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Let $B_t$ be a one dimensional Brownian motion. For $a, b > 0$ let $T_x = \inf\{t \in \mathbb{R}_+: B_t = x\}$ and $T_{-a \wedge b} = T_{-a} \wedge T_b$.

I now want to show the Laplace transform of $T_{a, b}$ is

$\mathbb{E}[\exp(-\lambda T_{a, b})] = \frac{cosh(\frac{a-b}{2}\sqrt{2\lambda})}{cosh(\frac{a+b}{2}\sqrt{2\lambda})}$

I know the usual way to this is to take a variant of the exponential martingale $M_t$, then use optional stopping to show that for some stopping time $T$

$E[M_T] = E[M_0] = c$

where $c \in \mathbb{R}$. This is the way it was presented in this answer for example.

My main problem is that I'm having trouble dealing with the two different values $-a$ and $b$. I tried working with the martingale

$M_t^{\lambda, c} = \sinh(\lambda (B_t + c))\exp(-\frac{\lambda^2}{2}t)$

but because the $c$ is part of it, the expected value at $t = 0$ is not the same for $-a$ and $b$. Then I tried using the fact, that I know that

$P(T_b < T_a) = \frac{-a}{b-a} \quad P(T_a < T_b) = \frac{b}{b-a}$

as well as

$\mathbb{E}[\exp(-\lambda T_{a})] = \exp(a\sqrt(2\lambda))$

to calculate something similar, but I feel like I keep getting stuck.

Hamilton
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1 Answers1

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For simplicity, let $T:=T_{-a,b}$ and let :

  • $A := \mathbb{E} ( e^{-\frac{\lambda^2}{2} T} \mathbb{1}_{ T_{-a}<T_b} )$
  • $B := \mathbb{E} ( e^{-\frac{\lambda^2}{2} T} \mathbb{1}_{ T_{-a}>T_b} )$

By the same exponential martingale method, you imply that: $$1= e^{-\lambda a} A+e^{\lambda b}B$$ and $$1= e^{\lambda a} A+e^{-\lambda b}B$$ While the transform Laplace of $T$ at $t= \lambda^2/2$ is, by definition: $$\mathbb{E}( e^{-\frac{\lambda^2}{2} T} )=A+B $$

So if our formula is correct, the work is done by now. $\square$

Paresseux Nguyen
  • 7,331