Answer to Question #181935 in Statistics and Probability for ANJU JAYACHANDRAN

Question

For normal distribution with mean zero and variance 2

σ show that :E(X)=root of (2/pi) sigma

 

Answer 1

This is an elegant problem of symmetry of the normal probability distribution. 

First of all, let $Y=X−μ. \text{ So } D(Y)=D(X)=σ$ . Notice that the figure of the normal probability distribution which Y follows is symmetrical with respect to its expectation E(Y)=0. The probability distribution of |Y| can be viewed as the result of "adding" the probability of +y and −y together. Intuitively, the figure of probability distribution of |Y| is made by folding the left half into the right and they plus together so that each point of the right half doubles its original value.

$E(|X-\mu|)=E(|Y|)=\int_0^{\infty} 2\dfrac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}xdx}$

Let $t=\dfrac{x}{\sigma}$ . SO we have-

$E(|Y|)=\sigma\sqrt{\dfrac{2}{\pi}}\int_0^{\infty}e^{-\frac{t^2}{2}tdt}$

               $=\sigma\sqrt{\dfrac{2}{\pi}}$