Thread: 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?

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

I(-y)= -I(y) due to the y^3.

Originally Posted by Duke

I(-y)= -I(y) due to the y^3.

Then the integral we may write

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

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

CB

=0

Originally Posted by Duke

=0

Yes.

CB