Marked Arrivals and Compound Poisson Processes

From Arrivals to Marked Arrivals

A Poisson process records how many events have occurred, but many applications attach a quantity to each event: a claim amount, a signed change, or a vector recording several types of impact. Summing those quantities produces a process whose path changes at arrival times. Its increments inherit an important structure from the Poisson arrivals, even when the attached quantities can be negative or multidimensional.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((N_t)_{t\ge0}\) be a Poisson process of rate \(\lambda>0\), with arrival times \(S_k=\inf\{t:N_t\ge k\}\), \(k\ge1\). Let \((Y_k)_{k\ge1}\) be iid random vectors in \(\mathbb{R}^d\), independent of the entire Poisson process. Write \(F\) for their common probability distribution. The scalar case is \(d=1\); in particular, the marks need not be nonnegative. The natural observed filtration is \(\mathcal{F}_t=\sigma(N_u,\mathbf{1}_{\{S_k\le u\}},Y_k\mathbf{1}_{\{S_k\le u\}}:0\le u\le t,\ k\ge1)\).

For deterministic times \(0\le s<t\), the increment is the sum of the marks attached to arrivals in \((s,t]\):

If there are no arrivals in the interval, this sum is zero. If marks are signed or vector-valued, an arrival need not increase every component, and several nonzero marks may cancel. These features change the path values, but not the basic increment structure.

Why the Increments Are Independent and Stationary

Proof. Take finitely many pairwise disjoint intervals \((a_\ell,b_\ell]\), \(1\le\ell\le r\), and define \(K_\ell=N_{b_\ell}-N_{a_\ell}\) and \(Z_\ell=C_{b_\ell}-C_{a_\ell}\). Independent increments of the Poisson process imply that the counts \(K_1,\ldots,K_r\) are independent, with \(K_\ell\) Poisson distributed with mean \(\lambda(b_\ell-a_\ell)\). Conditional on the count vector, \(Z_\ell\) is the sum of the \(K_\ell\) marks whose indices fall in interval \(\ell\). These index blocks are disjoint. Since the marks are iid and independent of the counts, the conditional distribution of each block sum depends only on its own count, and the block sums are conditionally independent. More explicitly, for Borel sets \(B_1,\ldots,B_r\subseteq\mathbb{R}^d\), let \(q_m(B)\) be the probability that the sum of \(m\) iid marks belongs to \(B\), with the sum for \(m=0\) equal to zero. Then

The middle equality follows from independence of the counts. This factorization proves independence of the interval increments. For one interval of length \(h\), its count has the Poisson distribution with mean \(\lambda h\). Given a count \(m\), its increment is a sum of \(m\) marks with the same common mark distribution, regardless of the interval's location. Consequently, its distribution depends on the interval only through \(h\), proving stationarity of increments. Finally, the defining empty sum gives \(C_0=0\). \(\square\)

The proof does not require the marks to be positive, scalar, or distinct. It uses two ingredients: disjoint time intervals contain independent Poisson counts, and disjoint blocks of iid marks are independent and have the same mark distribution. It does require the marks to be independent of the Poisson process; without that assumption, the marks selected by arrivals could have different distributions or be dependent across intervals.

Proof. Since the \(Y_k\) are iid, applying the same measurable function to each gives iid marks \(g(Y_k)\). They remain independent of the Poisson process because they are functions only of marks independent of that process. The preceding theorem applies to these \(\mathbb{R}^q\)-valued marks, so \(D_t\) starts at zero and has independent stationary increments. \(\square\)

A common case is recording only whether a mark meets a specified criterion: take \(g(y)=\mathbf{1}_{\{y\in A\}}\) for a Borel set \(A\). The recoded marks are zero or one, and their cumulative sum counts qualifying arrivals. More generally, a vector mark can be transformed by coordinate selection or another measurable rule without changing the independent stationary increment property.

Worked Constructions

Sample Paths and a Common Error

On every bounded time interval \([0,T]\), a Poisson process has finitely many arrivals almost surely because \(N_T\) is a finite-valued Poisson random variable. Between arrivals, \(C_t\) is constant; at each arrival, it changes by that arrival's mark. Thus its paths are right-continuous step functions with left limits. A zero mark produces no change in the observed path, and nonzero marks can cancel in the cumulative sum. Neither fact affects the theorem: its increments are still sums over disjoint blocks of iid marks.

Common pitfall: reusing the first mark. Suppose there is one arrival in each of two disjoint intervals and the marks are fair \(\pm1\). The correct increments are \(Z_1=Y_1\) and \(Z_2=Y_2\). A mistaken calculation might use \(Z_2=Y_1\) again, concluding that the increments are always equal. That would give \(\mathbb{P}(Z_1=Z_2)=1\). In fact, the second interval uses the next mark, \(Y_2\); the four pairs of signs each have probability \(\tfrac14\), so only two of the four pairs are equal and \(\mathbb{P}(Z_1=Z_2)=\tfrac12\). The correct indexing assigns each arrival its own mark and each interval its own block of marks.

It is also important to distinguish the count process from the cumulative mark process. \(N_t\) is nondecreasing, but \(C_t\) need not be: that property would require additional restrictions on the marks. Independent stationary increments alone do not imply monotone paths, and the proofs above impose no such restriction.