Follows now from Theorem 5.2. Q. E. D

Note that this result holds regardless of whether the matrix BE BT is non­singular or not. In the latter case the normal distribution involved is called “singular”:

Definition 5.2: Ann x 1 random vector Y has a singular Nn (д, E) distribution if its characteristic function is of the form yY (t) = exp(i ■ t т д – 21TE t) with E a singular, positive semidefinite matrix.

Because of the latter condition the distribution of the random vector Y in­volved is no longer absolutely continuous, but the form of the characteristic function is the same as in the nonsingular case – and that is all that matters. For example, let n = 2 and

*=й -=(0 "0.

where a 2 > 0 but small. The density of the corresponding N2(*, —) distribution

Подпись: f (У1. У2|а ) Подпись: exp( - y2/2) .. exP ( - y|/(2a2)) V2n a V2n Подпись: (5.5)

of Y = (Yb Y2)t is

Then lima4.0 f (yі, У2Іа) = 0 if y2 = 0, and lima40 f(yi, У2Іа) = то if y2 = 0. Thus, a singular multivariate normal distribution does not have a density.

In Figure 5.2 the density (5.5) for the near-singular case a2 = 0.00001 is displayed. The height of the picture is actually rescaled to fit in the box [-3, 3] x [-3, 3] x [-3, 3]. If we let a approach zero, the height of the ridge corresponding to the marginal density of Y1 will increase to infinity.

The next theorem shows that uncorrelated multivariate normally distributed random variables are independent. Thus, although for most distributions uncor – relatedness does not imply independence, for the multivariate normal distribu­tion it does.

Theorem 5.4: Let X be n-variate normally distributed, and let X1 and X2 be subvectors of components of X. If X1 and X2 are uncorrelated, that is, Cov(X1, X2) = O, then X1 and X2 are independent.

Подпись: E (X) = Подпись: *1 Iі2 Подпись: Var( X) Подпись: —11 —12 —21 — 22

Proof: Because X1 and X2 cannot have common components, we may with­out loss of generality assume that X = (XT, XT)T, X1 є Rk, X2 є Rm. Parti­tion the expectation vector and variance matrix of X conformably as

Then £12 = O and £21 = O because they are covariance matrices, and X1 and X2 are uncorrelated; hence, the density of X is

f (x) = f (x, X2)


-1 г


(xA (mA

£11 0


.W Vм2/.

0 £22



(V2^^det (£ e022)

_ exp (-i(xj – Ді)т£111(xj – мО)

= (V2n )V det(£„)

exp (-i(X2 – M2)T^221(X2 – М2))

X (V2n)V det(E22)

This implies independence of X1 and X2. Q. E.D.

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