For a real-valued standard Brownian motion $B= (B_t)_{t\geq 0}$ we define the stopping times

$ \tau_{a} := \inf \left\{ t> 0: B_t \leq a \right\},~a <0$,

$ \tau_{b} := \inf \left\{ t> 0: B_t \geq b \right\},~ b >0$,

$ \tau := \min (\tau_a , \tau_b). $

Prove for every $\lambda \in \mathbb{R}$

$$\mathbb{E} \left[\exp\left\{- \frac{\lambda^2 \tau}{2} \right\} \mathbf{1}_\left\{\tau = \tau_a \right\}\right] = \frac{\sinh (\lambda b)}{\sinh (\lambda(b-a))}. $$

Using Doob's optional sampling theorem, I get $$ \mathbb{E} \left[ \exp\left(\lambda B_{\tau}- \frac{1}{2}\lambda^2 \tau \right) \right] = 1$$ I'm having trouble with the dependence of $\lambda B_{\tau}$ and $\frac{1}{2}\lambda^2 \tau$. When I use Doob's theorem on $\tau_a$ or $\tau_b$ instead of on $\tau$, I obtain
$$\mathbb{E} \left[\exp\left\{- \frac{\lambda^2 \tau_a}{2} \right\} \right] = \exp(-\lambda a ) \quad \text{and} \quad \mathbb{E} \left[\exp\left\{- \frac{\lambda^2 \tau_b}{2} \right\} \right] = \exp(-\lambda b ).$$ Im not sure if that helps anything. From here i don't know how to proceed. I hope anybody can help me out here.


1 Answer 1



$$\begin{align*}1&= \mathbb{E} \exp \left( \lambda B_{\tau} - \frac{1}{2} \lambda^2 \tau \right) \\ &= \mathbb{E} \left[ 1_{\{\tau=\tau_a\}} \exp \left( \lambda B_{\tau} - \frac{1}{2} \lambda^2 \tau \right) \right] + \mathbb{E} \left[ 1_{\{\tau=\tau_b\}} \exp \left( \lambda B_{\tau} - \frac{1}{2} \lambda^2 \tau \right) \right] \\ &= \mathbb{E} \left[ 1_{\{\tau=\tau_a\}} \exp \left( \lambda a - \frac{1}{2} \lambda^2 \tau_a \right) \right] + \mathbb{E} \left[ 1_{\{\tau=\tau_b\}} \exp \left( \lambda b - \frac{1}{2} \lambda^2 \tau_b \right) \right] \\ &= e^{\lambda a} \mathbb{E} \left[ 1_{\{\tau=\tau_a\}} \exp \left(- \frac{1}{2} \lambda^2 \tau_a \right) \right] + e^{\lambda b} \mathbb{E} \left[ 1_{\{\tau=\tau_b\}} \exp \left(- \frac{1}{2} \lambda^2 \tau_b \right) \right]. \tag{1} \end{align*}$$

Replacing $\lambda$ by $-\lambda$ we find

$$1 = e^{-\lambda a} \mathbb{E} \left[ 1_{\{\tau=\tau_a\}} \exp \left(- \frac{1}{2} \lambda^2 \tau_a \right) \right] + e^{-\lambda b} \mathbb{E} \left[ 1_{\{\tau=\tau_b\}} \exp \left(- \frac{1}{2} \lambda^2 \tau_b \right) \right]. \tag{2}$$

This means that we have a system of linear equations $$\begin{align*} 1&=e^{\lambda a} u + e^{\lambda b} v\\ 1&= e^{-\lambda a} u + e^{-\lambda b} v \end{align*}$$ for the two unknown variables $$ u:=\mathbb{E} \left[ 1_{\{\tau=\tau_a\}} \exp \left(- \frac{1}{2} \lambda^2 \tau_a \right) \right] \quad \text{and} \quad v:= \mathbb{E} \left[ 1_{\{\tau=\tau_b\}} \exp \left(- \frac{1}{2} \lambda^2 \tau_b \right) \right].$$

Solve this linear system and you are done.

  • $\begingroup$ You're a hero. Thank you very much!! In retrospective it's obvious ;) $\endgroup$ Commented Nov 10, 2018 at 8:43
  • $\begingroup$ @ManuelSommer You are welcome. $\endgroup$
    – saz
    Commented Nov 10, 2018 at 9:26

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