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

≤ 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 k²^q = supE kqiT k⁴^q < ∞,

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

T ^,X /(^x2,i

N ÷ V'

^x3,i

—Ex2,i )(x2,i —Ex2,i)⁰

—EFzi X3,i) (x2,i —Ex2,i)

(^x2,i — Ex2,i)⁽^x3,i — EFzi ^χ3,i)0

(^x3,i — EFzi ^x3,iX^x^3,ii —EFzi ^x3,i¢ '

= ⅛∑ τ∑∑
i ts

/ (x2,it — Ex2,it)

× (x2,is — Ex2,is)⁰

(^x3,it — EFzi ^x3,it)

× (x2,is — Ex2,is)⁰

(X2,it — EX2,it) ∖

× (^x3,is EFzi ^x3,is¢

x3,it — EFz_i x3,it ₀

× (^x3,is EFzi ^x3,is¢ /

- ¹ , 1∙ ¹ pn - ^Г^γ22 ^Г23 A

= Ν∑Q∙T →^p^lNm N XEQiT -(^ Γ2₃ Γ33 J^,