Let $H_a$ be the first hitting time of $a$, $H_a = \inf\{t ≥ 0 : B_t = a\}$, where $B_t$ is a standard Brownian Motion Use Optional Stopping to compute the distribution of $B_{H_a∧H_b}$ for $ a< 0 < b$
The Optional Stopping theorem says the following:
Let $(X_t)t≥0$ be a uniformly integrable martingale with continuous sample paths. Let $S$ and $T$ be two stopping times with $S ≤ T$ . Then $X_S$ and $X_T$ are in $L^1$ and $X_S = E[X_T |F_S]$
If we stop the martingale then we get boundedness in $L^1$ and from there we can get uniform integrability, hence the theorem is applicable. It is trivial to see that $H_a∧H_b$ is a stopping time, too.
My problem is to find the distribution. Do we generally want to find $P(B_{H_a∧H_b}\geq x)=?$ or do we look at the mgf and try to guess the distribution (since there is a result about exponentiating a martingale)?
Thanks!