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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?

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    $\begingroup$ Do you mean $B$ instead of $M$? $\endgroup$
    – Stefan Hansen
    Dec 19, 2012 at 7:05
  • $\begingroup$ 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 $E(M_{\tau})=E(M_0)$. $\endgroup$
    – riotburn
    Dec 19, 2012 at 7:19
  • $\begingroup$ What is it exactly you're trying to find an expression for? $\endgroup$
    – Stefan Hansen
    Dec 19, 2012 at 7:21
  • $\begingroup$ 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. $\endgroup$
    – riotburn
    Dec 19, 2012 at 7:23

1 Answer 1

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

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  • $\begingroup$ 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 $P(B_{\tau}=b)$? $\endgroup$
    – riotburn
    Dec 19, 2012 at 7:48
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    $\begingroup$ 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$$0=-a(1-P(B_\tau=b))+bP(B_\tau=b).$$ $\endgroup$
    – Stefan Hansen
    Dec 19, 2012 at 7:50
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    $\begingroup$ You just blew my mind. Thank you so much! $\endgroup$
    – riotburn
    Dec 19, 2012 at 8:02
  • $\begingroup$ You're welcome :) $\endgroup$
    – Stefan Hansen
    Dec 19, 2012 at 8:03
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    $\begingroup$ Why does $E[\tau]=E[B_{\tau}^2]$? $\endgroup$
    – Rasputin
    Dec 14, 2018 at 16:40

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