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.
 A: "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}
