Seemingly unrelated equations SURE -...
Transcript of Seemingly unrelated equations SURE -...
Preliminaries. Kronecker Product
A (m × n), B (p × q)
A ⊗ B =
a11B . . . a1nB...
am1B amnB
(mp × nq)
• A1,A2 (m × n) B1, B2 (p × q)
Di = (Ai ⊗ B1) i = 1, 2
D1 + D2 = (A1 + A2) ⊗ B1
• C = (A ⊗ B)
αC = (αA ⊗ B) = (A ⊗ αB)
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Preliminaries. Kronecker Product
• Ci = (Ai ⊗ Bi) i = 1, 2
C1C2 = A1A2 ⊗ B1B2
• C = (A ⊗ B)
C′ = (A′ ⊗ B′)
• C = (A ⊗ B)
C−1 =(A−1 ⊗ B−1
)
•
tr (A ⊗ B) = tr (A) tr (B)
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Preliminaries. Kronecker Product
yt = Γxt + ǫt
This is a system of Seemingly Unrelated Regression Equations
(SURE).
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Seemingly Unrelated Regression Equations
The econometric specification is called seemingly unrelated regression
equations (SURE) because the individual regression equations have
no structural relationship in the sense that yt does not appear as an
RHS variable in the i−th (i 6= j) equation.
The OLS estimator fails to take into account cross-equation
information that can be exploited to improve estimator efficiency.
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Seemingly Unrelated Regression Equations
Hypothesis m equations
y1t = x′
1tβ1 + ε1t
...
ymt = x′
mtβm + εmt
xit
βi
(Ki × 1)
(Ki × 1)i = 1, . . . , m
E (εit |X1, . . . ,Xm ) = 0 i = 1, . . . , m
E(ε2it |X1, . . . ,Xm
)= σii i = 1, . . . , m
E (εitεit−k |X1, . . . ,Xm ) = 0 k 6= 0 i = 1, . . . , m
E (εitεjt |X1, . . . ,Xm ) = σij 6= 0 i, j = 1, . . . , m i 6= j
E (εitεjt−k |X1, . . . ,Xm ) = 0 k 6= 0
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Seemingly Unrelated Regression Equations
ǫt =
ε1t
ε2t
...
εmt
ǫt is a vector white noise
E (ǫtǫ′
t |X1, . . . ,Xm ) = Σ = σij i, j = 1, . . . , m
y1 = X1β1 + ǫ1
...
ym = Xmβm + ǫm
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Seemingly Unrelated Regression Equations
yi
Xi
ǫi
(T × 1)
(T × Ki)
(T × 1)
i = 1, . . . , m
E(ǫi |X1, . . . ,Xm
)= 0
E(ǫiεi′ |X1, . . . ,Xm
)= σiiIT
E(ǫiǫj′ |X1, . . . ,Xm
)= σijIT
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Seemingly Unrelated Regression Equations
y1
...
ym
=
X1 · · · 0...
. . ....
0 · · · Xm
β1
...
βm
+
ǫ1
...
ǫm
Y = Zδ + ǫ
where
Z =
X1 · · · 0...
. . ....
0 · · · Xm
δ =
β1
...
βm
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Seemingly Unrelated Regression Equations
E (ǫǫ′ |Z ) = Σ ⊗ IT
E (ǫǫ′ |Z ) =
E(ǫ1ǫ1′ |Z
)· · · E
(ǫ1ǫm′ |Z
)
.... . .
...
E(ǫmǫ1′ |Z
) ... E (ǫmǫm′ |Z )
=
σ11IT · · · σ1mIT
.... . .
...
σm1IT · · · σmmIT
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Seemingly Unrelated Regression Equations
Let
Σ ⊗ IT = Ω
the GLS estimator is:
βGLS =(X′Ω−1X
)−1X′Ω−1y
in this case:
δGLS =(Z′Ω−1Z
)−1Z′Ω−1Y
=(Z′ (Σ ⊗ IT )
−1Z
)−1
Z′ (Σ ⊗ IT )−1
Y
=(Z′
(Σ−1 ⊗ IT
)Z
)−1Z′
(Σ−1 ⊗ IT
)Y
This the Aitken-Zellner Estimator .
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Feasible GLS - Aitken-Zellner Estimator
Estimate of Σ
σij =ǫ
i′ǫ
j
T
ǫi = yi − yi = yi − XiβiOLS
δFGLS =(Z′
(Σ−1 ⊗ IT
)Z
)−1
Z′
(Σ−1 ⊗ IT
)Y
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Particular Cases
Two particular cases:
• Σ diagonal. There is no relation among the equations.
• Xi = X.
then δOLS = δGLS and V ar(δOLS |Z
)= V ar
(δGLS |Z
).
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Σ diagonal
Σ diagonal Assume that m = 2.
y1 = X1β1 + ǫ1
y2 = X2β2 + ǫ2
E(ǫ1ǫ1′ |X1,X1
)= σ11IT
E(ǫ2ǫ2′ |X1,X1
)= σ22IT
E(ǫ1ǫ2′ |X1,X1
)= σ12IT = 0
σ12 = 0
Σ−1 =
σ−111 0
0 σ−122
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Σ diagonal
δGLS =(Z′
(Σ−1 ⊗ IT
)Z
)−1Z′
(Σ−1 ⊗ IT
)Y
δ =
X1 0
0 X2
′
σ−111 IT 0
0 σ−122 IT
X1 0
0 X2
−1
×
X1 0
0 X2
′
σ−111 IT 0
0 σ−122 IT
y1
y2
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Σ diagonal
δGLS =
σ−111 X′
1X1 0
0 σ−122 X′
2X2
−1
σ−111 X′
1 0
0 σ−122 X′
2
y1
y2
=
(X′
1X1)−1
X′
1y1
(X′
2X2)−1
X′
2y2
= δOLS.
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same explanatory variables
Every regression function contains the same explanatory variables,
Xi = X.
Z =
X 0 · · · 0
0 X...
.... . . 0
0 · · · 0 X
Z = Im ⊗ X
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same explanatory variables
δGLS =(Z′
(Σ−1 ⊗ IT
)Z
)−1Z′
(Σ−1 ⊗ IT
)Y
=[(Im ⊗ X)′
(Σ−1 ⊗ IT
)(Im ⊗ X)
]−1(Im ⊗ X)′
(Σ−1 ⊗ IT
)Y
=[(
Σ−1 ⊗ X′)(Im ⊗ X)
]−1 (Σ−1 ⊗ X′
)Y
=[(
Σ−1 ⊗ (X′X))]−1 (
Σ−1 ⊗ X′)Y
=(Σ⊗ (X′X)
−1) (
Σ−1 ⊗ X′)Y
=[Im ⊗ (X′X)
−1X′
]Y
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same explanatory variables
δGLS =
(X′X)−1X′ 0 · · · 0
0 (X′X)−1
X′ 0...
. . ....
0 0 · · · (X′X)−1
X′
Y1
Y2
...
Ym
δGLS =
(X′X)−1
X′Y1
(X′X)−1
X′Y2
...
(X′X)−1
X′Ym
=
δOLS
V ar (δGLS |Z) =[(Im ⊗ X)′
(Σ−1 ⊗ IT
)(Im ⊗ X)
]−1
= Σ⊗ (X′X)−1
= V ar (δOLS |Z) .
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SURE
yt(I×1)
= β0(I×1)
+ B1(I×[k−1])
zt([k−1]×1)
+ ut(I×1)
β0 =
β01
...
β0I
B1 =
β′
11
...
β′
1I
yt =[
β0 B1
]
1
zt
+ ut
B =[
β0 B1
]=
β′
1
...
β′
I
β′
i =(β0i, β
′
1i
)′
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SURE
xt =
1
zt
yt = Bxt + ut
β = vec(B′
)
= vec[
β1 · · · βI
]
=
β1
...
βI
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SURE
In the SURE model where the explicative are the same in all
equations and one is the constant the true value of β0 and B1 is
obtained solving the following problem for each of the I equations:
E (yit |xt ) = arg min E
[yit − fi (x)]2
where fi (x) = β0i + z′β1i. We obtain
β0 = µy − B1µz
B1 = ΣyzΣ−1zz
The error variance is given by:
Ω = Σyy − ΣyzΣ−1zz Σzy
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SURE
OLS estimators of β0, B1,and Ω
X =
1 z′1
1 z′2...
...
1 z′T
Z = II ⊗ X
(X′X) = (II ⊗ X′X)
(X′X)−1
=(II ⊗ (X′X)
−1)
β =[II ⊗ (X′X)
−1X′
]Y
X′X =
T∑
z′t∑
zt
∑ztz
′
t
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SURE
A B
C D
−1
=
A−1 + A−1BE−1CA−1 −A−1BE−1
−E−1CA−1 E−1
E = D − CA−1B
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SURE
(X′X)−1
=
T∑
z′t∑
zt
∑(ztz
′
t)
−1
[(X′X)
−1]
11=
1
T+
1
T
∑z′t
(∑(ztz
′
t) −1
T
∑zt
∑z′t
)−1 ∑
zt1
T[(X′X)
−1]
21=
− 1
T
(∑(ztz
′
t) −1
T
∑zt
∑z′t
)−1 ∑
zt
[(X′X)
−1]
22=
(∑(ztz
′
t) −1
T
∑zt
∑z′t
)−1
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SURE
Σzz =1
T
∑(ztz
′
t) −1
T 2
∑zt
∑z′t
T Σzz =∑
(ztz′
t) −1
T
∑zt
∑z′t
(X′X)−1
X′ =1
T
1 + z′(Σ−1
zz
)z −z′Σ−1
zz
−Σ−1zz z Σ−1
zz
1 1 · · · 1
z1 z2 · · · zt
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SURE
β =[II ⊗ (X′X)
−1X′
]Y
β0 = y − B1z
B1 = ΣyzΣ−1zz
Ω = Σyy − ΣyzΣ−1zz
Σzy
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SURE
If
ut |Z ∼ IIDN (0,Ω)
then:
1. the exact distribution of β is given by:
β |Z ∼ N(β,Ω⊗ (X ′X)
−1)
2. the test statistic to verify the null hypothesis H0 : β0 = 0 is
given by:
T − I − K + 1
I
[1 + z′Σ−1
zz z]−1
β′
0Ω−1β0 |Z ∼ FI,T−I−K+1
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SURE
If
(u′
t,x′
t)′
are IID with respect to t and the LLN and the CLT hold
X′X
T
p→ E (xx′)
√T
(Im ⊗ X′)u
T
d→ N [0, Ω ⊗ E (xx′)]
The asymptotic distribution of β e data da√
T(β − β
)d→ N
[0, Ω ⊗ E (xx′)
−1]
where
E (xx′)−1
=
1 + µ′
zΣ−1zz µz −µ′
zΣ−1zz
−Σ−1zz µz Σ−1
zz
Eduardo Rossi c© - Econometria dei mercati finanziari 29
SURE
The test statistic for H0 : β0 = 0
T[1 + µ′
zΣ−1zz µz
]−1β′
0Ω−1β0
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SURE
Linear restrictions with two equations: I = 2
H0 :
β01
β02
=
0
0
Rβ =
1 0 · · · 0 0 · · · 0
0 0 · · · 0 1 · · · 0
β01
...
β02
...
=
0
0
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SURE
β′
0
[R
(Ω ⊗ (X′X)
−1)
R′
]−1
β0
= β′
0
R
σ11 σ12
σ21 σ22
⊗
1T
(1 + z′Σ−1
zz′z)
− 1T z′Σ−1
zz
− 1T Σ−1
zzz 1
T Σ−1zz
R′
−1
β0
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SURE
β′
0
[R
(Ω ⊗ (X′X)
−1)
R′
]−1
β0 = β′
0
bσ11
T
[1 + z′Σ−1
zzz] bσ12
T
[1 + z′Σ−1
zzz]bσ21
T
[1 + z′Σ−1
zzz] bσ22
T
[1 + z′Σ−1
zzz]
= T[1 + z′Σ−1
zzz]−1
β′
0
σ11 σ12
σ21 σ22
−1
β0
= T[1 + z′Σ−1
zzz]−1
β′
0Ω−1β0
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SURE
Theorem: Let x be the vector, (m × 1), x ∼ N (0, Σ). Let A be a
(m × m) matrix A ∼ Wm (n, Σ), n degrees of freedom with n ≥ m
with x and A indipendenti
(n − m + 1)
mx′A−1x ∼ Fm,n−m+1
The test statistic
x =[1 + z′Σ−1
zz z]−1/2
β0
A = T Ω
n = T − K
m = I
n − m + 1 = T − K − I + 1[1 + z′Σ−1
zzz]−1
β′
0Ω−1β0
T − I − K + 1
I|z ∼ FI,T−I−K+1
Eduardo Rossi c© - Econometria dei mercati finanziari 34
SURE
I = 2
z1t = α1 + β1zmt + u1t
z2t = α2 + β2zmt + u2t t = 1, . . . , T
E (ut |zmt ) = 0
z1t
z2t
=
α1
α2
+
β1
β2
zmt +
u1t
u2t
B =
α1 β1
α2 β2
Eduardo Rossi c© - Econometria dei mercati finanziari 35
SURE
B′ =
α1 α2
β1 β2
Y = Xδ + u
δ = vec (B′)
X =
1 zm1
......
1 zmT
Eduardo Rossi c© - Econometria dei mercati finanziari 36
SURE
X ′X =
1
... 1
zm
... zmT
1 zm1
......
1 zmT
=
T∑
zmt∑
zmt
∑z2mt
=
T Tzm
Tzm T(P
z2
mt
T
)
= T
1 zm
zm
(σ2
m + z2m
)
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