I've been working on a list of normal distribution exercises that so far had been pretty straightforward, mostly with applications and consultations on tables. This last one has got me, so far.
If $X$ is $N\left( {\mu ,{\sigma ^2}} \right)$ distributed, show that ${\rm E}\left( {\left| {X – \mu } \right|} \right) = \sigma \sqrt {{2 \over \pi }} $.
My first idea was to go from the definition of expected value on a normal distribution: $${\rm E}\left( X \right) = {1 \over {\sigma \sqrt {2\pi } }}\int_{ – \infty }^\infty {x{e^{ – {{{{(x – \mu )}^2}} \over {2{\sigma ^2}}}}}dx} $$ and then trying to plug in the $|X – \mu |$ factor where originally there was only an $x$, but I have no clue if I'm allowed to do that.
Anyway, following this line of thought I ended up with a non-convergent integral:
$$\eqalign{
& {\rm E}[|X – \mu |] = {1 \over {\sigma \sqrt {2\pi } }}\int_{ – \infty }^\infty {\left| {x – \mu } \right|{e^{ – {{{{(x – \mu )}^2}} \over {2{\sigma ^2}}}}}dx} \cr
& t = {{x – \mu } \over {\sqrt 2 \sigma }},{\rm{ }}dt = {1 \over {\sqrt 2 \sigma }}dx \cr
& {\rm E}[|X – \mu |] = {1 \over {\sqrt 2 \sigma \pi }}\int_{ – \infty }^\infty {2\sigma |t|{e^{{t^2}}}} \sqrt 2 \sigma dt \cr
& {\rm E}[|X – \mu |] = {{2\sigma } \over {\sqrt \pi }}\int_{ – \infty }^\infty {|t|{e^{{t^2}}}} \cr} $$
I'm not worried about getting a full answer, I'd love to get insight to try better approaches right now. Thank you.
Best Answer
$$\int_{\Bbb R}|t|e^{-t^2}dt=\int_0^\infty 2te^{-t^2}dt=1.$$