Random Vectors 1. STA442/2101 Fall See last slide for copyright information. 1 / 30

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1 Random Vectors 1 STA442/2101 Fall See last slide for copyright information. 1 / 30

2 Background Reading: Renscher and Schaalje s Linear models in statistics Chapter 3 on Random Vectors and Matrices Chapter 4 on the Distribution 2 / 30

3 Overview 1 Definitions and Basic Results 2 3 / 30

4 Random Vectors and Matrices A random matrix is just a matrix of random variables. Their joint probability distribution is the distribution of the random matrix. Random matrices with just one column (say, p 1) may be called random vectors. 4 / 30

5 Expected Value The expected value of a matrix is defined as the matrix of expected values. Denoting the p c random matrix X by [X i,j ], E(X) = [E(X i,j )]. 5 / 30

6 Immediately we have natural properties like E(X + Y) = E([X i,j ] + [Y i,j ]) = [E(X i,j + Y i,j )] = [E(X i,j ) + E(Y i,j )] = [E(X i,j )] + [E(Y i,j )] = E(X) + E(Y). 6 / 30

7 Moving a constant through the expected value sign Let A = [a i,j ] be an r p matrix of constants, while X is still a p c random matrix. Then ([ p ]) E(AX) = E a i,k X k,j k=1 [ ( p )] = E a i,k X k,j k=1 [ p ] = a i,k E(X k,j ) k=1 = AE(X). Similar calculations yield E(AXB) = AE(X)B. 7 / 30

8 Variance-Covariance Matrices Let X be a p 1 random vector with E(X) = µ. The variance-covariance matrix of X (sometimes just called the covariance matrix), denoted by cov(x), is defined as { cov(x) = E (X µ)(x µ) }. 8 / 30

9 cov(x) = E { (X µ)(x µ) } cov(x) = E X 1 µ 1 X 2 µ 2 ( ) X 1 µ 1 X 2 µ 2 X 3 µ 3 X 3 µ 3 = E (X 1 µ 1 ) 2 (X 1 µ 1 )(X 2 µ 2 ) (X 1 µ 1 )(X 3 µ 3 ) (X 2 µ 2 )(X 1 µ 1 ) (X 2 µ 2 ) 2 (X 2 µ 2 )(X 3 µ 3 ) (X 3 µ 3 )(X 1 µ 1 ) (X 3 µ 3 )(X 2 µ 2 ) (X 3 µ 3 ) 2 = = E{(X 1 µ 1 ) 2 } E{(X 1 µ 1 )(X 2 µ 2 )} E{(X 1 µ 1 )(X 3 E{(X 2 µ 2 )(X 1 µ 1 )} E{(X 2 µ 2 ) 2 } E{(X 2 µ 2 )(X 3 E{(X 3 µ 3 )(X 1 µ 1 )} E{(X 3 µ 3 )(X 2 µ 2 )} E{(X 3 µ 3 ) 2 } V ar(x 1) Cov(X 1, X 2 ) Cov(X 1, X 3 ) Cov(X 1, X 2 ) V ar(x 2 ) Cov(X 2, X 3 ) Cov(X 1, X 3 ) Cov(X 2, X 3 ) V ar(x 3 ). So, the covariance matrix cov(x) is a p p symmetric matrix with variances on the main diagonal and covariances on the off-diagonals. 9 / 30

10 Matrix of covariances between two random vectors Let X be a p 1 random vector with E(X) = µ x and let Y be a q 1 random vector with E(Y) = µ y. The p q matrix of covariances between the elements of X and the elements of Y is { cov(x, Y) = E (X µ x )(Y µ y ) }. 10 / 30

11 Adding a constant has no effect On variances and covariances cov(x + a) = cov(x) cov(x + a, Y + b) = cov(x, Y) These results are clear from the definitions: cov(x) = E { (X µ)(x µ) } cov(x, Y) = E { (X µ x )(Y µ y ) } Sometimes it is useful to let a = µ x and b = µ y. 11 / 30

12 Analogous to V ar(a X) = a 2 V ar(x) Let X be a p 1 random vector with E(X) = µ and cov(x) = Σ, while A = [a i,j ] is an r p matrix of constants. Then cov(ax) = E {(AX Aµ)(AX Aµ) } = E {A(X µ) (A(X µ)) } = E {A(X µ)(x µ) A } = AE{(X µ)(x µ) }A = Acov(X)A = AΣA 12 / 30

13 The Distribution The p 1 random vector X is said to have a multivariate normal distribution, and we write X N(µ, Σ), if X has (joint) density [ 1 f(x) = exp 1 ] Σ 1 2 (2π) p 2 2 (x µ) Σ 1 (x µ), where µ is p 1 and Σ is p p symmetric and positive definite. 13 / 30

14 Σ positive definite Positive definite means that for any non-zero p 1 vector a, we have a Σa > 0. Since the one-dimensional random variable Y = p i=1 a ix i may be written as Y = a X and V ar(y ) = cov(a X) = a Σa, it is natural to require that Σ be positive definite. All it means is that every non-zero linear combination of X values has a positive variance. And recall Σ positive definite is equivalent to Σ 1 positive definite. 14 / 30

15 Analogies (Multivariate normal reduces to the univariate normal when p = 1) Univariate Normal } f(x) = { 1 σ exp 1 (x µ) 2 2π 2 σ 2 E(X) = µ, V ar(x) = σ 2 (X µ) 2 σ χ 2 (1) 2 1 f(x) = exp { 1 Σ 2 1 (2π) p 2 2 (x µ) Σ 1 (x µ) } E(X) = µ, cov(x) = Σ (X µ) Σ 1 (X µ) χ 2 (p) 15 / 30

16 More properties of the multivariate normal If c is a vector of constants, X + c N(c + µ, Σ) If A is a matrix of constants, AX N(Aµ, AΣA ) Linear combinations of multivariate normals are multivariate normal. All the marginals (dimension less than p) of X are (multivariate) normal, but it is possible in theory to have a collection of univariate normals whose joint distribution is not multivariate normal. For the multivariate normal, zero covariance implies independence. The multivariate normal is the only continuous distribution with this property. 16 / 30

17 An easy example If you do it the easy way Let X = (X 1, X 2, X 3 ) be multivariate normal with µ = and Σ = Let Y 1 = X 1 + X 2 and Y 2 = X 2 + X 3. Find the joint distribution of Y 1 and Y / 30

18 In matrix terms Y 1 = X 1 + X 2 and Y 2 = X 2 + X 3 means Y = AX ( Y1 Y 2 ) = ( ) X 1 X 2 X 3 Y = AX N(Aµ, AΣA ) 18 / 30

19 You could do it by hand, but > mu = cbind(c(1,0,6)) > Sigma = rbind( c(2,1,0), + c(1,4,0), + c(0,0,2) ) > A = rbind( c(1,1,0), + c(0,1,1) ); A > A %*% mu # E(Y) [,1] [1,] 1 [2,] 6 > A %*% Sigma %*% t(a) # cov(y) [,1] [,2] [1,] 8 5 [2,] / 30

20 A couple of things to prove (X µ) Σ 1 (X µ) χ 2 (p) X and S 2 independent under normal random sampling. 20 / 30

21 Recall the square root matrix Covariance matrix Σ is real and symmetric matrix, so we have the spectral decomposition Σ = PΛP = PΛ 1/2 Λ 1/2 P = PΛ 1/2 I Λ 1/2 P = PΛ 1/2 P PΛ 1/2 P = Σ 1/2 Σ 1/2 So Σ 1/2 = PΛ 1/2 P 21 / 30

22 Square root of an inverse Positive definite Positive eigenvalues Inverse exists PΛ 1/2 P PΛ 1/2 P = PΛ 1 P = Σ 1, so ( Σ 1 ) 1/2 = PΛ 1/2 P. It s easy to show ( Σ 1 ) 1/2 is the inverse of Σ 1/2 Justifying the notation Σ 1/2 22 / 30

23 Now we can show (X µ) Σ 1 (X µ) χ 2 (p) Where X N(µ, Σ) Y = X µ N (0, Σ) Z = Σ 1 2 Y N (0, Σ 1 2 ΣΣ 1 2 ) = N (0, Σ Σ 2 Σ 2 Σ 1 2 = N (0, I) ) So Z is a vector of p independent standard normals, and p Y Σ 1 Y = Z Z = Zi 2 χ 2 (p) j=1 23 / 30

24 X and S 2 independent Let X 1,..., X n i.i.d. N(µ, σ 2 ). X = X 1. X n N ( µ1, σ 2 I ) Y = X 1 X. X n 1 X X = AX 24 / 30

25 Y = AX In more detail 1 1 n 1 n 1 n 1 n 1 n 1 1 n 1 n 1 n n 1 n 1 1 n 1 n 1 n 1 n 1 n 1 n X 1 X 2. X n 1 X n = X 1 X X 2 X. X n 1 X X 25 / 30

26 The argument Y = AX = Y is multivariate normal. X 1 X. X n 1 X X Cov ( X, (X j X) ) = 0 (Exercise) So X and Y 2 are independent. = So X and S 2 = g(y 2 ) are independent. Y 2 X 26 / 30

27 Leads to the t distribution If Z N(0, 1) and Y χ 2 (ν) and Z and Y are independent, then T = Z t(ν) Y/ν 27 / 30

28 Random sample from a normal distribution i.i.d. Let X 1,..., X n N(µ, σ 2 ). Then n(x µ) N(0, 1) and σ (n 1)S 2 σ 2 = (X µ) σ/ n χ 2 (n 1) and These quantities are independent, so T = = n(x µ)/σ (n 1)S 2 /(n 1) σ 2 n(x µ) t(n 1) S 28 / 30

29 Multivariate normal likelihood For reference L(µ, Σ) = n i=1 1 Σ 1 2 (2π) p 2 = Σ n/2 (2π) np/2 exp n 2 { exp 1 } 2 (xi µ) Σ 1 (x i µ) { tr( ΣΣ } 1 ) + (x µ) Σ 1 (x µ), where Σ = 1 n n i=1 (x i x)(x i x) is the sample variance-covariance matrix. 29 / 30

30 Copyright Information This slide show was prepared by Jerry Brunner, Department of Statistics, University of Toronto. It is licensed under a Creative Commons Attribution - ShareAlike 3.0 Unported License. Use any part of it as you like and share the result freely. The L A TEX source code is available from the course website: brunner/oldclass/appliedf17 30 / 30