Edgeworth expansions for a sample sum from a finite set of independent random variables

Transcription

1 E l e c t r o n i c J o u r n a l o f P r o b a b i l i t y Vol ), Paper no. 52, pages Journal URL Edgeworth expansions for a sample sum from a finite set of independent random variables Zhishui Hu Department of Statistics and Finance, University of Science and Technology of China, Hefei, , China huzs@ustc.edu.cn John Robinson School of Mathematics and Statistics, The University of Sydney, SW 2006, Australia, johnr@maths.usyd.edu.au Qiying Wang School of Mathematics and Statistics, The University of Sydney, SW 2006, Australia, qiying@maths.usyd.edu.au URL: Abstract Let {X 1,,X } be a set of independent random variables, and let S n be a sum of n random variables chosen without replacement from the set {X 1,,X } with equal probabilities. In this paper we give a one-term Edgeworth expansion of the remainder term for the normal approximation of S n under mild conditions. Key words: Edgeworth expansion, finite population, sampling without replacement. AMS 2000 Subject Classification: Primary 60F05, 60F15; Secondary:62E20. Submitted to EJP on October 16, 2006, final version accepted September 24, This research is supported in part by an Australian Research Council ARC) discovery project. 1402

2 1 Introduction and main results Let {X 1,, X } be a set of independent random variables, µ = EX, 1. Let R = R 1,, R ) be a random vector independent of X 1,, X, such that PR = r) = 1/! for any permutation r = r 1,, r ) of the numbers 1,,, and put S n = n j=1 X R j, 1 n, that is for a sum of n random variables chosen without replacement from the set {X 1,, X } with equal probabilities. In the situation that X = µ, 1, are nonrandom) real numbers, the sample sum S n has been studied by a number of authors. The asymptotic normality was established by Erdös and Rényi 1959) under quite general conditions. The rate in the Erdös and Rényi central limit theorem was studied by Bielis 1969) and later Höglund 1978). An Edgeworth expansion was obtained by Robinson 1978), Bicel and van Zwet 1978), Schneller 1989), Babu and Bai 1996) and later Bloznelis 2000a, b). Extensions to U-statistics and, more generally, symmetric statistics can be found in andi and Sen 1963), Zhao and Chen 1987, 1990), Koic and Weber 1990), Bloznelis and Götze 2000, 2001) and Bloznelis 2003). In contrast to rich investigations for the case X = µ, 1, are nonrandom) real numbers, there are only a few results concerned with the asymptotics of general S n discussed in this paper. von Bahr 1972) showed that the distribution of S n / V ars n may be approximated by a normal distribution under certain mild conditions. The rate of the normal approximation has currently been established by Zhao, Wu and Wang 2004), in which the paper improved essentially earlier wor by von Bahr 1972). Along the lines of Zhao, Wu and Wang 2004), this paper discusses Edgeworth expansions for the distribution of S n / V ars n. Throughout the paper, let γ 12 = 1 EX EX 2, =1 α j = 1 EX ) j, =1 β j = 1 EX j ), =1 for j = 1, 2, 3, 4, and p = n/, q = 1 p, b = 1 pα 2. Theorem 1. Suppose that α 1 = 0 and β 2 = 1. Then, for all 1 n <, sup PS n / nb x) G n x) x C 1n + nq) 1) + 3 nq log nq) exp { } nq δ, 1) where C is an absolute constant, G n x) = Φx) β 3 3pγ p 2 α 3 6 nb 3/2 Φ x), with Φx) being a standard normal distribution, 1n = nb) 1 α 4 + nb2 ) EX pex ) 4, =1

3 and where L 0 = 1 δ = 1 sup δ 0 b/9l 0 ) t 16 nb 1 =1 Ee itx E X 3 and δ 0 is so small that 192δ δ 0 1 cos1/16). Property 1) improves essentially a result of Mirahmedov 1983). The related result in Theorem 1 of Mirahmedov 1983) depends on max 1 EX 4. ote that it is frequently the case that 1 =1 EX4 is bounded, but max 1 EX 4 tends to. Also note that Corollary 1 of Mirahmedov 1983) requires lim t Ee itx ǫ < 1. This condition is quite restrictive since it taes away the most interesting case that the X are all degenerate. When X = µ, 1, are nonrandom) real numbers, it is readily seen that α 2 = 1, b = q, α 3 = β 3 = γ 12 = 1 =1 µ3,, 1 + nq) 1 3 nq) 1 1 µ 4. =1 In this case, the property 1) reduces to sup PS n / nq x) G 1n x) x C nq) 1 1 µ nq log nq) exp { } nq δ, =1 where G 1n = Φx)+ p q 6 1 nq =1 µ3 Φ x), which gives one of main results in Bloznelis 2000a, b). We next give a result complementary to Theorem 1. The result is better than Theorem 1 under certain conditions such as some of the X s are non-degenerate random variables and q is close to 0. Theorem 2. Suppose that α 1 = 0 and β 2 = 1. Then, for all 1 n, sup PS n / nb x) G n x) C 2n + 3 n log n exp { } n δ 1, 2) x where C is an absolute constant, G n x), L 0 and δ 0 are defined as in Theorem 1, 2n = nb 2 ) 1 β 4 and 1 δ 1 = 1 sup δ 0 b/9l 0 ) t 16 nb Ee itx. =1 In the next section, we prove the main results. Throughout the paper we shall use C, C 1, C 2,... to denote absolute constants whose value may differ at each occurrence. Also, IA) denotes the indicator function of a set A, A) denotes the number of elements in the set A, denotes =1, and denotes =1. The symbol i will be used exclusively for

4 2 Proofs of Theorems Let µ = EX and Ψt) = E exp{its n / nb}. Recall α 1 = 0 and β 2 = 1. As in 4) of Zhao, Wu and Wang 2004), Ψt) = [B n p)] 1 Eρ ψ, t)dψ, 3) ψ π nq where B n p) = 2πnqG n p), G n p) = 2πC n pn q n, X = X pµ and { ρ ψ, t) = q exp ipψ ipµ } { } t iqψ + p exp + itx. nq nb nq nb The main idea of the proofs is outlined as follows. We first provide the expansions and the basic properties for Eρ ψ, t) in Lemmas 1 4. In Lemma 5, the idea in von Bahr 1972) is extended to give an expansion of Ψt) for the case n/ 1/2. The proofs of Theorems 1 and 2 are finally completed by virtue of the classical Esseen s smoothing lemma. In the proofs of Lemmas 1 4, we assume that 1n < 1/16 and nq > 256, where 1n is defined as in Theorem 1. Throughout this section, we also define, hψ, t) = Eρ ψ, t) and gψ, t) = where fψ, t) = A 0 ψ 3 + 3A 1 ψt 2 + A 2 t 3, with 1 + i3 fψ, t) 6 ) e ψ2 +t 2 )/2, A 0 = q p pq, A 1 = 1 2pα 2) pq, A 2 = β 3 3pγ p 2 α 3 pb p 1/2 b 3/2. 4) Lemma 1. For ψ nq) 1/4 /4 and t 1/4 1n /4, we have where s 2 = ψ 2 + t 2. hψ, t) gψ, t) C[ 1n + nq) 1 ]s 4 + s 8 )exp{ s 2 /3}, 5) Proof. Define a sequence of independent random vectors U, V ), 1, by the conditional distribution given X as follows: PU = p/ pq, V = pµ / pb X ) = q, PU = q/ pq, V = X / pb X ) = p. Let W = ψu + tv. As in Lemma 1 of Zhao, Wu and Wang 2004), tedious but simple calculations show that EW = 0, EW 2 = ψ2 + t 2 ), = fψ, t), EW 4 8 EW 3 Eψ 4 U 4 + t4 V 4 ) 82 ψ 4 /nq) + t 4 1n ). 6) 1405

5 Furthermore, if we let B 2 n = EW 2, L j = E W j/ B j n, j = 3, 4, then L 2 3 = E W 3) 2 /B 6 n L 4 8nq) 1 + 1n ), 7) and whenever ψ nq) 1/4 /4 and t 1/4 1n /4, s := ψ 2 + t 2 L 1/4 4 ow, by recalling that W are independent r.v.s and noting that [ 8ψ 4 /nq) + t 4 1n ) ] 1/4 1/4 L 4 /2. 8) hψ, t) = Ee i P W / = Ee i s P W /B n, it follows from 7)-8) and the classical result see, for example, Theorem 8.6 in Bhattacharya and Ranga Rao 1976)) that, for ψ nq) 1/4 /4 and t 1/4 1n /4, hψ, t) 1 + i3 ) 6 fψ, t) e s2 /2 = hψ, t) 1 + i3 s 3 ) EW 3 e s2 /2 6B 3 n CL 4 + L 2 3)s 4 + s 8 )e s2 /3 C[ 1n + nq) 1 ]s 4 + s 8 )exp{ s 2 /3}. This proves 5) and hence completes the proof of Lemma 1. Lemma 2. For ψ nq) 1/4 /4 and t 1/4, we have d hψ, t) d gψ, t) C1 + ψ 12 ) nq) 1 ) + 1n e ψ 2 /4. 9) Proof. We first show that if ψ nq) 1/4 /4 and t 1/4, then Λψ, t) := d hψ, t) + t + i ) 6 d fψ, t) hψ, t) C1 + ψ 4 ) nq) 1 ) + 1n hψ, t). 10) To prove 10), define U, V ) and W = ψu + tv as in Lemma 1. Recall that hψ, t) = E exp{i W / }. It is readily seen that d hψ, t) = i hψ, t) I, 11) [ where I = E exp{iw / ] 1 [ } E V exp{iw / ] }. Recall nq > 256. It follows from 19) and 20) in Zhao, Wu and Wang 2004) that for ψ nq) 1/4 /4 and t 1/4, [ E exp{iw / ] 1 } = 1 + θ1 1 ψ 2 + t 2 EV 2 ), 12) 1406

6 where θ 1 1 and 1 ψ 2 +t 2 EV 2) 1/4. This, together with Taylor s expansion of eix, yields that recall EV = 0) I i EV W EV W 2 3 3/2E V W 3 + ψ2 + t 2 EV 2 1 EV W + 1 ) 2 EV W 2. 13) As in the proof of 6), for t 1/4, E V W 3 C1 + ψ 3 ) EU 4 + V 4 ) C1 + ψ 3 ) 2 [ nq) 1 ] + 1n, 14) ψ 2 + EV 2 ) EV W C1 + ψ 3 ) ψ 2 + EV 2 ) EV W 2 C1 + ψ4 ) 1 + EV 2 )EU2 + EV 2 ) C1 + ψ 3 ) 1 + EV 4 C1 + ψ 3 ) 2 [nq) 1 + 1n ], 15) C1 + ψ 4 ) [ C1 + ψ 4 ) pq) 1/ EV 2 ) E V 3 + E U 3) pq) 1/2 1 + EV 2 ) + E V 3 + EV 2 E V 3) 1 + EV 4 + )] EV 4 C1 + ψ 4 ) 5/2 [ nq) 1 + 1n ], 16) where, in the proof of 16), we have used the estimates: V V 4 and ow 10) follows from 11), 13)-16) and EV W = t, EV 2 E V 3 EV 2 )1/2 EV 4 EV 4. EV W 2 = 2A 1ψt + A 2 t 2 ) = 3 d fψ, t). 17) We next complete the proof of Lemma 2 by virtue of 10) and Lemma 1. We first notice that, by 6), for all ψ and t, fψ, t) 1 E W 3 1 ) 1/2 EW 2 EW 4 3 ψ 2 + t 2 ) 3/2 1n + nq) 1 ) 1/2, 18) and similarly by 17), for all ψ and t, d fψ, t) 3 E V W 2 9 ψ 2 + t 2 ) 3/2 1n + nq) 1 ) 1/2. 19) 1407

7 It follows from 18) and Lemma 1 that for ψ nq) 1/4 /4 and t 1/4, Therefore, by noting hψ, t) e ψ2 +t 2 )/2 C 1n + nq) 1 ) 1/2 e ψ2 /4. 20) d gψ, t) simple calculations show that d hψ, t) d gψ, t) = tgψ, t) i 6 d fψ, t) e ψ2 +t 2 )/2, Λψ, t) + t hψ, t) gψ, t) d fψ, t) hψ, t) +t 2 )/2 e ψ2 C1 + ψ 12 ) nq) 1 + 1n ) e ψ 2 /4, where we have used 5), 10), 19) and 20). The proof of Lemma 2 is now complete. Lemma 3. Assume that ψ nq) 1/4 /4. Then, for 1/4 1n /4 t ) 1 /16, where hψ, t) C 1n e ψ2 +t 2 )/4, 21) = 1 µ 3 / nb + 1 E X pµ 3 / nb 3/2 ). Assume that nq) 1/4 /4 ψ π nq. Then, hψ, t) Cnq) 4, 22) for all t δ 0 ) 1, where δ 0 is so small that 192δ δ 0 1 cos1/16). If in addition t 1/4, then we also have d hψ, t) Cnq) 4. 23) Proof. The proof of this lemma follows directly from an application of Lemmas 1-3 in Zhao, Wu and Wang2004). The choice of δ 0 can be found in the proof of Lemma 2 in Zhao, Wu and Wang2004). We omit the details. Lemma 4. Assume that n/ 1/2 and 2n nq) 1 /25, where 2n is defined as in Theorem 2. Then, for t 1 15 nb 3/2 /L 0, where L 0 = 1 E X 3 is defined as in Theorem 1. hψ, t) exp{ Ct 2 }, 24) 1408

8 Proof. We only need to note that the condition 2n nq) 1 /25 implies that 5qα 2 5qβ 1/2 4 b = 1 VarX ) + qα 2, 1 that is, VarX ) 4/5)b. Then 24) is obtained by repeating the proof of Lemma 4 in Zhao, Wu and Wang2004). Lemma 5. Assume that n/ 1/2 and 2n 1. Then, for u 1 16 min{ n/β 4 ) 1/4, 1 8 b } n/l 0, E exp{ius n / n} exp{ bu 2 /2} 1 + i3 u 3 b 3/2 6 A 2 Cn 1 β 4 u 2 + u 4 + u 6 b) exp{ 0.3 bu 2 }, 25) where L 0 = 1 E X 3 is defined as in Theorem 1. Proof. Write, for 1, f u) = E exp{iux / n}, b u) = exp{u 2 /2n)}f u) 1 and B j = 1) j+1 /j) =1 bj u) for 1 j n. As in von Bahr 1972), we have exp{u 2 /2}E exp{ius n / n p j B j ) i j n} = C P i j!,n, n j=1 ji, 26) j i j 0, 1 j n j=1 ) where C,n,r = { r ) / n r p r n) ), r n; 0, r > n. In view of 28) of Zhao, Wu and Wang 2004), for r > 0, C,n,r 1, and for n 4 and r n, To prove 25) by using 26), we need some preliminary results. C,n,r 1 r 2 /n. 27) Write β j = EX j, j = 2, 3, 4. Recall that 1 β 2 = 1. We have that β 4 1 and by Taylor s expansion, for u 1 16 n/β 4) 1/4, exp{u 2 /2n)} = n u n 2u4 + θ 4 n 3u6, where θ 4 1/24, f u) = 1 + iu µ u2 n 2n β 2 iu3 6n 3/2β 3 + θ 5u 4 n 2 β 4, where θ 5 1/24. ow, by noting that µ β 1/ β 4, β 2 β 1/ β 4 and β 3 β 3/ β 4, we obtain that, for u 1 16 n/β 4) 1/4, b u) = exp{u 2 /2n)}f u) 1 = iu n µ + u2 2n 1 β 2) + iu3 6n 3/23µ β 3 ) + R 1 u). 28) 1409

9 where R 1 u) 1 + β 4 )u 4 /n 2. Furthermore, by noting that β 1/4 4 u / n β 4 ) 1/4 u / n 1/8 since n/ 1/2 and u 1 16 n/β 4) 1/4, we have b 2 u) = u2 n µ2 + iu3 n 3/2µ 1 β 2 ) + R 2 u), 29) b 3 iu3 u) = n 3/2µ3 + R 3u), 30) b j u) 31/2)j β 4 )u 4 /n 2, for j 4, 31) where R 2 u) 31 + β 4 )u 4 /n 2 and R 3 u) 41 + β 4 )u 4 /n 2. Recalling that µ = 1 β 2) = 0, it follows from 28) 31) that, for u 1 16 n/β 4) 1/4, pb 1 = p p 2 B 2 = p 2 /2) p 3 B 3 = p 3 /3) b u) = iu3 6 n β 3 + θ 6 β 4 u 4 /n, 32) b 2 u) = u2 p 2 α 2 + iu3 p 2 n γ 12 + θ 7 β 4 u 4 /n, 33) b 3 u) = ip2 u 3 3 n α 3 + θ 8 β 4 u 4 /n, 34) p j B j b j u) 6β 4u 4 1/2) j 4 /n, for j 4. 35) where θ 6 2, θ 7 3 and θ 8 3. By virtue of 35), it is readily seen that, for u 1 16 n/β 4) 1/4, n p j B j 12β 4 u 4 /n, 36) j=4 oting that α 3 + β 3 + γ 12 3L 0 and recalling that 2n = nb 2 ) 1 β 4 1, it follows easily from 32) 34) and 36) that, for u 1 16 min{ n/β 4 ) 1/4, 1 8 b } n/l 0, n p j B j = 1 2 p α 2 u 2 u 3 L 0 u 4 β + θ θ 10 n n j=1 = 1 2 p α 2 u 2 + θ 11 bu 2, 37) where θ 9 3, θ and θ Also, if we let Lu) = 3 j=1 pj B j pα 2 u 2 /2, we have Lu) 0.2bu 2, Lu) + i u3 b 3/2 6 A 2 8β 4u 4 /n, 38) where A 2 is defined as in 4). As in the proof of 6), we may obtain A 2 2 = ) 2 1 EV 3 2 EV 2 EV 4 1n 17β 4 /nb 2 ). 1410

10 This together with 38) yields, for u 1 16 n/β 4) 1/4, [ u L 2 3 b 3/2 u) 6 A n β 4u 4] 2 u 6 bβ 4 /n + 128β4u 2 8 /n 2 β 4 u 4 + u 6 b)/n. 39) We are now ready to prove 25) by using 26). Rewrite 26) as exp{u 2 /2}E exp{ius n / n} = I 1 + I 2 + I 3, 40) where I 1 = n p j B j ) i j C P i j!,n, n j=1 ji, j I 2 = I 3 = j=1 i j 0, 1 j 3 j=1 i j 0, 1 j 3 j=1 3 p j B j ) i j i j! 3 p j B j ) i j, i j! C,n, P 3j=1 jij 1 ), where the summation in the expression of I 1 is over all i j 0, j = 1, 2, 3 and i j > 0 for at least one j = 4,, n. As in Mirahmedov 1983), it follows from 36)-37) that { 3 } I 1 exp p j B j exp j=1 { n j=4 n { n } p j B j exp p j B j j=4 j=1 Cn 1 β 4 u 4 exp{pα 2 u 2 / bu 2 }. As for I 2, it follows easily from 27) and 37) that I 2 Cn 1 Cn 1 i j 0, 1 j 3 j=1 i j 0, 1 j 3 j=1 } ) p j B j 1 3 p j B j i j 3 ) 2 ji j i j! 3 i 2 j pj B j i j i j! j=1 { 3 } 3 Cn 1 exp p j B j p j B j + p j B j 2) j=1 j=1 Cn 1 β 4 u 2 + u 4 )exp{pα 2 u 2 / bu 2 }. 1411

11 We next estimate I 3. Recalling that b = 1 pα 2 and noting that I 3 = e P 3 j=1 pj B j, we have I 3 e u2 /2 1 + i3 u 3 b 3/2 ) 6 A 2 e bu2 /2 e bu2 /2 e Lu) 1 Lu) + e bu2 /2 Lu) i3 u 3 b 3/2 6 A 2 [ 1/2)L 2 u)e Lu) + 8β 4 u 4 /n ] e bu2 /2 C n 1 β 4 u 4 + u 6 b)e 0.3 bu2, where Lu) = 3 j=1 pj B j pα 2 u 2 /2 and we have used 38)-39). Combining 40) and all above facts for I 1 -I 3, we obtain E exp{ius n/ ) n} exp{ bu 2 /2} 1 + i3 u 3 b 3/2 6 A 2 exp{ u 2 /2} I 1 + I 2 ) + I 3 e u2 /2 1 + i3 u 3 b 3/2 ) 6 A 2 C n 1 β 4 u 2 + u 4 + u 6 b) exp{ 0.3bu 2 }, e bu2 /2 which implies 25). The proof of Lemma 5 is now completed. After these preliminaries, we are now ready to prove the theorems. Proof of Theorem 1. Without loss of generality, assume that nq > 256 and 1n < 1/16. Write T 1 = 1n + nq) 1 and g n t) = 1 + i3 t 3 A ) 2 6 exp{ t 2 /2}, where A 2 is defined as in Lemma 1. We shall prove, i) if t 1/4, then Ψt) gn t) C t T 1 ; 41) ii) if t δ 0 ) 1, where δ 0 and are defined as in Lemma 3, then Ψt) g n t) C T t 8 )e t2 /4 + C nq) 3 ; 42) iii) if δ 0 ) 1 t T, then Ψt) gn t) C T 1 e t2 /4 + 3 nq exp { nq δ }, 43) where δ is defined as in Theorem 1. ote that A 2 /4 by 1n 1/16 and the last second inequality of 39). We have m sup x G nx) C1 + 1/2 A 2 ) 2C. So, by virtue of 41)-43) and Esseen s smoothing 1412

12 lemma, simple calculations show that supps n / nb x) G n x) x + t 1/4 1/4 t T 1 + T 1 t T Ψt) g n t) t C 1n + nq) 1 ) + 3 nq lognq) exp { nq δ }, where T 1 = min{δ 0 ) 1, T }, which implies 1) and hence Theorem 1. We next prove 41)-43). Throughout the proof, we write s 2 = ψ 2 + t 2. + CmT 1 Consider 42) first. ote that g n t) = 1 2π gψ, t)dψ. It is readily seen that Ψt) g n t) II 1 + II 2 + II 3 + II 4, 44) where II 1 = [B n p)] 1 II 2 = [B n p)] 1 II 3 = [B n p)] 1 II 4 = ψ nq) 1/4 /4 nq) 1/4 /4 ψ π nq ψ nq) 1/4 /4 hψ, t) gψ, t) dψ, 1 + [B n p)] 1 2π) 1/2 hψ, t) dψ, fψ, t) 6 fψ, t) 1 + ) e s2 /2 dψ, 6 ) e s2 /2 dψ. To estimate II j, j = 1, 2, 3, 4, we first recall that, by 18), for all ψ and t, and by virtue of Stirling s formula, fψ, t) 3s 3 1n + nq) 1 ) 1/2 s 3, 45) In view of 45) and 46), it is readily seen that By using 22), we have 1 2π/B n p) 1 + 1/nq. 46) II 3 + II 4 C nq) t 6 )e t2 /3. 47) As for II 1, if t min{ 1/4 1n /4, δ ) 1 }, Lemma 1 implies that if 1/4 1n II 2 C nq) 3. 48) II 1 C 1n + nq) 1 )1 + t 8 )e t2 /4 ; 49) /4 t δ ) 1, then it follows from 21) and 45) that fψ, t) II 1 hψ, t) dψ ψ nq) 1/4 /4 ψ nq) 1/4 /4 ) e s2 /2 dψ C 1n e t2 /4 + C t 3 )e t2 /2 C 1n e t2 /4. 50) 1413

13 Taing 47)-50) into 44), we obtain the required 42). Secondly we prove 41). Recall that g n t) = 1 2π gψ, t)dψ. As in 44), we have d Ψt) dg nt) III 1 + III 2 + III 3 + III 4, 51) where III 1 = [B n p)] 1 d hψ, t) d gψ, t) ψ nq) 1/4 /4 dψ, III 2 = [B n p)] 1 d hψ, t) nq) 1/4 /4 ψ π nq dψ, III 3 = [B n p)] 1 t fψ, t) t + ψ nq) 1/4 / d fψ, t) ) e s2 /2 dψ, III 4 = [B n p)] 1 2π) 1/2 t fψ, t) t d fψ, t) ) e s2 /2 dψ. By 18)-19) and 46), we have that for t 1/4 By 9), 23) and 46), we have that for t 1/4 III 3 + III 4 C 1n + nq) 1 ). 52) III 1 + III 2 C 1n + nq) 1 ). 53) Taing these estimates into 51), we obtain for t 1/4, Ψt) g n t) t sup d Ψx) dg nx) dx dx C t 1n + nq) 1 ), which yields 41). x 1/4 Finally we prove 43). We first notice that 1n 1/16nb). Indeed, if α 2 1/4, then 1n 1 EX pµ ) 2) 2 /nb 2 ) 1 2pα 2 ) 2 /nb 2 ) 1/16nb), and if α 2 > 1/4, then 1n 1 µ 2 )2 /nb) = α2 2 /nb) 1/16nb). This, together with the fact that 9 nb 3/2 1 E X 3, 1414

14 implies that if δ 0 ) 1 t T, then δ 0 b/9l 0 ) t / nb 16 nb and hence hψ, t) 2 1 2pq 1 E cosψ/ nq + tx / ) ) nb) { exp 2pq 1 E cosψ/ nq + tx / ) } nb) { exp 2pq 1 1/)E exp{iψ/ nq + itx / )} nb} { exp 2pq 1 1/) E exp{itx / )} nb} exp{ 2nq δ }. 54) We also note that 2 1/2 1n and this together with 45) implies that, for δ 0 ) 1 t T, fψ, t) ) g n t) e s2 /2 dψ C1 + t 3 )e t2 /2 C 1n e t2 /4. 55) Combining 54) and 55) and using the estimate 46), we obtain that, for δ 0 ) 1 t T, Ψt) g n t) [B n p)] 1 hψ, t) dψ + g n t) ψ π nq C 1n e t2 /4 + 3 nq exp { nq δ }, which yields 43). The proof of Theorem 1 is complete. Proof of Theorem 2. Without loss of generality, assume 2n 1. We first prove the property 2) for n/ 1/2 and 2n nq) 1 /25. Write T = nb 2 )/β 4, T1 = b1/2 16 min{ n/β 4 ) 1/4, 1 8 b } n/l 0 and T 2 = 15 1 nb 3/2 /L 0. As in the proof of Theorem 1, it follows from Esseen s smoothing lemma that sup x PS n / nb x) G n x) + + Ψt) g n t) + C 2n t T1 T1 t T 2 T2 t T t = Λ 1n + Λ 2n + Λ 3n + C 2n, say. 56) By virtue of Lemma 5, simple calculations show that Λ 1n C 2n. Recall 1n 17 2n. Applying Lemma 4 and similar arguments as in the proof of 50), we obtain that Λ 2n C 2n and also g nt) T2 t T t C 2n. Therefore, to prove 2), it remains to show that, for T2 t T, Ψt) 3 n exp { n δ 1 }. 57) In fact, by using 3) in Zhao, Wu and Wang 2004), for q > 0, Ψt) = E exp{its n / nb} = 2πG n p)) 1 π π e inψ q + pe iψ+itx j / nb ) dψ, 1415

15 where G n p) = 2πC n pn q n. This, together with the fact that δ 0 b/5l 0 ) t / nb 16 nb whenever T 2 t T, implies that q + p Ee itx j / nb ) Ψt) 2πG n p)) 1 { 2πG n p)) 1 exp p Ee itx j / nb 1 )} 3 n exp{ nδ 1 }, where we have used the inequality π/2 nqg n p) < 1 see, for instance, Lemma 1 in Höglund1978)). This proves 57) for q > 0. If q = 0, then = n hence by the independence of X, Ψt) = Ee itx j/ { nb exp Ee itx j / nb 1 )} exp{ nδ 1 }. This implies that 57) still holds for q = 0. We have now completed the proof of 57) and hence 2) for n/ 1/2 and 2n nq) 1 /25. ote that β 4 1, 1n 17 2n and b q 1/2 if n/ 1/2. We have that 1n + nq) n, whenever n/ 1/2 or 2n > nq) 1 /25. Based on this fact, by using a similar argument to that above and that in the proof of Theorem 1, we may obtain 2) for n/ 1/2 or 2n > nq) 1 /25, as well. The details are omitted. The proof of Theorem 2 is now complete. References [1] Babu, G. J. and Bai, Z. D. 1996). Mixtures of global and local Edgeworth expansions and their applications. J. Multivariate. Anal MR [2] von Bahr, B. 1972). On sampling from a finite set of independent random variables. Z. Wahrsch. Verw. Geb MR [3] Bhattacharya, R.. and Ranga Rao, R. 1976). ormal approximation and asymptotic expansions. Wiley, ew Yor. MR [4] Bicel, P. J. and von Zwet, W. R. 1978). Asymptotic expansions for the power of distribution-free tests in the two-sample problem. Ann. Statist MR [5] Bielis, A. 1969). On the estimation of the remainder term in the central limit theorem for samples from finite populations. Studia Sci. Math. Hungar in Russian. MR [6] Bloznelis, M. 2000a). One and two-term Edgeworth expansion for finite population sample mean. Exact results, I. Lith. Math. J. 403) MR [7] Bloznelis, M. 2000b). One and two-term Edgeworth expansion for finite population sample mean. Exact results, II. Lith. Math. J. 404) MR

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