0
$\begingroup$

I am interested in the probability density of $$ \tau =\inf\{t\geq 0: \vert W(t)\vert = 1\} $$ where $W(t)$ is a standard Wiener process. I have two approaches in mind:

  • First approach: I could compute the Laplace transform and invert it to obtain the density. To do so I apply the optional stopping theorem to the martingale $\exp(\lambda W(t)-\frac{\lambda^2}{2}t)$ to obtain $$ \mathbb{E}\left[\exp\left(\lambda W(\tau)-\frac{\lambda^2}{2}\tau\right)\right]= 1\\ \Leftrightarrow \mathbb{E}\left[\exp\left(-\lambda -\frac{\lambda^2}{2}\tau\right)1_{\{W(\tau)=-1\}}+\exp\left(\lambda -\frac{\lambda^2}{2}\tau\right)1_{\{W(\tau)=1\}}\right]= 1\\ \Leftrightarrow \mathbb{E}\left[\mathbb{E}\left[\exp\left(-\lambda -\frac{\lambda^2}{2}\tau\right)1_{\{W(\tau)=-1\}}+\exp\left(\lambda -\frac{\lambda^2}{2}\tau\right)1_{\{W(\tau)=1\}}\Big\vert \tau\right]\right]= 1\\ \Leftrightarrow \mathbb{E}\left[\exp\left(-\lambda -\frac{\lambda^2}{2}\tau\right)\mathbb{E}\left[1_{\{W(\tau)=-1\}}\vert \tau\right]+\exp\left(\lambda -\frac{\lambda^2}{2}\tau\right)\mathbb{E}\left[1_{\{W(\tau)=1\}}\Big\vert \tau\right]\right]= 1 $$ Because of symmetry we have $$ \mathbb{E}\left[1_{\{W(\tau)=1\}}\vert \tau\right] = \mathbb{E}\left[1_{\{W(\tau)=-1\}}\vert \tau\right]=\frac{1}{2} $$ and so we obtain $$ \mathbb{E}\left[\exp\left(-\frac{\lambda^2}{2}\tau\right)\right] = \frac{2}{\exp(-\lambda)+\exp(\lambda)} $$ or equivalently $$ \mathbb{E}\left[\exp\left(-\lambda\tau\right)\right] = \frac{2}{\exp(-\sqrt{2\lambda})+\exp(\sqrt{2\lambda})}. $$ The idea would now possibly be to compute the Taylor series of this expression and try to find functions which are mapped to the summands under the Laplace transform.
  • Second approach: The second approach would rely on the joint distribution of the running maximum and minimum of a Brownian motion: $$ \mathbb{P}(\tau > t) = \mathbb{P}(\max_{0\leq s\leq t} W(s)<1,\min_{0\leq s\leq t} W(s)>-1). $$ However, I do not know this joint distribution and could not find it anywhere. I would like to see a derivation of this joint probability distribution.

Furthermore, I am interested in what the standard approach to obtain the density is.

$\endgroup$
  • $\begingroup$ The joint probability of the running maximum and minimum can be obtained from Lévy's triple law which you can find, for instance, in the monograph by Schilling & Partzsch on Brownian motion (Section 6.5). Alternatively, you can have a look at the "Handbook of Brownian motion" by Borodin. However, there is no "nice" representation for the density of the joint distribution. (...I'm not sure what makes you belive that it is possible to compute the density of $\tau$ explicitly...) $\endgroup$ – saz Mar 2 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.