INFERENCE BASED ON OLS

Econometría Avanzada

CAN Estimators

Henceforth, we consider a parameter space Θand a “true” parameter

θ0 2Θ.

The estimator θn of the r 1 parameter vectorθ0 is CAN if,

n1/2(θn θ0) d

!Nr(0,D),

and it is usually said that,

θn

asy Nr(θ0,AsyVar(θn)),

and

AsyVar(θn) =

1

nD

CAN Estimators

Henceforth, we consider a parameter space Θand a “true” parameter

θ0 2Θ.

The estimatorθn of the r 1 parameter vectorθ0 is CAN if,

n1/2(θn θ0)!d Nr(0,D),

and it is usually said that,

θn

asy Nr(θ0,AsyVar(θn)),

and

AsyVar(θn) =

1

nD

CAN Estimators

Henceforth, we consider a parameter space Θand a “true” parameter

θ0 2Θ.

The estimatorθn of the r 1 parameter vectorθ0 is CAN if,

n1/2(θn θ0) d

!Nr(0,D),

and it is usually said that,

θn

asy Nr(θ0,AsyVar(θn)),

and

AsyVar(θn) =

1

nD

CAN Estimators

Henceforth, we consider a parameter space Θand a “true” parameter

θ0 2Θ.

The estimatorθn of the r 1 parameter vectorθ0 is CAN if,

n1/2(θn θ0) d

!Nr(0,D),

and it is usually said that,

θn

asy Nr(θ0,AsyVar(θn)),

and

AsyVar(θn) = 1

nD

CAN Estimators

Henceforth, we consider a parameter space Θand a “true” parameter

θ0 2Θ.

The estimatorθn of the r 1 parameter vectorθ0 is CAN if,

n1/2(θn θ0) d

!Nr(0,D),

and it is usually said that,

θn

asy Nr(θ0,AsyVar(θn)),

and

Example:Given W1, ...,Wn scalar r.v. withE(W1) =µand

V(W1) =σ2,the sample mean µ_n =En(W)satis…es

n1/2(µ_n µ)!d N 0,σ2 .

Thus,

AsyVar(µ_n) = σ 2

Example:Given W1, ...,Wn scalar r.v. withE(W1) =µand

V(W1) =σ2,the sample mean µ_n =En(W)satis…es

n1/2(µ_n µ)!d N 0,σ2 .

Thus,

AsyVar(µ_n) = σ

2

The matrix D can usually be consistently estimated from the sample, byDn say, and the corresponding estimator of the asymptotic

variance of θn is,

\

AsyVar(θn) = 1

nDn.

Therefore,

\

AsyVar(θn) 1/2(θn θ0) d

The matrix D can usually be consistently estimated from the sample,

byDn say, and the corresponding estimator of the asymptotic

variance of θn is,

\

AsyVar(θn) =

1

nDn. Therefore,

\

Notation

We assume that W1, ...,Wn,with Wi = (Yi,Zi0)0 is a random sample

of W= (Y,Z0)0 (i.e. are independent copies of the random vector

W), Y is scalar, andZis a p 1 valued vector. We assume that:

Assumption A1: V(Z)exists and is non singular.

De…ningZ= 1

Z , A1 can alternatively be written as:

Notation

We assume that W1, ...,Wn,with Wi = (Yi,Zi0)0 is a random sample

of W= (Y,Z0)0 (i.e. are independent copies of the random vector

W), Y is scalar, andZis a p 1 valued vector. We assume that:

Assumption A1: V(Z)exists and is non singular.

De…ningZ= 1

Z , A1 can alternatively be written as:

Notation

We assume that W1, ...,Wn,with Wi = (Yi,Zi0)0 is a random sample

of W= (Y,Z0)0 (i.e. are independent copies of the random vector

W), Y is scalar, andZis a p 1 valued vector. We assume that:

Assumption A1: V(Z)exists and is non singular.

De…ningZ= 1

Z , A1 can alternatively be written as:

Notation

We assume that W1, ...,Wn,with Wi = (Yi,Zi0)0 is a random sample

of W= (Y,Z0)0 (i.e. are independent copies of the random vector

W), Y is scalar, andZis a p 1 valued vector. We assume that:

Assumption A1: V(Z)exists and is non singular.

De…ningZ= 1

Z , A1 can alternatively be written as:

The best linear predictor of Y given Zis the linear projector:

L(Y_j1,Z) = β₀+Z0β

= Z0β,

where,

β = β0

β =

E(Y) E(Z)0β

V(Z) 1C(Z,Y) =E ZZ

0 1

E(ZY).

Therefore,U =Y Z0βsatis…es that,E(U) =E(ZU) =0,or

more compactly E(ZU) =0.

TheOLS estimator of βis its sample analog:

β_n =En ZZ0

1

The best linear predictor of Y given Zis the linear projector:

L(Y_j1,Z) = β₀+Z0β = Z0β,

where,

β = β0

β =

E(Y) E(Z)0β

V(Z) 1C(Z,Y) =E ZZ

0 1

E(ZY).

Therefore,U =Y Z0βsatis…es that,E(U) =E(ZU) =0,or

more compactly E(ZU) =0.

TheOLS estimator of βis its sample analog:

β_n =En ZZ0

1

The best linear predictor of Y given Zis the linear projector:

L(Y_j1,Z) = β₀+Z0β = Z0β,

where,

β = β0

β =

E(Y) E(Z)0β

V(Z) 1C(Z,Y) =E ZZ

0 1

E(ZY).

Therefore,U =Y Z0βsatis…es that,E(U) =E(ZU) =0,or more compactly E(ZU) =0.

TheOLS estimator of βis its sample analog:

β_n =En ZZ0

1

The best linear predictor of Y given Zis the linear projector:

L(Y_j1,Z) = β₀+Z0β = Z0β,

where,

β = β0

β =

E(Y) E(Z)0β

V(Z) 1C(Z,Y) =E ZZ

0 1

E(ZY).

Therefore,U =Y Z0βsatis…es that,E(U) =E(ZU) =0,or

more compactly E(ZU) =0.

TheOLS estimator of β is its sample analog:

Consistency

Theorem (Consistency)Under A1,theOLS estimator of βis consistent.

PROOF:

β_n =β+En ZZ0

1

En(ZU)

so

β_n = β+ En ZZ0

1

| {z }

p

!E ZZ0 1 LLN & Slutsky

En(ZU)

| {z }

p

!0

LLN

| {z }

Consistency

Theorem (Consistency)Under A1,theOLS estimator of βis

consistent.

PROOF:

β_n =β+En ZZ0 1

En(ZU)

so

β_n = β+ En ZZ0

1

| {z }

p

!E ZZ0 1 LLN & Slutsky

En(ZU)

| {z }

p

!0

LLN

| {z }

Consistency

Theorem (Consistency)Under A1,theOLS estimator of βis

consistent. PROOF:

β_n =β+En ZZ0

1

En(ZU)

so

β_n = β+ En ZZ0

1

| {z }

p

!E ZZ0 1 LLN & Slutsky

En(ZU)

| {z }

p

!0

LLN

| {z }

Consistency

Theorem (Consistency)Under A1,theOLS estimator of βis

consistent. PROOF:

β_n =β+En ZZ0

1

En(ZU)

so

β_n = β+ En ZZ0 1

| {z }

p

!E ZZ0 1 LLN & Slutsky

En(ZU)

| {z }

p

!0

LLN

Asymptotic Normality

Assumption A2 E ZZ0U2 exists.

Theorem (Asymptotic Normality)Under A1 and A2, β

n is CAN

with

AsyVar β

n =

1

nE ZZ

0 1 _{E ZZ}0_U2 _{E ZZ0} 1_.

proof:

n1/2 β_n β =En ZZ0

1

Asymptotic Normality

Assumption A2 E ZZ0U2 exists.

Theorem (Asymptotic Normality)Under A1 and A2, β

n is CAN

with

AsyVar β_n =

1

nE ZZ

0 1 _{E ZZ}0_U2 _{E ZZ0} 1_.

proof:

n1/2 β_n β =En ZZ0

1

Asymptotic Normality

Assumption A2 E ZZ0U2 exists.

Theorem (Asymptotic Normality)Under A1 and A2, β

n is CAN

with

AsyVar β_n =

1

nE ZZ

0 1 _{E ZZ}0_U2 _{E ZZ0} 1_.

proof:

n1/2 β_n β =En ZZ0

1

n1/2 β

n β = En ZZ

0 1

| {z }

p

!E ZZ0 1

by LLN

n1/2En(ZU)

| {z }

d

!Np+1 0,E ZZ0U2

by the CLT

| {z }

d

!Np+1 0,E ZZ0 1

Assumption A2’(Homoskedasticity)

E ZZ0U2 _{=E ZZ}0 _{E U}2 _{with E U}2 _<∞.

Corollary: Under A1 and A2’, β

n is CAN with

AsyVar β_n = 1

nE U

2 _{E ZZ0} 1_.

We shall call

Assumption A2’(Homoskedasticity)

E ZZ0U2 _{=E ZZ}0 _{E U}2 _{with E U}2 _<∞. Corollary: Under A1 and A2’, β

n is CAN with

AsyVar β_n = 1

nE U

2 _{E ZZ0} 1

.

We shall call

Assumption A2’(Homoskedasticity)

E ZZ0U2 _{=E ZZ}0 _{E U}2 _{with E U}2 _<∞.

Corollary: Under A1 and A2’, β

n is CAN with

AsyVar β_n = 1

nE U

2 _{E ZZ0} 1

.

We shall call

AsyVar estimates

Consider OLS residuals Uni =Yi Zi0β_n.

IHomoskedasticity: AsyVar β_n is estimated by:

^

AsyVar β

n =

1

nEn U 2

n En ZZ0

1

.

IHeteroskedasticity: AsyVar β_n is estimated by:

\

AsyVar β_n = 1

n En ZZ

0 1

En ZZ0Un2 En ZZ0

1

AsyVar estimates

Consider OLS residuals Uni =Yi Zi0β_n.

IHomoskedasticity: AsyVar β_n is estimated by:

^

AsyVar β

n =

1

nEn U

2

n En ZZ0 1

.

IHeteroskedasticity: AsyVar β_n is estimated by:

\

AsyVar β_n = 1

n En ZZ

0 1

En ZZ0Un2 En ZZ0

1

AsyVar estimates

Consider OLS residuals Uni =Yi Zi0β_n.

IHomoskedasticity: AsyVar β_n is estimated by:

^

AsyVar β

n =

1

nEn U 2

n En ZZ0

1

.

IHeteroskedasticity: AsyVar β_n is estimated by:

\

AsyVar β = 1 En ZZ0

1

En ZZ0Un2 En ZZ0

1

Next, we provide a theorem that justi…es the consistency of AsyVar^ β

n

Theorem: Under A1, ifE U2 exists,

PROOF: Since

Un =U Z0(β_n β)

En Un2 = En U2 + β β

n

0

En ZZ0 β β

n

+2 En(ZU)0 β β

n

By the LLN and consistency of β_n

En Un2 = En U2

| {z } p

!E(U2₎

+ β β

n

| {z }

=op(1)

0

En ZZ0

| {z } p

!E(ZZ0)

β β

n

| {z }

=op(1) +2 En(ZU)0

| {z }

p

!00

β β

n

| {z }

=op(1) = E U2

| {z } =σ2

PROOF: Since

Un =U Z0(β_n β)

En Un2 = En U2 + β β

n

0

En ZZ0 β β

n

+2 En(ZU)0 β β

n

By the LLN and consistency of β_n

En Un2 = En U2

| {z } p

!E(U2₎

+ β β

n

| {z }

=op(1)

0

En ZZ0

| {z } p

!E(ZZ0)

β β

n

| {z }

=op(1) +2 En(ZU)0

| {z }

p

!00

β β

n

| {z }

=op(1) = E U2

| {z } =σ2

PROOF: Since

Un =U Z0(β_n β)

En Un2 = En U2 + β β

n

0

En ZZ0 β β

n

+2 En(ZU)0 β β

n

By the LLN and consistency of β_n

En Un2 = En U2

| {z } p

!E(U2₎

+ β β

n

| {z }

=op(1)

0

En ZZ0

| {z } p

!E(ZZ0)

β β

n

| {z }

=op(1) +2 En(ZU)0

| {z }

p

!00

β β

n

| {z }

=op(1) = E U2

| {z } =σ2

PROOF: Since

Un =U Z0(β_n β)

En Un2 = En U2 + β β

n

0

En ZZ0 β β

n

+2 En(ZU)0 β β

n

By the LLN and consistency of β_n

En Un2 = En U2

| {z } p !E(U2₎

+ β β

n

| {z }

=op(1) 0

En ZZ0

| {z } p !E(ZZ0)

β β

n

| {z }

=op(1)

+2 En(ZU)0

| {z }

p !00

β β

n

| {z }

=op(1)

Assumption A3 E_kZ_k4 <∞ andE_jU_j4 <∞.

Next, we provide a theorem that justi…es the consistency of

\

AsyVar β_n

Theorem. UnderA1,A2 and A3,

En ZZ0Un2 =E ZZ0U2 +op(1)

PROOF: Since,

En ZZ0Un2 = En ZZ0U2 +En ZZ0 β β_n 0

ZZ0 β β_n

+2 En ZZ0 β β_n 0

Assumption A3 E_kZ_k4 <∞ andE_jU_j4 <∞.

Next, we provide a theorem that justi…es the consistency of \

AsyVar β_n

Theorem. UnderA1,A2 and A3,

En ZZ0Un2 =E ZZ0U2 +op(1)

PROOF: Since,

En ZZ0Un2 = En ZZ0U2 +En ZZ0 β β_n 0

ZZ0 β β_n

+2 En ZZ0 β β_n 0

Assumption A3 E_kZ_k4 <∞ andE_jU_j4 <∞.

Next, we provide a theorem that justi…es the consistency of

\

AsyVar β_n

Theorem. UnderA1,A2 and A3,

En ZZ0Un2 =E ZZ0U2 +op(1)

PROOF: Since,

En ZZ0Un2 = En ZZ0U2 +En ZZ0 β β_n 0

ZZ0 β β_n

+2 En ZZ0 β β_n 0

Assumption A3 E_kZ_k4 <∞ andE_jU_j4 <∞.

Next, we provide a theorem that justi…es the consistency of

\

AsyVar β_n

Theorem. UnderA1,A2 and A3,

En ZZ0Un2 =E ZZ0U2 +op(1)

PROOF: Since,

En ZZ0Un2 = En ZZ0U2 +En ZZ0 β β_n

0

ZZ0 β β_n

+2 En ZZ0 β β_n 0

by the LLN and consistency of β_n,

En ZZ0Un2 E ZZ0U2 En ZZ0U2 E ZZ0U2

+ β β

n 2

EnkZk4

+2 β β_n En kZk3jUj

= op(1),

after noticing that, by Cauchy-Schwartz inequality,

E ZZ0U2 E_kZ_k4 1/2 E_jU_j4 1/2 <∞,

and by Hölder’s inequality:

E kZ_k3_jU_j hE _kZ_k4 i3/4 hE_jU_j4i1/4< ∞,

Observe that there is always a trade-o¤ between assumed moments of

by the LLN and consistency of β_n,

En ZZ0Un2 E ZZ0U2 En ZZ0U2 E ZZ0U2

+ β β

n

2

EnkZk4

+2 β β_n En kZk3jUj

= op(1),

after noticing that, by Cauchy-Schwartz inequality,

E ZZ0U2 E_kZ_k4

1/2

EjU_j4 1/2

< ∞,

and by Hölder’s inequality:

E kZ_k3_jU_j hE _kZ_k4 i3/4 hE_jU_j4i1/4< ∞,

Observe that there is always a trade-o¤ between assumed moments of

by the LLN and consistency of β_n,

En ZZ0Un2 E ZZ0U2 En ZZ0U2 E ZZ0U2

+ β β

n

2

EnkZk4

+2 β β_n En kZk3jUj

= op(1),

after noticing that, by Cauchy-Schwartz inequality,

E ZZ0U2 E_kZ_k4 1/2

EjU_j4 1/2

< ∞,

and by Hölder’s inequality:

E kZ_k3_jU_j hE _kZ_k4 i3/4 hE_jU_j4i1/4< ∞,

Observe that there is always a trade-o¤ between assumed moments of

by the LLN and consistency of β_n,

En ZZ0Un2 E ZZ0U2 En ZZ0U2 E ZZ0U2

+ β β

n

2

EnkZk4

+2 β β_n En kZk3jUj

= op(1),

after noticing that, by Cauchy-Schwartz inequality,

E ZZ0U2 E_kZ_k4 1/2

EjU_j4 1/2

< ∞,

and by Hölder’s inequality:

Detour: Recall the basic properties of a norm

kAB_{k k}A_{k k}B_k

kA+B_{k k}A_k+_kB_k

Recall Hölder’s inequality:

E(_kXY_k) (E_kX_kp)1/p(E_kY_kq)1/q

with

1

p +

1

q =1

we have used 1/p=3/4 but alternative choices could be selected

Detour: Recall the basic properties of a norm

kAB_{k k}A_{k k}B_k

kA+B_{k k}A_k+_kB_k

Recall Hölder’s inequality:

E(_kXY_k) (E_kX_kp)1/p(E_kY_kq)1/q

with

1

p +

1

q =1

we have used 1/p=3/4 but alternative choices could be selected

Detour: Recall the basic properties of a norm

kAB_{k k}A_{k k}B_k

kA+B_{k k}A_k+_kB_k

Recall Hölder’s inequality:

E(_kXY_k) (E_kX_kp)1/p(E_kY_kq)1/q

with

1

p +

1

q =1

we have used 1/p=3/4 but alternative choices could be selected

Detour: Recall the basic properties of a norm

kAB_{k k}A_{k k}B_k

kA+B_{k k}A_k+_kB_k

Recall Hölder’s inequality:

E(_kXY_k) (E_kX_kp)1/p(E_kY_kq)1/q

with

1

p +

1

q =1

we have used 1/p=3/4 but alternative choices could be selected

Detour: Recall the basic properties of a norm

kAB_{k k}A_{k k}B_k

kA+B_{k k}A_k+_kB_k

Recall Hölder’s inequality:

E(_kXY_k) (E_kX_kp)1/p(E_kY_kq)1/q

with

1

p +

1

q =1

we have used 1/p=3/4 but alternative choices could be selected

Detour: Recall the basic properties of a norm

kAB_{k k}A_{k k}B_k

kA+B_{k k}A_k+_kB_k

Recall Hölder’s inequality:

E(_kXY_k) (E_kX_kp)1/p(E_kY_kq)1/q

with

1

p +

1

q =1

we have used 1/p=3/4 but alternative choices could be selected

Why do not we use AsyVar\ β_n in all occasions?

Because the con…dence interval will be usually wider and hypothesis

tests more ine¢ cient than those obtained with AsyVar^ β

n under

homoskedasticity.

Why do not we use AsyVar\ β_n in all occasions?

Because the con…dence interval will be usually wider and hypothesis tests more ine¢ cient than those obtained with AsyVar^ β

n under

homoskedasticity.

Why do not we use AsyVar\ β_n in all occasions?

Because the con…dence interval will be usually wider and hypothesis

tests more ine¢ cient than those obtained with AsyVar^ β

n under

homoskedasticity.

GLS

Assumption A4 E(Y_jZ) =Z0βand V(YjZ) =σ2(Z).

Assumption A5 E ZZ0/σ2(Z) is nonsingular.

The GLS estimator of βis:

βGLS

n = En

ZZ0

σ2(Z) 1

En

ZY

GLS

Assumption A4 E(Y_jZ) =Z0βand V(YjZ) =σ2(Z).

Assumption A5 E ZZ0/σ2(Z) is nonsingular.

The GLS estimator of βis:

βGLS

n = En

ZZ0

σ2(Z) 1

En

ZY

GLS

Assumption A4 E(Y_jZ) =Z0βand V(YjZ) =σ2(Z). Assumption A5 E ZZ0/σ2(Z) is nonsingular.

The GLS estimator of βis:

βGLS

n = En

ZZ0

σ2(Z) 1

En

ZY

Theorem. Under A4 and A5, the GLS estimator is CAN with

AsyVar βGLS_n = 1

n E

ZZ0

σ2(Z) 1

.

PROOF:

βGLS_n = β+ En

ZZ0

σ2(Z) 1

| {z }

=Op(1)

En

ZU

σ2(Z)

| {z }

=op(1)

whereU =Y Z0β,and,

n1/2 βGLS

n β = En

ZZ0

σ2(Z) 1

| {z }

p

!E _σZZ02₍Z) 1

n1/2En

ZU

σ2(Z)

| {z }

d

Theorem. Under A4 and A5, the GLS estimator is CAN with

AsyVar βGLS_n = 1

n E

ZZ0

σ2(Z) 1

.

PROOF:

βGLS_n =β+ En

ZZ0

σ2(Z) 1

| {z }

=Op(1)

En

ZU

σ2(Z)

| {z }

=op(1)

whereU =Y Z0β,and,

n1/2 βGLS

n β = En

ZZ0

σ2(Z) 1

| {z }

p

!E _σZZ02₍Z) 1

n1/2En

ZU

σ2(Z)

| {z }

d

Theorem. Under A4 and A5, the GLS estimator is CAN with

AsyVar βGLS_n = 1

n E

ZZ0

σ2(Z) 1

.

PROOF:

βGLS_n =β+ En

ZZ0

σ2(Z) 1

| {z }

=Op(1)

En

ZU

σ2(Z)

| {z }

=op(1)

whereU =Y Z0β,and,

n1/2 βGLS

n β = En

ZZ0

σ2(Z) 1

| {z }

n1/2En

ZU

σ2(Z)

TheAsyVar βGLS_n is estimated by:

\

AsyVar βGLS

n =

1

n En

ZZ0

σ2(Z) 1

A serious problem consists of estimating σ2( ) without knowing its

TheAsyVar βGLS_n is estimated by:

\

AsyVar βGLS

n =

1

n En

ZZ0

σ2(Z) 1

A serious problem consists of estimating σ2( ) without knowing its

Theorem. Under A4 and A5,

AsyVar β_n AsyVar βGLS_n is p.s.d.

Theorem. Under A4 and A5,

AsyVar β_n AsyVar βGLS_n is p.s.d.

Feasible GLS

In some occasions we know the functional form of σ2( ),i.e.

σ2(Z) = σ2_γ(Z) with γ a given vector of parameters.

Notice that we can write,

U2 = σ2_γ(Z) +error,

and we can estimate γ substitutingU2 by the OLS residuals. Let γn

Feasible GLS

In some occasions we know the functional form of σ2( ),i.e.

σ2(Z) = σ2_γ(Z) with γ a given vector of parameters.

Notice that we can write,

U2 = σ2_γ(Z) +error,

and we can estimate γ substitutingU2 by the OLS residuals. Let γn

Feasible GLS

In some occasions we know the functional form of σ2( ),i.e.

σ2(Z) = σ2_γ(Z) with γ a given vector of parameters.

Notice that we can write,

U2 = σ2_γ(Z) +error,

and we can estimate γsubstituting U2 by the OLS residuals. Let γn

The Feasible GLS is,

βFGLS_n = "

En

ZZ0

σ2_γ

n(Z)

!# 1

En

ZY

σ2_γ

n(Z) !

.

Under suitable regularity conditions,

βFGLS_n = βGLS_n +op n 1/2 ,

The Feasible GLS is,

βFGLS_n =

" En

ZZ0

σ2_γ

n(Z)

!# 1 En

ZY

σ2_γ

n(Z)

!

.

Under suitable regularity conditions,

βFGLS_n = βGLS_n +op n 1/2 ,

The Feasible GLS is,

βFGLS_n =

" En

ZZ0

σ2_γ

n(Z)

!# 1 En

ZY

σ2_γ

n(Z)

!

.

Under suitable regularity conditions,

βFGLS_n = βGLS_n +op n 1/2 ,

Presentation of results: standard errors, con…dence

intervals and t ratios

Theorem. UnderA1,A2,and A3 and ifE ZZ0U2 is nonsingular,

\

AsyVar β_n

1/2

β_n β !d Np+1(0,Ip+1).

Under A1, A20 and ifE U2 _>0,

^

AsyVar β

n

1/2

β

n β d

Presentation of results: standard errors, con…dence

intervals and t ratios

Theorem. UnderA1,A2,and A3 and ifE ZZ0U2 is nonsingular,

\

AsyVar β_n

1/2

β_n β !d Np+1(0,Ip+1).

Under A1, A20 and ifE U2 > 0,

^

AsyVar β

n 1/2

β

n β

d

We call standard error of the coe¢ cient β_j to the estimate of

r

AsyVar β_jn .

We have then the standard errors,

SE β_jn =AsyVar^ β_jn 1/2

under homoskedasticity.

SE β_jn =AsyVar\ β_jn 1/2

under heteroskedasticity.

We call standard error of the coe¢ cient β_j to the estimate of r

AsyVar β_jn .

We have then the standard errors,

SE β_jn =AsyVar^ β_jn

1/2

under homoskedasticity.

SE β_jn =AsyVar\ β_jn 1/2

under heteroskedasticity.

We call standard error of the coe¢ cient β_j to the estimate of r

AsyVar β_jn .

We have then the standard errors,

SE β_jn =AsyVar^ β_jn 1/2

under homoskedasticity.

SE β_jn =AsyVar\ β_jn

1/2

under heteroskedasticity.

We call standard error of the coe¢ cient β_j to the estimate of r

AsyVar β_jn .

We have then the standard errors,

SE β_jn =AsyVar^ β_jn 1/2

under homoskedasticity.

SE β_jn =AsyVar\ β_jn 1/2

under heteroskedasticity.

SALIDA 1

Dependent Variable: Y Method: Least Squares Included observations: 899

White Heteroskedasticity-Consistent Standard Errors & Covariance

Variable Coe¢ cient Std. Error t-Statistic Prob.

C -17766.84 1424.957 -12.46833 0.0000

LOG(Z1) 2116.400 157.2934 13.45511 0.0000

Z2 311.4864 47.75287 6.522883 0.0000

Z3 292.3367 87.12519 3.355363 0.0008

Z4 9.276353 16.01485 0.579235 0.5626

Z5 7.441149 16.33873 0.455430 0.6489

Z6 -886.3335 220.5673 -4.018426 0.0001

These results are reported as follow,

b

Y = 17,766.8

(1424.9)

+2,116.4

(157.2)

log(Z1) +311.5

(47.8) Z2+292.3(87.1) Z3

+9.28

(16.0) Z4+(7.4416.3) Z5 (886.3220.6) Z6+(246.9119.2) Z6Z2

Therefore, an approximate con…dence interval at the 95% for β_j is

given by,

β_nj 1.96 h

\

AsyVar β

n

i1/2

jj

Under homoskedasticity, the con…dence interval is,

β_nj 1.96 h

^

AsyVar β

n

i1/2

These results are reported as follow,

b

Y = 17,766.8

(1424.9)

+2,116.4

(157.2)

log(Z1) +311.5

(47.8) Z2+292(87.1.3) Z3

+9.28

(16.0) Z4+(716.44.3) Z5 (886220..63) Z6+(246119..29) Z6Z2

Therefore, an approximate con…dence interval at the 95% for β_j is

given by,

β_nj 1.96

h \

AsyVar β

n

i1/2

jj

Under homoskedasticity, the con…dence interval is,

β_nj 1.96 h

^

AsyVar β

n

i1/2

These results are reported as follow,

b

Y = 17,766.8

(1424.9)

+2,116.4

(157.2)

log(Z1) +311.5

(47.8) Z2+292(87.1.3) Z3

+9.28

(16.0) Z4+(716.44.3) Z5 (886220..63) Z6+(246119..29) Z6Z2

Therefore, an approximate con…dence interval at the 95% for β_j is

given by,

β_nj 1.96 h

\

AsyVar β

n

i1/2

jj

Under homoskedasticity, the con…dence interval is,

Signi…cance testing of the parameter β_j consists of testing:

H0 : β_j =0 versus H1 :β_j 6=0.

Under suitable conditions (theorems in this section),

t = βnj SE β_nj

Signi…cance testing of the parameter β_j consists of testing:

H0 : β_j =0 versus H1 :β_j 6=0.

Under suitable conditions (theorems in this section),

t = βnj

SE β_nj

We call t ratio (or t test) for testing the hypothesis

H0 :b0β₀+b1β₁+....+bpβ_p =c

with b0,b1, ...,bp,c known constants, to the statistic,

t = b0βn0+b1βn1+....+bpβnp c SE b0β_n0+b1β_n1+....+bpβ_np

!d N(0,1) under H0,

with

SE b0β_n0+b1β_n1+....+bkβ_np

= 8 > > > > > < > > > > > : ∑p i=0∑

p j=0bibj

h

\

AsyVar β_n

i

ij

1/2

or

∑k

i=0∑kj=0bibj

h

^

AsyVar β_n

i

ij

1/2

We call t ratio (or t test) for testing the hypothesis

H0 :b0β₀+b1β₁+....+bpβ_p =c

with b0,b1, ...,bp,c known constants, to the statistic,

t = b0βn0+b1βn1+....+bpβnp c

SE b0β_n0+b1β_n1+....+bpβ_np

!d N(0,1) under H0,

with

SE b0β_n0+b1β_n1+....+bkβ_np

= 8 > > > > > < > > > > > : ∑p i=0∑

p j=0bibj

h

\

AsyVar β_n

i

ij

1/2

or

∑k

i=0∑kj=0bibj

h

^

AsyVar β_n

i

ij

1/2

We call t ratio (or t test) for testing the hypothesis

H0 :b0β₀+b1β₁+....+bpβ_p =c

with b0,b1, ...,bp,c known constants, to the statistic,

t = b0βn0+b1βn1+....+bpβnp c SE b0β_n0+b1β_n1+....+bpβ_np

!d N(0,1) under H0,

with

SE b0β_n0+b1β_n1+....+bkβ_np

= 8 > > > > > < > > > ∑p i=0∑

p j=0bibj

h \ AsyVar β n i ij 1/2

or

h i 1/2

Testing

notation and some things worth knowing

We focus on tests with large sample justi…cation.

These apply to a wide variety of models, often under general

conditions, and useful …nite-sample justi…cation can be given only in special circumstances.

Testing

notation and some things worth knowing

We focus on tests with large sample justi…cation.

These apply to a wide variety of models, often under general

conditions, and useful …nite-sample justi…cation can be given only in special circumstances.

Testing

notation and some things worth knowing

We focus on tests with large sample justi…cation.

These apply to a wide variety of models, often under general

conditions, and useful …nite-sample justi…cation can be given only in special circumstances.

Let

θ r 1=

θ1 θ2

q 1

s 1.

True value:

θ0 = θ01 θ02 .

Consider the null hypothesis

H0 : θ01 =0 Composite if q <r

Simple if q =r. .

(in the composite case θ02 are nuisance parameters)

Alternative hypothesis:

Let

θ

r 1= θ1 θ2

q 1

s 1.

True value:

θ0 = θ01 θ02 .

Consider the null hypothesis

H0 : θ01 =0 Composite if q <r

Simple if q =r. .

(in the composite case θ02 are nuisance parameters)

Alternative hypothesis:

Let

θ

r 1= θ1 θ2

q 1

s 1.

True value:

θ0 = θ01 θ02 .

Consider the null hypothesis

H0 : θ01 =0 Composite if q <r

Simple if q =r. .

(in the composite case θ02 are nuisance parameters)

Alternative hypothesis:

Let

θ

r 1= θ1 θ2

q 1

s 1.

True value:

θ0 = θ01 θ02 .

Consider the null hypothesis

H0 : θ01 =0 Composite if q <r

Simple if q =r. .

(in the composite case θ02 are nuisance parameters)

Alternative hypothesis:

Let

θ

r 1= θ1 θ2

q 1

s 1.

True value:

θ0 = θ01 θ02 .

Consider the null hypothesis

H0 : θ01 =0 Composite if q <r

Simple if q =r. .

(in the composite case θ02 are nuisance parameters)

Alternative hypothesis:

Example: when testing on the mean, the variance is typically a nuisance parameter.

There is no real loss of generality in H0,since we can reparametrize

our problem,θ01 could be Mβ m,say.

De…nition: For a test statisticτˆn suppose that we reject H0 when

ˆ

τn >c.

Then

Πc

(θ01) = Pr(τˆn >cjθ01)

Example: when testing on the mean, the variance is typically a nuisance parameter.

There is no real loss of generality inH0,since we can reparametrize

our problem,θ01 could be Mβ m,say.

De…nition: For a test statisticτˆn suppose that we reject H0 when

ˆ

τn >c.

Then

Πc

(θ01) = Pr(τˆn >cjθ01)

Example: when testing on the mean, the variance is typically a nuisance parameter.

There is no real loss of generality inH0,since we can reparametrize

our problem,θ01 could be Mβ m,say.

De…nition: For a test statisticτnˆ suppose that we reject H0 when

ˆ

τn >c.

Then

Πc

(θ01) = Pr(τˆn >cjθ01)

Example: when testing on the mean, the variance is typically a nuisance parameter.

There is no real loss of generality inH0,since we can reparametrize

our problem,θ01 could be Mβ m,say.

De…nition: For a test statisticτˆn suppose that we reject H0 when

ˆ

τn >c.

Then

Πc

(θ01) = Pr(τnˆ >cjθ01)

Example: when testing on the mean, the variance is typically a nuisance parameter.

There is no real loss of generality inH0,since we can reparametrize

our problem,θ01 could be Mβ m,say.

De…nition: For a test statisticτˆn suppose that we reject H0 when

ˆ

τn >c.

Then

Πc

(θ01) = Pr(τˆn >cjθ01)

De…nition: (Consistency) The test in the previous de…nition is consistent i¤

Πc₍

θ01) ! 1 as n!∞, 8θ01 6=0, 8c >0.

Unfortunately, in a given problem there are any number of statistics which have similar null distributions (whenH0 is true) and give

consistent tests.

How do we choose between them?

Consider a sequence of local alternatives (“Pitman” alternatives):

H1n :θ01 =δn 1/2 (1)

for a …xed q 1 vectorδ.The choice of δ determines the direction

of departure from H0,e.g.

δ= (1,0,. . .,0)0.

De…nition: (Consistency) The test in the previous de…nition is consistent i¤

Πc_(θ01)

! 1 as n_!∞, ₈θ01 6=0, 8c >0.

Unfortunately, in a given problem there are any number of statistics which have similar null distributions (whenH0 is true) and give

consistent tests.

How do we choose between them?

Consider a sequence of local alternatives (“Pitman” alternatives):

H1n :θ01 =δn 1/2 (1)

for a …xed q 1 vectorδ.The choice of δ determines the direction

of departure from H0,e.g.

δ= (1,0,. . .,0)0.

De…nition: (Consistency) The test in the previous de…nition is consistent i¤

Πc_(θ01)

! 1 as n_!∞, ₈θ01 6=0, 8c >0.

Unfortunately, in a given problem there are any number of statistics which have similar null distributions (whenH0 is true) and give

consistent tests.

How do we choose between them?

Consider a sequence of local alternatives (“Pitman” alternatives):

H1n :θ01 =δn 1/2 (1)

for a …xed q 1 vectorδ.The choice of δ determines the direction

of departure from H0,e.g.

δ= (1,0,. . .,0)0.

De…nition: (Consistency) The test in the previous de…nition is consistent i¤

Πc_(θ01)

! 1 as n_!∞, ₈θ01 6=0, 8c >0.

Unfortunately, in a given problem there are any number of statistics which have similar null distributions (whenH0 is true) and give

consistent tests.

How do we choose between them?

Consider a sequence of local alternatives (“Pitman” alternatives):

H1n :θ01 =δn 1/2 (1)

for a …xed q 1 vectorδ.The choice of δ determines the direction

of departure from H0,e.g.

δ= (1,0,. . .,0)0.

De…nition: (Consistency) The test in the previous de…nition is consistent i¤

Πc_(θ01)

! 1 as n_!∞, ₈θ01 6=0, 8c >0.

Unfortunately, in a given problem there are any number of statistics which have similar null distributions (whenH0 is true) and give

consistent tests.

How do we choose between them?

Consider a sequence of local alternatives (“Pitman” alternatives):

H1n :θ01 =δn 1/2 (1)

for a …xed q 1 vectorδ.The choice of δ determines the direction of departure from H0,e.g.

De…nition: (Consistency) The test in the previous de…nition is consistent i¤

Πc_(θ01)

! 1 as n_!∞, ₈θ01 6=0, 8c >0.

Unfortunately, in a given problem there are any number of statistics which have similar null distributions (whenH0 is true) and give

consistent tests.

How do we choose between them?

Consider a sequence of local alternatives (“Pitman” alternatives):

H1n :θ01 =δn 1/2 (1)

for a …xed q 1 vectorδ.The choice of δ determines the direction

of departure from H0,e.g.

De…nition: Consider two statistics τn1ˆ , τn2ˆ which reject H0 when

ˆ

τni >c and where for some r.v. X

ˆ

τni !d X under H0 i =1,2.

If under H1n

Πc

1(θ01) =_nlim

!∞Pr(τˆn1 >cjθ01)>Π c

2(θ01) =_nlim

!∞Pr(τˆn2>cjθ01)

for all c >0 and for all δ,then we say thatτˆn1 is (asymptotically)

moree¢ cient than τˆn2.

IfΠc₁(θ01) =Πc

De…nition: Consider two statistics τˆn1, τˆn2 which reject H0 when

ˆ

τni >c and where for some r.v. X

ˆ

τni d

!X under H0 i =1,2.

If under H1n

Πc

1(θ01) =_nlim

!∞Pr(τn1ˆ >cjθ01)>Π

c

2(θ01) =_nlim

!∞Pr(τn2ˆ >cjθ01)

for all c >0 and for all δ,then we say thatτˆn1 is (asymptotically)

moree¢ cient than τˆn2.

IfΠc₁(θ01) =Πc

De…nition: Consider two statistics τˆn1, τˆn2 which reject H0 when

ˆ

τni >c and where for some r.v. X

ˆ

τni d

!X under H0 i =1,2.

If under H1n

Πc

1(θ01) =_nlim

!∞Pr(τˆn1 >cjθ01)>Π c

2(θ01) =_nlim

!∞Pr(τˆn2>cjθ01)

for all c >0 and for all δ,then we say thatτn1ˆ is (asymptotically)

moree¢ cient than τn2ˆ .

IfΠc₁(θ01) =Πc

De…nition: Consider two statistics τˆn1, τˆn2 which reject H0 when

ˆ

τni >c and where for some r.v. X

ˆ

τni d

!X under H0 i =1,2.

If under H1n

Πc

1(θ01) =_nlim

!∞Pr(τˆn1 >cjθ01)>Π c

2(θ01) =_nlim

!∞Pr(τˆn2>cjθ01)

for all c >0 and for all δ,then we say thatτˆn1 is (asymptotically)

moree¢ cient than τˆn2.

De…nition: (Noncentral χ2_q) X has a non-central χ2_q distribution,

with noncentrality parameter

Λ=

q

∑

1 λ2_j

and we write

X χ2q(Λ)

if

X =

q

∑

1

(uj +λj)2, uj NID(0, 1)

that is

X =

q

∑

1

De…nition: (Noncentral χ2_q) X has a non-central χ2_q distribution,

with noncentrality parameter

Λ=

q

∑

1 λ2_j

and we write

X χ2q(Λ)

if

X =

q

∑

1

(uj +λj)2, uj NID(0, 1)

that is

X =

q

∑

1

De…nition: (Noncentral χ2_q) X has a non-central χ2_q distribution,

with noncentrality parameter

Λ=

q

∑

1 λ2_j

and we write

X χ2q(Λ)

if

X =

q

∑

1

(uj +λj)2, uj NID(0, 1)

that is

X =

q

∑

1

De…nition: (Noncentral χ2_q) X has a non-central χ2_q distribution,

with noncentrality parameter

Λ=

q

∑

1 λ2_j

and we write

X χ2q(Λ)

if

X =

q

∑

1

(uj +λj)2, uj NID(0, 1)

that is

X =

q

∑

1

Note that if you have a vector

X Np(µ,Σ)

since

Σ 1/2

(X µ) Np(0,I)

or

Σ 1/2_{X N}

p(Σ 1/2µ,I)

then

X0Σ 1X χ2_p(Λ)

with noncentrality parameter

Note that if you have a vector

X Np(µ,Σ)

since

Σ 1/2

(X µ) Np(0,I)

or

Σ 1/2_{X N}

p(Σ 1/2µ,I)

then

X0Σ 1X χ2_p(Λ)

with noncentrality parameter

Note that if you have a vector

X Np(µ,Σ)

since

Σ 1/2

(X µ) Np(0,I)

or

Σ 1/2_{X N}

p(Σ 1/2µ,I)

then

X0Σ 1X χ2_p(Λ)

with noncentrality parameter

Note that if you have a vector

X Np(µ,Σ)

since

Σ 1/2

(X µ) Np(0,I)

or

Σ 1/2_{X N}

p(Σ 1/2µ,I)

then

X0Σ 1X χ2_p(Λ)

with noncentrality parameter

Note that if you have a vector

X Np(µ,Σ)

since

Σ 1/2

(X µ) Np(0,I)

or

Σ 1/2_{X N}

p(Σ 1/2µ,I)

then

X0Σ 1X χ2_p(Λ)

with noncentrality parameter

Testing linear restrictions

Let Mbe a q (p+1)matrix withq p+1 such that

rank(M) =q, and a q 1 vectorm. The hypothesis to be tested is:

H0 :Mβ=mversus H1 :Mβ6=m,

which can also be written as:

H0 :θ1 =0 versusH1 :θ16=0,

Testing linear restrictions

Let Mbe a q (p+1)matrix withq p+1 such that

rank(M) =q, and a q 1 vectorm. The hypothesis to be tested is:

H0 :Mβ=mversus H1 :Mβ6=m,

which can also be written as:

H0 :θ1 =0 versusH1 :θ16=0,

Testing linear restrictions

Let Mbe a q (p+1)matrix withq p+1 such that

rank(M) =q, and a q 1 vectorm. The hypothesis to be tested is:

H0 :Mβ=mversus H1 :Mβ6=m,

which can also be written as:

H0 :θ1 =0 versusH1 :θ16=0,

REMARK: In what follows, for the sake of notational convenience, we remove the subindex “0” from θ1.Then, θ1 refers to the “true”

parameters, and generic values are denoted by Latin letters (b,t,etc).

We can always rearrange the components of β such that,

Mβ = M1β(1)+M2β(2) where M1 is a nonsingular q q matrix.

β

(p+1) 1 =

0 B B @

β(1)

q 1

β(2)

(p+1 q) 1 1 C C

REMARK: In what follows, for the sake of notational convenience, we remove the subindex “0” from θ1.Then, θ1 refers to the “true”

parameters, and generic values are denoted by Latin letters (b,t,etc).

We can always rearrange the components of β such that,

Mβ = M1β(1)+M2β(2) where M1 is a nonsingular q q matrix.

β

(p+1) 1 =

0 B B @

β(1)

q 1

β(2)

(p+1 q) 1 1 C C

REMARK: In what follows, for the sake of notational convenience, we remove the subindex “0” from θ1.Then, θ1 refers to the “true”

parameters, and generic values are denoted by Latin letters (b,t,etc).

We can always rearrange the components of β such that,

Mβ = M1β(1)+M2β(2) where M1 is a nonsingular q q matrix.

β

(p+1) 1

= 0 B B @

β(1)

q 1

β(2)

(p+1 q) 1

1 C C

Thus, underH0,

M1β(1)+M2β(2)=m=)

β(1) =M₁1

h

m M2β(2)

i

Thus, underH0,

M1β(1)+M2β(2)=m=)

β(1) =M₁1 h

m M2β(2) i

Example: suppose q =2:

2β₁+β₂ = 1 β₁+β₂+2β₃ = 0

so,

M= 2 1 0

1 1 2 andm=

1 0 so

Mβ=M1β(1)+M2β(2)

where

M1= 2 1

1 1 , β

(1)

= β1

β₂ , M2 =

0

2 , β

(2) =β₃

so,

β(1)=M₁1

h

m M2β(2)

i

is

β₁ β₂ =

2 1 1 1 1 1 0 0

2 β3 =

1+2β₃

Example: suppose q =2:

2β₁+β₂ = 1

β₁+β₂+2β₃ = 0

so,

M= 2 1 0

1 1 2 and m= 1 0

so

Mβ=M1β(1)+M2β(2)

where

M1= 2 1

1 1 , β

(1)

= β1

β₂ , M2 =

0

2 , β

(2) =β₃

so,

β(1)=M₁1

h

m M2β(2)

i

is

β₁ β₂ =

2 1 1 1 1 1 0 0

2 β3 =

1+2β₃

Example: suppose q =2:

2β₁+β₂ = 1

β₁+β₂+2β₃ = 0

so,

M= 2 1 0

1 1 2 and m=

1 0

so

Mβ=M1β(1)+M2β(2)

where

M1= 2 1

1 1 , β

(1)

= β1

β₂ , M2 =

0

2 , β

(2) =β₃

so,

β(1)=M₁1

h

m M2β(2)

i

is

β₁ β₂ =

2 1 1 1 1 1 0 0

2 β3 =

1+2β₃

Example: suppose q =2:

2β₁+β₂ = 1

β₁+β₂+2β₃ = 0

so,

M= 2 1 0

1 1 2 and m=

1 0 so

Mβ=M1β(1)+M2β(2)

where

M1=

2 1 1 1 , β

(1)

= β1

β₂ , M2 =

0 2 , β

(2)

=β₃

so,

β(1)=M₁1

h

m M2β(2)

i

is

β₁ β₂ =

2 1 1 1 1 1 0 0

2 β3 =

1+2β₃

Example: suppose q =2:

2β₁+β₂ = 1

β₁+β₂+2β₃ = 0

so,

M= 2 1 0

1 1 2 and m=

1 0 so

Mβ=M1β(1)+M2β(2)

where

M1=

2 1

1 1 , β

(1)

= β1

β₂ , M2 =

0

2 , β

(2) =β₃

so,

β(1)=M11

h

m M2β(2) i

is

β₁ β₂ =

2 1 1 1 1 1 0 0

2 β3 =

1+2β₃

Example: suppose q =2:

2β₁+β₂ = 1

β₁+β₂+2β₃ = 0

so,

M= 2 1 0

1 1 2 and m=

1 0 so

Mβ=M1β(1)+M2β(2)

where

M1=

2 1

1 1 , β

(1)

= β1

β₂ , M2 =

0

2 , β

(2) =β₃

so,

β(1)=M11 h

m M2β(2)

i

IRestricted OLS:

(restricted means we imposeH0)

~_β

n =arg min

b₂Rp+1E

n

h

Y Z0b 2i

s.t.Mb=m.

Arranging the Z0 correspondingly

Z0 = Z(1)0 Z(2)0 ,

the model is written

Y =Z0β+U =Z(1)0β(1)+Z(2)0β(2)+U

and replacing (2), the restricted OLS estimator of β(2),~β

(2)

n ,is the

OLS estimator in the linear model:

h

Y Z(1)0M₁1mi= hZ(2)0 Z(1)0M₁1M2iβ(2)+U,

and the restricted estimator of β(1) is

~_β(1)

n =M

1 1

h

m M2~β

(2)

n

i

IRestricted OLS:

(restricted means we imposeH0)

~_β

n =arg min b₂Rp+1E

n

h

Y Z0b 2

i

s.t.Mb=m.

Arranging the Z0 correspondingly

Z0 = Z(1)0 Z(2)0 ,

the model is written

Y =Z0β+U =Z(1)0β(1)+Z(2)0β(2)+U

and replacing (2), the restricted OLS estimator of β(2),~β

(2)

n ,is the

OLS estimator in the linear model:

h

Y Z(1)0M₁1mi= hZ(2)0 Z(1)0M₁1M2iβ(2)+U,

and the restricted estimator of β(1) is

~_β(1)

n =M

1 1

h

m M2~β

(2)

n

i

IRestricted OLS:

(restricted means we imposeH0)

~_β

n =arg min

b₂Rp+1E

n

h

Y Z0b 2 i

s.t.Mb=m.

Arranging the Z0 correspondingly

Z0 = Z(1)0 Z(2)0 ,

the model is written

Y =Z0β+U =Z(1)0β(1)+Z(2)0β(2)+U

and replacing (2), the restricted OLS estimator of β(2),~β

(2)

n ,is the

OLS estimator in the linear model:

h

Y Z(1)0M₁1mi= hZ(2)0 Z(1)0M₁1M2iβ(2)+U,

and the restricted estimator of β(1) is

~_β(1)

n =M

1 1

h

m M2~β

(2)

n

i

IRestricted OLS:

(restricted means we imposeH0)

~_β

n =arg min

b₂Rp+1E

n

h

Y Z0b 2 i

s.t.Mb=m.

Arranging the Z0 correspondingly

Z0 = Z(1)0 Z(2)0 ,

the model is written

Y =Z0β+U =Z(1)0β(1)+Z(2)0β(2)+U

and replacing (2), the restricted OLS estimator of β(2),~β

(2)

n ,is the

OLS estimator in the linear model:

h

Y Z(1)0M₁1mi= hZ(2)0 Z(1)0M₁1M2iβ(2)+U,

and the restricted estimator of β(1) is

~_β(1)

n =M

1 1

h

m M2~β

(2)

n

i

IRestricted OLS:

(restricted means we imposeH0)

~_β

n =arg min

b₂Rp+1E

n

h

Y Z0b 2 i

s.t.Mb=m.

Arranging the Z0 correspondingly

Z0 = Z(1)0 Z(2)0 ,

the model is written

Y =Z0β+U =Z(1)0β(1)+Z(2)0β(2)+U

and replacing (2), the restricted OLS estimator of β(2),~β

(2)

n ,is the

OLS estimator in the linear model: h

Y Z(1)0M₁1mi= hZ(2)0 Z(1)0M₁1M2

i

β(2)+U,

and the restricted estimator of β(1) is

~_β(1)

n =M

1 1

h

m M2~β

(2)

n

i