Distribution of quadratic forms of normal variables

Asked Feb 05 '15 at 13:48

Active Feb 05 '15 at 13:54

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Suppose $X\sim \bf{N}_m(\bf{\mu}, \bf{\Sigma})$ where the covariance matrix $\bf \Sigma$ is positive definite. Show that the distribution of $X^T{\bf{B}}X \sim \chi^2(\bf{tr(B\Sigma)})$where $\bf B$ satisfies

$\bf{B\Sigma B}=\bf{B}$
$\bf\mu^TB\mu=0$

Now, previously I had done this result, when $\Sigma$ was replaced by $\bf{I}$ and $B$ was idempotent. This one is where I am stuck. I have found that $\bf B\Sigma$ and $\bf \Sigma B$ are idempotent. Now, I tried to find some $U\sim \bf{N_{\text{tr}(B\Sigma)}}(0, I)$ such that $U^TU=X^T{\bf{B}}X$ then we would be done, but I can't seem to find such an $U$. Can someone help me? Thanks.

edited Feb 05 '15 at 13:54

asked Feb 05 '15 at 13:48

shadow10

2

If this is a homework problem, please add a `self-study` tag. – Richard Hardy Feb 05 '15 at 13:51
You might make some progress with your strategy by writing $\Sigma=TT^\prime$ and considering $T^\prime B T$. – whuber Feb 05 '15 at 17:29
@whuber I have tried these days to proceed with your idea but nothing has come my way, could you tell me how to do it? Thanks a lot. – shadow10 Feb 08 '15 at 13:54

Distribution of quadratic forms of normal variables

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