The main finding of this note is an improvement of the ChenGoldsteinShao proof of the BerryEsseen bound for the combinatorial central limit theorem. A bound of the correct order in terms of thirdmoment type quantities with a small explicit constant is obtained. Moreover, our approach does not need to use a truncation step as in ChenGoldsteinShao. An example is also given to illustrate the optimality of the bound.
On the Berry-Esseen bound for a combinatorial central limit theorem Th`anh Lˆe Vˇan∗ Abstract The main finding of this note is an improvement of the Chen-Goldstein-Shao proof of the Berry-Esseen bound for the combinatorial central limit theorem. A bound of the correct order in terms of third-moment type quantities with a small explicit constant is obtained. Moreover, our approach does not need to use a truncation step as in Chen-Goldstein-Shao. An example is also given to illustrate the optimality of the bound. Key Words and Phrases: Berry-Esseen bound, combinatorial central limit theorem, zero-bias coupling, Stein’s method. 2010 Mathematics Subject Classifications: 60F05, 60D05. 1 Introduction and result Let n ≥ 2 and A = {aij , 1 ≤ i, j ≤ n} be an array of real numbers. In this note, we study the combinatorial central limit theorem, that is, the central limit theorem for random variables of the form n aiπ(i) , Y = YA = (1.1) i=1 where π is a random permutation with the uniform distribution over the symmetric group of all permutations of {1, . . . , n}. The central limit theorem for YA were proved by Wald and Wolfowitz [17] when the factorization aij = bi cj holds, and by Hoeffding [11] for general arrays. Bounds on the error in the normal approximation were later considered by a number of authors. Ho and Chen [10] used a concentration inequality approach and Stein’s method for exchangeable pairs [16], which yield the optimal rate only under condition that supij |aij | ≤ C. Bolthausen [1] also used Stein’s method with an inductive approach, which obtained a bound of the correct order in terms of third-moment type quantities, but with an unspecified constant. Goldstein [7] employing the zero bias version of Stein’s method obtained bounds with an explicit constant, but in terms of supi,j |ai,j |. Recently, Chen, Goldstein ∗ Department of Mathematics, Vinh University, Nghe An 42118, Vietnam. A part of research of the second author is also supported by the Vietnam Institute for Advanced Study in Mathematics (VIASM) and the Vietnam National Foundation of Sciences and Technology Development (NAFOSTED). Email: levt@vinhuni.edu.vn 1 and Shao [5, Theorem 6.2] used the zero bias variation of Ghosh [6] on the inductive method in Bolthausen [1] to prove a bound depending on a third moment type quantity of the matrix, but with an unspecified constant like Bolthausen [1]. In this note, we give an improvement of the Chen-Goldstein-Shao proof and obtain a bound of the correct order in terms of third-moment type quantities with a constant c = 90. Moreover, our approach do not need to use the truncation step as in [1, 5, 6]. We also give an example to illustrate the optimality of the bound. As far as we are aware, on bound depending on a third moment type quantity of the matrix, the best absolute constant is c = 447 which was obtained very recently by Chen and Fang [4]. (Neammanee and Suntornchost [15] obtained a constant c = 198. However, Chen and Fang [4] showed that the proof in [15] is incorrect.) Both in [4] and [15], the authors used the concentration inequality approach and method of exchangeable pairs, and considered the case where the elements of A are independent random variables. 2 We denote the mean and variance of YA by µA and σA , and use the following notation. ai. = 1 n n aij , 1 ≤ i ≤ n, a.j = j=1 1 n It is known that n aij , 1 ≤ j ≤ n, and a.. = i=1 n ai. = i=1 aij . i,j=1 a.π(i) , (1.2) i=1 n 2 = σA n n µA = na.. = and 1 n2 n 1 1 (a2 − a2i. − a2.j + a2.. ) = (aij − ai. − a.j + a.. )2 . n − 1 i,j=1 ij n − 1 i,j=1 (1.3) From (1.2), we automatically get that n (aiπ(i) − ai. − a.π(i) + a.. ). YA − EYA = (1.4) i=1 We also denote WA = (YA − µA )/σA and βA = n i,j=1 |aij − ai. − a.j + a.. |3 3 σA . (1.5) 1 x Throughout this note, Z is the standard normal random variable, Φ(x) = √ exp(−t2 /2)dt −∞ 2π is the distribution function of Z. For n ≥ 1, let Sn denote the symmetric group of all permutations of {1, . . . , n}, and let π denote a random permutation with the uniform distribution over Sn . For a set S, the indicator function of S is denoted by 1(S) and the cardinal of S is denoted by |S|. The following theorem will be proved in this note. Theorem 1.1. We have sup |P (WA ≤ x) − Φ(x)| ≤ x∈R 2 90βA . n (1.6) Let aij = aj for 1 ≤ i ≤ m and aij = 0 for m < i ≤ n, where 1 ≤ m < n and A = {a1 , . . . , an } is a set. Then YA = X1 + · · · + Xm , where {X1 , . . . , Xm } is a random sampling without replacement of size m from A. In this case, it is easy to see that with a ¯= m(n − m)[(n − m)2 + m2 ] βA = n n4 (V arX)3/2 n j=1 |aj − a ¯ |3 ≤ n j=1 aj /n, m(n − m) n j=1 |aj − a ¯ |3 n2 (V arX)3/2 . Therefore, we have the following corollary. Corollary 1.2. Let X = X1 + · · · + Xm , where {X1 , . . . , Xm } is a random sampling without replacement of size m from A = {a1 , . . . , an }. Then n 90m(m − n) j=1 |aj − a ¯ |3 X − EX sup P √ ≤ x − Φ(x) ≤ . n2 (V arX)3/2 V arX x∈R H¨ oglund [12] proved this corollary with the same bound but without an explicit constant by Fourier analysis. More recently, Goldstein [8] obtained the Wasserstein distance to the normal distribution, and Hu, Robinson, and Wang [13] proved the Cram´er-type large deviations for X. 2 Proof In view of (1.4) and (1.5), we may replace aij by aij − ai. − a.j + a.. σA 2 = 1. and assume a.. = ai. = a.j = 0, σA It was shown by Goldstein and Reinert [9] that for any mean zero random variable W with finite variance σ 2 , there exists a random variable W ∗ which satisfies EW f (W ) = σ 2 Ef (W ∗ ) for all absolutely continuous f with E|W f (W )| < ∞. We say that such a W ∗ has the W -zero biased distribution. Before present the proof of the theorem, we recall the zero-bias coupling construction for Y = YA in Goldstein [8] as follows. Choose I † , J † , K † , L† independently of the remaining random variables, with distribution P (I † = i, J † = j, K † = k, L† = l) = (aik + ajl − ail − ajk )2 . 4n2 (n − 1) For 1 ≤ i, j ≤ n, let τij be the permutation which transposes i and j. Set if L† = π(I † ), K † = π(J † ), πτπ−1 (K † ),J † π † = πτπ−1 (L† ),I † if L† = π(I † ), K † = π(J † ), πτπ−1 (K † ),I † τπ−1 (L† ),J † otherwise , and π ‡ = π † τI † ,J † . Then {π † (I † ), π † (J † )} = {π ‡ (I † ), π ‡ (J † )} = {K † , L† }. Let I = {I † , J † , π −1 (K † ), π −1 (L† )}, 3 and let Y † and Y ‡ be random variables given by (1.1) with π replaced by π † and π ‡ , respectively. Then, with U is the uniform distribution on [0, 1] independent of the remaining random variables, Goldstein [8] showed that Y ∗ = U Y † + (1 − U )Y ‡ (2.1) Y = S + T, Y † = S + T † , Y ‡ = S + T ‡ , (2.2) has Y -zero biased distribution, and where aiπ(i) , T † = aiπ(i) , T = S= i∈I / i∈I aiπ† (i) and T ‡ = i∈I aiπ‡ (i) . (2.3) i∈I Let 2 ≤ l ≤ 4, and D = {dij , 1 ≤ i, j ≤ n − l} be the (n − l) × (n − l) array formed by removing the l rows R ⊂ {1, . . . , n} and l columns C ⊂ {1, . . . , n} from A. Let E = {eij , 1 ≤ i, j ≤ n − l} be a matrix with eij = (dij − di. − d.j + d.. )/σD . 2 It follows that ei. = e.j = e.. = µE = 0, σE = 1, βE = βD . For x > −1 and 0 < α < 1, let hx,α (ω) be the function which is 1 for ω ≤ x and then drops linearly to the value 0 at x + α and is 0 for ω ≥ x + α. Let g(ω) = (ωf (ω)) , where f = fx,α be the unique bounded solution of the Stein equation f (ω) − ωf (ω) = h(ω) − Eh(Z). For an arbitrary random variable X and a ≤ b, we will use the following simple fact b−a P (a ≤ X ≤ b) ≤ √ + 2 sup |P (X ≤ x) − Φ(x)|. 2π x∈R (2.4) The proof of the following lemma is easy, and will be presented in Appendix. Lemma 2.1. Assume that n > 25000 and βA 1 < . n 160 (2.5) 2 σD ≥ 0.93, and βE = βD ≤ 1.153βA . (2.6) Then Proof of Theorem 1.1. For β > 0, set 2 M (β, n) = A ∈ Rn×n : ai. = a.j = 0, σA = 1, βA ≤ β , δ(β, n) = sup |P (WA ≤ x) − Φ(x)|, x ∈ R, A ∈ M (β, n) . 4 If A ∈ M (β, n), then −A ∈ M (β, n) and |P (W−A ≤ x) − Φ(x)| = |P (WA ≥ −x) − (1 − Φ(−x))| = |P (WA < −x) − Φ(−x)|. Therefore δ(β, n) = sup |P (WA ≤ x) − Φ(x)|, x ≥ 0, A ∈ M (β, n) . We then follow the computation in Chen and Shao [2, p. 246] to get δ(β, n) ≤ sup x≥0 1 − (1 − Φ(x)) ≤ 0.55. 1 + x2 (2.7) It suffices to prove that sup β>0 nδ(β, n) ≤ 90 for all n ≥ 2. β (2.8) If 2 ≤ n ≤ 25000, then for all β > 0 and A ∈ M (β, n), it easy to see that (see, e.g., [5, p. 171]) βA (n − 1)3/2 1 ≥ > . n n2 160 (2.9) Combining (2.7) and (2.9), we see that (2.8) holds for 2 ≤ n ≤ 25000. Assuming that n > 25000 and (2.8) holds for all m ≤ n − 1, we will prove that it also holds for n. Fix β > 0, by (2.7), we may assume that β/n < 1/160. Let A ∈ M (β, n) be arbitrary, we will prove that sup |P (WA ≤ x) − Φ(x)| ≤ x≥0 Firstly, we consider x such that 90βA . n (2.10) (1 + x)βA 1 ≥ . In this case, we have n 84 |P (YA ≤ x) − Φ(x)| = |P (YA > x) − (1 − Φ(x))| ≤ max{P (YA > x), 1 − Φ(x)} 1 E(YA + 1)2 , } (1 + x)2 1 + x 2 1 = max{ , } (1 + x)2 1 + x 1.05 89βA ≤ ≤ . 1+x n ≤ max{ (1 + x)βA 1 < . Let h = hx,α be defined earlier and f be the n 84 † solution to the Stein’s equation for this h. Let YA , YA‡ , YA∗ , S, T , T † , T ‡ be written as in (2.1), (2.2) It remains to consider the case 5 and (2.3), we have |Eh(YA ) − Eh(Z)| = |Ef (YA ) − EYA f (YA )| = |E(f (YA ) − f (YA∗ ))| ≤ |E(YA f (YA ) − YA∗ f (YA∗ ))| + |h(YA∗ ) − h(YA )| = E (S + T )f (S + T ) − (S + U T † + (1 − U )T ‡ )f (S + U T † + (1 − U )T ‡ ) 1 + E |YA∗ − YA | α := R1 + R2 . 1 I[x,x+α] (YA + r(YA∗ (2.11) − YA ))dr 0 Let I = (I † , J † , π −1 (K † ), π −1 (L† ), π(I † ), π(J † ), K † , L† ) and B = σ(I, U ). By (2.3), we see that T, T † , T ‡ are measurable with respect to B. Therefore T g(S + u)du R1 = E U T † +(1−U )T ‡ T E B g(S + u)du = E (2.12) U T † +(1−U )T ‡ T E I g(S + u)du (by the independence of U from {I, S}). = E U T † +(1−U )T ‡ Let R = I = {I † , J † , π −1 (K † ), π −1 (L† )}, C = {π(I † ), π(J † ), K † , L† }, l = |I|. Denote YD = n−l i=1 diθ(i) , where θ is a random permutation with the uniform distribution over Sn−l . Since S = i∈I / diπ(i) and π is chosen uniformly from Sn , we have L(S|I = i) = L(YD ) for all i. (2.13) E {I=i} g(S + u) = Eg(YD + u) ≤ 4 for all i and u. (2.14) By (2.13) and Lemma 2.2, Since (2.14) holds for all i, E I g(S + u) ≤ 4 for all u. (2.15) By Theorem 6.1 of Goldstein [8] and that n > 25000, E|YA∗ − YA | ≤ 28 4 βA 8+ + n−1 n − 1 (n − 1)2 ≤ 8.01βA . n (2.16) Combining (2.12), (2.15) and (2.16), we obtain R1 ≤ 4E|T − U T † − (1 − U )T ‡ | = 4E|YA∗ − YA | 32.04βA ≤ . n (2.17) Now we bound R2 . By the computation in [5, p. 173-174], R2 = 1 E |YA − YA∗ | α 1 P (S ∈ [x − qr , x + α − qr ]|B)dr , 0 6 (2.18) where qr = rU T † + r(1 − U )T ‡ + (1 − r)T. Since qr is measurable with respect to B for all r, it follows from (2.18) that 1 1 E |YA − YA∗ | sup P (S ∈ [x − u, x + α − u]|B)dr α 0 u∈R 1 = E |YA − YA∗ | sup P (S ∈ [x − u, x + α − u]|B) α u∈R 1 = E |YA − YA∗ | sup P (S ∈ [x − u, x + α − u]|I) α u∈R R2 ≤ (2.19) where the last equality we have used the independence of U from {S, I}. We have sup P (S ∈ [x − u, x + α − u]|I = i) u∈R = sup P (YD ∈ [x − u, x + α − u]) (by (2.13)) u∈R x − u − µD x + α − u − µD = sup P ≤ YE ≤ σD σD u∈R α + 2δ(βE , n − l) (by (2.4)) ≤ √ 2πσD α 180βE ≤√ + (by the inductive hypothesis) n−l 2πσD α 208βA ≤√ + (by Lemma 2.1). n 2πσD (2.20) As the last in (2.20) does not depend on i or u, it implies sup P (S ∈ [x − u, x + α − u]|I) ≤ √ u∈R α 208βA + . n 2πσD (2.21) Combining (2.16), (2.19) and (2.21), we obtain 208βA 1 α √ + E|YA∗ − YA | α n 2πσD 1 208βA 8.01βA α √ ≤ + α n n 2πσD 2 1667βA 3.32βA + . ≤ n αn2 From (2.11), (2.17) and (2.22), we obtain R2 ≤ sup |Ehx,α (YA ) − Ehx,α (Z)| ≤ x>−1 (2.22) 2 35.5βA 1667βA + . n αn2 (2.23) 71βA . n (2.24) √ Now, let α = 19 2πβA /n, we have from (2.23) that sup |Ehx,α (YA ) − Ehx,α (Z)| ≤ x>−1 For x ≥ 0, then x − α > −1. It thus follows from (2.24) that sup |Ehx−α,α (YA ) − Ehx−α,α (Z)| ≤ x≥0 71βA . n (2.25) By (2.24), (2.25) and the definition of h, we have for all x ≥ 0 P (YA ≤ x) − Φ(x) ≤ Ehx,α (YA ) − Ehx,α (Z) + Ehx,α (Z) − Φ(x) ≤ |Ehx,α (YA ) − Ehx,α (Z)| + P (x < Z ≤ x + α) 71βA α 90βA ≤ +√ = , n n 2π 7 (2.26) and P (YA ≤ x) − Φ(x) ≥ Ehx−α,α (YA ) − Ehx−α,α (Z) + Ehx−α,α (Z) − Φ(x) ≥ −|Ehx−α,α (YA ) − Ehx−α,α (Z)| − P (x − α < Z ≤ x) 71βA α 90βA ≥− −√ =− . n n 2π (2.27) 90βA for all x ≥ 0, i.e., (2.10) holds. n Taking the supremum over A ∈ M (β, n) and then taking supremum over β > 0, we conclude that Combining (2.26) and (2.27), we have |P (YA ≤ x) − Φ(x)| ≤ (2.8) holds for n. It remains to present and prove Lemma 2.2 which we have used above. The proof will be presented in Appendix. Lemma 2.2. Let x > −1 and 0 < α < 1 and the inductive hypothesis in the proof of Theorem 1.1 holds. If n > 25000, and (1 + x)βA /n < 1/84, (2.28) Eg(YD + u) ≤ 4 for all u. (2.29) then Example 2.3. Let 1 ≤ k, m < n and A = {a1 , . . . , an } be a set with aj = 0 or 1 and |{aj : aj ∈ A, aj = 1}| = k. Let A = (aij )n×n such that aij = aj for 1 ≤ i ≤ m, 1 ≤ j ≤ n and aij = 0 for m < i ≤ n, 1 ≤ j ≤ n. Then WA is the Hypergeometric distribution with parameters m, k, n. Let p = k/n and f = m/n, we have |aij − ai. − a.j + a.. | = (1 − f )|aj − p| f |aj − p| if 1 ≤ i ≤ m, if m < i ≤ n. 2 = n2 p(1 − p)f (1 − f )/(n − 1), and It is easy to show that µA = mp, σA np(1 − p)f (1 − f )[p2 + (1 − p)2 ][f 2 + (1 − f )2 ] βA = 3 n σA = 1 (n − 1)[p2 + (1 − p)2 ][f 2 + (1 − f )2 ] ≤ . nσA σA By Theorem 2.2 in Lahiri and Chatterjee [14], there exists a constant c0 > 0 such that sup P x∈R WA − µA c0 c0 βA ≤ x − Φ(x) ≥ ≥ . σA σA n Thus, the bound in (1.6) is optimal. A Appendix In this Section, we will prove Lemma 2.1 and Lemma 2.2. 8 2 Proof of Lemma 2.1. Firstly, we estimate σD . By (6.50) and (6.51) in [5], we get n−l |di. |3 ≤ i=1 n−l l 2 βA , (n − l)3 l 2 βA 5l2 βA 3 , |d | . ≤ .. (n − l)3 (n − l)4 |d.j |3 ≤ j=1 (A.1) Therefore n−l n−l d2i. ≤ (n − l)1/3 ( i=1 2/3 |di. |3 )2/3 ≤ l4/3 βA , (n − l)5/3 |d.j |3 )2/3 ≤ l4/3 βA . (n − l)5/3 i=1 and n−l n−l d2.j ≤ (n − l)1/3 ( j=1 2/3 i=1 From (1.3), we have 2 σD = n−l n−l 1 n−l−1 j=1 i=1 i,j d2.j + (n − l)2 d2.. . d2i. − (n − l) d2ij − (n − l) It thus follows that 2 σD ≥ n i,j=1 a2ij − {i∈R}∪{j∈C} a2ij − (n − l) i d2i. − (n − l) n−l−1 n−1 1 ≥ − n−l−1 n−l−1 2/3 a2ij + {i∈R}∪{j∈C} 1 2l4/3 n2/3 ≥1− 2n + n−l−1 (n − l)2/3 βA n j d2.j 2/3 l4/3 βA l4/3 βA + (n − l)2/3 (n − l)2/3 2/3 ≥ 0.93, where we have used the fact that a2ij ≤ (2nl − l2 )1/3 {i∈R}∪{j∈C} |aij |3 2/3 2/3 2/3 ≤ (2nl)1/3 βA ≤ 2n1/3 βA . {i∈R}∪{j∈C} We now prove the second half of (2.6). It follows from the first half of (2.6) and (A.1) that n−l −3 βD = σD |dij − di. − d.j + d.. |3 i,j=1 n−l −3 ≤ 1012 σD (|dij |3 /1002 + |di. + d.j − d.. |3 ) i,j=1 ≤ 101 2 100 ≤ 93 100 93 3/2 3/2 n−l (|dij |3 /1002 + 16(|di. |3 + |d.j |3 ) + 4|d.. |3 ) i,j=1 2 1012 2 52l + 101 βA ≤ 1.153βA , 1002 (n − l)2 where in the second inequality, we have used the fact that |x + y|3 = | x x |x|3 + ··· + + y|3 ≤ 1012 + |y|3 . 100 100 1002 9 Proof of Lemma 2.2. We will use the following forms which given by Chen and Shao [3, p. 2025]: 1 2 α N hx,α = Ehx,α (Z) = Φ(x) + √ se−(x+α−αs) /2 ds, 2π 0 √ ω 2 /2 if ω ≤ x, √2πe 2 Φ(ω)(1 − N hx,α ) 2πeω /2 (1 − Φ(ω))N h x,α fx,α (ω) = 2 2 −αeω /2 1+(x−ω)/α se−(x+α−αs) /2 ds if x < ω ≤ x + α, 0 √ 2 2πeω /2 (1 − Φ(ω))N hx,α if ω > x + α, (A.2) (A.3) and √ 2 2π(1 + ω 2 )eω /2 Φ(ω) + ω (1 − N hx,α ) √ 2 2π(1 + ω 2 )eω /2 (1 − Φ(ω)) − ω N hx,α + rx,α (ω) gx,α (ω) = √ 2π(1 + ω 2 )eω2 /2 (1 − Φ(ω)) − ω N hx,α if ω < x, if x ≤ ω ≤ x + α, (A.4) if ω > x + α, where rx,α (ω) = −ωeω 2 x+α /2 1+ ω x−ω x − s −s2 /2 e ds + 1 + . α α Chen and Shao [3] also proved that 0 ≤ rx,α (ω) ≤ 1 for x ≤ ω ≤ x + α and 0≤ √ 2π(1 + ω 2 )eω 2 /2 (1 − Φ(ω)) − ω ≤ 2 for all ω ≥ 0. 1 + ω3 (A.5) We have 0 ≤ f (ω) ≤ 1, |f (ω)| ≤ 1 (see Lemma 2.5 in [5]). Therefore, for all ω g(ω) = f (ω) + ωf (ω) ≤ 1 + |ω|. (A.6) For −1 < x ≤ 3, using (A.4)-(A.6), it is not hard to prove that gx,α (ω) ≤ 4 for all ω, so that (2.29) holds. For x > 3, using (A.2)-(A.5) and the fact that 1 − Φ(x) ≤ e−x √ 2 /2 √ /x 2π, we have 2 2π(x2 − 4x + 5)e(x−2) /2 Φ(x − 2) + x − 2 (1 − Φ(x)) 2 5 1 ≤ (x − 4 + )e−2x+2 + √ (x − 2)e−x /2 < 0.015, x x 2π g(x − 2) ≤ and √ 2 2πxex /2 Φ(x)(1 − Φ(x)) x2 2x ≤ + ≤ 1. 2 2 1+x (1 + x )(1 + x3 ) xfx,α (x) ≤ (A.7) (A.8) Furthermore, we have g ≥ 0, g(ω) ≤ 2(1 − Φ(x)) ≤ 2(1 − Φ(3)) for ω ≤ 0, g(ω) ≤ 2/(1 + 33 ) + 1(x < ω < x + α) for ω ≥ x and g is increasing for 0 ≤ ω < x (see Chen and Shao [3, p. 2025]). Therefore Eg(YD + u) = Eg(YD + u)1(YD + u ≤ 0) + Eg(YD + u)1(0 < YD + u ≤ x − 2) +Eg(YD + u)1(YD + u ≥ x) + Eg(YD + u)1(x − 2 < YD + u < x) ≤ 2(1 − Φ(3)) + g(x − 2) + 2/(1 + 33 ) + P (x < YD + u < x + α) +Eg(YD + u)1(x − 2 < YD + u < x) ≤ 1.09 + Eg(YD + u)1(x − 2 < YD + u < x). 10 (A.9) We estimate Eg(YD + u)1(x − 2 < YD + u < x) as follows. x Eg(YD + u)1(x − 2 < YD + u < x) = −g(ω)dP (ω < YD + u < x) x−2 x = g(x − 2)P (x − 2 < YD + u < x) + P (ω < YD + u < x)g (ω)dω x−2 x ω − u − µD x − u − µD g (ω)dω < YE < σD σD x−2 x x−ω √ ≤ 0.015 + + 2δ(βE , n − l) dg(ω) (by (2.4)) 2πσD x−2 x x−ω 180βE √ ≤ 0.015 + + dg(ω) (by the inductive hypothesis) n−l 2πσD x−2 x 1 180βE g(x− ) + √ (x − ω)dg(ω) ≤ 0.015 + n−l 2πσD x−2 x 207.6βA 1 ≤ 0.015 + (1 + x) + √ g(ω)dω (by the first half of (2.1)) n 2πσD x−2 1 100 1/2 ≤ 2.91 (by the second half of (2.1) and (2.28)). ≤ 0.015 + 2.472 + √ 2π 93 ≤ 0.015 + P (A.10) The conclusion (2.29) follows from (A.9) and (A.10). References [1] Bolthausen, Erwin. 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Probab. 42 (2005), no. 3, 661–683. [8] Goldstein, Larry. L1 bounds in normal approximation. Ann. Probab. 35 (2007), no. 5, 1888–1930. 11 [9] Goldstein, Larry; Reinert, Gesine. Stein’s method and the zero bias transformation with application to simple random sampling. Ann. Appl. Probab. 7 (1997), no. 4, 935–952. [10] Ho, Soo Thong; Chen, Louis H. Y. An Lp bound for the remainder in a combinatorial central limit theorem. Ann. Probab. 6 (1978), no. 2, 231–249. [11] Hoeffding, Wassily. A combinatorial central limit theorem. Ann. Math. Statist. 22, (1951), 558– 566. [12] H¨ oglund, Thomas. Sampling from a finite population: a remainder term estimate. Scand. J. Statist. 5 (1978), 69–71. [13] Hu, Zhishui; Robinson, John; Wang, Qiying. Cram´er-type large deviations for samples from a finite population. Ann. Statist. 35 (2007), no. 2, 673–696. [14] Lahiri, Soumendra N.; Chatterjee, Arindam. A Berry-Esseen theorem for hypergeometric probabilities under minimal conditions. Proc. Amer. Math. Soc. 135 (2007), no. 5, 1535–1545 (electronic). [15] Neammanee, Kritsana; Suntornchost, J. A uniform bound on a combinatorial central limit theorem. Stoch. Anal. Appl. 23 (2005), no. 3, 559–578. [16] Stein, Charles. A bound for the error in the normal approximation to the distribution of a sum of dependent random variables. Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability Vol. II: Probability theory, pp. 583–602. Univ. California Press, Berkeley, Calif., 1972. [17] Wald, Abraham.; Wolfowitz, Jacob. Statistical tests based on permutations of the observations. Ann. Math. Statist. 15, (1944), 358–372. 12 [...]... Soumendra N.; Chatterjee, Arindam A Berry- Esseen theorem for hypergeometric probabilities under minimal conditions Proc Amer Math Soc 135 (2007), no 5, 1535–1545 (electronic) [15] Neammanee, Kritsana; Suntornchost, J A uniform bound on a combinatorial central limit theorem Stoch Anal Appl 23 (2005), no 3, 559–578 [16] Stein, Charles A bound for the error in the normal approximation to the distribution of a. .. Goldstein, Larry; Shao, Qi-Man Normal approximation by Stein’s method Probability and its Applications (New York) Springer, Heidelberg, 2011 xii+405 pp [6] Ghosh, Subhankar Lp bounds for a central limit theorem with involutions Available at http://arxiv.org/abs/0905.1150 [7] Goldstein, Larry Berry- Esseen bounds for combinatorial central limit theorems and pattern occurrences, using zero and size biasing J Appl... [2] Chen, Louis H Y.; Shao, Qi-Man A non-uniform Berry- Esseen bound via Stein’s method Probab Theory Related Fields 120 (2001), no 2, 236–254 [3] Chen, Louis H Y.; Shao, Qi-Man Normal approximation under local dependence Ann Probab 32 (2004), no 3A, 1985–2028 [4] Chen, Louis H Y.; Fang, Xiao On the error bound in a combinatorial central limit theorem Available at http://arxiv.org/abs/1111.3159 [5] Chen,... Probab 42 (2005), no 3, 661–683 [8] Goldstein, Larry L1 bounds in normal approximation Ann Probab 35 (2007), no 5, 1888–1930 11 [9] Goldstein, Larry; Reinert, Gesine Stein’s method and the zero bias transformation with application to simple random sampling Ann Appl Probab 7 (1997), no 4, 935–952 [10] Ho, Soo Thong; Chen, Louis H Y An Lp bound for the remainder in a combinatorial central limit theorem Ann... Probab 6 (1978), no 2, 231–249 [11] Hoeffding, Wassily A combinatorial central limit theorem Ann Math Statist 22, (1951), 558– 566 [12] H¨ oglund, Thomas Sampling from a finite population: a remainder term estimate Scand J Statist 5 (1978), 69–71 [13] Hu, Zhishui; Robinson, John; Wang, Qiying Cram´er-type large deviations for samples from a finite population Ann Statist 35 (2007), no 2, 673–696 [14] Lahiri,... approximation to the distribution of a sum of dependent random variables Proceedings of the Sixth Berkeley Symposium on Mathematical Statistics and Probability Vol II: Probability theory, pp 583–602 Univ California Press, Berkeley, Calif., 1972 [17] Wald, Abraham.; Wolfowitz, Jacob Statistical tests based on permutations of the observations Ann Math Statist 15, (1944), 358–372 12 ... ω)dg(ω) ≤ 0.015 + n−l 2πσD x−2 x 207.6 A 1 ≤ 0.015 + (1 + x) + √ g(ω)dω (by the first half of (2.1)) n 2πσD x−2 1 100 1/2 ≤ 2.91 (by the second half of (2.1) and (2.28)) ≤ 0.015 + 2.472 + √ 2π 93 ≤ 0.015 + P (A. 10) The conclusion (2.29) follows from (A. 9) and (A. 10) References [1] Bolthausen, Erwin An estimate of the remainder in a combinatorial central limit theorem Z Wahrsch Verw Gebiete 66 (1984), no...We estimate Eg(YD + u)1(x − 2 < YD + u < x) as follows x Eg(YD + u)1(x − 2 < YD + u < x) = −g(ω)dP (ω < YD + u < x) x−2 x = g(x − 2)P (x − 2 < YD + u < x) + P (ω < YD + u < x)g (ω)dω x−2 x ω − u − µD x − u − µD g (ω)dω < YE < σD σD x−2 x x−ω √ ≤ 0.015 + + 2δ(βE , n − l) dg(ω) (by (2.4)) 2πσD x−2 x x−ω 180βE √ ≤ 0.015 + + dg(ω) (by the inductive hypothesis) n−l 2πσD x−2 x 1 180βE ... Subhankar Lp bounds for a central limit theorem with involutions Available at http://arxiv.org/abs/0905.1150 [7] Goldstein, Larry Berry- Esseen bounds for combinatorial central limit theorems and... H Y An Lp bound for the remainder in a combinatorial central limit theorem Ann Probab (1978), no 2, 231–249 [11] Hoeffding, Wassily A combinatorial central limit theorem Ann Math Statist 22,... Shao, Qi-Man Normal approximation under local dependence Ann Probab 32 (2004), no 3A, 1985–2028 [4] Chen, Louis H Y.; Fang, Xiao On the error bound in a combinatorial central limit theorem Available