October 3rd, Linear Algebra & Properties of the Covariance Matrix


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1 Linear Algebra & Properties of the Covariance Matrix October 3rd, 2012
2 Estimation of r and C Let rn 1, rn, t..., rn T be the historical return rates on the n th asset. rn 1 rṇ 2 r n =. r T n n = 1, 2,..., N.
3 Estimation of r and C The expected return is approximated by T ˆ r n = 1 T t=1 r t n and the covariance is approximated by ĉ mn = 1 T 1 or in matrix/vector notation Ĉ = 1 T 1 T (rn t ˆ r n )(rm t ˆ r m ), t=1 T (r t ˆ r)(r t ˆ r) an outer product. t=1
4 Importance of C The expected return r is often estimated exogenously (e.g. not statistically estimated). But statistical estimation of C is an important topic, is used in practice, and is the backbone of many finance strategies. A major part of Markowitz theory was the assumption for C to be a covariance matrix it must be symmetric positive definite (SPD).
5 Real Matrices/Vectors Let s focus on real matrices, and only use complex numbers if necessary. A vector x R N and a matrix A R M N are x 1 a 11 a a 1N x 2 x =. and A = a 21 a a 2N..... a M1 a M2... a MN x N
6 Matrix/Vector Multiplication The matrix A times a vector x is a new vector a 11 a a 1N x 1 a 21 a a 2N x 2 Ax = a M1 a M2... a MN x N a 11 x 1 + a 12 x a 1N x N a 21 x 1 + a 22 x a 2N x N =. RM. a M1 x 1 + a M2 x a MN x N Matrix/matrix multiplication is an extension of this.
7 Matrix/Vector transpose Their transposes are x 1 x x 2 =. x N a 11 a a 1N and A a 21 a a 2N =..... a M1 a M2... a MN Note that (Ax) = x A. = (x 1, x 2,..., x N ) a 11 a a M1 a 12 a a M2 = a 1N a 2N... a MN
8 Vector Inner/outer Product The vector x R N inner product with another vector y R N is x y = y x = x 1 y 1 + x 2 y x N y n. There is also the outer product, x 1 y 1 x 1 y 2... x 1 y N xy x 2 y 1 x 2 y 2... x 2 y N =..... yx. x N y 1 x N y 2... X N Y N
9 Matrix/Vector Norm The norm on the vector space R N is, and for any x R N we have x = x x. Properties of the norm x 0 for any x, x = 0 iff x = 0, x + y x + y (triangle inequality), x y x y with equality iff x = ay for some a R (CauchySchwartz), x + y 2 = x 2 + y 2 if x y = 0 (Pythagorean thm).
10 Linear Independence & Orthogonality Two vectors x, y are linear independent if there is no scalar constant a R s.t. y = ax. These two vectors are orthogonal if x y = 0. Clearly, orthogonality linear independent.
11 Span A set of vectors x 1, x 2,..., x n is said to span a subspace V R N if for any y V there are constants a 1, a 2,..., a n such y = a 1 x 2 + a 2 x a n x n. We write, span(x 1,..., x n ) = V. In particular, a set x 1, x 2,..., x N will span R N iff they are linearly independent: span(x 1, x 2,..., x N ) = R N x i linearly independent of x j for all i, j. If so, then x 1, x 2,..., X N are a basis for R N.
12 Orthogonal Basis Definition A basis of R N is orthogonal if all its elements are orthogonal, span(x 1, x 2,..., x N ) = R N and x i x j = 0 i, j.
13 Invertibility Definition A square matrix A R N N is invertible if for any b R N there exists x s.t. Ax = b. If so, then x = A 1 b. A is invertible if any of the following hold: A has rows that are linearly independent A has columns that are linearly independent det(a) 0 there is no nonzero vector x s.t. Ax = 0. In fact, these 4 statements are equivalent.
14 Eigenvalues Let A be an N N matrix. A scalar λ C is an eigenvalue of A if Ax = λx for some vector x = re(x) + 1 im(x) C N. If so x is an eigenvector. By 4th statement of previous slide, A 1 exists zero is not an eigenvalue.
15 Eigenvalue Diagonalization Let A be an N N matrix with N linearly independent eigenvectors. There is an N N matrix X and an N N diagonal matrix Λ such that A = X ΛX 1, where Λ ii is an eigenvalue of A, and the columns of X = [x 1, x 2,..., x N ] are eigenvectors, Ax i = Λ ii x i.
16 Real Symmetric Matrices A matrix is symmetric if A = A. Proposition A symmetric matrix has real eigenvalues. pf: If Ax = λx then λ 2 x 2 = λ 2 x x = x A 2 x = x A Ax = Ax 2, which is real and nonnegative (x = re(x) 1 im(x) i.e the conjugate transpose). Hence, λ is cannot be imaginary or complex.
17 Real Symmetric Matrices Proposition Let A = A be a real matrix. A s eigenvectors form an orthogonal basis of R N, and the diagonalization is simplified: i.e. X 1 = X. A = X ΛX, Basic Idea of Proof: Suppose that Ax = λx and Ay = αy with α λ. Then λx y = x A y = x Ay = αx y, and so it must be that x y = 0.
18 Skew Symmetric Matrices A matrix is skew symmetric if A = A. Proposition A skew symmetric matrix has purely imaginary eigenvalues. pf: If Ax = λx then λ 2 x 2 = λ 2 x x = x A 2 x = x A Ax = Ax 2 which is less than zero. Hence, λ 2 < 0 must be purely imaginary.
19 Positive Definitness Definition A matrix A R N N is positive definite iff x Ax > 0 x R N. It is only positive semidefinite iff x Ax 0 x R N.
20 Positive Definitness Symmetric Proposition If A R N N is positive definite, then A = A, i.e. it is symmetric positive definite (SPD). pf: For any x R N we have 0 < x Ax = 1 x (A + A 2 } {{ } )x + x ( A A } {{ } )x, sym. skewsym. but if A A 0 then there exists eigenvector x s.t. (A A )x = αx where α C \ R, which contradicts the inequality.
21 Eigenvalue of an SPD Matrix Proposition If A is SPD, then it has all positive eigenvalues. pf: By definition for any y R N \ {0} it holds that y Ay > 0. In particular, for any λ and x s.t. Ax = λx, we have λ x 2 = x Ax > 0 and so λ > 0. Consequence: for A SPD there s no x s.t. Ax = 0, and so A 1 exists.
22 So Can/How Do We Invert C? The invertibility of the covariance matrix is very important: we ve seen a need for it in the efficient frontiers and Markowitz problem, if it s not invertible then there are redundant assets, we ll need it to be invertible when we discuss multivariate Gaussian distributions. If there is no riskfree, our covariance matrix is practically invertible by definition: ( ) 0 < var w i r i = w i w j C ij = w Cw i i,j where w R N \ {0} is any allocation vector (w may not sum to 1). Hence, C is SPD C 1 exists.
23 So How Do We Know Ĉ is Covariance Matrix? Recall, Ĉ = 1 T 1 T t=1 (r t r)(r t r). Each summand is symmetric, (r t ˆ r)(r t ˆ r) is symmetric, and sum of symmetric matrices is symmetric. But each summand is not invertible because there exists a vector y s.t. y (r t ˆ r) = 0, which means that zero is an eigenvalue, (r t ˆ r) (r t ˆ r) y = 0. } {{ } =0
24 So How Do We Know Ĉ is Covariance Matrix? If span(r 1 ˆ r, r 2 ˆ r,..., r T ˆ r) = R N then Ĉ is invertible. So need T N. In order for Ĉ to be close C will need T N. Ĉ is also SPD: ( y Ĉy = y 1 T 1 ) T (r t ˆ r)(r t ˆ r) y t=1 = 1 T 1 T t=1 y (r t ˆ r)(r t ˆ r) y = 1 T 1 because at least 1 t s.t. y (r t ˆ r) 0. T (y (r t ˆ r)) 2 > 0, t=1
25 A Simple Covariance Matrix C = (1 ρ)i + ρu where I is the identity and U ij = 1 for all i, j. U1 = N1 Ux = 0 if x 1 = 0 where 1. = (1, 1,..., 1). Hence, for any x we have Cx = (1 ρ)ix + ρux = (1 ρ)x + ρ(1 x)1, which means x is an eigenvector iff 1 x = 0 or x = a1. The eigenvalues of C are 1 ρ + Nρ and 1 ρ, and the eigenvectors are 1 and the N 1 vectors orthogonal to 1, respectively.
26 Positive Definite (Complex Hermitian Matrices) Let C be a square matrix, i.e. C C N N. C is positive definite if z Cz > 0 z C N \ {0} where C N is Ndimensional complex vectors, and z is conjugate transpose, C is positive semidefinite if z = (z r + iz i ) = z r iz i. z Cz 0 z C N \ {0} If C positive definite and real, then it must also be symmetric i.e. C nm = C mn m, n.
27 Gershgorin Circles Proposition All eigenvalues of the matrix A lie in one of the Gershgorin circles, λ A ii j i A ij for at least 1 i N, where λ is an eigenvalue of A. pf: Let λ be an eigenvalue and let x be an eigenvector. Choose i so that x i = max j x j. We have j A ijx j = λx j. W.L.O.G. we can take x i = 1 so that j i A ijx j = λ A ii, and taking absolute values yields the result, λ A ii = A ij x j j i A ij x j A ij. j i j i
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