馮德成, 陳彩龍, 蔣文君
(西北師范大學(xué)數(shù)學(xué)與統(tǒng)計(jì)學(xué)院,甘肅蘭州730070)
本文用 X1,X2,… 或 S1,S2,… 表示定義在固定的概率空間(Ω,F(xiàn),P)上的隨機(jī)變量序列,X+=max(0,-X),I(A)表示集合A上的示性函數(shù).
定義1 設(shè)S1,S2,…是L1隨機(jī)變量序列,如果對(duì)任意的j=1,2,…都有
E{(Sj+1- Sj)f(S1,…,Sj)}≥0 (1)其中f是任一使上述期望存在且對(duì)每個(gè)變?cè)墙档暮瘮?shù),則稱{Sn,n≥1}為一個(gè)半鞅.此外,如果f是非負(fù)的,則稱{Sn,n≥1}為一個(gè)半下鞅.
上述定義首先是由Newman和Wright提出來的,此后有很多學(xué)者對(duì)半(下)鞅進(jìn)行了研究(如文獻(xiàn)[1]-[9]等).
而本文將文獻(xiàn)[1]中關(guān)于半鞅的極大值不等式的結(jié)論進(jìn)行了改進(jìn)和推廣,并得到了該不等式的一些應(yīng)用.
定理1 設(shè){Sn,n≥1}是半鞅,h(·)是一個(gè)不減函數(shù),{ck,k≥1}是一正數(shù)序列,若函數(shù)g(·)滿足g(0)=0,且對(duì)?x,y∈R,有g(shù)(y)-g(x)≥(y-x)h(x)且(ck-ck+1)g(Sk)≥0,k≥1,則對(duì)?ε > 0,有
證明: 令m(·)為R上的非負(fù)不減函數(shù),且m(0)=0,定義Sn'=max(c1g(S1),cng(Sn)),S0'=0.由于
由Sn'的定義知,S1'<0則m(Si')=0,而且當(dāng) Si' > Si-1'有 Si'=cig(Si),m(Si')=m(Si-1'),因此有
又因?yàn)?m(·)是 R上非負(fù)不減函數(shù),Si'>Si-1',則
設(shè)
又因?yàn)?ck- ck+1)g(Sk)≥0,k≥1,則
因?yàn)?g(·)滿足g(0)=0,且對(duì) ?x,y∈R,有g(shù)(y)-g(x)≥(y-x)h(x),h(Si)m(Si')是關(guān)于S1,…,Si的不減函數(shù),由{Si,i≥ 1}是半鞅,則ciE[(g(Si)- g(Si-1))m(Sn')]令 m(r)=I{t≥ ε},則
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注 若令g(·)是一個(gè)凸函數(shù),h(·)是函數(shù)g(·)的左導(dǎo)數(shù)函數(shù),即
則h(·)是不減函數(shù),且滿足
g(x)-g(x)≥(y-x)h(x)
此時(shí)定理1則為文獻(xiàn)[1]中的定理2.1.
定理2 設(shè){Sn,n≥1}是半鞅,h(·)是一個(gè)不減函數(shù),{ck,k≥1}是一正數(shù)序列,若函數(shù)g(·)滿足g(0)=0,且對(duì)?x,y∈R,有g(shù)(y)-g(x)≥(y-x)h(x)且(ck-ck+1)g(Sk)≥0,對(duì)p>1,?k≥1,都有E(g(Sk))p< ∞,則
其中
證明 由定理1,得
又因?yàn)?a≥0,b≥0,有alogb≤aloga++be+,則
由(9)得到(7).
推論1 設(shè){Sn,n≥1}是半鞅,h(·)是一個(gè)不減函數(shù),{ck,k≥1}是一正數(shù)序列,若函數(shù)g(·)滿足g(0)=0,且對(duì)?x,y∈R,有g(shù)(y)-g(x)≥(y-x)h(x)且(ck-ck+1)g(Sk)≥0,對(duì)p>1,?k≥1,都有E(g(Sk))p< ∞,則
其中p>1,
證明 由定理2,取ck≡1,則得(10)和(11).
定理3 設(shè){Sn,n≥1}是半鞅,h(·)是一個(gè)不減函數(shù),{ck,k≥1}是一正數(shù)序列,若函數(shù)g(·)滿足g(0)=0,且對(duì)?x,y∈R,有g(shù)(y)-g(x)≥(y-x)h(x)且(ck-ck+1)g(Sk)≥0,?k≥1,對(duì)0<p<1有
證明
由定理1,得
則(12)成立.
推論2 條件同定理1,則對(duì)k≥1,0<p<1,n≥1,有
證明 由定理3,取ck≡1,則得(13).
本文給出了半鞅的另一種形式的Chow型不等式,推廣了文獻(xiàn)[1]的結(jié)論.在此基礎(chǔ)上,得到了半鞅的其它極大值不等式,并進(jìn)行了一些應(yīng)用.
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