8
$\begingroup$

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?

  • 0
    I'm looking how to solve $E(\tau )$ in general. The stopping time defined in the original question is a practice question for my final.2012-12-19

1 Answers 1

12

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. $

  • 0
    @Rasputin That follows from the line just above.2018-12-15