# Strong Convergence Properties and Strong Stability for Weighted Sums of AANA Random Variables

Authors: Zhiyong Chen; Shuhe Hu; Jimin Ling; Xuejun Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The Khintchine-Kolmogorov-type convergence theorem and three-series theorem for AANA random variables are established. By using these convergence theorems, we obtain convergence results for AANA sequences, which extend the corresponding ones for independent sequences and NA sequences. In addition, we study the strong stability for weighted sums of AANA random variables and obtain some new results, which extend some earlier ones for NA random variables.


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## Body

## 1. Introduction

Firstly, let us recall some definitions.Definition 1 (cf. Wu [1]).
A sequence{Xn,n≥1}of random variables is said to be stochastically dominated by a random variableXif there exists a constantC such that
(1)P(|Xn|>x)≤CP(|X|>x)
for allx≥0andn≥1.Definition 2 (cf. Chow and Teicher [2]).
A sequence{Yn,n≥1}of random variables is said to be strongly stable if there exist two constant sequences{bn,n≥1}and{dn,n≥1}with0<bn↑∞such that
(2)bn-1Yn-dn⟶0a.s.Definition 3 (cf. Wu [1]).
A functionl(x)>0(x>0)is said to be quasimonotonically increasing function if there existx0>0and constantC>0with∀t≥x≥x0such thatl(t)≥Cl(x). A functionl(x)>0(x>0)is said to be quasimonotonically decreasing function if there existx0>0and constantC>0with∀t≥x≥x0such thatl(t)≤Cl(x).Definition 4 (cf. Wu [1]).
A real-valued functionl(x), positive and measurable on(0,∞), is said to be slowly varying if
(3)limx→∞l(λx)l(x)=1
for eachλ>0.Definition 5 (cf. Joag-Dev and Proschan [3]).
A finite collection of random variablesX1,X2,…,Xnis said to be negatively associated (NA, in short) if for every pair of disjoint subsetsA1, A2of{1,2,…,n},
(4)Cov{f(Xi:i∈A1),g(Xj:j∈A2)}≤0,
wheneverfandgare coordinatewise nondecreasing such that this covariance exists. An infinite sequence{Xn,n≥1}is NA if every finite subcollection is NA.Definition 6 (cf. Chandra and Ghosal [4]).
A sequence{Xn,n≥1}of random variables is called asymptotically almost negatively associated (AANA) if there exists a nonnegative sequenceq(n)→0asn→∞such that
(5)Cov{f(Xn),g(Xn+1,Xn+2,…,Xn+k)}≤q(n)[Var(f(Xn))Var(g(Xn+1,Xn+2,…,Xn+k))]1/2
for alln,k≥1and for all coordinatewise nondecreasing continuous functionsfandgwhenever the variances exist.Obviously, the family of AANA sequences contains NA (in particular, independent) sequences (takingq(n)=0, n≥1) and some more sequences of random variables which do not much deviates from being NA. An example of an AANA sequence which is not NA was introduced by Chandra and Ghosal [5].Since the concept of AANA random variables was introduced by Chandra and Ghosal [4], many applications have been found. For example, Chandra and Ghosal [4] derived the Kolmogorov-type inequality and the strong law of large numbers of Marcinkiewicz-Zygmund, Chandra and Ghosal [5] obtained the almost sure convergence of weighted averages, Ko et al. [6] studied the Hájek-Rènyi type inequality, and Wang et al. [7] established the law of the iterated logarithm for product sums. Yuan and An [8] established Rosenthal-type inequalities for maximum partial sums of AANA sequences. Wang et al. [9] studied some convergence properties for AANA sequence. Wang et al. [10] generalized and improved the results of Ko et al. [6] and studied the large deviation and Marcinkiewicz-type strong law of large numbers for AANA sequences. Yang et al. [11] investigated the complete convergence of moving average process for AANA sequence. Hu et al. [12] and Shen and Wu [13, 14] studied strong convergence property for weighted sums of AANA sequence. Wang et al. [15, 16] and Shen et al. [17] obtained some results on complete convergence for AANA sequence, and so forth.In this paper, we mainly study convergence results for AANA random variables, and the strong stability for weighted sums of AANA random variables, which extend the corresponding ones for independent sequences and NA sequences without necessarily adding extra conditions. The techniques used in the paper are the truncated method, the Khintchine-Kolmogorov-type convergence theorem and three-series theorem for AANA random variables.Throughout this paper, letI(A)be the indicator function of the setA, andXn(c)=-cI(Xn<-c)+XnI(|Xn|≤c)+cI(Xn>c)for somec>0.an=O(bn)denotes that there exists a positive constantCsuch that|an/bn|≤C. The symbolCrepresents a positive constant which may be different in various places. The main results of this paper depend on the following lemmas.Lemma 7 (cf. Yuan and An [8]).
Let{Xn,n≥1}be a sequence of AANA random variables with mixing coefficients{q(n),n≥1}, and letf1,f2,…be all nondecreasing (or nonincreasing) continuous functions; then{fn(Xn),n≥1}is still a sequence of AANA random variables with mixing coefficients{q(n),n≥1}.Lemma 8 (cf. Wang et al. [9]).
Let1<p≤2and{Xn,n≥1}be a sequence of AANA random variables with mixing coefficients{q(n),n≥1}. Assume thatEXn=0for alln≥1and∑n=1∞‍q2(n)<∞; then there exists a positive constantCpdepending only onpsuch that
(6)E(max1≤j≤n|∑i=1j‍Xi|p)≤Cp∑i=1n‍E|Xi|p
for alln≥1, whereCp=2p[22-pp+(6p)p(∑n=1∞‍q2(n))(p-1)].By Lemmas7 and 8, we can get the following Khintchine-Kolmogorov-type convergence theorem and three series theorem for AANA sequences, which can be applied to prove the main results of the paper. The proofs are standard, so we omit them.Corollary 9 (Khintchine-Kolmogorov-type convergence theorem).
Let{Xn,n≥1}be a sequence of AANA random variables with mixing coefficients{q(n),n≥1}and∑n=1∞q2(n)<∞. If
(7)∑n=1∞‍VarXn<∞,
then∑n=1∞(Xn-EXn)converges almost surely.Corollary 10 (three-series theorem for AANA random variables).
Let{Xn,n≥1}be a sequence of AANA random variables with mixing coefficients{q(n),n≥1}and∑n=1∞‍q2(n)<∞. Assume that for somec>0,
(8)∑n=1∞‍P(|Xn|>c)<∞,∑n=1∞‍EXn(c)converges,∑n=1∞‍VarXn(c)<∞.
Then,∑n=1∞Xnconverges almost surely.Remark 11.
Since NA implies AANA, Corollaries9 and 10 extend corresponding results for NA random variables (see Matula [18]) to AANA random variables without adding any extra condition.Lemma 12 (cf. Wu [19] or Shen [20]).
Let{Xn,n≥1}be a sequence of random variables which is stochastically dominated by a random variableX. For anyα>0andb>0, the following two statements hold:
(9)E|Xn|αI(|Xn|≤b)≤C1[E|X|αI(|X|≤b)+bαP(|X|>b)],E|Xn|aI(|Xn|>b)≤C2E|X|αI(|X|>b),
whereC1andC2are positive constants.Lemma 13 (cf. Wu [1]).
Leth(x)>0be a slowly varying function; then for anyδ>0,xδh(x)is a quasimonotonically increasing function andx-δh(x)is a quasimonotonically decreasing function.


## 2. Strong Convergence Properties of Weighted Sums for AANA Sequence

Theorem 14.
Let{Xn,n≥1}be a sequence of AANA random variables with∑n=1∞‍q2(n)<∞. Assume that{gn(x),n≥1}is a sequence of even functions defined onR. For eachn≥1,gn(x)is a positive and nondecreasing function in(0,∞)and satisfies one of the following conditions:(i)
for some0<r≤1,xr/gn(x)is a nondecreasing function in(0,∞);(ii)
for some1<r≤2,x/gn(x)andgn(x)/xrare nonincreasing functions in(0,∞); furthermore, assume thatEXn=0for eachn≥1.
For any positive number sequence{an,n≥1}withan↑∞ such that
(10)∑n=1∞‍Egn(Xn)gn(an)<∞,
then ∑n=1∞Xn/anconverges a.s., and
(11)an-1∑i=1n‍Xi⟶0a.s.,asn⟶∞.Proof.
For eachn≥1, denote
(12)Xn(an)=-anI(Xn<-an)+XnI(|Xn|≤an)+anI(Xn>an).
By Lemma 7, we can see that, for fixedn≥1,{Xn(an)}is still a sequence of AANA random variables. So by Corollary 10 in order to prove (11), we need only to prove the convergence of three series of (8), wherec=1.
Firstly, we prove that∑n=1∞‍P(|Xn/an|>1)<∞under condition (i) or (ii).
For eachn≥1, ifgn(x)satisfies condition (i), noting that{gn(x),n≥1}is a sequence of positive and nondecreasing even function in(0,+∞). Combining Markov’s inequality with (10), it follows that
(13)∑n=1∞‍P(|Xnan|>1)≤∑n=1∞‍P(gn(Xn)>gn(an))≤∑n=1∞‍Egn(Xn)gn(an)<∞.
Ifgn(x)satisfies condition (ii), it is easy to prove that (13) also holds.
Secondly, we will show∑n=1∞‍Var(Xn(an)/an)<∞.
Ifgn(x)satisfies (i), when|x|≤an, we have(|x|r/gn(x))≤(anr/gn(an)), which implies that
(14)|x|ranr≤gn(x)gn(an),x2an2≤(gn(x))2/r(gn(an))2/r.
Note that{gn(x),n≥1}is a sequence of positive and nondecreasing functions in(0,+∞), so0≤(gn(x)/gn(an))≤1when|x|≤an. Consequently,
(15)x2an2≤(gn(x)gn(an))2/r≤gn(x)gn(an),for0<r≤1.
On the other hand, ifgn(x)satisfies condition (ii), then we can also get that
(16)x2an2≤(gn(x)gn(an))2/r≤gn(x)gn(an),for1<r≤2.
Therefore, whether even functiongn(x)satisfies condition (i) or (ii), we can obtain
(17)Var(Xn(an)an)≤E(Xn(an)an)2=E(Xn2an2I(|Xn|≤an))+EI(|Xn|>an)≤E(gn(Xn)gn(an)I(|Xn|≤an))+E(gn(Xn)gn(an)I(|Xn|>an))=Egn(Xn)gn(an).
Therefore, it follows from (10) that
(18)∑n=1∞‍Var(Xn(an)an)≤∑n=1∞‍Egn(Xn)gn(an)<∞.
Finally, we prove that∑n=1∞‍E|Xn(an)/an|<∞.
Ifgn(x)satisfies condition (i), when|x|≤an, we have(|x|/an)≤(|x|r/anr), for0<r≤1. It follows that
(19)|E(Xn(an)an)|≤E(|Xn|anI(|Xn|≤an))+EI(|Xn|>an)≤E(|Xn|ranrI(|Xn|≤an))+E(gn(Xn)gn(an)I(|Xn|>an))≤E(gn(Xn)gn(an)I(|Xn|≤an))+E(gn(Xn)gn(an)I(|Xn|>an))=Egn(Xn)gn(an).
Ifgn(x)satisfies condition (ii), then by the fact thatEXn=0andx/gn(x)is a nonincreasing function in(0,∞), we get
(20)|E(Xn(an)an)|≤|E(XnanI(|Xn|≤an))|+EI(|Xn|>an)=|E(XnanI(|Xn|>an))|+EI(|Xn|>an)≤E(|Xn|anI(|Xn|>an))+EI(|Xn|>an)≤E(gn(Xn)gn(an)I(|Xn|>an))+E(gn(Xn)gn(an)I(|Xn|>an))≤2Egn(Xn)gn(an).
Therefore, whethergn(x)satisfies condition (i) or (ii), it also follows from (10) that
(21)∑n=1∞‍E|Xn(an)an|<∞.
The proof of Theorem 14 is completed by (13), (18), and (21).Corollary 15.
Let{Xn,n≥1}be a sequence of AANA random variables with∑n=1∞‍q2(n)<∞, and let{an,n≥1}be a sequence of positive numbers withan↑∞. There exists some0<p≤2such that
(22)∑n=1∞‍E|Xn|panp<∞.
If1<p≤2, we further assume thatEXn=0. Then, (11) holds.Proof.
We takegn(x)=|x|r,x∈(-∞,∞). If0<r≤1, one can find thatxr/gn(x)≡1forx≥0. So condition (i) of Theorem 14 is satisfied. If1<r≤2, we have thatx/gn(x)=x1-randgn(x)/xr≡1forx≥0. Thus, condition (ii) of Theorem 14 is satisfied. Consequently, the desired result (11) follows from Theorem 14 immediately.Remark 16.
If takingr=1in (i) andr=2in (ii), Theorem 14 and Corollary 15 extend the corresponding ones for NA random variables (see Gan [21]) to AANA random variables.Theorem 17.
Let1<α<2and{Xn,n≥1} be a sequence of AANA random variables with∑n=1∞‍q2(n)<∞and identical distribution
(23)P(|X1|>x)={L(x)x-α,x≥11,x<1,
whereL(x)is a slowly varying function. Let{an,n≥1}and{bn,n≥1}be sequences of positive constants satisfying0<bn↑∞. Denotecn=bn/anfor eachn≥1. Assume that
(24)∑n=1∞‍P(|Xn|>cn)<∞;
then
(25)bn-1∑k=1n‍akXk⟶0a.s.,asn⟶∞.Proof.
Since (23) and (24) imply thatck≥1for all sufficiently largek. Without loss of generality, we assumeck≥1for allk≥1.
By Borel-Cantelli Lemma, it is easily seen that (24) implies that
(26)∑k=1n‍akXkI(|Xk|>ck)=o(bn)a.s.
Denote
(27)Yk=-ckI(Xk<-ck)+XkI(|Xk|≤ck)+ckI(Xk>ck),k≥1;
thus,{Yk,k≥1}is still AANA from Lemma 7. It is easy to check that
(28)∑k=1n‍akXk=∑k=1n‍ak(Yk-EYk)+∑k=1n‍akEYk+∑k=1n‍akck(I(Xk<-ck)-I(Xk>ck))+∑k=1n‍akXkI(|Xk|>ck).
In order to show that bn-1∑k=1n‍akXk→0a.s., we only need to show that the first three terms above areo(bn)oro(bn)a.s.
ByCrinequality, Theorem  1b in [22, page 281] (or see Adler [23]) and (24), we can get
(29)∑k=1∞‍Var(Ykck)≤∑k=1∞‍ck-2EYk2≤3∑k=1∞‍ck-2E[ck2I(|Xk|>ck)+Xk2I(|Xk|≤ck)]=3∑k=1∞‍P(|Xk|>ck)+3∑k=1∞‍ck-2EXk2I(|Xk|≤ck)≤C+6∑k=1∞‍ck-2∫0ck‍tP(|Xk|>t)dt≤C+C∑k=1∞‍L(ck)ck-α≤C+C∑k=1∞‍P(|Xk|>ck)<∞.
It follows from Corollary 9 and Kronecker’s lemma that
(30)∑k=1n‍ak(Yk-EYk)=o(bn)a.s.
By (24) again,
(31)∑k=1∞‍E|akck(I(Xk<-ck)-I(Xk>ck))bk|≤∑k=1∞‍E(I(Xk<-ck)+I(Xk>ck))=∑k=1∞‍P(|Xk|>ck)<∞,
which implies that
(32)∑k=1∞‍akck(I(Xk<-ck)-I(Xk>ck))bkconvergesa.s.
By Kronecker’s lemma, it follows that
(33)∑k=1n‍akck(I(Xk<-ck)-I(Xk>ck))=o(bn)a.s.
By Theorem  1b in [22, page 281] (or see Adler [23]) and (24) again, we have
(34)∑k=1∞‍|akEYkbk|≤∑k=1∞‍ck-1[ckP(|Xk|>ck)+E|Xk|I(|Xk|>ck)]=∑k=1∞‍P(|Xk|>ck)+∑k=1∞‍ck-1E|Xk|I(|Xk|>ck)=2∑k=1∞‍P(|Xk|>ck)+∑k=1∞‍ck-1∫ck∞‍P(|Xk|>t)dt≤C+C∑k=1∞‍ck-1∫ck∞‍L(t)t-αdt≤C+C∑k=1∞‍L(ck)ck-α≤C+C∑k=1∞‍P(|Xk|>ck)<∞,
which implies that
(35)∑k=1∞‍akEYkbkconverges.
By Kronecker’s Lemma, it follows that
(36)∑k=1n‍akEYk=o(bn).
Hence, the desired result (25) follows from (26)–(36) immediately.Remark 18.
Theorem17 generalizes and extends the corresponding one for NA random variables (see Wang et al. [24]) to AANA random variables.Theorem 19.
Let1<r<2and{Xn,n≥1}a sequence of mean zero AANA random variables with∑n=1∞‍q2(n)<∞, which is stochastically dominated by a random variableX. Let{an,n≥1}be a sequence of positive constants satisfyingAn≐∑k=1n‍ak↑∞. Denotecn=An/anfor eachn≥1. Assume that
(37)E|X|r<∞,N(n)≐Card{i:ci≤n}=O(nr),n≥1;
then
(38)An-1∑k=1n‍akXk⟶0a.s.,asn⟶∞.Proof.
LetN(0)=0and denote
(39)Xn(cn)=-cnI(Xn<-cn)+XnI(|Xn|≤cn)+cnI(Xn>cn),n≥1.
It follows by (37) that
(40)∑i=1∞‍P(Xi≠Xi(ci))=∑i=1∞‍P(|Xi|>ci)=∑j=1∞‍∑ci≤j<ci+1‍P(|Xi|>ci)≤C∑j=1∞‍∑j-1<ci≤j‍P(|X|>j-1)=C∑j=1∞‍(N(j)-N(j-1))P(|X|>j-1)=C∑j=1∞‍(N(j)-N(j-1))∑n=j∞‍P(n-1<|X|≤n)=C∑n=1∞‍∑j=1n‍(N(j)-N(j-1))P(n-1<|X|≤n)≤C∑n=1∞‍nrP(n-1<|X|≤n)≤CE|X|r<∞.
By the equality above and Borel-Cantelli lemma, we can getP(Xi≠Xi(ci),i.o.)=0. Therefore, in order to prove (38), we only need to prove that
(41)An-1∑i=1n‍aiXi(ci)⟶0a.s.,n⟶∞.
ByCrinequality, Lemma 12, and (37) again,
(42)∑k=1∞‍Var(akXk(ck)Ak)≤∑k=1∞‍ck-2E(Xk(ck))2≤3∑k=1∞‍ck-2E[ck2I(|Xk|>ck)+Xk2I(|Xk|≤ck)]≤C∑k=1∞‍P(|X|>ck)+C∑k=1∞‍ck-2EX2I(|X|≤ck)≤C+C∑j=1∞‍∑j-1<ck≤j‍ck-2EX2I(|X|≤ck)≤C+C∑j=1∞‍∑j-1<ck≤j‍ck-2EX2I(|X|≤j)≤C+C∑j=2∞‍(N(j)-N(j-1))(j-1)-2×∑k=1j‍EX2I(k-1<|X|≤k)≤C+C∑k=2∞‍∑j=k∞‍(N(j)-N(j-1))×(j-1)-2EX2I(k-1<|X|≤k)≤C+C∑k=2∞‍∑j=k∞‍N(j)((j-1)-2-j-2)×EX2I(k-1<|X|≤k)≤C+C∑k=2∞‍∑j=k∞‍jr-3EX2I(k-1<|X|≤k)≤C+C∑k=2∞‍kr-2E|X|rk2-rI(k-1<|X|≤k)=C+C∑k=2∞‍E|X|rI(k-1<|X|≤k)≤C+CE|X|r<∞.
Hence, by the inequality above, Corollary 9 and Kronecker’s lemma, we have
(43)An-1∑i=1n‍ai(Xi(ci)-EXi(ci))⟶0a.s.,n⟶∞.
In order to prove (41), it suffices to prove that
(44)An-1∑i=1n‍aiEXi(ci)⟶0n⟶∞.
Notice thatEXn=0for eachn≥1, we have
(45)|EXnI(|Xn|≤cn)|=|EXnI(|Xn|>cn)|.
It follows from Lemma 12 and (37) that,
(46)∑k=1∞‍|akEXk(ck)Ak|≤∑k=1∞‍[P(|Xk|>ck)+ck-1|EXkI(|Xk|≤ck)|]=∑k=1∞‍[P(|Xk|>ck)+ck-1|EXkI(|Xk|>ck)|]≤∑k=1∞‍P(|Xk|>ck)+∑k=1∞‍ck-1E|Xk|I(|Xk|>ck)≤C∑j=1∞‍P(|X|>ck)+C∑k=1∞‍ck-1E|X|I(|X|>ck)≤C+C∑j=1∞‍∑ck≤j<ck+1‍ck-1E|X|I(|X|>ck)≤C+C∑j=1∞‍∑j-1<ck≤j‍ck-1E|X|I(|X|>j-1)≤C+C∑j=2∞‍(N(j)-N(j-1))(j-1)-1×∑k=j-1∞‍E|X|I(k<|X|≤k+1)≤C+C∑k=1∞‍∑j=2k+1‍N(j)((j-1)-1-j-1)×E|X|I(k<|X|≤k+1)≤C+C∑k=1∞‍∑j=2k+1‍jr-2E|X|I(k<|X|≤k+1)≤C+C∑k=1∞‍kr-1E|X|I(k<|X|≤k+1)≤C+C∑k=1∞‍E|X|rI(k<|X|≤k+1)≤C+CE|X|r<∞.
By Kronecker’s lemma, we can get (44) immediately. The proof is complete.


## 3. Strong Stability for Weighted Sums of AANA Sequence

Theorem 20.
Let{Xn,n≥1}be a sequence of AANA random variables with∑n=1∞‍q2(n)<∞, which is stochastically dominated by a random variableX. Let{an,n≥1}and{bn,n≥1}be two sequences of positive numbers withcn=bn/anandbn↑∞. DenoteN(x)=Card{n:cn≤x}, x>0. If the following conditions are satisfied:(i)
EN(|X|)<∞;(ii)
∫0∞‍tp-1P(|X|>t)(∫t∞‍y-(p+1)N(y)dy)dt<∞, for somep∈[1,2],then there existdn∈R,n=1,2,…, such that
(47)bn-1∑i=1n‍aiXi-dn⟶0a.s.Proof.
For eachi≥1, denote
(48)Xi(ci)=-ciI(Xi<-ci)+XiI(|Xi|≤ci)+ciI(Xi>ci).
By Definition 1 and conditions (i), we can obtain
(49)∑i=1∞‍P(Xi≠Xi(ci))=∑i=1∞‍P(|Xi|>ci)≤C∑i=1∞‍P(|X|>ci)≤C∑i=1∞‍∫0∞‍I(ci≤t)dP(|X|≤t)=C∫0∞‍N(t)dP(|X|≤t)=CEN(|X|)<∞.
By Borel-Cantelli lemma for any sequence{dn,n≥1}⊂R, with probability 1, the sequences{bn-1∑i=1naiXi-dn}and{bn-1∑i=1naiXi(ci)-dn}converge on the same set and to the same limit. We will prove thatbn-1∑i=1nai(Xi(ci)-EXi(ci))→0a.s., which implies (6) withdn=bn-1∑i=1naiEXi(ci). According to Lemma 7,{ai(Xi(ci)-EXi(ci)),i≥1}is a sequence of AANA random variables with mean zero. ByCrinequality and Lemma 12, we have
(50)∑n=1∞‍E|an(Xn(cn)-EXn(cn))|pbnp≤C∑n=1∞‍cn-pE(|Xn|pI(|Xn|≤cn))+C∑n=1∞‍cn-pcnpP(|Xn|>cn)≤C∑n=1∞‍cn-p[cnpP(|X|>cn)+E(|X|pI(|X|≤cn))]+C∑n=1∞‍P(|X|>cn)≤CEN(|X|)+C∑n=1∞‍cn-p∫0cn‍tp-1P(|X|>t)dt.
Notice that
(51)∑n=1∞‍cn-p∫0cn‍tp-1P(|X|>t)dt=∫0∞‍tp-1P(|X|>t)∑n:cn≥t‍cn-pdt≤C∫0∞‍tp-1P(|X|>t)(∫t∞‍y-(p+1)N(y)dy)dt,
where the last inequality follows from the fact that fort>0(52)∑n:cn≥t‍cn-p=limu→∞∑n:t≤cn≤u‍cn-p=limu→∞∫tu‍y-pdN(y)=limu→∞(u-pN(u)-t-pN(t)+p∫tu‍y-(p+1)N(y)dy),limu→∞¯(u-pN(u))≤limu→∞¯(p∫u∞‍y-(p+1)N(y)dy)≤p∫t∞‍y-(p+1)N(y)dy.
By (50), (51) and condition (i), (ii), we can get that
(53)∑n=1∞‍E|an(Xn(cn)-EXn(cn))|pbnp<∞.
Therefore, it follows from (53) and Corollary 15 that
(54)bn-1∑i=1n‍ai(Xi(ci)-EXi(ci))⟶0,a.s.
The proof is complete.Corollary 21.
Suppose that the conditions of Theorem20 are satisfied andEXn=0for eachn≥1. If∫1∞‍EN(|X|/s)ds<∞, thenbn-1∑i=1naiXi→0a.s.Proof.
According to the proof of Theorem20, we need only to prove that
(55)bn-1∑i=1n‍aiEXi(ci)⟶0,asn⟶∞.
Notice thatEXn=0for eachn≥1; then
(56)|EXi(ci)|=|-E(XiI(|Xi|>ci))-E(ciI(Xi<-ci))+E(ciI(Xi>ci))|=|E((Xi-ci)I(Xi>ci))+E((Xi+ci)I(Xi<-ci))|≤E((|Xi|+ci)I(|Xi|>ci)),(57)∑i=1∞‍ai|EXi(ci)|bi≤∑i=1∞‍ci-1E((|Xi|+ci)I(|Xi|>ci))=∑i=1∞‍P(|Xi|>ci)+∑i=1∞‍ci-1E(|Xi|I(|Xi|>ci))≤CEN(|X|)+∑i=1∞‍ci-1∫ci∞‍P(|Xi|>t)dt=CEN(|X|)+∫1∞‍∑i=1∞‍P(|Xi|>sci)ds≤CEN(|X|)+C∫1∞‍EN(|X|s)ds<∞.
By Kronecker’s lemma, we can get (55) immediately.Theorem 22.
Let{Xn,n≥1}be a sequence of AANA random variables with mean zero and∑n=1∞‍q2(n)<∞, which is stochastically dominated by a random variableX. Let{an,n≥1}and{bn,n≥1}be two sequences of positive numbers withcn=bn/anandbn↑∞. DenoteN(x)=Card{n:cn≤x},x>0. If the following conditions are satisfied:(i)
EN(|X|)<∞;(ii)
∫1∞‍EN(|X|/s)ds<∞;(iii)
max1≤j≤ncjp∑i=n∞ci-p=O(n), for somep∈[1,2],then
(58)bn-1∑i=1n‍aiXi⟶0a.s.Proof.
By (49), condition (i), and Borel-Cantelli lemma, it suffices to provebn-1∑i=1∞aiXi(ci)→0,a.s.So we need only to prove
(59)bn-1∑i=1n‍ai(Xi(ci)-EXi(ci))⟶0a.s.,(60)bn-1∑i=1n‍aiEXi(ci)⟶0,asn⟶∞.
We can get (60) from the proof of Corollary 21. In the following, we prove (59). Putε0=0andεn=max1≤j≤ncjforn≥1. According to Lemma 7,{ai(Xi(ci)-EXi(ci)),i≥1}is a sequence of AANA random variables with mean zero. ByCrinequality and Lemma 12,
(61)∑n=1∞‍E|an(Xn(cn)-EXn(cn))|pbnp≤C∑n=1∞‍cn-pE(|Xn|pI(|Xn|≤cn))+C∑n=1∞‍P(|Xn|>cn)≤C∑n=1∞‍cn-p[cnpP(|X|>cn)+E(|X|pI(|X|≤cn))]+C∑n=1∞‍P(|X|>cn)≤CEN(|X|)+C∑n=1∞‍cn-pE(|X|pI(|X|≤cn)).
It is easy to see that
(62)∑n=1∞‍cn-pE(|X|pI(|X|≤cn))≤∑n=1∞‍cn-pE(|X|pI(|X|≤εn))=∑n=1∞‍cn-p∑j=1n‍E(|X|pI(εj-1<|X|≤εj))≤∑j=1∞‍P(εj-1<|X|≤εj)εjp∑n=j∞‍cn-p≤∑j=1∞‍jP(εj-1<|X|≤εj)≤C∑n=1∞‍P(|X|>εn-1)≤C(1+∑n=1∞‍P(|X|>cn))≤C(1+EN(|X|))<∞.
Therefore,
(63)∑n=1∞‍E|an(Xn(cn)-EXn(cn))|pbnp<∞
follows from condition (i), (61) and (62). By Corollary 15 and (63), we can obtain (59) immediately. The proof is complete.Theorem 23.
Let{Xn,n≥1}be a sequence of AANA random variables with∑n=1∞‍q2(n)<∞, which is stochastically dominated by a random variableX. Let{an,n≥1}and{bn,n≥1}be two sequences of positive numbers withcn=bn/anandbn↑∞. DefineN(x)=Card{n:cn≤x},R(x)=∫x∞‍N(y)y-3dy,x>0. If(i)
N(x)<∞for anyx>0;(ii)
R(1)=∫1∞‍N(y)y-3dy<∞;(iii)
E(X2R(|X|))<∞,then there existdn∈R,n=1,2,…, such that
(64)bn-1∑i=1n‍aiXi-dn⟶0a.s.Proof.
According to Lemma7,{Xi(ci),i≥1}and{Xi(ci)/ci,i≥1}are sequences of AANA random variables.
SinceN(x)is nondecreasing, then for anyx>0(65)R(x)≥N(x)∫x∞‍y-3dy=12x-2N(x),
which implies thatEN(|X|)≤2E(X2R(|X|))<∞. Therefore,
(66)∑i=1∞‍P(Xi≠Xi(ci))=∑i=1∞‍P(|Xi|>ci)≤C∑i=1∞‍P(|X|>ci)≤CEN(|X|)<∞.
By Borel-Cantelli lemma for any sequence{dn,n≥1}⊂R, with probability 1, the sequences{bn-1∑i=1naiXi-dn}and{bn-1∑i=1naiXi(ci)-dn}converge on the same set and to the same limit. We will prove thatbn-1∑i=1nai(Xi(ci)-EXi(ci))→0a.s., which implies the theorem withdn=bn-1∑i=1naiEXi(ci). ByCrinequality and Lemma 12,
(67)∑n=1∞‍Var(anXn(cn)bn)≤∑n=1∞‍cn-2E(Xn(cn))2≤3∑n=1∞‍P(|Xn|>cn)+3∑n=1∞‍cn-2E(Xn2I(|Xn|≤cn))≤C∑n=1∞‍P(|X|>cn)+C∑n=1∞‍cn-2[cn2P(|X|>cn)+E(X2I(|X|≤cn))]≤CEN(|X|)+C∑n=1∞‍cn-2E(X2I(|X|≤cn))≐CEN(|X|)+CI,(68)I=∑n:cn≤1‍cn-2E(X2I(|X|≤cn))+∑n:cn>1‍cn-2E(X2I(|X|≤cn))≐I1+I2.
SinceN(1)=Card{n:cn≤1}≤2R(1)<∞, following from (65) and condition (ii), then we haveI1<∞.
(69)I2=∑n:cn>1‍cn-2E(X2I(|X|≤cn))=∑k=2∞‍∑(k-1)<cn≤k‍cn-2E(X2I(|X|≤cn))≤∑k=2∞‍(N(k)-N(k-1))(k-1)-2E(X2I(|X|≤k))≤∑k=2∞‍(N(k)-N(k-1))(k-1)-2E(X2I(|X|≤1))+∑k=2∞‍(N(k)-N(k-1))×(k-1)-2E(X2I(1<|X|≤k))≐I21+I22.
To proveI2<∞, we need to prove thatI21<∞andI22<∞:
(70)I21≤C∑k=2∞‍(N(k)-N(k-1))∑j=k-1∞‍j-3=C∑j=1∞‍j-3∑k=2j+1‍(N(k)-N(k-1))≤C∑j=1∞‍(j+1)-3N(j+1)≤C∫1∞‍y-3N(y)dy<∞.
SinceN(x)is nondecreasing andR(x)is nonincreasing, then
(71)I22=∑k=2∞‍(N(k)-N(k-1))(k-1)-2×∑m=2k‍E(X2I(m-1<|X|≤m))=∑m=2∞‍E(X2I(m-1<|X|≤m))×∑k=m∞‍(N(k)-N(k-1))(k-1)-2≤∑m=2∞‍E(X2I(m-1<|X|≤m))×∑k=m∞‍N(k)((k-1)-2-k-2)≤C∑m=2∞‍E(X2I(m-1<|X|≤m))×∑k=m∞‍∫kk+1‍N(x)x-3dx=C∑m=2∞‍R(m)E(X2I(m-1<|X|≤m))≤C∑m=2∞‍E(X2R(|X|)I(m-1<|X|≤m))≤CE(X2R(|X|))<∞.
By (67)–(71), we can get that
(72)∑n=1∞‍Var(anXn(cn)bn)<∞.
Therefore, it follows from Corollary 9 and Kronecker’s lemma that
(73)bn-1∑i=1n‍ai(Xi(ci)-EXi(ci))⟶0a.s.
Takingdn=bn-1∑i=1naiEXi(ci),n≥1, we can get (64). The proof is complete.Remark 24.
Since NA implies AANA, Theorem20 extends corresponding result for NA random variable (see Wang et al. [24]) to AANA random variables without adding any extra condition.
Similar to the proof of Corollary21, we can get the following corollary.Corollary 25.
Let the conditions of Theorem23 be satisfied andEXn=0for eachn≥1. If∫1∞‍EN(|X|/s)ds<∞, thenbn-1∑i=1naiXi→0a.s.Corollary 26.
Let{Xn,n≥1}be a sequence of AANA random variables with∑n=1∞‍q2(n)<∞, which is stochastically dominated by a random variableX. Let{an,n≥1}and{bn,n≥1}be two sequences of positive numbers withcn=bn/anandbn↑∞. Letf(x)=xrh(x), whereh(x)>0is a slowly varying function asx→∞,1<r<2. DefineN(x)=Card{n:cn≤x},x>0. If(i)
N(n)=O(f(n))for eachn≥1;(ii)
Ef(|X|)<∞.Then there existdn∈R,n=1,2,…, such that
(74)bn-1∑i=1n‍aiXi-dn⟶0a.s.Proof.
It is easy to verify that conditions (i)–(iii) of Theorem23 hold under the conditions of Corollary 25. So Corollary 25 is true by Theorem 23.Corollary 27.
Suppose that the conditions of Corollary26 are satisfied. IfEXn=0for eachn≥1, thenbn-1∑i=1naiXi→0a.s.Proof.
According to Corollary26, we need only to prove (60). ByEXn=0for eachn≥1and Lemma 12, we have
(75)∑i=1∞‍ai|EXi(ci)|bi≤∑i=1∞‍ci-1[ciEI(|Xi|>ci)+|EXiI(|Xi|≤ci)|]≤∑i=1∞‍ci-1[E|Xi|I(|Xi|>ci)+|EXiI(|Xi|>ci)|]≤2∑i=1∞‍ci-1E|Xi|I(|Xi|>ci)≤C∑i=1∞‍ci-1E|X|I(|X|>ci)=C∑k=1∞‍∑(k-1)<ci≤k‍ci-1E|X|I(|X|>ci>k-1)≤C∑k=2∞‍(k-1)-1(N(k)-N(k-1))×∑j=k∞‍E|X|I(j≤|X|<j+1)≤C∑j=2∞‍∑k=2j‍N(k)((k-1)-1-k-1)×E|X|I(j≤|X|<j+1).
Sincer>1, we can takeδ>0such thatr-δ>1. By Lemma 13 and differential mean value theorem, we can obtain
(76)∑k=2j‍N(k)((k-1)-1-k-1)≤C∑k=2j‍krh(k)((k-1)-1-k-1)=C∑k=2j‍kr-δ(kδh(k))((k-1)-1-k-1)≤Cjδh(j)∑k=2j‍kr-δ-2≤Cjδh(j)∫2j‍xr-δ-2dx≤Cjr-1h(j).
It is easily seen thatxr-1h(x)is a quasimonotonically increasing function by Lemma 13. Hence, we have by (75) and (76) that
(77)∑i=1∞‍ai|EXi(ci)|bi≤C∑j=2∞‍jr-1h(j)E|X|I(j≤|X|<j+1)≤C∑j=2∞‍E(|X|r-1h(|X|)|X|I(j≤|X|<j+1))=C∑j=2∞‍E(|X|rh(|X|)I(j≤|X|<j+1))≤CEf(|X|)<∞.
By Kronecker’s lemma, we can get (60) immediately. The proof is complete.



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*Source: 295041-2013-12-08.xml*