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. Well, with some linear transformation ( \(\displaystyle (a^T, b^T)^T*Y=(X_1, X_2)\)) and question a), 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 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.
Thank you in advance for taking time to answer my question.
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. Well, with some linear transformation ( \(\displaystyle (a^T, b^T)^T*Y=(X_1, X_2)\)) and question a), 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 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.
Thank you in advance for taking time to answer my question.
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