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By Lemma 5.6 we have
where
and
Um = ;Bm + Vm + e 1+
m
m
Bm = m 1=2 ES1(2 ; ES2) ; m 1=2 (ES1) 2 (4.7)
Vm = ;Z1(2 ; ES2)+ Z1Z2 + Z 2
1
m 1=2
where >0and m denotes a remainder term. Set V 0 m = Vm + e 1+
m . Thus
By Lemma 6.3,
+(2Z1 + Z2)ES1 (4.8)
P (Um y) =P (V 0 m y + Bm): (4.9)
sup P (V
y2R
0 m y) ; (y) ; m;1=2 (y)p(y) = o(e m): (4.10)
It remains to derive an expansion for Bm. By Lemma 7.1, bearing in mind that (m=n) = O(m ;1=2 ),
we have ES1 = (m=n) + o(m ;1 )+O(m ;1 ml) ES2 =1+o(m ;1=2 ) so that
m 1=2 ES1 = qm + o(qm + m ;1=2 )+O(m ;1=2 ml) m 1=2 ES1(2 ; ES2) = qm + o(qm):
If m = o(n 2 =(2 +1) ), then qm ! 0, and
If m Kn 2 =(2 +1) , then qm K +1=2 ,andwe obtain
(4.9) - (4.12) imply (2.13), (2.15) of Theorem 2.1.
Bm = qm + o(qm + m ;1=2 )+O(m ;1=2 ml): (4.11)
Bm = K +1=2 + o(1): (4.12)
Proof of Theorem 3.1. This follows the lines of that of Theorem 2.1. Relations (4.9), (4.10) remain
valid. Recall that qm =(m=n) 2 m 1=2 , and under Assumption m, qm = O(1). To expand Bm we use
(7.15) and (7.16) with k = 2 to deduce
= e(1`2)=( c1
c0
Bm = qm + qm(m=n) 2 + o(m ;1=2 )+O(m ;1=2 ml) (4.13)
) 2 2 `
(
9 2 )2 = d(1`4);e(1`2)e(2`2);e 2 (1`2) = c2
c0
4
25 ;(c1
c0
) 2 22
243
where e(k ` ) and d(k ` ) are de ned in (7.5) and (7.17). If m = o(n 4=5 ), then qm = o(1)sothat
Bm = qm + o(m ;1=2 )+O(m ;1=2 ml), and (3.13) follows from (4.9), (4.10). If m Kn 4=5 ,then
(m=n) 2 K 5=2 m ;1=2 and thus Bm = K 5=2 + m ;1=2 K 5 + o(m ;1=2 )+O(m ;1=2 ml): Therefore
from (4.9), (4.10) it follows that
sup P (Um y) ; y + K
y2R
5=2 + m ;1=2 K 5 ; m ;1=2 (y + K 5=2 )p(y + K 5=2 )
which implies (3.15).
= o(m ;1=2 )+O(m ;1=2 ml)
( `
2 )4