Compound Poisson Moments and Cumulants

Why Jump Moments Matter

A compound Poisson sum adds a random number of independent jumps. Its expected value and variance depend not only on how many jumps occur on average, but also on the sizes of the jumps. For higher-order cumulants, an important distinction appears: the cumulants of the compound sum are determined by the raw moments of one jump, not by that jump's cumulants.

This distinction is easy to miss because the cumulant generating function of a compound Poisson sum has a particularly simple form. The key is to identify exactly which quantity from the jump law appears in that form and to impose conditions that ensure the moments and cumulants being calculated are finite.

Setup and Notation

Definition: Let $N\sim\operatorname{Poisson}(\lambda)$, where $\lambda\geq 0$, and let $X_1,X_2,\ldots$ be independent and identically distributed random variables, independent of $N$. The compound Poisson sum is $$ S=\sum_{j=1}^{N}X_j, $$ with $S=0$ when $N=0$. The $X_j$ are the jumps. When the relevant moments exist, write $m_r=E[X_1^r]$ for the $r$th raw jump moment.

Jumps may be positive, negative, or zero. In particular, $S$ need not be a count. The case $\lambda=0$ gives $N=0$ almost surely and $S=0$ almost surely; the formulas below include this case.

For a random variable $Y$ with $E|Y|^r<\infty$, its $r$th cumulant can be defined from its characteristic function $\phi_Y(u)=E[e^{iuY}]$ by $$ \kappa_r(Y)=\frac{1}{i^r}\left.\frac{d^r}{du^r}\log\phi_Y(u)\right|_{u=0}. $$ The logarithm is taken on a neighborhood of zero where a continuous branch exists and equals zero at $u=0$. Such a neighborhood exists because $\phi_Y(0)=1$ and $\phi_Y$ is continuous. The first cumulant is the mean, and the second cumulant is the variance. When a moment generating function exists near zero, the same cumulants are the derivatives at zero of its logarithm, the cumulant generating function.

The raw moment $m_r=E[X_1^r]$ is not, in general, the $r$th cumulant $\kappa_r(X_1)$. For example, $m_2=\operatorname{Var}(X_1)+(E[X_1])^2$, whereas $\kappa_2(X_1)=\operatorname{Var}(X_1)$.

Mean and Variance by Conditioning

Theorem: If $E[X_1^2]<\infty$, then $$ E[S]=\lambda E[X_1]=\lambda m_1, \qquad \operatorname{Var}(S)=\lambda E[X_1^2]=\lambda m_2. $$

To derive these formulas, condition on the event $N=n$. A sum of $n$ independent jumps has conditional mean $nm_1$ and conditional variance $n(m_2-m_1^2)$. Using $E[N]=\operatorname{Var}(N)=\lambda$, the laws of total expectation and total variance give

E[S]=E[E[S\mid N]]=E[Nm_1]=\lambda m_1,

\begin{aligned} \operatorname{Var}(S) &=E[\operatorname{Var}(S\mid N)] +\operatorname{Var}(E[S\mid N])\\ &=E[N(m_2-m_1^2)]+\operatorname{Var}(Nm_1)\\ &=\lambda(m_2-m_1^2)+\lambda m_1^2\\ &=\lambda m_2. \end{aligned}

The variance formula includes the variation in the number of jumps as well as the variation in jump sizes. That is why it contains $E[X_1^2]$, not just $\operatorname{Var}(X_1)$.

Higher Cumulants from the Characteristic Function

Theorem: Let $r\geq 1$ be an integer and suppose $E|X_1|^r<\infty$. Then $S$ has finite absolute moments through order $r$, and for every $1\leq k\leq r$, $$ \kappa_k(S)=\lambda E[X_1^k]=\lambda m_k. $$

Here $m_k$ is the raw $k$th moment of a single jump. The result holds whether or not the jump moment generating function exists near zero. To see that the cumulants are well-defined, first note that for $n\geq1$ and $r\geq1$, $$ \left|\sum_{j=1}^n X_j\right|^r \leq n^{r-1}\sum_{j=1}^n |X_j|^r. $$ Thus $E[|S|^r\mid N=n]\leq n^r E|X_1|^r$; for $n=0$, the conditional moment is zero. Since a Poisson random variable has finite moments of every positive integer order, averaging over $N$ gives $E|S|^r<\infty$.

Now let $\phi_{X_1}$ denote the characteristic function of one jump. Conditional on $N=n$, independence gives the characteristic function $\phi_{X_1}(u)^n$ for the sum. Averaging over the Poisson probabilities,

\begin{aligned} \phi_S(u) &=\sum_{n=0}^{\infty}e^{-\lambda}\frac{\lambda^n}{n!} \phi_{X_1}(u)^n\\ &=\exp\{\lambda(\phi_{X_1}(u)-1)\}. \end{aligned}

Near zero, the logarithm branch that is zero at zero therefore satisfies $\log\phi_S(u)=\lambda(\phi_{X_1}(u)-1)$. The finite absolute $r$th moment permits differentiation of $\phi_{X_1}$ through order $r$: for each $k\leq r$, differentiation under the expectation is justified by the integrable bound $|X_1|^k\leq 1+|X_1|^r$. Consequently,

\left.\frac{d^k}{du^k}\phi_{X_1}(u)\right|_{u=0} =i^k E[X_1^k].

Differentiating the logarithm of $\phi_S$ $k$ times and applying the definition of cumulants now yields $$ \kappa_k(S)=\frac{\lambda}{i^k} \left.\frac{d^k}{du^k}\phi_{X_1}(u)\right|_{u=0} =\lambda E[X_1^k]. $$ This proves the result. $\square$

If the moment generating function $M_{X_1}(t)=E[e^{tX_1}]$ is finite on an open interval containing zero, conditioning on $N$ also gives $M_S(t)=\exp\{\lambda(M_{X_1}(t)-1)\}$ there. The logarithm of this expression is $\lambda(M_{X_1}(t)-1)$, whose $k$th derivative at zero is $\lambda E[X_1^k]$. The characteristic-function argument shows why the cumulant formula does not require this stronger moment-generating-function condition.

Worked Calculations

Worked Example: Constant Jumps

Suppose $N\sim\operatorname{Poisson}(3)$, and every jump has size $2$, so $X_j=2$ almost surely. Then $m_r=2^r$. The compound sum is $S=2N$, and its first three cumulants are

\kappa_1(S)=3(2)=6,\qquad \kappa_2(S)=3(2^2)=12,\qquad \kappa_3(S)=3(2^3)=24.

Thus $E[S]=6$, $\operatorname{Var}(S)=12$, and the third cumulant is $24$. The jump itself has $\kappa_1(X_1)=2$ and $\kappa_2(X_1)=\kappa_3(X_1)=0$, because it is constant. Multiplying those jump cumulants by $3$ would not give the cumulants of $S$. The compound-sum formula uses the jump raw moments instead.

Worked Example: Exponential Jumps

Let $N\sim\operatorname{Poisson}(4)$, and let the jumps be independent exponential random variables with rate $2$. For an exponential variable with rate $\beta$, $E[X_1^r]=r!/\beta^r$. Here this gives $$ m_1=\frac12,\qquad m_2=\frac{2}{2^2}=\frac12,\qquad m_3=\frac{6}{2^3}=\frac34. $$ Therefore the compound sum has $$ E[S]=4\left(\frac12\right)=2,\qquad \operatorname{Var}(S)=4\left(\frac12\right)=2,\qquad \kappa_3(S)=4\left(\frac34\right)=3. $$

For comparison, the first three cumulants of one exponential jump are $1/2,1/4,1/4$. Multiplying them by $4$ would give $2,1,1$, which is correct for the mean but not for the variance or third cumulant of the compound sum. In particular, $m_2=1/2$ includes both the jump variance $1/4$ and the square of its mean $1/4$.

Worked Example: Normal Jumps with Zero Third Jump Cumulant

Let $N\sim\operatorname{Poisson}(2)$, and let each jump have distribution $N(1,4)$, with variance $4$. Write $X_1=1+Z$, where $E[Z]=0$, $E[Z^2]=4$, and $E[Z^3]=0$. Then $$ m_1=1,\qquad m_2=E[(1+Z)^2]=1+4=5, $$ and $$ m_3=E[(1+Z)^3] =1+3E[Z]+3E[Z^2]+E[Z^3] =1+12=13. $$ It follows that $$ E[S]=2(1)=2,\qquad \operatorname{Var}(S)=2(5)=10,\qquad \kappa_3(S)=2(13)=26. $$

A normal jump has third cumulant zero, but the compound sum has third cumulant $26$. There is no contradiction: the compound-sum third cumulant depends on $E[X_1^3]$, which is $13$, rather than on the third cumulant of one jump.

Common Mistake and Practical Use

A common incorrect argument is: “Cumulants add for independent sums, so the $r$th cumulant of a compound Poisson sum is the Poisson mean times the $r$th cumulant of one jump.” That reasoning treats the random number of jumps as if it were fixed. Cumulants do add when the number of independent summands is fixed, but here the number of summands is itself random. The conditioning calculation or characteristic-function derivation accounts for that randomness and gives $\kappa_r(S)=\lambda E[X_1^r]$.

The distinction matters in applications that model event counts and event sizes together: for example, total claim amounts, aggregate medical costs, or accumulated rainfall from random storms. The jump law describes the magnitude of an individual event, while $\lambda$ controls the event rate. The compound cumulants combine both features, and the formulas can be used to check a proposed model or calculate the variability and asymmetry of its total.

Key takeaway: When the required jump moments are finite, the $r$th cumulant of a compound Poisson sum is the Poisson rate times the $r$th raw moment of one jump. In particular, the variance is $\lambda E[X_1^2]$, not $\lambda\operatorname{Var}(X_1)$.

Check Your Understanding

Use the formulas and conditions developed above to answer each question.

If $N\sim\operatorname{Poisson}(5)$ and every jump equals $3$, calculate the mean, variance, and third cumulant of the compound sum.
Why does the variance formula involve the second raw jump moment rather than only the variance of one jump?
For jumps with a finite third absolute moment, state the formula for the compound sum's third cumulant and identify the jump quantity it uses.
Explain why zero third cumulant for the jump distribution does not necessarily imply zero third cumulant for the compound sum.
What condition on the jump distribution is sufficient to calculate the compound sum's cumulants through order four using the characteristic-function argument?