I have two questions.

Let $(X_t)_{t\geq 0}$ be a Lévy process with Lévy measure $\nu$. The jump process $\Delta X=\left(\Delta X_t\right)_{t\geq 0}$ is defined by

$\Delta X_t=X_t-X_{t-}$, for every $t\geq0$, with $X_{t-}$ left limit in $t$.

For every $0\leq t <\infty$ e $A \in \mathcal{B}(\mathcal{R}-\{0\})$, let

$N(t,A)(w)= \#\left\{ 0 \leq s \leq t: \Delta X_t(w) \in A \right\}$ if $w \in \Omega_0$


$N(t,A)(w)=0$ if $w \in \Omega_0^c$, with $\Omega_0$ a measurable set with probability $1$ such that $t\longrightarrow X_t(w)$ is cadlag for every $w \in \Omega_0$.

Let $\nu(\cdot)=\mathbb{E}\left[N(1,\cdot)\right]$ be the intensity measure. We say that $A$ is bounded below if $0 \notin \overline{A}$.


1.If $A$ is bounded below, then $\left(N(t,A)\right)_{t\geq 0}$ is a Poisson process of intensity $\nu(A)$.

2.If $A_1, \dots , A_m \in \mathcal{B}(\mathcal{R}-\{0\})$ are disjoint e bounded below then the random variable $N(t,A_1) \dots N(t,A_m)$ are indpendent.

So my questions are:

i) How can I prove the second statement?

ii) How can I deduce from this theorem that $\nu(A)<\infty$ for all $A$ bounded below?


1 Answer 1


Your questions are covered by, say, Theorem 3 in Section 4 of Lalley's notes. The proof (given there) is somewhat involved.

  • $\begingroup$ Thank you very much! $\endgroup$
    – Joegin
    Commented Mar 18, 2023 at 10:37

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