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-1√NT (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 [σ^A-1 - σ2 (A1 + θTF1)-1]-1 Q
= θτ√NT [A-1F1A-1A2 - A-1F2 + θTR2] '
× [σV A-1F1A-1 - σ2 θTR1] -1
×θτ√NT [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
× [σ2 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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