By Lemma 2,
nN ∑i Dx,T XiXiDχ,τ
=Op(1),
nN Pi Dχ,τzizi) ( nN Pi zizi) -1 (nN Pi ziziDχ,T)
and by definition,
Gx,T Dx-,1T = Jx-,T1 = O(1) ,
as (N, T →∞) . Thus,
θTGχ,T ©A3 — CB-1C0} Gχ,T = Op (θT) = Op (1). (89)
Next, consider
√NTGχ,T {θT [(A4 + A5) - CB3-1 (B4 + B5)] }
θ2 √tg d-√ *d Dχ,τxi(ui1+zi) Λ
T x,τ x,Tj— (NPiDχ,τχizi)(N Pi ziz0) √NP Pizi (ui+zi))ʃ
By Lemma 2, under the local alternatives to random effects (Assumption 11),
√1N Pi Dx,Txi(ui + vi)
nN Pi Dx,T xi¾i)( nN Pi ziz0) N √1N
P ´ = Op(1).
∑i zi (Ui + Vi) ) J
By definition,
Thus,
θT √ΤGχ,T D-,T = O
√ntgxt ©θT £(A4 + A5)-CB-1 (B4 + B5)]}=Op{√0 = op(1). (90)
Substituting (89) and (90) into (88), we have
√nt(bg - β) = [Gx,TA1Gx,T + op(1)] 1[√NTGx,Ta2 + op(1)]
= √NT (bw - β) + Op(1).
The last equality results from Lemma 2(a), (b) and Theorem 5. ¥
Part (b)
Similarly to Part (a), we can easily show that under the assumptions given
in Part (b), the denominator in (88) is
NT X X √ Gx,T (xit xi) (xit xi) Gx,T + op (1) . (91)
Consider the second term of the numerator of (88):
θ2√TGχ,T I
√1N Pi xi(zi + zi)
— (nN ∑i Xizi)(NN ∑i zizi)
1( √1N Pi zi (zi + zi
)´ . (92)
56
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