Question

Suppose X is a normal random variable with mean 0 and variance 4. Then the mean of the absolute value of X is

• A. $\frac{1}{\sqrt{2\pi }}$
• B. $\frac{2\sqrt{2}}{\sqrt{\pi }}$
• C. $\frac{2\sqrt{2}}{\pi }$
• D. $\frac{2}{\sqrt{\pi }}$

Answer 1

The correct option is B $\frac{2\sqrt{2}}{\sqrt{\pi }}$

For normal distribution:
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{(x-\mu {)}^2}{2{\sigma }^2}}=\frac{1}{2\sqrt{2\pi }}{e}^{\frac{-x^2}{8}}$

$\mu -\text{mean},\sigma -\text{standard deviation}$ & given $\mu =0,{\sigma }^2=4$

mean of absolute value of x = E(|x|) =
$={\int }_{-\infty }^{\infty }|x|f\left(x\right)dx=2{\int }_0^{\infty }|x|f\left(x\right)dx$
$(\because \text{even function})$

Now put $\frac{x^2}{8}=y$

$\Rightarrow xdx=4dy$

$=\frac{1}{2\pi }{\int }_0^{\infty }{e}^{-y}\left(4dy\right)$

$=\frac{2\sqrt{2}}{\sqrt{\pi }}{\int }_0^{\infty }{e}^{-y}dy$

$=\frac{2\sqrt{2}}{\sqrt{\pi }}$