Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

(Jacob Rumans) #1
9.8. The Distributions of Certain Quadratic Forms 559

variables. We would like to treat this problem generally, but limitations of space
forbid this, and we find it necessary to restrict ourselves to some special cases; see,
for instance, Stapleton (2009) for discussion.
Recall from linear algebra that a symmetric matrixAisidempotentifA^2 =A.
In Section 9.1, we have already met some idempotent matrices. For example, the
matrixI−n^1 Jof Example 9.8.1 is idempotent. Idempotent matrices possess some
important characteristics. Supposeλis an eigenvalue of an idempotent matrixA
with corresponding eigenvectorv. Then the following identity is true:


λv=Av=A^2 v=λAv=λ^2 v.

Henceλ(λ−1)v= 0 .Sincev = 0 ,λ= 0 or 1. Conversely, if the eigenvalues
of a real symmetric matrix are only 0s and 1s then it is idempotent; see Exercise
9.8.10. Thus the rank of an idempotent matrixAis the number of its eigenvalues
which are 1. Denote the spectral decomposition ofAbyA=Γ′ΛΓ,whereΛis a
diagonal matrix of eigenvalues andΓis an orthogonal matrix whose columns are
the corresponding orthonormal eigenvectors. Because the diagonal entries ofΛare
0or1andΓis orthogonal, we have


trA=trΛΓΓ′=trΛ=rank(A);

i.e., the rank of an idempotent matrix is equal to its trace.

Theorem 9.8.4.LetX′=(X 1 ,...,Xn),whereX 1 ,...,Xnare iidN(0,σ^2 ).Let
Q=σ−^2 X′AXfor a symmetric matrixAwith rankr.ThenQhas aχ^2 (r)distri-
bution if and only ifAis idempotent.


Proof:By Theorem 9.8.2, the mgf ofQis


MQ(t)=

∏r

i=1

(1− 2 tλi)−^1 /^2 , (9.8.14)

whereλ 1 ,...,λrare thernonzero eigenvalues ofA. Suppose, first, thatAis
idempotent. Thenλ 1 =···=λr=1andthemgfofQisMQ(t)=(1− 2 t)−r/^2 ;
i.e.,Qhas aχ^2 (r) distribution. Next, supposeQhas aχ^2 (r) distribution. Then
fortin a neighborhood of 0, we have the identity


∏r

i=1

(1− 2 tλi)−^1 /^2 =(1− 2 t)−r/^2 ,

which, upon squaring both sides, leads to


∏r

i=1

(1− 2 tλi)=(1− 2 t)r,

By the uniqueness of the factorization of polynomials,λ 1 =···=λr= 1. HenceA
is idempotent.

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