1. ## Normal Distribution

If X is a normal random variable with mean $\mu$ and variance $\sigma^2$, show that:

$E(|X-\mu|)=\sqrt{\frac{2}{\pi}}\sigma$

2. aha!

its easier than i thought. make that substitution and you will get an integral you can actually evaluate directly.

we started from
$2 \int_{\mu}^{\infty} (x-\mu) \frac{1}{\sqrt{2\pi\sigma^2}} \exp \left( \frac{-(X-\mu)^2}{2\sigma^2} \right) dx$

sub $v=(X-\mu)^2$
so $dx = \frac{dv}{2\sqrt{v}}$

To get

$2 \int_{0}^{\infty} \frac{\sqrt{v}}{2\sqrt{v}} * \frac{1}{\sqrt{2\pi\sigma^2}} * \exp(\frac{-v}{2\sigma^2}) dv$

$= \frac{1}{\sqrt{2\pi\sigma^2}} \int_{0}^{\infty} \exp(\frac{-v}{2\sigma^2}) dv$

you can evaluate that integral using standard methods. hopefully the right answer drops out!
(reminder: $e^{-\infty} = 0$ and $e^0 = 1$

edit i did it to check:
Evaluating the integral gives

$\frac{2\sigma^2}{\sqrt{2\pi\sigma^2}} \left( 0 - - 1 \right)$

Which simplifies to the required answer.

phew!

not bad for an economist, if i do say so myself....even if it took about 10 tries