J-1
Tθ2 (CD) √T
Jx,T1
I Gχ,τAιGχ,τ + -√T-Dχ,τA3Dχ,

-1
1 1 ∏r≈. J J JxT „
√Niπ? ''ntgx∙t A2 + √ D
(A4 + A5)
By Lemma 2(d), (b), and (d),
B5 = Op (1),
√NTGχ,τA2 = Op (1),
Dx,T A5 = Op (1) ,
respectively. Under the fixed effect assumption (Assumption 10), Lemma 4
implies that
Dx,T A4 = Op (1) ,
as (N, T →∞) . Since
1J -T1
r- 2 , -XT = o (1) ,
√NtθT Tτ
and
1
= op (1)
Tθ2 (C0Dx,T
JxT JxT
Gχ,τ AιGχ,τ + -√=-Dχ,τ A3Dχ,τ -√=∙
(as shown in Part (a)), we have
(B4 + B5) - C0 μθ2A1 + A3^ μθ2A2 + (A4 + A5)^ = B4 + op (1) . (98)
But, according to Lemma 4,
B4 = N X ZiUi →p lZΞλ.
i
(99)
Therefore, (94), (97), (98) and (99) imply
γg →p γ + (lZξ1z) 1 lZΞλ,
as (N, T →∞) . ¥
Proof of Theorem 9
Using the notation in (87), we can express the Hausman test statistic by
HMNT
[(A1 + θTF1¢ 1 √NT (A2 + θTF2) - Af1√NτA2] 0
× hσVA1 1 - σ2 (A1 + θ2TF1) i
× [(A1 + θTF1)-1 √NT (A2 + θTF2) - Af 1√NτA2] .
60
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