Write
N X 12,i,N I2 ,i,N
i
II1,1 |
II1,21 |
0 |
0 |
II1,32 |
II1,33 |
II1,z |
∖ |
II1,21 |
II21,21 |
0 |
0 |
II21,32 |
II21,33 |
II21,z | |
0 |
0 |
0 |
0 |
0 |
0 |
0 | |
0 |
0 |
0 |
0 |
0 |
0 |
0 | |
II10 ,32 |
II20 1,32 |
0 |
0 |
II32,32 |
II32,33 |
II32,z | |
II1,33 |
II20 1,33 |
0 |
0 |
II30 2,33 |
II33,33 |
II33,z | |
II10,z |
II201,z |
0 |
0 |
II302,z |
II30 3,z |
IIz,z |
/ |
(53)
By Assumption 7(i), as N →∞,
II1,1
II10,21
II1,21
II21,21
ΓΘ1,Θ1
Γ0Θ1,Θ21
ΓΘ1,Θ21
ΓΘ21,Θ21
(54)
By the weak law of large numbers (WLLN) and Assumption 7, as N →∞,we
have
(1132,32 1132,33 II32,z
II30 2,33 II33,33 II33,z
II30 2,z II303,z IIz,z

Γ g32 ,g32 |
+Γ + ɪ μ32,μ32 |
Γ g32,g33 |
+Γ + ɪ μ32,μ32 |
Γ |
Γ0 g32 ,g33 |
+ Γ0 ʃ μ32,μ32 |
Γ g33,g33 |
+Γ μ33,μ33 |
Γg33,z . (55) |
Γ0 g32,z |
Γ0 g33,z |
Γz,z |
From Assumption 7 and WLLN with the assumption that Eg32,i (zi) = Eg33,i (zi)
= 0, it follows that
II1,32
II21,32
II1,33
II21,33
→p
rθ1,μ32
rθ21,μ32
rθ1,μ33
rθ21,μ33
(56)
as N →∞. In addition, by WLLN with Assumption 5,
μ ⅛. ) →p μ 0 ), (57)
as N →∞. The results (53)-(57) indicate that
N X I2,i,NI2,i,N →p ξ2, (58)
i
as N →∞. Finally, by Lemma 12, (52) and (58) imply (49).
Proof of (50): Notice that
N ∑D1T 1∑(xι,it - Exι,it)
it
41
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