E l e c t ro n ic
Jo ur
n a l o
f P
r o
b a b il i t y
Vol. 10 (2005), Paper no. 38, pages 1260-1285. Journal URL
http://www.math.washington.edu/∼ejpecp/
Limit Theorems for Self-normalized Large Deviation1
Qiying Wang
School of Mathematics and Statistics The University of Sydney, Sydney
NSW 2006, Australia E-mail: qiying@maths.usyd.edu.au URL: http://www.maths.usyd.edu.au/u/qiying/
Abstract. LetX, X1, X2,· · · be i.i.d. random variables with zero mean and £nite varianceσ2. It is well known that a £nite exponential moment assump-tion is necessary to study limit theorems for large deviaassump-tion for the standard-ized partial sums. In this paper, limit theorems for large deviation for self-normalized sums are derived only under £nite moment conditions. In partic-ular, we show that, ifEX4 <∞, then
P(Sn/Vn ≥x)
1−Φ(x) = exp
½ −x
3EX3
3√nσ3 ¾ ·
1 +O³1 +√ x n
´¸ ,
forx≥0andx=O(n1/6), whereS
n =Pni=1Xi andVn = (Pni=1Xi2)1/2.
Key Words and Phrases: Cram´er large deviation, limit theorem, self-normalized sum. AMS Subject Classi£cation (2000): Primary 60F05, Secondary 62E20.
Submitted to EJP on September 24, 2004. Final version accepted on September 16, 2005.
1
Introduction and main results
Let X, X1, X2,· · · , be a sequence of non-degenerate independent and identically distributed (i.i.d.) random variables with zero mean. Set
Sn=
n X
i=1
Xi, Vn2 =
n X
i=1
Xi2, n≥1.
The self-normalized version of the classical central limit theorem states that, asn → ∞,
sup
x ¯
¯P(Sn≥xVn)− ©
1−Φ(x)ª¯ ¯→0,
if and only if the distribution ofXis in the domain of attraction of the normal law, whereΦ(x)
denotes the standard normal distribution function. This beautiful self-normalized central limit theorem was conjectured by Logan, Mallows, Rice and Shepp (1973), and latterly proved by Gine, G¨otze and Mason (1997). For a short summary of developments that have eventually led to Gine, G¨otze and Mason (1997), we refer to the Introduction of the latter paper.
The self-normalized central limit theorem is useful whenxis not too large or when the error is well estimated. There are two approaches for estimating the error of the normal approxi-mation. One approach is to investigate the absolute error in the self-normalized central limit theorem via Berry-Esseen bounds or Edgeworth expansions. This has been done by many re-searchers. For details, we refer to Slavova (1985), Hall (1988) and Bentkus and G ¨otze (1996) for the Berry-Esseen bounds, Wang and Jing (1999) for an exponential nonuniform Berry-Esseen bound, Hall (1987) as well as Hall and Jing (1995) for Edgeworth expansions. See also van Zwet (1984), Friedrich (1989), Bentkus, Bloznelis and G ¨otze (1996), Bentkus, G ¨otze and van Zwet (1997), Putter and van Zwet (1998) and Wang, Jing and Zhao (2000). Another approach is to estimate the relative errorP(Sn ≥xVn)/(1−Φ(x)). In this direction, Jing, Shao and Wang (2003) re£ned Shao (1999), Wang and Jing (1999) as well as Chistyakov and G ¨otze (2003), and obtained the following result: if0 < σ2 = EX2 < ∞, then there exists an absolute constant A >0such that
expn−A(1 +x)2∆˜n,x o
≤ P ¡
Sn≥xVn¢
1−Φ(x) ≤ exp
n
A(1 +x)2∆˜n,x o
, (1.1) for allx≥0satisfying∆˜n,x ≤1/A, where
˜
Jing, Shao and Wang (2003) actually established (1.1) for independent random variables that are not necessarily identically distributed. It follows from (1.1) that ifE|X|3 <∞, then
P(Sn ≥xVn)
1−Φ(x) = 1 +O(1) (1 +x)
3n−1/2σ−3E
|X|3, (1.2) for0≤x≤An1/6σ/(E|X|3)1/3.
Result (1.2) is useful in statistics because it provides not only the relative error but also a Berry-Esseen type rate of convergence. Indeed, as a direct consequence of this result, it has been shown in Jing, Shao and Wang (2003) that bootstrapped studentized t-statistics possess large deviation properties in the region 0 ≤ x ≤ o(n1/6) under only a £nite third moment condition. However, (1.1) as well as (1.2) does not capture the term withn−1/2 explicitly. This short has limited further applications of the self-normaized large deviation.
In this paper we investigate the limit theorems for self-normalized large deviation. Under £nite moment conditions, a leading term withn−1/2in (1.1) and (1.2) is obtained explicitly. THEOREM 1.1. Assume thatEX4 <∞. Then, forx≥0andx=O(n1/6),
P¡
Sn ≥xVn¢
1−Φ(x) = exp
½ −x
3EX3
3√nσ3 ¾ ·
1 +O³1 +√ x n
´¸
, (1.3)
P¡
Sn ≤ −xVn¢
Φ(−x) = exp
½
x3EX3
3√nσ3 ¾ ·
1 +O³1 +√ x n
´¸
. (1.4)
If in additionEX3 = 0, then, for|x|=O¡ n1/6¢
,
P¡
Sn≤xVn¢
−Φ(x) =O¡
n−1/2e−x2/2¢
. (1.5)
WriteLn,x = (1 +x)ρn+ ∆n,x, whereρn=E|X|3/(√nσ3)and
∆n,x = (1 +x)3σ−3n−1/2E|X|3I{|X|>√nσ/(1+x)}+ (1 +x)4σ−4n−1EX4I{|X|≤√nσ/(1+x)}.
Furthermore, we obtain the following bounds which re£ne (1.1) under £nite third moment con-dition in the region0≤x≤cρ−n1/2, wherecis an absolute constant.
THEOREM 1.2. Assume thatE|X|3 < ∞. Then, there exists a positive absolute constant c
such that, for0≤x≤c ρ−n1/2,
exp³− x
3EX3
3√nσ3 −ALn,x ´
≤ P ¡
Sn≥xVn¢
1−Φ(x) ≤exp
³ − x
3EX3
3√nσ3 +ALn,x ´
. (1.6)
REMARK 1.1. Similar results to those in Theorem 1.1 hold for the standardized mean under
much stronger conditions. For instance, it follows from Section 5.8 of Petrov (1995) that, for x≥0andx=O(n1/6),
P³Sn≥x√nσ´
1−Φ(x) = exp
½
x3EX3
6√nσ3 ¾ ·
1 +O³1 +√ x n
´¸ ,
P³Sn≤ −x√nσ´
Φ(−x) = exp
½ −x
3EX3
6√nσ3 ¾ ·
1 +O³1 +√ x n
´¸ .
hold only when Cram´er’s condition is satis£ed, i.e.,EetX <∞fortbeing in a neighborhood of zero. We also notice that there are different formulae for the self-normalized and standardized cases.
REMARK 1.2. It is readily seen that Ln,xn = o(x
3
n/√n) for xn → ∞ and xn = O(ρ−n1/2). Hence−x33√EX3
nσ3 in (1.6) provides a leading term in this case. However it remains an open
prob-lem for more re£ned results.
This paper is organized as follows. The proofs of main results will be given in Section 3. Next section we present two auxiliary theorems that will be used in the proofs of main results. The proofs of these auxiliary theorems will be postponed to Sections 4 and 5 respectively. Without loss of generality, throughout the paper, we assume σ2 = EX2 = 1 and denote by A, A1, A2,· · · andc, C, C1, C2,· · · absolute positive constants, which may be different at each occurrence. If a constant depends on a parameter, sayu, then we writeA(u). In addition to the notation forLn,x and∆n,xde£ned in Theorem 1.2, we always let
ρn =n−1/2E|X|3 and Ψn(x) = h
1−Φ(x)iexp³− x
3EX3
3√n ´
.
2
Two auxiliary theorems
THEOREM 2.1. Leth=x/Bn, whereBnis a sequence of positive constants with
|Bn2 −n| ≤ C1nEX2I{|X|≥√n/(1+x)}. (2.1)
Suppose thatηj :=ηn(x, Xj),1≤j ≤n,satisfy the conditions:
¯ ¯ ¯Ee
hηj−1− x
2
2B2 n
+ x
3
3B3 n
EX3¯¯
¯ ≤ C2n
−1∆
n,x, (2.2)
¯ ¯ ¯Eηje
hηj − x
Bn ¯ ¯
¯ ≤ C3x
2n−1/2ρn,
(2.3) ¯
¯ ¯Eη
2
jehηj−1 ¯ ¯
¯ ≤ C4x ρn, (2.4)
E|ηj|3ehηj ≤ C5E|X|3. (2.5)
Then, for2≤x≤cρ−1
n withcsuf£ciently small and for any|b| ≤x2/4, P³
n X
j=1
ηj ≥(x+bx−1)Bn´≤(1 + 2|b|x−2)e−bΨn(x) exp(ALn,x); (2.6)
for2≤x≤c1ρ−n1/2 withc1suf£ciently small, P³
n X
j=1
ηj ≥x Bn´≥ Ψn(x) exp(−ALn,x). (2.7)
THEOREM 2.2. Write, for1≤m≤n,
Tm =
1
√ n
m X
j=1
ζj and Λn,m =
1
n2 m−1 X k=1
n X j=k+1
ψk,j,
whereζj :=ζn(x, Xj)andψk,j :=ψn(x, Xk, Xj)satisfy the conditions:
¯ ¯ ¯Eζ
2 j −1 +
x
√nEX3¯¯
¯ ≤ C6∆n,x/(1 +x)
2, (2.8)
¯ ¯ ¯Eζ
3
j −EX3 ¯ ¯
¯ ≤ C7 √
n∆n,x/(1 +x)3, (2.9)
Eζj = 0, |ζj| ≤ C8√n/(1 +x), Eζj4 ≤ C9EX4I{|X|≤√n/(1+x)}, (2.10)
E(ψk,j|Xk) = E(ψk,j|Xj) = 0, f or k 6=j, (2.11) E|ψk,j| ≤ C10|x|, E|ψk,j|3/2 ≤ C11|x|3/2¡
E|X|3¢2. (2.12)
Then, for 2 ≤ x ≤ c ρ−1
n with c suf£ciently small, and for any constants sequence λn(x)
satisfying|λn(x)| ≤C∆n,x/(1 +x),
P³Tn+ Λn,n ≥x+λn(x) ´
≤ Ψn(x) ¡
1 +ALn,x ¢
+ A1¡
3
Proofs of main results
Proof of Theorem 1.1. Note thatLn,x ≍(1 +x)/√nforx ≥0andx =O(n1/6). Theorem 1.1 follows immediately from Theorem 1.2. We omit the details.
Proof of Theorem 1.2. Without loss of generality, assumex ≥ 2. If0 ≤x < 2, the results are direct consequences of the Berry-Esseen bound (cf. Bentkus and G ¨otze (1996))
¯ ¯
¯P(Sn≥xVn)− ©
1−Φ(x)ª¯ ¯
¯≤A ρn.
We £rst provide four lemmas. For simplicity of presentation, de£neτ = √n/(1 +x)and assume2≤x≤c ρ−n1/2 withcsuf£ciently small throughout the section except where we point out.
LEMMA 3.1. We have, P³Sn− x
2√n(V 2
n −n)≥x √
n´ ≥ Ψn(x) exp(−ALn,x); (3.1)
and for2≤x≤c ρ−1
n withcsuf£ciently small and for arbitrary|δ| ≤x2/4, P³Sn− x
2√n(V 2
n −n)≥ ¡
x+δx−1¢√n´ ≤ (1 + 2|δ|x−2)e−δΨn(x) exp(ALn,x). (3.2)
Proof. We £rst prove (3.2). Let h = x/√n and ηj = Xj − 2√xn(X2
j −1). It follows from (4.12)-(4.14) in Wang and Jing (1999) that
¯ ¯ ¯Eη1e
hη1
− √xn¯¯
¯ ≤ 16x
2n−1/2ρn, ¯
¯ ¯Eη
2 1ehη
1
−1¯¯
¯ ≤ 30x ρn, E|η1|3ehη1
≤ 30E|X|3. Thus (3.2) follows immediately from (2.6) in Theorem 2.1 withB2
n=nif we prove ¯
¯ ¯Ee
hη1
−1− x
2
2n + x3
3n3/2EX 3¯¯
¯ ≤ C n
−1∆
n,x. (3.3)
Without loss of generality, assume x ≤ ρ−1
n /16. By noting E|X|3 ≥ (EX2)3/2 = 1, we get h ≤ 1/4. This implies that hη1 = h2/2 +hX1 −(hX1)2/2 ≤ 1 and |hη1|kehη1
≤
sups≤1|s|kes ≤efork= 0,1, . . . ,4. Therefore, using Taylor’s expansion ¯
¯ ¯e
x −
k X
j=0 xj j!
¯ ¯ ¯≤
|x|k+1
(k+ 1)!e
we obtain that Eehη1
= Eh1 +hη1+12 (hη1)2+ 16(hη1)3+ 24θ (hη1)4iI(|X1|≤τ)
+E³1 +hη1+θ1
2 ´
I(|X1|>τ)
= 1 +1
2 E(hη1) 2I(
|X1|≤τ)+
1
6E(hη1) 3I(
|X1|≤τ)
+θ1
2 P(|X1|> τ) + θ
24E(hη1) 4I
(|X1|≤τ), (3.4)
where|θ| ≤eand|θ1| ≤e. SinceEX2 = 1, it is readily seen that ¯
¯
¯E(hη1) 2I(
|X1|≤τ)−h
2 +E(hX)3¯¯
¯ ≤ 8n
−1∆n,x, ¯
¯
¯E(hη1) 3I(
|X1|≤τ)−E(hX)
3¯¯
¯ ≤ 32n
−1∆n,x, E(hη1)4I(|X1|≤τ) ≤ 16n
−1∆n,x.
Taking these estimates back into (3.4), we obtain (3.3), and hence (3.2).
The proof of (3.1) is similar by using (2.7). We omit the details. The proof of Lemma 3.1 is now complete.
The next lemma is from Lemma 6.4 in Jing, Shao and Wang (2003).
LEMMA 3.2. Let{ξi,1≤i≤n}be a sequence of independent random variables withEξi = 0andEξ2
i <∞. Then Ph
n X
i=1
ξi ≥an4Dn+¡ n X
i=1
ξi2¢1/2oi≤8e−a2/2 (3.5)
whereDn =¡ Pni=1Eξ2 i
¢1/2
.
In Lemmas 3.3-3.4, we use the notation:
¯
Xi =XiI{|Xi|≤τ}, Sn¯ =
n X
i=1
¯
Xi, V¯n2 =
n X
i=1
¯
Xi2, Bn2 =
n X
i=1 EX¯i2 Sn(i) =Sn−Xi, Vn(i) = (Vn2−Xi2)1/2.
LEMMA 3.3. We have P¡
Sn≥xVn ¢
Proof. LetΩn = (1−x−1/2,1 +x−1/2). Recalling2≤x ≤cρ−n1/2 withcsuf£ciently small, it follows from (2.27)-(2.29) in Shao (1999) that
P ¡
Sn≥xVn, Vn2/n /∈Ωn ¢ ≤ P ¡
Sn ≥xVn, Vn2 ≥9n ¢
+P ©
Sn ≥xVn, n(1 +x−1/2)≤Vn2 ≤9n ª
+P©
Sn≥xVn, Vn2 ≤n(1−x−1/2)ª ≤ 4 exp¡
−x2/2−x/8 +Ax3ρn¢ ≤ AΨn(x).
This, together with (3.2), implies that P¡
Sn≥xVn¢
=P ¡
Sn≥xVn, Vn2/n ∈Ωn ¢
+P ¡
Sn ≥xVn, Vn2/n /∈Ωn ¢ ≤P
½
Sn− x
2√n(V 2
n −n)≥(x−x−1/4) √
n ¾
+P ¡
Sn≥xVn, Vn2/n /∈Ωn ¢ ≤AΨn(x) exp(ALn,x),
as required. The proof of Lemma 3.3 is complete. LEMMA 3.4. We have
P¡
Sn ≥xVn ¢
≤ Ψn(x) exp(ALn,x) +A1e−3x
2
; (3.7)
P¡
Sn ≥xVn¢
≤ Ψn(x) exp(ALn,x) +A1(xρn)3/2. (3.8)
Proof. As in Wang and Jing (1999), forx≥2,
P(Sn≥xVn) ≤ P( ¯Sn≥xVn¯ ) +
n X
i=1 P£
Sn(i) ≥(x2−1)1/2Vn(i)¤
P(|Xi|> τ).
= Kn+In, say. (3.9)
It follows easily from Lemma 3.3 that for alli P£
Sn(i)≥(x2−1)1/2Vn(i)¤
≤AΨn(x) exp(ALn,x).
This, together withP(|Xi|> τ)≤n−1∆
n,x, implies that
In view of (3.9) and (3.10), the inequalities (3.7) and (3.8) will follow if we prove Kn ≤ Ψn(x) exp(ALn,x) +A1e−3x
2
, (3.11)
Kn ≤ Ψn(x) exp(ALn,x) +A1(xρn)3/2, (3.12) for2≤x≤cρ−1
n withcsuf£ciently small. We £rst prove (3.11). Let D2
n = Pn
i=1EX¯i4 and ξi = ¯Xi2 − EX¯i2. By the inequality
(1 +y)1/2 ≥1 +y/2−y2 for anyy≥ −1and Lemma 3.2,
Kn = PhSn¯ ≥xnBn2 +
n X
i=1
( ¯Xi2−EX¯i2)o1/2i
≤ PhSn¯ ≥xBnn1 + 1 2B2
n n X
i=1
ξi− 1 B4
n ³Xn
i=1
ξi´2oi
≤ Ph¯¯ ¯
n X
i=1 ξi¯¯
¯≥ √
6xn4Dn+³
n X
i=1
ξi2´1/2oi
+PhSn¯ ≥xBnn1 + 1 2B2
n n X
i=1
ξi− 1 B4
n ³Xn
i=1
ξi´2o,
¯ ¯ ¯
n X
i=1 ξi
¯ ¯ ¯≤
√
6xn4Dn+ ³Xn
i=1 ξ2
i
´1/2oi
≤ 8e−3x2 +Kn,1, (3.13)
where, after some algebra (see, Jing, Shao and Wang (2003), for example),
Kn,1 ≤ P³ n X
j=1
ηj ≥xBn´,
withηi = ¯Xi−2−1xB−1
n ( ¯Xi2−EX¯i2) + 24x3Bn−3( ¯Xi4+ 16EX¯i4). As in the proof of Lemma 3.1, tedious but elementary calculations show that the inequalities (2.1)-(2.5) hold true forB2
n = Pn
i=1EX¯i2 and theηi de£ned above. Therefore it follows from (3.2) in Theorem 2.1 that
Kn,1 ≤Ψn(x) exp(ALn,x), (3.14)
for 2 ≤ x ≤ cρ−1
We next prove (3.12). Note that
It is easy to see that ζj satisfy the conditions (2.8)- (2.10), ψk,j satisfy the conditions (2.11)-(2.12) in Theorem 2.2 and |λn(x)| ≤ 3xn−1 + 3(1 + x)−1∆
After these preliminaries, we are now ready to prove Theorem 1.2. As is well-known (see Wang and Jing (1999) for example),
P(Sn≥xVn) ≥ P³Sn− x
2√n(V 2
n −n)≥x √
n´.
The left hand inequality of (1.6) follows from Lemma 3.1 immediately.
To prove the right hand inequality of (1.6), we use Lemmas 3.4. IfΨn(x)≥(xρn)1/2, then by (3.12),
P(Sn≥xVn)≤Ψn(x) exp(ALn,x) ³
1 +A1xρn´≤Ψn(x) exp ³
A2Ln,x ´
. (3.16) Recall we may assume that2≤x≤ρ−1
n /16. It is readily seen that ©
1−Φ(x)ª−3
exp
½
x3EX3 √
n ¾
≤(2π)3/2x3e2x2. This implies that, whenΨn(x)≤(xρn)1/2,
e−3x2 ≤A©
1−Φ(x)ª3exp
½ −x
3EX3 √n
¾
≤A xρnΨn(x).
Therefore, by (3.7),
P(Sn≥xVn)≤Ψn(x) exp(ALn,x) ³
1 +A1xρn´≤Ψn(x) exp(ALn,x). (3.17) Collecting the estimates (3.16) and (3.17), we get the right hand inequality of (1.6). The proof of Theorem 1.2 is now complete.
4
Proof of Theorem 2.1
The proof of Theorem 2.1 is based on the conjugate method. To employ the method, let ξ1,· · ·, ξnbe independent random variables withξj having distribution functionVj(u)de£ned by
Vj(u) = E n
ehηjI(η
j ≤u) o.
Eehηj, for j = 1,· · ·, n,
Also de£neM2 n(h) =
Pn
j=1V ar(ξj), Gn(t) =Pn
Pn
j=1(ξj −Eξj) Mn(h) ≤t
o
and Rn(h) = (x+bx
−1)B
n−Pnj=1Eξj
Since EX2 = 1, we have E|X|3 ≥ 1 and there exists a positive constant c0 such that EX2I
(|X|≥τ) ≤ 1/(4C1) for 2 ≤ x ≤ c0ρ−n1. Let c1 = 28max{1, C1, . . . , C5} and c2 =
min{c0, c−11}. Taking account of the conditions (2.1)-(2.5), it can be shown by a elementary method that, for2≤x≤c2ρ−1
n , Eehηj = 1 + x
2
2n − EX3
3
³ x √
n ´3
+O1n−1∆n,x
= exp
½ x2
2n − EX3
3
³ x √ n
´3
+O2n−1∆n,x ¾
, (4.1)
Eξj = Eηjehηj
.
Eehηj = √x
n +O3x
2n−1/2ρn, (4.2) V ar(ξj) = Eηj2ehηj
.
Eehηj−(Eξj)2 = 1 +O4x ρn, (4.3)
E|ξj|3 ≤ E|ηj|3ehηj
.
Eehηj ≤ 2C5n1/2ρn, (4.4)
where|Oj| ≤1/(2c2), forj = 1, . . . ,4.
Using (4.3)-(4.4) and (2.1), we get, for2≤x≤c2ρ−1 n ,
n/2 ≤ Mn2(h) = n+O4n x ρn ≤ 3n/2, (4.5) |Rn(h)| = ¯¯
¯
(x+bx−1)Bn−Pn j=1Eξj Mn(h)
¯ ¯ ¯ ≤ (3/2)|b|x−1+ 2(C
1+|O3|)x2ρn, (4.6) |hMn(h)−x| ≤ 1
x ¯
¯h2Mn2(h)−x2 ¯ ¯ ≤ x
½¯ ¯ ¯
M2 n(h)
n −1
¯ ¯ ¯+
¯ ¯ ¯
1
B2 n
− 1 n ¯ ¯ ¯M
2 n(h)
¾
≤ (4C1+|O4|)x2ρn. (4.7) LetAn(h) =Rn(h) +hMn(h),c3 = 1
4(6C1+ 2|Q3|+|O3|)−
1andc4 = min{c2, c3}. It follows from (4.6)-(4.7) that, for2≤x≤c4ρ−1
n ,
|An(h)−x| ≤ |Rn(h)|+|hMn(h)−x| ≤ (3/2)|b|x−1+¡
6C1+ 2|O3|+|O4|¢
x2ρn (4.8) and (recall|b| ≤x2/4),
where|O5−1| ≤2|b|x−2 and|O6| ≤A. Therefore, for2≤x≤c4ρ−n1, I2(h) = ex2/2©
1−Φ(x)ª
e−R2n(x)/2
³
O5 +O6x ρn´. (4.13) As forI1(h), by (4.4)-(4.5), integration by parts and Berry-Esseen theorem, we get
|I1(h)| ≤2 sup
v |
Gn(v)−Φ(v)| ≤4Mn−3(h)
n X j=1
E|ξj|3 ≤16C5ρn.
This implies that forx≥2,
I1(h) = O7x ρnex
2
/2©
1−Φ(x)ª
, (4.14)
where|O7| ≤A.
It follows easily from (4.10)-(4.11) and (4.13)-(4.14) that for2≤x≤c4ρ−1 n , P³
n X
j=1
ηj ≥δn(x)´ ≤ (1 + 2|b|x−2)e−bΨn(x) exp ³
A∆n,x ´
(1 +A1xρn)
≤ (1 + 2|b|x−2)e−bΨn(x) exp ³
ALn,x ´
.
This proves (2.6). Similarly, by lettingb = 0, it follows from (4.10)-(4.11) and (4.13)-(4.14) that
P³ n X
j=1
ηj ≥x Bn´
≥ Ψn(x) exp ©
−A∆n,x−R2n(h)/2 ªh
1−n|O6|+|O7|eR2n(h)/2
o x ρni ≥ Ψn(x) exp{−A∆n,x}
h
1−A1x ρn i ≥ Ψn(x) exp{−A2Ln,x},
for2≤x≤c4ρn−1/2, where we have used the fact that|Rn(h)| ≤1by (4.6), and also Rn2(h)≤4(C1+|O3|)2x4n−1(E|X|3)2 ≤8(C1+|O3|)2∆n,x
since ¡E|X|3¢2 ≤ 2©
E|X|3I(
|x|>τ) ª2
+ 2EX4I(
|x|≤τ). This proves (2.7) and hence complete
5
Proof of Theorem 2.2
The idea for the proof of Theorem 2.2 is similar to Proposition 5.4 in Jing, Shao and Wang (2002), but we need some different details. Throughout this section, we use the following notations:g(t, x) = Eeitζ1/√n
The proof of Theorem 2.2 is based on the following lemmas. LEMMA 5.1. If(x, t)∈Ωn(x, t), then
Proof. It follows from Taylor’s expansion ofeix that ¯
This proves (5.1).
On the other hand, by using (5.5), we get that for(x, t)∈Ωn(x, t)and|t|3 ≤(1 +C7)−1/ρn,
g(t, x) = 1−r1(t, x), (5.7)
where
r1(t, x) = t
2
2nEζ 2 1 +η |
t|3
6n3/2E|ζ1| 3,
|η| ≤1, having|r1(t, x)| ≤1/4and
¯
¯r1(t, x) ¯ ¯ 2
≤ t 4
2n2 ¡
E|ζ1|3¢4/3+ |t|
6
18n3 ¡
E|ζ1|3¢2 ≤ 5
9(1 +C7)|t|
3n−1ρn. (5.8) Therefore, it follows fromln(1 +z) = z+η1z2, whenever|z|< 1/2, where|η1| ≤ 1, that for
(x, t)∈Ωn(x, t)and|t|3 ≤(1 +C7)−1/ρn,
lng(t, x) = −t
2
2nEζ 2
1 +θ|t|3n−1ρn, |θ| ≤1 +C7, (5.9)
lngn(t, x) = −t2
2Eζ
2
1 +θ|t|3ρn = − t2
2 +r(t, x), (5.10)
where, by using (5.5) again,
|r(t, x)| ≤ t
2
2|Eζ
2
1 −1|+|θ| |t|3ρn ≤ t2/3. Also, we have
|r(t, x)| ≤ t
2
2|Eζ
2
1 −1|+|θ| |t|3ρn ≤A(t2+|t|3)(1 +x)ρn. These estimates, together with |ez −1| ≤ |z|e|z|, imply that for (x, t) ∈ Ω
n(x, t) and|t|3 ≤
(1 +C7)−1/ρn, ¯ ¯ ¯g
n(t, x) −e−t
2 2
¯ ¯
¯ ≤ e
−t2
2
¯ ¯ ¯e
r(t,x) −1¯¯
¯ ≤ A(1 +x)ρn(t 2+
|t|3)e−t2/6. (5.11) Now, (5.2) follows from (5.6) and (5.11).
If we instead the estimate ofr1(t, x)in (5.8) by |r1(t, x)|2 ≤ t
4
2n2 ¡
Eζ12¢2+ |t|
6
18n3 ¡
E|ζ1|3¢2 ≤2n−2t4, we can rewrite (5.9) and (5.10) as
lng(t, x) = g(t, x)−1 +θ1n−2t4, |θ1| ≤2,
lngn(t, x) = n©
g(t, x)−1ª
Therefore, by using (5.5) and (5.7), we obtain for(x, t)∈Ωn(x, t),
This implies (5.3). The proof of Lemma 5.1 is now complete. LEMMA 5.2. If(x, t)∈Ωn(x, t), then
Proof. It follows from (2.11), (2.12), (5.5) and Holder’s inequality that
Therefore, it follows from independence ofζj and Lemma 5.1 that This proves (5.12).
To prove (5.13) and (5.14), put
Λ∗n,m = Λn,n−Λn,m = Therefore, (5.13) follows easily from (5.12) and (5.15) withm =n. In view of independence ofζj, on the other hand, (5.15) implies that for any1≤m≤n, where we have used the estimate (recalling (2.12)):
E|Λn,meit(Tn+Λ∗n,m)| = E This gives (5.14). The proof of Lemma 5.2 is now complete.
LEMMA 5.3. LetF be a distribution function with the characteristic functionf. Then for all
and2K(s) = K1(s) +iK2(s)/(πs),
K1(s) = 1− |s|, K2(s) =πs(1− |s|)cot πs+|s|, f or|s|<1,
andK(s)≡0for|s| ≥1.
Proof of Lemma 5.3 can be found in Prawitz (1972).
LEMMA 5.4. It holds that if0≤x≤cρ−1
n withc≤ 16(1 +C6)−
1, then for anyy ∈R,
|I+(y)|, |I−(y)| ≤ Aρne−y2/2+A1(1 +x3/2)ρ2n, (5.18)
whereK1(s)is de£ned as in Lemma 5.3,
I+(y) = 1
T Z T
−T
e−iytK1¡t T
¢
Eeit(Tn+Λn,n)dt,
I−(y) = 1
T Z T
−T
e−iytK1¡ −Tt¢
Eeit(Tn+Λn,n)dt, T = 1
6(1 +C7)
−1/ρn.
Proof. We only prove (5.18) forI+(y). Without loss of generality, we assumeρ
n≤12−3. This assumption implies that1 +x≤ 1
3(1 +C6)−
1/ρnfor0≤x≤cρ−1
n withc≤ 16(1 +C6)− 1. Let I+ =I+(y)andT1 =ρ−1/3
n . We have that I+ =
Z T1
−T1
· · ·+
Z
T1≤|t|≤T
· · · := I1+I2.
It is easy to see that[12nt−2log|t|]≤ n−2if|t| ≥12andn ≥6. Hence, recalling|T1| ≥ 12 and√n ≥12/E|X|3 ≥12, by (5.14) withm= [12nt−2log|t|] + 2,
|I2| ≤ 1 T
Z
T1≤|t|≤T
¯
¯Eeit(Sn+Λn,n) ¯
¯dt≤A(1 +x3/2)ρ2n. (5.19) NotingK1(s) = 1− |s|, for|s|<1, we obtain|I1| ≤ |I11|+|I12|, where
I11 = 1
T Z T1
−T1
e−iytEeit(Sn+Λn,n)dt, I12= 2
T2 Z T1
0
t¯¯Eeit(Sn+Λn,n) ¯ ¯dt.
It follows from (5.14) withm = [12nt−2log|t|] + 2again that
|I12| ≤ 2 T2
nZ 12 0
tdt+
Z T1
12
t¯¯Eeit(Sn+Λn,n) ¯ ¯dt
o
On the other hand, noting that
it follows from Lemmas (5.1)-(5.2) (i.e, (5.2) and (5.13)) that |I11| ≤ 1
Collecting all these estimates, we conclude the proof of Lemma 5.4. LEMMA 5.5. The integral
and T1 = ρ−n1/3. Similar to (5.19), it follows that |J2| ≤ A(1 +x3/2)ρ2n. By using (5.3) and (5.13), we have
|J12| ≤ to (5.20), it can be easily shown that
|J13| ≤ AT−2
On the other hand, simple calculation shows that i
Therefore, it follows from all these estimates andx≤cρ−1 n that This also completes the proof of Lemma 5.5. LEMMA 5.6. For any|y| ≤A(1 +x),
Proof. We £rst note that
This, together with Taylor’s expansion andEζ4
1 ≤C9EX4I(|X|≤τ), implies that
Therefore, taking account of (2.8)-(2.9) and (5.27), we have ¯
This concludes (5.23).
We next prove (5.24)-(5.26). It can be easily seen that∆n,x ≤(1+x)3ρn−1, and forx≤cρ−n1 withcsmall enough,
These estimates imply that
This proves (5.24) and (5.25). On the other hand, we have ¯
This, together with (5.27)-(5.24), implies that ¯ This gives (5.26). The proof of Lemma 5.6 is now complete.
After these lemmas, we are now ready to prove Theorem 2.2. By using (5.17) and Lemmas 5.4-5.5, we obtain for2≤x≤cρ−1
n withcsuf£ciently small, P³Tn+ ∆n,n≥x+λn(x) Using the well-known inequality
we have forx≥2, ¯ ¯ ¯x
3©
1−Φ(x)ª − x
2 √
2πe
−x2
/2¯¯ ¯≤
1
√
2πe
−x2
/2
ande−x2
/2 ≤Ax©
1−Φ(x)ª
. Taking these estimates back into (5.31), we get for2≤x≤cρ−1 n
withcsuf£ciently small,
P³Tn+ ∆n,n ≥x+λn(x) ´
≤ ©
1−Φ(x)ªh
1−x
3EX3
3√n +ALn,x i
+ A1x3/2ρ3/2n . Now we conclude the result if we prove
1− x
3EX3
3√n +ALn,x ≤ exp ½
−x 3EX3
3√n ¾
h
1 +A1Ln,x i
. (5.32)
In fact, (5.32) is obvious forEX3 <0. In the case thatEX3 >0, if x3
EX3
3√n ≤2, then
1− x
3EX3
3√n +ALn,x ≤ exp ½
−x 3EX3
3√n ¾
+ALn,x
≤ exp
½ −x
3EX3
3√n ¾
h
1 +Ae2Ln,x i
.
On the other hand, it follows easily from the de£nition of∆n,xthat for2≤x≤cρ−n1, A(xρn+ ∆n,x)≤(1 +x3EX3/√n)/9,
by choosingcsuf£ciently small. Therefore, if x33EX√ 3
n ≥2, then
1− x
3EX3
3√n +ALn,x ≤0≤exp ½
−x 3EX3
3√n ¾
h
1 +ALn,x i
.
Collecting all these estimates, we get the desired (5.32). This also completes the proof of Theorem 2.2.
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