1 Introduction

Definition 1.1 Two random variables X and Y are said to be negatively quadrant dependent (NQD) if for any x,yR,

P(Xx,Yy)P(Xx)P(Yy).
(1.1)

A sequence of random variables { X n ,n1} is said to be pairwise NQD if X i and X j are NQD for all i,jN and ij.

The concept of pairwise NQD was introduced by Lehmann [1] and many applications have been found. See, for example, Matula [2], Wang et al. [3], Wu [4], Li and Wang [5], Gan [6], Huang et al. [7], Chen [8], and so forth. Obviously, the sequence of pairwise NQD random variables is a family of very wide scope, which contains a pairwise independent random variable sequence. Many known types of negative dependence such as negative upper (lower) orthant dependence and negative association (see Joag-Dev and Proschan [9]) have been developed on the basis of this notion. Among them the negatively associated class is the most important and special case of a pairwise NQD sequence. So, it is very significant to study probabilistic properties of this wider pairwise NQD class which contains negatively orthant dependent (NOD) random variables as special cases. The main purpose of this paper is to study strong convergence results for weighted sums of pairwise NQD random variables, which generalize the previous known results for negatively associated random variables and negatively orthant dependent random variables, such as those of Wang et al. [10, 11].

Throughout the paper, denote X + max(0,X), X max(0,X). C denotes a positive constant which may be different in various places. Let a n b n denote that there exists a constant C>0 such that a n C b n for sufficiently large n. The main results of this paper depend on the following lemmas:

Lemma 1.1 (Lehmann [1])

Let X and Y be NQD, then

  1. (i)

    EXYEXEY;

  2. (ii)

    P(X>x,Y>y)P(X>x)P(Y>y), for any x,yR;

  3. (iii)

    If f and g are both nondecreasing (or nonincreasing) functions, then f(X) and g(Y) are NQD.

Lemma 1.2 (Matula [2])

Let { A n ,n1} be a sequence of events defined on a fixed probability space (Ω,F,P).

  1. (i)

    If n = 1 P( A n )<, then P( A n ,i.o.)=0;

  2. (ii)

    If P( A k A m )P( A k )P( A m ), km and n = 1 P( A n )=, then P( A n ,i.o.)=1.

Lemma 1.3 (Wu [4])

Let { X n ,n1} be a sequence of pairwise NQD random variables. If n = 1 log 2 nVar( X n )<, then n = 1 ( X n E X n ) converges almost surely.

2 Properties for pairwise NQD random variables

In this section, we will provide some properties for pairwise NQD random variables.

Property 2.1 Let { X n ,n1} be a sequence of pairwise NQD random variables, for any ε>0, X n 0a.s. n = 1 P(| X n |>ε)<.

Proof ‘⇐’ If n = 1 P(| X n |>ε)<, then X n 0 a.s. follows immediately from the Borel-Cantelli lemma.

‘⇒’ Let X n 0 a.s., we can see that X n + 0 a.s., X n 0 a.s. Denote

E n (1)= ( X n + > ε / 2 ) , E n (2)= ( X n > ε / 2 ) ,

it follows that P( E n (j),i.o.)=0, j=1,2. By Lemma 1.1(iii), we can see that { X n + ,n1} and { X n ,n1} are both pairwise NQD. By Lemma 1.1(ii) and Lemma 1.2(ii), we have n = 1 P( E n (j))<, j=1,2. Therefore, n = 1 P(| X n |>ε) n = 1 P( X n + >ε/2)+ n = 1 P( X n >ε/2)<. □

Property 2.2 Let { X n ,n1} be a sequence of pairwise NQD random variables and { x n ,n1} be a sequence of positive numbers. Denote E i =( X i > x i ), F i =( X i < x i ) and G i =(| X i |> x i ) for i=1,2,, then n = 1 P( G n )= implies P( G n ,i.o.)=1.

Proof It is easily seen that G i = E i + F i , E i F i = and P( G i )=P( E i )+P( F i ) for each i1. Thus, n = 1 P( G n )= implies that n = 1 P( E n )=, or n = 1 P( F n )=. By Lemma 1.2(ii), we have P( E n ,i.o.)=1 or P( F n ,i.o.)=1, which implies that P( G n ,i.o.)=1. □

Property 2.3 Under the conditions of Property 2.2, n = 1 P( G n )=P( G n ,i.o.)=1.

3 Strong convergence properties for weighted sums of pairwise NQD random variables

In this section, we will provide some sufficient conditions to prove the strong convergence for weighted sums of pairwise NQD random variables.

Theorem 3.1 Let 1<α<2 and { X n ,n1} be a sequence of mean zero pairwise NQD random variables with identical distribution

P ( | X 1 | > x ) ={ L ( x ) x α , x 1 , 1 , x < 1 ,
(3.1)

where L(x) is a slowly varying function at infinity and L(cx)/L(x)1 as x for all c>0. Let { a n ,n1} and { b n ,n1} be sequences of positive constants satisfying 0< b n . Denote c 1 = b 1 / a 1 and c n = b n /( a n logn) for n2. Assume that

n = 1 P ( | X n | > c n ) <,
(3.2)

then

1 b n k = 1 n a k X k 0a.s.
(3.3)

Proof By the Borel-Cantelli lemma and Kronecker’s lemma, it is easily seen that (3.2) implies that

k = 1 n a k X k I ( | X k | > c k ) =o( b n )a.s.
(3.4)

Denote

Y k = c k I( X k < c k )+ X k I ( | X k | c k ) + c k I( X k > c k ),k1,

then { Y k ,k1} is still pairwise NQD from Lemma 1.1(iii). Since

k = 1 n a k X k = k = 1 n a k ( Y k E Y k ) + k = 1 n a k E Y k + k = 1 n a k c k ( I ( X k < c k ) I ( X k > c k ) ) + k = 1 n a k X k I ( | X k | > c k ) ,
(3.5)

in order to show 1 b n k = 1 n a k X k 0 a.s., we only need to show that the first three terms above are o( b n ) or o( b n ) a.s.

By C r inequality, Theorem 1b in Feller [[12], p.281] and (3.2), we can get

By Lemma 1.3 and Kronecker’s lemma, we have k = 1 n a k ( Y k E Y k )=o( b n ) a.s.

By (3.2) again,

k = 1 E | a k ( log k ) c k ( I ( X k < c k ) I ( X k > c k ) ) b k | k = 1 E ( I ( X k < c k ) + I ( X k > c k ) ) = k = 1 P ( | X k | > c k ) < ,

which implies that

k = 1 a k c k ( I ( X k < c k ) I ( X k > c k ) ) b k  convergesa.s.

By Kronecker’s lemma, it follows that

k = 1 n a k c k ( I ( X k < c k ) I ( X k > c k ) ) =o( b n )a.s.

By Theorem 1a in Feller [[12], p.281] and (3.2) again, we have

k = 1 | a k ( log k ) E Y k b k | k = 1 c k 1 [ c k P ( | X k | > c k ) + E | X k | I ( | X k | > c k ) ] = 2 k = 1 P ( | X k | > c k ) + k = 1 c k 1 c k P ( | X k | > t ) d t O ( 1 ) + C k = 1 c k 1 c k L ( t ) t α d t O ( 1 ) + C k = 1 L ( c k ) c k α O ( 1 ) + C k = 1 P ( | X k | > c k ) < ,

which implies that

k = 1 a k E Y k b k  converges.

By Kronecker’s lemma, it follows that

k = 1 n a k E Y k =o( b n ).

Hence, the desired result (3.3) follows from the statements above immediately. □

Theorem 3.2 Let 1r<2, { X n ,n1} be a sequence of pairwise NQD random variables with identical distribution and { a n ,n1} be sequences of positive numbers with A n j = 1 n a j . Denote c 1 =1 and c n = A n /( a n logn) for n2. Assume that

(3.6)
(3.7)

then

A n 1 i = 1 n a i X i 0a.s.,n.
(3.8)

Proof Let N(0)=0 and

Y n = c n I( X n < c n )+ X n I ( | X n | c n ) + c n I( X n > c n ),n1.

By (3.7), it is easily seen that c n as n (otherwise, there exist infinite subscripts i and some n 0 such that c i n 0 , hence, N( n 0 )=, which is contrary to N( n 0 ) n 0 r from (3.7)). It follows by (3.6) and (3.7) that

i = 1 P ( X i Y i ) = i = 1 P ( | X i | > c i ) = j = 1 c i j < c i + 1 P ( | X i | > c i ) j = 1 j 1 < c i j P ( | X 1 | > j 1 ) = j = 1 ( N ( j ) N ( j 1 ) ) P ( | X 1 | > j 1 ) = j = 1 ( N ( j ) N ( j 1 ) ) l = j P ( l 1 < | X 1 | l ) = l = 1 j = 1 l ( N ( j ) N ( j 1 ) ) P ( l 1 < | X 1 | l ) l = 1 l r P ( l 1 < | X 1 | l ) E | X 1 | r < .

By the inequality above and Borel-Cantelli lemma, we can see that P( X i Y i ,i.o.)=0. Therefore, in order to prove (3.8), we only need to prove

A n 1 i = 1 n a i Y i 0a.s.,n.
(3.9)

By (3.6) and (3.7) again,

i = 1 log 2 i Var ( a i Y i A i ) i = 1 c i 2 E Y i 2 C i = 1 P ( | X i | > c i ) + C i = 1 c i 2 E X 1 2 I ( | X 1 | c i ) C + C j = 1 j 1 < c i j c i 2 E X 1 2 I ( | X 1 | c i ) C + C j = 1 j 1 < c i j c i 2 E X 1 2 I ( | X 1 | j ) C + j = 2 ( N ( j ) N ( j 1 ) ) ( j 1 ) 2 k = 1 j E X 1 2 I ( k 1 < | X 1 | k ) C + k = 2 j = k ( N ( j ) N ( j 1 ) ) ( j 1 ) 2 E X 1 2 I ( k 1 < | X 1 | k ) C + k = 2 j = k N ( j ) ( ( j 1 ) 2 j 2 ) E X 1 2 I ( k 1 < | X 1 | k ) C + k = 2 j = k j r 3 E X 1 2 I ( k 1 < | X 1 | k ) C + k = 2 k r 2 E | X 1 | r k 2 r I ( k 1 < | X 1 | k ) = C + k = 2 E | X 1 | r I ( k 1 < | X 1 | k ) C + E | X 1 | r < .

Therefore, by the inequality above, Lemma 1.3 and Kronecker’s lemma, we have

A n 1 i = 1 n a i ( Y i E Y i )0,a.s.
(3.10)

In order to prove (3.9), it suffices to prove that

A n 1 i = 1 n a i E Y i 0,n.
(3.11)

It is easily seen that E | X 1 | r < for 1r<2 implies that E| X 1 |<, thus

lim i c i P ( | X i | > c i ) =0.
(3.12)

By the Lebesgue dominated convergence theorem and E X 1 =0, we have

E X i I ( | X i | c i ) =E X 1 I ( | X 1 | c i ) E X 1 =0,i.

Therefore,

|E Y i | c i P ( | X i | > c i ) + | E X i I ( | X i | c i ) | 0,as i,
(3.13)

which implies (3.11) by Toeplitz’s lemma. The proof is completed. □

Remark 3.1 In Theorem 3.2, the condition A n = j = 1 n a j can be relaxed to 0< A n when 1<r<2. It suffices to prove (3.11). In fact, it follows by (3.6) and (3.7) that