Multivariate normal distribution and marginal distribution

Hi everyone,

I have the following exercise:

Given $\displaystyle Y \sim \mathcal{N}_p(\mu,\Omega ) $,

a) Consider the following decomposition $\displaystyle Y=(Y_1,Y_2)^T, \mu=(\mu_1, \mu_2)^T, \Omega=( \Omega_{11}, \Omega_{12};\Omega_{21},\Omega_{22} )$ ( omega is supposed to be a matrix).

Show that conditional $\displaystyle Y_1 |(Y_2=y_2) $ is $\displaystyle \mathcal{N}_p ( \mu_1+\Omega_{12}\Omega_{22}^{-1}(y_2-\mu_2),\Omega_{11}-\Omega_{12}\Omega_{22}^{-1}\Omega_{21})$, where p is the dimension of $\displaystyle Y_1$.

This one, I have shown.

b) Let $\displaystyle a,b \in \mathbb{R}^n$. Find the conditional $\displaystyle X_1|X_2=x_2$ where $\displaystyle X_1=a^TY,X_2=b^TY$. In which case this distribution doesn't depend on $\displaystyle x_2$?

This one is causing me trouble. I stated by writing explicitly the product in $\displaystyle f_{X}(a^TY|b^TY=x_2)$ but it gets me nowhere.

Thank you in advance for taking time to answer my question.

Re: Multivariate normal distribution and marginal distribution

Hey sunmalus.

One suggestion I have is to use your results in a) and find the situation where the conditional distribution is free from any value of x2. (Think of when sigma_12 * sigma_22_inverse = 0)

Re: Multivariate normal distribution and marginal distribution

Well, with some linear transformation ( $\displaystyle (a^T, b^T)^T*Y=(X_1, X_2)$)I found the conditional distribution for b) but I have some atrocious matrix multiplication to do to find the exact form of my new matrix Omega in terms of a and b and the old Omega. I'm really wondering if there isn't another way. Plus my answer for last part is, as chiro said, when sigma_12 * sigma_22_inverse = 0. But this implies a lot of ugly sub cases... what am I missing, I don't think it should be as messy as what I've found.