Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
@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 –  The Bridge Aug 10 '11 at 12:07
    
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 –  ShawnD Mar 20 '12 at 16:26
1  
@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. –  Ilya Mar 22 '12 at 10:58
add comment

1 Answer

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 http://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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.