=∫K2(u)du∫fz2(z)dz+op(1)
(A.31)
The result of Theorem (ii) follows from (A.31).
Proof of Theorem 1 (iii)
The proof of Theorem (iii) consists of the two steps.
Step 1. Show that JT = Jt + op (1) under the alternative hypothesis (4).
Step 2. Show that JT=J+op(1) under the alternative hypothesis (4),
where J = E{[Fy^z(Qθ(xt)∣ zt)-θ]2fz(zt)} . The combination of Steps 1 and 2 yields
Theorem (iii).
Step 1: Show that Jt = Jt + op (1) under the alternative hypothesis.
We need to show that the results of Step 2 and Step 3 in the proof of Theorem (i) hold under
the alternative hypothesis. First, we show that the result of Step 2 in the proof of Theorem (i)
still holds under the alternative hypothesis. We can show that J2(Qθ - CT)= op(1) by the
same procedures as in (A.24). Thus we focus on showing thatJ2 (Qθ) = op (1). As in the proof
of Theorem (i), denote S(g) ≡ ∂F[g] / ∂g . By taking a Taylor expansion of
Fy∣z(qθ (xs )| zs ) around qθ (zs ), we have
J2(Qθ)=-
1
T(T-1)
TT
∑ ∑
t=1 s≠t

— z I
—s-∣{1(yt ≤ Qθ(xt)) -Fyz(Qθ(xt)∣zt)}
h )
× S (Qθ (xs . zs ))
1T
= -∑ {1(yt ≤ Qθ(xt)) — Fy∖z (Qθ (xt ))}S(Qθ (xs, zs)) f z(zt)
T t=1
1T
(A.32)
≡7∑ u,S(Qθ(x,,z,)).f (zt).
T t=1
where Qθ(xs,zs) is between Qθ(xs) and Qθ(zs). By using the same procedures as in
19
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