Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

9.9. The Independence of Certain Quadratic Forms 563

Finally, using (9.9.1) and (9.9.3), we have the identity

AB={Γ′ 11 Λ 11 }Γ 11 Γ′ 21 {Λ 22 Γ 21 }. (9.9.6)

LetUdenote the matrix in the first set of braces. Note thatUhas full column
rank, so its kernel is null; i.e., its kernel consists of the vector 0 .LetVdenote the
matrix in the second set of braces. Note thatVhas full row rank, hence the kernel
ofV′is null.
For the proof then, supposeAB= 0 .Then

U[Γ 11 Γ′ 21 V]= 0.

Because the kernel ofUis null this implies each column of the matrix in the brackets
is the vector 0 ; i.e., the matrix in the brackets is the matrix 0. This implies that

V′[Γ 21 Γ′ 11 ]= 0.

In the same way, because the kernel ofV′is null, we haveΓ 11 Γ′ 21 = 0. Hence, by
(9.9.5), the random vectorsW 1 andW 2 are independent. Therefore, by (9.9.2) and
(9.9.4),Q 1 andQ 2 are independent.
Conversely, ifQ 1 andQ 2 are independent, then

{E[exp{t 1 Q 1 +t 2 Q 2 }]}−^2 ={E[exp{t 1 Q 1 }]}−^2 {E[exp{t 2 Q 2 }]}−^2 , (9.9.7)

for (t 1 ,t 2 )inanopenneighborhoodof(0,0). Note thatt 1 Q 1 +t 2 Q 2 is a quadratic
form inXwith symmetric matrixt 1 A+t 2 B. Recall that the matrixΓ 1 is orthogonal
and hence has determinant±1. Using this and Theorem 9.8.2, we can write the left
side of (9.9.7) as

E−^2 [exp{t 1 Q 1 +t 2 Q 2 }]=|In− 2 t 1 A− 2 t 2 B|

= |Γ′ 1 Γ 1 − 2 t 1 Γ′ 1 Λ 1 Γ 1 − 2 t 2 Γ′ 1 (Γ 1 BΓ′ 1 )Γ 1 |

= |In− 2 t 1 Λ 1 − 2 t 2 D|, (9.9.8)

where the matrixDis given by

D=Γ 1 BΓ′ 1 =

[

D 11 D 12

D 21 D 22

]

, (9.9.9)

andD 11 isr×r. By (9.9.2), (9.9.3), and Theorem 9.8.2, the right side of (9.9.7)

canbewrittenas

{E[exp{t 1 Q 1 }]}−^2 {E[exp{t 2 Q 2 }]}−^2 =

{r

∏

i=1

(1− 2 t 1 λi)

}

|In− 2 t 2 D|. (9.9.10)

This leads to the identity

|In− 2 t 1 Λ 1 − 2 t 2 D|=

{r

∏

i=1

(1− 2 t 1 λi)

}

|In− 2 t 2 D|, (9.9.11)