Im wondering if anyone has this calculated, I cant seem to find it anywhere online. I am trying to find the variance of absolute value of BM. Here is my attempt:
First, $f_{\lvert W_t \rvert} (x)=\sqrt{\frac{2}{ \pi t}} e^{\frac{-x^2}{2t}}$
and so:
\begin{align*} Var(\lvert W_t \rvert) &= E( \lvert W_t \rvert^2) - E( \lvert W_t \rvert)^2\\ &= \sqrt{\frac{2}{\pi t}} \int_0^{\infty} x^2 e^{\frac{-x^2}{2t}} - \left(\sqrt{\frac{2t}{ \pi}} \right)^2 \\ \end{align*}
With @Did 's contribution: Solving the Integral using integration by parts $(\int u dv = uv=\int vdu)$:
let $u=x~~ du = 1~~ dv=xe^{\frac{-x^2}{2t}}~~v=-te^{\frac{-x^2}{2t}}$
Therefore: $$ \int_0^{\infty} x^2 e^{\frac{-x^2}{2t}}dx=[-txe^{\frac{-x^2}{2t}} \rvert^{\infty}_0 + t \int^{\infty}_0 e^{\frac{-x^2}{2t}}dx $$ $$ =0 +t \int^{\infty}_0 \frac{\sqrt{2\pi t}}{\sqrt{2\pi t}}e^{\frac{-x^2}{2t}} $$ $$ =\frac{t}{2} \sqrt{2 \pi t} $$
Therefore: $$ \sqrt{\frac{2}{\pi t}} \int_0^{\infty} x^2 e^{\frac{-x^2}{2t}} - \left(\sqrt{\frac{2t}{ \pi}} \right)^2\\ = \sqrt{\frac{2}{\pi t}} \frac{t}{2} \sqrt{2 \pi t} - \frac{2t}{\pi}\\ =t(1-\frac{2}{\pi}) $$