ABSTRACT

Proof. Since S is symmetric and positive definite there exists a nonsingular matrix C such that S=CC'. Let Y=C-1X. Obviously Y has a p-variate normal distribution with mean v=C-1µ and covariance matrix I (identity matrix). From (6.8) we get

where Bi=C'AiC. Since C is nonsingular, rank (Ai)=rank(Bi), i=1,…, k. Obviously the theorem will be proved if we show that Y'BiY, i=1,…, k, are independently

distributed noncentral chi-squares if and only if in which case

Let us suppose that Y'BiY, i=1,…, k, are independently distributed

as Then is distributed as noncentral chi-square

Since Y'Y is distributed as and (6.9) holds, it follows

from the uniqueness of the characteristic function that and

which proves the necessity part of the theorem. To prove the

sufficiency part of the theorem let us assume that Since Qi is a quadratic form in Y of rank pi (rank of Bi) by Theorem 1.5.8, Qi can be expressed as

where the Zij are linear functions of Y1,…, Yp. Let

be a vector of dimension Then

where ? is a diagonal matrix of dimension p×p with diagonal elements +1 or - 1. Let Z=AY be the linear transformation that transforms the positive definite quadratic form Y'Y to Z'? Z. Since

for all values of Y we conclude that A'? A=I. In other words, A is nonsingular. Thus Z'? Z is positive definite and hence ? =I, A'A=I. Since A is orthogonal and Y has a pvariate normal distribution with mean v and covariance matrix I, the components of Z are independently normally distributed with unit variance. So Qi(i=1,…, k) are independently distributed chi-square random variables with pi degrees of freedom and noncentrality parameter v'Biv, i=1,…, k (see

Exercise 6.1). But Y'Y is distributed as Therefore

Q.E.D. Theorem 6.2.2. Let X=(X1,…, Xp)

' be normally distributed with mean µ and positive definite covariance matrix S. Then X'AX is distributed as a noncentral chisquare with k degrees of freedom if and only if SA is an idempotent matrix of rank k, i.e., ASA=A.