1
$\begingroup$

I am looking for ways to prove that the moment generating function of $X'AX$ given that $X \sim N(\vec{\mu}, \vec{\Sigma})$ and $A$ is symmetric is defined as:

$$M_{X'AX}(\vec{t})= \frac{1}{|I-2tA\Sigma|^{\frac{1}{2}}}e^{-\frac{1}{2}\mu'[I-(I-2tA\Sigma)^{-1}]\Sigma^{-1}\mu} $$

I found similar texts stating this property but without a proof. But usually I have to show that $A \Sigma$ is symettric idempotent and would already know that $Y'AY$ is non central chi square.

$\endgroup$
  • 2
    $\begingroup$ See math.stackexchange.com/questions/442472/… $\endgroup$ – kjetil b halvorsen Mar 8 at 12:08
  • 1
    $\begingroup$ You should reference or link the texts where you saw this. $\endgroup$ – kjetil b halvorsen Mar 8 at 23:50
  • $\begingroup$ One answer is stats.stackexchange.com/questions/262604/… but there the mgf is very different in form. $\endgroup$ – kjetil b halvorsen Mar 8 at 23:57
  • $\begingroup$ Answered this question with Graybill as reference (Man, the epub on genlib sucks). You guys can PM me for the answer if needed. $\endgroup$ – Xorion 1997 May 14 at 14:42
  • $\begingroup$ @kjetilbhalvorsen thanks for the links but most of it are unhelpful and even deviated me from the answer. The answer just really required bruteforcing matrix multiplications and kerneling. $\endgroup$ – Xorion 1997 May 14 at 14:47
2
$\begingroup$

I am quite sure @kjetil b halvorsen's answer at What is the moment generating function of the generalized (multivariate) chi-square distribution? reduces to the expression of MGF in this post on simplification.

A direct proof is not difficult either when $\Sigma$ is assumed to be positive definite.

The proof simply relies on the fact that for a symmetric positive definite matrix $B$, we have from the multivariate normal density

$$\int_{\mathbb R^p}\exp\left[-\frac12(x-\mu)' B^{-1}(x-\mu)\right] dx=(2\pi)^{p/2}(\det B)^{1/2}$$

Or,

$$\int_{\mathbb R^p}\exp\left[-\frac12 x' B^{-1}x+\mu' B^{-1}x-\frac12 \mu' B^{-1}\mu\right] dx=(2\pi)^{p/2}(\det B)^{1/2}$$

Taking $b'=\mu' B^{-1}$, this is same as

$$\int_{\mathbb R^p}\exp\left[-\frac12x' B^{-1}x+b' x\right]dx=(2\pi)^{p/2}(\det B)^{1/2}\exp\left(\frac12 b' B b\right) \tag{*}$$

For symmetric $A$,

\begin{align} M_{X' A X}(t)&=E\left[e^{tX' AX}\right] \\&=\frac1{(2\pi)^{p/2}(\det \Sigma)^{1/2}}\int_{\mathbb R^p} \exp(tx' Ax)\cdot \exp\left[-\frac12(x-\mu)' \Sigma^{-1}(x-\mu)\right] dx \\\\&=\frac{\exp(-\frac12 \mu'\Sigma^{-1}\mu)}{(2\pi)^{p/2}(\det \Sigma)^{1/2}}\int_{\mathbb R^p} \exp\left[-\frac12 x'(I-2tA\Sigma)\Sigma^{-1}x+\mu'\Sigma^{-1}x\right] dx \end{align}

We have $(I-2tA\Sigma)\Sigma^{-1}=\Sigma^{-1}-2tA$, which is assumed positive definite (it is already symmetric) for sufficiently small $|t|$. The MGF is now precisely of the form $(*)$.

Taking $B=(\Sigma^{-1}-2tA)^{-1}=\Sigma(I-2tA\Sigma)^{-1}$ and $b'=\mu'\Sigma^{-1}$ finally gives

$$M_{X'AX}(t)=(\det(I-2tA\Sigma))^{-1/2}\exp\left\{-\frac12 \mu'[I-(I-2tA\Sigma)^{-1}]\Sigma^{-1}\mu \right\}\,,$$

whenever the MGF exists. This also generalizes @whuber's answer here.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.