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

where the partition is made conformable to the size of

00     0     0     0     0     0

x1,it, x21,it, x22,it, x31,it, x32,it, x33,it, zi .

We now consider each element in N Pi I₁,i,NTI'_{1 i N}T. For I1,1, notice that by

Lemma 15(a),

N X (xι,i - Exι,i) (xι,i — Exι,i)⁰

i

≤ N X IKi — EXι,ik2 ≤ N X T X kxι,it - Exι,itk2 = Op (1).

Since each element in the diagonal matrix D1T tends to zero, we have

I1,1 = N X D1T (Xι,i — Exι,i) (xι,i — Ex1,i)⁰ D1T

= D1T Oⁱ p(1)D1T=op(1).

Next we consider the second diagonal block of ɪ Pi Iι,i,NTI'ι i _NT. Define

_ √τ ( X2,i — Ex2,i ʌ

qiT = T     X3,i — EFzix3,i J;

QiT = qiT qi⁰T .

Then, by Lemma 15 (b) and (d), for some constant q>1,

supE kQiT k^2q = supE kqiT k^4q < ∞,

which verifies the condition (37) of Corollary 14. In consequence, from Corollary
14 and Assumption 8(i), we have

-   ¹           , 1∙   ¹ pn - ^{Г γ}22  ^Г23 A

= Ν∑Q∙T →^{p l}Nm N XEQiT -(^ Γ2₃ Γ33 J^,

ii

39

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