Brownian motion: first-hitting-time with double barrier

Question

Let $(B_t)_t$ be a standard ($B_0=0$) Brownian motion , and

$$ T_{a,b} = \inf\{t>0 : B_t \not\in(a,b)\} $$

where $a<0<b$. What is the expected first-passage time $\mathbf{E}[T_{a,b}]$?

Answer 1

Hint: By the optional stopping theorem, both

$$(B_{t \wedge T_{a,b}})_{t \geq 0} \tag{1}$$

and

$$(B_{t \wedge T_{a,b}}^2- (t \wedge T_{a,b}))_{t \geq 0} \tag{2}$$

are martingales. Use the dominated convergence theorem to conclude from (2) that $T_{a,b} \in L^1$. Then let $t \to \infty$ to get

$$\mathbb{E}T_{a,b} = \mathbb{E}(B_{T_{a,b}}^2) \qquad \text{and} \qquad \mathbb{E}(B_{T_{a,b}})=0. \tag{3}$$

Since the Brownian motion has continuous sample paths, we have

$$B_{T_{a,b}} = a 1_{\{T_{a,b}=a\}}+b 1_{\{T_{a,b}=b\}} = a 1_{\{T_{a,b}=a\}}+b (1-1_{\{T_{a,b}=a\}})$$

and therefore we can write $(3)$ as a system of linear equations (with two unknown variables). Solve it.

Solution: $$\mathbb{E}T_{a,b} = -ab.$$

Answer 2

First note that sets of the form $\{t\in\mathbf{R}_{+}, B_t = x\}$ for $x\in\mathbf{R}$ are non empty by limit properties of the brownian motion. I assume more generally that $B$ starts at $x$, not $0$, but that $a < x < b$. (I may have switch your $a,b$ with mine, but who cares.)

Let us first recall three "optional sampling" results (théorème d'arrêt in french) from the Revuz-Yor book :

This is theorem 3.2 p. 65) Let $\left( M_t \right)_{t\in\mathbf{R}}$ be a $\left( \mathscr{F}_t \right)_{t\in\mathbf{R}}$-martingale. Let $\sigma$ and $\tau$ be two bounded $\left( \mathscr{F}_t \right)_{t\in\mathbf{R}}$-stopping times such that $\sigma \leq \tau$. Then we have $\mathbf{E}[M_{\sigma}|\mathscr{F}_{\tau}] = M_{\tau}$. If $\left( M_t \right)_{t\in\mathbf{R}}$ is moreover uniformly integrable, then the family $\left( M_{\tau} \right)_{\tau}$ where $\tau$ runs through the set of all stopping times is uniformly integrable, and if $\sigma$ and $\tau$ are two not necessarily bounded $\left( \mathscr{F}_t \right)_{t\in\mathbf{R}}$-stopping times such that $\sigma \leq \tau$ then we have $\mathbf{E}[M_{\sigma}|\mathscr{F}_{\tau}] = \mathbf{E}[M_{\sigma}|\mathscr{F}_{\infty}] = M_{\tau}$, as $M_{\infty}$ exist by uniform integrability of $\left( M_t \right)_{t\in\mathbf{R}}$. Practically only a corollary of this theorem is often use, using only one bounded stopping time $\tau$ to say that $\mathbf{E}[M_{\tau}] = M_{0}$. In fact this corolaresque conclusion is true under other hypothesis : this is the second optimal sampling theorem :

This is proposition 3.5 p. 67 A càdlàg $\left( \mathscr{F}_t \right)_{t\in\mathbf{R}}$-adapted process is a martingale if and only if for every bounded stopping time $\tau$ the random variable $M_{\tau}$ is $\mathbf{L}^1$ and if $\mathbf{E}[M_{\tau}] = \mathbf{E}[M_{0}]$.

This is corolary 3.6 p. 67 If $\left( M_t \right)_{t\in\mathbf{R}}$ is a $\left( \mathscr{F}_t \right)_{t\in\mathbf{R}}$-martingale and if $\tau$ is a stopping time, the stopped process $\left( M_t^{\tau} = M_{t\wedge\tau} \right)_{t\in\mathbf{R}}$ is $\left( \mathscr{F}_t \right)_{t\in\mathbf{R}}$-martingale.

Note $\tau = \min(\tau_a,\tau_b)$ : a classic result states that $\tau$ is a stopping time for the filtration generated by $B$. The brownian motion $B$ is a martingale for its own filtration (!). Then by the corollary, the stopped process $B^{\tau}$ is a martingale. Now, this martingale is bounded : we have $B_t^{\tau(\omega)} (\omega) = B_{t\wedge\tau_a (\omega) \wedge\tau_b (\omega)} (\omega) \in [a,b]$. Indeed, fix $\omega\in\Omega$, if $t < \tau (\omega)$, the continuous function $u\mapsto B_t (\omega)$ has not touched $a$ and $b$ yet, and as its value in $0$ is $x\in]a,b[$, it must still be in $]a,b[$ at $t$, that is $B_t^{\tau(\omega)} (\omega) = B_{t\wedge\tau_a (\omega)\wedge\tau_b(\omega)} (\omega) = B_{t} (\omega) \in [a,b]$. If $t \geq \tau (\omega)$ then $t\wedge\tau_a (\omega)\wedge\tau_b (\omega)= \tau_a (\omega)\wedge\tau_b(\omega)$ and then according to who the infimum of $\tau_a (\omega)$ and $\tau_b(\omega)$ is, $B_{t\wedge\tau_a (\omega)\wedge\tau_b (\omega)}$ is equal to $a$ or $b$. So here again $B_t^{\tau(\omega)} (\omega)\in [a,b]$. Now the martingale $B^{\tau}$ beeing bounded implies that it is uniformly integrable (exercise !), and then the second (stronger somehow) part of the theorem allows us to conclude that even if $\tau$ is not necessarily bounded, we have $\mathbf{E}[B_{\tau}] = \mathbf{E}[B_{\tau}^{\tau}] = \mathbf{E}[B_0] = x$.

Now $$B_{\tau} = B_{\tau} I(\tau a < \tau_b) + B_{\tau} I(\tau a \geq \tau_b)$$

$$ = B_{\tau_b} I(\tau a < \tau_b) + B_{\tau_a} I(\tau a \geq \tau_b)$$ $$ = a I(\tau a < \tau_b) + b I(\tau a \geq \tau_b)$$ so that $$x = \mathbf{E}[B_{\tau}] = a \mathbf{E}{I(\tau a < \tau_b})] + b \mathbf{E}[I(\tau a \geq \tau_b)]$$ that is $$a \mathbf{P}\left[ \tau a < \tau_b \right] + b \mathbf{P}\left[ \tau a \geq \tau_b \right] = x $$ and as $$\mathbf{P}\left[ \tau a < \tau_b \right] + \mathbf{P}\left[ \tau a \geq \tau_b \right] = 1$$ because $\mathbf{P}$ is a measure, solving the system with unknowns $\mathbf{P}\left[ \tau a < \tau_b \right]$ and $\mathbf{P}\left[ \tau a \geq \tau_b \right]$ gives $$\mathbf{P}\left[ \tau a < \tau_b \right] = \frac{b-x}{b-a}$$ and $$\mathbf{P}\left[ \tau a \geq \tau_b \right] = \frac{x-a}{b-a} $$

Now, knowing this, are you able to calculate $\tau$'s expectation ? ;-)