I do not understand how to demonstrate that the MST holds in the following situation:
There is a Brownian Motion process $\{B(t)\}$ for $t > 0$ and stopping time defined as $T:= inf\{t : B(t) = -a \ $or$ \ b\}$. Suppose $B(0)=0$.
At time $T$, I know that $B(T)$ is either $-a$ or $b$. However, I do not know how to get $P(B(T)=-a)$ and $P(B(T)=-b)$ so as to get $E[T]$ without the MST. An intuitive response would be much appreciated.
Since $B_T$ takes only the values $-a$ and $b$, we have $$B_T = -a 1_{\{B_T=-a\}} + b 1_{\{B_T=b\}}.$$ Using that $\mathbb{E}(B_T)=0$ we get $$0 = \mathbb{E}(B_T)= -a \mathbb{P}(B_T=-a) + b \mathbb{P}(B_T=b). \tag{1}$$ On the other hand, we know that $$\mathbb{P}(B_T=-a) + \mathbb{P}(B_T=b) = 1. \tag{2}$$ Solving this system of linear equations allows you to compute $\mathbb{P}(B_T=-a)$ and $\mathbb{P}(B_T=b)$.
In order to calculate $\mathbb{E}(T)$ you have to use that $(B_t^2-t)_{t \geq 0}$ is a martingale and apply the optional stopping theorem.