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Econometrics - Introduction to Matrix Algebra
Yale-NUS, YSS2211
February 3, 2016
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So far: 1 regressor (i.e. y = f (x))
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Next: 2 regressors (i.e. y = f (x1, x2))
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Next: 2 regressors (i.e. y = f (x1, x2))
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Next: 2 regressors (i.e. y = f (x1, x2))
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Next: 2 regressors (i.e. y = f (x1, x2))
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Next: 2 regressors (i.e. y = f (x1, x2))
Note: last four slides it’s the same graph, just different anglesThe model is:y = α + β1x1 − β2x2and to be precise, the four previous slides represent:y = 2 + 2x1 − x2
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So far
y = α + βx + ε (1)
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So far
in more detail y1 = α + βx1 + ε1y2 = α + βx2 + ε2y3 = α + βx3 + ε3
...yN = α + βxN + εN
(2)
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So far
which in matrix algebra isy1y2y3...yN
=
ααα...α
+
x1βx2βx3β
...xNβ
+
ε1ε2ε3...εN
(3)
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So far
which in matrix algebra isy1y2y3...yN
=
111...1
α +
x1x2x3...xN
β +
ε1ε2ε3...εN
(4)
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That is,
~y = α + β~x + ~ε (5)
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Let’s add one more explanatory variable
For instance, wage depends on education (x1) and age (X2)y1 = α + β1x1,1 + β2x2,1 + ε1y2 = α + β1x1,2 + β2x2,2 + ε2y3 = α + β1x1,3 + β2x2,3 + ε3
...yN − α + β1x1,N + β2x2,N + εN
(6)
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Let’s add one more explanatory variable
We want to estimate α, β1, β2. Like in the case with only oneregressor, we will minimize the sum of squared errors.
y1 − α + β1x1,1 + β2x2,1 = ε1y2 − α + β1x1,2 + β2x2,2 = ε2y3 − α + β1x1,3 + β2x2,3 = ε3
...yN − α + β1x1,N + β2x2,N = εN
(7)
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Our problem
We want to estimate α, β1, β2. Like in the case with only oneregressor, we will minimize the sum of squared errors.
Minα,β1,β2
N∑i=1
ε2i = Minα,β1,β2
N∑i=1
(yi − α + β1x1,i + β2x2,i )2 (8)
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First Order Conditions
[α] Nα̂ + β̂1
N∑i=1
x1,i + β̂2
N∑i=1
x2,i =N∑i=1
yi
[β1] α̂
N∑i=1
x1,i + β̂1
N∑i=1
x21,i + β̂2
N∑i=1
x1,ix2,i =N∑i=1
x1,iyi
[β2] α̂N∑i=1
x2,i + β̂1
N∑i=1
x1,ix2,i + β̂2
N∑i=1
x22,i =N∑i=1
x2,iyi
(9)
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First Order Conditions
This is t-e-d-i-o-u-s!
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Solution: Matrix Algebra
y1y2y3...yN
=
ααα...α
+
x1,1β1x1,2β1x1,3β1
...x1,Nβ1
+
x2,1β2x2,2β2x2,3β2
...x2,Nβ2
+
ε1ε2ε3...εN
(10)
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Solution: Matrix Algebra
y1y2y3...yN
=
111...1
α +
x1,1x1,2x1,3
...x1,N
β1 +
x2,1x2,2x2,3
...x2,N
β2 +
ε1ε2ε3...εN
(11)
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Solution: Matrix Algebra
y1y2y3...yN
︸ ︷︷ ︸
y
=
1 x1,1 x2,11 x1,2 x2,11 x1,3 x2,1...1 x1,N x2,1
︸ ︷︷ ︸
X
αβ1β2
︸ ︷︷ ︸
β
+
ε1ε2ε3...εN
︸ ︷︷ ︸
ε
(12)
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Solution: Matrix Algebra
y = Xβ + ε (13)
Note, using Matrix properties,
y1 = 1× α + x1,1 × β1 + x2,1 × β2 + ε1
and generally, for any observation i in our sample,
yi = α + x1,i × β1 + x2,i × β2 + εi
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Our Problem is the same as above (see equation (7))
We want to estimate α, β1, β2. Like in the case with only oneregressor, we will minimize the sum of squared errors.
y1 − α + β1x1,1 + β2x2,1 = ε1y2 − α + β1x1,2 + β2x2,2 = ε2y3 − α + β1x1,3 + β2x2,3 = ε3
...yN − α + β1x1,N + β2x2,N = εN
but now it looks nicer:y − Xβ = ε. Our problem is to find the argument that minimizes
Min ε′ε = Min (y − Xβ)′ (y − Xβ)
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Vector multiplication
Why do we write
Min ε′ε = Min (y − Xβ)′ (y − Xβ)
instead of
Min ε2 = Min (y − Xβ)2 ?
This is a useful convention. We cannot square or cube matrices,but we can always pre-multiply them by themselves, orpost-multiply them by themselves
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Vector multiplication: What is ε′ε ?
ε′ε =
ε1ε2ε3...εN
T
ε1ε2ε3...εN
i.e.
ε′ε =(ε1 ε2 ε3 . . . εN
)︸ ︷︷ ︸1× N
ε1ε2ε3...εN
︸ ︷︷ ︸N × 1
= ε21+ε22+ε23+...+ε2n =N∑i=1
ε2i
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Note: ε′ε 6= εε′ !
εε′ =
ε1ε2ε3...εN
ε1ε2ε3...εN
T
i.e.
εε′ =
ε1ε2ε3...εN
︸ ︷︷ ︸N × 1
(ε1 ε2 ε3 . . . εN
)︸ ︷︷ ︸1× N
=
ε21 ε1ε2 ε1ε3 . . . ε1εNε2ε1 ε22 ε2ε3 . . . ε2εNε3ε1 ε3ε2 ε23 . . . ε3εN
......
.... . .
...εNε1 εNε2 εNε3 . . . ε2N
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Our Problem is the same as above (see equation (7))
We want to estimate α, β1, β2 in y = Xβ + ε. Like in the casewith only one regressor, we will minimize the sum of squarederrors. That is, our problem is to find the argument that minimizes
Minβ ε′ε = Min (y − Xβ)′ (y − Xβ)
where β is a vector of parameters: β =
αβ1β2
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OLS in Matrix Algebra
Minβ ε′ε = Min (y − Xβ)′ (y − Xβ)
= Min(y ′ − β′X ′
)(y − Xβ)
= Min(y ′y − β′X ′y − y ′Xβ + β′X ′Xβ
)= Min
(y ′y − 2β′X ′y + β′X ′Xβ
)where the last step is due to the fact that β′X ′y is a scalar:(1× k)(k × N)(N × 1), and we know λ′ = λ for any scalar λAlso note: y ′y is a scalar, and β′X ′Xβ is also a scalar
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Matrix Differentiation
Let c be a k × 1 vector: c =
c1...ck
Let β be a k × 1 vector of parameters: β =
β1...βk
c ′β = ?
c1β1 + c2β2 + ...+ ckβk
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Matrix Differentiation
Let c be a k × 1 vector: c =
c1...ck
Let β be a k × 1 vector of parameters: β =
β1...βk
c ′β = ? c1β1 + c2β2 + ...+ ckβk
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Matrix Differentiation
What is then ∂(c ′β)∂β ?
∂(c ′β)∂β =
∂(c1β1+c2β2+...+ckβk )
∂β1∂(c1β1+c2β2+...+ckβk )
∂β2...
∂(c1β1+c2β2+...+ckβk )∂βk
=
c1c2...ck
= c
Note: since c ′β = β′c ⇒ ∂(β′c)∂β = c
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OLS in Matrix Algebra: FOC [ ∂(β′c)∂β = c and ∂(c ′β)
∂β = c]
∂ (y ′y − 2β′X ′y + β′X ′Xβ)
∂β
= −2X ′y + X ′Xβ + (β′X ′X )′ =
= −2X ′y + X ′Xβ + (X ′Xβ) =
= −2X ′y + 2X ′Xβ = 0
⇒ X ′y = X ′Xβ
⇒ (X ′X )−1X ′y = β̂
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OLS in Matrix Algebra: FOC
Note: (X ′X )−1X ′y = β̂ looks very much likeσxyσ2x
= β̂
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