Hindawi Publishing Corporation Journal of Applied Mathematics Volume 2013, Article ID 656257, 9 pages http://dx.doi.org/10.1155/2013/656257 Research Article The Almost Sure Local Central Limit Theorem for the Negatively Associated Sequences Yuanying Jiang1,2 and Qunying Wu1 1 2 College of Science, Guilin University of Technology, Guilin 541004, China School of Statistics, Renmin University of China, Beijing 100872, China Correspondence should be addressed to Yuanying Jiang; jyy@ruc.edu.cn Received 13 May 2013; Accepted 18 June 2013 Academic Editor: Ying Hu Copyright © 2013 Y. Jiang and Q. Wu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In this paper, the almost sure central limit theorem is established for sequences of negatively associated random variables: π limπ → ∞ (1/ log π) ∑π=1 (πΌ(ππ ≤ ππ < ππ )/π)π(ππ ≤ ππ < ππ ) = 1, almost surely. This is the local almost sure central limit theorem for negatively associated sequences similar to results by CsaΜki et al. (1993). The results extend those on almost sure local central limit theorems from the i.i.d. case to the stationary negatively associated sequences. 1. Introduction Definition 1. Random variables π1 , π2 , . . . , ππ , π ≥ 2 are said to be negatively associated (NA) if for every pair of disjoint subsets π΄ 1 and π΄ 2 of {1, 2, . . . , π}, [2] for the moment inequalities, and Wu and Jiang [3] for Chover’s law of the iterated logarithm. Assume that {ππ , π ≥ 1} is a strictly stationary sequence of NA random variables with πΈπ1 = 0, 0 < πΈπ12 < ∞. Define ππ = ∑ππ=1 ππ , Cov (π1 (ππ , π ∈ π΄ 1 ) , π2 (ππ , π ∈ π΄ 2 )) ≤ 0, ππ2 := Var ππ , (1) (2) ∞ where π1 and π2 are increasing for every variable (or decreasing for every variable) such that this covariance exists. A sequence of random variables {ππ , π ≥ 1} is said to be NA if every finite subfamily of {ππ , π ≥ 1} is NA. Obviously, if {ππ , π ≥ 1} is a sequence of NA random variables and {ππ , π ≥ 1} is a sequence of nondecreasing (or nonincreasing) functions, then {ππ (ππ ), π ≥ 1} is also a sequence of NA random variables. This definition was introduced by the Joag-Dev and Proschan [1]. Statistical test depends greatly on sampling, and the random sampling without replacement from a finite population is NA, but it is not independent. NA sampling has wide applications such as those in multivariate statistical analysis and reliability theory. Because of the wide applications of NA sampling, the notions of NA random variables have received more and more attention recently. We refer to Joag-Dev and Proschan [1] for fundamental properties, Shao π2 := Var π1 + 2 ∑Cov (π1 , ππ ) . π=2 (3) (1) Newman [4] and MatuΕa [5] showed that NA stationary sequences satisfy the central limit theorem (CLT) under π2 > 0, that is, σ΅¨σ΅¨ σ΅¨σ΅¨ π σ΅¨ σ΅¨ sup σ΅¨σ΅¨σ΅¨π ( π < π₯) − Φ (π₯)σ΅¨σ΅¨σ΅¨ = π (1) . (4) σ΅¨σ΅¨ ππ π₯∈π σ΅¨σ΅¨ (2) Applying MatuΕa [6] and Wu’s [7] methods, we can easily show that NA sequences satisfy the almost sure central limit theorem (ASCLT), that is, 1 π 1 ∑ πΌ {ππ ≤ π₯ππ1/2 } = Φ (π₯) π → ∞ log π π π=1 lim a.s. ∀π₯ ∈ R, (5) where Φ(π₯) is the standard normal distribution function and πΌ{π΄} denotes the indicator of the event π΄. 2 Journal of Applied Mathematics The ASCLT was stimulated by Brosamler [8] and Schatte [9]. Both were concerned with the partial sum of independent and identically distributed (i.i.d.) random variables with more than the second moment. The ASCLT was extensively studied in the past two decades and an interesting direction of the study is to prove it for dependent variables. There are some results for weakly dependent variables such as πΌ, π, π-mixing and associated random variables. Among those results, we refer to Peligrad and Shao [10], MatuΕa [11], and Wu [7]. More general version of ASCLT was proved by CsaΜki et al. [12]. The following theorem is due to them. Theorem A. Let {ππ , π ≥ 1} be a sequence of i.i.d. random variables with πΈ|π1 |3 < ∞, let πΈπ1 = 0. ππ , ππ satisfy −∞ ≤ ππ ≤ 0 ≤ ππ ≤ ∞, π = 1, 2, . . . , (6) log π = π (log π) , ≤ ππ < ππ ) ∑ π πΌπ π π=1 ππ := ∑ (11) under certain conditions. In our considerations, we will need the following CoxGrimmett coefficient which describes the covariance structure of the sequence π’ (π) := sup ∑ π∈ππ:|π−π|≥π σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨Cov (ππ , ππ )σ΅¨σ΅¨σ΅¨ , σ΅¨ σ΅¨ π ∈ π ∪ {0} . 3/2 π=1 π π (ππ (12) We remark that for a stationary sequence of NA random variables ∞ π’ (π) = −2 ∑ Cov (π1 , ππ ) , and assume that π So we need to investigate the limit behavior of π ∈ π. (13) π=π+1 ππ π σ³¨→ ∞, (7) and then 1 π πΌ {ππ ≤ ππ < ππ } ∑ =1 π → ∞ log π ππ (ππ ≤ ππ < ππ ) π=1 lim a.s. (8) This result may be called almost sure local central limit theorem, while (5) may be called almost sure global central limit theorem. Hurelbaatar [13] extended (8) to the case of πmixing sequences and Weng et al. [14] derived an almost sure local central limit theorem for the product of partial sums of a sequence of i.i.d. positive random variables under some regular conditions. For more details, we refer to Berkes and CsaΜki [15] and FoΜldes [16]. Our concern in this paper is to give a common generalization of (8) to the case of NA sequences. In the next section we present the exact results, postponing some technical lemmas and the proofs to Section 3. Assume in the following that {ππ , π ≥ 1} is a strictly stationary sequence of NA random variables with πΈπ1 = 0, 0 < πΈπ12 < ∞. We consider the limit behavior of the logarithmic average ∞ ∑ π’ (π) < ∞, (14) π=1 and for some π½ > 1, 1/3 (log π) πππ 1≤π≤π ∑ 2 −π½ = π ((log π) (log log π) ) . (15) Then we have lim π→∞ ππ = 1, log π a.s., (16) where ππ is defined by (11). (9) with −∞ ≤ ππ ≤ 0 ≤ ππ ≤ ∞, where the terms in the sum above are defined to be 1 if their denominator happens to be 0. More precisely let {ππ , π ≥ 1} and {ππ , π ≥ 1} be two sequences of real numbers and put Remark 3. Let ππ = −∞ and ππ = π₯ππ1/2 in (6). By the central limit theorem (4), we have ππ = π(ππ /ππ1/2 < π₯) → Φ(π₯), obviously (15) satisfies; then (16) becomes (5), which is the almost sure global central limit theorem. Thus the almost sure local central limit theorem is a general result which contains the almost sure global central limit theorem. Remark 4. The condition (15) is satisfied with a wide range of ππ ; for example, if ππ := π (ππ ≤ ππ < ππ ) , πΌ {ππ ≤ ππ < ππ } { { , if ππ =ΜΈ 0, πΌπ := { ππ { if ππ = 0. {1, Theorem 2. Let {ππ , π ≥ 1} be a strictly stationary sequence of NA random variables with πΈπ1 = 0, πΈ|π1 |3 < ∞ and let π2 > 0. ππ , ππ satisfy (6). Assume that ππ =ΜΈ 0 2. Main Results 1 π πΌ {ππ ≤ ππ < ππ } ∑ lim π → ∞ log π ππ (ππ ≤ ππ < ππ ) π=1 By Lemma 8 of Newman [4], we have π’(0) < ∞ and limπ → ∞ π’(π) = 0. In the following, ππ ∼ ππ denotes ππ /ππ → 1, π → ∞. ππ = π(ππ ) denotes that there exists a constant π > 0 such that ππ ≤ πππ for sufficiently large π. The symbols π, π1 , π2 , . . ., stand for generic positive constants which may differ from one place to another. (10) π½ ππ = 0 or ππ ≥ π (log log π) 2/3 (log π) (17) Journal of Applied Mathematics 3 holds, then the condition (15) is satisfied. In fact, letting 0 < πΏ < 1, we have with some π½ > 1 and positive constant π, then 1 π 1 ∑ ππ = 1 a.s. π → ∞ log π π π=1 1/3 lim (log π) πππ 1≤π≤π ∑ ππ =ΜΈ 0 (24) The following Lemma 8 is obvious. 1−πΏ πΏ (log π) −π½ ≤ π ∑ (log log π) (log π) 1≤π≤π π (18) 1−πΏ (log π) ≤ π(log log π) (log π) ∑ π 1≤π≤π −π½ πΏ −π½ 2 Lemma 8. Assume that the nonnegative random sequence {ππ , π ≥ 1} satisfies (24) and the sequence {ππ , π ≥ 1} is such that, for any π > 0, there exists a π0 = π0 (π, π) for which (1 − π) ππ ≤ ππ ≤ (1 + π) ππ , (25) a.s. (26) Then we have also ≤ π(log log π) (log π) 2 π > π0 . 1 π 1 ∑ ππ = 1 π → ∞ log π π π=1 lim −π½ = π ((log π) (log log π) ) . In the given theorem below, we strengthen the condition (6) on ππ and ππ . Meanwhile, as a compensation, we do not need to impose restricting condition (15) on ππ . Theorem 5. Let {ππ , π ≥ 1} be a strictly stationary sequence of NA random variables with πΈπ1 = 0, πΈ|π1 |3 < ∞, and π2 > 0, and let ππ and ππ satisfy −π1 π1/2−πΌ ≤ ππ ≤ −π2 π1/2−πΌ , The following Lemma 9 is an easy corollary to the Corollary 2.2 in MatuΕa [5] under strictly stationary condition, which studies the rate of convergence in the CLT under negative dependence. It was also studied in Pan [17]. Of course this is the Berry-Esseen inequality for the NA sequence. Lemma 9. Let {ππ , π ∈ π} be a strictly stationary sequence of NA random variables with πΈπ1 = 0, πΈ|π1 |3 < ∞, π2 > 0 satisfying (14). Then one has (19) π3 π1/2−πΌ ≤ ππ ≤ π4 π1/2−πΌ , where 0 < πΌ < 1/7. Assume that (14) hold, and then we have (16). σ΅¨σ΅¨ σ΅¨σ΅¨ π σ΅¨ σ΅¨ sup σ΅¨σ΅¨σ΅¨π ( π < π₯) − Φ (π₯)σ΅¨σ΅¨σ΅¨ = π (π−1/5 ) . σ΅¨σ΅¨ σ΅¨ π π₯∈π σ΅¨ π (27) Lemma 10. If the conditions of Lemma 9 hold, ππ and ππ satisfy (19). Then one has π1σΈ π−πΌ ≤ ππ ≤ π2σΈ π−πΌ 3. Proofs (π ≥ π0 ) , (28) The following lemmas play important roles in the proof of our theorems. The proofs are given in the Appendix. where πΌ is as (19). Lemma 6. Assume that {ππ , π ≥ 1} are random variables such that Lemma 11. If the conditions of Lemma 9 hold, ππ and ππ satisfy (19), and πΌ is as (19). Assume that ππ = π3πΌ/2 , and then the following asymptotic relations hold: ππ ≥ 0, πΈππ = 1, π = 1, 2, . . . (20) and furthermore there exists ππ ≥ 0 such that π·π = ∞, π·π /π·π−1 → 1, and π Var ( ∑ ππ ππ ) ≤ π=1 ∑ππ=1 ππ ↑ 1 σ΅¨ σ΅¨ π (σ΅¨σ΅¨σ΅¨πππΌ σ΅¨σ΅¨σ΅¨ ≥ ππ ) = π (log π) , πππ π 1≤π<π≤π (29) (21) 1 1 = π (log π) , ππππ (π − π − ππΌ )1/5 1≤π<π≤π (30) 1 π ∑ ππ ππ = 1 π→∞π· π π=1 π<π−ππΌ 1 πππ π 1≤π<π≤π ∑ a.s. Remark 7. Let ππ = 1/π in (21), and then π·π = log π. Thus, if π<π−ππΌ ∑ −π½ ππ·π2 (log π·π ) , with some π½ > 1 and positive constant π, and then lim ∑ π<π−ππΌ (22) ∑ππ=1 1/π ∼ 1 2 −π½ Var ( ∑ ππ ) ≤ π(log π) (log log π) , π π=1 (23) (31) = π (log π) , 1 πππ π 1≤π<π≤π ∑ π<π−ππΌ π σ΅¨ σ΅¨σ΅¨ π −π −π π σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨Φ ( π π π ) − Φ ( π )σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π1/2 σ΅¨σ΅¨σ΅¨ (π − π − ππΌ )1/2 σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ π − ππ + ππ ππ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨(Φ ( π ) − Φ ( )) 1/2 1/2 σ΅¨σ΅¨ σ΅¨σ΅¨ πΌ π (π − π − π ) σ΅¨ σ΅¨ = π (log π) . (32) 4 Journal of Applied Mathematics The main point in our proof is to verify the condition (23). We use global central limit theorem with remainders and the following elementary inequalities: σ΅¨ σ΅¨ σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨Φ (π₯) − Φ (π¦)σ΅¨σ΅¨σ΅¨ ≤ π σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ , for every π₯, π¦ ∈ R Applying Lemma 9, (33), (35), and (36) and noting that ππ = π1/2 (log π)1/3 , we obtain π (ππ − ππ − ππ ≤ ππ < ππ ) + π (ππ ≤ ππ < ππ − ππ + ππ ) (33) with some constant π. Moreover, for each π > 0, there exists π1 = π1 (π), such that σ΅¨ σ΅¨σ΅¨ σ΅¨σ΅¨Φ (π₯) − Φ (π¦)σ΅¨σ΅¨σ΅¨ σ΅¨ σ΅¨ ≥ π1 σ΅¨σ΅¨σ΅¨π₯ − π¦σ΅¨σ΅¨σ΅¨ , σ΅¨ σ΅¨ for every π₯, π¦ ∈ R, |π₯| + σ΅¨σ΅¨σ΅¨π¦σ΅¨σ΅¨σ΅¨ ≤ π. π1 π ≤ Var (ππ ) = ≤ π2 π + (Φ ( ≤ (34) (35) for some constant π1 , π2 > 0 and sufficiently large π. ππ + ππ −ππ + ππ π − ππ 2ππ 1 1 + + π 1/5 = π + + π 1/5 ππ ππ ππ ππ π π ≤ π( π1/2 (log π) ππ π1/2 1 + + ) ≤ π ( π1/2 π1/2 π1/5 π1/2 π π 1 π ππππ π=1 π=1 ∑∑ , π = 1, 2, . . . = 1 (π (ππ ≤ ππ < ππ , ππ ≤ ππ < ππ ) − ππ ππ ) ππ ππ ≤ 1 (π (ππ − ππ ≤ ππ − ππ < ππ − ππ , ππ ππ 1/3 π π 1/3 π 1/3 (log π) πππ π=1 ≤ π∑ 1/3 2 −π½ = π ((log π) (log log π) ) , (39) π π−1 1 1 1/5 πππ π π π=1 π=1 π =∑ π=1 1 (π (ππ − ππ ≤ ππ − ππ < ππ − ππ ) − ππ ) ππ 1 (π (ππ − ππ − ππ < ππ < ππ − ππ + ππ ) ππ σ΅¨ σ΅¨ −ππ + π (σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ ≥ ππ )) 1 (π (ππ − ππ − ππ ≤ ππ < πl ) ππ σ΅¨ σ΅¨ + π (ππ ≤ ππ < ππ − ππ + ππ ) +π (σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ ≥ ππ )) . π 1 1 log π ≤ π ∑ 6/5 6/5 π ππ π=1 π π=1 π ππ 1 π−1 ∑ 1/3 π (log π) ≤ π∑ πππ π=1 2 −π½ = π ((log π) (log log π) ) . By Chebyshev’s inequality and the condition of (15) and (35), we obtain 1 (π (ππ − ππ ≤ ππ − ππ < ππ − ππ ) ππ ππ ×π (ππ ≤ ππ < ππ ) − ππ ππ ) ≤ 1 π π (log π) (log π) 1/2 1 ≤ π π ∑ ∑ 3/2 1/2 3/2 π=1 π ππ π=1 π π=1 π ππ ππ ≤ ππ < ππ ) − ππ ππ ) ≤ 1/3 (log π) π1/2 ∑∑ Cov (πΌπ , πΌπ ) = 1/2 ≤ π∑ (36) with some constant π. Let 1 ≤ π < π and ππ = π1/2 (log π)1/3 . If either ππ = 0 or ππ = 0, then obviously Cov(πΌπ , πΌπ ) = 0, and so we may assume that ππ ππ =ΜΈ 0; then, we have ≤ 1 ). π1/5 (38) (log π) ≤ ∑ 3/2 ∑ π π π1/2 π π=1 π=1 Proof of Theorem 2. First assume that ππ − ππ ≤ ππ + By the condition of (15), we have π 1/2 ππ − ππ + ππ π 1 ) − Φ ( π )) + π 1/5 ππ ππ π 1/3 Let {ππ , π ≥ 1} be a strictly stationary sequence of NA random variables with π2 > 0; we can immediately get ππ2 ∼ ππ2 , that is, ππ2 ππ π −π −π ) − Φ ( π π π )) ππ ππ ≤ (Φ ( π (37) π 1 σ΅¨ σ΅¨ π (σ΅¨σ΅¨σ΅¨ππ σ΅¨σ΅¨σ΅¨ ≥ ππ ) ππππ π=1 π=1 ∑∑ π ≤∑ π=1 π π 1 π Var (ππ ) 1 π 1 ≤ π∑ ∑ ∑ 2 2/3 πππ π=1 πππ ππ π π=1 π(log π) π=1 1/3 (log π) πππ π=1 ≤ π∑ 2 (40) −π½ = π ((log π) (log log π) ) . Hence (37)–(40) imply that π π−1 Cov (πΌπ , πΌπ ) 2 −π½ = π ((log π) (log log π) ) . ππ π=1 π=1 ∑∑ (41) Journal of Applied Mathematics 5 But Var(πΌπ ) = 0 if ππ = 0 and Var (πΌπ ) = we obtain 1 − ππ 1 ≤ ππ ππ if ππ =ΜΈ 0. lim π (−π₯ππ1/2 ≤ ππ < 0) (42) π→∞ = lim π (−π₯ππ ≤ ππ < 0) = Φ (0) − Φ (−π₯) , Thus π→∞ π Var (πΌπ ) π2 π=1 π→∞ = lim π (0 ≤ ππ < π₯ππ ) = Φ (π₯) − Φ (0) . 1/3 (log π) 1 ≤ ∑ 2 ≤ ∑ π ππ 1≤π≤π πππ 1≤π≤π ππ =ΜΈ 0 (43) ππ =ΜΈ 0 2 π→∞ Applying the almost sure central limit theorem for NA random variables (5), that is, −π½ 1 π 1 ∑ πΌ {ππ ≤ π₯ππ1/2 } = Φ (π₯) π → ∞ log π π π=1 = π ((log π) (log log π) ) . lim Equations (41) and (43) together imply that π πΌ 2 −π½ Var ( ∑ π ) = π ((log π) (log log π) ) , π π=1 as π → ∞. Lemma 8, and (48), we have πΌ {ππ < −π₯ππ1/2 } 1 π ∑ 1/2 ≤ π < 0) π → ∞ log π π π=1 ππ (−π₯ππ lim Hence applying Remark 7, our theorem is proved under the restricting condition (36). Now we drop the restricting condition (36). Fix π₯ > 0 and define = Μπ = min (π , π₯ππ1/2 ) , π π (45) = πΜπ = π (Μ ππ ≤ ππ < Μππ ) , where π is defined by (3). Clearly πΜπ ≤ ππ , and so assuming πΜπ =ΜΈ 0; then, also we have ππ =ΜΈ 0, and thus 1 πΌπ = πΌ {ππ ≤ ππ < ππ } ππ a.s. 1 π πΌΜπ ∑ = 1 a.s., π → ∞ log π π π=1 (51) πΌ {Μ ππ ≤ ππ < Μππ } { { , if πΜπ =ΜΈ 0, πΌΜπ = { πΜπ { if πΜπ = 0. {1, (52) Equations (46) and (50)–(51) together imply that (46) lim sup 1 (πΌ {ππ ≤ ππ < πΜπ } + πΌ {Μππ ≤ ππ < ππ }) + ππ π→∞ + πΌ {ππ ≥ π₯ππ1/2 } π (0 ≤ ππ < π₯ππ1/2 ) 1 π πΌπ 1 − Φ (π₯) ∑ ≤1+2 log π π=1 π Φ (π₯) − Φ (0) a.s. (53) On the other hand, if πΜπ =ΜΈ 0, then we have 1 πΌ {Μ ππ ≤ ππ < Μππ } πΜπ π (−π₯ππ1/2 ≤ ππ < 0) 1 − Φ (π₯) Φ (π₯) − Φ (0) where 1 πΌ {Μ ππ ≤ ππ < Μππ } Μ ππ πΌ {ππ < −π₯ππ1/2 } (50) lim 1 (πΌ {Μ ππ ≤ ππ < Μππ } ππ + a.s., Since πΜπ and Μππ satisfy (36), we get +πΌ {ππ ≤ ππ < πΜπ } + πΌ {Μππ ≤ ππ < ππ }) ≤ Φ (−π₯) Φ (0) − Φ (−π₯) πΌ {ππ > π₯ππ1/2 } 1 π lim ∑ 1/2 π → ∞ log π π=1 ππ (0 ≤ ππ < π₯ππ ) πΜπ = max (ππ , −π₯ππ1/2 ) , ≤ a.s. ∀π₯ ∈ R, (49) (44) = (48) lim π (0 ≤ ππ < π₯ππ1/2 ) ∑ 1 πΌ {ππ ≤ ππ < ππ } ππ . By (35) and the central limit theorem for NA random variables (4), that is, σ΅¨σ΅¨ σ΅¨σ΅¨ π σ΅¨ σ΅¨ sup σ΅¨σ΅¨σ΅¨π ( π < π₯) − Φ (π₯)σ΅¨σ΅¨σ΅¨ = π (1) , (47) ππ π₯∈π σ΅¨σ΅¨ σ΅¨σ΅¨ ≥ π − πΜπ 1 πΌ {Μ ππ ≤ ππ < Μππ } (1 − π ) πΜπ ππ ≥ πΌΜπ (1 − π (ππ < −ππ₯π1/2 ) + π (ππ > ππ₯π1/2 ) ), min{π (0 ≤ ππ < ππ₯π1/2 ) , π (−ππ₯π1/2 ≤ ππ < 0)} (54) 6 Journal of Applied Mathematics and by the central limit theorem, lim π (ππ < −ππ₯π π→∞ min {π (0 ≤ ππ < Applying Lemma 9, (33), and (35), we obtain 1/2 1/2 ) + π (ππ > ππ₯π 1/2 ππ₯π ) , π (−ππ₯π1/2 ≤ ) π (ππ − ππ − ππ < ππ−π−ππΌ < ππ − ππ + ππ ) − ππ ππ < 0)} 1 − Φ (π₯) . =1−2 Φ (π₯) − Φ (0) ≤ (Φ ( (55) ππ − ππ + ππ (π − π − + (Φ ( Applying Lemma 8, (51) and (54) imply that lim inf π→∞ 1 π πΌπ 1 − Φ (π₯) ≥1−2 ∑ log π π=1 π Φ (π₯) − Φ (0) a.s., 1 π πΌπ 1 − Φ (π₯) ≥ lim sup ∑ Φ (π₯) − Φ (0) π → ∞ log π π=1 π ≥ lim inf π→∞ ≥1−2 1 ≥ lim sup π→∞ 1 − Φ (π₯) Φ (π₯) − Φ (0) (57) 1 (π − π − ππΌ )1/5 )) (61) ). Cov (πΌπ , πΌπ ) = π (log π) , ππ 1≤π<π≤π ∑ πΌ 1 ∑ π log π π=1 π because π − ππΌ < π < π, ππΎ − (π − ππΌ )πΎ → 0 as π → ∞ for πΎ < 1, πΌ < 1. But Var(πΌπ ) = 0 if ππ = 0 and (58) Var (πΌπ ) = a.s. 1 − ππ 1 ≤ ππ ππ if ππ =ΜΈ 0. (64) Then π (59) This completes the proof of Theorem 2. Proof of Theorem 5. Let π < π − ππΌ , 1 ≤ π < π, and ππ = π3πΌ/2 ; we have Cov (πΌπ , πΌπ ) 1 (π (ππ − ππ ≤ ππ − ππ+ππΌ + ππ+ππΌ − ππ ππ ππ < ππ − ππ , ππ ≤ ππ < ππ ) − ππ ππ ) 1 (π (ππ − ππ ≤ ππ − ππ+ππΌ + ππ+ππΌ − ππ ππ ππ < ππ − ππ ) π (ππ ≤ ππ < ππ ) − ππ ππ ) (60) 1 (π (ππ − ππ − ππ < ππ − ππ+ππΌ < ππ − ππ + ππ ) ππ σ΅¨ σ΅¨ −ππ + π (σ΅¨σ΅¨σ΅¨ππ+ππΌ − ππ σ΅¨σ΅¨σ΅¨ ≥ ππ )) (63) π>π−ππΌ π 1 (π (ππ − ππ − ππ < ππ−π−ππΌ < ππ − ππ + ππ ) ππ σ΅¨ σ΅¨ −ππ + π (σ΅¨σ΅¨σ΅¨πππΌ σ΅¨σ΅¨σ΅¨ ≥ ππ )) . (62) On the other hand, a.s. 1 π πΌπ lim ∑ = 1 a.s. π → ∞ log π π π=1 ≤ (π − π − ππΌ )1/2 π<π−ππΌ Thus ≤ + ππ − ππ − ππ − Φ( Cov (πΌπ , πΌπ ) = π (log π) . ππ 1≤π<π≤π 1 π πΌπ ∑ log π π=1 π 1 π πΌπ ≥ lim inf ≥1 ∑ π → ∞ log π π π=1 ≤ 1 π1/5 ππ )) 1/2 π ∑ By the arbitrariness of π₯, let π₯ → ∞ in (57); we have ≤ ππ ) π1/2 ) − Φ( Hence Lemma 11, (60), and (61) imply that and hence 1+2 + π( (56) ππΌ )1/2 π Var (πΌπ ) 1 1 ≤ ≤ = π (log π) . ∑ ∑ 2 2π 5/2−πΌ π π π π π=1 1≤π≤π π=1 ∑ (65) ππ =ΜΈ 0 Noting that π πΌπ ) π π=1 Var ( ∑ π Var (πΌπ ) π2 π=1 =∑ (66) Cov (πΌπ , πΌπ ) Cov (πΌπ , πΌπ ) +2 ∑ , ππ ππ 1≤π<π≤π 1≤π<π≤π +2 ∑ π>π−ππΌ π<π−ππΌ thus (62)–(66) imply that π πΌπ ) = π (log π) , π π=1 Var ( ∑ as π σ³¨→ ∞. (67) Hence applying Remark 7, we have 1 π πΌπ ∑ = 1 a.s. π → ∞ log π π π=1 lim This completes the proof of Theorem 5. (68) Journal of Applied Mathematics 7 Appendix ∑ππ=1 Proof of Lemma 6. Let ππ = Proof of Lemma 10. Applying Lemma 9, (33), and noting the conditions of (19) and 0 < πΌ < 1/7, we get ππ ππ , and then ∀π > 0 ππ = π (ππ ≤ ππ ≤ ππ ) σ΅¨σ΅¨ π Var (ππ /π·π ) πΈπ σ΅¨σ΅¨σ΅¨ −π½ σ΅¨ ≤ π(log π·π ) . (A.1) π (σ΅¨σ΅¨σ΅¨ π − π σ΅¨σ΅¨σ΅¨ ≥ π) ≤ σ΅¨σ΅¨ π·π π·π σ΅¨σ΅¨ π2 ≤ (Φ ( π Let π < 1, ππ½ > 1, and ππ = inf{ππ , π·ππ ≥ exp(π )}, and then π·ππ ≥ exp(ππ ), π·ππ −1 < exp(ππ ), for π·π ∼ π·π−1 ; we get 1≤ π·ππ ∼ exp (ππ ) π·ππ −1 < 1 σ³¨→ 1, exp (ππ ) (A.2) π·ππ ∼ exp (ππ ) , π·ππ−1 ∼ ≥ (Φ ( exp (π ) exp ((π − 1)π ) ≥ π( π 1 1 = exp (ππ [1 − (1 − ) ]) ∼ exp (ππ ⋅ π ⋅ ) π π = exp (π ⋅ π On account of π < 1, then π·ππ−1 ∼ exp (π ⋅ ππ−1 ) σ³¨→ 1, π=1 Proof of Lemma 11. By Lemma 10, Chebyshev’s inequality, and noting the condition of 0 < πΌ < 1/7, we have 1 σ΅¨ σ΅¨ π (σ΅¨σ΅¨σ΅¨πππΌ σ΅¨σ΅¨σ΅¨ ≥ ππ ) πππ π 1≤π<π≤π π<π−ππΌ πΌ (A.5) ≤ (log π·π ) −π½ π·ππ πΈπππ log (π − ππΌ ) = π (logπ) . π1+πΌ π=1 π ≤ π∑ σ³¨→ 0 a.s. π·ππ (A.6) π·ππ 1 1 ππππ (π − π − ππΌ )1/5 1≤π<π≤π π<π−ππΌ π ∑ π ππ = π=1 = 1, π·ππ (A.7) π ≤ π∑ π=1 thus πππ π·ππ σ³¨→ 1, a.s. π·ππ −1 πππ−1 π·ππ π·ππ−1 ≤ ππ π·ππ ππ ≤ π σ³¨→ 1 π·π π·ππ π·ππ−1 1 1 ( ∑ π1−πΌ 1≤π<(π−ππΌ )/2 π(π − ππΌ − π)1/5 (A.8) Now for ππ−1 ≤ π < ππ , for π·π ↑ ∞, π·ππ /π·ππ−1 → 1, and by ππ ≥ 0, then ππ ↑, and we have 1 ←σ³¨ It proves (29). By Lemma 10 and 0 < πΌ < 1/7, we get ∑ Since πΈπππ (A.13) π<π−π 1 < ∞. ππ½ π=1 π ≤ π∑ By the Borel-Cantelli lemma, − π 1 Var (πππΌ ) 1 π−π 1 ≤ π∑ 1+πΌ ∑ 3πΌ ππππ π π π 1≤π<π≤π π=1 π=1 ∑ πΌ ∞ 1 πππ π 1 + 1/5 ) ≥ π1σΈ π−πΌ . π1/2 π ∑ as π σ³¨→ ∞, σ΅¨σ΅¨ π πΈπππ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ π σ΅¨σ΅¨ ≥ π) ∑ π (σ΅¨σ΅¨σ΅¨σ΅¨ π − π·ππ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ π·ππ π=1 ∞ (A.12) 1/2−πΌ Thus Lemma 10 immediately follows from (A.11) and (A.12). ∞ ≤ π∑ ππ ππ 1 ) − Φ ( 1/2 )) − π 1/5 1/2 π π π ). (A.4) π·ππ π1/2−πΌ 1 + 1/5 ) ≤ π2σΈ π−πΌ . 1/2 π π ππ = π (ππ ≤ ππ ≤ ππ ) (A.3) π π−1 (A.11) Applying Lemma 9, (34), and noting the conditions of (19) and 0 < πΌ < 1/7, we have that is, and thus π·ππ ≤ π( ππ ππ 1 ) − Φ ( 1/2 )) + π 1/5 π1/2 π π a.s. (A.9) + π ≤ π∑ π=1 + This completes the proof of Lemma 6. (A.10) π ≤ π∑ π=1 (π−ππΌ )/2<π<π−ππΌ π(π 1 − ππΌ − π)1/5 1 1 1 ( ∑ π1−πΌ (π − ππΌ )1/5 1≤π<(π−ππΌ )/2 π hence ππ σ³¨→ 1 a.s. π·π ∑ log π π1−πΌ+1/5 1 1 ) ∑ π − ππΌ 1≤π<(π−ππΌ )/2 π1/5 π 1 ≤ π∑ = π (log π) . π π=1 ) (A.14) 8 Journal of Applied Mathematics It proves (30). Applying Lemma 9, (33), (35), ππ = π3πΌ/2 , and noting the condition of 0 < πΌ < 1/7, we obtain Noting that 0 < πΌ < 1/7, we deduce 1 Σ3 ≤ π ∑ 1−(7/2)πΌ 1/2 π (π − ππΌ − π)1/2 1≤π<π≤π π π<π−ππΌ σ΅¨ σ΅¨ π −π −π π σ΅¨σ΅¨ 1 σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨Φ ( π π π ) − Φ ( π )σ΅¨σ΅¨σ΅¨ ∑ ππππ σ΅¨σ΅¨σ΅¨ π1/2 σ΅¨σ΅¨σ΅¨ (π − π − ππΌ )1/2 1≤π<π≤π πΌ 1 1−(7/2)πΌ π 1≤π≤π ≤π ∑ π<π−π −ππ 1 1 ( − ) πΌ )1/2 √ πππ π − π − π (π π 1≤π<π≤π ×( ≤π ∑ 1 (π − πΌ π<π−π (A.15) ππ 1 +π ∑ ππππ (π − π − ππΌ )1/2 1≤π<π≤π ≤π ∑ 1≤π≤π ππΌ )1/2 1 1 1 ) + ∑ πΌ 1/2 π (π − π ) 1≤π<(π−ππΌ )/2 π 1≤π<(π−ππΌ )/2 log π π3/2−(7/2)πΌ ∑ (A.17) πΌ π<π−π It proves (31). The proof of (32) is similar to the proof of (31). This completes the proof of Lemma 11. ππ 1 +π ∑ := Σ1 + Σ2 + Σ3 . ππππ (π − π − ππΌ )1/2 1≤π<π≤π π<π−ππΌ Acknowledgments The authors are very grateful to the academic editor, professor Ying Hu, and the two anonymous reviewers for their valuable comments and helpful suggestions, which significantly contributed to improving the quality of this paper. This work is jointly supported by National Natural Science Foundation of China (11061012,71271210), Project Supported by Program to Sponsor Teams for Innovation in the Construction of Talent Highlands in Guangxi Institutions of Higher Learning ((2011)47), the Guangxi China Science Foundation (2013GXNSFDA019001). Now applying the same procedure as before, we have −ππ 1 π ( + ) 1/2 1−πΌ πΌ πππ π (π − π − π ) π(π − π − ππΌ )1/2 π 1≤π<π≤π Σ1 ≤ π ∑ π<π−ππΌ ≤π ∑ 1≤π≤π 1 π3/2−πΌ References 1 1 1 ×( ) + ∑ ∑ πΌ 1/2 1/2 πΌ π π − π π (l − π ) 1≤π<(π−ππΌ )/2 1≤π<(π−ππΌ )/2 1 1 1 ∑ 3/2 π πΌ − π − ππΌ )1/2 (π 1≤π≤π 1≤π<π−π +π ∑ log π 1 ≤π ∑ = π (log π) , 2−πΌ π π 1≤π≤π 1≤π≤π ≤π ∑ Σ2 ≤ π ∑ 1 1≤π<π≤π π<π−ππΌ ≤π ∑ 1≤π≤π π1−πΌ π1/2+πΌ (π 1 π1−πΌ ( 1≤π≤π 1 π1−πΌ ( − ππΌ − π)1/2 1 (π − ππΌ )1/2 + ≤π ∑ 1 = π (log π) . π 1≤π≤π ≤π ∑ ππΌ )1/2+πΌ 1 (π − 1≤π<(π−ππΌ )/2 1 (π − ππΌ )1/2 1 ∑ π1/2+πΌ 1 ) 1/2 π 1≤π<(π−ππΌ )/2 π1/2−πΌ + ∑ 1 (π − ππΌ )1/2+πΌ 1/2 (π − ππΌ ) ) 1 = π (log π) . π 1≤π≤π ≤π ∑ (A.16) [1] K. Joag-Dev and F. Proschan, “Negative association of random variables, with applications,” The Annals of Statistics, vol. 11, no. 1, pp. 286–295, 1983. [2] Q. M. 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Pan, “On the convergence rate in the central limit theorem for negatively associated sequences,” Chinese Journal of Applied Probability and Statistics, vol. 13, no. 2, pp. 183–192, 1997. 9 Advances in Operations Research Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Advances in Decision Sciences Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Mathematical Problems in Engineering Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Journal of Algebra Hindawi Publishing Corporation http://www.hindawi.com Probability and Statistics Volume 2014 The Scientific World Journal Hindawi Publishing Corporation http://www.hindawi.com Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 International Journal of Differential Equations Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Volume 2014 Submit your manuscripts at http://www.hindawi.com International Journal of Advances in Combinatorics Hindawi Publishing Corporation http://www.hindawi.com Mathematical Physics Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Journal of Complex Analysis Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 International Journal of Mathematics and Mathematical Sciences Journal of Hindawi Publishing Corporation http://www.hindawi.com Stochastic Analysis Abstract and Applied Analysis Hindawi Publishing Corporation http://www.hindawi.com Hindawi Publishing Corporation http://www.hindawi.com International Journal of Mathematics Volume 2014 Volume 2014 Discrete Dynamics in Nature and Society Volume 2014 Volume 2014 Journal of Journal of Discrete Mathematics Journal of Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Applied Mathematics Journal of Function Spaces Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Optimization Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 Hindawi Publishing Corporation http://www.hindawi.com Volume 2014