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I was trying to follow the proof of Theorem 4.2.1 of F.A. Graybill's Theory and Application of the Linear Model (1976). Here is an excerpt of the proof:

$$\eqalignno{\displaystyle m_{\mathbf{Y'}\mathbf{Y}}(t)&=\mathscr{E}[e^{t(\mathbf{Y'}\mathbf{Y})}]\cr &=(2\pi)^{-n/2}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}e^{t(\mathbf{Y'}\mathbf{Y})}e^{-\frac{1}{2}(\mathbf{y}-\mathbf{\mu})'(\mathbf{y}-\mathbf{\mu})}d\mathbf{y}\cr &=(2\pi)^{-n/2}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}\exp(-\frac{1}{2}[(1-2t)\mathbf{y'}\mathbf{y}-2\mathbf{y'}\mathbf{\mu}+\mathbf{\mu'\mu}])d\mathbf{y} } $$

The integral exists for all $t$ such that $t<\frac{1}{2}$, and we use Theorem 1.10.1 to evaluate it. We obtain

$$m_{\mathbf{Y'}\mathbf{Y}}(t)=(1-t)^{-n/2}\exp\left(\dfrac{2t\lambda}{1-2t}\right),\, t<\frac{1}{2}$$

Theorem 1.10.1 states that if $a_0$ and $b_0$ are scalar constants, $\mathbf{a}$ an $n\times 1$ vector of constants, $\mathbf{b}$ an $n\times 1$ vector of constants, $\mathbf{A}$ an $n\times n$ symmetric matrix of constants, and $\mathbf{B}$ a positive definite matrix of constants, then $\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\cdots\int_{-\infty}^{\infty}(\mathbf{x'Ax+x'a}+a_0)e^{(\mathbf{x'Bx+x'b}+b_0)}d\mathbf{x}=\frac{1}{2}\pi^{n/2}\lvert B\rvert^{-1/2}e^{(1/4)\mathbf{b'B^{-1}b}-b_0}[tr(\mathbf{AB^{-1})-b'B^{-1}a}+\frac{1}{2}\mathbf{b'B^{-1}AB^{-1}b}+2a_0]$

I can follow the part where we substitute the following in Theorem 1.10.1: $\mathbf{x}=\mathbf{y}$, $\mathbf{B}=(\frac{1}{2}-t)I, \mathbf{b}=-\mathbf{\mu}$ and $b_0=\dfrac{\mathbf{\mu'\mu}}{2}$, but I can't follow how $\mathbf{A}$, $\mathbf{a}$ and $a_0$ were derived apart from the fact that $\mathbf{x'Ax+x'a}+a_0=1$.

Question: Can anybody help me work this out by giving me tips or examples where the equation in Theorem 1.10.1 is applied, or is there another approach?

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1 Answer 1

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The integral looked scary but when I got to really work through, it was not difficult at all. But it was lengthy. A minor correction was $\mathbf{B}=\frac{1}{2}(1-2t)I$. Now it should not be difficult to see that $\mathbf{A}=\mathbf{0}$, $\mathbf{a}=\mathbf{0}$, and $a_0=1$. From there, substitution is straightforward.

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