If X is a normal random variable with mean $\displaystyle \mu$ and variance $\displaystyle \sigma^2$, show that:
$\displaystyle E(|X-\mu|)=\sqrt{\frac{2}{\pi}}\sigma$
aha!
its easier than i thought. make that substitution and you will get an integral you can actually evaluate directly.
we started from
$\displaystyle 2 \int_{\mu}^{\infty} (x-\mu) \frac{1}{\sqrt{2\pi\sigma^2}} \exp \left( \frac{-(X-\mu)^2}{2\sigma^2} \right) dx$
sub $\displaystyle v=(X-\mu)^2$
so $\displaystyle dx = \frac{dv}{2\sqrt{v}}$
To get
$\displaystyle 2 \int_{0}^{\infty} \frac{\sqrt{v}}{2\sqrt{v}} * \frac{1}{\sqrt{2\pi\sigma^2}} * \exp(\frac{-v}{2\sigma^2}) dv$
$\displaystyle = \frac{1}{\sqrt{2\pi\sigma^2}} \int_{0}^{\infty} \exp(\frac{-v}{2\sigma^2}) dv$
you can evaluate that integral using standard methods. hopefully the right answer drops out!
(reminder: $\displaystyle e^{-\infty} = 0$ and $\displaystyle e^0 = 1$
edit i did it to check:
Evaluating the integral gives
$\displaystyle \frac{2\sigma^2}{\sqrt{2\pi\sigma^2}} \left( 0 - - 1 \right)$
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