Volume 2010, Article ID 972324, 9 pagesdoi:10.1155/2010/972324 Research Article Second Moment Convergence Rates for Uniform Empirical Processes You-You Chen and Li-Xin Zhang Department o
Trang 1Volume 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
Copyrightq 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{U1, U2, , U n } be a sequence of independent and identically distributed U0, 1-distributed random variables Define the uniform empirical process as α n t n −1/2n
i1 I{U i ≤ t} − t, 0 ≤
t ≤ 1, α n sup0≤t≤1|α n t| In this paper, we get the exact convergence rates of weighted infinite
series of Eαn2
I{α n ≥ εlog n 1/β}
1 Introduction and Main Results
Let{X, X n ; n ≥ 1} be a sequence of independent and identically distributed i.i.d. random variables with zero mean Set S nn
i1 X i for n ≥ 1, and log x lnx ∨ e Hsu and Robbins
1 introduced the concept of complete convergence They showed that
∞
n1
P {|S n | ≥ εn} < ∞, ε > 0 1.1
if EX 0 and EX2< ∞ The converse part was proved by the study of Erd¨os in 2 Obviously, the sum in1.1 tends to infinity as ε 0 Many authors studied the exact rates in terms of ε
cf 3 5 Chow 6 studied the complete convergence of E{|S n | − εn α}, ε > 0 Recently, Liu
and Lin7 introduced a new kind of complete moment convergence which is interesting, and got the precise rate of it as follows
Trang 22 Journal of Inequalities and Applications
Theorem A Suppose that {X, X n ; n ≥ 1} is a sequence of i.i.d random variables, then
lim
ε0
1
− log ε
∞
n1
1
n2ES2n I {|S n | ≥ εn} 2σ2 1.2
holds, if and only if EX 0, EX 2 σ2, and EX 2log|X| < ∞.
Other than partial sums, many authors investigated precise rates in some different
cases, such as U-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 U0, 1 random variables and E n t is the empirical distribution function of it Denote the uniform empirical process by α n t √nE n t − t,
0≤ t ≤ 1, and the norm of a function ft on 0, 1 by f sup0≤t≤1|ft| Let Bt, t ∈ 0, 1
be the Brownian bridge We present one result of Zhang and Yang12 as follows
Theorem B For any δ > −1, one has
lim
ε0 ε 2δ2
∞
n1
log nδ
n P
α n ≥ εlog n
EB2δ2
δ 1 . 1.3
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 < β ≤ 2, δ > 2/β − 1, one has
lim
ε0 ε βδ1−2
∞
n2
log nδ−2/β
n E α n2
I
α n ≥ εlog n1/β
βE B
βδ1
β δ 1 − 2 . 1.4
Theorem 1.2 For 0 < β ≤ 2, δ > 2/β − 1, one has
lim
ε0 ε βδ1−2
∞
n3
log log nδ−2/β
n log n E α n2
I
α n ≥ εlog log n1/β
βE B βδ1
β δ 1 − 2 . 1.5
Trang 3Remark 1.3 It is well known that P {B ≥ x} 2∞
k1−1k1
e −2k2x2
, x > 0 see Cs¨org˝o and
R´ev´esz13, page 43 Therefore, by Fubini’s theorem we have
EBβδ1 βδ 1
∞
0
x βδ1−1 P {B ≥ x}dx
2βδ 1
∞
0
x βδ1−1
∞
k1
−1k1
e −2k2x2dx
β δ 1Γ
β δ 1/2
2βδ1/2
∞
k1
−1k1 k −βδ1
1.6
Consequently, explicit results of1.4 and 1.5 can be calculated further
2 The Proofs
In order to proveTheorem 1.1, we present several propositions first
Proposition 2.1 For β > 0, δ > −1, one has
lim
ε0 ε βδ1
∞
n2
log nδ
n P
B ≥ εlog n1/β E B βδ1
δ 1 . 2.1
Proof We calculate that
lim
ε0 ε βδ1
∞
n2
log nδ
n P
B ≥ εlog n1/β
lim
ε0 ε βδ1
∞
2
log yδ
y P
B ≥ εlog y1/β
dy
β
∞
0
t βδ1−1 P {B ≥ t}dt
EBβδ1
δ 1 .
2.2
Proposition 2.2 For β > 0, δ > −1, one has
lim
ε0 ε βδ1
∞
n2
log nδ
n
P α n ≥ εlog n1/β
− P B ≥ εlog n1/β
0. 2.3
Trang 44 Journal of Inequalities and Applications
Proof Following4, set Aε expM/ε β , where M > 1 Write
∞
n2
log nδ
n
P α n ≥ εlog n1/β − P B ≥ εlog n1/β
n≤Aε
log nδ
n
P α n ≥ εlog n1/β
− P B ≥ εlog n1/β
n>Aε
log nδ
n
P α n ≥ εlog n1/β − P B ≥ εlog n1/β
: I1 I2.
2.4
It is wellknown that α n· → B· see Cs¨org˝o and R´ev´esz d 13, page 17 By continuous mapping theorem, we haveα n → B As a result, it follows that d
Δn: sup
x
|P{α n ≥ x} − P{B ≥ x}| −→ 0, as n −→ ∞. 2.5
Using the Toeplitz’s lemmasee Stout 14, pages 120-121, we can get limε0 ε βδ1 I1 0 For
I2, it is obvious that
I2≤
n>Aε
log nδ
n P
B ≥ εlog n1/β
n>Aε
log nδ
n P
α n ≥ εlog n1/β
: I3 I4.
2.6
Notice that Aε − 1 ≥ Aε, for a small ε Via the similar argument in 4 we have
ε βδ1 I3 ≤ ε βδ1
n>Aε
log nδ
n P
B ≥ εlog n1/β
≤ C
∞
M/2 1β y βδ1−1 P
B ≥ ydy −→ 0, as M −→ ∞.
2.7
From Kiefer and Wolfowitz15, we have
P {α n ≥ x} ≤ Ce −Cx2
Trang 5ε βδ1 I4 ≤ Cε βδ1
n>Aε
log nδ
n exp
−Cε2
log n2/β
≤ Cε βδ1
∞
√
Aε
log xδ
x exp
−Cε2
log x2/β
dx
≤ C
∞
C M/2 2/β y βδ1/2−1 e −y dy −→ 0, as M −→ ∞.
2.9
From2.6, 2.7, and 2.9, we get limε0 ε βδ1 I2 0.Proposition 2.2has been proved
Proposition 2.3 For β > 0, δ > 2/β − 1, one has
lim
ε0 ε βδ1−2
∞
n2
log nδ−2/β
n
∞
εlog n1/β 2yP
B ≥ ydy 2EB βδ1
δ 1β δ 1 − 2 . 2.10
Proof The calculation here is analogous to2.1, so it is omitted here
Proposition 2.4 For 0 < β ≤ 2, δ > 2/β − 1, one has
lim
ε0 ε βδ1−2
∞
n2
log nδ−2/β
n
∞
εlog n 1/β 2yP
α n ≥ ydy −
∞
εlog n 1/β 2yP
B ≥ ydy
0.
2.11
Proof Like4 andProposition 2.2, we divide the summation into two parts,
∞
n2
log nδ−2β
n
∞
εlog n 1/β 2yP
α n ≥ ydy −
∞
εlog n 1/β 2yP
B ≥ ydy
n≤Aε
log nδ−2β
n
∞
εlog n 1/β
2yP
α n ≥ ydy −
∞
εlog n 1/β
2yP
B ≥ ydy
n>Aε
log nδ−2/β
n
∞
εlog n 1/β 2yP
α n ≥ ydy −
∞
εlog n 1/β 2yP
B ≥ ydy
: J1 J2.
2.12
Trang 66 Journal of Inequalities and Applications
First, consider J1,
J1 ≤
n≤Aε
log nδ−2/β
n
∞
εlog n 1/β
2y P
α n ≥ y− PB ≥ y dy
≤
n≤A ε
log nδ
n
∞
0
2x ε P
α n ≥ x εlog n1/β − P B ≥ x εlog n1/β
dx
≤
n≤A ε
log nδ
n
log n −1/βΔ−1/4
n
0
2x ε P
α n ≥ x εlog n1/β
−P B ≥ x εlog n1/β
dx
∞
log n −1/βΔ−1/4
n
2x εP B ≥ x εlog n1/β
dx
∞
log n −1/βΔ−1/4
n
2x εP α n ≥ x εlog n1/β
dx
:
n≤A ε
log nδ
n J11 J12 J13.
2.13
Since n ≤ Aε means ε < M/ log n 1/β, it follows
log n2/β
J11≤log n2/β log n −1/βΔ−1/4
n
0
2x εΔn dx
≤log n2/βΔn
log n−1/β
Δ−1/4
n log n−1/β
M 1/β2
≤Δ1/4
n M 1/βΔ1/2
n
2
−→ 0, as n −→ ∞.
2.14
By Lemma 2.1 in Zhang and Yang 12, we have P{B ≥ x} ≤ 2e −2x2
For J12, it is easy to get
log n2/β
J12≤log n2/β ∞
εlog n 1/βΔ−1/4
n
log n−2/β
· 2yPB ≥ ydy
≤ C
∞
Δ−1/4 n
2y exp
−2y2 dy
≤ C exp −2Δ−1/2
n −→ 0, as n −→ ∞.
2.15
Trang 7In the same way, by the inequality P {α n ≥ x} ≤ Ce −Cx2
, we can get
log n2/β
J13 ≤ C exp −CΔ −1/2
n −→ 0, as n −→ ∞. 2.16
Put the three parts together, we get thatlog n 2/β J11 J12 J13 → 0 uniformly in ε as
n → ∞ Using Toeplitz’s lemma again, we have lim ε0 ε βδ1−2 J1 0.
In the sequel, we verify limε0 ε βδ1−2 J2 0 It is easy to see that
J2≤
n>Aε
log nδ−2/β
n
∞
εlog n 1/β 2xP {B ≥ x}dx
n>Aε
log nδ−2/β
n
∞
εlog n 1/β
2xP {α n ≥ x}dx
: J21 J22.
2.17
We estimate J22first, by noticing 0 < β ≤ 2 and 2.8, it follows
J22≤
n>Aε
log nδ−2/β
n
∞
n
2ε
log y1/β
P
α n ≥ εlog y1/β
βy
log y1/β−1
dy
≤ C
∞
Aε
log xδ−2/β
x
∞
x
ε2
log y2/β−1
y exp
−Cε2
log y2/β
dy dx
≤ C
∞
Aε
ε2
log y2/β−1
y exp
−Cε2
log y2/β
log yδ−2/β1
dy
≤ Cε2
∞
Aε
log yδ
y exp
−Cε2log y dy
≤ Cε2logδ Aε
Aε Cε2 ≤ Cε2−βδ.
2.18
Therefore, we get limε0 ε βδ1−2 J22 0 So far, we only need to prove lim0 ε βδ1−2 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}
∞
a
P {X ≥ x}dx, 2.19
Trang 88 Journal of Inequalities and Applications
for a > 0 Therefore, we have
Eαn2
I
α n ≥ εlog n1/β
ε2
log n2/β
P
α n ≥ εlog n1/β
εlog n∞ 1/β 2yP
α n ≥ ydy.
2.20
FromProposition 2.1– 2.4, we have
lim
ε0 ε βδ1−2
∞
n2
log nδ−2/β
n Eαn2I
α n ≥ εlog n1/β
lim
ε0 ε βδ1∞
n2
log nδ
n P
α n ≥ εlog n1/β
lim
ε0 ε βδ1−2
∞
n2
log nδ−2/β
n
∞
εlog n 1/β
2yP
α n ≥ ydy
βE B βδ1
β δ 1 − 2 .
2.21
a
Proof of Theorem 1.2 From2.19, we have
ε βδ1−2
∞
n3
log log nδ−2/β
n log n Eαn2
I
α n ≥ εlog log n1/β
ε βδ1−2∞
n3
log log nδ−2/β
n log n
∞
εlog log n 1/β 2yP
α n ≥ ydy
ε βδ1∞
n3
log log nδ
n log n P
α n ≥ εlog log n1/β
.
2.22
Via the similar argument inProposition 2.1and 2.2,
lim
ε0 ε βδ1
∞
n3
log log nδ
n log n P
α n ≥ εlog log n1/β EBβδ1
δ 1 . 2.23 Also, by the analogous proof ofProposition 2.3and 2.4,
lim
ε0 ε βδ1−2
∞
n3
log log nδ−2/β
n log n
∞
εlog log n 1/β 2yP
α n ≥ ydy 2EBβδ1
δ 1β δ 1 − 2 .
2.24 Combine2.22, 2.23, and 2.24together, we get the result ofTheorem 1.2
Trang 9This work was supported by NSFCNo 10771192 and ZJNSF No J20091364
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... n1/β0. 2.3
Trang 44 Journal of Inequalities and Applications
Proof... −Cx2
Trang 5ε βδ1 I4 ≤ Cε βδ1... I3 I4.
2.6
Notice that Aε − ≥ Aε, for a small ε Via the similar argument in 4 we have
ε βδ1 I3 ≤ ε βδ1