# expected value

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Oct 15th 2011, 01:01 PM
Duke
Re: expected value
so i need to integrate y^3*(e^-(y^2/2v))/(2piv)^0.5 between + and - infinity and show its 0. How to compute the integral?
• Oct 15th 2011, 01:43 PM
CaptainBlack
Re: expected value
Quote:

Originally Posted by Duke
so i need to integrate y^3*(e^-(y^2/2v))/(2piv)^0.5 between + and - infinity and show its 0. How to compute the integral?

You don't, you look at the symmetry/antisymmetry about y=0.

Call the integrand I(y) so:

I(y)=y^3*(e^-(y^2/2v))/(2piv)^0.5

Now what is I(-y) in relation to I(y)?

CB
• Oct 16th 2011, 01:22 AM
Duke
Re: expected value
I(-y)= -I(y) due to the y^3.
• Oct 16th 2011, 01:59 AM
CaptainBlack
Re: expected value
Quote:

Originally Posted by Duke
I(-y)= -I(y) due to the y^3.

Then the integral we may write

$\int_{-\infty}^{\infty} I(y)dy =\int_{-\infty}^0 I(y) dy + \int_{0}^{\infty} I(y) dy$= ..

or you may just know what that means for the integral.

CB
• Oct 16th 2011, 03:54 AM
Duke
Re: expected value
=0
• Oct 16th 2011, 05:49 AM
CaptainBlack
Re: expected value
Quote:

Originally Posted by Duke
=0

Yes.

CB
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12