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For large m, by (6.12), Tr(S 2 )=Var(Q)=2 jjtjj 2 =16: Thus, since e ;1
m
to prove (6.16).
log j (t)j ;C ;1 (e mjjtjj 2 ) ;1 jjtjj 2 =16 = ;C ;1 m =16
Lemma 6.2 Let (4.4), Assumptions f l m hold. Then, with Vm given by (4.8),
where p(y) is given by (2.14).
sup P (Vm y) ; (y) ; m
y2R
;1=2 (y)p(y) = o(e m)
Proof. We shall derive the second order expansion
m for some >0,
P (Vm y) = (y) ; m ;1=2 (y)ep(y)+o(e m) (6.18)
uniformly in y 2 R where
y y
ep(y) =a1 + a2 + a3
2! 2 ; 1
3!
and the coe cients a1a2a3 are de ned (c.f. (2.1.16), p. 15 of Taniguchi (1991)) by
(6.19)
Cumj(Vm) =1fj=2g + m ;1=2 aj + o(e m) j =1 2 3: (6.20)
In fact we shall show that (6.20) holds with m( e m) instead of e m. We rstshow that
a1 = ;1 a2 =0 a3 = ;2: (6.21)
Write Vm = ;P + m ;1=2 Q + R, where P = Z1(2 ; ES2) Q= Z1Z2 + Z 2
1 R=(2Z1 + Z2)ES1:
Since EP = ER = 0 and by (7.39), (7.38), EQ = EZ1Z2 + EZ 2 1 = ;1+o(1) we obtain
and therefore a1 = ;1. Now, by (6.22),
We showthat
and
which with (6.23) implies
and thus a2 =0.
By Lemma 7.1,
Cum1(Vm) EVm = m ;1=2 EQ = ;m ;1=2 + o(m ;1=2 ) (6.22)
Cum2(Vm) =E(Vm ; EVm) 2 = EV 2
m + o(m ;1=2 )
= EP 2 ; 2EP(m ;1=2 Q + R)+E(m ;1=2 Q + R) 2 + o(m ;1=2 ): (6.23)
where = e(2` ), and (7.38) implies
EP 2 =1+o( m) EPQ = O( m) EPR = o( m) (6.24)
Ejm ;1=2 Q + Rj i = O(
i
m) i =2 3 4 (6.25)
Cum2(Vm) =1+o( m) (6.26)
ES1 = O( m) ES2 =1+ (m=n) + o( m) (6.27)
EP 2 = EZ 2
1(2 ; ES2) 2 =(1+2 (m=n) )(1 ; (m=n) ) 2 + o( m)
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