What is the value of $E[|X|]$?

Question

Let X be a zero mean unit variance Gaussian random variable.What is the value of $E[|X|]$?

Answer 1

$$E(|X|) = \frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-x^2/2}dx= \frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}xe^{-x^2/2}dx=\sqrt{\frac{2}{\pi}}$$

Answer 2

Write the expectancy:

$$ E[|X|]=\int_{-\infty}^{+\infty} |x|\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}dx \\ =\int_{0}^{+\infty}x\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}dx + \int_{-\infty}^{0}(-x)\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}dx \\ =\int_{0}^{+\infty}x\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}dx + \int_{\infty}^{0}x\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}(-dx) \\ =2\int_{0}^{+\infty}x\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}dx \\ =\sqrt{\frac{2}{\pi}} $$

First line -> second line : Chasles relation and setting the absolute value to -x in the negative domain of integration.

Second -> third line : change of variable in the second integral y=-x