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Let $B$ be a standard Brownian motion and for $a>0$ and $b>0$, and set $$\sigma_{a,b} = \inf\{t\,:\, B_t + bt = a\}.$$

There are at least two ways to solve the following problem (the other one is using the scaling property and Laplace transform), but I want to use Girsanov theorem to prove that the density of $\sigma_{a,b}$ is equal to $$a\,\left(2\pi t^3\right)^{-\frac{1}{2}}\exp\left(-(a-bt)^2\,/\,2t\right)$$

What I have done so far is using the reflection principle: \begin{equation} \begin{split} \mathbb P(\sigma_{a,b}<t) &=& 2\,\mathbb P(B_t + bt \geq a) = 2\,\mathbb P(B_t \geq a - bt) =\\ &=& 2\int_{\frac{a-bt}{\sqrt{t}}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx \end{split} \end{equation}

Clearly, differentiating the above expression does not yield the desired density, and I'm also mindful that I have not used Girsanov simply because I do not see a proper way of applying it.

I have also thought about defining a new Brownian motion $\widetilde{B_t} = B_t + bt$ and then use Girsanov's theorem such that $\frac{d\mathbb P}{d\mathbb Q} = Z$ but I'm unsure how to proceed down this path.

Any help is welcome. Thanks in advance.

Adam
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  • You cannot apply the reflection principle as you did because $B_t+bt$ is no Brownian motion (or am I missing something obvious)? Since you want to use Girsanov theorem, this question is a duplicate of http://math.stackexchange.com/q/1053294/ – saz Feb 16 '15 at 15:29
  • @saz thanks for pointing out that we can't apply Girsanov to that process, I skipped over that a bit. Indeed, it is the same question apparently. I have seen your detailed work answering the other question (great work), but I wonder if that is the only way of using Girsanov to solve this problem (tbh I would like to think that there is a simpler alternative solution). – Adam Feb 16 '15 at 16:10
  • Well, you are welcome to provide an easier solution. – saz Feb 16 '15 at 16:23

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