王寬程, 楊英鐘

(閩南理工學(xué)院 信息管理學(xué)院, 福建 泉州 362700)

AANA序列是一類廣泛的負(fù)相依序列,文獻(xiàn)[1]介紹了AANA序列的定義并指出NA(相關(guān)定義可見文獻(xiàn)[2])序列是AANA序列,其混合系數(shù)滿足(q(n)=0,n≥1),文中還舉出了一些例子.因此研究AANA陣列的收斂性質(zhì)是有意義的.文獻(xiàn)[3]研究AANA序列的Hajek-Renyi型極大值不等式及其應(yīng)用,文獻(xiàn)[4]建立了AANA序列的Rosenthal型不等式;文獻(xiàn)[5]研究了AANA序列極大不等式和強(qiáng)大數(shù)定律.有關(guān)更多AANA序列的研究成果,可以參見文獻(xiàn)[6-9].本文利用截尾方法和矩不等式,研究AANA陣列在h-可積下的Lp收斂性和完全收斂性.由于AANA列比NA列更弱,因此本文所得的結(jié)論對(duì)NA隨機(jī)變量序列仍然成立.

由于r階Cesaro一致可積嚴(yán)格弱于r階一致可積.而h-可積是比Cesaro一致可積更弱的條件[10]]. 本文研究了AANA陣列在h-可積條件下Lr收斂性和完全收斂性 ,因此本文所得結(jié)論推廣了Cesaro一致可積條件下的相應(yīng)結(jié)論.

1 相關(guān)定義

定義1[1]稱{Xn;n∈N}為漸近幾乎負(fù)相依(簡(jiǎn)稱AANA)隨機(jī)變量序列,如果存在非負(fù)序列q(n)→0(n→0),對(duì)任意的n,k≥1都有

Cov(f(Xn),g(Xn+1,…,Xn+k))≤q(n)(Var(f(Xn))Var(g(Xn+1,…,Xn+k)))1/2.

式中:f和g是任何兩個(gè)使上述方差存在且對(duì)每個(gè)變?cè)鶠榉墙档倪B續(xù)函數(shù).稱{q(n);n∈N}為該序列的混合系數(shù).

稱隨機(jī)陣列{Xnk;1≤k≤n,n∈N}是行為AANA陣列,固定n,假設(shè)每一行內(nèi)的隨機(jī)變量列{Xnk}是AANA的.

定義2[10]稱陣列{Xnk;1≤k≤n,n∈N}關(guān)于常數(shù)陣列{ank;1≤k≤n,n∈N}h-可積的,若滿足下列條件:

2 主要結(jié)論

引理1[6]設(shè){Xn;n∈N}為AANA序列,并且混合系數(shù)是{q(n);n∈N},若{fn;n∈N}皆是單調(diào)非降(或者單調(diào)非增)連續(xù)函數(shù),那么{fn(xn);n∈N}仍然是AANA序列,其混合系數(shù)仍然是{q(n);n∈N}.

定理1 設(shè){Xnk,1≤k≤n,n≥1}是行為兩兩AANA陣列,{ank,1≤k≤n,n≥1}是常數(shù)陣列,{h(n),n≥1}是單調(diào)不減序列,且h(n)→∞(n→∞),1≤p<2,若滿足以下條件:

證明 對(duì)每個(gè)1≤k≤n(n≥1)令

由引理1可知Ynk及Znk仍為AANA陣列,且有|Ynk|≤h(n)和|Znk|≤|Xnk|I(|Xnk|>h(n)).由引理2及Jessen不等式、Cr不等式可知

由條件(1),(2)可得

定理1證畢.

定理2 設(shè){Xnk;1≤k≤n,n≥1}為零均值且E|Xnk|<∞的行為AANA的陣列,{ank}是常數(shù)陣列,{h(n)}是單調(diào)不減序列,且h(n)→∞(n→∞).設(shè)α>0,αp>1,0<δ<1,q>0,t>0為實(shí)數(shù),滿足:αp-α-q<0及α-q+1<0.如果下面3個(gè)條件成立:

(ⅰ) {Xnk}是關(guān)于常數(shù)陣列{ank}的h-可積,且max|ank|=O(n-q);

?ε>0,有

(1)

證明 對(duì)每個(gè)1≤k≤n(n≥1),令

Ynk=XnkI(|Xnk|≤

h(n))-h(n)I(Xnk<-h(n))+

h(n)I(Xnk>h(n)),

由引理1可知{Ynk;1≤k≤n,n≥1}仍為AANA陣列.

先證

因?yàn)?ω∈Dn,有

且?1≤i-h(n)或Xnj(ω)>-h(n).

若記a=#{i:1≤i≤n,Xni(ω)>h(n)},b=#{i:1≤i≤n,Xni(ω)<-h(n)}.

則易知a≤1,b≤1.

當(dāng)a=0,b=0時(shí),有?j(1≤j≤n),|anjXnj(ω)|≤|anj|h(n),所以|anjXnj(ω)|=|anjYnj(ω)|,從而

當(dāng)a=1,b=0時(shí),僅有某個(gè)i0,使得Xni0(ω)>h(n),但仍有|ani0Xni0(ω)|<εnα,而其余的i,都有aniXni(ω)=aniYni(ω).若1≤k≤i0-1,則Sk(ω)=Uk(ω);若i0≤k≤n,則

Yi0(ω)=h(n)

從而

同理可證當(dāng)a=0,b=1時(shí)及當(dāng)a=1,b=1時(shí)的情況,故式(2)成立.因此,有

所以要證式(1),只需證明:

先證式(3),

再證式(4),

要證式(5)成立,只需證

由引理2及條件(ⅱ)可得

定理2證畢.

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