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With finitely many nonzero marks $z_j$ and independent Poisson processes $N^j$ of rates ${\lambda }_j$, the displayed process is a Lévy process with no Gaussian component. Its Lévy measure is ${\sum }_j{\lambda }_j{\delta }_{z_j}$, its mean is $t(\gamma +{\sum }_j{\lambda }_j{z}_j)$, and its variance is $t{\sum }_j{\lambda }_j{z}_j^2$. Its paths have finitely many jumps on every bounded interval and deterministic slope $\gamma$ between jumps. Coincident mark sizes combine their rates; zero marks have no effect. The drift here uses the uncompensated finite-sum convention, which differs from the drift parameter in a compensated Lévy–Khintchine formula.

A stable equilibrium M-M-s queue has a Poisson process of departures with the same rate as its input. Its present population is also independent of past departures, a component of quasireversibility. The birth-death process satisfies detailed balance; reversed upward jumps have a constant arrival rate, so they are generated by an independent Poisson process.

In stationarity, the departure process of an $M/M/\infty$ queue is a Poisson process of rate $\lambda$. By time reversibility, the departures before any fixed time are independent of the occupancy at that time.

The classical risk model has surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, where $N_t$ is a Poisson process of rate $\lambda$, independent claim sizes are positive, and premiums flow in at constant rate $c$. The relative safety loading is $\rho =c/(\lambda \mathbb{E}{X}_1)-1$. Ruin is the first time surplus becomes negative.

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.

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 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.

A flow-level model treats each document transfer as a flow whose rate depends on the current occupancy and resource capacities. With independent Poisson processes for arrivals and independent unit-mean exponential distributions for sizes, a route's total completion rate equals its allocated aggregate service.

A target galaxy with geometric encounter cross section $\pi R^2$ in a population of number density $N$ has encounter rate $\Gamma =\pi R^{2}N⟨v_{\mathrm{rel}}⟩$. A Poisson process gives merger probability $1-e^{-\Gamma T}$, approximated by $\Gamma T$ only for rare encounters. Counts of distinct pairs across the whole population additionally require a factor of one-half.

Place recovery marks at each vertex according to independent rate-$\mu$ Poisson processes and infection arrows on each oriented nearest-neighbour edge according to independent rate-$\lambda$ Poisson processes. A vertex is infected at time $t$ exactly when a time-directed path from an initially infected vertex reaches it while following arrows and avoiding recovery marks.

For the standard fixed routing loss network with independent Poisson processes, the stationary occupancy law depends on independent holding-time distributions only through their means. The occupancy process alone need not be a Markov process when those holding times lack the memoryless property.

For unit-rate Poisson process arrivals $Z_i$, $X_p$ is finite almost surely when $p>1$. There are finitely many positive arrivals before time one; the expected contribution afterwards is ${\int }_1^{\infty }{z}^{-p}dz<\infty$, by the Poisson expectation identity proved on simple functions and extended by monotone convergence. Merging independent processes of rates $1$ and $\rho$ produces rate $1+\rho$; rescaling their arrivals by $1+\rho$ restores unit rate. Hence the sum of their inverse-power contributions has law $(1+\rho {)}^p{X}_p$. In particular two independent unit-rate inverse-square sums add to a variable with law $4X_2$.

A loss network admits a call only when all its required resources have enough free capacity. Rejected calls do not queue. Under fixed routing and independent Poisson processes, its exact occupancy law has a product-form stationary distribution of a loss network.

An M-D-1 queue has a Poisson process of arrivals, a deterministic service time $\tau$, and one work-conserving server. With arrival rate $\lambda$ and $\lambda \tau <1$, its equilibrium idle probability is $1-\lambda \tau$: the stationary completed work per unit time is $\lambda \tau$. Its output generally need not be a Poisson process.

An $M/M/s/K$ queue has arrivals forming a Poisson process of rate $\lambda$, $s$ servers with independent service times having an exponential distribution of rate $\mu$, and total system capacity $K\ge s$. Its occupancy is the finite birth-death process on $\{0,\dots ,K\}$ with

An open migration process is a continuous-time Markov chain of colony populations with external Poisson processes, state-dependent departure rates, and fixed probabilistic routing that eventually exits the system. Its traffic equations of an open migration process determine the total arrival rates, and its equilibrium has a product-form stationary distribution of an open migration process when the normalizing sums are finite.