for some finite constant M. But Assumption 4(i) and (iii) imply that the right-
hand side of the last inequality in (45) is finite. This completes the proof. ¥
7 Appendix B: Proofs of Main Results
Proof of Theorem 1
For the desired result, it is enough to prove that
{ω ∈ Ω∣αz (G, H) (ω) ≤ x} ∈ Z,
(46)
for all x ∈ R. Since the partition Π = {∏ι,..., ∏i, ...}of Ω generates the sigma
field Z , we have
sup |(Pz (G ∩ H))(ω) - (PzG)(ω)(PzH)(ω)∣
G∈G, H∈H
=
i∈I
sup
G∈G, H∈H
P (G ∩ H ∩ Πi) P (G ∩ Πi) P (H ∩ Πi)
P∏ P∏ P∏
1Πi (ω),
where 1Πi (ω) denotes the indicator function that equals one if ω ∈ Πi, and
otherwise equals zero. Let I = {1, 2, ..., i, ...} be the set of positive integers, and
let
Ix = i ∈ I |
sup
G∈G, H∈H
P (G ∩ H ∩ Πi) P (G ∩ Πi) P (H ∩ Πi)
P∏ P∏ P∏
≤x .
Then, we have (46) because
{ω ∈ Ω∣αz (G,H) (ω) ≤ x} = ⅛∏ ∈ Z. ¥
Before we start proving the lemmas and theorems in Section 4, we introduce
some additional lemmas that are used repeatedly below. Recall that wit =
x01,it,x02,it , x03,it ,zi0 . We also repeatedly use the diagonal matrix DT defined
in Section 4.
Lemma 16 Suppose that Assumptions 1-8 hold. Define Ξ = Ξ1 + Ξ2, where
Ξ1
Ξ2
Γ22,22 |
Γ22,31 Γ22,32 | |||||
diag |
0k1 , 0 |
0 k21 , Γ22,31 Γ0 Γ22,32 |
Γ31,31 Γ31,32 Γ031,32 Γ32,32 |
,0k33,0kz ; | ||
/ Γθι |
,Θ1 |
ΓΘ1,Θ21 0 |
0 |
rΘ1,μ32 |
rΘ1,μ33 |
0 |
Γ0 ΓΘ1 |
,Θ21 |
ΓΘ21,Θ21 0 |
0 |
r‰,μ32 |
γθ2∣,∕,33 |
0 |
0 |
00 |
0 |
0 |
0 |
0 | |
0 |
00 |
0 |
0 |
0 |
0 | |
Γ0 ΓΘ1 |
Γ0 0 |
0 |
Γ g32,g32 |
Γ g32 ,g33 |
Γ | |
,μ32 |
θ21,μ32 |
+Γ μ32,μ32 0 |
+Γ μ32,μ32 |
g32 ,z | ||
Γ0 |
Γ0 0 |
0 |
Γ0 g32,g33 |
Γ g33,g33 |
Γ | |
ΓΘ1 |
,μ33 |
θ21,μ33 |
+Γ0 μ32,μ32 |
+Γ μ33,μ33 |
g33,z | |
0 |
00 |
0 |
Γ0 g32,z |
Γ0 g33,z |
Γ |
/
37
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