Reachability for Markov process, continuous time - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T21:42:48Zhttp://mathoverflow.net/feeds/question/72370http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/72370/reachability-for-markov-process-continuous-timeReachability for Markov process, continuous timeIlya2011-08-08T16:07:28Z2012-10-17T05:22:00Z
<p>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 $1$. </p>
<p>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.</p>
<p>I also asked in on <a href="http://math.stackexchange.com/questions/56326/reachability-for-markov-process-continuous-time" rel="nofollow">MSE</a>.</p>
http://mathoverflow.net/questions/72370/reachability-for-markov-process-continuous-time/72427#72427Answer by Poldavian for Reachability for Markov process, continuous timePoldavian2011-08-09T02:48:15Z2011-08-09T02:48:15Z<p>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 <a href="http://mathoverflow.net/questions/72426" rel="nofollow">http://mathoverflow.net/questions/72426</a> (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). </p>
<p>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. </p>