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Lets assume that Brownian Motion starts from some point $x$ for which $a<x<b$ holds. Let $\tau=\inf\{t:B_t\not\in [a,b]\}$ be a stopping time. Now I want to prove that for $\theta>0$ ,an appropriate martingale and the Optional Sampling Theorem the following holds; $$\mathbb E[\exp(-\frac{1}{2}\theta^2\tau)|B_0=x]=\frac{\cosh(\theta(x-(a+b)/2))}{\cosh(\theta(b-a)/2)}$$

My first idea was to rewrite the RHS using $\exp$ expressions instead of $\cosh$ expressions but this is only a formal thing.

My second thought was that it has something to with the result presented here: The Laplace transform of the first hitting time of Brownian motion

So we use $M_t=\exp(\theta B_t-\frac{1}{2}\theta^2 t)$ as the "appropriate integral" and the Laplace transform of $H_a=\inf\{t: B_t>a\}$ but I do not see how these things should be connected, may you have some idea.

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Ulrich Otto
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    @ Ulrich Otto : Hi I think the first part of your question is unclear (at least to me), can I restate it this way : "I want to prove the result (formula with cosh) using the martingale property of Brownian motion and Optimal sampling theorem " ? Am I correct ? Best regards – TheBridge Nov 20 '13 at 10:00

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The hitting time $H_a$ is not obviously related to the bilateral hitting time $\tau$ hence I fail to see how to use your idea. Instead, a common approach is to consider the function $u$ defined on $(a,b)$ by $$ u(x)=E_x[\mathrm e^{-\theta^2\tau/2}], $$ where $E_x[\ ]=E[\ \mid B_0=x]$, and to decompose $\tau$ as the independent sum of the first hitting time $\sigma$ of $\{x-z,x+z\}$ and the remaining time $\tau-\sigma$ until $\tau$, where $z$ is positive and small enough to guarantee that $(x-z,x+z)\subseteq(a,b)$.

The Markov property at time $\sigma$ and the independence of $\tau-\sigma$ and $\sigma$ yield $$ u(x)=E_x[\mathrm e^{-\theta^2\sigma/2}]\,\tfrac12(u(x+z)+u(x-z)). $$ The martingale $(M_t)$ you summoned in your question, plus the fact that $B_{\sigma}$ is independent of $\sigma$ and uniform on $\{x-z,x+z\}$, yield $$ E_x[\mathrm e^{-\theta^2\sigma/2}]=\left(\mathrm e^{-\theta x}E_x[\mathrm e^{\theta B_{\sigma}}]\right)^{-1}=\cosh(\theta z)^{-1}, $$ thus, $$ 2\cosh(\theta z)u(x)=u(x+z)+u(x-z). $$ When $z\to0$, this reads $$ 2(1+\tfrac12\theta^2z^2)u(x)=2u(x)+u''(x)z^2+o(z^2), $$ hence $$ u''(x)-\theta^2u(x)=0. $$ This, together with the boundary condition that $u(a)=u(b)=1$, yields the expression you are after.

Did
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  • Thanks for your answer. I have one question concerning your second sentence. You mention this is a common approach but do you have some exmaples or sources where I can find similar problems where this method works? – Ulrich Otto Nov 21 '13 at 20:27
  • If I had the book by Revuz and Yor with me, I would check it. – Did Nov 21 '13 at 23:17
  • No problem. I accepted the answer. I have the book by Revuz and Yor with me so I will go through it. – Ulrich Otto Nov 22 '13 at 00:17