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Let $X$ be a $N_n(\mu, \Sigma)$ random variable, where both $\mu \in \mathbb{R}^n$ and the positive definite $n \times n$ matrix $\Sigma$ are known. Given an $n \times n$ matrix $A$, what is the most general result regarding the distribution of $X^TA X$? To state it in an operational way, what are the procedures to follow to compute the distribution of $X^TAX$ when we receive matrix $A$?

The result I know about is that $A\Sigma$ is idempotent if and only if $X^TAX$ follows a non-central chi-squared distribution $\chi_r^2(\frac{1}{2}\mu^TA\mu)$, where $r = \text{rank}(A)$. This result follows directly from the moment generating function. Namely, write $M(t)$ for the moment generating function $\mathbb{E}[e^{tX^TAX}]$, then :

\begin{equation*} M(t) = [\text{det}(I_n - 2t A\Sigma)]^{-\frac{1}{2}}\exp{\left\{ -\frac{1}{2}\mu^T [I_n - (I_n - 2tA\Sigma)^{-1}]\Sigma^{-1}\mu\right \}} \end{equation*}

So that, if $A\Sigma$ is idempotent, one might further simplify this $M(t)$ until one recognizes this as non-central chi-squared .

What do I do if $A\Sigma$ is not idempotent? It is not clear how to read a meaningful distribution from $M(t)$. Maybe I have to use the characteristic function and then inverse Fourier transform ?

Another route I hope to be useful is to use Cochran's theorem trying to get a decomposition of $A$. However, it is not clear how to do such a thing for a generic matrix $A$ .

温泽海
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1 Answers1

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It'll be a (possibly infinite) linear combination of non-central $\chi^2$'s and as you noted, numerically inverting the characteristic function is one way to practically compute the actual probabilities.

Some (old) refs:

Ruben 1962

Ruben 1963

Press 1966

Harville 1971

Zack Fisher
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  • I do not see why it would be linear combination of non-central chi squares. Care to elaborate? – 温泽海 Jul 29 '24 at 13:50
  • @温泽海 Added a few references. I couldn't remember the details now after so many years. But it could be interesting to re-read them sometime. – Zack Fisher Jul 29 '24 at 14:06