Jumps of Lévy process Let $X:=(X_t)_{t\geq0}$ be a Lévy process with triple $(b,A,\nu)$. Is there any known relation between the "distribution" of its jumps and  the Lévy measure $\nu$? E.g. can we express something like $\mathbb{P}[X$ has $n$ jumps in $[0,1]]$ or $\mathbb{P}[X$ has a jump of absolute value $>u$ in $[0,1]]$ for some $u>0$ in terms of $\nu$?
 A: Sample path properties are discussed e.g. in Sato's Lévy processes and infinitely divisible distributions, Section 21. For example, the following results are given there:


*

*Sample functions of $X$ are a.s. continuous if and only if $\nu=0$.

*Sample functions of $X$ are a.s. piecewise constant if and only if $X$ is compound Poisson or a zero process.

*If $\nu(\mathbb{R}^d)=\infty$, then a.s. jumping times are countable and dense in $[0,\infty)$; if $0<\nu(\mathbb{R}^d)<\infty$, then a.s. jumping times are countable in increasing order and the first jumping time has an exponential distribution with mean $1/\nu(\mathbb{R}^d)$. In this latter case, the process $\{J(t)\}$ of jumps in $[0,t)$ is a Poisson process with intensity measure $\nu(\mathbb{R}^d)$, so the number of jumps in $[0,t)$ has a Poisson distribution with mean $t\nu(\mathbb{R}^d)$.

*$T_u$, the first time the process jumps by more than $u$, has an exponential distribution with mean $1/c$ if $\int_{D(u,\infty)}\nu(dx)=c<\infty$, where $D(u,\infty)=\{x\in\mathbb{R}^d: u<||x||<\infty\}$.

A: Yes, there is a very strong relation between the (distribution of the) jumps of a Lévy process and its Lévy measure. In fact, the Lévy measure describes the jump behaviour of the corresponding Lévy process:
Define the jump counting measure
$$N([0,t] \times B) := |\{0 \leq s \leq t; \Delta X_s \in B\}| \tag{1}$$
where $\Delta X_s := X_s-X_{s-}$ denotes the jump height at time $s$. So, basically, $N([0,t] \times B)(\omega)$ gives the number of jumps of height $\in B$ during the time interval $[0,t]$ of the sample path $s \mapsto X_s(\omega)$. For a fixed Borel set $B$ such that $0 \notin \overline{B}$, set
$$N_t := N([0,t] \times B).$$
Then one can show that $(N_t)_{t \geq 0}$ is again a Lévy process; more precisely a Poisson process, and
$$\mathbb{E}(N_t) = t \cdot \nu(B). \tag{2}$$
This means that $\nu$ characterizes the jump behavior of the process $(X_t)_{t \geq 0}$. Some important consequences:


*

*Whenever $\nu(B)=0$, then the process $(X_t)_{t \geq 0}$ has almost surely no jumps of height $B$. For example for the Poisson process we have $\nu = \delta_1$; hence, $\nu(B) = 0$ whenever $1 \notin B$. Consequently, by the above considerations, the Poisson process can only have jumps of size $1$.

*The same argumentation shows that a Lévy process with Lévy measure $\nu=0$ does not have any jumps.

*If $B$ is such that $\nu(B)<\infty$, then in any finite time interval $[0,T]$ we have only finitely many jumps of size $B$.

*We have $$\mathbb{P}(N_t = n) = \exp(-t \nu(B)) \cdot \frac{(t \nu(B))^n}{n!};$$ note that the left-hand side equals the probability that $(X_s)_{s \geq 0}$ has $n$ jumps of size $B$ during the time interval $[0,t]$.


In fact, one can show that $(1)$ defines a so-called Poisson random measure and define stochastic integrals with respect to such (random) measures. This leads finally to the Lévy-Itô decomposition.
