Seemingly unrelated equations SURE -...

38
Seemingly unrelated equations SURE Eduardo Rossi University of Pavia

Transcript of Seemingly unrelated equations SURE -...

Seemingly unrelated equationsSURE

Eduardo Rossi

University of Pavia

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

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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

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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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SURE

|X ′X | = σ2m + z2

m − z2m

= σ2m

(X ′X)−1

=

1 + zm/σ2m −zm/σ2

m

−zm/σ2m 1/σ2

m

1T

(1 + z′Σ−1

zz z)

− 1T z′Σ−1

zz

− 1T Σ−1

zz z 1T Σ−1

zz

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