39.4. QUADRATIC FORMS 819

=

(1√

2πσ

)n ∫Rn

eσ2t xT Axσ2 e−

12x·xσ2 d−→x =

(1√

2πσ

)n ∫Rn

e−1

2σ2 xT (I−2σ2tA)xd−→x

Now let y ≡ UTx where U is an orthogonal matrix such that UT(I−2σ2tA

)U = D, a

diagonal matrix having all positive diagonal entries. We can get such a thing whenever |t|is small enough because then the expression I− 2σ2tA will have all positive eigenvalues.Now det(U) =±1 because it is orthogonal. Changing variables to y and using the changeof variables formula,

M (t) =

(1√

2πσ

)n ∫Rn

e−1

2σ2 yT UT (I−2σ2tA)Uyd−→y

=

(1√

2πσ

)n ∫Rn

e−1

2σ2 yT D(t)yd−→y

where D(t) is the diagonal matrix which has the positive eigenvalues λ2k (t) down the diag-

onal. Then the above expression splits into factors of the form

1√2πσ

∫R

e−1

2σ2 y2kλ

2k dyk

So let u = λ kyk and this becomes

1√2πσ

1λ k

∫R

e−1

2σ2 u2du =

1λ k

.

Hence,

M (t) =n

∏k=1

1λ k

=

(n

∏k=1

1

λ2k

)1/2

=

(1

det(D(t))

)1/2

=1

det(I−2σ2tA)1/2 .

Remember the determinant is the product of the eigenvalues of the matrix and I− 2σ2tAand D are similar so they have the same eigenvalues, namely the diagonal entries of D. SeeCorollary 11.4.5 to Schur’s theorem. This is stated as the following lemma.

Lemma 39.4.1 Let A be symmetric and let X1, · · · ,Xn be independent and n(0,σ2

). Then

the moment generating function of XT AX is M (t) = det(I−2σ2tA

)−1/2 for all |t| suffi-ciently small.

Now suppose you have two symmetric matrices A,B and independent n(0,σ2

)random

variables X1, · · · ,Xn. Then there are two random variables XT AX,XT BX and we wantto determine when these two are independent. Then using similar reasoning to the above,it follows that for |s| , |t| both small enough,

M (t,s)≡ E(exp(tXT AX+ sXT BX

))=

1

det(I−2σ2tA−2σ2sB)1/2

If AB = 0, then(I−2σ

2tA)(

I−2σ2sB)

= I−2σ2tA−2σ

2sB+4σ4tsAB

= I−2σ2tA−2σ

2sB