3
$\begingroup$

Let $X$ be a strong Markov process in the continuous time with a state space $\mathbb R^n$. Consider a reachability problem for this process, i.e. $$ v(x):=\mathsf P_x(X_t\in A\text{ for some }0\leq t<\infty) $$ for an poen set $A$. In the discrete time there is an well-known (Bellman) integral equation on $v(x)$. It is necessary condition on $v$, the actual solution is given through the supremum over all solutions of this equation over function bounded with $0$ and $1$.

I am interested if there are similar results in the continuous time. In one paper I've read that if $X$ is a diffusion process then $$ \mathcal Av(x) = 0,\text{ for }x\in A^c $$ and $v(x) = 1$ for $x\in\partial A$. Unfortunately, there were no strict conditions on the process $X$ as well as a strict proof of such a characterization.

I also asked in on MSE.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ @Gortaur : Hi I think you might have a look at a paper from Mijatovic and Pistorius "Continuously Monitored Barrier Options under Markov Processes" where if I remember well some results you might be interested in are either proved or referenced. Regards $\endgroup$
    – The Bridge
    Aug 10, 2011 at 12:07
  • $\begingroup$ Maybe this isn't what you're asking, but it seems to me that for a diffusion the fact that $\mathcal{A}v(x)=0$ doesn't really need a proof $\endgroup$
    – ShawnD
    Mar 20, 2012 at 16:26
  • 1
    $\begingroup$ @ShawnD: well, to claim that something in mathematics doesn't need it proof is rather strong, isn't it? For example, it might happen that $v(x) = 1_{\mathbb R^n\setminus \{0\}}(x)$ even for a diffusion. $\endgroup$
    – SBF
    Mar 22, 2012 at 10:58

1 Answer 1

Reset to default
1
$\begingroup$

This is probably a consequence of the Kolmogorov backward/forward equations, by noting that the function $v$ does not depend on time. See my answer to your other question for references on Kolmogorov equations https://mathoverflow.net/questions/72426 (I realize this is a bit self-promoting on MO, but I have just seen both questions and they are intimately connected, so I plan to answer both anyway).

But as you said, this is a bit folklore, and I haven't seen a rigorous treatment either - if you have found a source of rigorous proof of this, please also let me know. Intuitive it's quite clear.

Improve this answer
$\endgroup$

You must log in to answer this question.

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