# The law of absolute value of a standard Brownian motion

How can we easily compute $\mathbb{E} [ \left|W_t\right|]$, where $W = (W_t)_{t \geq 0}$ is the one dimensional standard Brownian motion (or wiener process)?

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## 1 Answer

The finite one-dimensional distribution of $W_t$ is normal with mean 0 and variance $t$. This means that $E[|W_t|]=\sqrt{\frac{2t}{\pi}}$ because $$\int_0^{\infty} 2 x \frac{1}{\sqrt{2 \pi t}} \exp \left( - \frac{1}{2 t} x^2 \right) \mathrm{d} x = \sqrt{\frac{2t}{\pi}}$$

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Hello, and if I want to compute $Var[|W_t|]$. How do I solve it?? – Salvattore Jul 2 '14 at 19:02
@Salvattore Use the second moment – dimebucker91 Mar 29 '15 at 3:36