If X-N(u,v) show that E((X-u)^3)=0. Hint: use symmetry.

$\displaystyle E(X^3-uX^2-2uX^2+2Xu^2+xu^2-u^3)$

$\displaystyle = 3u^3-3uv+2u^3+E(X^3)$

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- Oct 15th 2011, 04:34 AMDukeexpected value
If X-N(u,v) show that E((X-u)^3)=0. Hint: use symmetry.

$\displaystyle E(X^3-uX^2-2uX^2+2Xu^2+xu^2-u^3)$

$\displaystyle = 3u^3-3uv+2u^3+E(X^3)$ - Oct 15th 2011, 05:27 AMCaptainBlackRe: expected value
- Oct 15th 2011, 05:30 AMCaptainBlackRe: expected value
- Oct 15th 2011, 05:32 AMTKHunnyRe: expected value
1) Where did you use symmetry?

2) Are you SURE $\displaystyle E[X^{2}] = \nu$? - Oct 15th 2011, 06:10 AMDukeRe: expected value
- Oct 15th 2011, 06:17 AMTKHunnyRe: expected value
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? - Oct 15th 2011, 06:29 AMDukeRe: expected value
E(X) is not 0. E(X)=u. I thought parity referered to odd and even

- Oct 15th 2011, 07:08 AMTKHunnyRe: expected value
...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? - Oct 15th 2011, 10:19 AMCaptainBlackRe: expected value
- Oct 15th 2011, 10:23 AMCaptainBlackRe: expected value
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 - Oct 15th 2011, 10:41 AMDukeRe: expected value
The link suggest I integrate Y^3 * pdf which from wikipedia is some horrible function.

- Oct 15th 2011, 10:47 AMCaptainBlackRe: expected value
Nobody asked you to do the integral just to write it out.

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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 - Oct 15th 2011, 11:03 AMDukeRe: expected value
how does this help?

- Oct 15th 2011, 11:24 AMCaptainBlackRe: expected value
- Oct 15th 2011, 12:01 PMSpringFan25Re: Show that question
**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..