高斯随机变量绝对值的期望

【结论】
对于均值为 0 的正态分布， $E {| X |} = \sqrt{2 / π} σ$

\begin{aligned} E [| X |] = & \int_{- \infty}^{\infty} | x | \frac{1}{\sqrt{2 π} σ} \exp {- \frac{(x - μ)^{2}}{2 σ^{2}}} d x \\ = & (y = \frac{x - μ}{σ}) \int_{- \infty}^{\infty} | μ + σ y | \frac{1}{\sqrt{2 π}} \exp {- \frac{y^{2}}{2}} d y \\ = & \int_{- \frac{μ}{σ}}^{\infty} (μ + σ y) \frac{1}{\sqrt{2 π}} \exp {- \frac{y^{2}}{2}} d y - \int_{- \infty}^{- \frac{μ}{σ}} (μ + σ y) \frac{1}{\sqrt{2 π}} \exp {- \frac{y^{2}}{2}} d y \\ = & μ (\int_{- \frac{μ}{σ}}^{\infty} \frac{1}{\sqrt{2 π}} \exp {- \frac{y^{2}}{2}} d y - \int_{- \infty}^{- \frac{μ}{σ}} \frac{1}{\sqrt{2 π}} \exp {- \frac{y^{2}}{2}} d y) + \\ \frac{σ}{\sqrt{2 π}} (\int_{- \frac{μ}{σ}}^{\infty} y \exp {- \frac{y^{2}}{2}} d y - \int_{- \infty}^{- \frac{μ}{σ}} y \exp {- \frac{y^{2}}{2}} d y) \\ = & μ [Φ (\frac{μ}{σ}) - Φ (- \frac{μ}{σ})] + \frac{σ}{\sqrt{2 π}} {[\exp {- \frac{y^{2}}{2}} |}_{\infty}^{- \frac{μ}{σ}} - \exp {- \frac{y^{2}}{2}} |_{- \frac{μ}{σ}}^{- \infty}] \\ = & μ [2 Φ (\frac{μ}{σ}) - 1] + \frac{2 σ}{\sqrt{2 π}} \exp {- \frac{μ^{2}}{2 σ^{2}}} . \end{aligned}