Question about $E(|Z|)$ at Normal distribution

Question

$Z$ is a standard normal variable.
How do I calculate $E(|Z|)$?
($E(Z)=0$).

Thank you!

Answer 1

Write $|z| = -z \cdot 1_{(-\infty,0)}+ z \cdot 1_{(0,\infty)}$.

The distribution is symmetric about zero, so $E|z| = 2 \int_0^\infty z P(z) dz$, where $P(z) = {1 \over \sigma \sqrt{ 2 \pi }} e^{- { z^2 \over 2 \sigma^2 }}$.

Since $P'(z) = (- { 1 \over \sigma^2 }) z P(z)$, we see that $E|z| = - { 2 \sigma^2} \int_0^\infty P'(z) dz = { -2 \sigma^2}(\lim_{z \to \infty } P(z)-P(0)) = { 2 \sigma^2}P(0) = \sqrt{ 2 \over \pi} \sigma$.

Answer 2

$\int\left|x\right|\phi\left(x\right)dx=2\int_{0}^{\infty}x\phi\left(x\right)dx=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}xe^{-\frac{1}{2}x^{2}}dx$

Now exploit substitution $u=\frac{1}{2}x^{2}$

Answer 3

$$ E[ |Z| ] = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}}|z| e^{-z^2/2}dz = 2\int_{0}^{+\infty} \frac{1}{\sqrt{2\pi}}ze^{-z^2/2}dz$$

From here, this integral is straightforward to compute.