Wishart Distribution Wishart Distribution
I Consider as usual a random sample x1, x2,..., xN from Np(µ, Σ).
I Recall: the sample covariance matrix S satisfies 1 S = A, N − 1 where N X 0 A = (xα − x¯)(xα − x¯) α=1
Statistics 784 NC STATE UNIVERSITY 1 / 13 Multivariate Analysis Wishart Distribution
I Write 0 x1 0 x2 X = . . 0 xN Then I 0 (x1 − x¯) 0 (x2 − x¯) 1 X = = I − 1 10 X dev . N N N . N 0 (xN − x¯) and 1 A = X0 X = X0 I − 1 10 X dev dev N N N N
Statistics 784 NC STATE UNIVERSITY 2 / 13 Multivariate Analysis Wishart Distribution
I The Helmert matrix is the orthogonal matrix
√1 √−1 0 ... 0 2 2 √1 √1 √−2 ... 0 6 6 6 . . . . . ...... HN = . . . . 1 1 1 −(N−1) √ √ √ ... √ N(N−1) N(N−1) N(N−1) N(N−1) √1 √1 √1 ... √1 N N N N
Statistics 784 NC STATE UNIVERSITY 3 / 13 Multivariate Analysis Wishart Distribution
I If we partition HN as ! H(1) H = N N √1 √1 √1 ... √1 N N N N then 0 1 I = H0 H = H(1) H(1) + 1 10 N N N N N N N N
I So 1 0 I − 1 10 = H(1) H(1) N N N N N N and
1 0 0 A = X0 I − 1 10 X = X0H(1) H(1)X = H(1)X H(1)X N N N N N N N N
Statistics 784 NC STATE UNIVERSITY 4 / 13 Multivariate Analysis Wishart Distribution
I If Z = HN X then the rows of Z are uncorrelated, and hence independent, with the first n = N − 1 distributed as Np(0, Σ), and the last distributed as √ Np Nµ, Σ So we can write I n X 0 A = zαzα α=1
where z1, z2,..., zn is a random sample from Np(0, Σ).
I A is said to have the Wishart distribution Wp(Σ, n).
Statistics 784 NC STATE UNIVERSITY 5 / 13 Multivariate Analysis Wishart Distribution Distribution of A
I First assume that Σ = Ip. Z(1) (1)X I Then the elements of = HN are iid N(0, 1). I We can use Gram-Schmidt orthogonalization to write
Z(1) = WT0
where:
I T is (p × p) lower-triangular; I the columns of W are mutually orthogonal and normalized. 0 and A = Z(1) Z(1) = TT0.
Statistics 784 NC STATE UNIVERSITY 6 / 13 Multivariate Analysis Wishart Distribution
I Anderson shows that, when n ≥ p:
I The non-zero elements of T are independent; I The sub-diagonal elements are all distributed as N(0, 1); 2 2 I The diagonal element ti,i satisfies ti,i ∼ χn−i+1. 1 I That is, the joint density of these 2 n(n + 1) variables is
p ! p i Y 1 X X tn−i exp − t2 i,i 2 i,j i=1 i=1 j=1 p 1 p(n−2) 1 p(p−1) Y 1 2 2 π 4 Γ (n + 1 − i) 2 i=1
Statistics 784 NC STATE UNIVERSITY 7 / 13 Multivariate Analysis Wishart Distribution
I The result for general Σ is that the joint density is
p ! Y 1 tn−i exp − trace