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Let x, y > 0. Defint eh first passage time of a Brownian motion $W_t$ as $\tau_a$ = min{t $\ge$ 0: $W_t$ = a}. I need to show that E[$e^{-u\tau_x}$$1_{\tau_x < \tau_{-y}}$] = $\frac{sinh(y\sqrt{2u})}{sinh((x + y)\sqrt{2u}}$.

My method, and the only method that I will be able to understand, is to use the optional sampling theorem. I noted that $Z_t = e^{\theta W_t - \frac{1}{2}\theta^{2}t}$ is martingale and that the optional sampling theorem states that $E[Z_{\tau_{min\, {a, t}}}$] = 1. Applying this to the stopping time $ {\tau_x, \wedge \tau_{-y}}$ I have managed to show that as t --> $\infty$ $Z_(\tau_x\wedge\tau_{-y})\wedge t$ = $e^{-\theta y - \frac{1}{2} \theta^2 \tau_{-y}}$$1_{\tau_{-y} \, < \, \tau_x}$ + $e^{\theta x - \frac{1}{2} \theta^2 \tau_{x}}$$1_{\tau_{x} \, < \, \tau_{-y}}$. I can't figure where to go from here. I had a similar problem, but there was only one level involved, where here we have two: x and y.

I believe we should now take the expectation of the expression I derived and set it equal to one by optional sampling theorem, but I don't know what follows. Thanks.

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As you have already shown above, \begin{align*} 1 = E\Big(e^{-\theta y - \frac{1}{2}\theta^2 \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+E\Big(e^{\theta x - \frac{1}{2}\theta^2 \tau_{x} }1_{\tau_x < \tau_{-y}} \Big). \end{align*} Set $\theta = \sqrt{2u}$, we obtain that \begin{align*} 1 = E\Big(e^{-\sqrt{2u}y - u \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+E\Big(e^{\sqrt{2u} x - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big), \end{align*} that is, \begin{align*} e^{\sqrt{2u}y} = E\Big(e^{- u \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+e^{\sqrt{2u} (x+y)}E\Big(e^{ - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big). \end{align*} On the other hand, by setting $\theta = -\sqrt{2u}$, we obtain that \begin{align*} 1 = E\Big(e^{\sqrt{2u}y - u \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+E\Big(e^{-\sqrt{2u} x - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big), \end{align*} that is, \begin{align*} e^{-\sqrt{2u}y} = E\Big(e^{- u \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+e^{-\sqrt{2u} (x+y)}E\Big(e^{ - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big). \end{align*} Consequently, \begin{align*} e^{\sqrt{2u}y} - e^{-\sqrt{2u}y} = \big[e^{\sqrt{2u} (x+y)} - e^{-\sqrt{2u} (x+y)} \big]E\Big(e^{ - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big). \end{align*} Then, \begin{align*} E\Big(e^{ - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big) = \frac{\sinh (\sqrt{2u}y)}{\sinh \big(\sqrt{2u} (x+y) \big)}. \end{align*}

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  • $\begingroup$ From Wikipedia: $\sinh x = \frac {e^x - e^{-x}} {2}$, please show the final steps of your solution. $\endgroup$ – emcor Dec 24 '14 at 14:23
  • $\begingroup$ I have added one more intermediate step. $\endgroup$ – Gordon Dec 24 '14 at 14:29
  • $\begingroup$ How do you eliminate the first expectation in $e^{-\sqrt{2u}y} = E\Big(e^{- u \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+e^{-\sqrt{2u} (x+y)}E\Big(e^{ - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big). $? $\endgroup$ – emcor Dec 24 '14 at 16:40
  • $\begingroup$ It is eliminated by a difference with $e^{\sqrt{2u}y} = E\Big(e^{- u \tau_{-y} }1_{\tau_{-y}<\tau_x} \Big)+e^{\sqrt{2u} (x+y)}E\Big(e^{ - u \tau_{x} }1_{\tau_x < \tau_{-y}} \Big)$. $\endgroup$ – Gordon Dec 24 '14 at 16:59
  • $\begingroup$ This is correct, but I had already worked this out a while back, if you see my identical posting on math.stackexchange I mention that I had found the solution. Indeed I did it exactly as you did it. Despite the late answer, +1. $\endgroup$ – user7348 Dec 25 '14 at 2:52

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