Expectation of Brownian hitting time

Question

Let $a,b >0$ and $B_t$ be one dimensional Brownian motion starting in $0$. I am searching some way to calculate $$E(\tau_{-a}1_{\{\tau_{-a}<\tau_b\}}),$$ where $\tau_x$ denotes the hitting time of some point $x \in \mathbb{R}$. Of course I know that by symmetry this expectation equals $\frac{a^2}{2}$ if $a=b$. But in this "unsymmetrical" I don't know how to calculate this expectation. Does anybody have an idea?

Answer 1

An alternative approach (motivated by the tediousness of differentiating that ratio of hyperbolic sines) can be made using the martingale $B_t^3 - 3tB_t$. By optional stopping, $E[B_\tau^3-3\tau B_\tau] =0$. Therefore, writing $q$ for $E[\tau_{-a}1_{\{\tau_{-a}<\tau_b\}}]$ (and noting that $E[\tau_b 1_{\{\tau_b<\tau_{-a}\}}]=ab-q$): $$ (-a)^3{b\over a+b}+b^3{a\over a+b} =3(-a)q+3b(ab-q), $$ because $P[\tau_{-a}<\tau_b]=b/(a+b)$, as is well known. Solving for $q$ you get $$ E[\tau_{-a}1_{\{\tau_{-a}<\tau_b\}}]={ab(a+2b)\over 3(a+b)}. $$

Answer 2

We define $\tau = \tau_{-a}\wedge\tau_{b}$ which is a stopping time as the min of two stopping times. We recall the exponential (continuous) martingale $M$: \begin{equation*} \forall t \geq 0, \quad M_t = \exp\left(\lambda B_t - \frac{\lambda^2 }{2}t\right) \end{equation*} We assume that $\tau$ is finite a.s. ($H$). Thanks to the Doob's optional theorem, we know that the stopped process $\left\{M_{t\wedge\tau}\right\}_{t\geq 0}$ is a martingale. Therefore: \begin{equation*} E\left[M_{t\wedge\tau}\right] = E\left[M_0\right] = 1 \end{equation*} But: \begin{align*} E\left[M_{t\wedge\tau}\right] &= E\left[e^{\lambda B_{t\wedge\tau} - \frac{\lambda^2}{2}(t \wedge \tau)}\right]\\ &= E\left[e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}1_{\tau \leq t}\right] + E\left[e^{\lambda B_{t} - \frac{\lambda^2}{2}t}1_{\tau \geq t}\right] \tag{1} \end{align*} On the event $\{\tau \leq t\}$, the function $t \mapsto e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}1_{\tau \leq t}$ is increasing and positive a.s.. Hence by Beppo-Levi: \begin{align*} \lim_{t\to + \infty} E\left[e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}1_{\tau \leq t}\right] &= E\left[e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}1_{\tau <+\infty}\right] \\ &= E\left[e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}\right] \text{ a.s. Thanks to $H$} \end{align*} On the event $\{\tau \geq t\}$, we know that $B_{t\wedge\tau} \leq \max\{a,b\}$ a.s.. Then if $\lambda > 0$, we have: \begin{equation*} \forall t \geq 0, \quad 0 \leq e^{\lambda B_{t} - \frac{\lambda^2}{2}t}1_{\tau \geq t} \leq e^{\lambda\max\{a,b\}} \end{equation*} By domination, we have: \begin{equation*} \lim_{t\to + \infty} E\left[e^{\lambda B_{t} - \frac{\lambda^2}{2}t}1_{\tau \geq t} \right] = 0 \quad \text{a.s. Thanks to $H$} \end{equation*} Gathering all the limits, we have: \begin{equation} E\left[e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}\right] = 1 \text{ a.s.} \end{equation} By continuity of the Brownian motion, we have $\forall \omega \in \Omega$, $B_{\tau}(\omega) = -a1_{\tau (\omega) = \tau_{-a}(\omega)} + b1_{\tau (\omega) = \tau_{b}(\omega)}$. Hence, \begin{equation} E\left[e^{\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}\right] = e^{-\lambda a}E\left[e^{- \frac{\lambda^2}{2}\tau_{-a}}1_{\tau = \tau_{-a}}\right] + e^{\lambda b}E\left[e^{- \frac{\lambda^2}{2}\tau_b}1_{\tau = \tau_{b}}\right] = 1 \end{equation} Given that $-B$ is still a Brownian motion, we can repeat all steps from above using the "modified" exponential martingale \begin{equation*} \tilde{M}_t = \exp\left(-\lambda B_t - \frac{\lambda^2 }{2}t\right) \end{equation*} This gives us: \begin{equation} E\left[e^{-\lambda B_{\tau} - \frac{\lambda^2}{2}\tau}\right] = e^{\lambda a}E\left[e^{- \frac{\lambda^2}{2}\tau_{-a}}1_{\tau = \tau_{-a}}\right] + e^{-\lambda b}E\left[e^{- \frac{\lambda^2}{2}\tau_b}1_{\tau = \tau_{b}}\right] = 1 \end{equation} Thus, we have a linear system: \begin{cases} e^{\lambda a}E\left[e^{- \frac{\lambda^2}{2}\tau_{-a}}1_{\tau = \tau_{-a}}\right] + e^{-\lambda b}E\left[e^{- \frac{\lambda^2}{2}\tau_b}1_{\tau = \tau_{b}}\right] = 1\\ e^{-\lambda a}E\left[e^{- \frac{\lambda^2}{2}\tau_{-a}}1_{\tau = \tau_{-a}}\right] + e^{\lambda b}E\left[e^{- \frac{\lambda^2}{2}\tau_b}1_{\tau = \tau_{b}}\right] = 1 \end{cases} It's quiet straightforward to solve it and get: \begin{cases} E\left[e^{- \frac{\lambda^2}{2}\tau_{-a}}1_{\tau = \tau_{-a}}\right] = \frac{\sinh(\lambda b)}{\sinh(\lambda (b+a))}\\ E\left[e^{- \frac{\lambda^2}{2}\tau_{b}}1_{\tau = \tau_{b}}\right] = \frac{\sinh(\lambda a)}{\sinh(\lambda (b+a))}\\ \end{cases}

As it seems that you are only interested in the first quantity, you can derivate both sides w.r.t $\mu = \lambda^2$ and take a decreasing sequence $\lambda$ which tends to $0$ to get $E(\tau_{-a}1_{\{\tau_{-a}<\tau_b\}})$.

Some remarks:

• In fact, we can show the hypothesis $H$ by taking the limit in $(1)$ and choosing a decreasing $\lambda_n$ to $0$.
• To be rigorous, we need to justify the derivative under the integral sign.