Absorbed Brownian motion probability

Question

Let $X(t)$ is a Brownian motion, and $S(t)$ is an abosrbed Brownian with two boundaries at $a$ and $b$.

I have defined $S(t)$ as

\begin{align*} S(t) = \begin{cases} a, & \text{ for $\left(m(t) \leq a\right) \cap \left( M(t) < b \right)$ } \\ b, & \text{ for $\left( M(t) \geq b \right) \cap \left(m(t) > a \right)$ } \\ X(t), & \text{ otherwise }, \end{cases} \end{align*} where

\begin{align*} & m({t}) = \min_{0 \leq u \leq t} \left \{ X(u) \right \}, \\ & M({t}) = \max_{0 \leq u \leq t} \left \{ X(u) \right \}. % & \Pr \Big(X(t) > y, \min_{0 \leq u \leq t} X(u) > 0, \\ % & \max_{0 \leq u \leq t} X(u) < R | X(0) = x \Big) = B(x,y). \end{align*}

Based on the definition, I tried to determine the probability of $S(t)$ in regard to $X(t)$ as

\begin{align*} \Pr(S(t) = a) = \Pr(m(t) \leq a, M(t) < b), \\ \Pr(S(t) = b) = \Pr(m(t) > a, M(t) \geq b). \\ \end{align*}

However, I have a feeling that my definition is wrong but I could not find where the mistake is.

Could you guys please give me some comments about my problems, especially about the probability of $S(t)$ based on $X(t)$. Thank you very much.

Answer 1

Your definition of $S(t)$ is probably incorrect. Hint: What if the brownian motion hits a at time 1 and b at time 2 and is now at 0 at time 3? What is S(t) as per your definition? (It is 0 (do you see why?), but that should not be. (and can you fix it?))

Answer 2

I think this might be of help. Consider the stopping times $$\tau_{a,t}=\inf\{s \in [0,t]:X_s\leq a\}$$ $$\tau_{b,t}=\inf\{s \in [0,t]:X_s\geq b\}$$ and set $\inf\{\emptyset\}=\infty$. We have $$S_{t}=a\mathbb{I}_{\{\tau_{a,t}<\tau_{b,t}\}}+b\mathbb{I}_{\{\tau_{a,t}>\tau_{b,t}\}}+X_t\mathbb{I}_{\{\tau_{a,t}=\infty\}\cap\{\tau_{b,t}=\infty\}}$$ So $S_t$ yields $a$ if $X_t$ hit $a$ first, yields $b$ if $X_t$ hit $b$ first, and yields $X_t$ if it stayed in $(a,b)$, that is, it never hit neither $a$ nor $b$. Therefore we have $$P(S_t=a)=P(\tau_{a,t}<\tau_{b,t})$$ and similarly for $P(S_t=b)$.