Stat 5101 Lecture Notes - School of Statistics

5.2. THE MULTIVARIATE NORMAL DISTRIBUTION 1435.2.1 The Density of a Non-Degenerate Normal RandomVectorHow about the density of the multivariate normal distribution? First wehave to say that it may not have a density. If the variance parameter M is notpositive definite, then the random vector will be concentrated on a hyperplane(will be degenerate) by Theorem 5.6, in which case it won’t have a density.Otherwise, it will.Another approach to the same issue is to consider that X will have supporton the whole of R n only if the transformationis invertible, in which case its inverse isand has derivative matrixg(z) =a+Bzh(x) =B −1 (x−a)∇h(x) =B −1Thus we find the density of X by the multivariate change of variable theorem(Corollary 1.6 of Chapter 1 of these notes)f X (x) =f Z [h(x)] · ∣∣det ( ∇h(x) )∣ ∣ .(= f Z B −1 (x − a) ) · ∣∣det ( B −1)∣ ∣ .∣∣det ( B −1)∣ ∣=exp ( − 1(2π) n/2 2 [B−1 (x − a)] ′ B −1 (x − a) )∣∣det ( B −1)∣ ∣=exp ( − 1(2π) n/2 2 (x − a)′ (B −1 ) ′ B −1 (x − a) )Now we need several facts about matrices and determinants to clean this up.First, (B −1 ) ′ B −1 = M −1 , where, as above, M = var(X) because of two factsabout inverses, transposes, and products.• The inverse of a transpose is the transpose of the inverse.Hence (B −1 ) ′ =(B ′ ) −1• The inverse of a product is the product of the inverses in reverse order,that is, (CD) −1 = D −1 C −1 for any invertible matrices C and D.Hence (B ′ ) −1 B −1 =(BB ′ ) −1 = M −1 .Second, |det ( B −1) | = det(M) −1/2 because of two facts about determinants,inverses, and products.• The determinant of an inverse is the multiplicative inverse (reciprocal) ofthe determinant.Hence det ( B −1) = det(B) −1 .