Introduction Eigen Values and Eigen Vectors An Application Matrix Calculus Optimal Portfolio. Portfolios. Christopher Ting.

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1 Portfolios Christopher Ting Christopher Ting : christopherting@smu.edu.sg : : LKCSB 5036 November 4, 2016 Christopher Ting QF 101 Week 12 November 4, /51

2 Table of Contents 1 Introduction 2 Eigen Values and Eigen Vectors 3 An Application 4 Matrix Calculus 5 Optimal Portfolio Christopher Ting QF 101 Week 12 November 4, /51

3 Pre-U Scalar and Vector A scalar a is a single number or one dimension. N: Natural numbers Z: Integers Q: Rational numbers R: Real numbers A vector a is a k 1 list of numbers arranged in a column. a 1 a 2 a =. Rk. a k Christopher Ting QF 101 Week 12 November 4, /51

4 Pre-U Matrix A matrix A is a k r rectangular array of numbers arranged in k rows and r columns. A 11 A 12 A 1r A 21 A 22 A 2r.... A k1 A k2 A kr Regular letters are used for scalars, lower case bold letters for vectors, and upper case bold letters for matrices. By convention, a i is the element of a vector in the i-th row, and A ij refers to the element of a matrix in the i-th row and j-th column. Christopher Ting QF 101 Week 12 November 4, /51

5 Matrix Transpose The transpose of matrix is the matrix obtained by rotating each row into a column (clockwise) with the first element as pivot, i.e., for each row i, A 1i A 2i [ Ai1 A i2 A ir]. Notation-wise, we write the transpose of A as A or A, and [ Aij ] = [ Aji ]. A ri Question If A is a k r matrix, what is the length and width of A? Christopher Ting QF 101 Week 12 November 4, /51

6 Matrix Addition Two matrices A and B are addable only when they are of the same order (rows column): A + B = [A ij ] + [B ij ] = [A ij + B ij ]. Matrix addition follows the commutative and associative laws. A + B = B + A. ( A + B ) + C = A + ( B + C ). Christopher Ting QF 101 Week 12 November 4, /51

7 Inner or Dot Product If a and b are both k 1, then the sum of element-wise products is called the inner product or dot product. a b = k a h b h. h=1 The result of an inner product is a scalar. Dot product is commutative: a b = b a. In Python, inner product is coded as numpy.inner(a, b). Christopher Ting QF 101 Week 12 November 4, /51

8 Linear Combination A vector can be written as a linear combination: a a 2 a = = a a a 0 k. a k The vectors with 1 in the i-th row and the rest 0 are called the basis vectors, and denoted by e i. Compactly, k the column vector is expressed as a = a i e i ; and the row vector a as a = k a i e i. i=1 i=1 Christopher Ting QF 101 Week 12 November 4, /51

9 Linear Dependence and Independence 1 A set of vectors a i, i = 1,2,...,n are said to be linearly dependent if there exist scalars c 1,c 2,...,c n such that c 1 a 1 + c 2 a 2 + c n a n = c i a i = o. (1) 2 A set of vectors a i, i = 1,2,...,n are said to be linearly independent if they are not linearly dependent. 3 Hence, a i, i = 1,2,...,n are linearly independent iff equation (1) has only the trivial solution: c i = 0, for all i = 1,2,...,n. Christopher Ting QF 101 Week 12 November 4, /51

10 Einstein s Notation In Einstein s notation, the inner product is written as a b = e i a i b j e j = a i b j e i e j = a i b j δ i j = a ib i, where δ i is called the Kronecker delta, which is defined as j e i e j = δ i j := { 1 when i = j; 0 otherwise. When a b = 0, we say that a and b are orthogonal. Christopher Ting QF 101 Week 12 November 4, /51

11 Inspirational Digression: General Relativity Space-Time Curvature = Energy-Momentum tensor R µν 1 2 Rg µν = 8πG c 4 T µν Christopher Ting QF 101 Week 12 November 4, /51

12 Matrix Multiplication Multiplication by a scalar: If c is a non-zero and finite scalar, c A = Ac = [ A ij c ]. If A is k r and B is r s so that the number of columns of A equals the number of rows of B, we say that A and B are conformable, and the matrix product AB can be defined. a 1 a 1 a b 1 a 1 b 2 a 1 b s 2 [ ] a 2 AB = b1 b 2 b s = b 1 a 2 b 2 a 2 b s..... a k b 1 a k b 2 a k b s a k Christopher Ting QF 101 Week 12 November 4, /51

13 Matrix Multiplication Christopher Ting QF 101 Week 12 November 4, /51

14 Square Matrix and Trace A matrix is said to be square when r = k. A square matrix is said to symmetric when A = A. The trace of a k k square matrix A is defined as k tra := A ii = A ij δ ij, i=1 i.e., the sum of its diagonal elements. Christopher Ting QF 101 Week 12 November 4, /51

15 Properties of Trace For square matrices A and B and a real and finite constant c, we have tr ( ca ) = c tra; tra = tra; tr ( A + B ) = tra + trb; tri k = k. For example, the trace of the matrix [ 3 ] 4 Tr(A) = 7 9 = = 12. Christopher Ting QF 101 Week 12 November 4, /51

16 Theorem Commutativity of Trace For k r A and r k B, we have tr ( AB ) = tr ( BA ). Proof: a 1 b 1 a 1 b 2 a 1 b k tr ( AB ) a 2 = tr b 1 a 2 b 2 a 2 b k... a k b 1 a k b 2 a k b k = a i b i = b i a i = tr ( BA ). We have applied the fact that the dot product is commutative. Christopher Ting QF 101 Week 12 November 4, /51

17 Rank of a Matrix 1 The rank of a matrix is defined as the number of its linearly independent rows, which is equal to the number of its linearly independent columns, i.e., row rank = column rank. 2 The rank of a matrix A is given by the maximum number of linearly independent rows (or columns). For example, [ 3 ] 4 rank 7 9 = 2, [ 3 ] 6 rank 2 4 = 1. Christopher Ting QF 101 Week 12 November 4, /51

18 Properties of Rank A matrix with a rank equal to its dimension is a matrix of full rank. A matrix that is not full rank is known as a short rank matrix, and is singular (non-invertible). Four important properties: rank(a) = rank ( A ). rank(ca) = rank ( A ), where c is a constant. rank(ab) min ( rank(a), rank(b) ) rank ( A A ) = rank ( AA ) = rank (A). Christopher Ting QF 101 Week 12 November 4, /51

19 Rank and Inverse Matrix The rank of the k r matrix (r k) A = [ a 1 a 2 a r ] is the number of linearly independent columns a j, and is written as ranka. We say that A has full rank if ranka = r. A square k k matrix A is said to be non-singular if it is has full rank, e.g. ranka = k. If A is non-singular then there exists a unique k k matrix A 1 called the inverse of A that satisfies AA 1 = A 1 A = I k. Christopher Ting QF 101 Week 12 November 4, /51

20 Properties of Matrix Inverse For non-singular A and C, ( A 1 ) ( = A ) 1 ; ( ) 1 AC = C 1 A 1 ; ( ) 1 A + C = A 1 ( A 1 + C 1) 1 C 1 ; A 1 ( A + C ) 1 = A 1 ( A 1 + C 1) 1 A 1. If A is an orthogonal matrix, then A 1 = A. Christopher Ting QF 101 Week 12 November 4, /51

21 Determinant of Square Matrix Let A be a general k k matrix. Let (j 1,j 2,...,j k ) denote a permutation of (1,2,...,k). There are k! permutations. There is a unique count of the number of inversions of the indices of such permutations relative to the natural order (1,2,...,k), and let ɛ (j1,j 2,...,j k ) = +1 if this count is even and ɛ (j1,j 2,...,j k ) = 1 if the count is odd. Then the determinant of A is defined as deta := π ɛ π A 1j1 A 2j2 A kjk = ξ j 1j 2 j k A 1j1 A 2j2 A kjk, where { ξ j 1j 2 j k 1 if (j1,j 2,...,j k ) is even; := 1 if (j 1,j 2,...,j k ) is odd. Christopher Ting QF 101 Week 12 November 4, /51

22 Properties of Determinant For example, if k = 2, then the two permutations of (1,2) are (1,2) and (2,1), for which ɛ (1,2) = 1 and ɛ (2,1) = 1. Thus, deta = ɛ (1,2) A 11 A 22 + ɛ (2,1) A 21 A 12 = A 11 A 22 A 21 A 12. If A is non-singular, deta 0. If A is orthogonal, then deta = ±1. If A is triangular (upper or lower), then deta = k i=1 A ii. Some other properties of a k k square matrix A include deta = deta ; det ( ca ) = c k deta; det ( AB ) = deta detb; det ( A 1) = ( deta ) 1 ; [ ] A B det = detddet ( A BD 1 C ), if detd 0. C D Christopher Ting QF 101 Week 12 November 4, /51

23 Determinant of a 3 3 Matrix a 11 a 12 a 13 det(a) = a 21 a 22 a 23 a 31 a 32 a = ( ) a 11 a 22 a 33 + a 21 a 32 a 13 + a 31 a 12 a ( ) a 13 a 22 a 31 + a 23 a 32 a 11 + a 33 a 12 a a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 a 11 a 12 a 13 a 21 a 22 a 23 Christopher Ting QF 101 Week 12 November 4, /51

24 Calculating the Inverse of a 2 2 Matrix The inverse of a 2 2 non-singular matrix [ ] [ ] a b 1 d b is. c d ad bc c a The expression in the denominator, ad bc, is the determinant of the matrix. If the matrix is [ ] 2 1, 4 6 then the inverse will be 1 8 [ ] = [ ] 8. As a check, multiply the two matrices together and it should give the identity matrix I. Christopher Ting QF 101 Week 12 November 4, /

25 Eigenvalues The characteristic equation of a k k square matrix A is det ( A λi k ) = 0. It is a polynomial of degree k in λ, so it has exactly k roots, which are not necessarily distinct and may be real or complex. They are called the latent roots or characteristic roots or eigenvalues of A. When the eigenvalues are real, they are written in descending order λ 1 λ 2 λ k. We also write λ min (A) = λ k and λ max (A) = λ 1. Christopher Ting QF 101 Week 12 November 4, /51

26 Eigenvectors If λ i is an eigenvalue of A, then A λ i I k is singular, i.e., there exists a non-zero vector h i such that ( A λi I k ) hi = o. The vector h i is called a latent vector or characteristic vector or eigenvector of A corresponding to λ i. It is a fundamental result of linear algebra that an equation Mv = o has a non-zero solution v if and only if the determinant det(m)is zero. Hence, f (λ) := det ( A λi ) = 0. Christopher Ting QF 101 Week 12 November 4, /51

27 Calculating Eigenvalues: An Example Let A be the 2 2 matrix A = [ ] Then the characteristic equation is A λi 2 = 0. That is [ ] [ ] λ = [ ] 5 λ 1 = = (5 λ)(4 λ) 2 = λ 2 9λ λ The solutions are λ = 6 and λ = 3. The characteristic roots are also known as. Christopher Ting QF 101 Week 12 November 4, /51

28 A Simple PageRank Algorithm Transition matrix of the directed graph of 4 web sites: A = Christopher Ting QF 101 Week 12 November 4, /51

29 A Simple PageRank Algorithm (Cont d) Denote by x 1,x 2,x 3, and x 4 the importance of the four sites. Analyzing the situation at each node, we get the system: x 1 = 1 x x 4 x 2 = 1 3 x 1 x 3 = 1 3 x x x 4 x 4 = 1 3 x x 2 It is equivalent to Ax = x, where x := [x 1 x 2 x 3 x 4 ]. Christopher Ting QF 101 Week 12 November 4, /51

30 Finding the Eigenvector The PageRank algorithm involves finding the eigenvector corresponding to the eigenvalue of 1! x 1 x 2 x 3 x 4 = 1 Since x 2 = x 1 /3, we substitute it in x 4 to find that x 4 = x 1 /2. In turn, we find that x 3 = x x x 1 2 = 3x 1 4. Christopher Ting QF 101 Week 12 November 4, /51 x 1 x 2 x 3 x 4.

31 Finding the Eigenvector (Cont d) So the eigenvector is v := x 1 x 2 x 3 x 4 1 = x 1/3 1 3/4. 1/2 To find x 1, we impose the condition that the sum of eigenvector s entries is equal to 1. Therefore ( x ) = 1. 2 Christopher Ting QF 101 Week 12 November 4, /51

32 Finding the Eigenvector (Cont d) So x 1 = Finally, the unique eigenvector is Therefore, the PageRank is, in declining order of importance, Site 1, Site 3, Site 4, and Site 2. Christopher Ting QF 101 Week 12 November 4, /51

33 Final Result So x 1 = Finally, the unique eigenvector is Therefore, the PageRank is, in declining order of importance, Site 1, Site 3, Site 4, and Site 2. Reference: PageRank Algorithm - The Mathematics of Google Search...some of you may have heard of quants but at Google, they re just called employees, because they re all quants... James Simon Christopher Ting QF 101 Week 12 November 4, /51

34 Definition of Vector Differentiation Let x be a column k-vector. Consider the function g(x) = g(x 1,x 2,...,x k ) : R k R. The vector derivative is g(x) x 1 g(x) x g(x) = x 2.. g(x) x k and [ x g(x) = g(x) x 1 x 2 g(x) ] g(x) x k Christopher Ting QF 101 Week 12 November 4, /51

35 Basic Properties For constant vector a and matrix A, x ( a x ) = x ( x a ) = a, x ( a x ) = a x ( Ax ) = A ( x Ax ) = ( A + A ) x x 2 x x ( x Ax ) = A + A Christopher Ting QF 101 Week 12 November 4, /51

36 Assets There are n assets whose expected returns are denoted by µ i := E ( r i ), i = 1,2,...,n. The covariance between asset i and asset j is denoted as σ ij. Arrange the covariances into a n n matrix: σ 11 σ 12 σ 1n Σ := [ ] σ 21 σ 22 σ 2n σ ij = σ n1 σ n2 σ nn The diagonal element σ ii is the variance of asset i. Note that the covariance matrix Σ is symmetric. Christopher Ting QF 101 Week 12 November 4, /51

37 Investment For each dollar invested, a fraction w i is invested in asset i. It must be that n w i = 1. The weights are arranged as a n-vector w. The portfolio s expected return and variance are, respectively, i=1 µ p := E ( ) n r p = w i E ( ) r i = wi µ i = w µ; σ 2 p := n i=1 j=1 i=1 n w i w j σ ij = w i Σ i j w j = w Σw. Christopher Ting QF 101 Week 12 November 4, /51

38 Numerical Illustration Suppose there are two assets. µ 1 = 5% and µ 2 = 8% per annum. The covariance is a 2 by 2 matrix. [ σ11 σ 12 Σ := σ 21 σ 22 ] = [ Given that the variance of asset 1 is , its volatility is = 25% per annum. The volatility of asset 2 is. The portfolio s expected return and variance are, respectively, µ p = 0.05w w 2 σ 2 p = w w 1w w 2 w w 2 2 ] Christopher Ting QF 101 Week 12 November 4, /51

39 Optimization Minimize half the portfolio variance under two constrains: n w i E ( ) r i i=1 = E ( r p ). n w i = 1. i=1 Constrained optimization with Lagrange multipliers λ and ψ: ( min L = 1 n n n w i w j σ ij λ w 1,w 2,...,w n 2 i=1 j=1 ) w i µ i µ p ψ i=1 ( n w i 1 i=1 ) Christopher Ting QF 101 Week 12 November 4, /51

40 In Matrix Form The Lagrangian L is L = 1 2 w Σw λ ( w µ µ p ) ψ(w 1 1). The first-order conditions with respect to w are Σw λµ ψ1 = 0 w µ = µ p w 1 = 1 Christopher Ting QF 101 Week 12 November 4, /51

41 Solution of First FOC The first FOC gives the solution for the weight vector w = Σ 1( λµ + ψ1 ). But what are the values of the Lagrange multipliers λ and ψ? To solve for λ and ψ, substitute the optimal weight vector above into the last two FOC s, µ w = µ Σ 1( λµ + ψ1 ) = µ p 1 w = 1 Σ 1( λµ + ψ1 ) = 1 Christopher Ting QF 101 Week 12 November 4, /51

42 Solution of Second and Third FOCs Let A := µ Σ 1 µ B := µ Σ 1 1 C := 1 Σ 1 1 The last two FOCs can be written as [ ][ ] [ A B λ µp = B C ψ 1 ] Solving these two linear equations, we obtain λ = Cµ p B AC B 2, ψ = A Bµ p AC B 2 Christopher Ting QF 101 Week 12 November 4, /51

43 Optimal Weight Vector and Portfolio Variance The optimal weight is then solved as Σ 1( ( Cµp B ) µ + ( ) ) A Bµ p 1 w = AC B 2. The portfolio variance is a quadratic function of the mean portfolio return µ p : V ( ) r p = w Σw = Cµ2 p 2Bµ p + A AC B 2 C = AC B 2 µ2 p 2B AC B 2 µ A p + AC B 2. Christopher Ting QF 101 Week 12 November 4, /51

44 Global Minimum Variance Portfolio The global minimum variance portfolio is obtained by minimizing V ( r p ) with respect to µp. dv ( ) r p = 2C dµ p AC B 2 µ p 2B AC B 2. The results of the first-order condition are µ = B C, σ2 = 1 C, w = Σ 1 1 C. Christopher Ting QF 101 Week 12 November 4, /51

45 Numerical Example For the two assets, compute the inverse matrix Σ 1 = [ ] = ( 0.01) ( 0.01) [ ] = [ ] Christopher Ting QF 101 Week 12 November 4, /51

46 Values of A,B, and C So the three scalars A, B, and C are A = µ Σ 1 µ = [ ][ ][ ] = ; B = µ Σ 1 1 = [ ][ ][ ] = C = 1 Σ 1 1 = Christopher Ting QF 101 Week 12 November 4, /51

47 Global Minimum Variance Portfolio Let s consider the global minimum variance portfolio. µ = B C = , σ2 = 1 C = So the minimum volatility is σ = 20.21%. The weight vector for w for the global minimum variance portfolio is w = Σ 1 1 C = [ ][ ] 1 = 1 [ ] 70.11% 29.89% Christopher Ting QF 101 Week 12 November 4, /51

48 Efficient Frontier Christopher Ting QF 101 Week 12 November 4, /51

49 Takeaways Quantitative (mathematics and programming) skills provide a competitive advantage. Ideas of eigenvalue and eigenvector appear in Google search engine! Investment optimization: Obtain highest possible return with minimal risk. Quants way of deriving the efficient frontier Christopher Ting QF 101 Week 12 November 4, /51

50 Week 12 Assignment 1. What are the eigenvalues and eigenvectors of a diagonal matrix? Let A = (A) Show that the characteristic equation is (λ 1)(λ + 3) 2 = 0. (B) Find the eigenvector corresponding to λ 1 = 1. (C) Find the eigenvector corresponding to λ = 3. Christopher Ting QF 101 Week 12 November 4, /51

51 Week 12 Additional Exercise 1. Consider the transformation of Cartesian coordinates (x, y) to the polar coordinates (r, ϕ): (A) Show that x = r cosϕ, y = r sinϕ x x r ϕ [ ] cosϕ r sinϕ J(r,ϕ) = y y = sinϕ r cosϕ r ϕ (B) Show that detj = r. 2. Apply the Jacobian detj in Problem 1 to show that I := exp ( x2 2 ) dx = 2π. Christopher Ting QF 101 Week 12 November 4, /51