Show ... is chi square distributed

Hendrik

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Nov 28, 2014
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8
Hi,

I cant figure out how to prove the following:
Let \(\displaystyle B\, :\, k\, \times\, r\) have full column rank and \(\displaystyle Y\, \sim\, N_k (\mu,\, V).\) Show that:

. . .\(\displaystyle (Y\, -\, \mu )'\, B\, (B' VB)^{-1}\, B'\, (Y\, -\, \mu)\, \sim\, \chi_r^2\)

What I got so for:
Let \(\displaystyle Z\, =\, V^{-1/2}\, (Y\, -\, \mu).\) Then \(\displaystyle Z\, \sim\, N(0,\, I).\) Thus, \(\displaystyle Z'\, Z\, =\, \left( V^{-1/2}\, (Y\, -\, \mu) \right)' \, V^{-1/2}\, (Y\, -\, \mu)\, \sim\, \chi_k^2\) Is this even true by the way?


This is quite similar to what is given. The difference is caused by the \(\displaystyle B\) matrices. I don't know how to work with those. Also, the number of degrees of freedom should equal \(\displaystyle r\) (the rank of \(\displaystyle B\)) and not \(\displaystyle k\). Why?

In \(\displaystyle B\, (B' V B)^{-1}\, B'\) I recognize \(\displaystyle B\, (B' B)^{-1}\, B'\) which is a projection matrix, but I don't know if and how to do something with that.

I hope someone can help me out!
 
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Until now no one has replied to my thread. If something is not clear enough, please let
me know and I will give additional information (At least, If I can..).
 
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