SF2940: Probability theory Lecture 8: Multivariate Normal Distribution


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1 SF2940: Probability theory Lecture 8: Multivariate Normal Distribution Timo Koski Timo Koski Matematisk statistik / 1
2 Learning outcomes Random vectors, mean vector, covariance matrix, rules of transformation Multivariate normal R.V., moment generating functions, characteristic function, rules of transformation Density of a multivariate normal RV Joint PDF of bivariate normal RVs Conditional distributions in a multivariate normal distribution Timo Koski Matematisk statistik / 1
3 PART 1: Mean vector, Covariance matrix, MGF, Characteristic function Timo Koski Matematisk statistik / 1
4 Vector Notation: Random Vector A random vector X is a column vector X 1 X 2 X =. = (X 1,X 2,...,X n ) T X n Each X i is a random variable. Timo Koski Matematisk statistik / 1
5 Sample Value Random Vector A column vector x = x 1 x 2. x n = (x 1,x 2,...,x n ) T We can think of x i is an outcome of X i. Timo Koski Matematisk statistik / 1
6 Joint CDF, Joint PDF The joint CDF (=cumulative distribution function) of a continuous random vector X is F X (x) = F X1,...,X n (x 1,...,x n ) = P (X x) = = P (X 1 x 1,...,X n x n ) Joint probability density function (PDF) f X (x) = n x 1... x n F X1,...,X n (x 1,...,x n ) Timo Koski Matematisk statistik / 1
7 Mean Vector µ X = E [X] = E [X 1 ] E [X 2 ]. E [X n ] a column vector of means (=expectations) of X., Timo Koski Matematisk statistik / 1
8 Matrix, Scalar Product If X T is the transposed column vector (=a row vector), then is a n n matrix, and XX T X T X = is a scalar product, a real valued R.V.. n Xi 2 i=1 Timo Koski Matematisk statistik / 1
9 Covariance Matrix of A Random Vector Covariance matrix C X := E [(X µ X )(X µ X ) T] where the element (i,j) is the covariance of X i and X j. C X (i,j) = E [(X i µ i )(X j µ j )] Timo Koski Matematisk statistik / 1
10 Remarks on Covariance X and Y are independent Cov(X,Y) = 0. The converse implication is not true in general, as shown in the next example. Let X N(0,1) and set Y = X 2. Then Y is clearly functionally dependent on X. But we have Cov(X,Y) = E [(X Y)] E [X] E [Y] = E [ X 3] 0 E [Y] = E [ X 3] = 0. The last equality holds, since one has g(x) = x 3 φ(x), so that g( x) = g(x). Hence E [ X 3] = + g(x)dx = 0, c.f., in the sequel, too. Timo Koski Matematisk statistik / 1
11 A Quadratic Form We see that = E = x T C X x = n n i=1j=1 [ n i=1 n n i=1j=1 x i x j C X (i,j). x i x j E [(X i µ i )(X j µ j )] n x i x j (X i µ i )(X j µ j ) j=1 ] ( ) Timo Koski Matematisk statistik / 1
12 Properties of a Covariance Matrix Covariance matrix is nonnegative definite, i.e., for all x we have x T C X x 0 Hence detc X 0. The covariance matrix is symmetric C X = CX T Timo Koski Matematisk statistik / 1
13 Properties of a Covariance Matrix The covariance matrix is symmetric C X = C T X since C X (i,j) = E [(X i µ i )(X j µ j )] = E [(X j µ j )(X i µ i )] = C X (j,i) Timo Koski Matematisk statistik / 1
14 Properties of a Covariance Matrix A covariance matrix is positive definite, x T C X x > 0 for all x = 0 iff (i.e. C X is invertible). detc X > 0 Timo Koski Matematisk statistik / 1
15 Properties of a Covariance Matrix Proposition Pf: By ( ) above = E x T C X x = x T E x T C X x 0 [ (X µ X )(X µ X ) T] x [ ] [ ] x T (X µ X )(X µ X ) T x = E x T w w T x where we have set w = (X µ X ). Then by linear algebra x T w = w T x = n i=1 w ix i. Hence ( [ ] n ) 2 E x T ww T x = E i x i 0. i=1w Timo Koski Matematisk statistik / 1
16 Properties of a Covariance Matrix In terms of the entries c i,j of a covariance matrix C = (c i,j ) n,n, i=1,j=1 there are the following necessary properties. 1 c i,j = c j,i (symmetry). 2 c i,i = Var(X i ) = σ 2 i 0 (the elements in the main diagonal are the variances, and thus all elements in the main diagonal are nonnegative). 3 c 2 i,j c i,i c j,j (CauchySchwartz inequality). Timo Koski Matematisk statistik / 1
17 Coefficient of Correlation The Coefficient of Correlation ρ of X and Y is defined as ρ := ρ X,Y := Cov(X,Y) Var(X) Var(Y), where Cov(X,Y) = E [(X µ X )(Y µ Y )]. This is normalized For random variables X and Y, 1 ρ X,Y 1 Cov(X,Y) = ρ X,Y = 0 does not always mean that X,Y are independent. Timo Koski Matematisk statistik / 1
18 Special case: Covariance Matrix of A Bivariate Vector X = (X 1,X 2 ) T. ( σ 2 C X = 1 ρσ 1 σ 2 ρσ 1 σ 2 σ2 2 where ρ is the coefficient of correlation of X 1 and X 2, and σ1 2 = Var(X 1), σ2 2 = Var(X 2). C X is invertible iff ρ 2 = 1, for proof we note that detc X = σ1 2 σ2 2 ( 1 ρ 2 ) ), Timo Koski Matematisk statistik / 1
19 Special case: Covariance Matrix of A Bivariate Vector if ρ 2 = 1, the inverse exists and Λ 1 = ( σ 2 Λ = 1 ρσ 1 σ 2 ρσ 1 σ 2 σ2 2 ( 1 σ1 2σ2 2 (1 ρ2 ) ), σ 2 2 ρσ 1 σ 2 ρσ 1 σ 2 σ 2 1 ), Timo Koski Matematisk statistik / 1
20 Y = BX+b Proposition X is a random vector with mean vector µ X and covariance matrix C X. B is a m n matrix. If Y = BX+b, then EY = Bµ X +b C Y = BC X B T Pf: For simplicity of writing, take b = µ = 0. Then C Y = EYY T = EBX(BX) T = [ = EBXX T B T = BE XX T] B T = BC X B T Timo Koski Matematisk statistik / 1
21 Moment Generating and Characteristic Functions Definition Moment generating function of X is defined as ψ X (t) def = Ee ttx = Ee t 1X 1 +t 2 X 2 + +t n X n Definition Characteristic function of X is defined as ϕ X (t) def = Ee ittx = Ee i(t 1X 1 +t 2 X 2 + +t n X n ) Special cases: take t 1 = 1,t 2 = t 3 =... = t n = 0, then ϕ X (t) = ϕ X1 (t 1 ). Timo Koski Matematisk statistik / 1
22 PART 2: Def I of a multivariate normal distribution We recall first some of the properties of univariate normal distribution Timo Koski Matematisk statistik / 1
23 Normal (Gaussian) Onedimensional RVs X is a normal random variable if where µ is real and σ > 0. Notation: X N(µ, σ 2 ) f X (x) = 1 σ 2π e 1 2σ 2(x µ)2 Properties: E(X) = µ, Var(X) = σ 2 Timo Koski Matematisk statistik / 1
24 Normal (Gaussian) Onedimensional RVs f X (x) x f X (x) x µ = 2, σ = 1/2, (b) µ = 2, σ = 2 (a) Timo Koski Matematisk statistik / 1
25 Central Moments Normal (Gaussian) Onedimensional RVs X N(0, σ 2 ). Then E [X n ] = { 0 n is odd (2k)! 2 k k! σ2k n = 2k, k = 0,1,2,.... Timo Koski Matematisk statistik / 1
26 Linear Transformation X N(µ X, σ 2 ) Y = ax +b is N(aµ X +b,a 2 σ 2 ) Thus Z = X µ X σ X N(0,1) and ( X µx P(X x) = P σ X or x µ X σ X ( F X (x) = P Z x µ ) ( ) X x µx = Φ σ X σ X ) Timo Koski Matematisk statistik / 1
27 Normal (Gaussian) Onedimensional RVs X N(µ, σ 2 ) then the moment generating function is [ ψ X (t) = E e tx] = e tµ+1 2 t2 σ 2, and the characteristic function is ϕ X (t) = E as found in previous Lectures. [ e itx] = e itµ 1 2 t2 σ 2 Timo Koski Matematisk statistik / 1
28 Multivariate Normal Def. I Definition An n 1 random vector X has a normal distribution iff for every n 1vector a the onedimensional random vector a T X has a normal distribution. We write X N(µ,Λ), when µ is the mean vector and Λ is the covariance matrix. Timo Koski Matematisk statistik / 1
29 Consequences of Def. I (1) An n 1 vector X N(µ,Λ) iff the onedimensional random vector a T X has a normal distribution for every nvector a. Now we know that (take B = a T in the preceding) [ ] Ea T X = a T µ, Var a T X = a T Λa Timo Koski Matematisk statistik / 1
30 Consequences of Def. I (2) Hence, if Y = a T X, then Y N ( a T µ,a T Λa ) and the moment generating function of Y is [ ψ Y (t) = E e ty] = e tat µ+ 2 1t2 a TΛa. Therefore ψ X (a) = Ee atx = ψ Y (1) = e at µ+ 1 2 at Λa. Timo Koski Matematisk statistik / 1
31 Consequences of Def. I (3) Hence we have shown that if X N(µ,Λ), then ψ X (t) = Ee ttx = e tt µ+ 1 2 tt Λt. is the moment generating function of X. Timo Koski Matematisk statistik / 1
32 Consequences of Def. I (4) In the same way we can find that ϕ X (t) = Ee ittx = e itt µ 1 2 tt Λt. is the characteristic function of X N(µ,Λ). Timo Koski Matematisk statistik / 1
33 Consequences of Def. I (5) Let Λ be a diagonal covariance matrix with λ 2 i s on the main diagonal, i.e., λ λ Λ = 0 0 λ , λ 2 n Proposition If X N(µ,Λ), then X 1,X 2,...,X n are independent normal variables. Timo Koski Matematisk statistik / 1
34 Consequences of Def. I (6) Pf: Λ is diagonal, the quadratic form becomes a single sum of squares. ϕ X (t) = e itt µ 1 2 tt Λt = = e i n i=1 µ it i 1 2 n i=1 λ2 i t2 i = e iµ 1t λ2 1 t2 1e iµ 2t λ2 2 t2 2 e iµ n t n 2 1 λ2 n t2 n is the product of the characteristic functions of X i N ( µ i, λ 2 ) i, which are thus seen to be independent N ( µ i, λ 2 ) i. Timo Koski Matematisk statistik / 1
35 Kac s theorem: Thm in LN Theorem X = (X 1,X 2,,X n ) T. The componentsx 1,X 2,,X n are independent if and only if φ X (s) = E [ e is X ] = n i=1 φ Xi (s i ), where φ Xi (s i ) is the characteristic function for X i. Timo Koski Matematisk statistik / 1
36 Further properties of the multivariate normal X N(µ,Λ) Every component X k is onedimensional normal. To prove this we take a = (0,0,..., }{{} 1,0,...,0) T position k and the conclusion follows by Def. I. X 1 +X 2 + X n is onedimensional normal. Note: The terms in the sum need not be independent. Timo Koski Matematisk statistik / 1
37 Properties of multivariate normal X N(µ,Λ) Every marginal distribution of k variables ( 1 k < n is normal. To prove this we consider any k variables X i1,x i2...x ik and then take a such that a j = 0 for j = i 1,...i k and then apply Def. I. Timo Koski Matematisk statistik / 1
38 Properties of multivariate normal Proposition X N(µ,Λ) and Y = BX+b. Then ( Y N Bµ+b,BΛB T). Pf: ψ Y (s) = E = e stb E E [ ] [ ] e st Y = E e st (b+bx) = [ e st BX ] = e stb E [ e (BT s) T X [ ] ( ) e s) T (BT X = ψ X B T s. ] Timo Koski Matematisk statistik / 1
39 Properties of multivariate normal X N(µ,Λ) ) ψ X (B T s = e s) T (BT µ+ 2(B 1 T s) T Λ(B T s). ( B T s) T µ = s T Bµ, ( ) T ( ) B T s Λ B T s = s T BΛB T s, e (BT s) T µ+ 1 2(B T s) T Λ(B T s) = e s T Bµ+ 1 2 st BΛB T s Timo Koski Matematisk statistik / 1
40 Properties of multivariate normal ) ψ X (B T s = e st Bµ+ 1 2 st BΛB Ts. ) ψ Y (s) = e stb ψ X (B T s = e stb e st Bµ+ 2 1sT BΛB T s which proves the claim as asserted. ψ Y (s) = e st (b+bµ)+ 1 2 st BΛB Ts, Timo Koski Matematisk statistik / 1
41 PART 3: Multivariate normal, Def. II: characteristic function, DEF III: density Timo Koski Matematisk statistik / 1
42 Multivariate normal, Def. II: char. fnctn Definition A random vector X with mean vector µ and a covariance matrix Λ is N(µ,Λ) if its characteristic function is ϕ X (t) = Ee ittx = e itt µ 1 2 tt Λt. Timo Koski Matematisk statistik / 1
43 Multivariate normal, Def. II implies Def. I We need to show that the onedimensional random vector Y = a T X has a normal distribution. [ ϕ Y (t) = E e ity] ] = E [e it n i=1 a i X i = = E [ e itat X ] = ϕ X (ta) = = e itat µ 1 2 t2 a T Λa and this is the characteristic function of N ( a T µ,a T Λa ). Timo Koski Matematisk statistik / 1
44 Multivariate normal, Def. III: joint PDF Definition A random vector X with mean vector µ and an invertible covariance matrix Λ is N(µ,Λ), if the density is f X (x) = 1 (2π) n/2 det(λ) e 1 2 (x µ) T Λ 1 (x µ) Timo Koski Matematisk statistik / 1
45 Multivariate normal It can be checked by a computation that e itt µ 2 1tTΛt = e itt x 1 R n (2π) n/2 det(λ) e 1 2 (x µ) TΛ 1 (x µ) dx (complete the square) Hence Def. III implies the property in Def. II. The three definitions are equivalent, in the case inverse of the covariance matrix exists. Timo Koski Matematisk statistik / 1
46 PART 4: Bivariate normal with density Timo Koski Matematisk statistik / 1
47 Multivariate Normal: the bivariate case As soon as ρ 2 = 1, the matrix ( σ 2 Λ = 1 ρσ 1 σ 2 ρσ 1 σ 2 σ2 2 ), is invertible, and the inverse is Λ 1 = 1 σ 2 1 σ2 2 (1 ρ2 ) ( σ 2 2 ρσ 1 σ 2 ρσ 1 σ 2 σ 2 1 ), Timo Koski Matematisk statistik / 1
48 Multivariate Normal: the bivariate case ρ 2 = 1, and X = (X 1,X 2 ) T, then f X (x) = = 1 2π detλ e 1 2 (x µ X ) T Λ 1 (x µ X ) 1 2πσ 1 σ 2 1 ρ 2 e 1 2 Q(x 1,x 2 ) Timo Koski Matematisk statistik / 1
49 Multivariate Normal: the bivariate case where Q(x 1,x 2 ) = [ (x1 ) 1 (1 ρ 2 ) µ 2 1 2ρ(x ( ) ] 1 µ 1 )(x 2 µ 2 ) x2 µ σ 1 σ 2 σ 1 For this, invert the matrix Λ and expand the quadratic form! σ 2 Timo Koski Matematisk statistik / 1
50 ρ = Timo Koski Matematisk statistik / 1
51 ρ = Timo Koski Matematisk statistik / 1
52 ρ = Timo Koski Matematisk statistik / 1
53 Conditional densities for the bivariate normal Complete the square of the exponent to write where f X,Y (x,y) = f X (x)f Y X (y) f X (x) = f Y X (y) = 1 e 1 2σ 2 (x µ 1 ) 2 1 σ 1 2π 1 e 1 2 σ 2 (y µ 2 (x)) 2 2 σ 2 2π µ 2 (x) = µ 2 + ρ σ 2 σ 1 (x µ 1 ), σ 2 = σ 2 1 ρ 2 Timo Koski Matematisk statistik / 1
54 Bivariate normal properties E(X) = µ 1 Given X = x, Y is Gaussian Conditional mean of Y given X = x: µ 2 (x) = µ 2 + ρ σ 2 σ 1 (x µ 1 ) = E(Y X = x) Conditional variance of Y given X = x: Var(Y X = x) = σ2 2 ( 1 ρ 2 ) Timo Koski Matematisk statistik / 1
55 Bivariate normal properties Conditional mean of Y given X = x: µ 2 (x) = µ 2 + ρ σ 2 σ 1 (x µ 1 ) = E(Y X = x) Conditional variance of Y given X = x: Var(Y X = x) = σ2 2 ( 1 ρ 2 ) Check Section and Exercise By this is seen that the conditional mean of Y given X variable in a bivariate normal distribution is also the best LINEAR predictor of Y based on X, and the conditional variance is the variance of the estimation error. Timo Koski Matematisk statistik / 1
56 Marginal PDFs Timo Koski Matematisk statistik / 1
57 Proof of conditional pdf Consider f X,Y (x,y) f X (x) = σ 1 2π 2πσ 1 σ 2 1 ρ 2 e 1 2 Q(x,y)+ 1 2σ 1 2 (x µ 1 ) 2 Timo Koski Matematisk statistik / 1
58 Proof of conditional pdf 1 2 Q(x,y)+ 1 2σ1 2 (x µ 1 ) 2 = 1 2 H(x,y), Timo Koski Matematisk statistik / 1
59 Proof of conditional pdfs H(x,y) = [ (x ) 1 2 (1 ρ 2 ) µ1 2ρ(x µ ( ) ] 1)(y µ 2 ) y 2 µ2 + σ 1 σ 2 σ 1 ( x µ1 σ 1 ) 2 σ 2 Timo Koski Matematisk statistik / 1
60 Proof of conditional pdf H(x,y) = ρ 2 (x µ 1 ) 2 (1 ρ 2 ) σ1 2 2ρ(x µ 1)(y µ 2 ) σ 1 σ 2 (1 ρ 2 + (y µ 2) 2 ) σ2 2(1 ρ2 ) Timo Koski Matematisk statistik / 1
61 Proof of conditional pdf H(x,y) = ( ) 2 y µ 2 ρ σ 2 σ 1 (x µ 1 ) σ 2 2 (1 ρ2 ) Timo Koski Matematisk statistik / 1
62 Conditional pdf 1 1 ρ 2 σ 2 2π e f X,Y (x,y) = f X (x) 2 1 (y µ 2 ρ σ 2 σ1 (x µ 1 )) 2 σ 2 2(1 ρ2 ) This establishes the bivariate normal properties claimed above. Timo Koski Matematisk statistik / 1
63 Bivariate normal properties : ρ Proposition (X,Y) bivariate normal ρ = ρ X,Y Proof: E [(X µ 1 )(Y µ 2 )] = E(E([(X µ 1 )(Y µ 2 )] X)) = E((X µ 1 )E [Y µ 2 ] X)) Timo Koski Matematisk statistik / 1
64 Bivariate normal properties : ρ = E((X µ 1 )E [(Y µ 2 )] X)) = E(X µ 1 )[E(Y X) µ 2 ] [ = E((X µ 1 ) µ 2 + ρ σ ] 2 (X µ 1 ) µ 2 σ 1 = ρ σ 2 σ 1 E(X µ 1 )((X µ 1 )) Timo Koski Matematisk statistik / 1
65 Bivariate normal properties : ρ = ρ σ 2 σ 1 E(X µ 1 )(X µ 1 ) = ρ σ 2 σ 1 E(X µ 1 ) 2 = ρ σ 2 σ 1 σ 2 1 = ρσ 2 σ 1 Timo Koski Matematisk statistik / 1
66 Bivariate normal properties : ρ In other words we have checked that ρ = E [(X µ 1)(Y µ 2 )] σ 2 σ 1 ρ = 0 bivariate normal X,Y are independent. Timo Koski Matematisk statistik / 1
67 PART 5: Generating a multivariate normal variable Timo Koski Matematisk statistik / 1
68 Standard Normal Vector: definition Z N(0,I) is a standard normal vector. I is the n n identity matrix. f Z (z) = 1 (2π) n/2 det(i) e 1 2 (z 0) T I 1 (z 0) = 1 (2π) n/2e 1 2 zt z Timo Koski Matematisk statistik / 1
69 Distribution of X = AZ+b X = AZ+b, Z is standard Gaussian, then X = N (b,aa T) (follows by a rule in the preceding) Timo Koski Matematisk statistik / 1
70 Multivariate Normal: the bivariate case If ( σ 2 Λ = 1 ρσ 1 σ 2 ρσ 1 σ 2 σ2 2 ), then Λ = AA T, where A = ( σ1 0 ρσ 2 σ 2 1 ρ 2 ), Timo Koski Matematisk statistik / 1
71 Standard Normal Vector X N(µ X,Λ), and A is such that Λ = AA T (An invertible matrix A with this property exists always, if Λ is positive definite (we need the symmetry of Λ, too.) Then Z = A 1 (X µ X ) is a standard Gaussian vector. Proof: We give the first idea of his proof, a rule of transformation. Timo Koski Matematisk statistik / 1
72 Rule of transformation If X has density f X (x), Y = AX+b, A is invertible, then f Y (y) = Note that if Λ = AA T, then so that deta = detλ. 1 deta f ( X A 1 (y b) ) detλ = deta deta T = deta deta = deta 2, Timo Koski Matematisk statistik / 1
73 Diagonalizable Matrices An n n matrix A is orthogonally diagonalizable, if there is an orthogonal matrix P (i.e., P T P =PP T = I) such that where Λ is a diagonal matrix. P T AP = Λ, Timo Koski Matematisk statistik / 1
74 Diagonalizable Matrices Theorem If A is an n n matrix, then the following are equivalent: (i) A is orthogonally diagonalizable. (ii) A has an orthonormal set of eigenvectors. (iii) A is symmetric. Since covariance matrices are symmetric, we have by the theorem above that all covariance matrices are orthogonally diagonalizable. Timo Koski Matematisk statistik / 1
75 Diagonalizable Matrices Theorem If A is a symmetric matrix, then (i) Eigenvalues ofaare all real numbers. (ii) Eigenvectors from different eigenspaces are orthogonal. That is, all eigenvalues of a covariance matrix are real. Timo Koski Matematisk statistik / 1
76 Diagonalizable Matrices Hence we have for any covariance matrix the spectral decomposition C = n λ i e i ei T, (1) i=1 where Ce i = λ i e i. Since C is nonnegative definite, and its eigenvectors are orthonormal, 0 e T i Ce i = λ i e T i e i = λ i, and thus the eigenvalues of a covariance matrix are nonnegative. Timo Koski Matematisk statistik / 1
77 Diagonalizable Matrices Let now P be an orthogonal matrix such that P C X P = Λ, and X N(0,C X ), i.e., C X is a covariance matrix and Λ is diagonal (with the eigenvalues of C X on the main diagonal). Then if Y = P T X, we have that Y N(0,Λ). In other words, Y is a Gaussian vector and has independent components. This method of producing independent Gaussians has several important applications. One of these is the principal component analysis. Timo Koski Matematisk statistik / 1
78 Timo Koski Matematisk statistik / 1
SF2940: Probability theory Lecture 8: Multivariate Normal Distribution
SF2940: Probability theory Lecture 8: Multivariate Normal Distribution Timo Koski 24.09.2014 Timo Koski () Mathematisk statistik 24.09.2014 1 / 75 Learning outcomes Random vectors, mean vector, covariance
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