If X-N(u,v) show that E((X-u)^3)=0. Hint: use symmetry.
$\displaystyle E(X^3-uX^2-2uX^2+2Xu^2+xu^2-u^3)$
$\displaystyle = 3u^3-3uv+2u^3+E(X^3)$
Oh, so $\displaystyle E[X^{2}] = \nu$ IF $\displaystyle E[X] = 0$. That might be related to Symmetry.
As for $\displaystyle E[X^{3}]$, think really hard about the "Parity" that CaptainBlack is talking about. We know already that $\displaystyle E[X] = 0$. Hmmm... More Symmetry?
...and since we don't care about X, but $\displaystyle X-\mu$, we did get somewhere.
"Parity" has other meanings.
In this case, if you cube a negative number, what do you get? Positive or Negative?
If you cube two numbers, does their relationship change? In other words, if a > b, for a, b > 0, what can we say about a^3 vs. b^3? Same or different?
This thread is going nowhere because you are not taking much notice of what people are telling you, now:
Go back to post #3 Where you were told that
$\displaystyle E[(X-\mu)^3)]=E(Y^3)$
and that $\displaystyle Y \sim N(0,v)$
Now write out $\displaystyle E(Y^3)$ as an integral and post what you get here.
As a further hint you will find the definition of the expectation of a function of a continuous RV in Definition 1 >>here<<
CB
Nobody asked you to do the integral just to write it out.
================================================== =
The normal density function:
$\displaystyle f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{- \frac{(x-\mu)^2}{2\sigma^2}}$
Making this specific to the RV $\displaystyle Y$:
$\displaystyle f(y)=\frac{1}{\sqrt{2\pi v}}e^{-\frac{y^2}{2v}}$
CB
edit: removed, for some reason i didn't notice the earlier 10 replies on the thread, if anyone used my answer i think there was a sloppy change in my definition of f(x), be warned..