# Conditional Variance

• January 14th 2011, 01:21 PM
reflex
Conditional Variance
Hi,

In this proof http://onlinecourses.science.psu.edu/stat414/node/118 I don't understand why the VAR(Y|X) (VAR(Y|X=x) is constant and can be moved outside the integral, isn't it being integrated over x?

Thanks!
• January 14th 2011, 05:17 PM
Guy
It is assumed that $\sigma_{Y|X}$ is constant, i.e. it does not depend on $x$. See (3) at the beginning of the notes.

Now, (3) follows immediately if $(X, Y)$ is bivariate normal i.e. you can prove (3). If, on the other hand, we aren't assuming a bivariate normal, then the proof is fine and (3) is the key assumption that lets you pull it out of the integral.
• January 14th 2011, 05:37 PM
reflex
Thanks, I see (the notation confused me).

How generally would I go about proving (3) knowing it's bivariate normal?
• January 15th 2011, 07:07 AM
Guy
Quote:

Originally Posted by reflex
Thanks, I see (the notation confused me).

How generally would I go about proving (3) knowing it's bivariate normal?

I actually happen to have a proof sitting on my computer already; it derives the conditional distribution of Y|X:

http://i56.tinypic.com/2j0yw0m.jpg

Some explanation may be in order if you haven't seen something like this before. Essentially the idea is to show that the conditional pdf is proportional (for fixed x, letting y vary) to the pdf of something you know. Then, because they both have to integrate to 1, you conclude that they are equal. (3) follows from this because $\sigma^2 _Y (1 - \rho^2)$ doesn't depend on $X$.
• January 15th 2011, 07:34 AM
reflex
Wow, thanks! I wasn't expecting the whole proof!

Thanks again!