3
$\begingroup$

Let

  • $E$ be a normed $\mathbb R$-vector space;
  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
  • $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
  • $(X_t)_{t\ge0}$ be an $E$-valued càdlàg $(\mathcal F_t)_{t\ge0}$-Lévy process on $(\Omega,\mathcal A,\operatorname P)$;
  • $\tau^B_0:=0$ and $$\tau^B_n:=\inf\underbrace{\left\{t>\tau^B_{n-1}:\Delta X_t\in B\right\}}_{=:\:I^B_n}$$ for $n\in\mathbb N$ and $B\in\mathcal B(E)$;
  • $$\pi_t(B):=\sum_{\substack{s\in[0,\:t]\\\Delta X_s\ne0}}1_B\left(\Delta X_s(\omega)\right)=\left|\left\{s\in(0,t]:0\ne\Delta X_s\in B\right\}\right|$$ for $B\in\mathcal B(E)$ and $t\ge0$.

Let $k\in\mathbb N$ and $B_1,\ldots,B_k\in\mathcal B(E)$ be disjoint. How can we show that $\left(\pi_1(B_1),\ldots,\pi_k(B_k)\right)$ is independent? (Maybe we need to assume that $0\not\in\overline{B_i}$ for all $i\in\{1,\ldots,k\}$.)

Since $B_1,\ldots,B_k\in\mathcal B(E)$ are disjoint, it clearly holds $$I^{B_i}_m\cap I^{B_j}_n=\emptyset\tag1$$ for all $i,j\in\{1,\ldots,k\}$ and $m,n\in\mathbb N$. But how do we need to proceed?

I don't know if it is useful, but we may note the following: If $B\in\mathcal B(E)$ with $0\not\in\mathcal B(E)$, then $\pi_t(B)<\infty$ and $$\pi_t(B)=\sum_{n\in\mathbb N}1_{[0,\:t]}\left(\tau^B_n\right)\tag1$$ (since $\tau^B_n\xrightarrow{n\to\infty}\infty$, the sum on the right-hand side is finite) for all $t\ge0$. Moreover, we can show that $\tau^B_1$ is exponentially distributed and if we define $\tilde\Omega:=\left\{\tau^B_1<\infty\right\}$, $\tilde{\operatorname P}[A]:=\operatorname P\left[A\mid\tilde\Omega\right]$ for $A\in\tilde{\mathcal A}:=\left.\mathcal A\right|_{\tilde\Omega}$ and $\tilde\tau^B_n:=\left.\tau^B_n\right|_{\tilde\Omega}$ for $n\in\mathbb N$, then $\left(\tilde\tau^B_n-\tilde\tau^B_{n-1}\right)_{n\in\mathbb N}$ is independent and identically distributed on $\left(\tilde\Omega,\tilde{\mathcal A},\tilde{\operatorname P}\right)$.

$\endgroup$
1
  • $\begingroup$ What is $\Delta X_t$? $\endgroup$
    – 温泽海
    Commented Apr 1, 2022 at 12:12

1 Answer 1

1
$\begingroup$

The jumps of a L'evy process are either countable and dense, or form a Poisson point process. The assumptions of the OP imply we are in the latter case, so the required independence is a basic property of Poisson processes. See section 21 in the book
Ken-Iti, S., 1999. Lévy processes and infinitely divisible distributions. Cambridge university press.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .