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



Observe that for any conformable matrices P and Q, we have

(P + Q)-1 - P-1 = -P-1QP-1 + (P + Q)-1 QP-1QP-1.

Using this fact, we write

(A1 + θ2τF1)-1 - A-1 = -θTA-1F1A-1 + θTR1,         (100)

where R1 = (A1 + θ2τF1)-1 F1A-1F1A-1. Define

Q = (A1 + θTF1 )-1 NT (A2 + θTF2) - A-1 NTA2.

Then,

Q

= (A1 + θTF1)-1 NT (A2 + θTF2) - A-1NT (A2 + θTF2)

+A-1 √NT θT F2

= {(A1 + θTF1)-1 - A-11 √NT {A2 + θTF2} + A-1 NTθTF2

= -A-1 (θTF1) A-1 √NT {A2 + θTF2} + A-1 √NTθTF2

+θT R1 √NT {A2 + θT F2}

= -θT √NT [A-1F1A-1A2


-a-1f2] -θT √NT


θT A-1F1A-1F2

- θT R1 {A2 + θT F2j∙


= -θT√NT A-1F1 A-1A2 - A-1 F2] - θT√NTr2,            (101)
where
R2 = A-1F1A-1F2 - R1 {A2 + θTF2} .

In view of (100) and (101) , we now can rewrite the Hausman statistic as

HMnt = Q^A-1 - σ2 (A1 + θTF1)-1]-1 Q

= θτNT [A-1F1A-1A2 - A-1F2 + θTR2] '

× [σV A-1F1A-1 - σ2 θTR1] -1

×θτNT [A-1F1A-1A2 - A-1F2 + θTR2] ;

or equivalently,

HMnt

= θτ √NT


G1


J-TDχ,τF1Dχ,τ Jχ-T) G1Gχ,τ
-G1 Jx-1 Dχ,τ F2 + θ2 G-1τ R2


A2


× 2 G1 J'  ' F D J ' ) G1 + σ2 θT (⅛ r1g⅛) ] "ɪ

×θτ √NT


G1


Jχ,TDχ,τF1Dχ,τ J-T) G1Gχ,τA2
G1 Jx-T Dχ,τ F2 + θT G-1τ R2

61



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