Thread: 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!

It is assumed that is constant, i.e. it does not depend on . See (3) at the beginning of the notes.

Now, (3) follows immediately if 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.

Thanks, I see (the notation confused me).

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

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:



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 doesn't depend on .

Wow, thanks! I wasn't expecting the whole proof!

Thanks again!