Cramér-type large deviations for samples from a nite population

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1 Cramér-type large deviations for samples from a nite population by Zhishui Hu, John Robinson and Qiying Wang USTC, University of Sydney and University of Sydney January 0, 006 Abstract. Cramér-type large deviations for means of samples from a nite population are established under wea conditions. The results are comparable to results for the so-called self-nomalized large deviation for independent random variables. Cramér-type large deviations for nite population Student t -statistic are also investigated. Key Words and Phrases: Cramér large deviation, moderate deviation, nite population. Short title: Cramér-type large deviation for nite populations. AMS 000 Subject Classi cation: Primary 60F05, Secondary 6E0. 1 Introduction and results Let X 1, X,..., X n be a simple random sample drawn without replacement from a nite population {a} = {a 1,, a }, where n <. Denote µ = EX 1, σ = varx 1, S n = n X, p = n/, q = 1 p, ω = pq. Under appropriate conditions, the nite central limit theorem [see Erdös and Rényi 1959] states that P S n nµ xσω may be approximated by 1 Φx, where Φx is the distribution function of a standard normal variate. The absolute error of this normal approximation, via Berry-Esseen bounds and Edgeworth expansions, has been widely investigated in the literature. We only refer to Bielis 1969 and Höglund 1978 for the rates in the Erdös and Rényi central limit theorem; Robinson 1978, Bicel and van Zwet 1978, Babu and Bai 1996 as well as Bloznelis 000a, b for the Edgeworth expansions. Extensions to U-statistics and, more 1

2 generally, symmetric statistics can be found in andi and Sen 1963, Zhao and Chen 1987, 1990, Koic and Weber 1990 as well as Bloznelis and Götze 000, 001. In this paper we shall be concerned with the relative error of P S n nµ xσω to 1 Φx. In this direction, Robinson 1977 derived a large deviation result that is similar to the type for sums of independent random variables in Petrov 1975, Chapter VIII. However, to mae the main results in Robinson 1977 applicable, it essentially requires the assumption that 0 < p 1 p p < 1. This ind of condition not only taes away a major dif culty in proving large deviation results but also limits its potential applications. The aim of this paper is to establish a Cramér-type large deviation for samples from a nite population under wea conditions. In a reasonably wide range for x, we show that the relative error of P S n nµ xσω to 1 Φx is only related to E X1 µ 3 /σ 3 with an absolute constant. We also obtain a similar result for the so-called nite population Student t-statistic de ned by t n = n X µ/ˆσ q, where X = S n /n and ˆσ = n j=1 X j X /n 1. It is interesting to note that the results for both nite population standardized mean and Student t-statistic are comparable to the so-called self-nomalized large deviation for independent random variables, which has been recently developed by Jing, Shao and Wang 003. Indeed, Theorems 1.1 and 1.3 below can be considered as analogous to Theorem.1 by Jing, Shao and Wang 003 in the independent case. The Berry-Esseen bounds and Edgeworth expansions for the Student t-statistic have been investigated in Babu and Singh 1985, Rao and Zhao 1994 and Bloznelis 1999, 003. We now state our main ndings. Theorem 1.1. There is an absolute constant A > 0 such that exp { A1 + x 3 β 3 /ω } P S n nµ xσω 1 Φx for 0 x 1/Aω σ/ max a µ, where β 3 = σ 3 E X 1 µ 3. exp { A1 + x 3 β 3 /ω }, 1 The following result is a direct consequence of Theorem 1.1, and provides a Cramér-type large deviation result for samples from a nite population. Theorem 1.. There exists an absolute constant A > 0 such that P S n nµ xσω = 1 + O11 + x 3 β 3 /ω, 1 Φx

3 and P S n nµ xσω Φ x = 1 + O11 + x 3 β 3 /ω, 3 for 0 x 1/A min { ω σ/ max a µ, ω /β 3 1/3}, where O1 is bounded by an absolute constant. In particular, if ω /β 3, then, for any 0 < η 0, P S n nµ xσω P S n nµ xσω 1, 1, 4 1 Φx Φ x uniformly in 0 x η min { ω σ/ max a µ, ω /β 3 1/3}. Results and 3 are useful because they provide not only the relative error but also a Berry-Esseen rate of convergence. Indeed, by the fact that 1 Φx e x / /1 + x for x 0, we may obtain P S n nµ xσω Φx A1 + x e x / β 3 /ω, for x 1/A min { ω σ/ max a µ, ω /β 3 1/3}. This provides an exponential nonuniform Berry-Esseen bound for samples from a nite population. Remar 1.1. We do not have any restriction on the {a} in Theorems 1.1 and 1.. Indeed, for any {a}, µ = 1 a, σ = 1 a µ, E X 1 µ 3 = 1 a µ 3. Removing the trivial case that all a are the same, we always have max a µ > 0, σ > 0 and E X 1 µ 3 <. Remar 1.. Háje 1960 proved that if 0 < p 1 p p < 1, then S n nµ/σω 0, 1 if and only if ω σ/ max a µ. Theorems 1.1 and 1. therefore provide reasonably wide ranges for x to mae the results hold true. To be more precise, as an example, consider a = α, where α > 1/3. In this special case, simple calculations show that min { ω σ/ max a µ, ω /β 3 1/3} pq 1/6, which implies that Theorem 1. holds true for x being in the best range 0, o[pq 1/6 ]. The following Theorem 1.3 provides a relative error P t n D x to 1 Φx, which is only related to E X 1 µ 3 /σ 3 with an absolute constant as in Theorem 1.1. Cramér-type large deviation results for the Student t-statistic may be obtained accordingly as in Theorem 1.. We omit the details. 3

4 Theorem 1.3. There is an absolute constant A > 0 such that exp { A1 + x 3 β 3 /ω } P t n x 1 Φx exp { A1 + x 3 β 3 /ω }, 5 for all 0 x 1/Aω σ/ max a µ, where β 3 is de ned as in Theorem 1.1. This paper is organized as follows. Major steps of the proofs of Theorems are given in Section. As a preliminary, in a general setting, Section 3 provides a Berry-Esseen bound for the associated distribution of P S n nµ x related to the conjugate method. Proofs of three propositions used in the main proofs are offered in Sections 4-6. Throughout the paper we shall use A, A 1, A,... to denote absolute constants whose values may differ at each occurrence. We also write b = x/ω, V n = n X, V 1n = V n n and V n = n [ X 1 EX 1 ], and, when no confusion arises, denotes, and denotes. The symbol i will be used exclusively for 1. Proofs of theorems Without loss of generality, we assume µ = 0 and σ = 1. Otherwise, it suf ces to consider that {X 1, X,..., X n } is a simple random sample drawn without replacement from a nite population {a } = {a 1 µ/σ,, a µ/σ}, where n <. Proof of Theorem 1.1. When 0 x, property 1 follows from the Berry-Esseen bound for samples from a nite population see, Höglund 1978, for example: P S n xω 1 Φx A β 3 /ω. When x 1/A ω / max a, property 1 follows from the following Proposition.1 with ξ = 0, ξ 1 = 0 and h = 0. Proposition.1 will be proved in Section 4. Proposition.1. There exists an absolute constant A > 0 such that, for all 0 ξ 1/, ξ 1 36 and x 1/A ω / max a, P bs n ξ b qv 1n + ξ 1 b 4 q V n x exp { } Ax 3 β 3 /ω, 6 1 Φx 4

5 and P bs n ξ b qv 1n + ξ 1 b 4 q V n x + h 1 Φx [ h x ] exp { h + Ax 3 β 3 /ω }, 7 where h is an arbitrary constant which may depend on x with h x /5. Remar.1. The restrictions for ξ and ξ 1 in proposition.1 may be reduced to more general 0 ξ A 0 and ξ 1 A 1, where A 0 and A 1 are two absolute constants. Proof of Theorem 1.. This follows immediately from Theorem 1.1. Proof of Theorem 1.3. When 0 x 4, property 5 follows from the Berry-Esseen bound for nite population Student t-statistic. See, Bloznelis 1999, for example. ext, assume 4 x 1/Aω / max a. Without loss of generality, assume that A 8 and n 4. ote that max a 1 since a =. It is readily seen that x 0 x 1 = [ 1 + x q 1/n ] 1/ 1 x /n, 8 where x 0 = x n 1/ /n + x q 1 1/. It follows from 8 that x/ x 0 3x/ and x 0 x x 3 β 3 /ω. Hence, by noting 1 Φx xφ x/1 + x for x 0 see, for example, Revuz and Yor1999, p30, we have log 1 Φx 0 = 1 Φx which yields that x0 x Φ t 1 Φt dt x0 x 1 + t dt x x x 0 x 3 β 3 /ω, t exp{ x 3 β 3 /ω } 1 Φx 0 1 Φx exp{x3 β 3 /ω }. 9 We are now ready to prove Theorem 1.3. As is well-nown, for x 0, P t n x = P S n /V n x 0 q. ote that b 0 x 0 q Vn x 0 + b 0qV n / x 0 + b 0qV n n/, where b 0 = x 0 /ω. It follows from 6, 8 and 9 that, for 4 x 1/Aω / max a, P S n x 0 qvn P b 0 S n b 0qV n n/ x 0 1 Φx 0 exp{ Ax 3 0β 3 /ω } 1 Φx exp { A 1 x 3 β 3 /ω }, 5

6 which implies the rst inequality of 5. In view of the following Propositions. and.3, the second inequality of 5 may be obtained by a similar argument to that in the proof of 5.13 in Jing, Shao and Wang 003, and the details are omitted. The proofs of Propositions. and.3 will be given in Section 5 and Section 6 respectively. Proposition.. There exists an absolute constant A > 0 such that P S n x qv n 1 Φx exp{a x 3 β 3 /ω } + Ae 4x, for x 1/A ω / max a. Proposition.3. There exists an absolute constant A > 0 such that P S n x qv n 1 Φx exp{ax 3 β 3 /ω } + A xβ 3 /ω 4/3, for x 1/Aω / max a. 3 Preliminaries The main aim of this section is to derive a Berry-Esseen bound for the associated distribution of P S n x related to the conjugate method. The result and several related lemmas are established in a general setting, and will be used in the proofs of the propositions. For z = x + iy, de ne, Kz = log βz with βz = pe qz + qe pz, 10 where p, q > 0 and p + q = 1. Consider a sequence of constants {b} = {b 1,, b } with b = 0, and let K, K x = u b + α u, where α u is the solution of the equation and K be the values of Kx, K x and K x evaluated at K u b + α = Throughout the section we assume that C 0 > 0 is a given constant and u C 0 / max b. ote that, for any real u with u C 0 / max b, K ub + α is negative when α < C 0 and positive when α > C 0, and it is strictly monotone in the range C 0 α C 0, by virtue of 13 and 14 below. It is readily seen that 11 has a unique solution α = α u, and C 0 α C 0. 6

7 We continue to assume that X 1, X,..., X n is a random sample without replacement from {b}, where n <, and continue to use the notation S n = n X, p, q and ω in Section 1. De ne H n x; u = Ee usn IS n x/ee usn, = pq as and assume C > 0 a constant depending only on C 0, which may differ at each occurrence. The main result in this section is as follows. Theorem 3.1. We have x m sup H n x; u Φ x σ C pq 1/ b 3/ b 3/, 1 where m = b K, σ = b K / b K K. Theorem 3.1 provides an extension of the classical result for samples from a nite population given by Höglund Its proof will be given after ve lemmas. Our rst lemma summarizes some basic properties of Kz. Lemma 3.1. We have K 0 = 0, pqe t K x < 0 < K x pqe t, for 0 < x t; 13 Furthermore, if x 1/16, then pq e 3t < K x < pq e 3t, for x t; 14 K x + iy 3/ e 5t pq, for x t and y π/. 15 Kx/ pq x / 1/ x 3, 16 K x/ pq x x, 17 K x/ pq 1 q px 8 x. 18 Proof. The proof of Lemma 3.1 is straightforward and the details are omitted. To introduce the following lemmas, we write, for 1, η = u b + α and ξ = v b + y 0, 19 where ν and y 0 are two real variables speci ed later. By using Lemma 3.1, it is readily seen that e C 0 βη e C 0, η C 0, K pqe C 0 and pqe 6C 0 K pqe 6C

8 The property 0 will be used heavily in the lemmas below. In the remainder of this section, we also de ne ρu, v, y 0 = βη + iξ. Lemma 3.. There exist 0 < ε 0 π/8 and δ 0 > 0 depending only on C 0, such that, for y 0 ε 0 and v < δ 0 b / b 3, { ρu, v, y 0 = exp K + i ξ K ξk / } 1 + R, 1 where βz is de ned as in 10 and R C pq 1 ξ 3 exp 4 Also, for ε 0 y 0 π and v < δ 0 b / b 3, ρu, v, y 0 e C 0 where Proof. We rst prove 1. De ne ξ K. βη + iξ C exp { [ K ε 0K /4 ]}, D 1 = { : vb π/4} and D = { : vb > π/4}. It suf ces to show that there exist 0 < ε 0 π/8 and γ 1 > 0 depending only on C 0 such that, if y 0 ε 0 and v < γ 1 b / b 3, then I 1 := βη + iξ βη D 1 D { = exp K + i ξ K ξk / } 1 + R1, 3 where R 1 C pq ξ 3 exp 1 4 ξ K, and I := [ ] βη + iξ βη + iξ βη D 1 D D C pq { ξ 3 exp K ξk /4 }. 4 Indeed, it follows from 3-4 that { ρu, v, y 0 = exp K + i ξ K ξk / } 1 + R, 8

9 where as required. { R R 1 + I exp K + ξ C pq ξ 3 exp 1 4 We next give the proofs of 3 and 4. K ξ K, / } Recall we assume that ε 0 π/8. If D 1, then ξ < π/ since y 0 π/8. This fact, together with 15, 0 and Taylor s formula: for x, y R, 1 Kx + iy = Kx + iyk x y K x/ iy 3 1 t K x + itydt/, implies that, whenever D 1, Kη + iξ K iξ K + ξ K / ξ 3 max x C 0 y <π/ 0 K x + iy /6 e 10C 0 pq ξ 3. Therefore, { βη + iξ = exp K + i ξ K ξ D 1 K / + L 1 }, 5 D 1 where L 1 e 10C 0 pq D 1 ξ 3. On the other hand, if D, then ξ π/8 since y 0 π/8. This, together with 0, yields that, whenever D, and hence iξ K ξ K / [8/π + 4/π]e 6C 0 pq ξ 3, βη = exp D { = exp { D K }, where L [8/π + 4/π]e 6C 0 pq D ξ 3. D K + i ξ K ξ K / + L }, 6 Recalling b = 0, if we choose ε 0 and δ 0 so small that 4C 1 max{ε 0, δ 0 }e 6C 0 1/4, where C 1 = max{8/π + 4/π, e 4C 0 }e 6C 0, then for y 0 ε 0 and v < δ 0 b / b 3, L 1 + L C 1 pq ξ 3 4C 1 pq y v 3 b 3 4C 1 max{ε 0, γ 1 } pq y0 + v b 4C 1 max{ε 0, γ 1 }e 6C 0 ξ K 1/4 ξ K, 7 9

10 by using 0. Combining 5-7, where which yields 3. { I 1 = exp K + i ξ K ξk / } 1 + R1, R 1 = e L 1 +L 1 L 1 + L e L 1 + L C pq 1 ξ 3 exp ξ 4 K, As for 4, by noting from 5-7 that, for any 0 D, βη + iξ βη e C 0 βη + iξ βη D 1 D { 0 } D 1 D { e C 0 exp K ξ / } + L 1 + L K e C 0 exp { K ξ K /4 }, 8 since e C 0 βη e C 0, we have I βη j + iξ j βη j βη + iξ j D D 1 ow 4 follows from 9 and βη + iξ βη = { e C 0 exp K ξk /4 } iξ 1 0 j D D {j} βη βη j + iξ j βη j. 9 β η + itξ dt e C 0 pq ξ C pq ξ 3, for D, where we have used the estimates: ξ π/8 for D, and for all 0 t 1, β η + itξ = pq e qη +itξ e pη +itξ e C 0 pq. 30 This proves 4 and also completes the proof of 1. We next prove. As in 7 of Robinson 1977, we obtain βη + iξ = e K [1 K 1 cos ξ ] exp K K 1 cos ξ = exp { K K [1 cosy 0 L 1 ] }, 31 10

11 where L 1 = cosξ cosy 0. ote that L 1 1 cosvb + sinvb v b /+ vb. It follows from 0 that, for any given δ 0 > 0, if v < δ 0 b / b 3, then L1 K pq e 6C 0 L1 pq e 6C 0 [δ 0/ 3/ b b 3 + δ 0 b b / ] b 3 pq e 6C 0 δ 0/ + δ 0 e 1C 0 δ 0/ + δ 0 K, 3 where we have used the fact that, by Hölder s inequality, b /3 b 3 1/3 and b 1/3 b 3 /3. 33 By taing δ 0 = min{γ 1, γ }, where γ 1 is de ned as in the proofs of 3-4 and γ is a constant satisfying e 1C 0 γ / + γ 1 cos ε 0 /4, it follows easily from 31-3, and K 0 K 0 1 cos ξ 0 C [recall 0] for any 1 0, that if v < δ 0 b / b 3 and ε 0 y 0 π, then 0 βη + iξ { [ exp K K 1 cos ξ ] } + K 0 K 0 1 cos ξ 0 { [ C exp K ε 0K /4 ]}, for any 1 0, where we have used the well-nown facts: 1 cosy 0 1 cosε 0 ε 0/ ε 4 0/4 ε 0/3, since 0 < ε 0 π/8. This proves the second inequality of. The rst inequality of holds true since βη 0 + iξ 0 e C 0 for each 1 0. The proof of Lemma 3. is now complete. Lemma 3.3. Let ε 0 and δ 0 be de ned as in Lemma 3.. Suppose that v < δ 0 b / b 3. Then, for y 0 ε 0, d ρu, v, y 0 dv and for ε 0 y 0 π, i b K b ξ K ρu, v, y0 C pq b ξ exp d ρu, v, y 0 C pq b exp dv { K ξ K /4 } ; 34 { K ε 0K /4 }

12 Proof. ote that d ρu, v, y 0 dv = i b j β η j + iξ j βη + iξ, j j=1 where i = 1. The property 35 follows immediately from and 30. We next prove 34. De ne D 1 and D as in Lemma 3.. We may write d ρu, v, y 0 dv = i D 1 b K η + iξ ρu, v, y 0 + i D b β η + iξ j βη j + iξ j. By virtue of 8, it suf ces to show that II := i b K η + iξ i b K + b ξ K D 1 D 1 C pq b ξ, 36 and β η + iξ K βη + iξ C pq ξ, for D. 37 In fact, as in the proof of Lemma 3., by using the Taylor s formula of K x + iy, II b K η + iξ K iξ K + b ξ K D D 1 1/ max K x + iy b ξ + e 6C 0 pq b ξ x <C 0 D y <π/ D 1 C pq b ξ, where we have used 15 and the fact that ξ > π/8 when D. This proves 36. The property 37 follows from ξ > π/8 for D, and hence β η + iξ K βη + iξ = pq eqη e iξ 1 pe η + q e C 0 pq ξ 8e C 0 /π pq ξ. The proof of Lemma 3.3 is complete. Lemma 3.4. There exists a δ 1 > 0 depending only on C 0, such that for v < δ 1 b / b 3, 1/ { Ee u+ivsn = G n p 1 K exp K + ivm 1 } v σ 1 + R, 38 1

13 where G n p = π n p n q n, m and σ are de ned as in Theorem 3.1 and R C v 3 pq b 3 + 1/ω e v σ /4. In particular, by letting v = 0 in 38, 1/ { } Ee u Sn = G n p 1 K exp K 1 + O 1 /ω, 39 where O 1 C 1 and C 1 is a constant depending only on C 0. Proof. As in Erdos and Renyi1959, for any α, Ee u+ivsn = πg n p 1 π π q + pe u+ivb+α+iθ e nα+iθ dθ. Let α be the solution of 11, y 0 = ψ/ω, and η and ξ as in 19. Some algebra shows that Ee u+ivsn = πω G n p 1 + ρu, v, ψ/ω dψ ψ ε 0 ω ε 0 ω < ψ <πω = III 1 + III, say, 40 where ε 0 is de ned as in Lemma 3.. Let δ 1 = min{δ 0, e 3C 0 ε 0 / }, where ε 0 and δ 0 are de ned as in Lemma 3.. We will show that, for v < δ 1 b / b 3, III C/ω Gn p 1 K III 1 = G n p 1 K 1/ { exp K 1 } 4 v σ 1/ { exp K + ivm 1 v σ, 41 } 1 + R 1, 4 where R 1 C v 3 pq b 3 + 1/ω e v σ /4. Then 38 follows easily from The proof of 41 is straightforward by 0 and Lemma that and hence for v < δ 1 b / b 3, Indeed, it follows from e 6C 0 ω K e 6C 0 ω, 43 v σ v b K e 6C 0 pqv b δ 1e 6C 0 pq b 3 / b 3 δ 1e 6C 0 ω ε 0 By and Lemma 3., it is readily seen that III C G n p 1 exp C G n p 1 exp [ K ε 0 C/ω Gn p 1 K ] K /4 [ K 1 4 v σ ε 0 K 13 K /, 44 ] /8 1/ exp { K 1 4 v σ },

14 as required. We next prove 4. ote that ξ K = v b K since K = 0, b K { gψ, v := ψ + vω K } K ω = ξ K v σ 45 and e gψ,v/ dψ = πω / K 1/. It follows from 45 and Lemma 3. that, for v < δ 1 b / b 3, 1/ { III 1 = G n p 1 K exp K + ivm 1 } v σ 1 + R, 46 where R is de ned as in 1 and R e gψ,v/ dψ + e 3C 0 R e gψ,v/ dψ := L 3 + L 4. ψ ε 0 ω ψ ε 0 ω By 0, 43 and Hölder s inequality, ω b K K e 3C 0 It follows easily that ψ ε 0 ω ψ 3 e gψ,v/4 dψ C b K 1 + v ω 1/ e 6C 0 pq 1/ b 1/. 47 b K K 3 C This, together with the de nitions of R and gψ, v, implies that [1 + pq ω v 3 b 3 ]. L 4 C pq e v σ ψ ε /4 ξ 3 e gψ,v/4 dψ 0 ω 4C pq e v σ /4 v 3 b 3 + ω 3 ψ 3 e gψ,v/4 dψ ψ ε 0 ω C v 3 pq b 3 + 1/ω e v σ /4. 48 As for L 3, by noting from 47 that, for v < δ 1 b / b 3, vω b K K δ 1 e 6C 0 pq 1/ b 3// b 3 ε 0 ω /, it is readily seen [recall 43] that L 3 ψ ε 0 ω / exp e 6C 0 ψ / dψ C/ω. 49 Taing the estimates 48 and 49 bac into 46, we obtain the required 4. The proof of Lemma 3.4 is now complete. 14

15 Lemma 3.5. If v < min{pq b 1/, δ 1 b / b 3 }, then d [ ] e ivm Ee u+ivsn + vσ dv e 1 v σ Ee us n { } C exp K b 3/ b, 50 where δ 1, m, σ and K are de ned as in Lemma 3.4. Proof. Let ε 0 be de ned as in Lemma 3.. By 40, we have d [ ] e ivm Ee u+ivsn + vσ dv e 1 v σ Ee us n πω G n p 1 J 1 + J + J 3 + J 4, 51 where J 1 = dρu, v, ψ/ω i m b ξ K ρu, v, ψ/ω dψ, ψ ε 0 ω dv J = b ξ K vσ ρu, v, ψ/ω e ivm dψ, ψ ε 0 ω J 3 = v σ ρu, v, ψ/ω e ivm dψ πω G n p e 1 v σ Ee us n, ψ ε 0 ω [ d e ivm ρu, v, ψ/ω ] J 4 = dψ. dv ε 0 ω ψ πω De ne gψ, v as in 45. Similarly to the proof of 48, it follows from Lemma 3.3 that J 1 C pq e K b ξ ψ ε e gψ,v/4 dψ 0 ω C pq e K v b 3 + ω b ψ e gψ,v/4 dψ ψ ε 0 ω C pq e K v b 3 + Cω b [ 1 + v pq b] C e K b 3 / b, 5 since v pq b 1/, where we have used the estimate: b b b 3 by 33. Also, by noting where g 1 ψ, v = ψ + vω J b ξ K = vσ b K + g 1 ψ, v, ω b K K b K ω, it follows from 1 in Lemma 3. that e K ψ ε 0 ω + C pq ψ ε 0 ω 15 g 1 ψ, v e gψ,v/ dψ ξ 3 g 1 ψ, v e gψ,v/4 dψ. 53

16 Since g 1ψ, v e gψ,v/ dψ = 0, and g 1 ψ, v e 3C 0 g 1/ ψ, v by 43, as in the proof of 49, we have g 1 ψ, v e gψ,v/ dψ g 1 ψ, v e gψ,v/ dψ C/ω. ψ >ε 0 ω ψ ε 0 ω On the other hand, as in the proof of 48, pq ξ 3 g 1 ψ, v e gψ,v/4 dψ C v 3 pq b 3 + 1/ω. ψ ε 0 ω Taing these estimates bac into 53, and noting b K e 6C 0 pq b e 6C 0 ω b 3 / b, and also b K e 6C 0 1/ pq b 1/, by 0 and 33, we have that for v pq b 1/, b K J C e K v 3 pq b 3 + 1/ω ω C e K v 3 pq 3/ b 3/ + 1 b 3 / b C e K b 3 / b. 54 As for J 3, by using 39 and 4, we obtain that for v pq b 1/, J 3 C v σ ω K 1/[ v 3 pq ] b 3 + 1/ω e K C e K b 3 / b, 55 where we have used 43, σ e6c 0 pq b since 0, and some routine calculations. Finally we estimate J 4. In fact, by using, 35 and 43, and noting m = b K pqe 4C 0 b since 0, we have d ρu, v, ψ/ω J 4 + m ρu, v, ψ/ω dψ ε 0 ω ψ πω dv C pq e K b e cω dψ ε 0 ω ψ πω C pq e K b /ω C e K b 3 / b. 56 Combining and noting [Lemma 1 in Höglund1978] π/ πω G n p < we obtain the required 50. The proof of Lemma 3.5 is complete.. 16

17 We are now ready to prove Theorem 3.1. Let T = δ b / b 3, where δ = min{δ 0, δ 1 } with that δ 0 and δ 1 are de ned as in Lemmas 3. and 3.4. De ne fv = Ee u+ivsn /Ee usn and gv = e ivm v σ /. ote that fv and gv are characteristic functions of the random variable with distribution function H n x; u and the normal random variable with mean m and variance σ, respectively. By Esseen s smoothing inequality, sup x x m H n x; u Φ σ T Recalling b = 0 and 0, it is readily seen that T σ = b b K / K K v 1 fv gv dv + 1/T σ. 58 > e 6C 0 pq b b K / K e 6C 0 pq b. 59 This, together with 58, implies that 1 will follow if we prove T T v 1 fv gv dv C pq 1/ b 3/ b 3/. 60 Without loss of generality, we assume ω suf ciently large so that O 1 /ω 1/, where O 1 is de ned as in 39. Otherwise 60 is trivial by the fact 1/ b 3 / b 3/. For O 1 /ω 1/, it follows from Lemma 3.4 that fv gv exp v σ/ R O 1 /ω 1 + O 1 /ω C v 3 pq b 3 + 1/ω e v σ /4. 61 This, together with 59, implies that v 1 fv gv dv C pq 1/ b 3 / b 3/ + C/ω T 1 v T C pq 1/ b 3 / b 3/, 6 where T 1 = min{pq b 1/, T }. In the following, we let f 1 v = e ivm fv and g 1 v = e ivm gv = exp{ 1 v σ }. 17

18 By 39 and Lemma 3.5, for v T 1, f 1v g 1v = [Ee usn ] 1 d [ e ivm Ee u+ivsn ] 1/ dv C G np K 1 + O 1 /ω b 3 / b C b 3 / b, where we have used O 1 /ω 1/ and the fact that, by 43 and 57, G n p K 1/ e C 0 G n pω C. + vσ e 1 v σ Ee us n This, together with the fact that fv gv = f 1 v f v v sup 0tv f 1t g 1t, implies that v T 1 v 1 fv gv dv Cpq 1/ b 3 / b 3/. 63 ow 60 follows from 6 and 63. The proof of Theorem 3.1 is complete. 4 Proof of Proposition.1 Roughly speaing, the proof of Proposition.1 is based on the conjugate method and an application of Theorem 3.1 to the b speci ed in 64 below. We need some preliminaries rst. Let 0 < λ, 0 θ 1 and θ 1 7. De ne, for = 1,,, b = λba θb q a 1 [ + θ 1 b 4 q a 1 1 ] a j Since a = 0 and a =, it is readily seen that max a 1 and b = 0. Also, when b max a 1/18, we have that, bβ 3 1/18, max b 1/3, 65 b λ b 5b 3 qβ 3, 66 b 3 9b 3 β So, recalling b = x/ω, hold true if 0 x 1/18 ω / max a. De ne Kz as in 10. We still use the notation K, K and K denote the values of Kz, K z and K z evaluated at z = b + α, where α is the solution of the equation K b + α = As shown in the solution of 11, if 65 holds true, then α is unique and α 1/3. We establish four lemmas before the proof of Proposition.1. 18

19 Lemma 4.1. If 0 x 1/18 ω / max a, then { α min 1/3, / } b, α 9/8 b 3 β Proof. The inequality that α 1/3 has been proved above. By noting b + α 1/16 by 65, it follows from 17, 68 and b = 0 that α = [ K b + α /pq b + α ] b + α = b + α b + α /. This yields α / b, and hence the rst result of 69 follows. Furthermore, by using Hölder s inequality, b 1/3 and 67, α 4/ b 4 9/8b 3 β 3, which implies the second result of 69. The proof of Lemma 4.1 is complete. Lemma 4.. If 0 x 1/18 ω / max a, then K λ x / 4 x 3 β 3 /ω, 70 b K λ x 4 x 3 β 3 /ω, 71 K ω 41 x, 7 b K 6 x, 73 b K λ x 1 x 3 β 3 /ω. 74 Proof. We prove 70. The others are similar and omitted. Applying 16 with x = b + α and using Hölder s inquality, [ K 1 pq b + α ] pq b 3 + α This, together with b = 0, and 69, implies that K λ x / ] [K 1 pq b + α + 1 pq b λ b + 1 ωα 4 b 3 ω β 3 = 4x 3 β 3 /ω, as required. 19

20 Let Y j, j = 1,,..., n be a random sample of size n without replacement from {b 1, b,..., b } de ned by 64, T n T n λ, θ, θ 1 = n Y, m m λ, θ, θ 1 = b K, σ σλ, θ, θ 1 = b K b K / K, and H nu = E expt nit n u/e expt n. Lemma 4.3. There exists an absolute constant λ 0 > 0 such that, for x λ 0 ω / max a, exp{λ x / Ax 3 β 3 /ω } E expt n exp{λ x / + Ax 3 β 3 /ω }. 76 Proof. Without loss of generality, assume λ 0 min{1/18, 1/8C 1 + 4}, where C 1 is de ned as in 39. Recall that max b 1/3 by 65. It follows from Lemma 3.4 with C 0 = 1/3, u = 1 and v = 0 that E expt n = G n p 1 K 1/ exp{ where G n p = π n p n q n and R C 1 /ω. By Stirling s formula, p n q n = πω n 1/ 1 + O ω, K }1 + R, 77 where O 1/6. This, together with ω x max a /λ 0 18 recall max a 1, implies that j=1 ω 1 G np R = 1 + O 3 ω 1, 78 where O 3 C On the other hand, it follows from 7 that K 1/ ω = 1 + O 4 b, 79 where O 4 8. From 78-79, for x λ 0 ω / max a, exp{ A 1 x 3 β 3 /ω } K 1/ G n p R exp{a 1 x 3 β 3 /ω }, 80 where A 1 = C and we have used the fact that 1/ω + b x 3 β 3 /ω since b = x/ω and β 3 1. ow 76 follows easily from 70, 77 and 80. The proof of Lemma 4.3 is complete. Lemma 4.4. There exists an absolute constant λ 1 > 0 such that, for x λ 1 ω / max a, m λ x 4 x 3 β 3 /ω, 81 σ λ x x 3 β 3 /ω, 8 0

21 If in addition 1 λ, then := sup y where uy = y σ + m and C is de ned as in Theorem 3.1. Also, for all y satisfying m y + σ, Hnuy Φy 1 Cβ 3 /ω 1/4, 83 P T n y 1/ exp{ m σ } E expt n. 84 Proof. Without loss of generality, assume λ 1 min{1/18, 1/5C}, where C is de ned as in Theorem 3.1. Then 81 and 8 follow from by a simple calculation. If 1 λ, by noting β 3 /ω xβ 3 /ω min{1/18, 1/50C} since β 3 max a, it follows easily from that pq b 4x /5 and pq 1/ b 3 / b 3/ 1β 3 /ω 1/4C. 85 By 85 and Theorem 3.1 with C 0 = 1/3 and u = 1 recall max b 1/3, which implies 83. Cpq 1/ b 3 / b 3/ 1 Cβ 3 /ω 1/4, We next prove 84. In fact, by 83 and the conjugate method, for all y satisfying m y + σ, / P Tn y = e m E expt n = e m σ y m /σ y e u dh nu e xσ dh n uy dh nuy e m σ P 0, 1 1/ exp{ m σ }, where 0, 1 is a standard normal random variable and we have used the fact that P 0, 1 > 3/4. This proves 84 and also completes the proof of Lemma 4.4. After these preliminaries, we are now ready to prove Proposition.1. 1

22 In addition to the previous notation, we further let T 1n = T n 1, ξ, ξ 1, m 1 = m 1, ξ, ξ 1, σ 1 = σ 1, ξ, ξ 1, ε = x + h m 1 /σ 1 and H 1n u = E exp{t 1n }IT 1n u/e exp{t 1n }. ote that bs n ξ b qv 1n + ξ 1 b 4 q V n = T 1n. It follows from the conjugate method that, P bs n ξ b qv 1n + ξ 1 b 4 q V n x + h = P T 1n x + h = E exp{t 1n } x +h = E exp{t 1n } e x h = E exp{t 1n } e x h e t dh 1n t [ ] e tσ 1 dh 1n σ 1 t + ε + m 1 0 L + R 86 where L = R = 0 0 e tσ 1 dφt + ε, ] } e tσ 1 d {H 1n [σ 1 t + ε + m 1 Φt + ε. We next estimate E exp{t 1n }, L and R for 0 ξ 1/, ξ 1 36, h x /5 and x ηω / max a, where we assume η suf ciently small such that η min{1/18, λ 0, λ 1 }, with λ 0 and λ 1 de ned as in Lemmas 4.3 and 4.4. This η chosen guarantees that Lemmas hold true, and since β 3 max a, Clearly, by Lemma 4.3, β 3 /ω xβ 3 /ω η/ 1/ exp { x / A x 3 β 3 /ω } E exp{t1n } exp { x / + A x 3 β 3 /ω }. 88 In order to estimate L, we note that L = 1 e σ t 1 t+ε dt π 0 = e ε / e ε +σ t 1 t dt π 0 := e ε / π L 1. 89

23 Write ψt = { 1 Φt } /Φ t = e t / e y / dy. It is readily seen that, t 3/4 tψt 1 for t, and ψ t = tψt 1 t for t > On the other hand, ψ{ε + σ } = L 1, and by virtue of 81-8 and 87, and if in addition h x /5, ε h/σ 8x β 3 /ω 91 ε + σ x 3 h /x + 41 x β 3 /ω x/3. 9 Using 90-9, it follows from Taylor s expansion that, for h x /5 and x ηω / max a, L 1 = ψ{ε + σ } = ψx + ψ θ { ε + σ x }, [where θ x/3, 5x/3] = ψx 1 + τ + O 5 x β 3 /ω, where τ 9 h /x and O Therefore, taing account of 89, we get for h x /5 and x ηω / max a, L = e x / { 1 Φx } e ε / 1 + τ + O 5 x β 3 /ω. 93 As for R, by 83 and integration by parts, [ ] R sup H 1n σ 1 t + m 1 Φt 4Cβ 3 /ω. t This, together with 90, implies that for x, where O 6 3 πc. R = O 6 x β 3 /ω e x / { 1 Φx }, 94 Combining 86, 88 and 93-94, it is readily seen that for any h x /5 and x ηω / max a, This proves 7. P bs n ξ b qv 1n + ξ 1 b 4 q V n x + h 1 Φx [ h x ] exp { h + Ax 3 β 3 /ω }. 3

24 Similarly, by letting h = 0, it follows from 86, 88, 91 and that if, in addition, x ω /β 3, then P bs n ξ b qv 1n + ξ 1 b 4 q V n x 1 Φx exp { Ax 3 β 3 /ω ε / } [ } ] 1 { O 5 + O 6 e ε / x β 3 /ω exp { } [ ] A 1 x 3 β 3 /ω 1 A xβ 3 /ω exp { A 3 x 3 β 3 /ω }, 95 by choosing η suf ciently small. From 95, the property 6 will follow if we prove that, for x ω /β 3 and x ηω / max a, P bs n ξ b qv 1n + ξ 1 b 4 q V n x exp { } A x 3 β 3 /ω Φx We will prove 96 by using 84. Let λ = 1 + 8xβ 3 /ω, θ = λξ and θ 1 = λξ 1. ote that, 1 λ 3/ by 87, 0 θ 3/4 since 0 ξ 1/ and θ 1 7 since ξ By virtue of 81-8, 87 and x ω /β 3, we have m λ x +4 x 3 β 3 /ω, σ x x3 β 3 /ω and m λx σ λλ 1x 8x 3 β 3 /ω 0. ow, by 84 with y = λx and Lemma 4.3, for x ω /β 3 and x ηω / max a, P bs n ξ b qv 1n + ξ 1 b 4 q V n x = P Tn λx 1 exp{ m σ }E exp{t n} 1 exp{ x / x Ax 3 β 3 /ω } 1 Φx exp{ A 1 x 3 β 3 /ω }, which implies 96. The proof of Proposition.1 is now complete. 4

25 5 Proof of Proposition. By the inequality 1 + y 1/ 1 + y/ y for any y 1, P S n x qv n = P S n x nq 1 + V P S n x [ nq 1 + V 1n n P V1n 36x n V 1n ] n n n n X 1 + 5p a 4 1/ + P S n x nq 1 + V n 1n n 36x X n 1 + 5p a 4 := R 1n + R n, say. 97 ote that R n = P bs n 1 b qv 1n + 36 b 4 q V n x h 0, where, whenever x 1/18 ω / max a, h 0 = 180p x4 a n x4 EX 1 n n 3x3 β 3 ω, and also 0 h 0 x /5. It follows from Proposition.1 with ξ = 1/, ξ 1 = 36 and h = h 0 that there exists an absolute constant A > 18 such that, for all x 1/A ω / max a, R n 1 Φx exp{ax 3 β 3/ω }. 98 This, together with 97, implies that Proposition. will follow if we prove, for all x > 0, R 1n e 4x. 99 Theorem.1 of de la Pena, Klass and Lai 004 will be used to prove 99. To use the theorem, let Y i = Xi 1, A = n Y and B = n + 4p a 4 1/. It follows from Y 5

26 Theorem 4 of Hoeffding1963 also see Lemma 6. below that, for any λ R, E exp {λ A λ } B { = exp λ p a 4 where we have used the inequality e x x } E exp { n } λy λ Y { exp λ p }[ ] n a 4 E exp{λy 1 λ Y1 } { exp λ p }[ a EλY 1 IλY 1 1/ { = exp λ p }[ a 4 1 EλY 1 IλY 1 1/ { exp λ p }[ ] n a λ EY1 { exp λ p } a 4 + λ ney1 { = exp λ p a 4 + λ p } a 1 1, 1 + xix 1/. This yields that two random variables A and B > 0 satisfy the condition 1.4 in de la Pena, Klass and Lai 004. ow, by noting EB EB 6p a 4 and applying Theorem.1 of de la Pena, Klass and Lai 004, we have by letting t = x. Similarly, n P V 1n 6x X 1 + 5p 1/ a 4 P A 6x B + EB e 6xt/ E exp ta/ B + EB e 6xt/ +t e 4x, 100 n P V 1n 6x X 1 + 5p 1/ a 4 e 4x. 101 By virtue of 100 and 101, we obtain 99. The proof of Proposition. is now complete. 6 Proof of Proposition.3 Throughout the section, let ε j, 1 j be iid random variables with P ε 1 = 1 = 1 P ε 1 = 0 = p, which are also independent of all other random variables, and B = j=1 ε j 6 ] n ] n

27 p. By the inequality 1 + y 1/ 1 + y/ y for any y 1 again, we have P S n x qv n = P S n x nq 1 + V n n 1/ n P S n x nq 1 + V n n V n n n n = P ε a x nq 1 + ε pg + x n = P ε a 1 ε a 1 n ν ν j x h B = 0 1 j B = 0 n = P T + Λ x h B = 0, 10 where h = xpq a 1 /n, T = ε pg, Λ = x n where, for all j = 1,,, ν j = ε j pa j 1 and g j = 1 j ν ν j, a j xa j 1 x1 p + a nq n n j 1 1 a 1, and where, in the proof of 10, we have used the fact that a = 0, a = and ε p = ε 1 p + p = ε p1 p + pq. We need the following lemmas before the proof of Proposition.3. Lemma 6.1. For any random variable Z with E Z <, E Z B = 0 = 1 B n p where B n p = πω P B = 0 and πω πω EZe itb /ω dt, π/b n p 1 + ω. 104 Proof. ote that B = j=1 ε j n is an integer and for any integer, π π e it dt = { π if = 0 0 if 0. The proof of 103 is now obvious. The estimate for B n p follows from P B = 0 = n p n q n and Stirling s formula. 7

28 Lemma 6.. Let the population {C} consist of values c 1,, c. Let X 1,, X n denote a random sample without replacement from {C} and let Ỹ1,, Ỹn denote a random sample with replacement from {C}. Then for any continuous and convex function fx, n 1 Ef n X + 1 n Ef 1 jn X X Xj n Ef Ỹ. 105 Ef 1 jn Ỹ Ỹ j. 106 Proof. 105 is Theorem 4 of Hoeffding1963. We next prove 106. As in the proof of Theorem 4 in Hoeffding1963, for any function f, there exists a function ḡ f x 1,, x n which is symmetric in x 1,, x n such that Ef Ỹ Ỹ j = Eḡ f X 1,, X n. 107 By noting Ef 1 jn 1 jn Ỹ Ỹ j = 1 n 1,..., n=1 as in 6.6 of Hoeffding1963, ḡ f can be written as [ n n f c j c j ], ḡ f x 1,, x n = [ p, i1,, i, r 1,, r f r j x ij where the sum j=1 j=1 j=1 r j x i j ], 108 is taen over the positive integers, i 1,, i, r 1,, r such that = 1,,..., n, j=1 r j = n and i 1,..., i are all different and do not exceed n. The coef cients p are non-negative and do not depend on the function f. In particular, when f = x, n ḡ x x 1,, x n = K 0 x + K 1 x x j, jn since ḡ f is symmetric on x 1,..., x n, where K 0 and K 1 are constants. Since E n Ỹ Ỹ j = K 0 E X + K 1 E X Xj 1 jn by 107 and 109, we have that nn 1 K 0 n c c = + K 1nn 1 1 holds true for any c 1,, c R, and hence K 0 = n 1 1 jn j=1 c c and K 1 = 1. On the other hand, by letting f = 1 in 107 and 108, p, i1,, i, r 1,, r =

29 By virtue of , it follows from the Jensen s inequality that, for any continuous and convex function fx, n 1 Ef n X + 1 X Xj = Ef ḡ x X 1,, X n 1 jn Eḡ f X 1,, X n = Ef Ỹ Ỹ j. 1 jn This yields 106 and hence completes the proof of Lemma 6.. Lemma 6.3. i. We have E 1 j E ν E j=1, 1 j ν ν j 3/ B = 0 ν j 3/ B = 0 3/ ν ν j B = 0 An β 3, 111 An β 3, 11 An β ii. If η, 1, are iid random variables with P η = 1 = 1 P η = 0 = mt, 0 mt 1, independent of all other rv s, then E E 1 j 1 j 3/ η η j ν ν j B = 0 3/ η 1 η j ν ν j B = 0 Am tn β 3, 114 Amtn β Proof. We rst prove 113. ote that ν ν j = ε j ε a j 1a 1 + p ε a 1 1 j 1 j + p a j 1a 1. 1 j By the c r -inequality, we have E 1 j 3/ ν ν j B = 0 4I 1 + 4I + I 3, 116 9

30 where Since a =, I 1 = E 1 j ε j ε a j 1a 1 3/ B = 0, I = p 3/ E ε a 1 3/ B = 0, I 3 = p 3 a j 1a 1 3/. 1 j I 3 p 3 a 1 3/ p 3 3/ a 4 p 3 a 3 n β Recall that X 1, X,..., X n is a random sample without replacement from {a} = {a 1,, a }. Suppose that Y 1, Y,..., Y n is a random sample with replacement from {a}. ote that Y j are iid random variables with EfY 1 = 1 fa for any f.. It follows from Lemma 6. and the classical results for iid random variables that I = p 3/ n E X 1 3/ p 3/ E p 3/ E n n Y 1 3/ Y 1 EY 1 3/ + p 3/ ney1 1 3/ n 4p 3/ E Y 1 EY 1 3/ + p 3 a 1 3/ 16p 5/ a p 3 a 1 3/ 18p 5/ a 3 18 n β Similarly, it follows from Lemma 6. and the classical results for U-statistics that 1 3/I1 = 1 3/E Xj 1X 1 3/ E 1 jn 1 jn Y 1Yj 1 3/ + p 3/ E n X 1 3/ A n E Y n β 3 A 1 n β Combining , we obtain the required 113. We next prove 11. ote that, by a =, ν j = ε a 1 ε j a j 1 ν j=1, pa 1 j=1, ε j a j 1 + pε p a 1. j=1 30

31 By the c r -inequality, we have E ν j=1, ν j 3/ B = 0 4I 4 + I 5 + I 6, where, as in the proofs of , I 4 = = p I 5 = p I 6 = Eε a 1 j=1, a 1 3/ E A nn 1 1 This yields 11. ε j a j 1 3/ B = 0 j=1, a 1 3/ j=1 ε j a j 1 3/ j=1, j=1, ε j = n 1 a j 1 3/ A n β 3, a 1 3/ E ε j a j 1 3/ B = 0 A n β3, a 1 E 3 pε p 3/ B = 0 A p The proof of 111 is simple. Indeed, 1 j E ν ν j 3/ B = 0 a 6 A n β 3. E A a 3 ε 1 pε p 3/ B = 0 A 1 p a 3 = A1 n β 3. We nally prove 114 and 115. By 113 and the c r inequality, it suf ces to prove E 1 j 3/ η mtη j mtν ν j B = 0 E 1 j 3/ η mtν ν j B = 0 Am tn β 3, 10 Amtn β In fact, recalling that η are iid random variables with Eη 1 = mt, independent of all other random variables, it follow from conditional expectation arguments and moment results for 31

32 degenerate U-statistics and 111 that E A 1 j 1 j Am t 3/ η mtη j mtν ν j B = 0 E η mtη j mtν ν j 3/ B = 0 1 j A m t n β 3. E ν ν j 3/ B = 0 This proves 10. Similarly, it follows from conditional expectation arguments and moment results for partial sums and 11 that E 1 j 3/ η mtν ν j B = 0 = E η mtν A mt A mt n β 3, ν E j=1, j=1, 3/ ν j B = 0 3/ ν j B = 0 which implies 11. The proof of Lemma 6.3 is now complete. To introduce the following lemmas, we de ne ft = Ee ittn+λn B = 0, f 1 t = Ee ittn B = 0, f t = EΛ n e ittn B = 0, and for = 1,,, g t, ψ = E exp{iε ptg + ψ/ω }. We also use the notation = xβ 3 /ω. Lemma 6.4. If t 1/18 1, then for x 1/18ω / max a and any 0 mt 1, [ ] ft A1 + tx m 1/ t e mtt /4 + ω e 1/40mtω + A t 3/ mt + A t m 4/3 t 4/3. 1 3

33 Proof. De ne {η, = 1,, } as in Lemma 6.3 ii. Furthermore, let T1 = η ε pg, T = 1 η ε pg, Λ 1 = x η n η j ν ν j, Λ = x η n 1 η j ν ν j, 1 j Λ 3 = x n 1 j 1 η 1 η j ν ν j. 1 j ote that T + Λ = T1 + T + Λ 1 + Λ + Λ It follows from 13, e it 1 t and e it 1 it t 3/, that ft = Ee itt 1 +T +Λ 1 +Λ +Λ 3 B = 0 Ee itt 1 +T +Λ +Λ 3 B = 0 + t E Λ 1 B = 0 Ee itt 1 +T +Λ 3 B = 0 + t EΛ e itt 1 +T +Λ 3 B = 0 +8 t 3/ E Λ 3/ B = 0 + t E Λ 1 B = 0 := Ξ 1 t, x + Ξ t, x + Ξ 3 t, x + Ξ 4 t, x. 14 We rst estimate Ξ 3 t, x and Ξ 4 t, x. By Lemma 6.3 ii, we obtain that, E Λ 3/ B = 0 Ax 3/ mtn 1 β3 Amt, and, by Hölder s inequality, E Λ 1 B = 0 [ E Λ 1 3/ B = 0 ] /3 Am 4/3 t 4/3. These facts yield that Ξ 3 t, x + Ξ 4 t, x A t 3/ mt + A t m 4/3 t 4/3. 15 ext we estimate Ξ 1 t, x. Write B1 = η ε p, B = 1 η ε p, and B = { : η = 1}, B c = { : η = 0}. 16 ote that, given η 1,, η, T1, B1 σ{ε, B}, T, Λ 3 B σ{ε, B c }, 33

34 and B = B 1 + B. It follows that T 1 and B 1 are independent of T, Λ 3, B, given η 1,, η, and hence by Lemma 6.1, 1 Ξ 1 t, x = E exp {itt1 + T + Λ B n p 3 + iψb /ω } dψ ψ πω EE η exp {itt1 + iψb 1/ω } dψ ψ πω = E E η exp {iη ε ptg + ψ/ω } dψ, 17 ψ πω where E η denotes the condition expectation given η, = 1,,. Let ε be an independent copy of ε. ote that, by Taylor s expansion of e iz, E E η exp {iη ε ptg + ψ/ω } = E E η exp {iη ε ε tg + ψ/ω } = E exp {iη ε ε tg + ψ/ω } 1 1/tg + ψ/ω Eη Eε ε + 1/6 tg + ψ/ω 3 Eη 3 E ε ε 3 1 pq mt tg + ψ/ω + pq/3 mt tg + ψ/ω 3. This, together with that fact that g = 0 and for x 1/18ω / max a, pq g 1 xβ 3 /ω and pq g 3 5β 3 /ω, 18 yields that for t < 1/18 1, ψ < 3/8ω and x 1/18ω / max a, Jt, ψ := EE η exp {iη ε ptg + ψ/ω } E E η exp {iη ε ptg + ψ/ω } 1/ { exp pq/mt tg + ψ/ω + pq/6mt } tg + ψ/ω 3 { exp pq/mt t g mtψ / +pq/3mt } tg 3 + /3mt ψ 3 /ω { exp pq/mt t g + pq/3mt } tg 3 mtψ /4 { } exp 1/mtt 1 xβ 3 /ω 5/3 t β 3 /ω mtψ /4 exp{ mtt /4 mtψ /4}. 19 To estimate Jt, ψ for 3/8ω ψ πω, we rst note that E E η exp {iη ε ptg + ψ/ω } = E exp {iη ε ε tg + ψ/ω } = 1 pq + pq E cos [ η tg + ψ/ω ] = 1 pq mt + pq mt cos tg + ψ/ω

35 De ne D = { : g } and D c = { : g > }. It is readily seen that, for D, t < 1/18 1 and 3/8ω ψ πω, 3 64 tg + ψ/ω π or 1 64 π tg + ψ/ω 3 64 and hence cos tg + ψ/ω cos3/64 < On the other hand, it follows from 18 that, for x 1/18ω / max a, 4pq 1 D c 4x β 3 ω D c D c g pq xβ 3 /ω pq 1, where D c denotes the number of D c. Thus D c / and D = D c /. By virtue of 130 and all above facts, we obtain that for t < 1/18 1, 3/8ω ψ πω and x 1/18ω / max a, Jt, ψ E E η exp {iη ε ptg + ψ/ω } 1/ D D { exp pq mt [ 1 cos ] } tg + ψ/ω exp { 1/40 mt ω }. 131 Combining 17, 19 and 131, it follows that, for t < 1/18 1 and x 1/18ω / max a, Ξ 1 t, x Amt 1/ e mtt /4 + Aω e 1/40mtω. 13 Finally, we estimate Ξ t, x. ote that Λ = x n j B ν c j B ν, where B and B c is de ned in 16. Similarly to 17, Ξ t, x = t B n p 4 t x n 4 t x n 4 t x mt n ψ πω E ψ πω E ψ πω 1j E Λ e itt 1 +T +Λ 3 +iψ B /ω dψ [ ] E η ν j E η ν exp {itt1 + iψb1/ω } dψ j B c B [ 1j ] 1 η j η E ν j E ν Ω j t, ψ dψ E ν j E ν EΩ j t, ψdψ, 133 ψ πω where Ω j t, ψ = l j, E η exp {iη l ε l p tg l + ψ/ω }. 35

36 As in the proof of 13 with minor modi cations, we have that, for t < 1/18 1, x 1/18ω / max a, and for all 1 j, ψ πω EΩ j t, ψdψ Amt 1/ e mtt /4 + Aω e 1/40mtω. This, together with 133 and the fact that 1 j E ν j E ν pq a + 1 = 16ω 4, yields that, for x 1/18ω / max a and t < 1/18 1, Ξ t, x A tx e mtt /4 + ω e 1/40mtω. 134 Taing estimates 15, 13 and 134 into 14, we obtain 1. The proof of Lemma 6.4 is now complete. Lemma 6.5. Suppose that x 1/18ω / max a. i. If t 1/18 1 and ψ πω, then l=1, j, for all 1 j, and g l t, ψ e t +ψ /4 + e 1/40ω, 135 d g t, ψ 4 t + ψ e t +ψ /4 + e 1/40ω. 136 dt ii. If t 1/18 1/3 and ψ < 1/18 1/3, then and if in addition t 1/4, then where g t, ψ gt, ψ A 4/3 e t +ψ /4, 137 d g t, ψ dt d gt, ψ dt A 4/3 1 + ψ 6 e ψ /4, 138 gt, ψ = e t +ψ / {1 + g t, ψ 1 + t + ψ }. Proof. By letting mt = 1 in 19 and 131, together with minor modi cations, we obtain 135. ote that, under the conditions of part ii, s := t + ψ 1/64 1/3 and g t, ψ 1 + t + ψ / s + s 3,

37 by 18 and Taylor s expansion of e iz. 137 follows easily from some routine calculations. See, for example, Lemma 10.1 of Jing, Shao and Wang 003 with minor modi cations. We next prove 136 and 138. ote that d g t, ψ dt where g t, ψ = [g t, ψ] 1 dg t,ψ, and dt dgt, ψ dt = g t, ψ g t, ψ, = tgt, ψ + dg t, ψ dt + t e t +ψ /. 140 Simple calculations show that d g t, ψ d gt, ψ dt dt J 1 + J + J 3 e t +ψ /, 141 where J 1 = g t, ψ g t, ψ gt, ψ, J = g t, ψ + t g t, ψ 1 + t + ψ /, J 3 = g t, ψ dg t, ψ. dt By the inequality e iz 1 iz z /, it is readily seen that, for any t and ψ, and g t, ψ 1 pq/tg + ψ/ω, 14 dg t, ψ + pq g tg + ψ/ω pq/ g tg + ψ/ω. 143 dt Since pqg by 18, it follows from 14 that, if ψ < 1/18 1/3 and t 1/4, then g t, ψ 1 1/4 and hence [g t, ψ] 1 = 1 + θ 1 pq tg + ψ/ω, 144 where θ 1 < 1. In view of 143 and 144, it follows from 18 again that J 3 [g t, ψ] 1 1 dg t, ψ dt pq g tg + ψ/ω 3 8pq 1 + ψ 3 g 4 + g /ω 3 4/3 1/3 8pq 1 + ψ 3 g 3 + g 3 /3 /ω ψ 3 4/3, 37

38 Similarly, by recalling g = 0, we have g t, ψ + t pq g 1 + pq g tg + ψ/ω ψ 3. which, together with 137 and 139, implies that J 1 A 1 + ψ 3 4/3 e ψ /4 and J A 1 + ψ 6 4/3. Taing the estimates of J 1, J and J 3 into 141, we obtain 138. Similarly, by noting that dg t, ψ pq g tg + ψ/ω dt t pq g + ψ pq g 1/ 4 t + ψ, 145 it follows from 135 that d g t, ψ dt j=1, gj t, ψ dg t, ψ dt 4 t + ψ e t +ψ /4 + e 1/40ω, which implies 136. The proof of Lemma 6.5 is now complete. Lemma 6.6. Suppose that x 1/18ω / max a. Then, for t 1/18 1/3, f 1 t e t / A min{ t, 1} 1 + t 6 e t /4 + ω 6, 146 and f 1 t gt, 0 A min{ t, 1} 4/3 1 + t 6 e t /4 + ω 6, 147 where gt, ψ is de ned as in Lemma 6.5. where Proof. We only prove follows from 147 and 139 with ψ = 0. First assume t 1/4. By Lemma 6.1, we have 1 f 1 t = g t, ψdψ = II 1 t + II t + II 3 t + II 4 t, 148 B n p ψ πω II 1 t = 1 gt, ψdψ, π 1 II t = B n p 1 gt, ψdψ 1 gt, ψdψ, π B n p ψ 1/18 1/3 II 3 t = 1 g t, ψ gt, ψ dψ, B n p ψ 1/18 1/3 II 4 t = 1 g t, ψdψ. B n p 1/18 1/3 ψ πω 38

39 In view of 104, 135, 137 and 139, it is readily seen that II t + II 3 t + II 4 t A 4/3 1 + t 6 e t /4 + A ω In order to estimate II 1 t, write g m t, 0 = Eε p m e itg ε p, m = 1,, 3. We rst note that, by Taylor s expansion of e iz, g t, ψ = g t, 0 + iψ g 1 ψ t, 0 ω ω g t, 0 + i3 ψ 3 g 3 6ω 3 t, 0 + R t, ψ, 150 where R t, ψ 1/4ψ/ω 4 E ε p 4 1/4 pq ψ 4 /ω 4, and g t, 0 = pq + itg Eε p 3 + R 1 t, 151 where R 1 t t g E ε p 4 / pqt g /. By virtue of and the fact that g = 0, e ψ / dψ = π and ψ e ψ / = 0, = 1, 3, we have II 1 t gt, 0 = = 1 π 1 π gt, ψ gt, 0e ψ / dψ [ g t, ψ g t, 0 + ψ /] e ψ +t / dψ = Rt, 15 where Rt 1 π A ω R t, ψ + ψ ω R1 t e ψ +t / dψ 1 + t pq g e t / A 1 4/3 1 + t e t /. Combining 148, 149 and 15, we obtain 147 for t 1/4. ext assume t 1/4. ote that f 1 t gt, 0 = t 0 f 1s g s, 0ds. It suf ces to show that, for t 1/4, f 1t g t, 0 A 4/3 + A ω We continue to use the decomposition of f 1 t in 148. In view of 136 and 138, II 3t + II 4t A 4/3 + A ω 6, for t 1/4. It follows easily from 140, 145 and 149 that, II t A 4/3 + A ω 6, 39

40 for t 1/4. In order to estimate II 1t, we rst note that, as in , dg t, ψ dt dg t, 0 dt = iψ dg 1 t, 0 ω dt ψ ω dg t, 0 + R dt t, ψ, 154 where R t, ψ 1/6 ψ /ω 3 tg E ε p 4 1/6 pq g t ψ 3 /ω 3, and dg t, 0 dt = ig Eε p 3 + R 1t, 155 where R 1 t t g E ε p 4 / pq t g /. It follows from , g = 0, pq g and ψe ψ / = 0 that, for t 1/4, Υ := dg t, ψ dg t, 0 e ψ / dψ dt dt Aω R 1 t + A R t, ψ e ψ / dψ Aω pq g + Aω 3 pq g A. 156 Therefore, by 140, 15 and 156, we have that, for t 1/4, II 1t g t, 0 = 1 π dgt, ψ dt dgt, 0 e ψ / dt t II 1 t gt, 0 + Υ π e t / A 4/3. dψ Combining 148 and all above estimates for II t, = 1,, 3, 4, we obtain 153. The proof of Lemma 6.6 is now complete. Lemma 6.7. Suppose that x 1/18ω / max a. Then, for t 1/18 1/3, f t A1 + t e t /4 + ω 6, 157 ft f 1 t A t 3/ + A t 1 + t e t /4 + ω Proof. We rst prove 157. Write ε = ε ptg +ψ/ω. ote that, by 18, Eν = 0, a = and Taylor s expansion of e iz, E ν e iε Eν e itg ε p 1e iε pψ/ω + Eν e iε pψ/ω 1 tg a + 1Eε p + ψ /ω a + 1Eε p t pq g 3 1/3 a 3 /3 + ψ ω 6 t β 3 ω + ψ ω 6 t + ψ β 3 ω. 40

41 This, together with Lemma 6.1, 135 and the independence of ε, implies that x f t = E ν n ν j e i ε l dψ B n p which yields 157. x n ψ πω ψ πω 1 j 1 j E ν e iε E νj e iε j l=1, j, A xn 1 + t β3 ω e t /4 + ωe 3 1/40ω A1 + t e t /4 + ω 6, g l t, ψ dψ By virtue of 157 and 113, the proof of 158 is simple. Indeed, by 113, we have and hence ft f 1 t it f t = Ee ittn e itλn 1 itλ n B = 0 t 3/ E Λ n 3/ B = 0 A t 3/ x 3/ β3/n A t 3/, ft f 1 t ft f 1 t it f t + t f t A t 3/ + A t 1 + t e t /4 + ω 6, as required. The proof of Lemma 6.7 is now complete. Lemma 6.8. Suppose that x 1/18ω / max a. There exists an absolute constant A such that, for all y 4x, P T + Λ y B = 0 1 Φy + A x e y / + A 4/3. Proof. ote that Lemmas 6.4, 6.6 and 6.7 are similar to Lemmas in Jing, Shao and Wang 003. The proof of Lemma 6.8 is similar to Lemma 10.5 of Jing, Shao and Wang 003 with some routine modi cations. We omit the details. We are now ready to prove Proposition.3. ote that max a ω, h = xpq a 1 /n x max a β 3 /n, 41

42 and x h x. It follows from 10 and Lemma 6.8 that P S n x qv n P T + Λ x h B = 0 1 Φx h + Ax e x h / + A 4/3 1 Φx + A1 + x e x /+x + A 4/3 1 Φx1 + Ax e x + A 4/3 1 Φx exp{ax 3 β 3 /ω } + A 4/3, xβ 3 /ω where we have used the result: Φx Φx h hφ x h h e x h / e x /+x. This yields Proposition.3. Acnowledgments. The authors would lie to than the referees and the Editor for their valuable comments, which have led to this much improved version of the paper. This research is supported in part by ARC DP REFERECES Babu, G. J. and Bai, Z. D Mixtures of global and local Edgeworth expansions and their applications. J. Multivariate. Anal Babu, G. J. and Singh, K Edgeworth expansions for sampling without replacement from nite populations. J. Multivariate. Anal Bicel, P. J. and van Zwet, W. R Asymptotic expansions for the power of distributionfree tests in the two-sample problem. Ann. Statist Bielis, A On the estimation of the remainder term in the central limit theorem for samples from nite populations. Studia Sci. Math. Hungar in Russian. Bloznelis, M A Berry-Esseen bound for nite population student s statistic. Ann. Probab Bloznelis, M. 000a. One and two-term Edgeworth expansion for nite population sample mean. Exact results, I. Lith. Math. J Bloznelis, M. 000b. One and two-term Edgeworth expansion for nite population sample mean. Exact results, II. Lith. Math. J

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44 Sinica. Ser. A Zhao, L.C. and Chen, X. R ormal approximation for nite population U-statistics. Acta Math. Appl. Sinica Zhishui Hu John Robinson, Qiying Wang Department of Statistics and Finance School of Mathematics and Statistics F07 University of Science and Technology of China University of Sydney SW 006 Hefei 3006, China Australia huzs@ustc.edu.cn johnr@maths.usyd.edu.au qiying@maths.usyd.edu.au 44