(N\) claims.

Theorem: Under the compound aggregate assumptions above,

$$ E[S^2] = E[N]E[X^2] + E[N(N-1)](E[X])^2. $$

Here \(X\) denotes a generic claim severity having the common distribution of the \(X_j\). The formula requires the displayed expectations to be finite.

To see why the two terms appear, first fix the claim count at \(N=n\). Squaring the sum produces \(n\) squared-severity terms and \(n(n-1)\) cross-products. Independence and identical distribution give \(E[X_j^2]=E[X^2]\) and \(E[X_jX_k]=(E[X])^2\) for distinct \(j\) and \(k\).

\begin{aligned} S^2 &=\left(\sum_{j=1}^{n}X_j\right)^2\\ &=\sum_{j=1}^{n}X_j^2+\sum_{\substack{j,k=1\\j\ne k}}^{n}X_jX_k,\\ E[S^2\mid N=n] &=nE[X^2]+n(n-1)(E[X])^2. \end{aligned}

Taking expectations over \(N\) replaces \(n\) by its mean in the first term and \(n(n-1)\) by its expectation in the second. In particular, when \(N=0\), both terms are zero, as they should be because the aggregate loss is then zero.

Find the severity moments.
Calculate \(E[X]\) and \(E[X^2]\) from the stated severity model. For a continuous severity with density \(f_X\), these are integrals against \(f_X\).

Find the frequency moments.
Calculate \(E[N]\) and \(E[N(N-1)]\). Do not substitute \(\operatorname{Var}(N)\) for the factorial moment.

Apply the formula and check.
Compute \(E[S^2]\), then use \(E[S]=E[N]E[X]\) to obtain \(\operatorname{Var}(S)\). Check the variance against the equivalent expression derived below.

The check follows by rearranging the second-moment result. Since \(E[N(N-1)]=\operatorname{Var}(N)+(E[N])^2-E[N]\), and \(E[S]=E[N]E[X]\), subtraction of \((E[S])^2\) gives

\operatorname{Var}(S) = E[N]\operatorname{Var}(X) + \operatorname{Var}(N)(E[X])^2.

This is an equivalent way to check the aggregate variance, not a replacement for identifying the requested quantity. A common error is to report \(E[S^2]\) when the question asks for \(\operatorname{Var}(S)\), or to omit the cross-product term in \(E[S^2]\).

Worked Example: Binomial Count and Two-Point Severity

A hypothetical portfolio has \(N\sim\operatorname{Binomial}(2,1/2)\) claims during a period. Claim severities are independent of \(N\), with \(P(X=1)=1/2\) and \(P(X=3)=1/2\), in thousands of dollars. Find \(E[S^2]\) and \(\operatorname{Var}(S)\), in squared thousands of dollars. Use the second-moment formula and check the variance.

Method and assumptions. This is a compound binomial model with independent, identically distributed discrete severities. The severity values are measured in thousands of dollars, so the moments of \(S\) are in the corresponding powers of that unit.

First calculate the severity moments:

E[X]=\tfrac12(1)+\tfrac12(3)=2, \qquad E[X^2]=\tfrac12(1^2)+\tfrac12(3^2)=5.

For the claim count, a binomial count with parameters \(2\) and \(1/2\) has \(E[N]=2(1/2)=1\). Its factorial moment is \(E[N(N-1)]=2(1/2)^2=1/2\). Therefore,

E[S^2] =1(5)+\tfrac12(2^2) =5+2 =7.

The aggregate mean is \(E[S]=E[N]E[X]=1(2)=2\). Thus,

\operatorname{Var}(S)=E[S^2]-(E[S])^2=7-2^2=3.

Check. The severity variance is \(5-2^2=1\), and the binomial count variance is \(2(1/2)(1/2)=1/2\). The equivalent variance expression gives \(1(1)+(1/2)(2^2)=1+2=3\), matching the result. The common mistake avoided here is using \(E[N]\) in place of \(E[N(N-1)]\) for the cross-products.

Continuous Severity: Integrate Before Aggregating

For a continuous severity, calculate its raw moments from its density before applying the aggregate formula. In particular, \(E[X^2]\) is not \((E[X])^2\); confusing these quantities understates the aggregate second moment unless the severity is constant.

Worked Example: Poisson Count and Uniform Severity

A hypothetical aggregate loss has \(N\sim\operatorname{Poisson}(2)\) claims. Independently, each claim severity is uniform on \([0,4]\) thousand dollars. Find \(E[S^2]\) and \(\operatorname{Var}(S)\), in squared thousands of dollars.

Method and assumptions. The count is Poisson with parameter \(2\), and the severities are independent and identically distributed, independent of the count. A uniform \([0,4]\) severity has density \(f_X(x)=1/4\) for \(0\le x\le4\).

Integrate to obtain the severity moments:

E[X]=\int_0^4 x\frac14\,dx =\frac14\left[\frac{x^2}{2}\right]_0^4 =2,

E[X^2]=\int_0^4 x^2\frac14\,dx =\frac14\left[\frac{x^3}{3}\right]_0^4 =\frac{16}{3}.

For a Poisson random variable with parameter \(2\), \(E[N]=2\) and \(E[N(N-1)]=2^2=4\). Substitution gives

E[S^2] =2\left(\frac{16}{3}\right)+4(2^2) =\frac{32}{3}+16 =\frac{80}{3}.

The aggregate mean is \(E[S]=2(2)=4\), so

\operatorname{Var}(S)=\frac{80}{3}-4^2 =\frac{80}{3}-\frac{48}{3} =\frac{32}{3}.

Check. The uniform severity variance is \(16/3-2^2=4/3\), and the Poisson count variance is \(2\). The equivalent variance expression gives \(2(4/3)+2(2^2)=8/3+8=32/3\). Both calculations agree. Using \((E[X])^2=4\) instead of \(E[X^2]=16/3\) would give an incorrect first term.

Worked Example: Geometric Count and Discrete Severity

A hypothetical model has a claim count with geometric probabilities \(P(N=n)=\frac13(\frac23)^n\), for \(n=0,1,2,\ldots\). Each severity is \(2\) thousand dollars with probability \(3/4\) and \(5\) thousand dollars with probability \(1/4\), independently of the count and of other severities. Find \(E[S^2]\) and \(\operatorname{Var}(S)\), in squared thousands of dollars.

Method and assumptions. The stated geometric distribution is supported on nonnegative integers, including zero. Its mean is \((1-p)/p=2\) for \(p=1/3\), and its variance is \((1-p)/p^2=6\). Hence

E[N(N-1)] =\operatorname{Var}(N)+(E[N])^2-E[N] =6+2^2-2 =8.

For severity,

E[X]=\frac34(2)+\frac14(5)=\frac{11}{4}, \qquad E[X^2]=\frac34(2^2)+\frac14(5^2)=\frac{37}{4}.

The aggregate second moment is therefore

E[S^2] =2\left(\frac{37}{4}\right)+8\left(\frac{11}{4}\right)^2 =\frac{37}{2}+\frac{121}{2} =79.

The aggregate mean is \(E[S]=2(11/4)=11/2\). It follows that

\operatorname{Var}(S) =79-\left(\frac{11}{2}\right)^2 =79-\frac{121}{4} =\frac{195}{4} =48.75.

Check. The severity variance is \(37/4-(11/4)^2=27/16\). The equivalent variance expression gives \(2(27/16)+6(11/4)^2=54/16+726/16=780/16=195/4\). The result agrees. The key trap is to use the specified geometric convention: this count includes zero and has mean \(2\), not a geometric count supported on \(1,2,\ldots\).

Common Checks and Interpretation

A reliable calculation keeps the raw moments and variances distinct. For example, the cross-product contribution uses \((E[X])^2\), whereas the squared-severity contribution uses \(E[X^2]\). The frequency term likewise uses \(E[N(N-1)]\), not simply the count variance. Once the second raw moment is found, subtract the square of the aggregate mean to obtain the variance.

The units provide a useful reasonableness check: if claim amounts are measured in thousands of dollars, \(E[S]\) is measured in thousands of dollars, while \(E[S^2]\) and \(\operatorname{Var}(S)\) are measured in squared thousands of dollars. Also, \(E[S^2]\) must be at least \((E[S])^2\), since variance cannot be negative. A result violating this inequality signals an arithmetic or formula error.

These identities are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. On a compound-model problem, write down the count and severity assumptions, compute their required moments separately, and keep the second-moment calculation distinct from the final variance calculation.

Check Your Understanding

For each problem, apply the second-moment formula and distinguish the raw second moment from the variance.

A hypothetical count is \(N\sim\operatorname{Binomial}(3,0.2)\), and each claim severity is a constant \(4\) thousand dollars. Find \(\operatorname{Var}(S)\), in squared thousands of dollars. (Answer: \(7.68\))
A hypothetical count is Poisson with parameter \(1.5\), and severities are uniform on \([0,2]\) thousand dollars. Find \(E[S^2]\), in squared thousands of dollars. (Answer: \(4.25\))
A count has geometric probabilities \(P(N=n)=\frac12(\frac12)^n\), for \(n=0,1,2,\ldots\). Severities are \(1\) and \(4\) thousand dollars with equal probabilities. Find \(E[S^2]\), in squared thousands of dollars. (Answer: \(21\))
Exactly two claims occur. Each severity is uniform on \([0,3]\) thousand dollars, independently. Find \(E[S^2]\), in squared thousands of dollars. (Answer: \(10.5\))
A hypothetical count has \(E[N]=4\) and \(\operatorname{Var}(N)=5\). Independent, identically distributed severities have \(E[X]=3\) and \(E[X^2]=11\). Find \(\operatorname{Var}(S)\). (Answer: \(53\))

What You'll Learn