Calculating absolute value of Brownian motion

Question

I am trying to calculate $\mathbb{E}\mid B_{t} \, \mid$ in terms of $t$, but am really stuck. Any toughts?

Answer 1

Hint: This is the expected value of the absolute value of a $\mathcal{N}(0,t)$ random variable. For any $t > 0$, the density of $B_t$ is $f_t(x) = \frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}$, so $$\Bbb{E}\left[|B_t|\right] = \int_{-\infty}^{\infty}|x|f_t(x)\, dx = 2\int_{0}^{\infty}x\cdot\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}} \, dx,$$ since the integrand is even.

To compute this integral, try a substitution like $u=\frac{x^2}{2t}$ (so $du = \frac{x}{t}\, dx$).