Research Article Complete Convergence for Weighted Sums of Sequences of Negatively Dependent Random Variables

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1 Journal of Probability and Statistics Volume 2011, Article ID , 16 pages doi: /2011/ Research Article Complete Convergence for Weighted Sums of Sequences of Negatively Dependent Random Variables Qunying Wu College of Science, Guilin Uversity of Technology, Guilin , China Correspondence should be addressed to Qunying Wu, Received 30 September 2010; Accepted 21 January 2011 Academic Editor: A. Thavaneswaran Copyright q 2011 Qunying 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. Applying to the moment inequality of negatively dependent random variables the complete convergence for weighted sums of sequences of negatively dependent random variables is discussed. As a result, complete convergence theorems for negatively dependent sequences of random variables are extended. 1. Introduction and Lemmas Defition 1.1. Random variables X and Y are said to be negatively dependent ND if P X x, Y y P X x P Y y 1.1 for all x, y R. A collection of random variables is said to be pairwise negatively dependent PND if every pair of random variables in the collection satisfies 1.1. It is important to note that 1.1 implies that P X>x,Y>y P X >x P Y>y 1.2 for all x, y R. Moreover, it follows that 1.2 implies 1.1, and, hence, 1.1 and 1.2 are equivalent. However, 1.1 and 1.2 are not equivalent for a collection of 3 or more random variables. Consequently, the following defition is needed to define sequences of negatively dependent random variables.

2 2 Journal of Probability and Statistics Defition 1.2. The random variables X 1,...,X n are said to be negatively dependent ND if, for all real x 1,...,x n, P n Xj x j P n Xj >x j n P X j x j, n P X j >x j. 1.3 An infite sequence of random variables {X n ; n 1} is said to be ND if every fite subset X 1,...,X n is ND. Defition 1.3. Random variables X 1,X 2,...,X n, n 2, are said to be negatively associated NA if, for every pair of disjoint subsets A 1 and A 2 of {1, 2,...,n}, cov f 1 X i ; i A 1,f 2 Xj ; j A 2 0, 1.4 where f 1 and f 2 are increasing in every variable or decreasing in every variable, provided this covariance exists. A random variables sequence {X n ; n 1} is said to be NA if every fite subfamily is NA. The defition of PND is given by Lehmann 1, the concept of ND is given by Bozorga et al. 2, and the defition of NA is introduced by Joag-Dev and Proschan 3. These concepts of dependence random variables have been very useful in reliability theory and applications. First, note that by letting f 1 X 1,X 2,...,X n 1 I X1 x 1,X 2 x 2,...,X n 1 x n 1, f 2 X n I Xn x n and f 1 X 1,X 2,...,X n 1 I X1 >x 1,X 2 >x 2,...,X n 1 >x n 1, f 2 X n I Xn >x n, separately, it is easy to see that NA implies 1.3. Hence, NA implies ND. But there are many examples which are ND but are not NA. We list the following two examples. Example 1.4. Let X i be a binary random variable such that P X i 1 P X i for i 1, 2, 3. Let X 1,X 2,X 3 take the values 0, 0, 1, 0, 1, 0, 1, 0, 0, and 1, 1, 1, each with probability 1/4. It can be verified that all the ND conditions hold. However, P X 1 X 3 1, X P X 1 X 3 1 P X Hence, X 1, X 2,andX 3 are not NA. In the next example X X 1,X 2,X 3,X 4 possesses ND, but does not possess NA obtained by Joag-Dev and Proschan 3. Example 1.5. Let X i be a binary random variable such that P X i 1.5 fori 1, 2, 3, 4. Let X 1,X 2 and X 3,X 4 have the same bivariate distributions, and let X X 1,X 2,X 3,X 4 have joint distribution as shown in Table 1.

3 Journal of Probability and Statistics 3 Table 1 X 1,X 2 0, 0 0, 1 1, 0 1, 1 Marginal 0, , X 3,X 4 1, , marginal It can be verified that all the ND conditions hold. However, P X i 1, i 1, 2, 3, 4 >P X 1 X 2 1 P X 3 X 4 1, 1.6 violating NA. From the above examples, it is shown that ND does not imply NA and ND is much weaker than NA. In the papers listed earlier, a number of well-known multivariate distributions are shown to possess the ND properties, such as a multinomial, b convolution of unlike multinomials, c multivariate hypergeometric, d dirichlet, e dirichlet compound multinomial, and f multinomials having certain covariance matrices. Because of the wide applications of ND random variables, the notions of ND random variables have received more and more attention recently. A series of useful results have been established cf. Bozorga et al. 2,Ami 4, Fakoor and Azarnoosh 5, Nili Sa et al. 6, Klesov et al. 7, and Wu and Jiang 8. Hence, the extending of the limit properties of independent or NA random variables to the case of ND random variables is highly desirable and of considerable sigficance in the theory and application. In this paper we study and obtain some probability inequalities and some complete convergence theorems for weighted sums of sequences of negatively dependent random variables. In the following, let a n b n a n b n denote that there exists a constant c>0 such that a n cb n a n cb n for sufficiently large n, andleta n b n mean a n b n and a n b n. Also, let log x denote ln max e,x and S n n X j. Lemma 1.6 see 2. Let X 1,...,X n be ND random variables and {f n ; n 1} a sequence of Borel functions all of which are monotone increasing or all are monotone decreasing. Then {f n X n ; n 1} is still a sequence of ND r. v. s. Lemma 1.7 see 2. Let X 1,...,X n be nonnegative r. v. s which are ND. Then E n X j n EX j. 1.7

4 4 Journal of Probability and Statistics In particular, let X 1,...,X n be ND, and let t 1,...,t n be all nonnegative or non-positive real numbers. Then E exp n t j X j n E exp t j X j. 1.8 Lemma 1.8. Let {X n ; n 1} be an ND sequence with EX n 0 and E X n p <, p 2. Then for B n n EX2 i, { } E S n p c p E X i p B p/2 n, 1.9 { } E max S i p c p log p n E X i p B p/2 n, i n where c p > 0 depends only on p. Remark 1.9. If {X n ; n 1} is a sequence of independent random variables, then 1.9 is the classic Rosenthal inequality 9. Therefore, 1.9 is a generalization of the Rosenthal inequality. Proof of Lemma 1.8. Let a > 0, X i min X i,a, and S n n X i. It is easy to show that {X i ; i 1} is a negatively dependent sequence by Lemma 1.6. Notingthat ex 1 x /x 2 is a nondecreasing function of x on R and that EX i EX i 0, tx i ta, we have e tx E e tx i 1 tex i E i 1 tx i t 2 X 2 i 1 e ta 1 ta a 2 EX 2 i t 2 X 2 i e ta 1 ta a 2 EX 2 i exp{ e ta 1 ta a 2 EX 2 i }. Here the last inequality follows from 1 x e x, for all x R. Note that B n n EX2 i and {X i ; i 1} is ND, we conclude from the above inequality and Lemma 1.7 that, for any x>0andh>0, we get n e hx E e hs n e hx E e hx i e hx { exp hx e ha 1 ha n E e hx i } a 2 B n. 1.12

5 Journal of Probability and Statistics 5 Letting h ln xa /B n 1 /a > 0, we get e ha 1 ha a 2 B n x a B n xa a ln 1 x 2 B n a Putting this one into 1.12, we get furthermore { e hx E e hs n x exp a x } xa a ln B n Putting x/a t into the above inequality, we get P S n x P X i >a P S n x P X i >a e hx Ee hs n P X i > x { } x 2 exp t t ln 1 t tb n P X i > x t e t 1 x2. t tb n 1.15 Letting X i take the place of X i in the above inequality, we can get P S n x P S n x P X i > x t e t 1 x2 t tb n P X i < x t e 1 t x2. t tb n 1.16 Thus P S n x P S n x P S n x P X i < x t 2e t 1 x t tb n to Multiplying 1.17 by px p 1, letting t p, and integrating over 0 <x<, according E X p p 0 x p 1 P X x dx, 1.18

6 6 Journal of Probability and Statistics we obtain E S n p p p 0 x p 1 P S n x dx 0 x p 1 P X i x p dx 2pe p x p 1 p p 1 E X i p pe p p/2 u p/2 1 pb n 1 u p du p p p 1 E X i p p p/2 1 e p B 2, p 2 0 B p/2 n, 0 p 1 x2 dx pb n 1.19 where B α, β 1 0 xα 1 1 x β 1 dx x α 1 1 x α β dx, α, β > 0 is Beta function. Letting 0 c p max p p 1,p 1 p/2 e p B p/2,p/2, we can deduce 1.9 from From 1.9, we can prove 1.10 by a similar way of Stout s paper 10, Theorem Lemma Let {X n ; n 1} be a sequence of ND random variables. Then there exists a positive constant c such that, for any x 0 and all n 1, 2 1 P max X k >x P X k >x cp max k >x Proof. Let A k X k >x and α n 1 P n A k 1 P max X k >x. Without loss of generality, assume that α n > 0. Note that {I Xk >x EI Xk >x ; k 1} and {I Xk < x EI Xk < x ; k 1} are still ND by Lemma 1.6.Using 1.9, weget 2 E I Ak EI Ak 2 E I Xk >x EI Xk >x I Xk < x EI Xk < x 2 2 2E I Xk >x EI Xk >x 2E I Xk < x EI Xk < x c P A k Combing with the Cauchy-Schwarz inequality, we obtain P A k n P A k, A j E I Ak I n A j E I Ak EI Ak I n A P A j k P n A j

7 Journal of Probability and Statistics 7 E 2 I Ak EI Ak EI n A j 1/2 1 α n P A k c 1 αn α n α n α n 1/2 P A k 1 α n P A k 1 c 1 αn α n P A k 1 α n P A k Thus α 2 n P A k c 1 α n, 1.23 that is, 2 1 P max X k >x P X k >x cp max k >x Main Results and the Proofs The concept of complete convergence of a sequence of random variables was introduced by Hsu and Robbins 11 as follows. A sequence {Y n ; n 1} of random variables converges completely to the constant c if P X n c >ε <, for all ε>0. In view of the Borel- Cantelli lemma, this implies that Y n 0 almost surely. Therefore, complete convergence is one of the most important problems in probability theory. Hsu and Robbins 11 proved that the sequence of arithmetic means of independent and identically distributed i.i.d. random variables converges completely to the expected value if the variance of the summands is fite. Baum and Katz 12 proved that if {X, X n ; n 1} is a sequence of i.i.d. random variables with mean zero, then E X p t 2 < 1 p<2, t 1 is equivalent to the condition that nt P n 1 1 X i /n 1/p >ε <, for all ε>0. Recent results of the complete convergence can be found in Li et al. 13, LiangandSu 14, Wu 15, 16, and Sung 17. In this paper we study the complete convergence for negatively dependent random variables. As a result, we extend some complete convergence theorems for independent random variables to the negatively dependent random variables without necessarily imposing any extra conditions. Theorem 2.1. Let {X, X n ; n 1} be a sequence of identically distributed ND random variables and {a nk ;, n 1} an array of real numbers, and let r>1, p>2. If, for some 2 q<p, { N n, m 1 k 1; a nk m 1 1/p} m q r 1 /p, n,m 1, 2.1 EX 0 for 1 q r 1, 2.2 a 2 nk nδ for 2 q r 1 and some 0 <δ< 2 p, 2.3

8 8 Journal of Probability and Statistics then, for r 2, E X p r 1 < 2.4 if and only if n r 2 k P max a X i >εn 1/p <, ε > For 1 < r < 2, 2.4 implies 2.5, conversely, and 2.5 and n r 2 P max a nk X k > n 1/p decreasing on n imply 2.4. For p 2, q 2, we have the following theorem. Theorem 2.2. Let {X, X n ; n 1} be a sequence of identically distributed ND random variables and {a nk ;1 k n, n 1} an array of real numbers, and let r>1. If { N n, m 1 k; a nk m 1 1/2} m r 1, n,m 1, 2.6 EX 0, 1 2 r 1, a nk 2 r 1 O 1, 2.7 then, for r 2, E X 2 r 1 log X < 2.8 if and only if n r 2 k P max a X i >εn 1/2 <, ε > For 1 < r < 2, 2.8 implies 2.9, conversely, and 2.9 and n r 2 P max a nk X k > n 1/2 decreasing on n imply 2.8. Remark 2.3. Since NA random variables are a special case of ND r. v. s, Theorems 2.1 and 2.2 extend the work of Liang and Su 14, Theorem 2.1. Remark 2.4. Since, for some 2 q p, k N a nk q r 1 1asn implies that { N n, m 1 k 1; a nk m 1 1/p} m q r 1 /p as n, 2.10

9 Journal of Probability and Statistics 9 taking r 2, then conditions 2.1 and 2.6 are weaker than conditions 2.13 and 2.9 in Li et al. 13. Therefore, Theorems 2.1 and 2.2 not only promote and improve the work of Li et al. 13, Theorem 2.2 for i.i.d. random variables to an ND setting but also obtain their necessities and relax the range of r. Proof of Theorem 2.1. Equation To prove 2.5 it suffices to show that n r 2 k P max a ± X i >εn1/p <, ε >0, 2.11 where a max a, 0 and a max a, 0. Thus, without loss of generality, we can assume that a > 0 for all n 1, i n. For 0 <α<1/p small enough and sufficiently large integer K, which will be determined later, let X 1 X 2 n α I a X i < n α a X i I a X i n α n α I a X i >n α, a X i n α I n α <a X i <εn 1/p /K, X 3 X 4 a X i n α I εn 1/p /K<a X i < n α, a X X 1 X 2 X 3 a X i n α I a X i εn 1/p /K a X i n α I a X i εn 1/p /K, 2.12 S j nk k X j, j 1, 2, 3, 4; 1 k n, n 1. Thus S nk k a X i 4 S j nk.notethat max S nk > 4εn 1/p 4 max S j nk >εn. 1/p 2.13 So, to prove 2.5 it suffices to show that I j n r 2 P max S j nk >εn 1/p <, j 1, 2, 3,

10 10 Journal of Probability and Statistics For any q >q, a q r 1 q a r 1 1 a p <j 1 j q r 1 /p 1 a p <j 1 N n, j 1 N n, j j q r 1 /p N n, j j q r 1 /p j 1 q r 1 /p j 1 q q r 1 /p < Now, we prove that n 1/p max ES 1 nk 0, n i For 0 <q r 1 < 1, taking q<q <psuch that 0 <q r 1 < 1, by 2.4 and 2.15, we get n 1/p max ES 1 nk n 1/p E a X i I ax i n α n α P a X i >n α n 1/p E a X i q r 1 a X i 1 q r 1 I a X i n α n α αq r 1 E a X i q r n 1/p α αq r 1 0, n. ii For 1 q r 1, letting q<q <p,by 2.2, 2.4, and 2.15, weget n 1/p max ES 1 nk n 1/p E a X i I ax i >n α n α P a X i >n α n 1/p a X i q r 1 1 E a X i I a X i n α n α αq r 1 E a X i q r 1 n α 2.18 n 1/p α αq r 1 0.

11 Journal of Probability and Statistics 11 Hence, 2.16 holds. Therefore, to prove I 1 < it suffices to prove that Ĩ 1 n r 2 P max S 1 nk ES 1 nk >εn 1/p <, ε > Note that {X 1 ;1 i n, n 1} is still ND by the defition of X 1 and Lemma 1.6.Usingthe Markov inequality and Lemma 1.8, we get for a suitably large M, which will be determined later, Ĩ 1 n r 2 M/p E max S 1 nk M ES 1 nk n r 2 M/p log M n n E X 1 M E X 1 M/ Ĩ 11 Ĩ 12. Taking M>max 2,p r 1 1 αq / 1 αp, then r 2 M/p αm αq r 1 < 1, and, by 2.15,weget Ĩ 11 n r 2 M/p log M n E a X i M I a X i n α n Mα P a X i >n α n r 2 M/p log M n E a X i q r 1 n α M q r 1 n α M q r 1 E a X i q r 1 n r 2 M/p αm αq r 1 log M n 2.21 <. i For q r 1 < 2, taking q<q <psuch that q r 1 < 2 and taking M>max 2, 2p r 1 / 2 2αp αpq r 1,from 2.15 and r 2 M/p αm Mαq r 1 /2 < 1, we have Ĩ 12 [ n r 2 M/p log M n E a X i q r 1 n α 2 q r 1 I a X i n α n 2α αq r 1 E a X i q r 1 n r 2 M/p αm Mαq r 1 /2 log M n ] M/ <.

12 12 Journal of Probability and Statistics ii For q r 1 2, taking q<q <pand M>max 2, 2p r 1 / 2 pδ, where δ is defined by 2.3, weget,from 2.3, 2.4, 2.15,andr 2 M/p δm/2 < 1, [ Ĩ 12 n r 2 M/p log M n a 2 r 1 n2α αq E a X i q r 1 n r 2 M/p δm/2 log M n <. Since X 2 >εn 1/p a X i n α I n α <a X i <εn 1/p /K >εn 1/p ] M/2 there at least exist K indices k such that a nk X k >n α, we have P X 2 >εn 1/p 1 i 1 <i 2 < <i K n P a 1 X i1 >n α,a 2 X i2 >n α,...,a K X ik >n α that By Lemma 1.6, {a X i ;1 i n, n 1} is still ND. Hence, for q<q <pwe conclude P X 2 >εn 1/p 1 i 1 <i 2 < <i K n K P a j X ij >n α K P a X i >n α n αq r 1 E a X i q r 1 K 2.26 n αq r 1 K, via 2.4 and X 2 > 0 from the defition of X 2. Hence by 2.26 and by taking α>0 and K such that r 2 αkq r 1 < 1, we have I 2 n r 2 P X 2 >εn 1/p n r 2 αq r 1 K < Similarly, we have X 3 < 0andI 3 <.

13 Journal of Probability and Statistics 13 Last, we prove that I 4 <.LetY KX/ε. By the defition of X 4 P max S 4 nk Combing with 2.15, >εn 1/p P P X 4 n >εn 1/p a X i > εn1/p K P a X i > εn1/p K P 1 a p <j 1 Y > nj 1/p N n, j 1 N n, j P l Y p <l 1 l nj l/n N n, j 1 N n, j P l Y p <l 1 l n l n l q r 1 /p P l Y p <l 1. n and 2.1, we have 2.28 l q r 1 /p I 4 n r 2 P l Y p <l 1 n l n l n r 2 q r 1 /p l q r 1 /p P l Y p <l 1 l 1 l r 1 P l Y p <l 1 l E Y p r 1 E X p r 1 <. Now we prove Since maxa nj X j j j 1 max a X i max a X i, 1 j n 1 j n 1 j n 2.30 then from 2.5 we have n r 2 P max anj X j >n 1/p < j n

14 14 Journal of Probability and Statistics Combing with the hypotheses of Theorem 2.1, P max anj X j >n 1/p 0, n j n Thus, for sufficiently large n, P max anj X j >n 1/p < j n By Lemma 1.6, {a nj X j ;1 j n, n 1} is still ND. By applying Lemma 1.10 and 2.1, we obtain P a nk X k >n 1/p 4CP max a nkx k >n 1/p Substituting the above inequality in 2.5, weget n r 2 P a nk X k >n 1/p < So, by the process of proof of I 4 <, E X p r 1 n r 2 P a nk X k >n 1/p < Proof of Theorem 2.2. Let p 2, α<1/p 1/2, and K>1/ 2α. Using the same notations and method of Theorem 2.1, we need only to give the different parts. Letting 2.7 take the place of 2.15, similarly to the proof of 2.19 and 2.26, we obtain n 1/2 maxes 1 n 1/2 α 2α r 1 0, n nk Taking M>max 2, 2 r 1, we have Ĩ 11 n 1 1 2α M/2 r 1 log M n< For r 1 1, taking M>max 2, 2 r 1 / 1 2α 2α r 1,weget Ĩ 12 n 1 1 2α r 1 2α M/2 r 1 log M n<. 2.39

15 Journal of Probability and Statistics 15 For r 1 > 1, EX 2 < from 2.8. Letting M>2 r 1 2,bytheHölder inequality, [ ] M/2 Ĩ 12 n r 2 M/2 log M n a 2 n2α 2α r 1 E a X i 2 r 1 1/ r 1 r 2/ r 1 n r 2 M/2 log M n a 2 r 1 1 M/ n 1 M/2 r 1 r 1 log M n<. By the defition of K, I 2 n 1 r 1 2αK 1 < Similarly to the proof 2.31, we have I 4 l 1 l 1 l n 1 l r 1 P l Y 2 <l 1 l r 1 log lp l Y 2 <l 1 E Y 2 r 1 log Y E X 2 r 1 log X 2.42 <. Equation Using the same method of the necessary part of Theorem 2.1, we can easily get E X 2 r 1 log X n r 2 P a nk X k >n 1/2 < Acknowledgments The author is very grateful to the referees and the editors for their valuable comments and some helpful suggestions that improved the clarity and readability of the paper. This work was supported by the National Natural Science Foundation of China , the Support Program the New Century Guangxi China Ten-Hundred-Thousand Talents Project , and the Guangxi China Science Foundation 2010GXNSFA Professor Dr. Qunying Wu engages in probability and statistics.

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