Third Moments And Cumulants Of Aggregate Loss

Why Third Cumulants Matter

The mean and variance of an aggregate loss describe its center and spread, but they do not describe how its tail is balanced around its center. The third cumulant captures that asymmetry: a positive value generally indicates a longer or heavier upper side, while a negative value generally indicates a longer or heavier lower side. For aggregate losses, the third cumulant reflects both claim-size variation and claim-count variation.

Let $N$ be the number of claims, and let $X_1,X_2,\ldots$ be the individual claim amounts. The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. The results below assume that $N$ is independent of the claim amounts, that the claim amounts are independent and identically distributed, and that the relevant third moments are finite. No interest or mortality assumptions are involved: the examples concern undiscounted aggregate losses over a fixed period.

Definition: For a random variable $Z$ with a finite third moment, its first three cumulants are $\kappa_1(Z)=E[Z]$, $\kappa_2(Z)=\operatorname{Var}(Z)$, and $\kappa_3(Z)=E[(Z-E[Z])^3]$. Write $\mu=E[X]$, $\sigma^2=\operatorname{Var}(X)$, and $\tau_3=E[(X-\mu)^3]$ for a single claim amount $X$. Its raw moments are $m_r=E[X^r]$; in particular, $\tau_3=m_3-3m_1m_2+2m_1^3$.

The third cumulant is the third central moment. This differs from the raw third moment $E[Z^3]$: the raw moment is measured about zero, whereas the third cumulant is measured about the mean. That distinction is central when combining frequency and severity information.

The Random-Sum Cumulant Formula

Write $\kappa_r(N)$ for the $r$th cumulant of the claim count. Under the assumptions above, the aggregate’s first three cumulants are:

\begin{aligned} \kappa_1(S)&=\kappa_1(N)\mu,\\ \kappa_2(S)&=\kappa_1(N)\sigma^2+\kappa_2(N)\mu^2,\\ \kappa_3(S)&=\kappa_1(N)\tau_3 +3\kappa_2(N)\mu\sigma^2 +\kappa_3(N)\mu^3. \end{aligned}

The first two expressions give the familiar compound-aggregate mean and variance. The third expression separates three sources of the aggregate’s third cumulant: severity asymmetry, interaction between frequency variation and severity variation, and the frequency’s own third cumulant.

Why the Third-Cumulant Formula Holds

The formula follows by conditioning on $N$ and expressing aggregate loss as a centered sum plus a count-related shift. Given $N$, define $A=S-N\mu$ and $B=\mu(N-E[N])$. Then $S-E[S]=A+B$. Conditional on $N$, $A$ is a sum of $N$ independent centered severities, so $E[A\mid N]=0$, $E[A^2\mid N]=N\sigma^2$, and $E[A^3\mid N]=N\tau_3$.

Expand the cube and take conditional expectations. The terms containing $A$ to the first power have conditional expectation zero. Therefore,

\begin{aligned} E[(S-E[S])^3] &=E[A^3]+3E[BA^2]+E[B^3]\\ &=E[N]\tau_3 +3\mu\sigma^2 E[N(N-E[N])] +\mu^3 E[(N-E[N])^3]. \end{aligned}

Since $E[N(N-E[N])]=\operatorname{Var}(N)=\kappa_2(N)$, $E[N]=\kappa_1(N)$, and $E[(N-E[N])^3]=\kappa_3(N)$, this is exactly the stated formula for $\kappa_3(S)$. The derivation also shows why the middle term is present: the conditional variance of aggregate loss grows with $N$, and so it interacts with the count’s deviation from its mean.

Definition: If $S$ has a finite positive variance, its skewness is $\gamma_S=\kappa_3(S)/\kappa_2(S)^{3/2}$. Skewness is dimensionless. It is undefined when the aggregate variance is zero.

For a Poisson count with parameter $\lambda$, $\kappa_1(N)=\kappa_2(N)=\kappa_3(N)=\lambda$. Substituting into the general formula gives $\kappa_3(S)=\lambda(\tau_3+3\mu\sigma^2+\mu^3)=\lambda E[X^3]$. This convenient result uses the raw third moment of severity. It is a Poisson special case, not the general formula for an arbitrary frequency model.

Worked Example: Poisson Frequency and Two-Point Severity

A hypothetical portfolio has a Poisson claim count with parameter $\lambda=2$ per period. Each claim is independently either $1,000 or $3,000, each with probability $0.5$. Calculate the aggregate mean, variance, third cumulant, and skewness. Use thousands of dollars for the calculations. Interest: none; losses are undiscounted. Mortality: not modeled.

Method and severity quantities. Let $X$ denote a claim amount in thousands of dollars. Its mean and raw moments are $m_1=0.5(1)+0.5(3)=2$, $m_2=0.5(1^2)+0.5(3^2)=5$, and $m_3=0.5(1^3)+0.5(3^3)=14$. Thus $\mu=2$, $\sigma^2=m_2-\mu^2=5-4=1$, and $\tau_3=m_3-3\mu m_2+2\mu^3=14-30+16=0$.

For $N\sim\operatorname{Poisson}(2)$, all three frequency cumulants equal 2. The aggregate mean is $2(2)=4$ thousand dollars. Its variance is $2(1)+2(2^2)=2+8=10$ thousand dollars squared. Its third cumulant is $2(0)+3(2)(2)(1)+2(2^3)=0+12+16=28$ thousand dollars cubed.

The skewness is $28/10^{3/2}=28/31.6228\ldots=0.8854$, rounded to four decimals. The positive skewness indicates asymmetry toward larger aggregate losses. As a check, the Poisson shortcut gives $\lambda m_3=2(14)=28$, agreeing with the general formula.

Worked Example: Binomial Frequency and the Importance of Count Variation

For a hypothetical coverage, the claim count is $N\sim\operatorname{Binomial}(3,0.4)$. Each claim is independently either $1,000 or $2,000, with equal probability. Find the aggregate variance, third cumulant, and skewness. Use thousands of dollars. Interest: none; losses are undiscounted. Mortality: not modeled.

Method and frequency cumulants. For a binomial count with parameters $n,p$, the first three cumulants are $np$, $np(1-p)$, and $np(1-p)(1-2p)$. Here they are $1.2$, $3(0.4)(0.6)=0.72$, and $3(0.4)(0.6)(0.2)=0.144$, respectively.

For the severity, $\mu=0.5(1)+0.5(2)=1.5$, $m_2=0.5(1)+0.5(4)=2.5$, and $m_3=0.5(1)+0.5(8)=4.5$. Hence $\sigma^2=2.5-1.5^2=0.25$, and $\tau_3=4.5-3(1.5)(2.5)+2(1.5^3)=0$.

The aggregate variance is $1.2(0.25)+0.72(1.5^2)=0.3+1.62=1.92$ thousand dollars squared. Its third cumulant is $1.2(0)+3(0.72)(1.5)(0.25)+(0.144)(1.5^3)=0+0.81+0.486=1.296$ thousand dollars cubed.

Therefore the skewness is $1.296/1.92^{3/2}=1.296/2.6604\ldots=0.4871$, rounded to four decimals. The severity itself has zero third cumulant, but the aggregate does not: count variation creates both nonzero terms. The common incorrect calculation $\kappa_3(S)=E[N]\tau_3=0$ omits those terms. The correction is to include the frequency-variance and frequency-third-cumulant contributions.

Worked Example: Negative Binomial Frequency with Asymmetric Severity

A hypothetical portfolio has a negative binomial claim count $N$, defined as the number of failures before $r=2$ successes when the success probability is $p=0.5$. Each claim amount, in thousands of dollars, is 1 with probability 0.5, 2 with probability 0.25, and 4 with probability 0.25. Calculate the aggregate mean, variance, third cumulant, and skewness. Interest: none; losses are undiscounted. Mortality: not modeled.

Method and count cumulants. For this negative binomial parameterization, with $q=1-p$, the first three cumulants are $rq/p$, $rq/p^2$, and $rq(1+q)/p^3$. Thus $\kappa_1(N)=2$, $\kappa_2(N)=4$, and $\kappa_3(N)=12$.

The severity moments are $m_1=0.5(1)+0.25(2)+0.25(4)=2$, $m_2=0.5(1)+0.25(4)+0.25(16)=5.5$, and $m_3=0.5(1)+0.25(8)+0.25(64)=18.5$. Therefore $\mu=2$, $\sigma^2=5.5-2^2=1.5$, and $\tau_3=18.5-3(2)(5.5)+2(2^3)=1.5$.

The aggregate mean is $2(2)=4$ thousand dollars. Its variance is $2(1.5)+4(2^2)=3+16=19$ thousand dollars squared. Its third cumulant is $2(1.5)+3(4)(2)(1.5)+12(2^3)=3+36+96=135$ thousand dollars cubed.

The skewness is $135/19^{3/2}=135/82.8191\ldots=1.6301$, rounded to four decimals. All three contributions to the third cumulant are positive here, so the aggregate has a pronounced positive skew. As a check on the severity calculation, $E[(X-2)^3]=0.5(-1)^3+0.25(0)^3+0.25(2)^3=1.5$, matching the cumulant obtained from the raw moments.

Interpreting the Result and Avoiding Traps

The formula shows why aggregate skewness cannot usually be inferred from severity skewness alone. Even symmetric severity can produce an asymmetric aggregate if the frequency distribution contributes nonzero terms. Conversely, the sign and size of the aggregate third cumulant depend on the combined terms; they are not determined solely by whether claim amounts are positive.

Do not substitute raw moments into the general formula. The first term uses the severity’s third cumulant $\tau_3$, not its raw third moment $m_3$. The raw-moment shortcut $\lambda m_3$ applies when the count is Poisson.
Do not omit count variation. If $\kappa_2(N)$ or $\kappa_3(N)$ is nonzero, the corresponding terms may contribute even when $\tau_3=0$.
Keep units consistent. If claim amounts are expressed in thousands of dollars, the mean, variance, and third cumulant have units of thousands, thousands squared, and thousands cubed. Skewness has no units.
Use the correct definition of frequency cumulants. In particular, $\kappa_3(N)$ is the third central moment, not $E[N^3]$.

These calculations are useful when comparing aggregate models or interpreting why one aggregate distribution may have a heavier upper side than another. They summarize shape; they do not, by themselves, determine tail probabilities or replace a full aggregate distribution calculation.

Key takeaway: For an independent random sum, aggregate skewness is obtained by first combining frequency and severity through $\kappa_3(S)=\kappa_1(N)\tau_3+3\kappa_2(N)\mu\sigma^2+\kappa_3(N)\mu^3$, then dividing by $\kappa_2(S)^{3/2}$. The formula requires finite third moments and the stated independence assumptions.

Check Your Understanding

For each problem, calculate the requested quantity using the stated frequency model and the random-sum formula. Amounts are in thousands of dollars unless stated otherwise.

A hypothetical aggregate has $N\sim\operatorname{Poisson}(3)$, and every claim is exactly 2. Find the aggregate third cumulant and skewness. (Answer: $\kappa_3(S)=24$, skewness $=0.5774$.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.25)$. Claim amounts are 1 and 3 with equal probability. Find the aggregate mean, variance, and third cumulant. (Answer: mean $=2$, variance $=4$, $\kappa_3(S)=7.5$.)
Let $N$ be the number of failures before one success, with success probability $0.5$. Claim amounts are 1 and 2 with equal probability. Find the aggregate variance, third cumulant, and skewness. (Answer: variance $=4.75$, $\kappa_3(S)=22.5$, skewness $=2.1734$.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(5,0.4)$. Its claim severity has mean 2, variance 3, and third central moment 1. Find the aggregate mean, variance, third cumulant, and skewness. (Answer: mean $=4$, variance $=10.8$, $\kappa_3(S)=25.52$, skewness $=0.7190$.)

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