Econ 424 Review of Matrix Algebra

Transcription

1 Econ 424 Review of Matrix Algebra Eric Zivot January 22, 205

2 Matrices and Vectors Matrix 2 A = ( ) = of rows, = of columns Square matrix : = Vector x ( ) = 2.

3 Remarks R is a matrix oriented programming language Excel can handle matrices and vectors in formulas and some functions Excel has special functions for working with matrices. There are called array functions. Must use to evaluate array function ctrl-shift-enter

4 Transpose of a Matrix Interchange rows and columns of a matrix Example A = x = A 0 = transpose of A ( ) ( ) A 0 = x 0 = h 2 3 i

5 Rfunction t(a) Excel function TRANSPOSE(matrix) ctrl-shift-enter

6 Symmetric Matrix AsquarematrixA is symmetric if Example A = 2 2 A = A 0 A 0 = 2 2 Remark: Covariance and correlation matrices are symmetric

7 Basic Matrix Operations Addition and Subtraction (element-by-element) = = = =

8 Scalar Multiplication (element-by-element) =2=scalar 3 = 2 = ( ) =

9 Matrix Multiplication (not element-by-element) A (3 2) = B (2 3) = Note: A and B are comformable matrices: of columns in = ofrows in A B (3 2) = (2 3) Remark: In general, A B 6= B A

10 Example A = A B = B = = Roperator A%*%B Excel function MMULT(matrix, matrix2) ctrl-shift-enter

11 Identity Matrix The dimensional identity matrix has all diagonal elements equal to, and all off diagonal elements equal to 0 Example I 2 = 0 0 Remark: The identity matrix plays the roll of in matrix algebra 0 I 2 A = = 2 +0 = = A

12 Similarly 2 0 A I 2 = = = A Rfunction diag(n) creates dimensional identity matrix

13 Matrix Inverse Let A = square matrix. A = inverse of A satisfies ( ) A A=I AA =I Remark: A is similar to the inverse of a number: =2 = 2 =2 2 = = 2 2=

14 Rfunction solve(a) Excel function MINVERSE(matrix) ctrl-shift-enter

15 Representing Systems of Linear Equations Using Matrix Algebra Consider the system of two linear equations + = 2 = The equations represent two straight lines which intersect at the point = 2 3 = 3

16 Matrix algebra representation: 2 = or A z = b where A = 2 z = and b =

17 We can solve for by multiplying both sides by A A A z = A b = I z = A b = z = A b or = 2 Remark: AslongaswecandeterminetheelementsinA we can solve for the values of and in the vector z Since the system of linear equations has a solution as long as the two lines intersect, we can determine the elements in A provided the two lines are not parallel.

18 There are general numerical algorithms for finding the elements of A and programs like Excel and R have these algorithms available. However, if A is a (2 2) matrix then there is a simple formula for A Let A = Then where A = det(a) det(a) = =0

19 Let s apply the above rule to find the inverse of A in our example: A = = 2 2 Notice that A A = Our solution for is then = 0 0 so that = 2 3 and = 3 z = A b = = =

20 In general, if we have linear equations in unknown variables we may write the system of equations as = = 2. = = which we may then express in matrix form as or = 2. A ( ) x ( ) = b ( )

21 The solution to the system of equations is given by x = A b where A A = I and I is the ( ) identity matrix. If the number of equations is greater than two, then we generally use numerical algorithms to find the elements in A

22 Representing Summation Using Matrix Notation X = = x ( ) = 2. ( ) =.

23 Then Equivalently x 0 = ³ 2 = = 0 x = ³ 2. = =. X = X =

24 Sum of Squares X = 2 = x 0 x = ³ 2 = = 2. X 2 =

25 Sums of cross products X = = x 0 y = ³ 2 2. = = = y 0 x X =

26 R function t(x)%*%y, t(y)%*%x crossprod(x,y) Excel function MMULT(TRANSPOSE(x),y) MMULT(TRANSPOSE(y),x) ctrl-shift-enter

27 Portfolio Math with Matrix Algebra Three Risky Asset Example Let denote the return on asset = and assume that and are jointly normally distributed with means, variances and covariances: Portfolio x Portfolio return = [ ] 2 =var( ) cov( )= = share of wealth in asset + + = = + +

28 Portfolio expected return = [ ]= + + Portfolio variance 2 =var( )= Portfolio distribution ( ) 2

29 Matrix Algebra Representation R = x = Portfolio weights sum to = Σ = x 0 =( ) = = + + =

30 Digression on Covariance Matrix Using matrix algebra, the variance-covariance matrix of the return vector R is defined as var(r) =cov(r) =[(R )(R ) 0 ]=Σ Because R has elements, Σ is the matrix Σ =

31 For the case =2,wehave à [(R ) (R ) 0 ]= ( ) 2 à ( = ) 2! ( )( 2 2 ) ( 2 2 )( ) ( 2 2 ) 2 à [( = ) 2! ] [( )( 2 2 )] [( 2 2 )( )] [( 2 2 ) 2 ] Ã! Ã! var( ) cov( = 2 ) 2 = 2 cov( 2 ) var( 2 ) = Σ!

32 Portfolio return = x 0 R =( ) = + + = R 0 x Portfolio expected return = x 0 =( ) = + + = 0 x

33 Excel formula MMULT(transpose(xvec),muvec) ctrl-shift-enter Rformula crossprod(x,mu) t(x)%*%mu

34 Portfolio variance 2 =var(x 0 R)=[ ³ x 0 R x 0 2 ]=[ ³ x 0 (R ) 2 ] = [x 0 (R )x 0 (R )] = [x 0 (R )(R ) 0 x]= = x 0 [(R )(R ) 0 ]x = x 0 Σx =( ) =

35 Excel formulas MMULT(TRANSPOSE(xvec),MMULT(sigma,xvec)) MMULT(MMULT(TRANSPOSE(xvec),sigma),xvec) ctrl-shift-enter Note: x 0 Σx = x 0 Σ x = x 0 (Σx) Rformulas t(x)%*%sigma%*%x Portfolio distribution ( 2 )

36 Covariance Between 2 Portfolio Returns 2 portfolios Portfolio returns Covariance x = y = x 0 = y 0 = = x 0 R = y 0 R cov( )=x 0 Σy = y 0 Σx

37 Derivation cov( )=cov(x 0 R y 0 R) = [(x 0 R x 0 ])(y 0 R y 0 ])] = [x 0 (R )y 0 (R )] = [x 0 (R )(R ) 0 y] = x 0 [(R )(R ) 0 ]y = x 0 Σy

38 Excel formula MMULT(TRANSPOSE(xvec),MMULT(sigma,yvec)) MMULT(TRANSPOSE(yvec),MMULT(sigma,xvec)) ctrl-shift-enter Rformula t(x)%*%sigma%*%y

39 Bivariate Normal Distribution Let and be distributed bivariate normal. The joint pdf is given by exp 2( 2 ) ( ) = Ã 2 q 2 Ã! 2 2 ( )( )! 2 + where [] = [ ]= sd() = sd( )= and = cor( )

40 Define Ã! Ã! Ã 2 X = x = Σ = 2 Then the bivariate normal distribution can be compactly expressed as where (x) = = Ã 2 det(σ) 2 2 (x )0 Σ (x )!! det(σ) = = = 2 2 ( 2 ) We use the shorthand notation X ( Σ)

41 Derivatives of Simple Matrix Functions Result: LetA be an symmetric matrix, and let x and y be an vectors. Then x x0 y = x Ax= x x0 Ax = x 0 y. x 0 y (Ax) 0. (Ax) 0 x 0 Ax. x 0 Ax = y = A =2Ax

42 We will demonstrate these results with simple examples. Let A = Ã For the firstresultwehave! x = Ã 2! x 0 y = y = Ã 2! Then x x0 y = x 0 y 2 x 0 y = ( ) 2 ( ) = Ã 2! = y

43 Next, consider the second result. Note that and Ax = Ã!Ã 2! = Ã ! (Ax) 0 =( ) Then x Ax= ( ) 2 ( ) Ã =! = A

44 Finally, consider the third result. We have x 0 Ax = ³ 2 Ã!Ã 2 Then ³ x x0 Ax= ³ Ã!Ã! =2 =2Ax 2! = = Ã !