Compound Poisson process

A Compound Poisson process has the formwhere $N$ is a Poisson process and the independent identically distributed marks $Y_n$ are independent of $N$. It is a Lévy process with a finite Lévy measure.

Let $Y$ be a Compound Poisson process with nonzero jump marks and positive rate, independent of a càdlàg process $Z$. Conditional on the path of $Z$, its discontinuities form a countable set. The Poisson jump times have continuous distributions and avoid that set almost surely. At every jump of $Y$, the sum $Y+Z$ therefore has the same nonzero jump. In particular a continuous Lévy process cannot have a positive-rate independent compound Poisson component. Applying this to each truncation of its Lévy measure proves that its jump measure is zero.

For nondecreasing compound Poisson input $U$, let $W={\sup }_{t\ge 0}(U(t)-Ct)$ and $W_{\delta }={\sup }_{k\ge 0}(U(k\delta )-Ck\delta )$. The second supremum is bounded above by the first. For each $t$, the next grid point is at most $\delta$ later, its input is at least $U(t)$, and it offers at most an extra $C\delta$ service. Hence $W\le W_{\delta }+C\delta$. When the workload is finite almost surely, these inequalities give stationary-tail brackets. If its logarithmic exponential tail rate exists, adding the bounded shift $C\delta$ does not change that rate.

Let $Y_t={\sum }_{j=1}^{N_t}{X}_j$ for a rate-$\lambda$ Poisson process and independent and identically distributed nonnegative marks independent of that process, with $\varphi (q)=\mathbb{E}{e}^{-q{X}_1}$. The Laplace transform of a nonnegative random variable for the increment over $[s,t]$ is $\exp [\lambda (t-s)(\varphi (q)-1)]$. That increment is independent of the past of $Y$, so the displayed nonnegative process has mean one and satisfies $\mathbb{E}[Z_t\mid \sigma (Y_u:u\le s)]=Z_s$. No finite first moment of the marks is needed for this martingale when $q\ge 0$.

For independent Poisson processes with rates $r_{+},r_{-}\ge 0$, their difference is a Lévy process with characteristic function $\exp (t[r_{+}(e^{iu}-1)+r_{-}(e^{-iu}-1)])$. When the combined rate is positive, its jumps occur at rate $r_{+}+r_{-}$, with direction probabilities proportional to the two rates. Equal unit rates give $\exp (2t(\cos u-1))$ and a rate-two Compound Poisson process with Rademacher distribution jumps. Its paths are integer-valued, càdlàg, and of finite variation on compact intervals.

At unit jump rate, this is $X_t={\sum }_{k=1}^{N_t}{\xi }_k$, where $N$ is a rate-one Poisson process and the independent marks are uniform signs. Equivalently it is the difference of two independent rate-$1/2$ Poisson processes. It is a Lévy process with Characteristic function $\exp \{t(\cos \theta -1)\}$, mean zero, variance $t$, and Lévy characteristic exponent $1-\cos \theta$ in the negative-exponent convention. A general total jump rate rescales time.

For the unit-rate continuous-time symmetric simple random walk, the processes $n^{-\alpha }{X}_{nt}$ have three finite-dimensional regimes. For $0<\alpha <1/2$ no probability limit is possible, because the one-time characteristic functions tend to zero off the origin and are discontinuous there. At $\alpha =1/2$ the limit is standard Brownian motion. For $\alpha >1/2$ it is the identically zero Lévy process. This follows from $n(1-\cos (n^{-\alpha }\theta ))\sim ({\theta }^2/2)n^{1-2\alpha }$ and independent increments. Thus a nonzero limit uniquely determines the exponent, whereas merely asking for a Lévy limit does not.

For a strictly positive jump rate $\lambda$, a Compound Poisson process $X_t={\sum }_{j=1}^{N_t}{Y}_j$ is a martingale for its natural filtration exactly when $\mathbb{E}|Y_1|<\infty$ and $\mathbb{E}{Y}_1=0$. The marks are iid and independent of the rate-$\lambda$ Poisson process.

Necessity of mark integrability follows from the positive-probability event $N_t=1$: on that event $X_t$ has the mark law, so an infinite first absolute mark moment would make $X_t$ nonintegrable. With integrable marks, independent increments giveThe martingale condition is therefore precisely zero mark mean. For arbitrary integrable marks, $X_t-\lambda t\mathbb{E}{Y}_1$ is the compensated martingale.

Symmetric unit-sign marks give a martingale of variance $\lambda t$, whereas deterministic unit marks give $N_t$, whose compensated version is $N_t-\lambda t$. The strict-rate assumption avoids a degenerate exception: at $\lambda =0$ the process is identically zero and is a martingale regardless of a hypothetical unused mark law.

For a vector-valued Compound Poisson process, its coordinate processes are independent exactly when its Lévy measure is supported on the union of the coordinate axes. A jump can then change only one coordinate. Splitting the marked Poisson process by the axis of its mark gives independent coordinate jump processes. Conversely, independent finite-activity coordinates have no simultaneous nonzero jumps almost surely, excluding Lévy mass away from the axes.

For a positive rate and two-dimensional marks $(g_1(U),g_2(U))$, with $U$ uniform on $[0,1]$, the condition is $g_1(U)g_2(U)=0$ almost surely. If both functions are continuous, it is equivalent toA nonzero value of the continuous product would persist on an interval of positive uniform probability. Without continuity only the almost-everywhere condition is necessary.

For example, marks $((1/2-U{)}^{+},(U-1/2{)}^{+})$ give independent coordinates by thinning into the two half-intervals. Marks $(U,1-U)$ produce common jumps in both coordinates almost surely, so the coordinates are dependent even though a single scalar uniform generates the marks. The positive-rate qualification is needed: a zero-rate process has independent constant coordinates irrespective of its unused mark functions.