o f P r o b a b i l i t y Vol. 10 (2005), Paper no. 38, pages Journal URL ejpecp/

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1 E l e c t r o i c J o u r a l o f P r o b a b i l i t y Vol , Paper o. 38, pages Joural URL ejpecp/ Limit Theorems for Self-ormalized Large Deviatio 1 Qiyig Wag School of Mathematics ad Statistics The Uiversity of Sydey, Sydey NSW 006, Australia qiyig@maths.usyd.edu.au URL: Abstract. Let X, X 1, X, be i.i.d. radom variables with zero mea ad ite variace σ. It is well kow that a ite expoetial momet assumptio is ecessary to study limit theorems for large deviatio for the stadardized partial sums. I this paper, limit theorems for large deviatio for selformalized sums are derived oly uder ite momet coditios. I particular, we show that, if EX 4 <, the P S /V x 1 Φx } [ = exp x3 EX x ] O, σ 3 for x 0 ad x = O 1/6, where S = X i ad V = X i 1/. Key Words ad Phrases: Cramér large deviatio, limit theorem, self-ormalized sum. AMS Subject Classi catio 000: Primary 60F05, Secodary 6E0. Submitted to EJP o September 4, 004. Fial versio accepted o September 16, This research is supported i part by ARC DP04517

2 1 Itroductio ad mai results Let X, X 1, X,, be a sequece of o-degeerate idepedet ad idetically distributed i.i.d. radom variables with zero mea. Set S = X i, V = Xi, 1. The self-ormalized versio of the classical cetral limit theorem states that, as, sup P S xv 1 Φx } 0, x if ad oly if the distributio of X is i the domai of attractio of the ormal law, where Φx deotes the stadard ormal distributio fuctio. This beautiful self-ormalized cetral limit theorem was cojectured by Loga, Mallows, Rice ad Shepp 1973, ad latterly proved by Gie, Götze ad Maso For a short summary of developmets that have evetually led to Gie, Götze ad Maso 1997, we refer to the Itroductio of the latter paper. The self-ormalized cetral limit theorem is useful whe x is ot too large or whe the error is well estimated. There are two approaches for estimatig the error of the ormal approximatio. Oe approach is to ivestigate the absolute error i the self-ormalized cetral limit theorem via Berry-Essee bouds or Edgeworth expasios. This has bee doe by may researchers. For details, we refer to Slavova 1985, Hall 1988 ad Betkus ad Götze 1996 for the Berry-Essee bouds, Wag ad Jig 1999 for a expoetial ouiform Berry-Essee boud, Hall 1987 as well as Hall ad Jig 1995 for Edgeworth expasios. See also va Zwet 1984, Friedrich 1989, Betkus, Blozelis ad Götze 1996, Betkus, Götze ad va Zwet 1997, Putter ad va Zwet 1998 ad Wag, Jig ad Zhao 000. Aother approach is to estimate the relative error P S xv /1 Φx. I this directio, Jig, Shao ad Wag 003 re ed Shao 1999, Wag ad Jig 1999 as well as Chistyakov ad Götze 003, ad obtaied the followig result: if 0 < σ = EX <, the there exists a absolute costat A > 0 such that } exp A 1 + x,x for all x 0 satisfyig,x 1/A, where P S xv 1 Φx } exp A 1 + x,x, 1.1,x = σ EX I X σ/1+x xσ 3 1/ E X 3 I X σ/1+x. 161

3 Jig, Shao ad Wag 003 actually established 1.1 for idepedet radom variables that are ot ecessarily idetically distributed. It follows from 1.1 that if E X 3 <, the P S xv 1 Φx for 0 x A 1/6 σ/e X 3 1/3. = 1 + O1 1 + x 3 1/ σ 3 E X 3, 1. Result 1. is useful i statistics because it provides ot oly the relative error but also a Berry-Essee type rate of covergece. Ideed, as a direct cosequece of this result, it has bee show i Jig, Shao ad Wag 003 that bootstrapped studetized t-statistics possess large deviatio properties i the regio 0 x o 1/6 uder oly a ite third momet coditio. However, 1.1 as well as 1. does ot capture the term with 1/ explicitly. This short has limited further applicatios of the self-ormaized large deviatio. I this paper we ivestigate the limit theorems for self-ormalized large deviatio. Uder ite momet coditios, a leadig term with 1/ i 1.1 ad 1. is obtaied explicitly. THEOREM 1.1. Assume that EX 4 <. The, for x 0 ad x = O 1/6, P } [ S xv = exp x3 EX x ] 1 Φx O, 1.3 σ 3 P } [ S xv x 3 EX x ] = exp Φ x O. 1.4 σ 3 If i additio EX 3 = 0, the, for x = O 1/6, P S xv Φx = O 1/ e x /. 1.5 Write L,x = 1 + xρ +,x, where ρ = E X 3 / σ 3 ad,x = 1 + x 3 σ 3 1/ E X 3 I X > σ/1+x} x 4 σ 4 1 EX 4 I X σ/1+x}. Furthermore, we obtai the followig bouds which re e 1.1 uder ite third momet coditio i the regio 0 x cρ 1/, where c is a absolute costat. THEOREM 1.. Assume that E X 3 <. The, there exists a positive absolute costat c such that, for 0 x c ρ 1/, exp x3 EX 3 3 σ AL 3,x where A is a absolute positive costat. P S xv 1 Φx 16 exp x3 EX 3 3 σ + AL 3,x. 1.6

4 REMARK 1.1. Similar results to those i Theorem 1.1 hold for the stadardized mea uder much stroger coditios. For istace, it follows from Sectio 5.8 of Petrov 1995 that, for x 0 ad x = O 1/6, P S x σ 1 Φx P S x σ Φ x } [ x 3 EX x ] = exp O, σ 3 } [ = exp x3 EX x ] O. σ 3 hold oly whe Cramér s coditio is satis ed, i.e., Ee tx < for t beig i a eighborhood of zero. We also otice that there are differet formulae for the self-ormalized ad stadardized cases. REMARK 1.. It is readily see that L,x = ox 3 / for x ad x = Oρ 1/. Hece x3 EX 3 3 σ 3 i 1.6 provides a leadig term i this case. However it remais a ope problem for more re ed results. This paper is orgaized as follows. The proofs of mai results will be give i Sectio 3. Next sectio we preset two auxiliary theorems that will be used i the proofs of mai results. The proofs of these auxiliary theorems will be postpoed to Sectios 4 ad 5 respectively. Without loss of geerality, throughout the paper, we assume σ = EX = 1 ad deote by A, A 1, A, ad c, C, C 1, C, absolute positive costats, which may be differet at each occurrece. If a costat depeds o a parameter, say u, the we write Au. I additio to the otatio for L,x ad,x de ed i Theorem 1., we always let ρ = 1/ E X 3 ad Ψ x = Two auxiliary theorems [ ] 1 Φx exp x3 EX 3 3 Throughout the sectio we assume that X, X 1, X,, are i.i.d. radom variables satisfyig EX = 0, EX = 1 ad E X 3 <. Two theorems i this sectio are established uder quite geeral settig, which will be iterestig i themselves. The proofs of these two theorems will be give i Sectios 4 ad 5 respectively.. 163

5 THEOREM.1. Let h = x/b, where B is a sequece of positive costats with B C 1 EX I X /1+x}..1 Suppose that η j := η x, X j, 1 j, satisfy the coditios: Ee hη j 1 x + x3 EX 3 C 1 B 3B 3,x,. Eη j e hη j x C3 x 1/ ρ,.3 B Eηj e hη j 1 C 4 x ρ,.4 E η j 3 e hη j C 5 E X 3..5 The, for x cρ 1 with c suf cietly small ad for ay b x /4, P η j x + bx 1 B 1 + b x e b Ψ x expa L,x ;.6 for x c 1 ρ 1/ with c 1 suf cietly small, P η j x B Ψ x exp A L,x..7 THEOREM.. Write, for 1 m, T m = 1 m ζ j ad Λ,m = 1 m 1 k=1 j=k+1 where ζ j := ζ x, X j ad ψ k,j := ψ x, X k, X j satisfy the coditios: Eζj 1 + ψ k,j, x EX 3 C6,x /1 + x,.8 Eζ 3 j EX 3 C7,x /1 + x 3,.9 Eζ j = 0, ζ j C 8 /1 + x, Eζ 4 j C 9 EX 4 I X /1+x},.10 The, for x c ρ 1 Eψ k,j X k = Eψ k,j X j = 0, for k j,.11 E ψ k,j C 10 x, E ψ k,j 3/ C 11 x 3/ E X 3..1 with c suf cietly small, ad for ay costats sequece λ x satisfyig λ x C,x /1 + x, P T + Λ, x + λ x Ψ x 3/. 1 + AL,x + A1 xρ

6 3 Proofs of mai results Proof of Theorem 1.1. Note that L,x 1 + x/ for x 0 ad x = O 1/6. Theorem 1.1 follows immediately from Theorem 1.. We omit the details. Proof of Theorem 1.. Without loss of geerality, assume x. If 0 x <, the results are direct cosequeces of the Berry-Essee boud cf. Betkus ad Götze 1996 P S xv 1 Φx } A ρ. We rst provide four lemmas. For simplicity of presetatio, de e τ = /1 + x ad assume x c ρ 1/ out. with c suf cietly small throughout the sectio except where we poit LEMMA 3.1. We have, P S x V x Ψ x exp A L,x ; 3.1 ad for x c ρ 1 with c suf cietly small ad for arbitrary δ x /4, P S x V x + δx δ x e δ Ψ x expa L,x. 3. Proof. We rst prove 3.. Let h = x/ ad η j = X j x X j 1. It follows from i Wag ad Jig 1999 that Eη 1 e hη 1 x 16 x 1/ ρ, Eη1e hη x ρ, E η 1 3 e hη 1 30 E X 3. Thus 3. follows immediately from.6 i Theorem.1 with B = if we prove Ee hη 1 1 x + x3 3 3/ EX3 C 1,x. 3.3 Without loss of geerality, assume x ρ 1 /16. By otig E X 3 EX 3/ = 1, we get h 1/4. This implies that hη 1 = h / + hx 1 hx 1 / 1 ad hη 1 k e hη 1 sup s1 s k e s e for k = 0, 1,..., 4. Therefore, usig Taylor s expasio e x k x j j! j=0 x k+1 k + 1! ex 0, for k 1, 165

7 we obtai that [ Ee hη 1 = E 1 + hη hη hη θ hη 4 1 ]I 4 X1 τ + E 1 + hη 1 + θ I 1 X1 >τ = Ehη 1 I X1 τ Ehη 1 3 I X1 τ + θ 1 P X 1 > τ + θ 4 Ehη 1 4 I X1 τ, 3.4 where θ e ad θ 1 e. Sice EX = 1, it is readily see that Ehη 1 I X1 τ h + EhX 3 8 1,x, Ehη 1 3 I X1 τ EhX 3 3 1,x, Ehη 1 4 I X1 τ 16 1,x. Takig these estimates back ito 3.4, we obtai 3.3, ad hece 3.. The proof of 3.1 is similar by usig.7. We omit the details. The proof of Lemma 3.1 is ow complete. The ext lemma is from Lemma 6.4 i Jig, Shao ad Wag 003. LEMMA 3.. Let ξ i, 1 i } be a sequece of idepedet radom variables with Eξ i = 0 ad Eξ i <. The [ P ξ i a 4D + where D = Eξ i 1/. ξ i 1/ }] 8e a / 3.5 I Lemmas , we use the otatio: X i = X i I Xi τ}, S = X i, V = S i = S X i, V i = V X i 1/. LEMMA 3.3. We have X i, B = E X i P S xv A Ψ x expa L,x

8 Proof. Let Ω = 1 x 1 /, 1 + x 1 /. Recallig x cρ 1/ with c suf cietly small, it follows from.7-.9 i Shao 1999 that P S xv, V / / Ω P S xv, V 9 + P S xv, 1 + x 1 / V 9 } + P S xv, V 1 x 1 / } 4 exp x / x/8 + Ax 3 ρ A Ψ x. This, together with 3., implies that P S xv = P S xv, V / Ω + P S xv, V / / Ω P S x V x x 1 /4 } + P S xv, V / / Ω AΨ x expa L,x, as required. The proof of Lemma 3.3 is complete. LEMMA 3.4. We have P S xv Ψ x expa L,x + A 1 e 3x ; 3.7 P S xv Ψ x expa L,x + A 1 xρ 3/. 3.8 Proof. As i Wag ad Jig 1999, for x, P S xv P S x V + It follows easily from Lemma 3.3 that for all i P [ S i ] x 1 1/ V i P Xi > τ. = K + I, say. 3.9 P [ S i ] x 1 1/ V i AΨ x expa L,x. This, together with P X i > τ 1,x, implies that I A,x Ψ x expa L,x

9 I view of 3.9 ad 3.10, the iequalities 3.7 ad 3.8 will follow if we prove for x cρ 1 K Ψ x expa L,x + A 1 e 3x, 3.11 K Ψ x expa L,x + A 1 xρ 3/, 3.1 with c suf cietly small. We rst prove Let D = E X i 4 ad ξ i = X i E X i. By the iequality 1 + y 1/ 1 + y/ y for ay y 1 ad Lemma 3., [ K = P S x B + X i E X } 1/ ] i [ P S xb ξ B i 1 }] ξ B 4 i [ 1/ }] P ξ i 6x 4D + + P [ S xb ξ i ξ B i 1 ξ B 4 i }, 1/ }] ξ i 6x 4D + 8e 3x + K,1, 3.13 where, after some algebra see, Jig, Shao ad Wag 003, for example, with η i = X i 1 xb 1 K,1 P η j xb, X i E X i + 4x 3 B 3 X i E X i 4. As i the proof of Lemma 3.1, tedious but elemetary calculatios show that the iequalities.1-.5 hold true for B = E X i ad the η i de ed above. Therefore it follows from 3. i Theorem.1 that ξ i K,1 Ψ x expa L,x, 3.14 for x cρ 1 desired with c suf cietly small. Take this estimate back ito 3.13, we get the 168

10 We ext prove 3.1. Note that V = = X i 1 + X k 1 X j 1 k j X i 1 + X k E X k X j E X j k j + 1E X j 1 X j 1 E X j. By the iequality 1 + y 1/ 1 + y/ y for ay y 1 agai, we have where [ K = P S x [ P S x = P X } 1/ ] i 1 V 1 V }] ζ j k=1 j=k+1 ζ j = X j x X j 1 + x 3/ X j 1 + ζ j = ζ j Eζ j 1x 3/ E X 1 1 = X j E X j x 1 + X j 1 E X j, + x X 4 3/ j E X j 4, ψ k,j = x X k E X k X j E X j, λ x = Eζ 1 ψ k,j x λ x, EX I X τ X j E X j = EXI X τ x EX I X τ + x E X 1 1 1x. EX I X τ It is easy to see that ζ j satisfy the coditios.8-.10, ψ k,j satisfy the coditios i Theorem. ad λ x 3x x 1,x 61 + x 1,x. Hece, i view of 3.15, 3.1 follows immediately Theorem.. This also completes the proof of Lemma

11 After these prelimiaries, we are ow ready to prove Theorem 1.. As is well-kow see Wag ad Jig 1999 for example, P S xv P S x V x. The left had iequality of 1.6 follows from Lemma 3.1 immediately. To prove the right had iequality of 1.6, we use Lemmas 3.4. If Ψ x xρ 1/, the by 3.1, P S xv Ψ x expa L,x 1 + A 1 xρ Recall we may assume that x ρ 1 /16. It is readily see that 1 Φx } 3 exp x 3 EX 3 } π 3/ x 3 e x. This implies that, whe Ψ x xρ 1/, e 3x A 1 Φx } 3 exp x3 EX 3 } A xρ Ψ x. Ψ x exp A L,x Therefore, by 3.7, P S xv Ψ x expa L,x 1 + A 1 xρ Ψ x expa L,x Collectig the estimates 3.16 ad 3.17, we get the right had iequality of 1.6. The proof of Theorem 1. is ow complete. 4 Proof of Theorem.1 The proof of Theorem.1 is based o the cojugate method. To employ the method, let ξ 1,, ξ be idepedet radom variables with ξ j havig distributio fuctio V j u de ed by }/ V j u = E e hη j Iη j u Ee hη j, for j = 1,,, Also de e M h = V arξ j, G t = P ξ j Eξ j M h } t ad 170 R h = x + bx 1 B M h Eξ j.

12 Sice EX = 1, we have E X 3 1 ad there exists a positive costat c 0 such that EX I X τ 1/4C 1 for x c 0 ρ 1. Let c 1 = 8 max1, C 1,..., C 5 } ad c = mic 0, c 1 1 }. Takig accout of the coditios.1-.5, it ca be show by a elemetary method that, for x c ρ 1, Ee hη j = 1 + x EX3 x 3 + O1 1,x 3 x = exp EX3 x } 3 + O 1,x, / Eξ j = Eη j e hη j Ee hη j = x + O 3 x 1/ ρ, 4. / V arξ j = Eηj e hη j Ee hη j Eξ j = 1 + O 4 x ρ, 4.3 / E ξ j 3 E η j 3 e hη j Ee hη j C 5 1/ ρ, 4.4 where O j 1/c, for j = 1,..., 4. Usig ad.1, we get, for x c ρ 1, / Mh = + O 4 x ρ 3/, 4.5 R h = x + bx 1 B Eξ j M h 3/ b x 1 + C 1 + O 3 x ρ, 4.6 hm h x 1 h M x h x M x h } M B h 4C 1 + O 4 x ρ. 4.7 Let A h = R h + hm h, c 3 = 1 4 6C 1 + Q 3 + O 3 1 ad c 4 = mic, c 3 }. It follows from that, for x c 4 ρ 1, ad recall b x /4, A h x R h + hm h x 3/ b x 1 + 6C 1 + O 3 + O 4 x ρ 4.8 x/ A h 3x/

13 After these prelimiaries, we ext give the proof of Theorem.1. Write δ x = x + bx 1 B. By the cojugate method, P η j > δ x = = Ee hη j = Ee hη j δ x e hu dp e hδx hmhv dg v + R h }, 0 Ee hη j e x b 0 ξ j u, e hmhv d [G v + R h } Φ v + R h }] + e hmhv dφ v + R h } 0 := I 0 h e x b I 1 h + I h It follows from 4.1 that, for x c 4 ρ 1, x } exp x3 3 EX3 A,x Next we estimate I h. We have I h = I 0 h = Ee hη j x } exp x3 3 EX3 + A,x e hmhv 1 v+rh dv π 0 = e R h/ π 0 e hmh+rhv 1 v dv := e R h/ π I 3 h. 4.1 Write ψx = 1 Φx } /Φ x = e x / / x e y dy. Clearly, ψa h} = I 3 h, ad for x, 1 x ψx 1 x ad ψ x = xψx 1 x. These estimates, together with 4.8 ad 4.9, imply that for x c 4 ρ 1, I 3 h = ψx + ψ θ A h x }, [where θ x/, 3x/] = ψx [ O 5 + O 6 x ρ ], 17

14 where O 5 1 b x ad O 6 A. Therefore, for x c 4 ρ 1, I h = e x / 1 Φx } e R x/ O 5 + O 6 x ρ As for I 1 h, by , itegratio by parts ad Berry-Essee theorem, we get I 1 h sup v This implies that for x, where O 7 A. G v Φv 4M 3 h E ξ j 3 16 C 5 ρ. I 1 h = O 7 x ρ e x / 1 Φx }, 4.14 It follows easily from ad that for x c 4 ρ 1, P η j δ x 1 + b x e b Ψ x exp A,x 1 + A 1 xρ 1 + b x e b Ψ x exp A L,x. This proves.6. Similarly, by lettig b = 0, it follows from ad that P η j x B Ψ x exp A,x Rh/ } [ 1 ] Ψ x exp A,x } [1 A 1 x ρ Ψ x exp A L,x }, } ] O 6 + O 7 e R h/ x ρ for x c 4 ρ 1/, where we have used the fact that R h 1 by 4.6, ad also R h 4C 1 + O 3 x 4 1 E X 3 8C 1 + O 3,x sice E X 3 E X 3 I x >τ } + EX 4 I x τ. This proves.7 ad hece complete the proof of Theorem

15 5 Proof of Theorem. The idea for the proof of Theorem. is similar to Propositio 5.4 i Jig, Shao ad Wag 00, but we eed some differet details. Throughout this sectio, we use the followig otatios: gt, x = Ee itζ 1/ ad Ω x, t = x, t : 1 + x C 6 1 /ρ, t 16 } 1 + C 7 1 /ρ. The proof of Theorem. is based o the followig lemmas. LEMMA 5.1. If x, t Ω x, t, the gt, x e t /6, 5.1 g t, x e t A 1 + x ρ t + t 3 e t /6, 5. [ g t, x e t 1 + gt, x 1 } ] + t A 1 + x ρ t 3 + t 8 e t / Proof. It follows from Taylor s expasio of e ix that I view of.8 ad.9, we have that gt, x 1 + t Eζ1 1 6 t 3 3/ E ζ Eζ C xρ ad E ζ C 7 E X Takig these estimates back ito 5.4, we obtai for x, t Ω x, t, This proves 5.1. gt, x 1 t + t Eζ t 3 3/ E ζ t + t 6 + t 6 e t /6. Usig 5.1, we have that for x, t Ω x, t ad t C 7 1 /ρ, g t, x e t e t /6 1 + C 7 ρ t 3 e t /

16 O the other had, by usig 5.5, we get that for x, t Ω x, t ad t C 7 1 /ρ, where gt, x = 1 r 1 t, x, 5.7 r 1 t, x = t Eζ 1 + η t 3 6 3/ E ζ 1 3, η 1, havig r 1 t, x 1/4 ad r 1 t, x t4 E ζ1 3 4/3 + t E ζ C 7 t 3 1 ρ. 5.8 Therefore, it follows from l1 + z = z + η 1 z, wheever z < 1/, where η 1 1, that for x, t Ω x, t ad t C 7 1 /ρ, where, by usig 5.5 agai, Also, we have l gt, x = t Eζ 1 + θ t 3 1 ρ, θ 1 + C 7, 5.9 l g t, x = t Eζ 1 + θ t 3 ρ = t rt, x t Eζ θ t 3 ρ t /3. + rt, x, 5.10 rt, x t Eζ θ t 3 ρ At + t xρ. These estimates, together with e z 1 z e z, imply that for x, t Ω x, t ad t C 7 1 /ρ, g t, x e t e t Now, 5. follows from 5.6 ad e rt,x 1 If we istead the estimate of r 1 t, x i 5.8 by we ca rewrite 5.9 ad 5.10 as A1 + xρ t + t 3 e t / r 1 t, x t4 Eζ 1 + t E ζ1 3 t 4, l gt, x = gt, x 1 + θ 1 t 4, θ 1, l g t, x = gt, x 1 } + θ 1 t

17 Therefore, by usig 5.5 ad 5.7, we obtai for x, t Ω x, t, l g t, x + t gt, x 1 } + t + 1 t 4 recallig t 1 6 ρ 1 /6. Also, we have l g t, x + t t Eζ t 3 6 E ζ t 4 t /3 t Eζ t 3 6 E ζ t 4 A1 + xρ t + t 4. Now, by usig e z 1 z z e z, we obtai that for x, t Ω x, t, } g t, x e t = e t e l g t,x+ t 1 [ = e t gt, x 1 } ] + t + r t, x, where r t, x 1 t l g t, x + t exp l g t, x + t } A 1 + x ρ t 4 + t 8 e t /3. This implies 5.3. The proof of Lemma 5.1 is ow complete. LEMMA 5.. If x, t Ω x, t, the EΛ, e itt Axρ t e t /6, 5.1 Ee itt+λ, g t, x Axρ t e t /6 + A 1 x 3/ ρ t 3/ 5.13 ad for ay 1 m, Ee itt+λ, 1 + A x t e m t /1 + A 1 m 1 x 3/ ρ t 3/ Proof. It follows from.11,.1, 5.5 ad Holder s iequality that Eψ 1, e itζ 1+ζ / = Eψ 1, e itζ 1/ 1 e itζ / 1 t E ψ1, ζ 1 ζ } t E ψ1, 3/} /3 E ζ1 3} /3 Axρ t. 176

18 Therefore, it follows from idepedece of ζ j ad Lemma 5.1 that EΛ, e itt Eψ1, e itζ 1+ζ / gt, x Axρ t e t /6. This proves 5.1. To prove 5.13 ad 5.14, put Λ,m = Λ, Λ,m = 1 1 k=m+1 j=k+1 By.1 ad e iz 1 iz z 3/, Ee itt+λ, Ee itt+λ,m iteλ,m e itt+λ,m ψ k,j ad Λ,m = 0, m. t 3/ E Λ,m 3/ A t 3/ m E ψ 1, 3/ Am 1 x 3/ ρ t 3/ Therefore, 5.13 follows easily from 5.1 ad 5.15 with m =. I view of idepedece of ζ j, o the other had, 5.15 implies that for ay 1 m, Ee itt+λ, g m t, x + Ax t g m t, x + Am 1 x 3/ ρ t 3/ where we have used the estimate recallig.1: 1 + Ax t e m t /6 + Am 1 x 3/ ρ t 3/ E Λ,m e itt+λ,m = E 1 m k=1 j=k+1 ψ k,j e itt+λ,m gt, x m E ψ 1, A x gt, x m. This gives The proof of Lemma 5. is ow complete. LEMMA 5.3. Let F be a distributio fuctio with the characteristic fuctio f. The for all y R ad T > 0 it holds that where lim F z 1 T z y + V.P. exp iyt 1 T T K t ftdt T lim F z 1 T z y V.P. exp iyt 1 T T K t ftdt, 5.17 T T V.P. T h = lim h 0 T T +, h 177

19 ad Ks = K 1 s + ik s/πs, K 1 s = 1 s, K s = πs1 s cot πs + s, for s < 1, ad Ks 0 for s 1. Proof of Lemma 5.3 ca be foud i Prawitz 197. LEMMA 5.4. It holds that if 0 x cρ 1 with c C 6 1, the for ay y R, where K 1 s is de ed as i Lemma 5.3, I + y, I y Aρ e y / + A x 3/ ρ, 5.18 I + y = 1 T I y = 1 T T T T T e iyt t K 1 Ee itt +Λ, dt, T e iyt K 1 t T Ee itt +Λ, dt, T = C 7 1 /ρ. Proof. We oly prove 5.18 for I + y. Without loss of geerality, we assume ρ 1 3. This assumptio implies that 1 + x 11 + C /ρ for 0 x cρ 1 I + = I + y ad T 1 = ρ 1/3. We have that I + = T1 + T 1 T 1 t T := I 1 + I. with c C 6 1. Let It is easy to see that [1t log t ] if t 1 ad 6. Hece, recallig T 1 1 ad 1/E X 3 1, by 5.14 with m = [1t log t ] +, I 1 Ee its+λ, dt A1 + x 3/ ρ T T 1 t T Notig K 1 s = 1 s, for s < 1, we obtai I 1 I 11 + I 1, where I 11 = 1 T T1 T 1 e iyt Ee its+λ, dt, I 1 = T T1 0 t Ee its+λ, dt. It follows from 5.14 with m = [1t log t ] + agai that I 1 T 1 0 tdt + T1 1 t Ee its+λ, } dt A1 + x 3/ ρ

20 O the other had, otig that 1 π e iyt t / dt = e y /, it follows from Lemmas i.e, 5. ad 5.13 that I 11 1 e iyt t / dt + 1 e t / dt + 1 T T t T 1 T Aρ e y / + A x 3/ ρ. T1 Collectig all these estimates, we coclude the proof of Lemma 5.4. LEMMA 5.5. The itegral J + y = i T t π V.P. e iyt K T T J y = i T π V.P. e iyt K t T T its+λ, dt Ee t, T 1 Ee its+λ, e t / dt its+λ, dt Ee t, T = C 7 1 /ρ, satisfy that, for ay y R ad 0 x cρ 1 with c 11 + C 6 6 1, J + y + 1 Φy L y A1 + x 3/ ρ 3/, 5.1 J y + 1 Φy L y A1 + x 3/ ρ 3/, 5. where K s is de ed as i Lemma 5.3 ad L y = EΦ y ζ } 1 Φy 1 Φ y. Proof. We oly prove 5.1. Similar to the proof of Lemma 5.4, we assume ρ 1 3, which implies 1 + x 11 + C /ρ for 0 x cρ 1 with c 11 + C Write J + = J + y [ ad f t, x = 1 + gt, x 1 } ] + t e t /. The J + ca be rewrite as J + = J 11 + J 1 + J 13 + J, where J 11 = i T1 π V.P. e iyt f t, x dt T 1 t, J 1 = i T1 } dt π V.P. Ee its+λ, f t, x t, J 13 = i π V.P. T1 J = i π V.P. T 1 e iyt T 1 e iyt T 1 t T t K 1 its+λ, dt }Ee T t, e iyt K t T 179 Ee its +Λ, dt t

21 ad T 1 = ρ 1/3. Similar to 5.19, it follows that J A1 + x 3/ ρ. By usig 5.3 ad 5.13, we have J 1 T1 Ee its+λ, g t, x dt T 1 t + } A 1 + x ρ + x 3/ ρ 3/. T1 T 1 g t, x f t, x dt t Notig that K s 1 As, for s 1/ cf., e.g., Lemma.1 i Betkus 1994, similar to 5.0, it ca be easily show that T1 J 13 AT t Ee its+λ, dt A1 + x 3/ ρ. T 1 O the other had, simple calculatio shows that i π V.P. e iyt f t, x dt t = 1 + Φy + L y. Therefore, it follows from all these estimates ad x cρ 1 that J Φy L y J Φy L y + J 1 + J 13 + J } f t, x dt + A 1 + x ρ + x 3/ ρ 3/ t A t T x ρ + x 3/ ρ 3/ A 1 + x 3/ ρ 3/. This also completes the proof of Lemma 5.5. LEMMA 5.6. For ay y A1 + x, L y EX3 y π 6 yx e y / where L y is de ed as i Lemma 5.5. For 0 x cρ 1 } A ρ +,x /1 + x } e y /, 5.3 with c suf cietly small, ad y 0 = x + λ x, where λ x C,x /1 + x, e y 0 / e x / A ρ, 5.4 Φy 0 Φx A ρ +,x /1 + x } e x / + A 1 ρ, 5.5 L y 0 + EX3 3 x e x / A ρ +,x /1 + x } e x / + A x ρ. 5.6 π 180

22 Proof. We rst ote that Φ y = y e y /, Φ 3 y = y 1 e y / π π 5.7 ad Φ 4 y A1 + y 3 e y /. 5.8 Usig 5.8, y A1 + x ad ζ 1 A /1 + x, it ca be easily see that for θ 1, Φ 4 y + θ ζ 1 A1 + y 3 exp y + y ζ } 1 A1 + x 3 e y /. This, together with Taylor s expasio ad Eζ 4 1 C 9 EX 4 I X τ, implies that Q := EΦ y ζ 1 Φy Eζ 1 Φ y + Eζ / Φ3 y 1 4 E ζ4 1 Φ 4 y + θ ζ 1 θ 1 A EX 4 I X τ 1 + x 3 e y /. Therefore, takig accout of.8-.9 ad 5.7, we have This cocludes 5.3. L y EX3 y π 6 yx L y xex3 e y / Φ y EX3 6 Φ3 y + Q + 1 Φ y Eζ1 1 αx EX / Φ3 y Eζ1 3 βex 3 + A ρ +,x /1 + x } e y /. ρ 6 π e y / ρ 6 π e y / We ext prove It ca be easily see that,x 1+x 3 ρ 1, ad for x cρ 1 with c small eough, y 0 x x λ x + λ x C + C,x x /3, x y 0 x x + x λ x x + C,x 3x /4. 181

23 These estimates imply that e y 0 / e x / 1 y 0 x e x /+ y0 x / A ρ 1 + x 3 e x /3 Aρ, 5.9 Φy 0 Φx y 0 x Φ x + y 0 x Φ x + θ y 0 x θ 1 A,x 1 + x e x / + ρ 1 + x7 e x y 0 x / A,x 1 + x e x / + A 1 ρ This proves 5.4 ad 5.5. O the other had, we have y y 0x x y0 x + x 6 y 0 x A,x. This, together with , implies that L y 0 + EX3 3 x e x / Aρ,x e y 0 / + A 1 ρ +,x /1 + x } e y 0 / π This gives 5.6. The proof of Lemma 5.6 is ow complete. A ρ +,x /1 + x } e x / + A x ρ. After these lemmas, we are ow ready to prove Theorem.. By usig 5.17 ad Lemmas , we obtai for x cρ 1 P T +, x + λ x [ ] I y J y 0 with c suf cietly small, 1 1 Φy 0 + L y 0 + Aρ e y 0 / + A 1 x 3/ ρ 3/, where y 0 = x + λ x. Furthermore, it follows from Lemma 5.6 that P T +, x + λ x 1 Φx EX3 3 x e x / + A ρ +,x /x e x / + A 1 x 3/ ρ 3/ π Usig the well-kow iequality 1 1 π x 1 e x / 1 Φx 1 e x /, x > 0, x 3 πx 18

24 we have for x, x 3 1 Φx } x e x / 1 e x / π π ad e x / Ax 1 Φx }. Takig these estimates back ito 5.31, we get for x cρ 1 with c suf cietly small, P T +, x + λ x Now we coclude the result if we prove 1 x3 EX AL,x exp 1 Φx }[ 1 x3 EX 3 x3 EX AL,x ] + A 1 x 3/ ρ 3/. } [ 1 + A 1 L,x]. 5.3 I fact, 5.3 is obvious for EX 3 < 0. I the case that EX 3 > 0, if x3 EX 3 3, the } 1 x3 EX A L,x exp x3 EX A L,x } exp x3 EX 3 [ Ae L,x]. O the other had, it follows easily from the de itio of,x that for x cρ 1, Axρ +,x 1 + x 3 EX 3 / /9, by choosig c suf cietly small. Therefore, if x3 EX 3 3, the 1 x3 EX A L,x 0 exp x3 EX 3 3 } [ 1 + A L,x]. Collectig all these estimates, we get the desired 5.3. This also completes the proof of Theorem.. REFERENCES Betkus, V., Blozelis, M. ad Götze, F A Berry-Essee boud for Studet s statistic i the o-i.i.d. case. J. Theoret. Probab. 9, Betkus, V. ad Götze, F The Berry-Essee boud for Studet s statistic. A. Probab. 4,

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