Let $(W_s)_{s \geq 0}$ be a Wiener process and $\tau$ be a random variable with an exponential distribution with parameter $1$. Suppose that $W$ and $\tau$ are independent. Determine the distribution of $W_{\tau}$.
My first (and only) idea was to compute the characteristic function of $W_{\tau}$. In order to do that I thought about expanding $e^{itW_{\tau}}$ into power series and computing the $n$-th moment of $W_\tau$ but I have no foggiest idea how to tackle this.
$\newcommand{\E}{\operatorname{E}}$ The characteristic function is \begin{align} s \mapsto {} & \E\left( e^{isW_\tau} \right) = \E\left( \E\left( e^{isW_\tau} \mid\tau \right) \right) = \E\left( e^{-\tau s^2/2} \right) =\int_0^\infty e^{-t s^2/2} f_\tau(t)\, dt \\[6pt] = {} & \int_0^\infty e^{-t s^2/2} e^{-t}\,dt = \int_0^\infty e^{-(1+s^2/2)t} \, dt = \frac 1 {1+\frac{s^2}2}. \end{align}
This can be recognized as the characteristic function of the Laplace distribution with expected value $0$ and scale parameter $1/\sqrt 2$, with density $$ f(x) = \frac 1 2 \sqrt 2\ e^{-|x|\sqrt 2}. $$