Laplace Transform of Stopping Time of Brownian Motion

Asked May 19 '18 at 02:35

Active May 19 '18 at 05:25

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Let W represent a standard Brownian Motion in one dimension. For some $b > 0$, let $S_b$ denote the first time that $|W_t| = a$. How do we show that the Laplace transform of $S_b$ is $$E[\exp(\lambda S_b)] = \cosh(s\sqrt{2\lambda})^{-1}$$

My intuition is that this can be obtained by applying optional stopping theorem to a martingale.

edited May 19 '18 at 05:25

Math1000

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asked May 19 '18 at 02:35

Francois Wassert

2

Any idea what kind of martingale you might want to consider? – saz May 19 '18 at 08:55
I'm not really sure? Some-type of exponential martingale? – Francois Wassert May 27 '18 at 03:01
Well, there aren't so many exponential martingales for Brownian motion, are there? Why not give it a try and see what you get? – saz May 27 '18 at 05:34
@saz I actually considered the martingale $e^{\theta{B_{t}}-(\theta^{2}t/2)}$ and using the optional stopping theorem and DCT, I concluded that $E_{0}[e^{\theta{B_{S_{b}}}-(\theta^{2}S_{b}/2)}]=1$ – user135520 Apr 27 '20 at 16:22
How does one obtain the $\cosh$? is it from $\exp(\theta{B}{S{b}})$? – user135520 Apr 27 '20 at 16:24
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@user135520 See this question and also this question – saz Apr 27 '20 at 16:28

Laplace Transform of Stopping Time of Brownian Motion

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