# Thread: expected value

1. ## expected value

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

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

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

2. ## Re: expected value

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

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

$= 3u^3-3uv+2u^3+E(X^3)$
Some words of explanation of what you are trying to do would be a great help.

CB

3. ## Re: expected value

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

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

$= 3u^3-3uv+2u^3+E(X^3)$
Consider the random variable $Y=X-u$ which is $\sim N(0,v)$

So:

$E((X-u)^3)=E(Y^3)$

Now the rest of the argument depends only on the parity of the normal density and of $y^3$.

CB

4. ## Re: expected value

1) Where did you use symmetry?

2) Are you SURE $E[X^{2}] = \nu$?

5. ## Re: expected value

Originally Posted by TKHunny
1) Where did you use symmetry?

2) Are you SURE $E[X^{2}] = \nu$?
That was a hint given in the question

No $v=E[X^2]-(E[X])^2$ I have just written down final simplification. How do you find E(X^3) in terms of u,v?

6. ## Re: expected value

Oh, so $E[X^{2}] = \nu$ IF $E[X] = 0$. That might be related to Symmetry.

As for $E[X^{3}]$, think really hard about the "Parity" that CaptainBlack is talking about. We know already that $E[X] = 0$. Hmmm... More Symmetry?

7. ## Re: expected value

E(X) is not 0. E(X)=u. I thought parity referered to odd and even

8. ## Re: expected value

...and since we don't care about X, but $X-\mu$, 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?

9. ## Re: expected value

Originally Posted by Duke
E(X) is not 0. E(X)=u. I thought parity referered to odd and even
Even and odd functions - Wikipedia, the free encyclopedia

CB

10. ## Re: expected value

Originally Posted by Duke
E(X) is not 0. E(X)=u. I thought parity referered to odd and even
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

$E[(X-\mu)^3)]=E(Y^3)$

and that $Y \sim N(0,v)$

Now write out $E(Y^3)$ 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

11. ## Re: expected value

The link suggest I integrate Y^3 * pdf which from wikipedia is some horrible function.

12. ## Re: expected value

Originally Posted by Duke
The link suggest I integrate Y^3 * pdf which from wikipedia is some horrible function.
Nobody asked you to do the integral just to write it out.

================================================== =

The normal density function:

$f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{- \frac{(x-\mu)^2}{2\sigma^2}}$

Making this specific to the RV $Y$:

$f(y)=\frac{1}{\sqrt{2\pi v}}e^{-\frac{y^2}{2v}}$

CB

13. ## Re: expected value

how does this help?

14. ## Re: expected value

Originally Posted by Duke
how does this help?
Have you done what was asked? Do you have it in front of you scribbled down on a piece of paper?

CB

15. ## Re: 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..

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