R E S E A R C H
Open Access
Complete convergence and complete
moment convergence for negatively
associated sequences of random variables
Qunying Wu and Yuanying Jiang
**Correspondence: [email protected]
College of Science, Guilin University of Technology, Guilin, 541004, P.R. China
Abstract
In this paper, we study the complete convergence and complete moment
convergence for negatively associated sequences of random variables withEX= 0,
Eexp(lnα|X|) <∞,
α
> 1. As a result, we extend some complete convergence and complete moment convergence theorems for independent random variables to the case of negatively associated random variables without necessarily imposing any extra conditions. Our results generalize corresponding results obtained by Gut and Stadtmüller (Stat. Probab. Lett. 81:1486-1492, 2011) and Qiu and Chen (Stat. Probab. Lett. 91:76-82, 2014).MSC: Primary 60F15
Keywords: negatively associated random variables; complete convergence; complete moment convergence
1 Introduction and main results
Definition . Random variablesX,X, . . . ,Xn,n≥, are said to be negatively associated (NA) if for every pair of disjoint subsetsAandAof{, , . . . ,n},
covf(Xi;i∈A),f(Xj;j∈A)
≤,
wherefandfare increasing for every variable (or decreasing for every variable) functions
such that this covariance exists. A sequence of random variables{Xi;i≥}is said to be NA if its every finite subfamily is NA.
By Joag-Dev and Proschan ( []), we have the following lemma.
Lemma .(Joag-Dev and Proschan, []) Let{Xi;i≥}be a sequence of NA random variables.
(i) If{fi;i≥}is a sequence of nondecreasing(or nonincreasing)functions,then
{fi(Xi);i≥}is also a sequence of NA random variables.
(ii) Increasing functions defined on disjoint subsets of a set of negatively associated random variables are negatively associated.
This definition was introduced by Joag-Dev and Proschan ( []). Statistical test de-pends greatly on sampling. The random sampling without replacement from a finite pop-ulation is NA, but is not independent. NA sampling has wide applications such as in mul-tivariate statistical analysis and reliability theory. Because of the wide applications of NA sampling, the limit behaviors of NA random variables have received more and more atten-tion recently. One can refer to: Joag-Dev and Proschan ( []) for fundamental prop-erties, Newman ( []) for the central limit theorem, Matula ( []) for the three series theorem, Shao ( []) for the moment inequalities.
The concept of complete convergence of a sequence of random variables was introduced by Hsu and Robbins ( []). In view of the Borel-Cantelli lemma, complete conver-gence implies almost sure converconver-gence. Chow ( []) first investigated the complete moment convergence, which is more exact than complete convergence. Thus, complete convergence and complete moment convergence are two of the most important problems in probability theory. Their recent results can be found in Wu ( [], []), Xu and Tang ( []), Guoet al.( []), Gut and Stadtmüller ( []), and Qiu and Chen ( []). In addition, Gut and Stadtmüller ( []) and Qiu and Chen ( []) obtained, respectively, complete convergence and complete moment convergence theo-rems for independent identically distributed sequences of random variable withEX= , Eexp(lnα|X|) <∞,α> . In this paper, based on Gut and Stadtmüller ( []) and Qiu and Chen ( []), we extend the complete convergence and complete moment the-orems for independent random variables to the negatively associated sequences of ran-dom variables without necessarily imposing any extra conditions, which extend the cor-responding results of Gut and Stadtmüller ( []) and Qiu and Chen ( []).
In the following, the symbolcstands for a generic positive constant which may differ from one place to another. Let anbn denote that there exists a constantc> such thatan≤cbnfor sufficiently largen,lnxmeansln(max(x, e)), andIdenotes an indicator function.
Theorem . Letα> ,{X,Xn;n≥}be a sequence of NA identically distributed random variables with partial sums Sn=
n
i=Xi,n≥.Suppose that
EX= , Eexplnα|X|<∞, (.)
then
∞
n=
explnαnln
α–n
n P
max
≤k≤n|Sk|>nβ
<∞ for allβ> . (.)
Conversely,if(.)holds for someβ> ,thenEexp(lnα|X/(β)|) <∞;furthermore,ifβ≤
/,thenEexp(lnα|X|) <∞,ifβ> /,thenEexp(( –λ)lnα|X|) <∞for anyλ> .
Theorem . Assume that the conditions of Theorem.and(.)hold.Then
∞
n=
explnαnln
α–n
n+q E
max
≤k≤n|Sk|–βn q
+<∞ for allβ> and all q> . (.)
Remark . By mimicking the analogous part in the proof of Theorem . in Qiu and Chen ( []), (.) and (.) imply, respectively,
∞
n=
explnαnln
α–n
n P
sup
k≥n Sk
k >β
<∞ for allβ>
and
∞
n=
explnαnln
α–n
n E
sup ≤k≤n
Sk
k –β
q
+
<∞ for allβ> and allq> .
Remark . Corresponding results of Gut and Stadtmüller ( []) and Qiu and Chen ( []) are the special cases of our Theorems . and . when{X,Xn;n≥}is i.i.d.
2 Proofs
The following two lemmas will be useful in the proofs of our theorems, and the first is due to Shao ( []).
Lemma .(Shao, [], Theorem ) Let{Xi; ≤i≤n}be a sequence of negatively associated random variables with zero means and finite second moments.Let Sk=
k i=Xi and Bn=
n
i=EXi.Then,for all y> ,a> and <θ< ,
Pmax ≤k≤nSk≥y
≤Pmax ≤k≤nXk>a
+
–θ exp
– y
θ
(ay+Bn)
+ ln
+ay
Bn
.
Lemma . For any random variable X andα> ,
Eexplnα|X|<∞ ⇔ ∞
n=
explnαnln
α–n
n P
|X|>n<∞.
Proof Letan≈bndenote that there exist constantsc> andc> such thatcan≤bn≤ canfor sufficiently largen. We have
∞
n=
explnαnln
α–n
n P
|X|>n
= ∞
n=
explnαnln
α–n
n ∞
j=n
Pj<|X| ≤j+
= ∞
j=
Pj<|X| ≤j+ j
n=
explnαnln
α–n
n
≈
∞
j=
explnαjEIj<|X| ≤j+
≈
∞
j=
Eexplnα|X|Ij<|X| ≤j+
≈Eexplnα|X|,
Proof of Theorem. Letβ> be arbitrary, set, forn≥,bn=βn/(lnαn), define, for ≤k≤n,
Xk=XkI{Xk≤bn}+bnI{Xk>bn}, Sn= n
k=
Xk,
Xk = (Xk–bn)I{bn<Xk≤n}, Xk = (Xk–bn)I{Xk>n}.
Obviously,Xk=Xk+X k+Xk andXkis increasing onXk, thus, by Lemma .(i),{Xk;k≥ }is also a sequence of NA random variables. Note that
max ≤k≤nSk>nβ
⊆max
≤k≤nSk>nβandXk≤bnfor allk≤n
∪max
≤k≤nSk>nβandbn<Xk≤nfor exactly onek≤nand
Xj≤bnfor allj=k
∪X k= for at least twok≤n
∪X k = for at least onek≤n
ˆ
=An∪Bn∪Cn∪Dn.
Therefore,
P
max ≤k≤nSk>nβ
≤P(An) +P(Bn) +P(Cn) +P(Dn). (.)
By condition (.),EX= , andEexp(lnα|X|) <∞,α> , we getEXI(X≤bn) = –EXI(X> bn) andEX<∞. It is well known thatEX<∞impliesEXI(|X|>bn)→,n→ ∞, and we setδ= –ˆ β–> , for sufficiently largen,
max ≤k≤n
ESk≤ max ≤k≤n
kEXI(X≤bn)+nbnEI(X>bn)
≤nE|X|I|X|>bn
+nb–n EXI|X|>bn
≤nEXI(|X|>bn) bn
=ln
α
n
β EX
I|X|>b
n
≤βδlnαn, (.)
so that, takingy= (n–δlnαn)β,a= b
n,θ= / in Lemma ., for sufficiently largen, we get
P(An)≤P
max ≤k≤nSk>nβ
≤Pmax ≤k≤n
Sk–ESk>n–δlnαnβ
exp
– (n–δln
α
n)β
(βn(nln–δαlnnαn)+nEX)
= exp
– β
( –δlnαn n )
β( –δlnαn
n ) +
EXlnαn
n
lnαn
≤exp–lnαn, (.)
from β(–δln
αn
n )
β(–δlnαn n )+EX
lnαn n
→ > asn→ ∞. By the Markov inequality, (.), and (lnn+
ln(β/) –αln lnn)α/lnα
n→ < –δ/ asn→ ∞, for sufficiently largen, (lnn+ln(β/) –
αln lnn)α≤( –δ/)lnαn, thus,
P|X|>bn
≤ Eexp(lnα|X|) exp(lnαb
n)
exp(lnn+ln(β/) –αln lnn)α
≤exp–( –δ/)lnαn, (.)
and, hence, by combining (.) and Lemma .(ii),max≤k≤n≤i≤k,i=k
XiandXkare NA random variable, we get
P(Bn)≤P
∃≤k≤nsuch that max ≤k≤n
≤i≤k,i=k
Xi>βn–n,Xk>bn
≤
n
k=
P
max ≤k≤n
≤i≤k,i=k
Xi>βn–n=βδn
P(Xk>bn). (.)
Similar to the proof of (.), we havemax≤k≤n|E
≤i≤k,i=kXi| ≤βδln
αn, so that, taking
y=βδ(n–lnαn),a= bn,θ= / in Lemma ., using the fact that β δ(–lnαn
n )
βδ(–lnαnn)+EX(n–)lnαn
n →
> asn→ ∞, for sufficiently largen, we get
P
max ≤k≤n
≤i≤k,i=k
Xi>βδn
≤P
max ≤k≤n
≤i≤k,i=k
Xi–EXi>βδn–lnαn
exp
– (n–ln
α
n)βδ
(βδn(lnn–αlnnαn)+ (n– )EX)
=exp
– β
δ( –lnαn n )
βδ( –lnαn n ) +
EX(n–)lnαn
n
δlnαn
≤exp–δlnαn.
Substituting the above inequality and (.) in (.), we obtain
P(Bn)nexp
–δlnαn– ( –δ/)lnαn
= exp–lnαn n (elnn)(δlnα–n)/
By (.),
P(Cn) = P
∃≤k<k≤nsuch thatXk = ,X k=
≤
≤k<k≤n
P(Xk>bn,Xk>bn)≤n
P|X|>b
n
nexp–( –δ/)lnαn
= nexp–( –δ/)lnαn=nexp– – ( –δ)lnαn
= exp–lnαn n
(elnn)(–δ)lnα–n
≤exp–lnαn. (.)
This, together with (.), (.), (.), and (.), shows
P
max ≤k≤nSk>βn
exp–lnαn+nP|X|>n. (.)
Because –Xkis decreasing onXk, by Lemma .(i),{–X, –Xk;k≥}is also a sequence of NA random variables. Obviously,{–X, –Xk;k≥}also satisfies the condition (.). There-fore, replacingXkby –Xkin (.), we get
P
max
≤k≤n(–Sk) >βn
exp–lnαn+nP|X|>n.
Thus,
P
max
≤k≤n|Sk|>βn
≤P
max ≤k≤nSk>βn
+P
max
≤k≤n(–Sk) >βn
exp–lnαn+nP|X|>n. (.)
From (.) and Lemma .,
∞
n=
explnαnln α–n
n P
max
≤k≤n|Sk|>βn
∞
n= lnα–n
n +
∞
n=
explnαnln
α–n
n P
|X|>n
<∞.
That is, (.) holds.
Conversely, if (.) holds, then combining withmax≤k≤n|Xk| ≤max≤k≤n|Sk|, it fol-lows that
∞
n=
explnαnln
α–n
n P
max
≤k≤n|Xk|> βn
<∞, (.)
it implies thatP(max≤k≤n|Xk|> βn)→,n→ ∞, hence, for sufficiently largen,
Pmax
≤k≤n|Xk|> βn
<
Obviously, NA implies pairwise negative quadrant dependent (PNQD) from their defini-tions. Thus, by Lemma . of Wu ( []),
–P
max
≤k≤n|Xk|> βn n
k=
P|Xk|> βn
≤cP
max
≤k≤n|Xk|> βn
,
from which, combining with (.), we have
nP|X|> βn≤cPmax
≤k≤n|Xk|> βn
.
Consequently, by (.),
∞
n=
explnαnln
α–n
n P
|X| β >n
<∞,
and, hence, we haveEexp(lnα|X/(β)|) <∞from Lemma .. Therefore, if <β≤/,
thenEexp(lnα|X|)≤Eexp(lnα|X/(β)|) <∞, ifβ> /, then for anyλ> ,
( –λ)lnαx
lnα(x/(β))→ –λ< , asx→+∞.
This implies that there exists a constantMsuch that for allx≥M, we have ( –λ)lnαx≤
lnα(x/(β)). Hence,
Eexp( –λ)lnα|X| = Eexp( –λ)lnα|X|I|X| ≤M
+Eexp( –λ)lnα|X|I|X|>M
≤c+EexplnαX/(β)I|X|>M EexplnαX/(β)
< ∞.
This completes the proof of Theorem ..
Proof of Theorem. Note that
∞
n=
explnαnln
α–n
n+q E
max
≤k≤n|Sk|–βn q
+
=βq
∞
n=
explnαnln α–n
n+q n
qxq–P
max
≤k≤n|Sk|–βn>βx
dx
+βq
∞
n=
explnαnln
α–n
n+q ∞
n
qxq–P
max
≤k≤n|Sk|–βn>βx
dx
∞
n=
explnαnln
α–n
n P
max
≤k≤n|Sk|>βn
+ ∞
n=
explnαnln α–n
n+q ∞
n xq–P
max
≤k≤n|Sk|>βx
Hence, by (.), in order to establish (.), it suffices to prove that
∞
n=
explnαnln
α–n
n+q ∞
n xq–P
max
≤k≤n|Sk|>βx
dx<∞. (.)
Letβ> be an arbitrary, set, forx≥n,bx=βx/(lnαx), define, for ≤k≤n,
Yk=XkI{Xk≤bx}+bxI{Xk>bx}, Un= n
k=
Yk,
Yk = (Xk–bx)I{bx<Xk≤x}, Yk = (Xk–bx)I{Xk>x}.
By similar methods to the proof of (.), we have
P
max ≤k≤nSk>xβ
≤P(Ax) +P(Bx) +P(Cx) +P(Dx), (.)
which leads to
Ax=
max
≤k≤nUk>xβ ,
Bx=
max
≤k≤nSk>xβandbx<Xk≤xfor exactly onek≤nandXj≤bxfor allj=k ,
Cx=
Yk = for at least twok≤n, Dx=
Yk = for at least onek≤n.
Using similar methods to those used in the proof of (.)-(.), forδ= –ˆ β–> andx≥n, we havemax≤k≤n|EUk| ≤βδln
α
x, and
P(Ax)exp
–lnαx,
P|X|>bx
exp–( –δ/)lnαx,
P(Bx)nexp
–δlnαx– ( –δ/)lnαx
=exp–lnαx n
x(δlnα–n)/≤exp
–lnαx,
P(Cx)≤nP
|X|>bx
exp–lnαxnexp–( –δ)lnαx ≤exp–lnαx,
P(Dx)≤nP(X>x)≤nP
|X|>x,
which, combining with (.), shows
P
max ≤k≤nSk>xβ
exp–lnαx+nP|X|>x.
ReplacingXkby –Xkin the above inequality, we have
Pmax
≤k≤n(–Sk) >xβ
Therefore,
Pmax
≤k≤n|Sk|>xβ
≤Pmax ≤k≤nSk>xβ
+Pmax
≤k≤n(–Sk) >xβ
exp–lnαx+nP|X|>x.
Hence, ∞
n xq–P
max
≤k≤n|Sk|>xβ
dx
∞
n
xq–exp–lnαxdx+
∞
n
xq–nP|X|>xdx
ˆ
=I+I. (.)
By the fact that (a+b)α≥aα+bαfor anya,b> andα> ,
∞
n=
explnαnln
α–n
n+q I
= ∞
n=
explnαnln
α–n
n+q ∞
nqtq–exp–(lnn+lnt)αdt
≤
∞
n=
explnαnln α–n
n+q n
qexp–lnαn ∞
tq–exp–lnαtdt
∞
n=
explnαnln
α–n
n+q n
qexp–lnα
n
= ∞
n= lnα–n
n <∞. (.)
By (.) and Lemma .,
∞
n=
explnαnln α–n
n+q I
= ∞
n=
explnαnln
α–n
n+q ∞
n
xq–P|X|>xdx
= ∞
n=
explnαnln α–n
n+q ∞
j=n j+
j
xq–P|X|>xdx
∞
n=
explnαnln
α–n
n+q ∞
j=n
P|X|>jjq–
= ∞
j=
P|X|>jjq– j
n=
explnαnln
α–n
n+q
∞
j=
explnαjln
α–j
j P
|X|>j<∞,
Conversely, (.) implies (.), that is, the conclusion was established. This completes
the proof of Theorem ..
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
QW conceived of the study, drafted, completed, read, and approved the final manuscript. YJ conceived of the study, and drafted and approved the final manuscript.
Authors’ information
Qunying Wu: Professor, Doctor, working in the field of probability and statistics. Yuanying Jiang: Associate professor, Doctor, working in the field of probability and statistics.
Acknowledgements
The authors are very grateful to the referees and the editors for their valuable comments and some helpful suggestions, which improved the clarity and readability of the paper. This work was supported by the National Natural Science Foundation of China (11361019) and the Support Program of the Guangxi China Science Foundation
(2015GXNSFAA139008).
Received: 14 January 2016 Accepted: 3 June 2016 References
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