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.