Thm/2 [ J2(Qθ) - J2(Qθ - Ct)]
= Op(CTh-m/2)
=op(1) .
(A.25)
[3] Thm/2 [J3(Qθ)- J3(Qθ -Ct)] = Op(1):
Noting that H3t (5, t, Qθ ) = 0 because of F(Qθ (xj )| Zj ) - θ = 0 for j = t, s, we have
J3(Qθ)-J3(Qθ-CT)
1 ʃ. ʃ. 1 (z -z A
=--1--∑ ∑ ɪ K z-÷- I
T(T -1) ⅛ ⅛ hm L h )
× {F(Qθ(xt)-Ct |z,)-θ}{F(Qθ(xs)-Ct∣Zs)-θ}
1TT1(zzI
= ^~∑ ∑ ∑ mK l^tv^ I CT S (Qθ ( Xt )) S (Qθ ( Xs ))
T(T- 1) t=1 s≠th Lh )
= CT1 ∑ S(Q(Xt ))S(Q( X ))f Z(z, ) (A.26)
T t=1
Thus, we have
EJ3( Qθ ) - J3( Qθ - ' )∣
1T
≤ ΛCT-∑ E∖fZ(z■,)
T t=1
≤ ΛCT1 ∑ Ef (z-,)∣ + ΛCT1 ∑ E∖f■ (z,)-f,(z,)∣
Tt=1 Tt=1
1T1T 2
≤ ΛCT-∑ Efz(Zt) + ΛCT-∑ E{f■ (Zt)-fz(Zt)}
Tt=1Tt=1
= O(CT2 ) (A.27)
Finally, we have
Th""-[J3(Qθ)- J3(Qθ -Ct)]
= Op(Thm/2CT2)
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