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 $W\le W_{\delta }+C\delta$. This supplies explicit stationary-tail brackets and shows that time discretization preserves the workload's exponential tail 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.

A compound Poisson process with rate $\lambda$ and mark $Y$ is a martingale exactly when $\mathbb{E}|Y|<\infty$ and $\mathbb{E}Y=0$.

The coordinates of a vector-valued Compound Poisson process are independent exactly when its Lévy measure is supported on the union of the coordinate axes. For marks $(g_1(U),g_2(U))$ with $U$ uniform on $[0,1]$ and continuous $g_i$, this is equivalent tofor every $y\in [0,1]$.