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.

$\endgroup$
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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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