Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Do you mean $B$ instead of $M$? – Stefan Hansen Dec 19 '12 at 7:05
Using M as any martingale such that the optional stopping theorem states: Let $\tau$ be a stopping time and $M_n$ be a martingale. If there is $K\in \mathbb(N)$ such that $\tau \le K$ almost surely then $E(M_{\tau})=E(M_0)$. – riotburn Dec 19 '12 at 7:19
What is it exactly you're trying to find an expression for? – Stefan Hansen Dec 19 '12 at 7:21
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. – riotburn Dec 19 '12 at 7:23
up vote 7 down vote accepted

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 $$ E[\tau]=E[B_\tau^2]=a^2P(B_\tau=-a)+b^2P(B_\tau=b)=ab. $$

share|cite|improve this answer
Thank you for the answer! The first part this question comes from asks to prove that $(B^2_t - t)$ is a martingale. I never made the connection nor what the usefulness of the stopping theorem was. Two questions. How do you get $P(B_{\tau}=-a) = 1 - P(B_{\tau})$? How do you get $P(B_{\tau}=b)$? – riotburn Dec 19 '12 at 7:48
Because $B_\tau\in \{-a,b\}$ a.s. means exactly that $P(B_\tau=-a)+P(B_\tau=b)=1$ and therefore $P(B_\tau=-a)=1-P(B_\tau=b)$. I got $P(B_\tau=b)$ by solving for it in the equation$$0=-a(1-P(B_\tau=b))+bP(B_\tau=b).$$ – Stefan Hansen Dec 19 '12 at 7:50
You just blew my mind. Thank you so much! – riotburn Dec 19 '12 at 8:02
You're welcome :) – Stefan Hansen Dec 19 '12 at 8:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.