Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID 972324,9 pages doi:10.1155/2010/972324

Research Article

Second Moment Convergence Rates for Uniform Empirical Processes

You-You Chen and Li-Xin Zhang

Department of Mathematics, Zhejiang University, Hangzhou 310027, China Correspondence should be addressed to You-You Chen, cyyooo@gmail.com Received 21 May 2010; Revised 3 August 2010; Accepted 19 August 2010 Academic Editor: Andrei Volodin

Copyright © 2010 Y.-Y. Chen and L.-X. Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let [U-i, U2,..., Un} be a sequence of independent and identically distributed U[0,1]-distributed random variables. Define the uniform empirical process as an(t) = n~1/2^n=1(I[U < t}- t),0 < t < 1, \\an\\ = sup0<K1|an(t)|. In this paper, we get the exact convergence rates of weighted infinite series of E\\an\\2I[\\an\\ > e(logn)1^}.

1. Introduction and Main Results

Let [X,Xn; n > 1} be a sequence of independent and identically distributed (i.i.d.) random variables with zero mean. Set Sn = ^X for n > 1, and log x = ln(x V e). Hsu and Robbins [1] introduced the concept of complete convergence. They showed that

{|S„|> en} < œ, £> 0 (1.1)

if EX = 0 and EX2 < to. The converse part was proved by the study of Erdos in [2]. Obviously, the sum in (1.1) tends to infinity as e \ 0. Many authors studied the exact rates in terms of e (cf. [3-5]). Chow [6] studied the complete convergence of E[|Sn| - ena}+, e> 0. Recently, Liu and Lin [7] introduced a new kind of complete moment convergence which is interesting, and got the precise rate of it as follows.

Theorem A. Suppose that {X,Xn; n > 1} is a sequence ofi.i.d. random variables, then

lim^ ^ 2

£\0 - log £^1 n2

fJ-2ES2nI{|Sn|> £n} = 2a2

holds, if and only if EX = 0, EX2 = a2, and EX2log+|X| < to.

Other than partial sums, many authors investigated precise rates in some different cases, such as ^-statistics (cf. [8, 9]) and self-normalized sums (cf. [10, 11]). Zhang and Yang [12] extended the precise asymptotic results to the uniform empirical process. We suppose U1, U2, ••• ,Un is the sample of U[0,1] random variables and En(t) is the empirical distribution function of it. Denote the uniform empirical process by an(t) = sJn(En(t) - t), 0 < t < 1, and the norm of a function f (t) on [0,1] by \\f || = sup0<i£1 |f (t)|. Let B(t), t e [0,1] be the Brownian bridge. We present one result of Zhang and Yang [12] as follows.

Theorem B. For any 6 > -1, one has

£\ £26+2I^{\\an\\> = ^

Inspired by the above conclusions, we consider second moment convergence rates for the uniform empirical process in the law of iterated logarithm and the law of the logarithm. Throughout this paper, let C denote a positive constant whose values can be different from one place to another. [x] will denote the largest integer < x. The following two theorems are our main results.

Theorem 1.1. For 0 <p < 2, 6> 2/p - 1, one has

£\ £«6+'-2§ E^tl{Na„\\> £(,og„)'") = B+gl. (1.4)

Theorem 1.2. For 0 <p < 2, 6> 2/p - 1, one has

^{„an,, > £(,og,og„)-) = ^. (1.5)

Remark 1.3. It is well known that P[\\B\\ > x} = 2 (-1)fc+1e-2fc2x2, x > 0 (see Csorgo and Revesz [13, page 43]). Therefore, by Fubini's theorem we have

Epf^ = p(6 + 1) x«6+1)-1P[\\B\\ > x}dx h

A TO TO

= 2p(S + 1) | x«6+1)-1Y (-1)fc+1e-2fc2x2 dx J0 k=1

= p(6 + 1)t(^(6 + 1)/2) »( fc+1 ^+1) 2P(s+1)/i 1) k .

E||Bf(g+1) 6 +1 '

Proposition 2.2. For ¡5 > 0, 6> -1, one has

Consequently, explicit results of (1.4) and (1.5) can be calculated further. 2. The Proofs

In order to prove Theorem 1.1, we present several propositions first. Proposition 2.1. For ¡5> 0, 6 > -1, one has

„m ,,6+1,-g (¡s^p {|BU £(1og }. (2.1)

«\0 n ^ >6 + 1

Proof. We calculate that

Km«K^ilonn6P{||B||> «(logn)1/5}

= l\«¡^^J^P{|B| > «(logy)1/5}dy

¡•TO

= p\ t5(6+1)-1P{|B|> t}dt 0

lim«5(6+1^{M> «(logn)1/5} - P{||B||> «(logn)1/5}|= 0. (2.3)

«\0 n=2 n

Proof. Following [4], set A(e) = [exp(M/£^)], where M> 1. Write

2^^g-^6!^{\an\\> £(logn)1/") - P{\\B\\> £(logn)1/")|

= 2 {\\an\\ > £(logn)1/^) - P{\\B\\> £(logn)1/^)|

n<A(£) n 1 1 J 1 Jl (2.4)

+ 2 il°gn6|P{\\an\\> £(log n)1/^) - p{\\B\\> £(log n)1/^ )|

n>A(£)

=: I1 +12.

It is wellknown that an(■) -— B(-) (see Csorgo and Revesz [13, page 17]). By continuous mapping theorem, we have \\an\\ -— \\B\\. As a result, it follows that

An := sup|P{\\an\\ > x}- P{\\B\\> x}|-— 0, as n -—to. (2.5)

Using the Toeplitz's lemma (see Stout [14, pages 120-121]), we can get lim£\o£^(6+1)I1 = 0. For I2, it is obvious that

I2 < X \\B\\> £(log n)1/^ + X M-P{\M> £(log n)1/"}

n>A(£) n 1 J n>A(£) n 1 J (2.6)

0og n)6n i„d„ , 2 (logn

=: I3 +14.

Notice that A(£) - 1 >\/A(e), for a small £. Via the similar argument in [4] we have

£^+1)I3 < £^(6+1) ^ {\\B\\ > £(log n)1/^)

n>A(£) n (2 7)

< C y«6+1)-1P{\\B\ > y}dy -- 0, as M —^ to.

J (M/2)1^

From Kiefer and Wolfowitz [15], we have

P{\\an\\> x}< Ce~Cxl. (2.8)

Journal of Inequalities and Applications Therefore,

0°g n

n>A(e)

exp j -C«2 (log n)2/5j

eP(6+1)I4 < C«5(6+1) ^

< ce^l expf-C«2 (logx)2/5}dx

\Aë) X 1 '

< CI y5(6+1)/2-1e~ydy 0, as M to.

' C(M/2)2/5

From (2.6), (2.7), and (2.9), we get lim«\0«5(6+1)l2 = 0. Proposition 2.2 has been proved. □ Proposition 2.3. For ¡5 > 0, 6 > 2/5 - 1, one has

lim«¡(6+1)-2£ T 2yP{|B| > y}dy = 2£'B|5(6+1) „ . (2.

«\0 n=2 n ¡«(logn)^ y ^ (6 + 1)(¡(6 + 1) - 2)) '

Proof. The calculation here is analogous to (2.1), so it is omitted here. Proposition 2.4. For 0 <p < 2, 6> 2/p - 1, one has

lim e5(6+1)-2V

«\0 n=2

(log n)

/"TO /"TO

2yP{||«n| > y}dy - 2yP{||B| > y}dy

* «(log n)1/5 J «(log n)1/5

«(log n)1/5

(2.11)

Proof. Like [4] and Proposition 2.2, we divide the summation into two parts,

(log n)

f TO f TO

2yP{HanH> y}dy - 2yP{||B| > y}dy

■'«(log n)1/5 •'«(log n)1/5

«(logn) 6-25

n<A(«)

n>A(«)

(log n)

f TO ^ TO

2yP{|an||> y}dy ^ 2yP{||B|| > y}dy

./«(logn)1/5 J «(logn)1/5

/" TO /" TO

2yP{|an|> y}dy - 2yP{||B| > y}dy

■'«(log n)1/5 •'«(log n)1/5

(2.12)

=: J1 + J2.

First, consider J1,

^ floen)6-/ r J1 < X --2y|^\an\\> y} - P{\\BH > y} |dy

n<A(£) n J £(log n)1/f

< 2 ^log n f 2(x + £)|p {\an\ > (x + £)(log n)V^ ) - p{\\B\\> (x + £)(log n)1/^)|dx

n<A(£) n J0 1 1 J 1 J 1

^ (log n)6 /r(logn)-1/pA-1/^ 17

< X M^i 2(x + £)|p{\\an\ > (x + £)(logn)

n<A(£) n \J0 11 '

-P{\\B\\ > (x + £)(logn)1/^|dx

+ 2(x + £)p(\\B\ > (x + £)(logn)1/^dx

J(log n)-1/^A-1/4 1

r to n

+ 2(x + £)p( \ \an \ \ > (x + £) (log n) 1/^dx

J(loB n)(-1/^)A-1/4 1 J ,

(log n)(-1/^ A-1/4

=: 2 (J11 + J12 + J13).

n<A(£) n

Since n < A(£) means £ < (M/ log n)1/^, it follows

,(log n)-1/^A-1/4

(2.13)

/•(log n) An

(log n) 2/(SJn < (log n) 2/?\ 2(x + £) Andx

< (log n) an( (log n )-1/^an1/4 + (log n)~1/pM1/^2 (2.14)

< (a-/4 + M1/^a-/2)2 -— 0, as n .

By Lemma 2.1 in Zhang and Yang [12], we have P{\\B\\ > x}< 2e 2x2. For J12, it is easy to get

(logn)J < (logn)2/n (logn)-2/P ■ 2yP{\\B\\ > y}dy

j£(log n)1/^ + A- 4

< C 2yexp{-2y2}dy (2.15)

< C expj-2a-1/2} -— 0, as n -—to .

In the same way, by the inequality P[\\an\\ > x}< Ce-Cx2, we can get

(logn)2/ph3 < C exp{-Ca-1/2} -— 0, as n -—to. (2.16)

Put the three parts together, we get that (log n)2/5(J11 + J12 + J13) — 0 uniformly in e as n —> to. Using Toeplitz's lemma again, we have lime\oe^(6+1)-2J1 = 0. In the sequel, we verify lime\0e/(6+1)-2 J2 = 0. It is easy to see that

„ (log^6-2/5 fTO J2 < Y --2xP[\\B\\> x}dx

n>A(e) n ' e(log n)1/5

„ (logn)6-2/5 fto (2.17)

+ V v 67--2xP[\an\> x}dx

n>A(e) n ' e(log n)1/5

= : J21 + J22.

We estimate J22 first, by noticing 0 < / < 2 and (2.8), it follows

Qogn)6-2/P f- - .1///D,, ww^V/n e

j22 < ^ (log "J f2«(log y)1/5P{ |an |> «(log y)1/5}-«- (log y)1/5-1 dy

n>A(«) n n 5y

fTO (log x)6-2/5 fTO «2 (log y)2/5-1 r ,, n25

< C J au^Ï-L -exp{ -C«^(log y 2/5}dydx

< C fto «^logyy)2/5-1 exp{-C«2(logy)2/5}(logy)6-2/5+1dy (2.18)

JA(«) y 1 '

< Cë2 i" (log y exp { -C«2 log y} dy

JA(«) y 1 '

' A(«) y

^l^^f» < c«2->6.

(A(«))C

Therefore, we get lim«\0£5(6+1)-2 J22 = 0. So far, we only need to prove lime\0£5(6+1)-2 J21 = 0. Use the inequality P{||B|| > x} < 2e-2x2 again and follow the proof of J22, we can get this result. The proof of the proposition is completed now. □

Proof of Theorem 1.1. According to Fubini's theorem, it is easy to get

EXI{X > a} = aP{X > a} + P{X > x}dx, (2.19)

for a> 0. Therefore, we have

E||a„||2i{||a„y > e(logn)1//}} ■ e2(logn)2/(Sp{\\an\\ > e(logn)1/p]

+ J^iognf» 2Vp{ ||an» > v)dv-

From Proposition 2.1- 2.4, we have

,5-2/p

œ /1 \6-2/p>

limm-E||an||2i{»an» > .(logn)1/p]

e\0 n^2 n I- >

Q°g n)

Km ep(6+1)-2jr °og f p 2yP{»an» > y}dy

e\0 „■? n Je (log n) p

= e^(6+1)E11an| > e(logn)1/p}

(log n) n=2 n Je(log n)'

pEpf0^

« P(6 + 1) - 2'

Proof of Theorem 1.2. From (2.19), we have

^+1)-2£3 (logn°lgognn E»an»24 »an» > e(loglogn)1/p}

■ " ,, 2yP{»«.!> y}dy (2.22)

«■3 n log n J e(log log n)1/p

(loglog n)

■3 n log n Je(log log n)'

^^gognfp {»an»> e(log log n)1/p }.

limeP(6+1)-2] f 2yP{»an» > y}dy ■

e\0 n log n Je(log log n)1/"

(2.20)

(2.21)

Via the similar argument in Proposition 2.1 and 2.2,

lim (loglog n)6 p {»an» > e || log log n^ } ■ EM^. (2.23)

e\0 n^3 n log n I" m - 11 6 6 11 > 6 + 1

Also, by the analogous proof of Proposition 2.3 and 2.4,

(loglogn)6-2/P p 2E||B||P(6+1)

=3 n log n J £(log log n )1/^ Ul * (6 + 1)03(6 + 1) - 2))

(2.24)

Combine (2.22), (2.23), and (2.24)together, we get the result of Theorem 1.2. □

Acknowledgment

This work was supported by NSFC (No. 10771192) and ZJNSF (No. J20091364).

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