Moment Generating Functions And Cumulant Composition

Why Generating Functions Compose

An aggregate loss is a random sum: the number of payments varies, and each payment has its own random amount. A useful way to describe the aggregate distribution is to combine the frequency and severity models through their generating functions. The key is to condition on the count first. That step also makes clear why the aggregate moment generating function (MGF) may have a narrower domain than the severity MGF alone.

Let $N$ be a nonnegative integer-valued claim count, and let $X_1,X_2,\ldots$ be independent, identically distributed, nonnegative severities, independent of $N$. Define the aggregate loss by $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. Write $M_X(t)=E[e^{tX}]$ for the severity MGF when finite, and $G_N(z)=E[z^N]$ for the probability generating function of the count.

Definition: The MGF of $S$ is $M_S(t)=E[e^{tS}]$, at real values of $t$ for which this expectation is finite. Its cumulant generating function is $K_S(t)=\log M_S(t)$ on that domain. For the count, define $K_N(s)=\log E[e^{sN}]$ wherever the expectation is finite. Since $G_N(z)=E[z^N]$, the count functions satisfy $G_N(z)=E[e^{(\log z)N}]$ for $z>0$ when the relevant expectation exists.

Deriving the Compound MGF

Conditional on $N=n$, the aggregate is the sum of $n$ independent severities. Independence gives a product of their MGFs. Averaging over the possible counts then gives the compound formula.

Condition on the count.
For a fixed $n$, independence implies $E[e^{tS}\mid N=n]=M_X(t)^n$, provided the severity MGF is finite.

Average over counts.
Taking expectation over $N$ gives $M_S(t)=E[M_X(t)^N]=G_N(M_X(t))$.

Take logarithms.
Where the expressions are finite, $K_S(t)=\log G_N(M_X(t))=K_N(K_X(t))$, where $K_X(t)=\log M_X(t)$.

The formula is not merely a formal substitution. If $M_X(t)$ is finite, then $M_X(t)>0$, and the conditional expectation is $M_X(t)^n$. Its average over $N$ is finite exactly when $G_N(M_X(t))$ is finite. Equivalently, the count MGF must be finite at $s=K_X(t)$. The compound domain therefore depends on both the severity and count distributions.

Theorem: For independent, identically distributed nonnegative severities and an independent count, at any $t$ where $M_X(t)<\infty$, the aggregate MGF is finite if and only if $G_N(M_X(t))<\infty$, and then $M_S(t)=G_N(M_X(t))$. If $P(N>0)>0$, a point where $M_X(t)=\infty$ also has $M_S(t)=\infty$. If $N=0$ almost surely, however, $S=0$ almost surely and $M_S(t)=1$ for every real $t$, regardless of whether $M_X(t)$ exists.

The last case matters: it is not correct to require severity-MGF finiteness at every aggregate-MGF point without qualification. When the count is always zero, severity is never used. For nonnegative losses, $M_X(t)$ is finite for every $t\leq 0$, but positive values of $t$ may exceed its domain. Even when the severity MGF is finite, the count distribution can impose an additional restriction.

Worked Examples

Worked Example: Binomial Count with Exponential Severity

A hypothetical policy covers at most two independent events. Let $N\sim\operatorname{Binomial}(2,0.4)$. Conditional on the count, let the event severities be independent exponential random variables with rate $2$ per dollar, independent of $N$. Find the aggregate MGF and its domain, then calculate the aggregate mean and variance.

Method and count function. For $N\sim\operatorname{Binomial}(2,0.4)$, the probability generating function is $G_N(z)=(0.6+0.4z)^2$. For an exponential severity of rate $2$, $M_X(t)=2/(2-t)$ for $t<2$. The count is bounded, so its PGF is finite for every finite argument. Because $P(N>0)=1-0.6^2=0.64>0$, the severity domain supplies the aggregate restriction $t<2$.

Compose the functions. Substituting the severity MGF into the count PGF gives $M_S(t)=[0.6+0.4(2/(2-t))]^2$, for $t<2$. At $t=0$, this is $[0.6+0.4]^2=1$, as an MGF must be. For $t\geq2$, the severity MGF is infinite and a positive count has positive probability, so the aggregate MGF is infinite.

Calculate the mean and variance. The count moments are $E[N]=2(0.4)=0.8$ and $\operatorname{Var}(N)=2(0.4)(0.6)=0.48$. An exponential variable with rate $2$ has $E[X]=1/2$ and $\operatorname{Var}(X)=1/4$. Thus $E[S]=E[N]E[X]=0.8(0.5)=0.4$. For a random sum, $\operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$, so $\operatorname{Var}(S)=0.8(0.25)+0.48(0.25)=0.20+0.12=0.32$.

Check. The MGF is defined at zero and equals one. The mean is $0.4$, and the variance includes both variation in severities and variation in the count. The common mistake avoided here is to use only $E[N]\operatorname{Var}(X)$, omitting the count-variation term.

Worked Example: A Geometric Count Restricts the Domain

A hypothetical portfolio has count probabilities $P(N=n)=0.75(0.25)^n$, for $n=0,1,2,\ldots$. Every event produces a fixed payment of $\$2$, so $X=2$. Find the aggregate MGF and its domain, and calculate the aggregate mean and variance.

Method and count function. Here $M_X(t)=e^{2t}$ for all real $t$. The count PGF is a geometric series: $G_N(z)=0.75\sum_{n=0}^{\infty}(0.25z)^n=0.75/(1-0.25z)$, provided $0.25z<1$. Since $z=e^{2t}>0$, the composition is finite exactly when $e^{2t}<4$, or $t<\log(4)/2=\log 2$.

Aggregate MGF. Therefore, $M_S(t)=0.75/(1-0.25e^{2t})$ for $t<\log 2$. At $t=0$, the value is $0.75/(1-0.25)=1$. As $t$ approaches $\log 2$ from below, the denominator approaches zero, and the MGF diverges. The severity MGF exists everywhere, but that alone does not ensure the aggregate MGF exists everywhere.

Moments. The geometric count has $E[N]=0.25/0.75=1/3$ and $\operatorname{Var}(N)=0.25/(0.75)^2=4/9$. Since $S=2N$, $E[S]=2/3$ and $\operatorname{Var}(S)=4(4/9)=16/9$.

Check. The fixed severity makes $S$ exactly twice the count, agreeing with both moments. The common mistake avoided here is to assume that a severity MGF existing for every real $t$ guarantees the same for the aggregate; the count PGF has its own convergence restriction.

Worked Example: The Always-Zero Count Exception

Suppose a hypothetical contract has $N=0$ with probability one. A severity, if an event were to occur, would have an exponential distribution with rate $1$. Determine the aggregate MGF domain and explain why the severity MGF does not restrict it.

Method. For an exponential severity of rate $1$, $M_X(t)=1/(1-t)$ for $t<1$, and it is infinite for $t\geq1$. But the count is always zero, so the defining sum for $S$ is empty on every outcome and $S=0$ with probability one.

It follows directly that $M_S(t)=E[e^{t\cdot0}]=1$ for every real $t$, and $K_S(t)=\log 1=0$ for every real $t$. In particular, the aggregate MGF remains finite for $t=1$ and for $t>1$, even though $M_X(t)$ is infinite there.

Check. The aggregate distribution is a point mass at zero, so its MGF is identically one. The common mistake avoided is applying the usual severity-domain restriction when $P(N>0)=0$; in this case no severity is ever included in the aggregate.

Cumulants and Their Composition

An MGF can be difficult to manipulate directly, while its logarithm often has a simpler composition. When the relevant MGFs are finite in a neighborhood of zero, derivatives of the cumulant generating function at zero give cumulants: the first is the mean, the second is the variance, and the third is the third cumulant.

Result: On points where the composition is finite, $K_S(t)=K_N(K_X(t))$. If the functions are differentiable around zero and the required moments exist, then $E[S]=E[N]E[X]$ and $\operatorname{Var}(S)=E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2$.

The variance formula follows by differentiating the composition. The chain rule gives $K_S'(t)=K_N'(K_X(t))K_X'(t)$. At zero, $K_X(0)=0$, $K_N'(0)=E[N]$, and $K_X'(0)=E[X]$, giving the mean. Differentiating once more gives $K_S''(t)=K_N''(K_X(t))(K_X'(t))^2+K_N'(K_X(t))K_X''(t)$. At zero, $K_N''(0)=\operatorname{Var}(N)$ and $K_X''(0)=\operatorname{Var}(X)$. Therefore the second derivative is precisely the stated variance formula. This derivation requires the cumulant functions and derivatives to exist around zero; it should not be applied where the necessary moments do not exist.

For completeness, if third moments exist, one more differentiation gives $\kappa_3(S)=E[N]\kappa_3(X)+3\operatorname{Var}(N)E[X]\operatorname{Var}(X)+\kappa_3(N)(E[X])^3$. This follows by differentiating $K_N(K_X(t))$ a third time and evaluating at zero. It illustrates the same composition principle: aggregate cumulants contain contributions from both severity variability and frequency variability.

Common Domain and Composition Traps

Checking only severity. A finite $M_X(t)$ does not guarantee a finite $M_S(t)$. The count function must also be finite at $M_X(t)$, or equivalently $K_N$ must be finite at $K_X(t)$.
Checking only the count formula. Substitution into $G_N$ presumes that $M_X(t)$ is finite, except in the special case $N=0$ almost surely. If $P(N>0)>0$ and the severity MGF is infinite, the aggregate MGF is infinite for nonnegative severities.
Ignoring the zero-count case. If $N=0$ almost surely, the aggregate is always zero. Its MGF exists for all real $t$, independently of the severity model.
Using moments without existence conditions. Differentiating at zero yields aggregate moments only when the required moment-generating or cumulant-generating functions are finite around zero and the derivatives exist.

Key takeaway: For an independent random sum, condition on $N$ to obtain $M_S(t)=G_N(M_X(t))$ wherever the composition is defined and finite; then $K_S(t)=K_N(K_X(t))$. If $P(N>0)>0$, aggregate finiteness requires a finite severity MGF as well as convergence of the count function. If $N=0$ almost surely, the aggregate MGF is identically one, even where the severity MGF is infinite.

Check Your Understanding

For each problem, identify the relevant composition or domain condition before calculating.

A count is Bernoulli with success probability $0.3$, and every event pays a fixed $\$100$. Find the aggregate mean and variance. (Answer: Mean $\$30$; variance $\$2{,}100$.)
A count has probabilities $P(N=0)=0.5$, $P(N=1)=0.3$, and $P(N=2)=0.2$. Every event pays $\$3$. Find the aggregate MGF and state its domain. (Answer: $M_S(t)=0.5+0.3e^{3t}+0.2e^{6t}$, finite for every real $t$.)
A count has $P(N=n)=0.8(0.2)^n$ for $n\geq0$, and every event pays $\$5$. Find the aggregate MGF domain. (Answer: $t<\log(5)/5$.)
A count is zero with probability $0.6$ and one with probability $0.4$. Each event severity is exponential with rate $4$. Give the aggregate MGF and its domain. (Answer: $M_S(t)=0.6+0.4\cdot4/(4-t)$ for $t<4$, and it is infinite for $t\geq4$.)
A count is zero almost surely, while the hypothetical severity is exponential with rate $1$. State the aggregate MGF for all real $t$, including $t\geq1$. (Answer: $M_S(t)=1$ for every real $t$.)

What You'll Learn