I was self-studying a book on stochastic calculus and got some difficulties dealing with the following question.
Suppose $W_t$ is a Brownian motion path and $T$ is a random hitting time. The stopped process is: $$X_t= \begin{cases} W_t & t < T \\ W_T & t \geq T \end{cases} $$ I have shown that $X_t$ is a martingale.
The question is: Suppose $W_0 = 0$, and $x_l<0<x_r$, and that $T$ is the first hitting time, which is $T=min\{t|W_t=x_l\ \text{or} \ W_t = x_r\}$. Use the fact that this stopped process is a martingale to find a formula for $Pr(W_T = x_l)$. Assumption here is $E[T]<\infty$.
Thanks in advance for any helpful insights.
Since $X_t$ is a martingale and $\mathbb ET<\infty$, we can apply the optional stopping theorem to conclude that $\mathbb EX_T=\mathbb EX_0=\mathbb EW_0=0$. Now, $X_T$ is a random variable that satisfies $\mathbb P(X_T\in\{x_l,x_r\})=1$. Thus, we have the system of equations $$ \mathbb P(X_T=x_l)+\mathbb P(X_T=x_r)=1,\qquad x_l\mathbb P(X_T=x_l)+x_r\mathbb P(X_T=x_r)=0, $$ where the second equation comes from expressing the expectation $\mathbb EX_T$ in two different ways. Solving the system of equations yields $$ \mathbb P(X_T=x_l)=\frac{x_r}{x_r-x_l}\qquad \mathbb P(X_T=x_r)=\frac{-x_l}{x_r-x_l}. $$ Thus, $\mathbb P(W_T=x_l)=x_r/(x_r-x_l)$.