1
$\begingroup$

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $E$ be a normed $\mathbb R$-vector space and $(X_t)_{t\ge0}$ be an $E$-valued càdlàg Lèvy process on $(\Omega,\mathcal A,\operatorname P)$.

How can we prove that there is a (unique) transition kernel $\pi$ from $(\Omega,\mathcal A)$ to $([0,\infty)\times E,\mathcal B([0,\infty)\times E))$ with$^1$ $$\pi(\omega,[0,t]\times B\setminus\{0\})=|\{s\in[0,t]:\Delta X_s(\omega)\}\|\tag1$$ for all $\omega\in\Omega$, $t\ge0$ and $B\in\mathcal B(E)$?

Clearly, for fixed $\omega\in\Omega$, we should be able to find a unique measure $\pi_\omega$ on $([0,\infty)\times E,\mathcal B([0,\infty)\times E))$ with $$\pi_\omega([0,t]\times B\setminus\{0\})=|\{s\in[0,t]:\Delta X_s(\omega)\}\|\tag2$$ for all $t\ge0$ and $B\in\mathcal B(E)$ by the classical measure-theoretic results, since $\mathcal B([0,\infty))$ is generated by the $\cap$-stable system $\{[0,t]:t\ge0\}$. The only thing I'm unsure about is whether we need to specify the values for $\{0\}\subseteq E$, since $\mathcal B(E)$ is clearly not generated by $\{B\setminus\{0\}:B\in\mathcal B(E)\}$ to obtain the uniqueness.

However, even when the latter issue is solved, I'm not sure how we obtain the $\mathcal A$-measurability of $\Omega\ni\omega\mapsto\pi_\omega(A)$ for every $A\in\mathcal B([0,\infty)\times E)$ ...


As usual, $x(t-):=\lim_{s\to t-}$ and $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$, where $x(0-):=x(0)$.

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.