How do you take the expectation of a stopping time with respect to a Brownian motion? The specific question is:
$$
\tau = \inf\{ t \ge 0: B(t) \in \{-a, b\}\}
$$
I understand the optional stopping theorem tells us that $E[M_\tau ] = E[M_0]$ but how do I use that to find the expectation?
Best Answer
We want to use the optional stopping theorem on the two martingales $(B_t)_{t\geq 0}$ and $(B_t^2-t)_{t\geq 0}$. Note that $\tau<\infty$ a.s. so $B_\tau \in \{-a,b\}$ a.s. and hence by the optional stopping theorem, we have $$ \begin{align*} 0&=E[B_0]=E[B_\tau]=-aP(B_\tau=-a)+bP(B_\tau=b)\\ &=-a(1-P(B_\tau=b))+bP(B_\tau=b) \end{align*} $$ which implies that $$ P(B_\tau=b)=\frac{a}{a+b},\quad P(B_\tau=-a)=\frac{b}{a+b}. $$ Using the optional stopping theorem on $(B_t^2-t)_{t\geq 0}$ we get that $$ 0=E[B_0^2-0]=E[B_\tau^2-\tau] $$ and hence $$ E[\tau]=E[B_\tau^2]=a^2P(B_\tau=-a)+b^2P(B_\tau=b)=ab. $$