Question

For normal distribution with mean zero and variance σ² show that:

E(|x|)= (√2/x)σ

Answer 1

The probability density function (PDF) for a normal $X\sim (\mu, \sigma^2 )$ is:

$$f_X(x)=\dfrac{1}{\sigma \sqrt{2\pi }}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}$$

For normal distribution with mean zero and variance σ²

$$f_X(x)=\dfrac{1}{\sigma \sqrt{2\pi }}e^{-\frac{1}{2}(\frac{x}{\sigma})^2}$$

$$E(|X|)=\displaystyle\int_{-\infin}^{\infin}|x|f_X(x)dx$$

$$=2\displaystyle\int_{0}^{\infin}x\dfrac{1}{\sigma \sqrt{2\pi }}e^{-\frac{1}{2}(\frac{x}{\sigma})^2}dx$$

Let $z=\dfrac{x}{\sigma}$

Then $dx=\sigma dz$

$$E(|X|)=\displaystyle\int_{-\infin}^{\infin}|x|f_X(x)dx$$

$$=2\displaystyle\int_{0}^{\infin}x\dfrac{1}{\sigma \sqrt{2\pi }}e^{-\frac{1}{2}(\frac{x}{\sigma})^2}dx$$

$$=\displaystyle\int_{0}^{\infin}\dfrac{2\sigma}{ \sqrt{2\pi }}ze^{-\frac{1}{2}z^2}dz$$

$$=-\dfrac{2\sigma}{ \sqrt{2\pi }}\big[e^{-\frac{1}{2}z^2}\big]\begin{matrix} \infin \\ 0 \end{matrix}=\sqrt{\dfrac{2}{\pi}}\cdot\sigma$$

$$E(|X|)=\sqrt{\dfrac{2}{\pi}}\cdot\sigma$$