Self-normalized Cramér type moderate deviations for the maximum of sums

Bernoulli 193, 2013, 1006 1027 DOI: 10.3150/12-BEJ415 arxiv:1307.6044v1 [math.st] 23 Jul 2013 Self-normalized Cramér type moderate deviations for the maximum of sums WEIDONG LIU 1, QI-MAN SHAO 2 and QIYING WANG 3 1 Department of Mathematics and Institute of Natural Sciences, Shanghai Jiao Tong University, Shanghai, China. E-mail: liuweidong99@gmail.com 2 Department of Mathematics, Hong Kong University of Science and Technology, Clear Water Bay, Kowloon, Hong Kong. E-mail: maqmshao@ust.hk 3 School of Mathematics and Statistics, University of Sydney, Australia. E-mail: qiying@maths.usyd.edu.au Let X 1,X 2,... be independent random variables with zero means and finite variances, and let S n = n Xi and V n 2 = n X2 i. A Cramér type moderate deviation for the maximum of the self-normalized sums max S k /V n is obtained. In particular, for identically distributed X 1,X 2,..., it is proved that max S k xv n/1 Φx 2 uniformly for 0<x on 1/6 under the optimal finite third moment of X 1. Keywords: independent random variables; maximum of self-normalized sums 1. Introduction and main results Let X 1,X 2,... be a sequence of independent non-degenerate random variables with zero means. Set n S n = and Vn 2 = Xj 2. j=1 X j The past decade has brought significant developments in the limit theorems for the socalled self-normalized sum, S n /V n. It is now well understood that the limit theorems for S n /V n usually require fewer moment assumptions than those for their classical standardized counterpart, and thus have much wider applicability. For examples, for identically distributed X 1,X 2,..., a self-normalized large deviation holds without any moment assumption Shao [11], and a Cramér type moderate deviation Shao [12], j=1 S n xv n lim =1, 1.1 n 1 Φx This is an electronic reprint of the original article published by the ISI/BS in Bernoulli, 2013, Vol. 19, No. 3, 1006 1027. This reprint differs from the original in pagination and typographic detail. 1350-7265 c 2013 ISI/BS

2 W. Liu, Q.-M. Shao and Q. Wang holds uniformly for x [0,on 1/6 provided that E X 1 3 <, whereas a finite momentgenerating condition of X 1 is necessary for a similar result for the standard sum S n / VarS n see, e.g., Linnik [9]. For more related results, we refer to de la eña, Lai and Shao [5] for a systematic treatment of the theory and applications of selfnormalization and Wang [13] for some refined self-normalized moderate deviations. As for the Cramér type moderate deviations for the maximum of self-normalized sums, namely for max S k /V n, Hu, Shao and Wang [7] were the first to prove that if X 1,X 2,... is a sequence of i.i.d. random variables with EX 4 1 <, then max S k xv n lim =2, 1.2 n 1 Φx uniformly for x [0,on 1/6. This contrasts with the moderate deviation result for the maximum of partial sums of Aleshkyavichene [1, 2], where a finite moment-generating condition is required. However, in view of the result given in 1.1, it is natural to ask whether a finite third moment suffices for 1.2. The main purpose of this paper is to provide an affirmative answer to this question. Indeed, we have the following more general result for independent random variables. Theorem 1. Assume that max k 1 E X k 2+r < and min k 1 EXk 2 >0, where 0<r 1. Then 1.2 holds uniformly in 0 x on r/4+2r. As in the moderate deviation result for self-normalized sum S n /V n, Theorem 1 is sharp in both the moment condition and the range in which the result 1.2 holds true. Examples can be constructed similarly as done by Chistyakov and Götze[4] and Shao[12]. Inparticular,forr=1andidenticallydistributed X 1,X 2,...,Theorem1establishes1.2 under the optimal finite third moment of X 1. Theorem 1 can be extended further; in fact, it is a direct consequence of Theorem 2 below. Set Bn 2 = n EX2 i, L n,r = n E X i 2+r and d n,r = B n /Ln,r 1/2+r, where 0 < r 1. Theorem 2. For 0<r 1, suppose that d n,r as n, and that max n j=k E X j 2+r n j=k E X j 2 τlr/2+r n,r d δ n,r for some δ,τ >0. 1.3 Then 1.2 holds uniformly in 0 x min{b n,od n,r }. Remark 1. For i.i.d. random variables with EX i =0 and E X i 3 <, Jing, Shao and Wang [8] proved that 1.1 can be refined as S n xv n 1 Φx =1+O11+x 3 E X 1 3 /EX 2 1 3/2

Self-normalized moderate deviation for maximum 3 uniformly in x [0,n 1/6 EX 2 1 1/3 /EX 2 1 1/2, where O1 is bounded by an absolute constant. We conjecture that a similar result holds for max S k /V n, that is, max S k xv n 1 Φx uniformly in x [0,n 1/6 EX 2 1 1/2 /E X 1 3 1/3. =2+O11+x 3 E X 1 3 /EX 2 1 3/2 This paper is organized as follows. The proof of the main theorems is given in the next section. The proofs of two technical propositions are deferred to Sections 3 and 4, respectively. Throughout the paper, A,A 1,... denotes absolute constants and C δ,τ denotes a constant depending only on δ and τ, which might be different at each appearance. 2. roofs of theorems roof of Theorem 1. Simple calculations show that if max E X k 2+r < and min k 1 k 1 EX2 k >0, then Bn 2 n, L n,r n, d n,r n r/4+2r and 1.3 holds for δ =1 and some τ >0, where the notation a n b n denotes 0<lim n a n /b n <lim n a n /b n <. Therefore, Theorem 1 follows immediately from Theorem 2. roof of Theorem 2. First note that for ǫ>0, 1 Bn 2 EX 2 k I X k ǫb n ǫ r d 1/2+r n,r whenever d n,r 0. That is, the Lindeberg condition is satisfied for the sequence X 1,X 2,... On the other hand, routine calculations show that, given d n,r 0, Vn/B 2 n 2 1 in probability. Given these facts, the invariance principle see Theorem 2 of Brown [3] and the continuous mapping theorem imply that max S k /V n D N0,1. This yields 1.2 uniformly for 0 x M, where M is an arbitrary constant. Thus, Theorem 2 will follow if we can prove lim lim M n Toward this end, let n,x = x2 B 2 n sup M x min{b n,od n,r} EX 2 i{ X i >B n /x}+ x3 B 3 n 0 max S k xv n 1 Φx 2 =0. 2.1 E X i 3 I{ X i B n /x},

4 W. Liu, Q.-M. Shao and Q. Wang and write n 0 n 0 x=max { k: } EXj 2 192Bnlogx e/x 2 2,1 k n. 2.2 j=k It can be readily seen that the condition 1.3, together with 0<x min{b n,od n,r } and d n,r, imply the existence of an absolute constant A such that n,x 0, and 0 x B n, n,x minδ 9/2,1/A, 2.3 n j=n 0+1 E X j 3 I{ X j B n /x} n j=n 0+1 E X j 2 B n x 1+δ 2.4 for all sufficiently large n, where δ is defined as in 1.3. The result 2.1 follows immediately from the following proposition. roposition 1. For all x 2 satisfying 2.3 and 2.4, we have max S k xv n 1 Φx =2+O1x min{1/4,δ/20} + 1/9 n,x, 2.5 where O1 is bounded by a constant C δ that depends only on δ. The main idea of the proof of roposition 1 is to use truncation and the maximum probability inequality and then apply a moderate deviation theorem of Sakhanenko [10] to the truncated variables. A suitable truncation level is ensured by using an inequality from Jing, Shao and Wang [8], page 2181. This avoids the conjugate argument of Hu, Shao and Wang [7], and makes it possible to prove the main result under an optimal moment assumption. It remains to prove roposition 1. In addition to the notation in the previous section, let γ =72 1 minδ,1, ε=max2 2/9 n,x,γx 1/2,γx δ/10, m=[x 2 /2], N 0 = and, for 1 l m, N l ={j 1,j 2,...,j l } {1,2,...,n}. Furthermore, write X i = X i I{ X i εb n /x}, and for 0 l m and, S N l k = S N l k = k,i/ N l Xi, k,i/ N l X i, V Nl n 2 = V Nl n 2 =,i/ N l X2 i,,i/ N l X 2 i, B Nl n 2 = BNl n 2 =,i/ N l E X 2 i,,i/ N l EX 2 i.

Self-normalized moderate deviation for maximum 5 Note that if s,t R 1, x 1, c 0 and s+t x c+t 2, then s x 2 1 1/2 c. Similar to the arguments in reported by Jing, Shao and Wang [8], page 2181, we have max S k xv n max S N0 k x V n N0 + + j 1=1 max j 1=1 max S k xv n, X j1 εb n /x S N0 k x V n N0 max SN1 k x 2 1Vn N1 X j1 εb n /x 2.6 and max S k xv n max S N0 k x V n N0 2.7 max S N1 k x 2 1 V n N1 X j1 εb n /x. j 1=1 Repeating 2.6 m-times gives max S k xv n max S N0 k x V n N0 { m n m+1 + Z k x+ X k εb n /x}, where Z k x= j 1=1 [ k ] X ji εb n /x j k =1 max S N k j 1 j n x 2 k V N k n. 2.8 Note that X k εb n /x x2 ε 2 B 2 n x2 ε 2 B 2 n EXkI{ X 2 k εb n /x} EX 2 ki{ X k B n /x}+ x3 ε 3 B 3 n ε 3 n,x ε 3/2 /16 1/16. E X k 3 I{ X k B n /x} 2.9

6 W. Liu, Q.-M. Shao and Q. Wang It follows from m=[x 2 /2] that [ n m+1 X k εb n /x] e x2. 2.10 This, together with 2.7 and 2.8, implies that roposition 1 will follow if we prove the following two propositions. roposition 2. For all 0 l m, all x/2 y x, and all x 2 satisfying 2.3 and 2.4, we have max SN l k y V N l n 2+C δ,τ ε 2 n,x +ε. 2.11 1 Φy roposition 3. For all x 2 satisfying 2.3 and 2.4, we have Indeed, noting that max SN 0 k x V n N0 =2+C δ,τ ε 2 n,x +ε. 2.12 1 Φx x /2 1 Φx 1 e 2π1+x2 e x2 x2 /2 2πx for x 1, we have that for 1 k m=[x 2 /2] and x 1, 1 Φ x 2 k 1 Φx 2e k/2. This, together with 2.8 2.11, implies that for all x 2 satisfying 2.3 and 2.4, max S k xv n { m e x2 +2 1 Φx+ {1 Φ { n } k } x 2 k} X j ǫb n /x {1+C δ,τ ǫ 2 n,x +ǫ} 2.13 21 Φx{1+C δ,τ ǫ 3 n,x +ǫ+x 1 } 21 Φx{1+C δ,τ x min{1/4,δ/20} + 1/9 n,x }. Similarly,by2.7,2.11and2.12,weobtainthatforallx 2satisfying2.3and2.4, max S k xv n 21 Φx{1 C δ,τ x min{1/4,δ/20} + 1/9 n,x}. 2.14 Combining 2.13 and 2.14, we obtain 2.5, and thus roposition 1. j=1

Self-normalized moderate deviation for maximum 7 Itremainstoproveropositions2and3,whichwegiveinSections3and4,respectively. The proof of Theorem 2 is now complete. 3. roof of roposition 2 Let b=y/b N l n. First, note that S N l k y V N l n max 2b max + max S N l k b V N l n 2 +y 2 ε 2 3.1 S N l k y V N l n, b V N l n y ε. Furthermore, we have max max S N l k y V N l n, b V N l n y ε + max S N l k y V N l n,b2 V N l n 2 >y 2 +εy S N l k y V N l n,b2 V N l n 2 <y 2 εy 3.2 and 2b max n =:I 1 +I 2 S N l k b2 V N l n 2 +y 2 ε 2 {2b S N l k b2 V N l n 2 +y 2 ε 2, E[ V N l n 2 V N l k 2 ] [ V N l n 2 V N l k 2 ] ε 2 /b 2 } n + {2b S N l k b2 V N l k 2 +b 2 E[ V N l n 2 V N l k 2 ]+y 2 2ε 2 } =:I 3 +I 4. 3.3 By 3.1 3.3, roposition 2 follows from the following Lemma 1. Lemma 1. Under the conditions of roposition 2, we have I 1 C δ,τ y 2 exp y 2 /2, 3.4 I 2 C δ,τ y 2 exp y 2 /2, 3.5 I 3 C δ,τ y 2 exp y 2 /2, 3.6 I 4 2[1 Φy][1+C δ,τ ε 2 n,x +ε]. 3.7

8 W. Liu, Q.-M. Shao and Q. Wang To prove Lemma 1, we start with some preliminaries. Note that γmax{x 1/2,x δ/10 } ε min{1/24,δ/72}, n,x ε/2 9/2. 3.8 This fact 3.8 is repeatedly used in the proof without further explanation. Define k 0 = 0,k T =n and,1 i<t, by { k =max k: EXj 2 2 1 ε 3 Bn }. 2 /x2 By the definition of, j= 1+1 j= 1+1 EX 2 j 2 1 ε 3 B 2 n /x2 and k i+1 j= 1+1 EX 2 j >2 1 ε 3 B 2 n /x2 3.9 for any 1 i<t. By 2.3 and 3.8, x 2 max EX2 k x 2 max [EX2 k{ X k >B n /x}+e X k 3 I{ X k B n /x} 2/3 ] B 2 n n,x + 2/3 n,x ε 3 B 2 n/4, which, together with 3.9, implies that Therefore, j= 1+1 T 1 T 14 1 ε 3 Bn 2 /x2 EX 2 j 4 1 ε 3 B 2 n/x 2. j= 1+1 EX 2 j B2 n, 3.10 which yields T 4x 2 /ε 3 +1. For 1 +1 j 1, define events { A j ={ S N l j y B N l n 2 1+ε/y}, C j = X k E X k εbn }. Nl /y k=j+1,k/ N l Note that k N l EX 2 k ε3 B 2 n /8 for all 0 l m=[x2 /2] by 3.10, and thus B 2 n B N l n 2 = EXk 2 k N l EX 2 k 1 ε 3 /8B 2 n 7 8 B2 n. 3.11 Applying the Chebyshev inequality, we have, for any 1 j and x/2 y x, C j 1 y2 k=j+1 EX2 k ε 2 B N l n 2 1 4ε/7 1/2. 3.12

Self-normalized moderate deviation for maximum 9 We are now ready to prove Lemma 1. roof of 3.4. It follows from 3.12 and the independence between C j and {A l,l j} that T ki I 1 A j 2 2 [ T [ T T j= 1+1 A ki 1+1+ j= 1+2 A ki 1+1,C ki 1+1+ A c 1,...,A c j 1,A j j= 1+2 ] A c 1,...,A c j 1,A j,c j S N l E S N l y B N l n 2 1+ε/y εb N l n /y D, where D ki = j=1 E X j I{ X j >εb n /x}. Taking t=y 1+ε/y/B N l n and noting tεb n /y+d ki 2ε+1, ] 3.13 we have S N l E S N l y B N l n 2 1+ε/y εb N l n /y D 9exp y 2 εy 9exp y 2 εy j=1,j/ N l Eexpt X j E X j j=1,j/ N l 9exp y 2 /2 εy/2+a n,x. 1+ EX2 j 2 t2 +8t 3 E X j 3 e 2tεBn/x Submitting this estimate into 3.13 and recalling T 4x 2 /ε 3 +1, x/2 y x and ε γx 1/2, we obtain I 1 4ε 3 x 2 +1exp y 2 /2 εy/2+a n,x C δ,τ y 2 e y2 /2. 3.14 This proves 3.4. roof of 3.5. For this part, let Y ki = j=k X2 i 1+1,j/ N l j, and define Ā j ={ S N l j y V N l n 2 Y ki,b 2 [ V N l n 2 Y ki ]<y 2 εy}, 1 j n.

10 W. Liu, Q.-M. Shao and Q. Wang From 3.12 and the independence between C j and {Āl,l j}, it follows that [ ] T I 2 Ā 1+1+ Āc k,...,āc i 1 j 1,Āj 2 [ T j= 1+2 Ā 1+1,C ki 1+1+ j= 1+2 Āc 1,...,Āc j 1,Āj,C j 3.15 T 2 S N l E S N l y V N l n 2 Y ki εb N l n /y D,b 2 [ V N l n 2 Y ki ]<y 2 εy =:2 T I 2,i, where, as before, D ki = j=1 E X j I{ X j >εb n /x}. Furthermore, for,...,t, I 2,i V N l n 2 Y ki <1 εb N l n 2 [y] + S N l E S N l y =:I 2,i,0 + I 2,i,k. B N l n 2 [1 k+1ε/y] εb N l n /y D ki, B N l n 2 [1 k+1ε/y]< V N l n 2 Y ki <B N l n 2 [1 kε/y] [y] Note that, for any t 1 0 and t 2 0, Eexpt 1 X k E X k +t 2 E X 2 k X 2 k 1+ 1 2 Et 1 X k E X k +t 2 E X 2 k X 2 k 2 +8t 3 1E X k 3 +8t 3 2E X k 6 e 2t1εBn/x+t2EX2 k 3.16 exp 1 2 t2 1E X 2 k + 1 2 4t 1t 2 +t 2 2εB n /xe X k 3 +8t 3 1 +8t 3 2ε 3 B 3 n/x 3 E X k 3 e 2t1εBn/x+t2max EX 2 k. Let t 1 =y 1 k+1ε/y/b N l n and t 2 =ε 1 y 2 /B N l n 2 in 3.16. Noting that we have for 1 k [x], t 1 εb N l n /y+d ε+1, I 2,i,k t 1 S N l E S N l +t 2 {E[ V N l n 2 Y ki ] [ V N l n 2 Y ki ]} y 2 k+1ε/y+t 2 kεb N l n 2 /y 2 ]

Self-normalized moderate deviation for maximum 11 exp y 2 +k+1ε/y t 2 kεb N l n 2 /y+2 exp k i 1,k/ N l Eexpt 1 X k E X k +t 2 E X 2 k X 2 k Eexpt 1 X k E X k Eexpt 2 E X k 2 X k 2 k= 1+1,k/ N l k=+1,k/ N l y 2 /2+2 1 k+1ε/y t 2 kεb N l n 2 /y+2 +At 1 t 2 +t 2 2 εb n/x+t 3 1 +t3 2 ε3 B 3 n /x3 n E X k 3 exp y 2 /2+2 1 k+1ε/y ky+aε 1 n,x +2 Aexp y 2 /2 y/2. Similarly, by 3.16 with t 1 =0, we have I 2,i,0 t 2 {E[ V N l n 2 Y ki ] [ V N l n 2 Y ki ]} t 2 εb N l n 2 ε 2 Aexp y 2 +Aε 1 n,x A 1 exp y 2 /2 y. Combining above inequalities yields I 2 A4x 2 /ε 3 +1e y2 /2 y A 1 y 2 e y2 /2. 3.17 The proof of 3.5 is now complete. roof of 3.6. Following the arguments in the estimates of I 1 and I 2, we have I 3 T T ki j= 1+1 ki j= 1+1 {2b S N l j b 2 V N l n 2 +y 2 ε 2, E[ V N l n 2 V N l j 2 ] [ V N l n 2 V N l j 2 ] ε 2 B N l n 2 /y 2 } {2b S N l j b 2 [ V N l n 2 Y ki ]+y 2 ε 2 }, E X k 2 X k 2 2 1 ε 2 Bn Nl 2 /y 2 k=+1,k/ N l 3.18

12 W. Liu, Q.-M. Shao and Q. Wang 2 =:2 T 2b S N l b 2 [ V N l n 2 Y ki ]+y 2 2ε, T I 3i. E X k 2 X k 2 2 1 ε 2 Bn Nl 2 /y 2 k=+1,k/ N l As in the proof of 3.16, it can be easily shown that for α 0, where and Ee b X j αb 2 X2 j exp{1/2 αb 2 E X j 2 +A j n,x }, 3.19 j n,x = x2 Bn 2 EXj 2 I{ X j εb n /x}+ x3 Bn 3 E X j 3 I{ X j εb n /x} ε 1 x 2 B 2 n EXj 2 I{ X j B n /x}+ x3 Bn 3 E X j 3 I{ X j B n /x} Ee αe X 2 j X 2 j b2 X2 j /2 exp{ 1 2 b2 E X 2 j +2α 2 B n /x+x 3 /B 3 nεe X j 3 e αmax EX 2 k }. 3.20 Next, let t satisfy te tmax EX 2 k = εb n 24x n j=+1 E X j 3. Clearly t exists. Furthermore, we have t x 2 /Bn. 2 Indeed, if tmax EXk 2 ε, then by 3.10 and recalling ε 1/24, If tmax EXk 2 ε, then t t ε/ max EX2 k 4ε 2 x 2 /B 2 n x2 /B 2 n. εb n 24e ε x n j=+1 E X j 3 εx2 30B 2 n n,x 1 15 7/9 n,x x2 /B 2 n x2 /B 2 n. Now it follows from 3.19 and 3.20 with α=t that I 3i b S N l 2 1 b 2 [ V N l n 2 Y ki ]+t E X k 2 X k 2 k=+1,k/ N l

Self-normalized moderate deviation for maximum 13 y 2 /2 2ε+2 1 tε 2 B N l n 2 /y 2 [ exp 2ε y 2 /2 ε2 B N l 2y 2 n 2 t ] 1 j=1,j/ N l Ee b Xj 2 1 b 2 X2 j j= 1+1,j/ N l Ee b Xj Aexp y 2 /2exp A 1 exp y 2 /2exp ε2 Bn 2t n Ee 2 1 b 2 X2 j +te X j 2 X j 2 j=+1,j/ N l j=+1 EX2 j ε 1 n,x ε2 Bn 2t 3x 2 y2n 2B N l n 2 +2t 2 B n /x+x 3 /B 3 nε 4x 2 y2n A 1 exp y 2 /2exp ε2 Bnt 2 4x 2 y2n j=+1 j=+1 EX2 j 4Bn 2 j=+1 EX2 j 4Bn 2 + y2ki j= 1+1 EX2 j 2B N l n 2 E X j 3 e tmax EX 2 k + ε2 x 2 12Bn 2t. 3.21 Note that when t 2δx2 logx B, tmax n 2 1 k EX 2 ε3 k < δ 2 logx by 3.10. Thus, by the definition of t, Now considering t 2δx2 logx B 2 n ε3 t εb n 24x 1+δ/2 n j=+1 E X j 3. and t 2δx2 logx B 2 n ε3, we have, by 3.21, I 3i Ay δ/3ε exp y 2 /2 +Ae y2 /2 exp y2 n j=+1 EX2 j ε 3 B 3 n 4Bn 2 144x 3+δ/2 n j=k E X. i+1 j 3 From the definition of n 0, n j=n 0+1 EX2 j 192B2 n x 2 logx and thus by 2.4 j=n 0+1 E X j 3 192τB3 nlogx x 3+δ. 3.22 For i<i 0, where i 0 =max{i: +1 n 0 }, we have n y 2 EXj 2 x2 EXj 2 /4 24B2 n logx. j= j=n 0

14 W. Liu, Q.-M. Shao and Q. Wang It now follows from 3.22, 3.8 and the fact T 4x 2 /ε 3 +1 that I 3 2 T I 3i 2ATy δ/3ε e y2 /2 +2Ae y2 /2 i 0 e 6logx +2Ae y2 /2 T +1 ε 3 B 3 n exp 144x 3+δ/2 n j=k E X i+1 j 3 3.23 A 1 4x 2 /ε 3 +1e y2 /2 y 6 +e Axδ/2 ε 3 /logx C δ,τ y 2 e y2 /2. This completes the proof of 3.6. roof of 3.7. For this result, we need the following moderate deviation theorem for the standardized sum due to Sakhanenko [10] also see Heinrich [6]. Lemma 2. Suppose that η 1,...,η n are independent random variables such that Eη j =0 and η j 1 for j 1. Write σn 2 = n j=1 Eη2 j and L n = n j=1 E η j 3 /σn. 3 Then there exists an absolute constant A>0 such that for all 1 x min{σ n,ln 1/3 }/A, where O1 is bounded by an absolute constant. To prove 3.7, write n j=1 η j xσ n =1+O1x 3 L n, 3.24 1 Φx ξ j =2b X j b 2 X2 j +b 2 E X j, 2 { } j E j = ξ k z, where z =2y 2 ε 2.,k/ N l Note that ξ j Eξ j 4ε+2ε 2 5ε, and by the non-uniform Berry Esseen bound, there exists an absolute constant A 0 such that for any 1 k n and c>0, n ξ j Eξ j cε j=k,j/ N l 1 Φt+ A n 0 j=k,j/ N l E ξ j Eξ j 3 1+t 3 s 3 n,k 1 2 1 t e s2 /2 ds+ 5A 0 2π c 1+t 3 t, 0

Self-normalized moderate deviation for maximum 15 where s 2 n,k = n j=k,j/ N l Varξ j and t=cε/s n,k. Because t 0 e s2 /2 ds t1+t 3 /2 for any t 0, we may choose c 0 10A 0 2π such that for all, n By virtue of 3.25, we obtain that j=k,j/ N l ξ j Eξ j c 0 ε 1/2. 3.25 I 4 =E 1 + E c 1,...,Ec k 1,E k k=2 2 E 1, ξ j Eξ j c 0 ε j=2,j/ N l +2 E c 1,...,Ec k 1,E k, ξ j Eξ j c 0 ε k=2 j=k+1,j/ N l n 2 ξ k Eξ k z c 0 ε D n,,k/ N l 3.26 where D n = n j=1,j/ N l Eξ j. Write z =z c 0 ε D n. It is not difficult to show that D n 2b E X j I{ X j εb n /x} 4ε 2 n,x, j=1,j/ N l s 2 n,1 = Varξ j =4b 2 EXj 2 +O1ε 1 n,x j=1,j/ N l j=1,j/ N l =4y 2 +O1ε 2 n,x, where O1 30. This yields that z s n,1 =y+o1[ε+ε 2 n,x /y], where O1 40. Therefore, by Lemma 2 with η j =ξ j Eξ j I 4 2[1 Φz /s n,1 ] [ 1+Az /s n,1 3 s 3 2[1 Φy][1+Aε+ε 2 n,x ], n,1 j=1,j/ N l E ξ j 3 ] 3.27

16 W. Liu, Q.-M. Shao and Q. Wang where we have used the fact that whenever xθ n 0, 1 Φx+θ n 1 Φx =1+O1xθ n. This proves 3.7, and also completes the proof of roposition 2. 4. roof of roposition 3 By roposition 2, it suffices to show that S k x V n 21 Φx1 C δ,τ ε 2 n,x +ε. 4.1 max Toward this end, let b = x/bn N0 throughout this section. Recall 3.8, which we use repeatedly in the proof without further explanation. Let n 0 be defined as in 2.2. It can be readily seen that max S k x V n 2b max S k b 2 V 2 n +x 2 n 0 k n n {2b S k b 2 V 2 k +b 2 E V n 2 V k 2 +x2 +ε} k=n 0 n {2b S k b 2 V 2 k +x 2 +ε, V n 2 V k 2 E V n 2 V k 2 εb2 n /x2 } k=n 0 =:I 5 I 6. To complete the proof of roposition 3, we only need to show the following lemma. Lemma 3. Under the conditions of roposition 3, we have 4.2 I 5 21 Φx1 C δ,τ ε 2 n,x +ε, 4.3 I 6 C τ,δ x 2 e x2 /2. 4.4 roof of 4.3. We have n I 5 {2b S k b 2 V 2 k +b 2 E V n 2 V k+x 2 2 +ε} n0 {2b S k b 2 V 2 k +b 2 E V n 2 V k+x 2 2 +ε} =:I 5,1 I 5,2. 4.5

Self-normalized moderate deviation for maximum 17 Write ξ j =2b X j b 2 X2 j +b 2 E X j, 2 { j } F j = ξ k y, where y=2x 2 +ε. As in the proof of 3.25, there exists a constant c 0 such that for all 0 k n 1, n j=k+1 ξ j Eξ j c 0 ε This, together with the independence of ξ j, yields that I 5,1 =F 1 + F c 1,...,Fc k 1,F k k=2 F 1,y ξ 1 y+4ε+ 2 +2 F 1,y ξ 1 y+4ε, 1/2. F c 1,...,F c k 1,F k,y ξ j Eξ j c 0 ε k=2 j=2 F c 1,...,F c k 1,F k,y k=2 k ξ j y+4ε, j=1 n 2 ξ k Eξ k y+c 0 +4ε+D n, j=k+1 k ξ j y+4ε j=1 ξ j Eξ j c 0 ε where D n = n j=1 Eξ j. Similarly to the proofs of 3.26 3.27, it follows from Lemma 2 with η j =ξ j Eξ j that n I 5,1 2 ξ k Eξ k y+c 0 +4ε+D n 21 Φx1 Aε+ε 2 n,x. On the other hand, similar to the proofs of 3.26 and 3.27, we have I 5,2 2 n0 ξ j 2x 2 +1 c 0 ε D n j=1 4.6

18 W. Liu, Q.-M. Shao and Q. Wang Cx 1 exp x2 2 x2n 2B n Cx 2 e x2 /2. j=n 0+1 EX2 j 4.7 This, together with 4.6, implies 4.3. roof of 4.4. Define k 0 =1, and k i =k i 1 +1 if EX2 k i 1 +1 >ε2 B 2 n /x6, and otherwise k i =max {k n: Let m satisfy k m 1 <n k m and define k j=k i 1 +1 EX 2 j } ε2b2 n x 6 +1. =k i for i<m, and k m =n. Because j= 1+1 EX2 j >ε2 B 2 n /x6 for i<m, we have m 1 Bn 2 j= 1+1 EX 2 j >m 1ε2 B 2 n /x6, whichimpliesthat m ε 2 x 6 +1.Furthermore,supposethat i 0 satisfies0 1 <n 0 0, where n 0 is defined as in 2.2. Set X k =X k I{ X k 16 1 εb n /x 3 }, ˆX k =X k I{16 1 εb n /x 3 < X k εb n /x}, Ẑ ki = Note that 2b X k 2ε. Simple calculations show that m ki I 6 {2b S k b 2 V 2 k +x 2 +ε, k= 1+1 V 2 n V 2 k E V 2 n V 2 k εb2 n /x2 } k i 1 k= 1+1 ˆX k. m ki 1 k= 1+1 {2b S k b 2 V 2 k +x 2 ε, V 2 n V 2 1+1 E V 2 n V 2 1+1 2 1 εb 2 n/x 2 } 4.8

Self-normalized moderate deviation for maximum 19 Noting that m 2b S ki 1 +Ẑ +EẐ b 2 V 2 ki +x 2 2ε, V n 2 V k 2 E V 2 i 1+1 n V k 2 i 1+1 2 1 εbn 2 /x2 m j + max 2b X k E X k ε 1+1 j 1 =:I 6,1 +I 6,2. σ 2 ni := k i 1 k= 1+1 k= 1+1 E X 2 k ε2 B 2 n x 6 and X k 16 1 εb n /x 3, it follows from m ε 2 x 6 +1 and Lévy s inequality that with t=2bx 2 /ε m ki 1 I 6,2 X k E X k ε/2b 2σ ni m k= 1+1 e tε/2b 2σni Ae x2 m exp{at 2 σ 2 ni } k i 1 k= 1+1 Ee t X k E X k 2A 1 ε 2 x 6 +1e x2 C τ,δ x 2 e x2 /2, 4.9 where we used the fact that ε γx 1/2. To estimate I 6,1, let t=24ε 1 x 2 B 2 n logx. Note that 2bEẐ 32x4 εb 2 n k i 1 k= 1+1 Similar to the estimate for I 3 in 3.4, we obtain I 6,1 m EX 2 k 32ε x 2 8ε. 2b S ki 1 +Ẑ EẐ b 2 V 2 ki 1 +x 2 18ε, m V 2 n V 2 1+1 E V 2 n V 2 1+1 2 1 εb 2 n /x2 b S ki 1 E S ki 1 +Ẑ EẐ b 2 V 2 ki 1 /2 +t V 2 n V 2 1+1 te V 2 n V 2 1+1 x2 /2+12logx 9ε

20 W. Liu, Q.-M. Shao and Q. Wang Ax 12 e x2 /2 Ax 12 e x2 /2 m m { ki 1 j=1 exp Recall, by the definition of, x 2 k i 1 Bn 2 j= 1+1 x 4 k i 1 Bn 4 j= 1+1 Ee b X j E X j 2 1 b 2 X2 j k i 1 j= 1+1 Ee b ˆX j E ˆX j +t X 2 j E X 2 j A n,x +A x2 1 j= 1+1 EX2 j B 2 n e 24εlogx +A x3 1 j=k E X i 1+1 j 3 e 24εlogx ε 1 logx B 3 n +A x4 1 j=k E X 4 i 1+1 j Bn 4 e 24εlogx ε 1 logx 2 +A x4 n j= E X j 4 Bn 4 e 24εlogx ε 1 logx. 2 EX 2 j ε2 x 4, E X 4 j εx3 B 3 n On the other hand, we have k i 1 j= 1+1 E X j 3 ε2 x 2 B 2 n k i 1 j= 1+1 ε 1 x 24ε γ 1 min{x δ/10+δ/3,x 3/2 } γ 1 x min{δ/2,3/2}, } n Ee t 2 X j E X j 2 j= EX 2 j ε 4 x 4. and by 2.2 2.4 and the inequality n j=n 0+1 EX2 j 192x 2 B 2 nlogx, for all i i 0, x 4 Bn 4 j= E X 4 j εb n /x j=n 0+1 ετb 2 n/x 2+δ Substituting these estimates into 4.10 gives E X j 3 j=n 0+1 EX 2 j C τ,δ εx δ logx. 4.10 I 6,1 C δ,τ ε 2 x 6 +1x 12 e x2 /2 expc τ,δ x 1/2 log 2 x+c τ,δ x δ/2 log 3 x C δ,τ x 2 e x2 /2. This proves 4.4, which also completes the proof of Lemma 3. 4.11

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