999精品在线视频,手机成人午夜在线视频,久久不卡国产精品无码,中日无码在线观看,成人av手机在线观看,日韩精品亚洲一区中文字幕,亚洲av无码人妻,四虎国产在线观看 ?

A LIMINF RESULT FOR LéVY’S MODULUS OF CONTINUITY OF A FRACTIONAL BROWNIAN MOTION

2019-04-13 03:59:32LIUYonghongWANGWeina
數學雜志 2019年2期

LIU Yong-hong,WANG Wei-na

(School of Mathematics and Computing Science,Guilin University of Electronic Technology;Guangxi Colleges and Universities Key Laboratory of Data Analysis and Computation,Guilin 541004,China)

Abstract:In this paper,we investigate functional limit for Lévy’s modulus of continuity of a fractional Brownian motion.By using large deviation and small deviation for Brownian motion,a liminf for Lévy’s modulus of continuity of a fractional Brownian motion is obtained,which extends the corresponding result of Brownian motion.

Keywords: fractional Brownian motion;Lévy’s modulus of continuity;liminf result

1 Introduction and Main Result

Let{X(t);t ≥ 0}be a standard α-fractional Brownian motion with 0< α <1 and X(0)=0.The{X(t);t≥0}has a covariance function

for s,t≥0,and representation

where

{X(t);t≥ 0}has stationary increments with E(X(s+t)?X(s))2=t2α,s,t≥ 0 and is a standard Brownian motion when

Let C0[0,1]be the space of continuous functions from[0,1]to R with value zero at the origin,endowed with usual normand

Then H is a Hilbert space with respect to the scalar product

Define a mapping I:C0[0,1]→ [0,∞]by

The limit set associated with functional laws of the iterated logarithm for{X(t);t≥0}is Kα,the subset of functions in C0[0,1]with the form

here the function g(x)ranges over the unit ball of L2(R1),and hence.The subset K of C0[0,1]is defined by

For 0

In[1],Monrad and Rootzén gave a Chung’s functional law of the iterated logarithm for fractional Brownian motion,as follows,for any f∈K,hf,fi<1,

where γ(f)is a constant satifying 2?1/2cα(1 ? hf,fi)?α≤ γ(f) ≤ 2?1/2Cα(1 ? hf,fi)?α,c,C denote the positive constants in(2.13)of[1].

Inspired by the arguments of Monrad and Rootzén,in the present paper,we obtain a liminf result for Lévy’s modulus of continuity of a fractional Brownian motion.The main result is stated as follows.

Theorem 1.1 For each f∈K with hf,fi<1,then

where b(f)is a constant satisfying

here c and C denote the positive constants in(2.13)of[1].

2 Some Lemmas

Our proofs are based on the following lemmas.

In order to prove(3.1)below,we need the following Lemma 2.1.

Lemma 2.1(see(3.14)of[2])Let{X(t);t≥0}be fractional Brownian motion as above,σ2(u)=E(X(t+u)? X(t))2,we have that for any ε>0,there exists a positive constant k0=k0(ε)such that

for any T,00.

In order to prove(3.2)below,we need the following Lemma 2.2 and Lemma 2.3.

Lemma 2.2(see Lemma 2.3 in[3])Let 0<α<1,0≤q0<1 and fix 0

where Ik=(skdk?1,skdr].Let 0< β

there is a constant C0>0 depending only on α such that uniformly in t,u,k,

Lemma 2.3 Let{Γ(t):t≥ 0}be a centred Gaussian process with stationary increments and Γ(0)=0.We assume that σ2(u)=E(Γ(t+u)? Γ(t))2.Let T>0,we have,for x large enough,any ε>0,

where C1>0 is a constant.

Proof This conclusion is from page 49 in[4].

3 The Proof of Theorem 1.1

We only need to show the following two claims:

3.1 The Proof of(3.1)

For any 0< ε<1,choose δ>0,such that.Then we have

By Proposition 4.2 in[1],we have for any δ>0 and n large enough

By Corollary 2.2 in[1],

Choose d>(η? 1)?1,then

which implies,by the Borel-Cantelli lemma,

On the other hand,for any δ>0,

By Lemma 2.1,we have that for any ε>0,there exists a positive k0=k0(ε)such that

By the Borel-Cantelli lemma,

By(3.3)–(3.5),we get

Remark that hnis ultimately strictly decreasing to 0,so for any small h,there is a unique n such that h ∈ (hn+1,hn].Let,s∈[0,1],t∈[0,1].We define

By the definition of in fimum,for any ε>0,there exists∈ (hn+1,hn]such that ξn≥ξ(h0n)? ε.

Noting that

By(3.6)–(3.9),we have

3.2 The Proof of(3.2)

Note that

then it is sufficient to show that

For r=1,2,3,···,we define

for 0 ≤ t≤ 1,dr=rr+(1?γ),sr=r?r,0< γ <1.Then{Zr(·)},r=1,2,···are independent and

where Yr(sr,·)is as in Lemma 2.2.

In order to prove(3.10),we need to prove that for any ε>0,

and

First of all,we prove(3.15).

Now using the argument as in Lemma 2.2,we have

where q,δ are as in Lemma 2.2.For any ε>0,we have,by Lemma 2.3,

Taking n sufficiently large such that,we get(3.15)by the definition of the sequence{hn:n≥1}

Second,we prove(3.14).

For any ε>0,choose δ>0,such that.Let β =2?1/2Cα(1 ? hf,fi)?α,then

By Proposition 4.2 in[1],we have for any δ>0 and n large enough,

By Corollary 2.2 in[1],

Thus

Similar to the proof of(3.15),we have the following estimate for I2,

主站蜘蛛池模板: 五月天综合婷婷| 2020国产精品视频| 国产菊爆视频在线观看| 国产无码精品在线| 免费国产一级 片内射老| 午夜视频免费一区二区在线看| 97在线公开视频| 欧美日本一区二区三区免费| 综合网久久| 热99精品视频| 亚洲高清国产拍精品26u| 55夜色66夜色国产精品视频| 国产精品林美惠子在线观看| 国产第一页免费浮力影院| 国产91线观看| 日韩在线影院| 2020精品极品国产色在线观看| 国产全黄a一级毛片| 99精品一区二区免费视频| 国产免费观看av大片的网站| 欧美成人亚洲综合精品欧美激情 | 一级成人a毛片免费播放| 全裸无码专区| 国产一级精品毛片基地| 中文字幕av无码不卡免费| 国产成人夜色91| 伊人精品视频免费在线| 亚洲一本大道在线| 精品久久久久久中文字幕女| 欧美日本激情| 乱人伦99久久| 无码日韩视频| 久久狠狠色噜噜狠狠狠狠97视色| 国产日本视频91| 欧美在线一二区| 亚洲娇小与黑人巨大交| 无码专区第一页| 久久精品丝袜| 免费 国产 无码久久久| 国产乱人视频免费观看| 日韩精品一区二区三区中文无码| 日韩在线欧美在线| 最新精品久久精品| 丁香婷婷久久| 人人妻人人澡人人爽欧美一区 | 性欧美精品xxxx| 97se亚洲综合在线天天| 亚洲丝袜第一页| 青青草91视频| 欧美精品1区2区| 欧美成在线视频| 亚洲天堂视频在线观看免费| 999精品色在线观看| 国内精品视频在线| 找国产毛片看| 国产农村妇女精品一二区| 日韩欧美高清视频| 青青草国产在线视频| 无码'专区第一页| 超碰aⅴ人人做人人爽欧美 | 波多野结衣无码中文字幕在线观看一区二区 | 先锋资源久久| 成人精品免费视频| 国产不卡国语在线| 看你懂的巨臀中文字幕一区二区| 精品国产香蕉在线播出| 国产一区二区人大臿蕉香蕉| 五月天香蕉视频国产亚| 亚洲高清日韩heyzo| 午夜视频在线观看免费网站| 99这里只有精品6| 久久久久亚洲精品无码网站| 日本a∨在线观看| 无码精品福利一区二区三区| 日韩福利在线视频| 精品人妻系列无码专区久久| 在线观看热码亚洲av每日更新| 欧美在线黄| 综合天天色| 欧美曰批视频免费播放免费| 国产黄色爱视频| 视频二区亚洲精品|