Large-N and Large-T Properties of Panel Data Estimators and the Hausman Test



for some finite constant M. But Assumption 4(i) and (iii) imply that the right-
hand side of the last inequality in (45) is finite. This completes the proof.
¥

7 Appendix B: Proofs of Main Results

Proof of Theorem 1

For the desired result, it is enough to prove that

{ω Ωαz (G, H) (ω) x} Z,

(46)


for all x R. Since the partition Π = {ι,..., ∏i, ...}of Ω generates the sigma
field
Z , we have

sup   |(Pz (G H))(ω) - (PzG)(ω)(PzH)(ω)

GG, HH

=

iI


sup

GG, HH


P (G H Πi)   P (G Πi) P (H Πi)

P∏      P∏   P∏


1Πi (ω),


where 1Πi (ω) denotes the indicator function that equals one if ω Πi, and
otherwise equals zero. Let
I = {1, 2, ..., i, ...} be the set of positive integers, and
let

Ix = i I |


sup

GG, HH


P (G H Πi)   P (G Πi) P (H Πi)

P∏      P∏   P∏


x .


Then, we have (46) because

{ω Ωαz (G,H) (ω) x} = ⅛∏ Z. ¥

Before we start proving the lemmas and theorems in Section 4, we introduce
some additional lemmas that are used repeatedly below. Recall that w
it =
x
01,it,x02,it , x03,it ,zi0 . We also repeatedly use the diagonal matrix DT defined
in Section 4.

Lemma 16 Suppose that Assumptions 1-8 hold. Define Ξ = Ξ1 + Ξ2, where

Ξ1


Ξ2


Γ22,22

Γ22,31 Γ22,32

diag

0k1 , 0

0

k21 ,     Γ22,31

Γ0

Γ22,32

Γ31,31 Γ31,32

Γ031,32 Γ32,32

,0k33,0kz ;

/ Γθι

1

ΓΘ121    0

0

rΘ132

rΘ133

0

Γ0

ΓΘ1

21

ΓΘ2121   0

0

r,μ32

γθ2,,33

0

0

00

0

0

0

0

0

00

0

0

0

0

Γ0

ΓΘ1

Γ0         0

0

Γ

g32,g32

Γ

g32 ,g33

Γ

,μ32

θ2132

μ32,μ32

0

μ32,μ32

g32 ,z

Γ0

Γ0         0

0

Γ0

g32,g33

Γ

g33,g33

Γ

ΓΘ1

,μ33

θ21,μ33

0

μ32,μ32

μ33,μ33

g33,z

0

00

0

Γ0

g32,z

Γ0

g33,z

Γ
z,z

/

37




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