Stat 5101 Lecture Notes - School of Statistics

150 Stat 5101 (Geyer) Course NotesLemma 5.14. Suppose X is partitioned as in (5.30a) and has variance matrix(5.30c), and suppose that M 22 is positive definite. Thenare uncorrelated.X 1 − M 12 M −122 X 2 and X 2And, we should note, by Theorem 5.13, if X is multivariate normal, thenX 1 − M 12 M −122 X 2 is independent of X 2 .Proof. Obvious, just calculate the covariancecov(X 1 − M 12 M −122 X 2, X 2 )=cov(X 1 ,X 2 )−M 12 M −122 cov(X 2, X 2 )= M 12 − M 12 M −122 M 22=0Every conditional of a normal random vector is also normal, but it is hard forus to give an explicit expression for the degenerate case. This is not surprising,because all our methods for finding conditional densities and degenerate normaldistributions don’t have densities. So here we will be satisfied with describingthe non-degenerate case.Theorem 5.15. Every condition distribution of a non-degenerate multivariatenormal distribution is non-degenerate (multivariate or univariate) normal.In particular, if X is partitioned as in (5.30a), has the multivariate normaldistribution with mean vector (5.30b) and variance matrix (5.30c), thenX 1 | X 2 ∼N(µ 1 +M 12 M −122 [X 2 − µ 2 ], M 11 − M 12 M −122 M 21). (5.33)Proof. First note that the conditional distribution is multivariate normal byLemma 5.10, because the joint density is the exponential of a quadratic, henceso is the conditional, which is just the joint density considered as a function ofx 1 with x 2 fixed renormalized.So all that remains to be done is figuring out the conditional mean andvariance. For the conditional mean, we use Lemma 5.14 and the commentfollowing it. Because of the independence of X 1 − M 12 M −122 X 2 and X 2 ,butE(X 1 − M 12 M −122 X 2 | X 2 )=E(X 1 −M 12 M −122 X 2)E(X 1 − M 12 M −122 X 2 | X 2 )=E(X 1 |X 2 )−M 12 M −122 X 2by linearity of expectations and functions of the conditioning variable behavinglike constants, andE(X 1 − M 12 M −122 X 2)=µ 1 −M 12 M −122 µ 2.