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!
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 $\displaystyle \sigma_{Y|X}$ is constant, i.e. it does not depend on $\displaystyle x$. See (3) at the beginning of the notes.
Now, (3) follows immediately if $\displaystyle (X, Y)$ 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.
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 $\displaystyle \sigma^2 _Y (1 - \rho^2)$ doesn't depend on $\displaystyle X$.