From Arrival Counts to a Jump-Sum Process
A Poisson process describes when events occur, but not how large their effects are. A compound Poisson process adds a random mark to each arrival and sums the marks observed by time \(t\). Conditioning on the number of arrivals turns questions about the resulting distribution into calculations with independent marks.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). The Brownian motion plays no role in the construction below; it is included only to keep the ambient stochastic-calculus notation fixed. Let \(N_t\) be a Poisson process of rate \(\lambda>0\), independent of \(B_t\). Let \(Y_1,Y_2,\ldots\) be independent, identically distributed real-valued random variables, independent of \(N\) and \(B\). Write \(T_k\) for the time of the \(k\)-th arrival of \(N\). Use the filtration generated by \(B\), \(N\), and the marks revealed at their arrival times, with the usual completion and right-continuous augmentation.
Write \(\phi_Y(u)=\mathbb{E}[e^{iuY_1}]\) for the characteristic function of one mark. No moment assumption is needed for this characteristic function: \(|e^{iuY_1}|=1\), so the expectation exists for every real \(u\). For a probability distribution \(F\), let \(F^{*n}\) denote the distribution of a sum of \(n\) independent marks with law \(F\); set \(F^{*0}=\delta_0\), the point mass at zero.
Distribution and Characteristic Function
Proof. Condition on the event \(N_t=n\). On that event, \(S_t=Y_1+\cdots+Y_n\), whose distribution is \(F^{*n}\); for \(n=0\), the sum is zero and its distribution is \(F^{*0}\). Since \(\mathbb{P}(N_t=n)=e^{-\lambda t}(\lambda t)^n/n!\), summing the conditional probabilities gives the claimed mixture. For the characteristic function, independence of the marks gives \(\mathbb{E}[e^{iu(Y_1+\cdots+Y_n)}]=\phi_Y(u)^n\), including \(n=0\), for which the value is \(1\). Therefore $$ \mathbb{E}[e^{iuS_t}] =\sum_{n=0}^{\infty}e^{-\lambda t}\frac{(\lambda t)^n}{n!}\phi_Y(u)^n =e^{-\lambda t}\exp\!\big(\lambda t\phi_Y(u)\big) =\exp\!\big(\lambda t(\phi_Y(u)-1)\big). $$ The series is absolutely convergent because \(|\phi_Y(u)|\le1\). At \(t=0\), the mixture has only its \(n=0\) term, so \(S_0=0\) and the characteristic function is \(1\), as the formula also gives. \(\square\)
The mixture describes the distribution even when the marks have no finite moments. It also identifies a possible atom at zero: the event of no arrivals contributes mass \(e^{-\lambda t}\) at zero. Other configurations of marks can contribute additional mass there, so this is not necessarily the entire atom.
The process has stationary independent increments as well. To see why, consider disjoint time intervals. Their Poisson arrival counts are independent, and the marks assigned to arrivals in disjoint intervals are disjoint blocks of the independent sequence \(Y_1,Y_2,\ldots\). Conditional on the counts, the sums over those blocks are independent and each has a distribution determined only by its interval's count. Averaging over the independent counts preserves independence. The distribution of an increment depends only on the interval length, since the Poisson count does. These facts describe this process itself; they are not assumptions needed to derive the one-time mixture above.
Mean and Variance
Proof. The second-moment assumption ensures both \(m\) and \(q\) are finite. Conditional on \(N_t=n\), independence of the marks gives $$ \mathbb{E}[S_t\mid N_t=n]=nm,\qquad \mathbb{E}[S_t^2\mid N_t=n]=nq+n(n-1)m^2. $$ For the second identity, expand the square: the \(n\) diagonal terms each have expectation \(q\), and the \(n(n-1)\) ordered cross terms each have expectation \(m^2\). Put \(a=\lambda t\). The Poisson probabilities give \(\mathbb{E}[N_t]=a\) and \(\mathbb{E}[N_t(N_t-1)]=a^2\): for the first, \(\sum_{n\ge1}n e^{-a}a^n/n!=a\); for the second, \(\sum_{n\ge2}n(n-1)e^{-a}a^n/n!=a^2\). Thus \(\mathbb{E}[S_t]=am\) and \(\mathbb{E}[S_t^2]=aq+a^2m^2\). Subtracting \((\mathbb{E}[S_t])^2=a^2m^2\) yields \(\operatorname{Var}(S_t)=aq\). If \(t=0\), then \(a=0\), \(S_t=0\), and both formulas still hold. \(\square\)
The variance involves the second raw moment \(q\), not just the variance of one mark. Randomness comes both from the mark sizes and from the number of marks that arrive. The formula can also be written \(\lambda t\big(\operatorname{Var}(Y_1)+m^2\big)\), since \(q=\operatorname{Var}(Y_1)+m^2\).
Worked Example: Fixed-Size Jumps
Let the arrival rate be \(\lambda=3\) and let every mark equal \(2\), so \(Y_k=2\) with probability one. Find the distribution, characteristic function, mean, variance, and the probability that \(S_1=4\).
Here \(S_t=2N_t\). For each integer \(n\ge0\), the event \(S_t=2n\) is exactly the event \(N_t=n\), so $$ \mathbb{P}(S_t=2n)=e^{-3t}\frac{(3t)^n}{n!}. $$ The characteristic function of a mark is \(\phi_Y(u)=e^{2iu}\); substituting it into the theorem gives \(\mathbb{E}[e^{iuS_t}]=\exp(3t(e^{2iu}-1))\). The mark moments are \(m=2\) and \(q=2^2=4\). Therefore \(\mathbb{E}[S_t]=3t\cdot2=6t\) and \(\operatorname{Var}(S_t)=3t\cdot4=12t\).
At \(t=1\), \(S_1=4\) requires \(N_1=2\), so $$ \mathbb{P}(S_1=4)=e^{-3}\frac{3^2}{2!}=\frac{9}{2}e^{-3}. $$ Check: since \(S_1=2N_1\), its mean is twice the Poisson mean \(3\), giving \(6\); its variance is four times the Poisson variance \(3\), giving \(12\). Both agree with the mark-moment calculation.
Worked Example: Symmetric Random Jumps
Let \(\lambda=2\), and let each mark be \(1\) or \(-1\), each with probability \(1/2\). Compute the characteristic function, mean, and variance at time \(t\), and check the mean and variance at \(t=1\).
The mark characteristic function is $$ \phi_Y(u)=\tfrac12e^{iu}+\tfrac12e^{-iu}=\cos u. $$ Thus \(\mathbb{E}[e^{iuS_t}]=\exp(2t(\cos u-1))\). The mark moments are \(m=\tfrac12(1)+\tfrac12(-1)=0\) and \(q=\tfrac12(1^2)+\tfrac12((-1)^2)=1\). The moment theorem gives \(\mathbb{E}[S_t]=2t\cdot0=0\) and \(\operatorname{Var}(S_t)=2t\cdot1=2t\).
At \(t=1\), condition on \(N_1=n\). Symmetry of the marks gives conditional mean zero, and the conditional second moment is \(nq+n(n-1)m^2=n\). Averaging gives mean zero and second moment \(\mathbb{E}[N_1]=2\), hence variance \(2-0^2=2\). This independent conditioning check agrees with the stated formulas.
Worked Example: Exponential Random Jumps
Let \(\lambda=3\), and let each mark have the exponential distribution with rate \(2\), supported on \(y>0\). Find the characteristic function, mean, variance, and distributional density away from zero at \(t=1\).
The mark density is \(2e^{-2y}\) for \(y>0\). Direct integration gives $$ \phi_Y(u)=\int_0^\infty 2e^{-2y}e^{iuy}\,dy=\frac{2}{2-iu}. $$ Also \(m=\tfrac12\) and \(q=\tfrac12\), the first two raw moments of an exponential variable with rate \(2\). The characteristic function of \(S_t\) is \(\exp(3t(2/(2-iu)-1))\); its mean is \(3t/2\), and its variance is \(3t/2\).
At \(t=1\), conditional on \(N_1=n\ge1\), the sum of the \(n\) marks has density \(2^n x^{n-1}e^{-2x}/(n-1)!\) for \(x>0\). Weighting by \(\mathbb{P}(N_1=n)=e^{-3}3^n/n!\), the unconditional density for \(x>0\) is $$ f_{S_1}(x)=e^{-3-2x}\sum_{n=1}^{\infty}\frac{3^n2^n x^{n-1}}{n!(n-1)!}. $$ There is also an atom of size \(e^{-3}\) at zero, from \(N_1=0\); positive marks ensure that no positive number of jumps can sum to zero.
Check: the probability of at least one arrival is \(1-e^{-3}\). Each conditional gamma density integrates to \(1\), so the mixture density integrates to \(\sum_{n\ge1}e^{-3}3^n/n!=1-e^{-3}\), matching that probability. The mean and variance at \(t=1\) are both \(3\cdot\tfrac12=\tfrac32\), as given by the moment theorem.
A Common Variance Mistake
A tempting but incorrect calculation is \(\operatorname{Var}(S_t)=\lambda t\operatorname{Var}(Y_1)\). It treats the number of jumps as fixed, discarding the variability in the Poisson count. For the fixed-size example, \(\operatorname{Var}(Y_1)=0\), so that calculation would say the compound sum has zero variance. But \(S_1=2N_1\), and \(N_1\) is random, so \(\operatorname{Var}(S_1)=4\operatorname{Var}(N_1)=4\cdot3=12\).
The correct formula is \(\lambda t\mathbb{E}[Y_1^2]\). In the fixed-size example, \(\mathbb{E}[Y_1^2]=4\), giving \(3\cdot4=12\) at \(t=1\). More generally, the identity $$ \lambda t\mathbb{E}[Y_1^2] =\lambda t\operatorname{Var}(Y_1)+\lambda t(\mathbb{E}[Y_1])^2 $$ shows explicitly the additional contribution from random arrival counts. A related detail is that a positive jump-mark mean makes \(\mathbb{E}[S_t]\) positive, but it does not change the variance formula into one based only on the centered marks.
The construction is a finite random sum, not a Brownian stochastic integral. In particular, the characteristic-function calculation uses conditioning on the Poisson count and independence of the marks; it does not use an Itô correction term or an ordinary chain rule. For distributions with heavy-tailed marks, the compound sum and its characteristic function are still defined, even though the mean or variance formulas require the corresponding moment assumptions.
Check Your Understanding
In each question, assume the arrival rate is positive and the marks are independent and identically distributed, independent of the Poisson count.
- If \(\lambda=4\) and \(Y_k=3\) for every \(k\), find the characteristic function, mean, and variance of \(S_t\).
- For marks taking values \(0\) and \(2\), each with probability \(1/2\), compute \(\phi_Y(u)\), \(\mathbb{E}[Y_1]\), and \(\mathbb{E}[Y_1^2]\).
- Explain why the distribution mixture includes the \(n=0\) term, and state its probability weight at time \(t\).
- Suppose \(\mathbb{E}[Y_1]=m\) and \(\mathbb{E}[Y_1^2]=q<\infty\). Derive \(\mathbb{E}[S_t^2]\) from the mean and variance formulas.
- For symmetric marks \(Y_k\in\{-1,1\}\) with equal probabilities, identify which mark moment makes the mean zero and which gives the variance of \(S_t\).