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UnB - Financial Econometrics I Otavio Medeiros
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The Matrix
Otavio R. de Medeiros
UnB
Programa de Pós-Graduação em Administração
Programa Multiinstitucional e Interregional de Pós-Graduação em Ciências Contábeis UnB-UFPB-UFRN
Financial Econometrics I
UnB - Financial Econometrics I Otavio Medeiros
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Matrices• A Matrix is a collection or array of numbers• Size of a matrix is given by number of rows
and columns R x C• If a matrix has only one row, it is a row
vector• If a matrix has only one column, it is a
column vector• If R = C the matrix is a square matrix
UnB - Financial Econometrics I Otavio Medeiros
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Definitions• Matrix is a rectangular array of real
numbers with R rows and C columns.
are matrix elements.
11 12 1
21 22 2
1 2
...
...A
...
n
n
m m mn
a a a
a a a
a a a
( 1, ; 1, )ija i m j n
UnB - Financial Econometrics I Otavio Medeiros
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Definitions
• Dimension of a matrix: R x C.• Matrix 1 x 1 is a scalar.• Matrix R x 1 is a column vector.• Matrix 1 x C is a row vector.• If R = C, the matrix is square.• Sum of elements of leading diagonal = trace.• Diagonal matrix : square matrix with all elements off the leading
diagonal equal to zero.• Identity matrix: diagonal matrix with all elements in the leading
diagonal equal to one.• Zero matrix: all elements are zero.
UnB - Financial Econometrics I Otavio Medeiros
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Definitions
• Rank of a matrix: is given by the maximum number of linearly independent rows or columns contained in the matrix, e.g.:
3 42
7 9
3 61
2 4
rank
rank
UnB - Financial Econometrics I Otavio Medeiros
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Matrix Operations
• Equality: A = B if and only if A and B have the same size and aij = bij i, j.
• Addition of matrices: A+B= C if and only if A and B have the same size and aij + bij = cij i, j.
2 4 1 2 1 6
3 5 4 1 7 6
UnB - Financial Econometrics I Otavio Medeiros
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Matrix operations
• Multiplication of a scalar by a matrix:
k.A = k.[aij], i.e. every element of the matrix is multiplied by the scalar.
UnB - Financial Econometrics I Otavio Medeiros
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Matrix operations• Multiplication of matrices: if A is m x n and B is n x p,
then the product of the 2 matrices is A.B = C, where C is a m x p matrix with elements:
• Example:
Note: A.BB.A
1
n
ij ik kjk
c a b
2 4 1 2 2 ( 1) 4 4 2 2 4 1 14 8
3 5 4 1 3 ( 1) 5 4 3 2 5 1 17 11
UnB - Financial Econometrics I Otavio Medeiros
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Transpose of a matrix
• matrix transpose: if A is m x n, then the transpose of A is n x m, i.e.:
11 12 1 11 21 1
21 22 2 12 22 2
1 2 1 2
... ...
... ...A ; A '
... ...
n m
n m
m m mn n n mn
a a a a a a
a a a a a a
a a a a a a
UnB - Financial Econometrics I Otavio Medeiros
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Properties of transpose matrices
• (A+B)+C=A+(B+C)
• (A.B).C=A(B.C)
i. (A')'=A
ii. (A+B)'=A'+B'
iii. (A.B)'=B'.A'
iv. If A is square and if A=A', then A is symmetrical.
UnB - Financial Econometrics I Otavio Medeiros
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• Propriedades:
PROJEÇÕES
1
( )́´
( )´ ´ ´
( )´ ´ ´
( )
-1 -1
-1 -1
-1 -1
-1 -1
A A
A B A B
AB B A
AA = A A = I
(A ) = A
AB B A
(A´) = (A )´
UnB - Financial Econometrics I Otavio Medeiros
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Square matrices :
• Identity matrix I:
Note: A.I = I.A = A, where A has the same size as I.
1 0 0 0
0 1 0 0I
0 0 1 0
0 0 0 1
UnB - Financial Econometrics I Otavio Medeiros
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Square matrices :
• Diagonal matrix:
1
2
0 ... 0
0 ... 0
0 0 ... n
UnB - Financial Econometrics I Otavio Medeiros
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Square matrices:
• Scalar matrix = diagonal matrix, when
n .
• Zero matrix: A + 0 = A; A x0 = 0.
UnB - Financial Econometrics I Otavio Medeiros
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• Trace of a matrix:
If A is m x n and B is n x m, then AB and BA are square matrices and tr(AB) = tr (BA)
1
(A)
( A) ( (A))
n
iii
tr a
tr c c tr
UnB - Financial Econometrics I Otavio Medeiros
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Determinants • matrix 3 x 3:
2 3 2
1 1 2
3 2 2
1 2 1 2 1 12 3 2
2 2 3 2 3 2
2(2 4) 3(2 6) 2(2 3)
4 12 2 6
UnB - Financial Econometrics I Otavio Medeiros
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Determinants
• Matrix 3 x 3:
2 3 2 2 3 2 2 3
1 1 2 1 1 2 1 1
3 2 2 3 2 2 3 2
2 1 2 3 2 3 2 1 2 2 1 3 2 2 2 3 1 2
4 18 4 6 8 6 6
UnB - Financial Econometrics I Otavio Medeiros
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Inverse matrix• The inverse of a square matrix A, named A-1, is the matrix
which pre or post multiplied by A gives the identity matrix.• B = A-1 if and only if BA = AB = I• Matrix A has an inverse if and only if det A 0 (i.e. A is
non singular).• (A.B)-1 = B-1.A-1
• (A-1)’=(A’)-1 if A é symmetrical and non singular, then A-1 is symmetrical.
• If det A 0 and A is a square matrix of size n, then A has rank n.
UnB - Financial Econometrics I Otavio Medeiros
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Steps for finding an inverse matrix
• Calculation of the determinant: Kramer’s rule or cofactor matrix.
• Minor of the element aij is the determinant of the
submatrix obtained after exclusion of the i-th row and j-th column.
• Cofactor is the minor multiplied by (-1)i+j,
UnB - Financial Econometrics I Otavio Medeiros
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Steps for finding an inverse matrix
• Laplace expansion: take any row or column and get the determinant by multiplying the products of each element of row or columns by its respective cofactor.
• Cofactor matrix: matrix where each element is substituted by its cofactor.
UnB - Financial Econometrics I Otavio Medeiros
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i. Adjunct matrix is the transpose of the cofactor matrix, i.e. adj A = C’.
ii. Inverse matrix: 1 1A A
Aadj
11 12 11 12
21 22 21 22
11 11 12 12
22 12
21 11
1
11 22 12 21 11 22 12 2
11 22 11 22
12 21 12 21A
21 12 21 12
22 11 22 11
A =
adj A = C'=
1A
m a c a
m a c aa a a aC
m a c aa a a a
m a c c
a c a c
a a
a a
adjA
a a a c a a a c
22 12
1 21 11
a a
a a
UnB - Financial Econometrics I Otavio Medeiros
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Example2 x 2 matrix :
-1
4 3A A 4 3 2 3 12 6 6
2 3
3 2 3 -3C A = C' =
3 4 -2 4
3 -3 0,5 0,51 1A A=
-2 4 0,33 0,66A 6
adj
adj
UnB - Financial Econometrics I Otavio Medeiros
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Example• 3 x 3 matrix :
1 2 1 2 1 1
2 2 3 2 3 22 3 2 2 4 1
3 2 2 2 2 31 1 2 det 6 2 2 5
2 2 3 2 3 23 2 2 4 2 1
3 2 2 2 2 3
1 2 1 2 1 1
2 2 4
6 6 64 2 2
6 6 61 5 1
6 6 6
A A cofactor matrix
Inverse
UnB - Financial Econometrics I Otavio Medeiros
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Matrix differentiation:The derivative of a scalar (1 x 1) w.r.t. a column vector (n x 1) is a column
vector the elements of which are the derivatives of the scalar w.r.t each
element of the column vector, i.e.
Let be a scaly
1
2
1
2
ar and a column vector . Hence:
( )
( )( )
( )
n
n
x
x
x
y
x
yy
x
y
x
x x
x
UnB - Financial Econometrics I Otavio Medeiros
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Matrix differentiation:1 1
2 2
1 1 2 2
1
Let be a column vector and a column vector . Hence, we can write:
´
The derivative of ´ w.r.t. vector will be:
( ´ )
( ´ )
n n
n n
a x
a x
a x
a x a x a x
x
a a x x
a x
a x x
a x
a x
x
1 1 2 2
11
1 1 2 22
2 2
1 1 2 2
( )
( ´ ) ( )
( ´ ) ( )
n n
n n
nn n
n n
a x a x a x
xa
a x a x a xa
x x
aa x a x a x
x x
a x
a
a x
UnB - Financial Econometrics I Otavio Medeiros
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Matrix differentiation:
11 12 1 1
12 22 2 2
1 2
2 211 1 12 1 2 13 1 3 1 1 22 2 23 2 3 2 2
Let A be a symmetric matrix and x a column vector .
Hence:
´ 2 2 2 2 2
n
n
n n nn n
n n n n
a a a x
a a a x
a a a x
a x a x x a x x a x x a x a x x a x x
A x
x Ax
2
111 1 12 2 1
12 1 22 2 22
1 1 2 2
The derivative of ´ w.r.t. vector will be:
( ´ )
2( )( ´ )
2( )( )2
2( )( ´ )
nn n
n n
n n
n n nn n
n
a x
xa x a x a x
a x a x a xx
a x a x a x
x
x Ax x
x Ax
x Axx´Ax
x
x Ax
11 12 1 1
21 22 2 2
1 2
2 ´
n
n
n n nn n
a a a x
a a a x
a a a x
Ax 2x A
UnB - Financial Econometrics I Otavio Medeiros
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Y X1 1 1 X' = 1 1 1 1 12 1 2 1 2 2 3 31 1 22 1 3 X'X = 5 112 1 3 11 27 |X'X| = 14
cofactor m. 27 -11 Adj(X'X) = 27 -11-11 5 -11 5
(X'X)^(-1)= 1,929 -0,79 X'Y = 8 B = 0,50-0,79 0,357 19 0,50
Since = 0.5 e = 0.5, the regression equation is yt = 0.5 + 0.5xt
Solution:
Linear regression, example 1: Perform a linear regression, given that the data for the dependent variable are 1, 2, 1, 2, 2 and for the independent variable are 1, 2, 2, 3, 3.
UnB - Financial Econometrics I Otavio Medeiros
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y = 0.5x + 0.5
R2 = 0,5833
0
0,5
1
1,5
2
2,5
3
0 0,5 1 1,5 2 2,5 3 3,5 4 4,5
UnB - Financial Econometrics I Otavio Medeiros
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• Linear regression, example 2: a firm manufacturing bikes is preparing a project and wishes to find out what is the relationship between bike sales and national income (GDP). In the last 5 years, bike sales increased by 5%, 9%, 5%, 6% and 10%, whereas GDP increased by 2,5%, 4%, 3%, 2,5%, 4%. What is the relationship between bike sales and GDP?
UnB - Financial Econometrics I Otavio Medeiros
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The past relationship between bike sales growth and GDP growth is given by: yt = -0.0204 + 2.826xt
Example 2:Solution 1
Y X5% 1 2,50% X' = 1 1 1 1 19% 1 4% 0,025 0,04 0,03 0,025 0,045% 1 3%6% 1 2,50% X'X = 5 0,16
10% 1 4% 0,16 0,01 |X'X| = 0,00115
cofactor m. 0,005 -0,16 Adj(X'X) = 0,005 -0,16-0,16 5 -0,16 5
(X'X)^(-1)= 4,652 -139,1 X'Y = 0,35 B = -0,020-139,13 4348 0,012 2,826
UnB - Financial Econometrics I Otavio Medeiros
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Hint: to avoid working with decimals, we can multiply y and x by 100. To find the correct final result, divide by 100. doesn’t change.
Example 2:Solution 2
Y X5 1 2,5 X' = 1 1 1 1 19 1 4,0 2,5 4 3 2,5 45 1 3,06 1 2,5 X'X = 5 16
10 1 4,0 16 53,5 |X'X| 11,5
cofactor m 54 -16 Adj(X'X) = 53,5 -16-16 5 -16 5
(X'X)^(-1)= 4,65 -1,391 X'Y = 35 B = -2,0435 -0,02043-1,39 0,4348 119 2,8261 2,82609
UnB - Financial Econometrics I Otavio Medeiros
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y = 2,8261x - 0,0204
R2 = 0,835
-15%
-10%
-5%
0%
5%
10%
15%
20%
25%
30%
-4% -2% 0% 2% 4% 6% 8% 10%
GDP growth
Sal
es g
row
th
Graph (Excel):
UnB - Financial Econometrics I Otavio Medeiros
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Goodness of fit:• A measure of the goodness of fit of a regression is
the coefficient of determination R2, which is defined as:
2
2 1
2 2
1 1
´1 1
( ) ( )
where:
= regression residuals
values of the dependent variable
= mean of the values of the dependent variable
n
ii
n n
i ii i
i
i
ee e
Ry y y y
e
y
y
UnB - Financial Econometrics I Otavio Medeiros
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Goodness of fit:
• When all the residuals are equal to nil, R2 = 1, meaning that the regression is perfect, with all data points located on the line.
• When
then R2 = 0, meaning that there is no regression.
• Hence, the range for R2 will be: 0 < R2 < 1
• Values of R2 close to 1 indicate a good regression, while low values of R2 indicate a bad or inexisting regression.
2
1
´ ( )n
ii
y y
e e
UnB - Financial Econometrics I Otavio Medeiros
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Calculation of R2 – Example 1:
52 2
1
1 1 1 1 1 0
2 1 2 2 1,5 0,50,5
1 1 2 1 1,5 0,50,5
2 1 3 2 2 0
2 1 3 2 2 0
1´ 0,5 ( 0,5) 0,5; (1 2 1 2 2) / 5 1,6
5 ii
y y
Y = XB + e e = Y XB
e
e e
52 2 2 2 2 2
1
2
1 1,6 0,6
2 1,6 0,4
; ( ) 0,6 0,4 0,6 0,4 0,4 1,21 1,6 0,6
2 1,6 0,4
2 1,6 0,4
0,51 0,5833
1,2
ii
y y y y
R