Moments and Cumulants of Compound Poisson Processes

From a Poisson Count to a Sum of Marks

A Poisson process records how many events have occurred. In many applications, each event also carries a random amount: a claim size, a dose, a batch quantity, or a jump in a cumulative total. The resulting process is a sum of marks, and its variability reflects both the random number of events and the sizes of their marks. Transform calculations make these two sources of randomness explicit.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N_t)_{t\ge0}$ be a Poisson process of rate $\lambda>0$. Let $Y_1,Y_2,\ldots$ be independent, identically distributed real-valued random variables, independent of the entire Poisson process. These are the marks. Define the compound Poisson process $(X_t)_{t\ge0}$, with state space $S=\mathbb{R}$ and continuous time $t\ge0$, by

X_t=\sum_{k=1}^{N_t}Y_k,

where the sum is zero when $N_t=0$. Use the filtration $\mathcal{F}_t=\sigma(N_u,Y_k\mathbf{1}_{\{S_k\le u\}}:0\le u\le t,\ k\ge1)$, completed by null events, where $S_k$ is the time of the $k$th Poisson arrival. The process starts at $X_0=0$. Write $m_r=\mathbb{E}[Y_1^r]$ when the indicated moment exists.

The characteristic function is available without any moment assumption on $Y_1$. For moment and cumulant formulas, additional assumptions are needed: the mean requires $\mathbb{E}|Y_1|<\infty$, the variance and covariance require $\mathbb{E}[Y_1^2]<\infty$, and the cumulant-generating-function formula below assumes that $\mathbb{E}e^{\theta Y_1}$ is finite for $\theta$ in an open interval containing zero.

The Transform and Its Cumulants

Theorem (Compound Poisson transform and cumulants): For real $u$, the characteristic function of $X_t$ is $\mathbb{E}e^{iuX_t}=\exp\{\lambda t(\mathbb{E}e^{iuY_1}-1)\}$. If the mark moment-generating function $M_Y(\theta)=\mathbb{E}e^{\theta Y_1}$ is finite in an open interval containing zero, then the cumulant-generating function is $K_t(\theta)=\log\mathbb{E}e^{\theta X_t}=\lambda t(M_Y(\theta)-1)$. Consequently, for every integer $r\ge1$, the $r$th cumulant is $\kappa_r(X_t)=\lambda t\,m_r$.

Proof. Condition on the count $N_t=n$. Given this count, $X_t$ is the sum of $n$ independent marks, so its conditional characteristic function is $(\mathbb{E}e^{iuY_1})^n$. Since $N_t$ has Poisson probabilities $e^{-\lambda t}(\lambda t)^n/n!$, summing over $n\ge0$ gives

\mathbb{E}e^{iuX_t} =\sum_{n=0}^{\infty}e^{-\lambda t}\frac{(\lambda t)^n}{n!} (\mathbb{E}e^{iuY_1})^n =e^{-\lambda t}\exp\{\lambda t\mathbb{E}e^{iuY_1}\} =\exp\{\lambda t(\mathbb{E}e^{iuY_1}-1)\}.

Under the stated exponential-moment assumption, the same conditioning argument with $e^{\theta X_t}$ gives $\mathbb{E}e^{\theta X_t}=\exp\{\lambda t(M_Y(\theta)-1)\}$. Taking the logarithm gives $K_t(\theta)$. The cumulant of order $r$ is, by definition, the $r$th derivative of $K_t$ at zero. Differentiating yields $K_t^{(r)}(0)=\lambda t M_Y^{(r)}(0)=\lambda t\mathbb{E}[Y_1^r]$. The derivatives are justified by finiteness of the moment-generating function on a neighborhood of zero, which permits differentiation under the expectation there. This proves both transform formulas and the cumulant identity. $\square$

The first cumulant is the mean, and the second is the variance. In particular, the second cumulant involves the second raw moment of a mark, not just its variance. The same fact follows directly from conditioning on the Poisson count, requiring only a finite second moment.

Moment formulas: If $\mathbb{E}|Y_1|<\infty$, then $\mathbb{E}X_t=\lambda t\,\mathbb{E}Y_1$. If $\mathbb{E}[Y_1^2]<\infty$, then $\operatorname{Var}(X_t)=\lambda t\,\mathbb{E}[Y_1^2]$. These formulas include $t=0$, when $X_0=0$.

To verify the variance formula directly, conditional on $N_t=n$, the sum has mean $n\mathbb{E}Y_1$ and variance $n\operatorname{Var}(Y_1)$. The law of total expectation gives $\mathbb{E}X_t=\mathbb{E}N_t\,\mathbb{E}Y_1=\lambda t\,\mathbb{E}Y_1$. The law of total variance gives

\begin{aligned} \operatorname{Var}(X_t) &=\mathbb{E}[\operatorname{Var}(X_t\mid N_t)] +\operatorname{Var}(\mathbb{E}[X_t\mid N_t])\\ &=\lambda t\,\operatorname{Var}(Y_1) +\lambda t\,(\mathbb{E}Y_1)^2 =\lambda t\,\mathbb{E}[Y_1^2]. \end{aligned}

Here we used $\mathbb{E}N_t=\operatorname{Var}(N_t)=\lambda t$, the standard Poisson count formulas. The second term is the variability in the number of marks multiplied by the squared mean mark; omitting it would ignore count variability.

Covariance Across Times

For $0\le s\le t$, the increment $X_t-X_s$ is the sum of marks arriving in $(s,t]$. It is independent of $X_s$: Poisson counts on disjoint intervals are independent, and the marks attached to the two sets of arrivals are independent. Thus $X_t=X_s+(X_t-X_s)$, with the two terms independent. When the marks have finite second moment, this gives the cross-time covariance.

Proposition (Covariance across times): If $\mathbb{E}[Y_1^2]<\infty$, then for all $s,t\ge0$, $$ \operatorname{Cov}(X_s,X_t)=\lambda\min(s,t)\,\mathbb{E}[Y_1^2]. $$

Proof. First suppose $s\le t$. By independent increments, $X_t=X_s+Z$, where $Z=X_t-X_s$ is independent of $X_s$. Therefore $\operatorname{Cov}(X_s,X_t)=\operatorname{Cov}(X_s,X_s+Z)=\operatorname{Var}(X_s)+\operatorname{Cov}(X_s,Z)=\operatorname{Var}(X_s)$. The moment formula gives $\operatorname{Var}(X_s)=\lambda s\,\mathbb{E}[Y_1^2]$. If $t\le s$, interchange $s$ and $t$, using symmetry of covariance. Both cases give the displayed expression. $\square$

Worked Calculations

Worked Example: Fixed-Size Marks

Let the event rate be $\lambda=\tfrac32$, let each mark be the fixed amount $Y_k=2$, and consider $t=\tfrac43$. Then $\lambda t=\tfrac32\cdot\tfrac43=2$. The mark transform is $M_Y(\theta)=e^{2\theta}$, so

K_t(\theta)=2(e^{2\theta}-1).

Differentiating gives $K_t'(\theta)=4e^{2\theta}$, $K_t''(\theta)=8e^{2\theta}$, and $K_t'''(\theta)=16e^{2\theta}$. At zero, the mean is $4$, the variance is $8$, and the third cumulant is $16$. More generally, the $r$th cumulant here is $2\cdot 2^r$, since $m_r=2^r$. Directly, $X_t=2N_t$, and $\mathbb{E}X_t=2\cdot2=4$, $\operatorname{Var}(X_t)=4\cdot2=8$. These direct calculations agree with the transform.

Worked Example: Two Possible Random Mark Sizes

Suppose $\lambda=2$ and each mark equals $1$ or $3$, each with probability $\tfrac12$. At time $t=1$, the mark moment-generating function is $M_Y(\theta)=\tfrac12e^\theta+\tfrac12e^{3\theta}$, so $K_1(\theta)=2(M_Y(\theta)-1)$. The first three mark raw moments are

m_1=\tfrac12(1)+\tfrac12(3)=2,\qquad m_2=\tfrac12(1^2)+\tfrac12(3^2)=5,\qquad m_3=\tfrac12(1^3)+\tfrac12(3^3)=14.

Therefore the first three cumulants are $\kappa_1=2(2)=4$, $\kappa_2=2(5)=10$, and $\kappa_3=2(14)=28$. Thus $\mathbb{E}X_1=4$ and $\operatorname{Var}(X_1)=10$. As a check on the variance, $\operatorname{Var}(Y_1)=5-2^2=1$, so conditioning on the count gives $2(1)+2(2^2)=10$ as well. This calculation shows why using only the mark variance would miss the count-variability contribution.

Worked Example: Covariance at Two Times

Use rate $\lambda=\tfrac12$ and marks equal to $1$ or $3$ with probability $\tfrac12$ each. Their first two raw moments are $\mathbb{E}Y_1=\tfrac12(1)+\tfrac12(3)=2$ and $\mathbb{E}[Y_1^2]=\tfrac12(1)+\tfrac12(9)=5$. At times $s=2$ and $t=5$, the covariance formula gives

\operatorname{Cov}(X_2,X_5) =\tfrac12\min(2,5)\cdot 5=5.

Independently, $\operatorname{Var}(X_2)=\tfrac12\cdot2\cdot5=5$, and $X_5-X_2$ is independent of $X_2$. Hence $\operatorname{Cov}(X_2,X_5)=\operatorname{Cov}(X_2,X_2+(X_5-X_2))=5+0=5$, verifying the result. The means are $\mathbb{E}X_2=\tfrac12\cdot2\cdot2=2$ and $\mathbb{E}X_5=\tfrac12\cdot5\cdot2=5$; these are not the covariance, because covariance measures joint fluctuation around the means.

Common Mistake: Using the Mark Variance Alone

A frequent incorrect calculation is $\operatorname{Var}(X_t)=\lambda t\,\operatorname{Var}(Y_1)$. It accounts for variability in mark sizes conditional on the count, but not variability in the count itself. For the two-point marks in the preceding example, $\operatorname{Var}(Y_1)=1$, $\lambda=2$, and $t=1$. The incorrect calculation gives $2\cdot1=2$. The correct calculation is $\lambda t\,\mathbb{E}[Y_1^2]=2\cdot5=10$, equivalently $2\operatorname{Var}(Y_1)+2(\mathbb{E}Y_1)^2=2+8=10$. Substitution into the total-variance decomposition confirms the correction.

For deterministic marks $Y_k=c$, the mark variance is zero, but $X_t=cN_t$ is still random whenever $c\ne0$ and $t>0$. Its variance is $\lambda t c^2$, entirely due to the random count. If $c=0$, or if $t=0$, the process value is identically zero and the formula also gives zero. These limiting cases are useful checks on the assumptions and the result.

Check Your Understanding

For each question, state which mark moments are needed and verify numerical answers by substitution.

If $\lambda=3$, $t=2$, and every mark equals $4$, find the mean, variance, and third cumulant of $X_t$.
If $\lambda=1$, $t=1$, and $Y_1$ is $0$ or $2$ with equal probabilities, calculate $\mathbb{E}X_t$ and $\operatorname{Var}(X_t)$.
For finite-second-moment marks, explain why $\operatorname{Cov}(X_s,X_t)=\operatorname{Var}(X_s)$ when $s\le t$.
What assumption allows the cumulant-generating function to be differentiated near zero? Which mean and variance formulas can still hold without it?
Identify the missing source of variation in the calculation $\operatorname{Var}(X_t)=\lambda t\,\operatorname{Var}(Y_1)$.