# Cauchy density function for Brownian motion

Let $$\{W(t):t\geq0\}$$ be a Brownian motion, and let $$\{\mathcal{F}_{t},t\geq0\}$$ be its natural filtration. Let $$\{W_{2}(t):t\geq0\}$$ be a Brownian motion, independent of $$\{W(t):t\geq0\}$$. Denote, for $$a>0$$, $$\tau_{a}=\inf\{t\geq0:W(t)=a\}.$$ Using that the probability density of the first hitting time of $$a>0$$ for Brownian motion is given by $$f(t)= \left\{ \begin{array}{ll} \displaystyle\frac{ae^{-a^2/2t}}{\sqrt{2\pi t^{3}}}&\text{if }t>0\\ 0&\text{otherwise}. \end{array} \right.$$ Show that the probability denisty of $$W_{2}(\tau_{a})$$ is given by the Cauchy density function $$f(y)=\frac{a}{\pi(a^{2}+y^{2})},\quad y\in\mathbb{R}.$$

First, I do not fully understand the question. Is the Cauchy density function the probability density function of the second Brownian motion at time $$\tau_{a}$$, when the first Brownian motion is stopped? The distribution of a Brownian motion should have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case.

"Is the Cauchy density function the probability function of the second Brownian motion at time $$\tau_a$$, when the first Brownian motion is stopped?" Yes.
"The probability of the a Brownian motion should only have a growing variance, while the mean remains zero. I do not understand the Cauchy density distribution in this case." As $$a$$ increases, so will $$\tau_a$$ overall; and although a Cauchy distribution doesn't have finite variance, the distribution does get more dispersed as $$a$$ increases. So there doesn't seem to be a conflict with your understanding that the variance of Brownian motion is increasing in time.
And the solution to the problem: \begin{align} f_{W_2(\tau_a)}(y)dy &=P[W_2(\tau_a)\in(y,y+dy)]\\ &=\int_{t=0}^\infty P[W_2(\tau_a)\in(y,y+dy)|\tau_a=t]f_{\tau_a}(t)dt\\ &=\int_{t=0}^\infty \frac{1}{\sqrt{2\pi t}}e^{-\frac{y^2}{2t}}dy\frac{1}{\sqrt{2\pi t^3}} ae^{-\frac{a^2}{2t}}dt\\ &= \left(\int_{0}^\infty \frac{a}{2\pi t^2}e^{-\frac{y^2+a^2}{2t}}dt\right)dy,\quad\text{let }s=\frac{1}{t},\\ &= \left(\int_\infty^0\frac{as^2}{2\pi}e^{-\frac{y^2+a^2}{2}s}\left(-\frac{1}{s^2}\right)ds \right)dy\\ &=\left( \int_0^\infty\frac{a}{2\pi}e^{-\frac{y^2+a^2}{2}s}ds \right)dy\\ & = \frac{a}{\pi(a^2+y^2)}dy. \end{align}