Suppose $D$ is a domain in $\mathbb{R}^d$, $x\in D$. $X_t$ is a Levy process with Lévy triplet ${\displaystyle (b,0,\mu )}$ . Can one give a brief proof for: $$ \mathbb{P}_x(X_{\tau_D^-}\in \partial D)=0, $$ where $\tau_D:=\inf\{t>0: X_t\in D^c \}$.

I think the above equation holds for a large class of Pure jump Markov processes, but I can't find the relevant literature now.


1 Answer 1


This does not seem to be true: take $X_t$ to be a symmetric stable process in $\mathbf{R}$ with index $\alpha > 1$, and $D = (-1, 1) \setminus \{0\}$. Then $X_t$ hits $0$ with positive probability.

Things get even worse when one thinks about non-symmetric processes: if $X_t$ is strictly stable with one-sided jumps and with index $\alpha > 1$, and $D = (-1, 1)$, then $X_t$ hits the boundary with positive probability, even though $D$ is extremely regular.

For one-dimensional Lévy processes exiting an interval this is precisely the question whether the process creeps or not. See, for example, Kyprianou's book Introductory Lectures on Fluctuations of Lévy Processes with Applications for further information.

In higher dimensions, if $D$ is regular enough, the answer should be similar. However, I doubt this is rigorously proved anywhere.


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