Now, the same expansion used for KI21 can be applied to KI22(fθ*, g):
ki 22(fθ* ,g) =
1 XX μ fθ* (xi) - g(xi) v b (x )
n M g(xi) f
1 X <fθ* (xi) - g(xi)A2 b, ʌ 1 1 τ
- 2n Σ —g(χi)— f fn(xi) = Jn1- 2J∙2∙
(96)
E Jin(fθ* ,g)) = E
K(u) (fθ* (x) — g(x)) g(x + hu)dxdu.
(97)
Applying the same steps used for S2n we can show that
lim sup E
n→∞
(Jin(fθ*,g)) ≤ У
(fθ* (x) — g(x)) g(x)dx = E (fθ* (x) — g(x))
lim E ( Jin(fθ* ,g)) = E (fθ* (x) — g(x)). (98)
n→∞
It follows that J1n(fθ*, g) = Op(1). Repeating the same steps once more for J2n(fθ*, g) we obtain:
E А XX μ fθ∙ (xi) - g(xi) ..^ A = E (t μ fθ- (xi) -g(xi) V2 fn^^d
n i=1 g(xi) g(xi)
=E
(fθ* (xi) - g(xi))2
g(xi)
fn(xi)dxi I = (( K(u) ( ) g^ g( g(x + hu)dxdu,
g(x)
lim sup E (J2n(fθ* ,g)) ≤ ( (fθ∙ (x) — g(x))2 dx
n→∞
lim E (J2n(fθ* ,g))= ( (fθ* (x) — g(x))2 dx > 0.
n→∞
Then also J2n(fθ* ,g) = Op(1). This implies that KI22(fθ* ,g) = Jni — 2 Jn2 = Op(1).
Then it is clear that given assumptions A1-A4, if h → 0,nh→∞then
(99)
(100)
(101)
KI22(fθ* ,g) →p E (fθ* (x) — g(x)) — 2 У (fθ* (x) — g(x))2 dx = E [lnfθ* — lng],
(102)
this implies that nh1/2 KTI 22 →p ∞, hence we need to rescale it by dn = n-1h-1/2 where dn → 0 as
This is embodied in assumption A6:
n →∞.
Finally we can put all terms together:
KI =
(ln
x
KTI22 ` ah1/2Cn
(103)
—~~ —~~ —~~
fTn(x) — lnfbθ(x))fTn(x)dx == KTI1
—-~.
— KTI2
(nh1/2) 1√2σιNι — (nh1/2) 1√2σ2N2 — 2 Cn
KI21(fb, fθ* ) + KI22(fθ* , g)J ,
(104)
since we showed that
32
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