first time for a diffusion process to hit a set.

-2

By definition a stopping time $\tau $ is a random variable such that $ \lbrace\tau \leq t\rbrace \in {\mathcal F}(t)$, $\forall t \geq 0$. I would like to show that $\tau :=\inf\lbrace t\geq 0| X\in E\rbrace$ , where $E$ is noempty closed subset or a noempty open set of $\mathbb R^n$, is a stoppping time. $\bf X_t$ is a diffusion process $d{\bf X}_t=b(t,{\bf X}_t)dt +\sigma(t,{\bf X}_t)d{\bf B}_t,$ ${\bf B}_t$ is a brownian motion.

flag	asked May 17 2010 at 0:20 Walter 22●3

I'm not the one that voted this down but perhaps it was because you haven't defined many of the symbols in your question. – jc May 17 2010 at 1:18

This is just the same question you(?) asked a while ago, adding the statement that X is a diffusion (and losing the comments). That's annoying. I posted a link to the page on PlanetMath answering your question. – George Lowther May 17 2010 at 1:26

If you can notice the first time that X enters E, then $\tau$ is a stopping time. – Steve Huntsman May 17 2010 at 3:47

George, sorry that i deleted the question, i thought i had badly posed the problem, i understand that it can be annoying. I did follow that link but i still dont understand the solutions. – Walter May 17 2010 at 14:05

Make sure to define everything you're talking about (say, what's F?). Also, if you're asking mathoverflow to prove something for you, you need to explain what you've tried, what you expect might or might not work, etc. – Scott Morrison♦ May 17 2010 at 15:02

show 1 more comment