Aggregate Skewness and Higher Cumulants

Why Aggregate Skewness Matters

The mean and variance describe an aggregate loss’s center and spread, but they do not describe how its probability is distributed around that center. An aggregate loss can have a substantial right tail even when its mean and variance are known. Skewness gives a standardized measure of asymmetry, while the fourth cumulant gives information about tail weight relative to a normal distribution. These measures help assess whether a symmetric description is plausible; they do not, by themselves, determine the full loss distribution.

Let \(S=\sum_{j=1}^{N}X_j\) be an aggregate loss, where \(N\) is the claim count and \(X_1,X_2,\ldots\) are claim severities. Unless stated otherwise, assume \(N\) is independent of the severities and the severities are independent and identically distributed. The formulas below concern cumulants of \(S\), not just those of an individual claim.

Definition: If a random variable \(W\) has a moment-generating function finite in a neighborhood of zero, its cumulant-generating function is \(K_W(t)=\log E[e^{tW}]\). Its \(r\)th cumulant is \(\kappa_r(W)=K_W^{(r)}(0)\). In particular, \(\kappa_1(W)=E[W]\), \(\kappa_2(W)=\operatorname{Var}(W)\), and \(\kappa_3(W)=E[(W-E[W])^3]\). The fourth cumulant is \(\kappa_4(W)=E[(W-E[W])^4]-3\operatorname{Var}(W)^2\).

Cumulants are convenient for sums: cumulant-generating functions add for independent random variables, so the \(r\)th cumulant of a sum of a fixed number of independent, identically distributed claims is that number times the severity’s \(r\)th cumulant. For a random claim count, the count’s variability also contributes. This is why aggregate cumulants are not generally just \(E[N]\) times the corresponding severity cumulants.

From Frequency and Severity to Aggregate Cumulants

Write \(m_r=E[X^r]\) for the \(r\)th raw moment of an individual severity, and write \(\kappa_{r,N}\) and \(\kappa_{r,X}\) for frequency and severity cumulants. For the first four severity cumulants, raw moments give

\begin{aligned} \kappa_{1,X}&=m_1,\\ \kappa_{2,X}&=m_2-m_1^2,\\ \kappa_{3,X}&=m_3-3m_2m_1+2m_1^3,\\ \kappa_{4,X}&=m_4-4m_3m_1-3m_2^2+12m_2m_1^2-6m_1^4. \end{aligned}

The aggregate mean and variance can also be written using severity raw moments: \(\kappa_1(S)=E[N]m_1\) and \(\kappa_2(S)=E[N]m_2+\operatorname{Var}(N)m_1^2\). For skewness and higher cumulants, it is useful to use the cumulants of both components directly.

Theorem: Under the independent-frequency and iid-severity compound-sum model, the first four aggregate cumulants are

\begin{aligned} \kappa_1(S)&=\kappa_{1,N}\kappa_{1,X},\\ \kappa_2(S)&=\kappa_{1,N}\kappa_{2,X}+\kappa_{2,N}\kappa_{1,X}^2,\\ \kappa_3(S)&=\kappa_{1,N}\kappa_{3,X} +3\kappa_{2,N}\kappa_{1,X}\kappa_{2,X} +\kappa_{3,N}\kappa_{1,X}^3,\\ \kappa_4(S)&=\kappa_{1,N}\kappa_{4,X} +4\kappa_{2,N}\kappa_{1,X}\kappa_{3,X} +3\kappa_{2,N}\kappa_{2,X}^2\\ &\quad+6\kappa_{3,N}\kappa_{1,X}^2\kappa_{2,X} +\kappa_{4,N}\kappa_{1,X}^4. \end{aligned}

These expressions require the relevant moments to exist; the cumulant-generating-function derivation applies when the generating functions are finite near zero. When using formulas based on moments alone, it is enough that the moments required by the calculation exist. If \(N\) is fixed at \(n\), its cumulants above the first are zero, and each aggregate cumulant reduces to \(n\kappa_{r,X}\).

Why the Compound-Sum Formulas Hold

Let \(K_X(t)\) and \(K_N(t)\) be the cumulant-generating functions of a severity and the claim count. Conditional on \(N\), the aggregate is a sum of \(N\) independent severities, so its conditional moment-generating function is \(M_X(t)^N\). Taking expectations over \(N\) gives

E[e^{tS}]=E[M_X(t)^N]=M_N(\log M_X(t)), \qquad K_S(t)=K_N(K_X(t)).

To see how the third-cumulant formula follows, expand \(K_X(t)\) through the cubic term as \(\kappa_{1,X}t+\kappa_{2,X}t^2/2+\kappa_{3,X}t^3/6\), and expand \(K_N(u)\) as \(\kappa_{1,N}u+\kappa_{2,N}u^2/2+\kappa_{3,N}u^3/6\). Substituting \(u=K_X(t)\), the coefficient of \(t^3/6\) is

\kappa_{1,N}\kappa_{3,X} +3\kappa_{2,N}\kappa_{1,X}\kappa_{2,X} +\kappa_{3,N}\kappa_{1,X}^3.

This is \(\kappa_3(S)\). Expanding the same composition through the fourth-order term gives the displayed expression for \(\kappa_4(S)\). The cross terms arise because the count itself is random: changing \(N\) changes the number of severities contributing to the aggregate.

Skewness and Excess Kurtosis

Definition: When \(\kappa_2(S)>0\), the aggregate skewness is \(\gamma_1(S)=\kappa_3(S)/\kappa_2(S)^{3/2}\). The aggregate excess kurtosis is \(\gamma_2(S)=\kappa_4(S)/\kappa_2(S)^2\); ordinary kurtosis is \(\gamma_2(S)+3\). Positive skewness indicates a longer or more influential right tail than left tail, while negative skewness indicates the reverse.

Skewness is dimensionless: multiplying every loss by a positive constant changes the cumulants but leaves \(\gamma_1\) unchanged. A value near zero is consistent with symmetry, but does not prove it; different nonsymmetric distributions can have zero third cumulant. The fourth cumulant can provide additional information, but it also cannot establish the distribution’s shape by itself.

A particularly useful case is Poisson frequency. If \(N\sim\operatorname{Poisson}(\lambda)\), every frequency cumulant equals \(\lambda\). Substitution into the compound-sum formulas gives a compact result for every order for which the severity moment exists:

\kappa_r(S)=\lambda E[X^r].

For example, the Poisson aggregate’s third cumulant uses the severity’s third raw moment, not its third central moment. Confusing these is a common source of error.

Worked Example: Poisson Frequency With Two Severity Amounts

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(4)\) claims. Each claim is independent of the count and has severity \(X\) equal to 1 or 3 thousand dollars, each with probability \(1/2\). Find the first four aggregate cumulants, skewness, and excess kurtosis. State what the skewness suggests about symmetry.

Method. Use the Poisson compound-sum identity with severity raw moments. The common trap is to use central severity moments in place of the raw moments required by this identity.

Step 1: Calculate severity raw moments.

\begin{aligned} m_1&=\tfrac12(1)+\tfrac12(3)=2,\\ m_2&=\tfrac12(1^2)+\tfrac12(3^2)=5,\\ m_3&=\tfrac12(1^3)+\tfrac12(3^3)=14,\\ m_4&=\tfrac12(1^4)+\tfrac12(3^4)=41. \end{aligned}

Step 2: Calculate aggregate cumulants. Since \(\lambda=4\), \(\kappa_r(S)=4m_r\). Thus \(\kappa_1(S)=8\), \(\kappa_2(S)=20\), \(\kappa_3(S)=56\), and \(\kappa_4(S)=164\), in thousand-dollar units raised to the respective cumulant order.

Step 3: Standardize.

\gamma_1(S)=\frac{56}{20^{3/2}}=\frac{56}{89.4427\ldots}=0.6261, \qquad \gamma_2(S)=\frac{164}{20^2}=\frac{164}{400}=0.4100.

The skewness is positive, so the aggregate is not well described as exactly symmetric on the basis of this diagnostic. The calculation does not determine how accurate any particular approximation would be. The check is consistent: the aggregate variance is \(20\), matching \(\lambda E[X^2]=4(5)\), and the calculated excess kurtosis uses the fourth cumulant, not the fourth central moment.

Worked Example: Random Claim Occurrence Adds Skewness

A hypothetical policy produces either no claim or one claim: \(N\sim\operatorname{Bernoulli}(0.5)\). If a claim occurs, its severity is 1 or 3 thousand dollars with equal probabilities, independently of \(N\). Calculate the aggregate’s first four cumulants and skewness. A useful trap to avoid is setting the aggregate third cumulant to zero just because the severity distribution is symmetric.

Step 1: Find the component cumulants. The severity has mean \(\kappa_{1,X}=2\), variance \(\kappa_{2,X}=1\), and third cumulant \(\kappa_{3,X}=0\). Its fourth central moment is 1, so \(\kappa_{4,X}=1-3(1)^2=-2\). For a Bernoulli variable with success probability \(p\), the first four cumulants are \(p\), \(p(1-p)\), \(p(1-p)(1-2p)\), and \(p(1-p)[1-6p(1-p)]\). At \(p=0.5\), these are \(0.5,\ 0.25,\ 0,\) and \(-0.125\).

Step 2: Apply the compound formulas.

\begin{aligned} \kappa_1(S)&=0.5(2)=1,\\ \kappa_2(S)&=0.5(1)+0.25(2^2)=0.5+1=1.5,\\ \kappa_3(S)&=0.5(0)+3(0.25)(2)(1)+0(2^3)=1.5,\\ \kappa_4(S)&=0.5(-2)+4(0.25)(2)(0)+3(0.25)(1^2)\\ &\quad+6(0)(2^2)(1)+(-0.125)(2^4)\\ &=-1+0+0.75+0-2=-2.25. \end{aligned}

Step 3: Calculate skewness and check. The variance is positive, so

\gamma_1(S)=\frac{1.5}{1.5^{3/2}}=0.8165

to four decimal places. Directly, \(S\) is 0 with probability \(0.5\), 1 with probability \(0.25\), and 3 with probability \(0.25\), in thousands. Around its mean of 1, the third central moment is \(0.5(-1)^3+0.25(0)^3+0.25(2)^3=1.5\), confirming the aggregate third cumulant. Random occurrence creates positive skewness even though claim severity itself is symmetric.

Worked Example: Fixed Count and Exponential Severity

A hypothetical contract produces exactly three independent claims. Each claim amount is exponentially distributed with mean 1 thousand dollars. Find the aggregate’s first four cumulants, skewness, and excess kurtosis. State why the frequency-variability terms do not appear.

Method. For an exponential severity with mean \(\theta\), the cumulants are \(\kappa_{r,X}=(r-1)!\theta^r\). Here \(\theta=1\), so the first four severity cumulants are \(1,1,2,6\). Because \(N=3\) is fixed, its first cumulant is 3 and all its higher cumulants are zero.

Step 1: Scale the severity cumulants by the fixed count.

\kappa_1(S)=3,\qquad \kappa_2(S)=3,\qquad \kappa_3(S)=6,\qquad \kappa_4(S)=18.

Step 2: Standardize.

\gamma_1(S)=\frac{6}{3^{3/2}}=\frac{2}{\sqrt{3}}=1.1547, \qquad \gamma_2(S)=\frac{18}{3^2}=2.0000.

The substantial positive skewness signals asymmetry in this aggregate. As a check, the aggregate is the sum of three independent exponential amounts with the same scale, so its cumulants are three times the individual cumulants. No frequency-variation contribution is present because the count cannot vary.

Common Exam Traps and Interpretation

Using a raw moment where a cumulant is required. For a general frequency distribution, use the compound formulas with frequency and severity cumulants. For Poisson frequency, the compact identity instead uses severity raw moments.
Forgetting count variability. A random \(N\) contributes terms involving \(\kappa_{2,N},\kappa_{3,N},\) and \(\kappa_{4,N}\). Only a fixed count makes those higher frequency cumulants vanish.
Using the wrong standardization. Skewness divides by variance to the power \(3/2\), not by variance. Excess kurtosis divides the fourth cumulant by variance squared.
Calling the fourth cumulant the fourth central moment. They differ by \(3\operatorname{Var}(S)^2\). Ordinary kurtosis is excess kurtosis plus 3.
Treating zero skewness as proof of symmetry. Skewness is one diagnostic, not a complete description of a distribution.

For a professional assessment, calculate the cumulants first, then interpret their standardized values. Positive aggregate skewness warns that the distribution is not symmetric, but it does not specify the tail probability at any particular threshold. A symmetric approximation should not be accepted merely because the aggregate mean and variance are available or because the third cumulant is small.

Check Your Understanding

For each problem, calculate the requested quantities and check whether the result supports a symmetric description.

A hypothetical portfolio has \(N\sim\operatorname{Poisson}(3)\), with every claim exactly 2 thousand dollars. Find \(\kappa_1(S),\kappa_2(S),\kappa_3(S)\), and \(\gamma_1(S)\). (Answer: \(6,\ 12,\ 24\); skewness \(0.5774\).)
Exactly four independent claims occur. Each claim is 0 or 2 thousand dollars with equal probability. Find the aggregate variance, third cumulant, fourth cumulant, and excess kurtosis. (Answer: \(\kappa_2(S)=4,\ \kappa_3(S)=0,\ \kappa_4(S)=-8\); excess kurtosis \(-0.5000\).)
A hypothetical policy has \(N\sim\operatorname{Bernoulli}(0.2)\), and each claim, if any, is exactly 5 thousand dollars. Calculate the aggregate variance, third cumulant, fourth cumulant, skewness, and excess kurtosis. (Answer: \(4,\ 12,\ 4\); skewness \(1.5000\); excess kurtosis \(0.2500\).)
Let \(N\sim\operatorname{Poisson}(2)\). Each hypothetical claim is 1 thousand dollars with probability \(0.75\) and 5 thousand dollars with probability \(0.25\). Find the first four aggregate cumulants and the aggregate skewness. (Answer: \(\kappa_1(S)=4,\ \kappa_2(S)=14,\ \kappa_3(S)=64,\ \kappa_4(S)=314\); skewness \(1.2218\).)
Can a calculated aggregate skewness of zero establish that the aggregate loss distribution is symmetric? Give a precise reason. (Answer: No. Zero third cumulant only means zero third central moment; a nonsymmetric distribution can also have zero skewness.)

What You'll Learn