Một số vấn đề lý thuyết xác suất trên không gian banach

8/2/2019 Mt s vn l thuyt xc sut trn khng gian banach

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Mc lc

Li m u i

1 Kin thc chun b 11.1 Bin ngu nhin vi gi tr trong khng gian Banach . . 11.2 Bin Rademacher, nguyn l Co . . . . . . . . . . . . . . 71.3 Cc bt ng thc i vi bin ngu nhin thc v Mar-

tingale . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Bt ng thc ng chu ca tch o . . . . . . . . . 12

2 Tng ca cc bin ngu nhin c lp 13

2.1 i xng ho v mt s bt ng thc ca tng cc binngu nhin c lp . . . . . . . . . . . . . . . . . . . . . 14

2.2 Tnh kh tch ca tng i lng ngu nhin c lp . . 232.3 S tp trung v dng iu ui . . . . . . . . . . . . . . 43

3 Lut mnh s ln 563.1 Pht biu chung cho nh l gii hn . . . . . . . . . . . 563.2 Cc lut s ln . . . . . . . . . . . . . . . . . . . . . . . 68

Kt lun 78

Ti liu tham kho 80

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Li ni u

L thuyt xc sut ra i vo th k 17 bi cc nh ton hc Php.Tuy nhin, phi n na u th k 20 mi thc s c mt c s vng

chc. K t mn khoa hc ny khng ngng pht trin. Ngy nay n tr thnh mt ngnh ton hc ln chim v tr quan trng, khng chc nhiu ng dng m cn l mt ngnh ton c tm l thuyt trnh cao.

L thuyt xc sut trong khng gian Banach l mt nhnh mi caton hc, nhm m rng nghin cu vector ngu nhin v hn chiu. l mt s khi qut t nhin ca qu trnh ngu nhin, cnhiu nh ton hc nghin cu nh A. Beck, B. Maurey, G.Pisier, J-P.

Kahane, J. Hoffmanm-Jorgensen, vi nhiu kt qu quan trng c tmthy . Do , ti chon ti "Mt s vn ca l thuyt xcsut trn khng gian Banach" lm ti lun vn tt nghipthc s ca mnh. Tm hiu vn ny, ti mong mun s nm bt cnhng kt qu c bn ca th h trc t c, v c gng rt ranhng kt lun, nhn xt ca ring mnh. T trang b cho mnh vnkin thc v phng php nghin cu c th i su vo lnh vc .Vi kh nng v thi gian c hn, nn ti ch dng li vic nghincu tng cc bin ngu nhin c lp cng lut s ln trong khng gianBanach. Vi l do , bn lun vn c chia lm ba chng.

Chng 1: Nhng kin thc chun b ca lun vn. Trong chngny, tc gi nu nhng khi nim c bn v bin ngu nhin trong khnggian Banach, chui bin Rademacher v bt ng thc ng chu. y lnhng kt qu quan trng nht nghin cu tng cc bin ngu nhintrong khng gian Banach cc chng sau.

Chng 2 Trnh by v tng ca cc bin ngu nhin c lp. yl mt trong hai chng chnh ca lun vn. Trong chng ny, c

chia thnh ba phn: Phn u xem xt phng php i xng ho trong

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nghin cu cc tnh cht ca tng cc bin ngu nhin c lp, vi ccbt ng thc quan trng nh bt ng thc co. Phn hai nghin cutnh kh tch ca tng cc bin ngu nhin c lp vi cc nh l quan

trng l nh l 2.11 v 2.11. Phn cui, quan trng nht vi vic sdng bt ng thc ng chu nh gi bin c ui, nh l 2.29.Chng 3 Trnh by v lut mnh s ln ca tng cc bin ngu

nhin trong khng gian Banach. Nghin cu s hi t hu chc chn catng cc bin ngu nhin c lp. Chng ny c chia lm hai phn:Phn u nu pht biu chung ca nh l gii hn vi kt qu quantrng nht l nh l 3.5; phn hai l p dng pht biu chung a racc lut s ln c th.

Qua y, tc gi xin c gi li cm n su sc n ngi thy, ngihng dn khoa hc ca mnh l GS. TSKH ng Hng Thng. Ngi a ra ti v hng dn tn tnh trong sut qu trnh nghin cuca tc gi. ng thi tc gi cng chn thnh cm n cc thy c trongkhoa Ton - C - Tin hc trng i hc Khoa hc T nhin, i hcQuc gia H Ni, to mi iu kin cho tc gi v kin thc, ti liuv th tc hnh chnh tc gi hon thnh bn lun vn ny.

H Ni, nm 2009Hc vin

T Cng Sn

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Chng 1

Kin thc chun b

Phn ny, ta s a ra mt s khi nim v kt qu cn dng trong phntip theo nh: Cc khi nim, tnh cht c bn lin quan ti bin ngunhin vi gi tr trong khng gian Banach; bt ng thc ng chu; btng thc co; v dy bin Rademacher vi cc tnh cht ca n.

Vi mc tiu chnh ca lun vn l nghin cu v tng cc bin ngunhin trong khng gian Banach. V vy, chng ny ch chng minh hai

bt ng thc v dy tng ring l bt ng thc Levy v bt ng thcOttavani-Kolmogorov.

1.1 Bin ngu nhin vi gi tr trong khng

gian Banach

y, trnh by mt s khi nim v tnh cht lin quan ti bin ngunhin nhn gi tr trong khng gian Banach nh: Khi nim v binngu nhin vi gi tr trong khng gian Banach, cc s hi t trong cabin ngu nhin trong khng gian Banach, tnh kh tch.

K hiu B l khng gian Banach trn R vi chun ., B l khnggian lin hp ca B.

Gi thit khng gian xc sut (, A,P) l y , v khng gian Btho mn iu kin: tn ti tp m c D l tp con ca hnh cu

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n v trong khng gian lin hp B sao cho :

x = supf

D

|f(x)| (vi x B).

Mt bin ngu nhin hoc vc t ngu nhin vi gi tr trong khng gianBanach B l mt nh x o c X t khng gian xc sut (, A,P)vo B, vi B c trang b trn mt i s sinh bi cc tp mca B.

Bin ngu nhin X vi gi tr trong B c gi l Radon, nu mi > 0, tn ti tp compac K() trong B sao cho

P

{X

K

} 1

.

Ni cch khc, o nh ca P qua X l mt o Radon trn (B, B).Hay tng ng vi X nhn hu ht cc gi tr trong mt khng giantuyn tnh ng, tch c. Hn na, iu ny li tng ng vi tnhcht: X l gii hn hu chc chn ca dy hm n gin:

i

xiIAi vi xi B, Ai A.

i vi bin ngu nhin X, khi o xc sut nh trn B = Xca P qua X c gi l phn phi xc sut ca X.

Khi phn phi ca bin ngu nhin Radon l hon ton c xcnh bi hnh chiu ca n; chnh xc hn, nu X,Y l cc bin ngunhin Radon sao cho mi f B: f(X) v f(Y) (nh l bin ngu nhinthc) c cng phn phi th X = Y.

Kt hp vi nh l trong trng hp thc, th ta c : cc phim hmc trng trn B:

E exp{if(X)} = B

exp{if(x)}d(x) f B

xc nh hon ton phn phi ca X.Nu bin ngu nhin X c phn phi tho mn X = (X) th ta

ni bin ngu nhin X l i xng.Ta ni, n hi t yu ti v k hiu n nu v ch nu

limndn = d

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vi mi lin tc v b chn trn B.Khng gian tt c cc o xc sut Radon trn (B, P(B)) cng

vi t p yu xc nh t s hi t yu trn l khng gian metric .

V vy kim tra dy n hi t yu ta cn ch ra (n) compac yu ngthi tt c cc gii hn c th l nh nhau.i vi iu kin u, mt tiu chun quan trng kim tra l nh

l Prokhonov:Tp (i)iI trong P(B) l compac tng i vi t p yu khi v ch khimi > 0, tn ti tp compac K trong B

i(K) 1 vi mi i I.

Dy (Xn) cc bin ngu nhin Radon vi gi tr trong B gi l hit n X nu dy phn phi Xn X. kim tra (Xn) hi t yun X ta cn kin tra f(Xn) hi t yu ti f(X) vi mi f B v dy(Xn) l cht theo ngha:

Mi > 0, tn ti tp compac K trong B vi

P(Xn > ) 1 vi mi n hoc n ln.

Dy (Xn) gi l hi t theo xc sut ti X nu vi mi > 0

limnP

{Xn X > } = 0.

Dy c gi l b chn theo xc sut ( hay bi chn ngu nhin) nu vimi > 0 tn ti A > 0 sao cho

supP{Xn > A} < .Ta c, nu (Xn) hi t theo xc sut th n hi t yu. iu ngc li

ng nu phn phi gii han tp trung ti mt im. Do , mun kimtra (Xn) hi t theo xc sut v 0 th ch cn kim tra (Xn) l cht vtt c cc gii hn c th bng 0.

Nu 0 < p , k hiuLp(B) = Lp(, A,P, B)

l khng gian tt c cc bin ngu nhin X (trn (, A,P)) nhn gi trtrong B, sao choXp kh tch

EXp = XpdP < , p < .3

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vX = esssupX < , p = .

Khi Lp(B) cng vi

Xp = (E

X

p)1/p l mt khng gian Banach

vi 1 p (v l khng gian vector metric vi 0 < p < 1 ).Nu (Xn) hi t ti X trong Lp(B), th ta ni (Xn) hi t trung bnh

ti X. V khi (Xn) hi t trung bnh ti X th n cng hi t theo xcsut ti X.

Dy (Xn) hi t hu chc chn ti X nu

P{ limn

Xn = X} = 1.

Dy (Xn) l b chn hu chc chn nu

P{supn

Xn < } = 1.

Khc vi s hi t theo xc sut, hi t hu chc chn khng metric hoc, v r rng hi t hu chc chn ko theo hi t theo xc sut.

phn tip theo, ta nu mt s nh ngha v tnh cht lin quann tnh kh tch.

Mt bin ngu nhin Radon X vi gi tr trong B gi l kh tchmnh nu X kh tch (tc thuc L1(B)).By gi, ta nh ngha mt loi kh tch khc l kh tch yu (kh

tch Pettis):Gi s, bin ngu nhin Radon X tho mn: vi mi f B, bin

ngu nhin thc f(X) kh tch. Nu ta xt ton t

T : B L1(, A,P)

nh ngha bi T(f) = f(X) th T l ton t b chn, v vy xc nhmt phim hm tuyn tnh lin tc trn B, n l mt phn t trongB. Nu phn t ny thc s thuc B th ta k hiu phn t l EXv khi ta ni X l kh tch yu (hay kh tch Pettis). Ni cch khc,nu tn ti a B sao cho vi mi f B, ta c Ef(X) = f(a) th X lkh tch yu, v vit EX = a.

Nu bin ngu nhin Radon X kh tch mnh th kh tch yu. ngthi ta c

EX EX

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y ta cng nhc li mt tnh cht quan trng c thit lp tbt ng thc Jensen v tnh c lp. Nu X l bin ngu nhin Radonvi gi tri trong B m EX = 0 ( tc Ef(X) = 0 vi mi f

D) ta ni

X c k vng khng hay X l quy tm.Vi F l mt hm li trong R+, X, Y l 2 bin ngu nhin c lp

ly gi tr trong B sao cho EF(X) < v nu Y c k vng 0.Ta c:

EF(X + Y) EF(X). (1.1)Tip theo ta s chng minh hai bt ng thc quan trng ca dy

tng ring ca dy bin ngu nhin c lp (c th tham kho cc chngminh ny trong [1])

nh l 1.1. (Bt ng thc Levy) Cho(Xi) l bin ngu nhin i xng

nhn gi tr trong B. Vi mik, tSk =k

i=1

Xi. Th vi mi s nguyn

N v t > 0 ta c:

P{maxkN

Sk > t} 2P{SN > t}

vP{max

iNXi > t} 2P{SN > t}.

Nu(Sk) hi t theo xc sut ti S, th ta c bt ng thc m rng sau

P{maxk

Sk > t} 2P{S > t}

v tng t, khi thaySi biXi.Hn na, nu tnh kh tch c m bo th vi mi p: 0 t

}

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Khi N

k=1{ = k} = {max

kNSk > t}.

Vy

P{SN > t} =N

k=1

P{Sn > t, = k}. ()

Li do, vi mi k th (X1, ...,Xk, Xk+1,...,XN) cng phn phi vi(X1,...,XN) v { = k} ch ph thuc vo X1,...,Xk nn ta cng c:

P{SN > t} =N

k=1

P{Sk Rk > t, = k} ()

y Rk = SN Sk. Cng v vi v (*) v (**) v dng bt ng thctam gic, ta c:

2P{SN > t} =N

k=1

P{ = k} = P{maxkN

Sk > t}.

c iu phi chng minh.Bt ng thc tip theo c cp ti l bt ng thc Ottavani-

Kolmogorov. Chng minh ca n c suy ra theo kiu chng minh cabt ng thc Levy.

nh l 1.2. Cho {Xi}iN l dy bin ngu nhin Borel c lp vi ccgi tr trong khng gian Banach tch c B. XtSk =

k

i=1Xi (k N)

th vi mis, t > 0

P{maxkN

Sk > s + t} P{SN > t}1 max

kNP(SN Sk > s)

Chng minh. Xt

= inf{k N : Sk > s + t} nu tn ti k nh th+ nu khng tn ti k nh vy

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Khi { = k} ch ph thuc vo X1, X2, . . . , X k vN

k=1{ = k} = {maxkN Sk > s + t}nn

Nk=1

P{ = k} = P{maxkN

Sk > s + t}.

Ta thy, khi

Sk

> s + t

SN

Sk

SN

Sk

> t.

Vy nn, khi = k v SN Sk s th SN > t.Cng vi tnh c lp ca { = k} v SN Sk nn:

P{SN > t} = P({SN > t} N

k=1

{ = k})

=

N

k=1 P{ = k, SN > t} N

k=1 P{ = k, SN Sk > s}=

Nk=1

P{SN Sk > s}P{ = k} infP{SNSk s}N

k=1

P{ = k}

= (1 maxP{SN Sk > s})P{max SN > s + t} pcm

1.2 Bin Rademacher, nguyn l Co

Vic nghin cu trc tip chui ngu nhin trong khng gian Banach lkh khn. V vy, nh mt bc trung gian ta s nghin cu cc kt qucho trng hp chui c bit dng

i

xii, vi i l cc bin ngu nhin

thc no . Nh l mt php nhng bin ngu nhin thc vo khnggian cc bin ngu nhin nhn gi tr trong khng gian Banach. Di

y, ta xem xt cc tnh cht ca chui dng trn.

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Ta gi dy (i) cc bin ngu nhin Bernulli c lp nhn gi tr1vi xc sut bng nhau v bng 1/2 l dy Rademacher, v chuidng ii (vi i B) l chui Rademacher.Cho (Xi) l mt dy cc bin ngu nhin trong B, gi (i) l dyRademacher c lp ca (Xi), khi : (Xi) l i xng nu v ch nu(Xi) cng phn phi vi (iXi).

y, ta nu mt s tnh cht (chng minh c trnh by [5] v[7]) ca chui Rademacher.

nh l 1.3. Vi 0 < p < khi tn ti cc hng s dng Ap vBp ph thuc vo p sao cho vi mi dy s thc hu hn(i) Ta c

Ap(2i )1/2 iip Bp(2i )1/2.c bit, p = 1 th

(

2i )1/2

2

ii1.

nh l 1.4. (Bt ng thc Co) Cho F : R+ R+ l li, khng gim.Mi dy hu hn (xi) trong khng gian Banach B v mi dy s thc

(i) sao cho vi mii, i 1 vi mii. Ta c:EF(

i

iixi) EF(

i

ixi) (1.2)

P(

i

iixi > t) 2P(

i

ixi > t). (1.3)

Cho nh x : R R l nh x co khi |(s) (t)| |s t| vimi s, t R. Nu h l mt nh trn tp T, chng ta t:

h(t)T = hT = suptT

h(t).

Ta c nh l sau:

nh l 1.5. Cho hm sF : R+ R+ l li v tng. Hn na

i : R

R, i

N

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l nh x co, sao cho i(0) = 0. Th vi mi tp T trongRN

EF(1

2

N

i=1 ii(ti)T) EF(N

i=1 itiT).Nu(xi)iN l cc im trong khng gian Banach, th

E( supf1

|N

i=1

if2(xi)|) 4E

N

i=1

ixixi

. (1.4)

Gi X l mt chui Rademacher trong B ta c mt bt ng thc

ui sau: P{X M + t} 2exp(t/82). (1.5) y M = M(X) l k hiu median ca X , = sup

f1(Ef2(X))1/2.

Ta cng c nh gi:

M 2EX v 2 EX2.(nh gi u tin l t bt ng thc Markov

P{X 2EX} EX2EX =

1

2.

nh gi th hai l hin nhin).

1.3 Cc bt ng thc i vi bin ngu

nhin thc v MartingaleMc tiu ca ta l nghin cu cc tnh cht ca tng cc bin ngu nhinvi gi tr trong khng gian Banach. Tuy nhin, hu ht u dn nvic nh gi cc bin c ui dng:

{|SN ESN| > t}.Mt khc, nu t

di = E(SN|Ai)9

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th di l hiu martingale thc, v

N

i=1 di = SN ESN.V vy, vn li chnh l vic nghin cu bin c ui ca cc martingalethc.

Phn ny, ta s xem xt nhng nh gi ui ca cc martingalethc (chng minh c trong [5]) phc v cho chng sau, c th:

Cho L1 = L1(, A,P) l khng gian tt c cc hm o c f trong sao cho E|f| < gi s ta xt h cc i s:

{, } = Ao A1 ... AN = Av E(f|Ai) l k vong c iu kin i vi Ai cho f trong L1 t

di = E(f|Ai) E(f|Ai1)gi l hiu martingale (v E(di|Ai1) = 0) ta c:

f

Ef =

N

i=1 di.Khi , ta c mt s bt ng thc v c lng ui cho martingalethc sau:

nh l 1.6. Cho f L1, f Ef =N

i=1

di l tng ca hiu martingale

tng ng i vi (Ai)iN. Gi sdi < t a = (N

i=1di2)

1

2 th

vi mit > 0:P{|f Ef| > t} 2exp{t2/2a2}.

nh l 1.7. Cho f trong L1 v f Ef =N

i=1di l tng ca hiu

martingale tng ng i vi (Ai)iN, ta = maxiN

di v

b (N

i=1E(d

2i |Ai1))

1/2

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tht > 0:

P

{|f

Ef

|> t

} 2exp

{

t2

2b

2(1

exp(at/b2))

}.

nh l 1.8. Cho 1 0

P{|f Ef| > t} 2exp{t2/Cqa2}Cp > 0 ch ph thuc vo q.

nh l 1.9. Cho (Xi) l dy hu hn ca cc bin ngu nhin thc clp c k vng khng, sao cho Xi a i. Th mi > 0 , tn ti sthc dng K()( ln), v ()sao cho mi t tho mn: t

K()b,

ta ()b2 vib = (iEX2i )1/2 ta c:P{

i

Xi > t} exp{(1 + )t2/2b2}.

nh l 1.10. Cho (Zi)i N l bin ngu nhin thc dng tht > 0

P{maxk

N

Zi > t} N

i=1(Zi > t)/(1 +

N

i=1(Zi > t)).

c bit, viP{maxkN

Zi > t} 1/2 th:

Ni=1

(Zi > t) 2P{maxiN

Zi > t}.

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1.4 Bt ng thc ng chu ca tch o

Phn cui ca chng ny, ta cp n mt bt ng thc rt quan

trng nghin cu tng i lng ngu nhin. l bt ng thcng chu cho tch o.

Cho mt khng gian xc sut (E, , ) v mt s nguyn N > 1.K hiu P l tch o N trn EN mt im x trong EN c h sx = (x1,...,xN) vi xi E, A l mt tp con ca EN. Chng ta t:

H(A,q,k) =

{x

E

N :

x1,...,xq

A sao cho card

{i

N : xi /

{x1i ,...,x

qi

}} k

}Khi , ta c bt ng thc ng chu c lng c ca H(A,q,k) vi o P (Chng minh c trong [6]).

nh l 1.11. ViA l tp o c vi o tch trong khng gianEN.Khi , c hng sK :

P(H(A,q,k)) 1 [K(ln( 1

P(A))

k+

1

q)]k.

viP l o xc sut ngoi.

c bit, viP(A) 12

vk q ta c:

P(H(A,q,k)) 1 (Koq

)k. (1.6)

Khi , vi dy bin ngu nhin (Xn)nN c lp nhn gi tr trongkhng gian o c E, th tn ti khng gian xc sut tch N sao chovi = (n)nN trong N, Xn() ch ph thuc vo n. Vy (1.6) suyra:

Khi P(X A) 12

v k q ta c:

P(X H(A,q,k)) 1 (Koq

)k. (1.7)

chng sau, ta s p dng nhn xt quan trng ny nh gi bin

c ui ca tng bin ngu nhin c lp.

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Chng 2

Tng ca cc bin ngu

nhin c lp

Chng ny, ta tm cch m rng cc kt qu ca tng cc bin ngunhin thc, cho trng hp bin ngu nhin nhn gi tr trong khnggian Banach. Chng hn v s hi t, v tnh kh tch, v cc nh gi

ca bin c ui.Tuy nhin, chng ta cn nhn mnh rng trong khng gian Banach

tng qut, thiu hn gi thit trc giao E(

Xi)2 =EX2i ; vi (Xi) l

dy cc bin ngu nhin c lp c k vng khng. V vy vic m rngny s l cng vic tng i kh khn. Do ta cn c nhng phngphp khc nghin cu. Phng php u tin c ni n l phngphp i xng ho, c trnh by trong phn mt; phng php thhai l phng php dng dy Rademacher cng c trnh by phn

mt v phn ba; phng php th ba l phng php nghin cu thngqua dy martigale thc phn ba; v cui cng v cng l quan trngnht l s dng bt ng thc ng chu, c ch r phn ba.

Vi tng nh trn, chng ny c chia lm ba phn: Phn mtnghin cu phng php i xng ho, v p dng n chng minhnh l Lvy- It-Nisio, bt ng thc co. Phn hai nghin cu tnh khtch ca tng cc i lng ngu nhin, bt u bng bt ng thcHoffmana-Jorgensen, sau l bt ng thc momen ca tng cc bin

ngu nhin c lp, v nhng kt lun v tnh kh tch. Phn ba vi p

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dng ca bt ng thc ng chu, nh gi c lng ui ca tngcc bin ngu nhin c lp v nh l m rng v tnh kh tch.

Cc kt qu chng ny dng nghin cu chng sau v cc

nh l gii hn.

2.1 i xng ho v mt s bt ng thc

ca tng cc bin ngu nhin c lp

tng c bn trong nghin cu tng cc bin ngu nhin c lp l

khi nim i xng ho. Nu X l bin ngu nhin bt k xc nh trn(, A,P); ta c th xy dng mt bin ngu nhin i xng theo nghaX = X X, xc nh trn ( , A A,P P) v c gi l ixng ho ca X. Vi X l bn sao c lp vi X (xy dng trn khnggian xc sut khc (, A,P)). Phn phi ca X v X-X l thc s linquan; chng hn, ta c mt s bt ng thc sau:

nh l 2.1. Vi mi t; a > 0; ta c

P{X a}P{X > t + a} P{ X > t}.Chng minh. T X X X X suy ra

{X a}{X > t + a} { X X > t}

cng vi tnh c lp, v cng phn phi ca X v X ta c pcm.

c bit, khi ta chon a sao cho P(

X

a)

1

2

th ta c

P{X > t + a} 2P{ X > t}.iu ny cng suy ra mt kt lun quan trng v mi lin h ca

tnh kh tch gia X v X rng:H qu 2.2.

EXp < E

Xp < (2.1)

vi mip > 0.

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Chng minh. p dng bt ng thc:

EX Xp E | X + X |p Cp(EXp + EXp)

ta c ngayEXp < EX Xp < .

Ngc li, t EX Xp < +0

pxp1P(X X x)dx <

+0

xp1P(X x)dx <

(theo bt ng thc trn). Vy ta c iu phi chng minh.

Nhn xt 2.3. Ta c th tng qut khng nh trn (vi vic chng minhtng t) cho hm f l hm kh vi, li, tng th

Ef(X) < Ef( X) < .Phng trnh (2.1) l cha thc s tt cho nhiu p dng khc, v

vy n c ci tin nh sau:

nh l 2.4. Vi mia, t > 0

inffD

P{| f(X) | a}P{X > t + a} P{X X > t}. (2.2)

Chng minh. Tht vy, ly sao cho X() > t + a, khi tn tih D cho: |h(X())| > t + a (do trong B th x = sup

fDf(x)).

T suy ra vi |h(X())| a thX() X() |h(X() X())|

= |h(X()) h(X())| > t + a a = t.V vy m

{ : |h(X())| a} { : X() X() > t}.Do ; vi X() > t + a

inffD P

{|f(X)| a} P{X X > t}.

15

19/83

Suy ra

I{X()>t+a} inff

DP{|f(X)| a} P{X() X > t}.

Ly tch phn theo , p dng nh l Fubini v v X, X cng phnphi nn

inffD

P{|f(X)| a}P{X > t + a}

= inffD

P{|f(X)| a}P{X > t + a} P{X X > t}.

Nhn xt 2.5. Tng t, ta chng minh c rng t, a > 0 thP{X > t + a} P{X X > t} + sup

fDP{|f(X)| > a}.

Chng minh. Tht vy, vi sao cho X() > t + a th tn ti h D h(X()) > t + a, t suy ra vi m X() X() < t th vimi f D ta c f(X() X()) < t. Vy

|h(X())

| h(X())

h(X()

X()) > a.

Nn{X() X > t}{ |f(X)| > a} = .

V th m

I{X()>t+a} P{X() X > t} + supfD

P{|f(X)| > a}.

Sau ly tch phn hai v theo ta c iu phi chng minh.

Thc t, dy i xng ( Xi) xy dng t (Xi) l rt tin li cho vicnghin cu (Xi). lm r iu , ta bt u bng nh l Levy-Ito-Nisio cho dy bin ngu nhin c lp khng nht thit i xng (c thchng minh trc tip nh l ny bng cch p dng tiu chun Kca Prokhorov v bt ng thc Ottavani-Kolmogorov, c trnh bytrong [2] v [7]).

16

20/83

nh l 2.6. Cho {Xi} l dy bin ngu nhin Borel c lp vi cc gitr trong khng gian Banach tch c B. XtSn =

n

i=1Xi (n 1). Cc

iu sau l tng ngi) (Sn) hi t h.c.c.ii) (Sn) hi t theo xc sut.iii) (Sn) hi t yu.

Chng minh. Trc ht ta chng minh cho trng hp i xng:Ta thy (i) (ii) (iii) l hin nhin.

Ta s chng minh (iii) (ii). chng minh iu ny ta chng minhXi

P

0.

Do (Xi) l compc tng i yu, nn tn ti dy con (Xik) Xik yu Xv vy suy ra mi hm tuyn tnh f th f(Xik)

yu f(X). Nhng f(Sn)hi t theo phn phi ti bin ngu nhin thc, nn vi mi > 0 vtn ti M > 0 sao cho

supP{|f(Sn)| > M} < 2.Do tnh i xng nn

supnPXP{|

ni=1

if(Xi)| > M}

= supnP{|f(

ni=1

iXi)| > M}

= supnP{|f(Sn)| > M} < 2

vi (i) l dy Rademacher c lp ca (Xi); vi mi n ta t

A = { : P{|n

i=1

if(Xi)| > M} < }

suy raP(A) = PX(A) > 1

v vy vi 18

thn

i=1f2

(Xi()) 8M2

.

17

21/83

t ta c:

supn

P

{

n

i=1 f2(Xi) 8M2} < suy ra

i

f2(Xi) < h.c.c.

Vy f(Xi)h.c.c 0.

Li v (Xi) l dy cht vi gii hn duy nht l 0, suy ra XiP 0.

Khi ta c (ii), v nu khng th (Sn) l dy cauchy theo xc sut nntn ti > 0 v dy tng (nk) Tk = Snk+1

Snk khng hi t theo xc

sut ti 0, nhng v k

Tk =i

Xi hi t yu nn ta li p dng l lun

trn chok

Tk th dn n TkP 0, iu ny l v l. Vy ta chng

minh c (ii).Ta chng minh (ii) (i)

Do SnP S, khi tn ti dy con (nk) sao cho

k P{SnK S > 2k} < .S dng bt ng thc Levy ta c

P{ maxnk1nnk

Sn Snk1 > 2k+1} 2P{Snk Snk1 > 2k+1}

2(P{Snk S > 2k} + P{Snk1 S > 2k1}).Theo b Borel-Cantelli ta c, dy (Sn) l dy c bn hu chc chn

nn cng hi t hu chc chn, vy ta c (i).Tip theo ta chng minh cho trng hp tng qut.

Tt nhin ta ch cn chng minh (iii) suy ra (i) l .Gi s (Sn) hi t yu ti bin ngu nhin S, trn khng gian xc

sut khc (, A,P); ta xt bn sao (Xi) ca (Xi) v xt Sn =n

i=1 Xn.

Khi (Sn) hi t yu ti S, l bin ngu nhin c cng phn phi vi

S.t

Sn = Sn Sn ;

S = S S nh ngha trn , vy (Sn) l i

xng. Li do tnh lin tc ca tch chp nn Sn yu S, theo chng minh18

22/83

trn th Sn h.c.c S. Khi tn ti trong sao choSn Sn() h.c.c S S() ().

(v nu ngc li th tn ti < 1 P(A) > 1/2 vi

A = { : P{Sn Sn()hi t } < }nhng theo nh l Fubini th P{Sn Snhi t } < 1).Ta thy rng (Sn(

)) l dy compact tng i trong B, hn na chm c trng , vi mi f trong B tho mn:

exp{if(Sn())} exp{if(S())}Suy ra f(Sn()) f(S()) v v vy Sn() hi t trong B nS(), cng vi (*) ta c iu phi chng minh.

Phng php i xng ho c minh ho r hn trong mnh sauy.

nh l 2.7. Cho F : R+ R+ l li, th mi dy hu hn(Xi) ccbnn c lp c k vng 0 trongB, sao cho EF(Xi) < i :

EF(1

2i

iXi) EF(i

Xi) EF(2i

iXi)

vi(i) l dy Rademecher c lp ca (Xi).

Chng minh. Xt X = X X v ly (i) l dy Rademecher c lp t(Xi) v (Xi). Khi v

i

Xi vi k vng 0, theo bt ng thc (1.1) ta

cEF(i Xi) EF(i Xi).

Li nh l Frubini vi ch (Xi) v (iXi) cng phn phi nn ta c

EF(

i

Xi) = EF(i

i Xi).S dng bt ng thc (1.1) cng vi tnh cht li v

i

i(Xi + Xi) c

k vng khng, ta c:

EF(i

i Xi) EF(2i

iXi).

19


23/83

VyF(

i

Xi) EF(2

i

iXi).

Bt ng thc cn li c chng minh tng t

EF(1

2

i

iXi) EF(12

i

i Xi) = EF(12

i

Xi) EF(i

Xi).

Phng php i xng ho v cc nh gi trn ch ra rng huht cc kt qu i vi bin ngu nhin i xng, c th di truyn chotrng hp chung. Vi l do ny, trong cc phn sau ta ch yu xt vi

cc bin ngu nhin i xng.Phn tip theo, ta s dng cc kt qu bit ca tng cc bin

Rademecher thu c cc kt qu cho tng cc bin ngu nhin clp tng qut. Vi t tng tng qut nh sau:

Xut pht t kt qu ca dy Rademecher, chng han dng

Ef(x11,...,xNN) Eg(x11,...,xNN) vi mi x1,...,xN Bsuy ra

Ef(X1()1,...,XN()N) Eg(X1()1,...,XN()N) vi mi .V vy, ly k vng hai v ta c

Ef(X11,...,XNN) Eg(X11,...,XNN)Khi , nu thm gi thit (Xi) i xng th ta c kt qu i vi ccbin (Xi):

Ef(X1,...,XN) Eg(X1,...,XN).Vi t tng trn, ta bt u vi vic m rng bt ng thc co chotrng hp tng qut.

nh l 2.8. Cho (Xi) l dy hu hn cc bin ngu nhin i xngnhn gi tr trong B, lyi vi l hai bin ngu nhin thc sao cho:

i = i(Xi) vi i : R R l i xng, v tng t i vii. Khi , nu |i| |i| a.s vi mi i, cho mi hm li, khng gimF : R+ R+ kh tch th

EF(i

iXi) EF(i

iXi).

20


24/83

Chng ta cng c, vi t > 0P(

iiXi > t) 2P(

iiXi > t).

Bt ng thc trong trng hp c bit p dng khi i = 1{XiAi} 1 i vi Ai c lp, i xng trong B ( trong trng hp c thAi = {x ai}) ta c:

EF(

i

Xi1{Xiai}) EF(

i

Xi).

Chng minh. Dy (Xi) cng phn phi nh (iXi) v bi gi thit i

xng ca i nn (iXi) v (iiXi) cng phn phi, sau theo nh lFubini ta c:

EF(

i

iXi) = EXEF(

i

iiXi).

Tng t , ta cng c

EF(

i

iXi) = EXEF(

i

iiXi).

Vi mi bi nguyn l co (nh l 1.4 ) thEF(

i

ii()Xi()) EF(

i

ii()Xi()).

Ly k vng theo X th

EXEF(

i

iiXi) EXEF(

i

iiXi).

Cng vi hai ng thc trn, ta c bt ng thc u.Bt ng thc th hai c thit lp tng t vi vic p dng bt ngthc (1.3).

Tht vy, do tnh cht cng phn phi v nh l Fubini ta c

P(

i

iXi > t) = PXP(

i

iiXi > t).

V

P(i

iXi > t) = PXP(i

iiXi > t).

21

25/83

Vi mi , theo bt ng thc (1.3) th

P(i

ii()Xi() > t) 2P(i

ii()Xi() > t).

Sau ly tch phn hai v theo ta c iu phi chng minh.

Nhn xt 2.9. Vi X, Y l hai bin ngu nhin i xng, c lp v hmF l hm li, tng; p dng vi 1 = 1, 2 = 0 v 1 = 1, 2 = 1 th ta c:

EFX + Y EFX.T bt ng thc th hai ca nh l cng vi biu din tch phn

ca k vng ta suy ra tnh cht sau:Nu F l hm khng gim, v kh vi th ta c:

EF(

i

iXi) 2EF(

i

iXi).

Nhn xt cui cng l xut pht t cu hi nu (Xi) khng i xngth bt ng thc ca nh l cn ng khng? Ta c kt qu nh sau:

Nu (Xi) khng nht thit i xng nhng c k vng khng cngvi cc gi thit khc nh trn nh l th

EF(i

iXi) EF(2i

iXi).

Chng minh. V

EF(

i

iXi) EF(

i

i Xi) EF(i

i Xi) EF(2i

iXi).

nh l 2.10. Cho F : R+ R+ l li v tng, vi dy(Xi) cc binngu nhin ty EF(Xi) < i.

EF

1

2supfD

i

i|f(Xi)|

EF(

i

iXi). (2.3)

KhiXi l bin ngu nhin i xng v c lp trong L2(B), chng ta c:

EsupfDi f2(Xi) supfDi Ef2(Xi) + 8Ei XiXi. (2.4)22


26/83

Chng minh. (2.3) l n gin khi p dng nh l 1.5, viT = {t = (f(Xi))iN; f = 1} v i(x) = |x| ta c

EF12

supfD

i

i|f(Xi)| EFsupfD|i if(Xi)|= EF(

i

iXi).

chng minh (2.4) ta xt c lng:

EsupfDi f2(Xi) E

supfD

|

i

f2(Xi) Ef2(Xi)| + supfD

i

Ef2(Xi)

= supfD

i

Ef2(Xi) + E

supfD

|

i

f2(Xi) Ef2(Xi)|

.

T nh l 2.7 suy ra:

E

supfD

|

i

f2(Xi) Ef2(Xi)|

2E

supfD

i

if2(Xi)

v, bi (1.4)

E

supfD

i

if2(Xi)

4E

i

iXiXi.

Cng vi tnh cht i xng ca (Xi) ta suy ra iu phi chng minh.

2.2 Tnh kh tch ca tng i lng ngu

nhin c lp

Phn ny ta xem xt tnh kh tch ca tng cc i lng ngu nhinc lp c gi tr trn khng gian Banach. iu dn n, cn c mt

23

27/83

nh gi cho cc bin c ui. y s a ra vi tng n gin,c bit l mt bt ng thc quan trng ca J.Hoffmann- Jorgensen vmt s h qu ca n.

nh l 2.11. Cho (Xi)iN l bin ngu nhin c lp nhn gi tr trongB. Xt

Sk =k

i=1

Xi k N.

Vis, t > 0 ta c

P{maxk 3t+s} (P{maxk t})2+P{maxi s}. (2.5)

Nu bin ngu nhin i xng th vi mit, s > 0

P{maxk 2t + s} 4(P{SN > t})2 + P{maxi s}. (2.6)

Chng minh. t = inf{j N : Sj > t} vi nh ngha ny th{ = j} ch ph thuc X1, X2,...,Xj v {max

kNSk > t} =

N

j=1{ = j}

(hp ri nhau).Trn { = j} th:Nu k < j ta c Sk t.Nu k j ta c

Sk Sj1 + Xj + Sk Sj t + Xj + Sk Sj.

Nn trong mi trng hp ta u c, trn { = j}

maxkN Sk t + max Xi + maxjkNSk Sj

Vy trn { = j} th:

{max Sk > 3t + s} {maxiN

Xi > s} { maxj 2t}.

Li do, tnh c lp ca { = j} v Sk Sj ta c

P{ = j, maxkN Sk > 3t + s}

24


28/83

P{ = j, maxiN

Xi > s} + P{ = j}P{ maxj 2t}.

V maxj 3t + s}

P{ = j, maxiN

Xi > s} + P{ = j}P{2maxkN

Sk > 2t}.

Ly tng theo j = 1...N v cng vi nhn xt

maxkN

{Sk > t} =N

j=1

{ = j} ( hp ri nhau) ta c :

P{maxkN Sk > 3t + s} =

P{maxkN

Sk > t, maxkN

Sk > 3t+s} = P{N

j=1

{ = j}, maxkN

Sk > 3t+s}

P{maxkN

Sk > s, maxiN

Xi > s} + (N

j=1

P{ = j})P{2maxkN

Sk > 2t}

= P{maxiN Xi > s} + (P{maxkN Sk > t})2.Vy ta c iu phi chng minh.

chng minh (2.6) ta c nhn xt, vi i = 1...N thSN Sj1 + Xj + SN Sj.

Vy trn { = j} th

SN t + maxi Xi + SN Sj.Suy ra

P{ = j, SN > 2t + s} P{ = j, max

iXi > s} + P{ = j}P{SN Sj > t}.

p dng bt ng thc Lvy cho bin ngu nhin i xng (nh l 1.1)ta c:

P{ = j, SN > 2t+s} P{ = j, max Xi > s}+2P{ = j}P{SN > t}.25

29/83

Ly tng theo j ta c

P{SN > 2t + s} = P{maxkN

Sk > t, SN > 2t + s}

P{maxkN

Sk > t, max Xi > s} + 2P{maxkN

Sk > t}P{SN > t}.

p dng bt ng thc Lvy (nh l 1.1) ln na, ta c iu phichng minh.

Nhn xt 2.12. T tng ca nh l trn l phn hoch khng gianthnh cc min nh gi trn tng min. Tc gi suy ngh rng nuta phn hoch khng gian mn hn ta s c nhng nh gi tt hn y, chng han ta t:

1 = inf{j N : Sj > t}2 = inf{j N : Xj > s}

sau nh gi trn tng min (1, 2) = (i, j). Tuy nhin, tc gi vncha gi quyt c vn ny.

Bt ng thc trn c s dng ch yu vi t = s, iu th vca chng c bt ngun t bnh phng xc sut v phi. iu nycn c tho lun tip phn sau.

Nh h qu ca bt ng thc trn, mnh tip theo l mt bck thut trc khi trnh by v tnh kh tch. Vi vic p dng biu dintch phn ca k vng, ta c:

nh l 2.13. Cho 0 0 : P{

maxkN

Sk

> t}

(2.4p)1

}th

EmaxkN

Skp 2.4pEmaxiN

Xip + 2(4to)p. (2.7)NuXi l bin ngu nhin i xng v

to = inf{t > 0 : P{SN > t} (8.3p)1}th

ESNp

2.3p

EmaxiN Xip

+ 2(3to)p

. (2.8)

26


30/83

Chng minh. u tin ta chng minh (2.7)Ta ly u > to v s dng (2.5)

EmaxkN

Skp = 4p+

0

P{maxkN

Sk > 4t}dtp

= 4p(

u0

+

+u

)P{maxkN

Sk > 4t}dtp

(4u)p + 4p

+

u (P{maxkN Sk > t})2dtp + 4p

+

u P{max Xi > t}dtp

(4u)p + 4pP{maxkN

Sk > u}+0

P{maxkN

Sk > t}dtp + 4pEmaxiN

Xip.

Do vic chon u > to nn 4pP{maxkN

Sk > t} 12

v vy nn:

EmaxkN Sk

p

(4u)p

+

1

2E

maxkN Skp

+ 4

pE

maxiN Xip

EmaxkN

Skp 2.4pEmaxiN

Xip + 2(4to)p

iu ny ng vi bt k u > to nn ta c iu phi chng minh.Tip theo ta chng minh (2.8)Tng t nh phn trn, ly u > to bi biu din tch phn ca k vngv (2.6) ta c

ESNp = 3p+0

P{SN > 3t}dtp = 3p(u

0

++u

)P{SN > 3t}dtp

(3u)p + 4.3p+u

(P{SN > t})2dtp + 3p+u

P{max Xi > t}dtp

(3u)p

+ 4.3

pP

{SN > u}

+

0

P

{SN > t}dtp

+ 3

pE

maxiN Xip

.

27


31/83

Do vic chon u > to nn 4.3pP{SN > u} 12

khi th

ESNp (3u)p + 4.3pP{SN > u}+0

P{SN > t}dtp

+3pEmaxiN

Xip (3u)p + 12ESNp + 3pEmax

iNXip

3pESNp 2(3u)p + 2.3pEmaxiN

Xip.

Do u bt k u > to nn ta c iu phi chng minh.

Nhn xt 2.14. By gi, ta p dng cch lm tng t cho cc hm khcca dy tng ring, c gng thu c nhng bt ng thc mi, vta c kt qu sau:

Cho (Xi)iN l bin ngu nhin c lp trong L1(B), t Sk =k

i=1

Xi, k N khi vi

to = inf{t > 0 : P{maxkN

Sk > t} 1/8}

th

E{ln(maxkN

Sk + 1)} 8E{ln(maxiN

Xi + 1)} + 2 ln(4to + 1).

Nu Xi l bin ngu nhin i xng v

to = inf{t > 0 : P{SN > t} 1/24}

thE ln(SN + 1) 6E ln(max

iNXi) + 2 ln(3to + 1).

Chng minh. Ta ch chng minh cng thc u, cng thc th hai tngt.

Vi u > to ta c

E ln(maxkN

Sk

+ 1) =

+

0

P

{max max

kN Sk

> 4t

}d ln(4t + 1)

28


32/83

u0

4

4t + 1dt +

+u

P{maxkN

Sk > 4t} 44t + 1

du.

Sau p dng bt ng thc (2.5) cng ch vi t > 0 th4

4t + 1 0 v (Zi ) l dy hu hn cc bin ngu nhindng trongLp chon > 0, v t

o = inf{t > 0,

i

P(Zi > t) }

th:

(1 + )1po + (1 + )1

i

+

o

P{Zi > t}dtp Emaxi

Zpi

po +

i

+o

P{Zi > t}dtp.

Chng minh. Bt ng thc v phi l tm thng v:

Emaxi

Zpi =

+0

P{maxi

Zi > t}dtp

o0

P{maxi

Zi > t}dtp++o

P{maxi

Zi > t}dtp po++o

i

P{Zi > t}dtp

( do{

maxi

Zi

> t}

= i {Zi > t} P{maxi Zi > t} i P{Zi > t}).29


33/83

By gi, ta chng minh bt ng thc tri:p dng bt ng thc ca nh l 1.10. Do (Zi)iN l bin ngu nhinc lp dng, nn vi mi t th

P{maxi

Zi > t} i

P{(Zi > t)}1 +i

P{(Zi > t)} .

Vi ch y =x

1 + xng bin trn (1, +) v

o = inf{t > 0,iP(Zi > t) }.

Nn ta cNu t o th

i

P(Zi > t) P{maxi

Zi > t} 1 +

.

Nu t > o, do Zi dng nn

iP{(Zi > t)} vy

P{maxi

Zi > t} i

P{(Zi > t)}1 +

.

V vy m

Emaxi

Zpi =

+

0P{max

iSi > t}dtp

o

0(1+)1dtp+

+

oi

P{(Zi > t)}1 +

dtp

= (1 + )1po + (1 + )1

i

+o

P{(Zi > t)}dtp.

Khng nh v s tng ng ca cc mmen c th hin mnh sau.

30


34/83

nh l 2.16. cho p, q > 0 v (Xi ) l dy hu hn cc bin ngu nhinc lp v i xng trong Lp(B).Khi tn ti hng sKpq ch ph thuc vo p,q sao cho

i

Xip Kpq maxi

Xip +

i

XiI{Xio}q

vi o = inf{t > 0,i

P{Xi > t} (8.3p)1} v k hiua Kpq b nghal K1pq b a Kpqb.Chng minh. Do bt ng thc tam gic

EXi + Yip Cp(EXip + EYip) (Cp 2p)ta c

E

i

Xip 2pE

i

XiI{Xio}p + 2pE

i

XiI{Xi>o}p.

Chng ta p dng (2.8) cho s hng th 2 v phi ca bt ngthc ny, vi nh ngha ca o th

to = inf{t > 0 : P{i

XiI{Xi>o} > t} (8.3p)1} = 0

( v mi t > 0 th ta c

{ :

i

XiI{Xi>o} > t} { :

i

XiI{Xi>o} > 0}

i { : Xi() > o}nn P{

i

XiI{Xi>o} > t} i

P{ : Xi() > o} (8.3p)1 ).Suy ra

E

i

XiI{Xi>o}p 2.3pEmaxi

Xip.

Quay li s hng u v p dng li nh l 2.13 ( bt ng thc (2.8))vi nhn xt

to (8.3p

Ei

XiI{Xio}q

)

1

q

31


35/83


36/83

Nhng theo bt ng thc Marcop v bt ng thc co, ta c

P{

iXiI{Xio} > (8.3qE

iXip) 1p}

E

i

XiI{Xio}p

8.3qEi

Xip E

i

Xip

8.3qEi

Xip 1

8.3q.

V vy to (8.3qEi

Xip)1

p nn

E

iXiI{Xio}q 2.3qqo + 2.3q.8

1

p3qp (E

iXip)

qp . ()

Tuy nhin theo b 2.15 th

Emaxi

Xip (1+)1po+1

1 +

i

+o

P{(Xi > t)}dtp (1+)1po.

Vy

V P() 2.3qq

p (1 + )q

p (Emaxi Xip)q

p + 2.3q.81

p3q

p (Ei

Xip)q

p .

Suy ra

i

XiI{Xio}q = (E

i

XiI{Xio}q)1

q

Cpq maxi

Xip + Cpq

i

Xip. ()

Hn na theo bt ng thc L-vy (nh l 1.1) th

Emaxi

Xip 2E

i

Xip.

Nn max

iXip C

i

Xip. ( )

T (**) v (***) ta c :

i

XiI{Xio}q Cpqi

Xip.

33

37/83

Sau li kt hp vi (***) ta c:

Cpqi Xip max

iXip + i XiI{Xio}

q.

Ta c iu phi chng minh.

By gi, chng ta tng kt v tnh kh tch cho tng cc bin ngunhin c lp nhn gi tr trn khng gian Banach, da trn cc btng thc v cc lp lun trn.

nh l 2.17. Cho (Xi)iN l dy bin ngu nhin c lp vi gi tr

trongB

t Sn =

ni=1 Xi (n 1) v 0 < p < th nu sup Sn < h.c.c, chng ta c s tng ng gia(i) E sup

nSnp < .

(ii) E supn

Xnp < .Hn na, nu(Sn) hi t h.c.c tiS, (i) v (ii) s tng ng vi(iii) ESp < .(iv) (Sn) hi t trongLp.

Chng minh. (i) suy ra (ii) l d dng, v t

Xn+1 = Sn+1 Snsuy ra

Xn+1 Sn+1 + Sn.Vy ta c

sup

n Xn

2sup

n Sn

.

Nn khi E supn

Snp < suy ra E supn

Xnp < . Tc (i) suy ra (ii).Chng minh (ii) suy ra (i) khi N c nh, bi nh l 2.13 vi

to = inf{t > 0 : P{maxkN

Sk > t} (2.4p)1}.

Ta cEmax

nN Sn

p

2.4pEmax

nN Xi

p + 2(4to)

p.

34

38/83

V supn

Sn nu tn ti M > 0 sao cho to M v khng ph thucvo N, cho N ra v cng ta c (i).

Vi gi thit v s hi t hu chc chn, u tin ta chng minh (i),(ii) tng ng vi (iii)(i) suy ra (iii) l hin nhin, do sup

nSnp

n

Xnp

Chng minh (iii) suy ra (ii):Ta i xng ho v p dng tnh cht 2.1 suy ra:

Ei

Xip < sau p dng bt ng thc Levy ta c c (ii) cho(

Xi) ri li p dng tnh cht (2.1).

Ta chng minh (iii) tng ng vi (iv)

(iv) suy ra (iii) l hin nhin, by gi ta chng minh (iii) suy ra (iv) thtvy:V

Sn Sp 2p(supn

Snp + Sp)

hn na do ESp < nn E supn

Snp < vy theo nh l lebesguev hi t b chn suy ra (iv).

Nhn xt 2.18. T nh l ny ta c th rt ra nhng nhn xt sau:Th nht: Nu (Xn) l dy bin ngu nhin c lp vi gi tr trongB v tha mn P{Xn < c} = 1 vi mi n, v Sn hcc S th

ESp < vi mi p > 0

(iu ny c c l v vi gi thit trn th E supn

Xnp ap < ).Th hai: T nh l 2.17 ta c

iIAi hi t hu chc chn th

Ei

IAi = i

P(Ai) < y tng ng cho phn c lp ca b Borel- Cantelli.

Vi t tng, c gng tm mt khng nh v mi quan h gia shi t hu chc chn v hi t trung bnh ging nh nh l Kolmogrov-Khinchin. Ta c kt qu sau:

H qu 2.19. Cho dy (Xn) cc bin ngu nhin c lp thuc Lp()

vi gi tr trong khng gian Banach, t Sn =

ni=1 Xi. Vip > 0:35

39/83

Nu(Sn) hi t trongLp thSn hi t hu chc chn.NuSn hi t hu chc chn tiS v mt trong hai iu sau tho mn:

i) Tn tic

P{Xn < c} = 1, vi min.ii) S thuc Lp.

Th(Sn) hi t trongLp.

Chng minh. Khng nh u tin, l do s hi t trung bnh ko theos hi t theo xc sut, tuy nhin p dng nh l Levy v s hi tu,suy ra (Sn) hi t hu chc chn.

Tip theo, ta chng minh khng nh th hai, gi s Snhcc

S Ta schng minhESn Sp 0.

Tht vy: vi iu kin (i)

P{Xn < c} = 1, vi mi nv p dng nh l 2.17 th

E supn

Snp <

v vy mSn Sp 2p sup

nSnp < .

Do gi thit, thSn Sp hcc 0.

Nn theo nh l hi t b chn lebesgue suy ra

ESn Sp 0.Vi iu kin (ii)

ESp < theo nh l 2.17 th

E supn

Snp < .

Sau lp li lp lun nh phn (i).V vy, ta c iu phi chng minh.

36

40/83

By gi, vi mc tiu m rng nh l 2.17. Ta c kt qu sau:

H qu 2.20. Cho (an) l mt dy cc s dng tng ti v cng, (Xi) l

bin ngu nhin c lp nhn gi tr trong B v tSn =n

i=1Xi vin > 1

th nusup{Snan

} < h.c.c. Khi , vi0 < p < ta c cc mnh sau tng ng:

i) E supn

(Snan

)p < .ii) E sup

n(Xnan )

p < .

Chng minh. Ta xt dy mi (Yi) ca dy bin ngu nhin vi gi tr

trong khng gian Banach l(B)(l(B) l tp cc dy x = (xn) vi chun sup x = sup

n{xn})

Bi php t

Yi = (0; ...; 0;Xiai

;Xi

ai+1; .....).

Vi i thnh phn u bng khng,v r rng Yi = Xiai vi i (do(ai) l n iu tng ).

ng thini

Yi = (X1a1

;X1 + X2

a2; ..;

X1 + ... + Xnan

;X1 + ... + Xn

an+1; ...).

Suy ra

ni

Yi = sup1in

Siai

.

V vy msup

n

ni

Yi = supn

Snan

< h.c.c.

p dng nh l 2.17 cho dy Yi trong l(B), ta c iu phi chngminh.

Lin quan n nh l 2.17, ta cn c hai khng nh sau:

37

41/83

nh l 2.21. Vi gi thit ca nh l2.17, cng vi iu kinSnhcc S

th cc khng nh ca nh l tng ng vi iu kin t > 0 ta c

i=1

EXnpI{Xi>t} < .

H qu 2.22. Cho p > 0, X1,...,Xn l dy bin ngu nhin c lp, khi

i=1

Xi hi t hu chc chn khi v ch khi vi mi a > 0 hai iu kin

sau c tho mn:

i)

i=1P{Xi a} < .

ii) i=1

XiI{Xia} hi t trung bnh cp p.

Chng minh hai khng nh ny c trnh by trong [7].By gi ta xt chui hi t hu chc chn S =

i

Xi cc bin ngu

nhin c lp, b chn u nhn gi tr trong B, i xng hoc c kvng khng .Chng ta s nghin cu tnh kh tch ca S. Vi gi thitc bit hn l Xi l i xng v Xi a < cho i. Vi mi N,ta t SN =

Ni=1

Xi. Bi (2.6) vi mi t > 0

P{SN > 2t + a} (2P{SN > t})2

( do P{maxiN

Xi > a} = 0) vi to xc nh dy

tn = 2n(to + a) a

khi p dng bt ng thc trn vi t := tn1

ta c

P{SN > tn} = P{SN > 2tn1 + a} (2P{SN > tn1})2.Lp li n-1 ln, ta thu c kt qu

P{SN > tn} 22n+12(P{SN > to})2n.Nu S =

i

Xi l hi t h.c.c, th s tn ti to sao cho

N, P{SN > to} 18

38

42/83

khi ta cP{SN > 2n(to + a)} 22n.

V vy, s tn ti > 0 (to + a) < 1.Khi , t

e.2n(to+a)P{SN > 2n(to + a)} e2n{(to+a)1}.

iu ny cha chng minh cht ch, nhng cng dn ta n phnon rng

supE exp(SN) < .V v vy theo b Fatou th: E exp(S) < .

Tip theo, ta s chng minh tnh cht trn, thng qua bt ng thcsau:

nh l 2.23. Cho (Xi)iN l bin ngu nhin c lp, i xng nhngi tr trong B.

Sn =k

i=1

Xi k N gi s rngXi a, cho mii N th mi, t > 0.

E exp

SN

exp(t) + 2 exp((t + a))P{

SN

> t}E exp

S

N.

Chng minh. t = inf{k N; Sk > t} ta vit

E exp SN

{N}

exp SNdP +N

i=1

{=k}

exp SNdP

exp(t) +N

i=1

{=k}exp SNdP.

Trn tp { = k} thSN Sk1 + Xk + SN Sk t + a + SN Sk.

Li do tnh c lp ca { = k} v SN Sk ta c:

{=k} exp SNdP exp((t + a)) {=k} exp SN SkdP39

43/83

= exp((t + a))P{ = k}E exp SN Sk.Li vi bt ng thc Jensen vi k vng khng ( theo (1.1)) ta c

E exp SN Sk E exp SN.V vy m

{=k}

exp SNdP exp((t + a))E exp(SN).P{ = k}

vi ch N

k=1{ = k} = {max

kNSk > t} cng vi bt ng thc Levy ta

c Nk=1

{ = k} = P{maxkN

Sk > t} 2P{SN > t}.

Sau ly tng theo k :

E exp S exp(t) + exp((t + a))E exp SNN

i=1

P{ = k}

= exp(t) + exp((t + a))E exp SNP{maxkN Sk > t}

exp(t) + 2 exp((t + a))E exp SNP{SN > t}.Ta c pcm.

T , ta c khng nh sau v tnh kh tch ca tng bin ngunhin c lp:

H qu 2.24. Cho (Xi) l mt dy i xng, c lp ( hoc c k vng

khng) b chn u v Sn = ki=1

Xi hi t h.c.c, th tn ti > 0

E exp S < .Chng minh. p dng nh l 2.23 vi t l gi tr sao cho: vi mi NP{SN t} (4e)1 v = (t + a)1, chng ta c ngay

supE exp SN 2exp(t) < .Trng hp khng i xng, ta dng phng php i xng ho v tnhcht 2.1 ta c iu phi chng minh.

40

44/83

T vic phn tch chng minh ca nh l, ta c th tng qut nhl trn nh sau:

H qu 2.25.Vi cc gi thit ca nh l2.23v vi mip > 0 th:

E exp SNp exp(tp)+2 exp(Cp(t+a)p)P{SN > t}E exp CpSNp.Vi

Cp =

1 nu 0 1.Chng minh. t = inf

{k

N;

Sk

> t

}ta vit

E exp SNp exp(tp) +N

i=1

{=k}

exp SNpdP.

Tuy nhin, trn { = k} th:SNp (t + a + SN Sk)p Cp(t + a)p + CpSn Skp.

V vy m{=k}

exp SNpdP exp{Cp(t + a)p}.

{=k}

exp CpSN SkpdP

= exp{Cp(t + a)p}.P{ = k}E exp CpSN Skp.Li do hm exp(Cpxp) l li, p dng bt ng thc 1.1 ta c

E exp Cp

SN

Sk

p

E exp Cp

SN

p.

T y ta c iu phi chng minh.

Ging nh t tng ca khng nh trn, ta thay hm m bi hmlu tha v vi l lun trong chng minh ca nh l 2.23. Ta c mtbt ng thc thuc dng bt ng thc Kolmogorov nh sau:

H qu 2.26. Vi gi thit ca nh l2.23, cho mit > 0 v1 p <

P{SN > t} 1

2p (1 22p(tp + Emax

Xi

p)

ESNp ).

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45/83

Chng minh. t = inf{k N; Sk > t}.Trn tp { = k} th

SNp

22p

(tp

+ maxi Xip

) + 2p

SN Skp

ta vit

ESNp tp + [22p(tp + max Xip) + 2pESNp]P{maxkN

Sk > t}.

Ri p dng bt ng thc Levy ta c iu phi chng minh.

Khi quan st bt ng thc ca nh l ta thy, c s hng cha

xc sut ca bin c ui. iu lm ta lin tng n cng thc tchphn ca k vng. C th ho nhn xt trn ta c c h qu sau:

H qu 2.27. Vi gi thit ca nh l 2.23v mi sao cho: exp a > 1/4 ta c

supNE exp SN 1 +

1 162 exp a4 exp a

.

Vi > 0 th1

1 + E exp SN 1 + 2ESN.E exp SN.

Chng minh. T bt ng thc ca nh l 2.23 ta c

E exp SN exp(t) + exp(2t + t/n + a)P{SN > t}E exp SN exp{( + 1/n)t}E exp SN

exp{t/n} + 2 exp{a} exp{t}P{SN > t}E exp SNLy tch phn hai v, gii phng trnh bc hai v cho n tin ra v cngta c khng nh th nht.

Cng t bt ng thc ca nh l 2.23 ta c :

E exp SN exp(t) + exp(2t + t)P{SN > t}E exp SNChia hai v cho exp(2t) ri ly tch phn hai v ta c khng nhhai.

42

46/83

Cho (Xi) l mt dy i xng, c lp (hoc c k vng khng), b

chn u v vi Sn =k

i=1Xi hi t h.c.c. Th ging nh dy h qu

2.24 ta c khng nh rng E exp S < vi > 0.Tuy nhin, khng nh ca h qu 2.24 khng tho mn mtkt qu bit ca trng hp thc, E exp |S| log+ |S| < vi > 0. m rng kt qu ny trong trng hp gi tr vc t, ta s s dngbt ng thc ng chu, th hin phn tip theo.

2.3 S tp trung v dng iu ui

Trong phn nay, chng ta ngin cu ch yu v tnh kh tch v bin cui ca tng bin ngu nhin c lp, da trn phng php ng chu(nh l 1.11) .

Nhng trc khi quay li tho lun vi tnh ng chu trong nghincu ny, chng ta s a ra mt s kt qu da trn cc bt ng thcmartingale. Mc d n khng mnh xem xt trong trng hptng qut cho tnh kh tch v cu hi v nh gi bin c ui. Tuynhin chng l nhng cng c n gin v hu ch trong nhiu tnhhung. Mt trong nhng p dng l s m rng cho s chiu v hntrong trng hp cc bt ng thc m nh bt ng thc Bernstein(nh l 1.7), Kolmogropv, Bennett...

u tin ta nhc li v bt ng thc Bennett.Ly (Xi) l mt dy ca cc bin ngu nhin thc k vng khng,

sao cho X i v nu b2 =EX2i t > 0

P

{i Xi > t} exp{t

a (

t

a+

b2

a2) ln(1 +

at

b2)}

(2.9)

Bt ng thc l loi in hnh ca dng iu ui ca tng cc binngu nhin c lp, dng iu ny ph thuc trn quan h ca t v t

sb2

a2, c th ta xt mt vi trng hp sau:

Nu t b2

a2hay 0 at

b2th t bt ng thc ln(1 + t) t 1

2t2 vi

43


47/83

0 t 1 (2.9) suy ra:

P

{i Xi > t} exp(t2

2b2

+at3

2b4

).

Nu t b2

2ath

at3

2b4 t

2

4b2v vy ta c nh gi sau:

P{

i

Xi > t} exp{t2

4b2}.

Nu t > 0 th ln(1 +at

b2

) > 0 nn t (2.9) suy ra:

P{

i

Xi > t} exp{ta

(ln(1 +at

b2) 1)}.

Nhn xt dn n vic s dng cc bt ng thc martingale trongnghin cu tng cc vc t ngu nhin c lp, l quan st n ginnhng cc k hu ch ca Yurinski.

Ly (Xi)i N l bin ngu nhin kh tch trong B, k hiu Ai l -

i s sinh bi X1, X2,...,Xi i N, Ao l - i s tm thng, talun vit SN =

Ni=1

Xi v t:

di = EAiSN EAi1SN

(di)iN l hiu martingale thc ( v EAi1di = 0 ) v

N

i=1 di = EANSN EAoSN = SN ESN.Chng ta c khng nh sau:

B 2.28. Gi s bin ngu nhin Xi l c lp, th h.c.c i N|di| X + ESN.

Hn na, nuXi trongL2() ta s c EAi1d2i EXi2.Chng minh. Ta c

(EAiEAi1)(SNSNXi) = di(EAiSNXiEAi1SNXi) = di.44

48/83

Li do bt ng thc tam gic

|di| (EAi + EAi1)(|SN SN Xi|) (EAi + EAi1)(Xi)= Xi + EXi.

V k vng c iu kin l hnh chiu trong L2 nn ta cng c btng thc th hai ca b .

ngha ca quan st pha trn l biu din lch ca tng SN ccvc t ngu nhin c lp Xi vi k vng ca n ESN thng qua hiumartingale thc di (

N

i=1di = SN ESN) vi di li c c lng

bi Xi .in hnh ca cch nhn ny l bt ng thc bnh phng martin-

gale. Sau y l h qu trc tip ca b 2.28 v tnh trc giao cahiu martingale

ESN ESN2 N

i=1

EXi2. (2.10)

T b 2.28 v cng vi cc bt ng thc martingale chng 1 cth c p dng xem xt tnh cht tp chung ca SN ESN.Tht vy sau y l mt vi pht biu cho iu ny, vn vi k hiu

a = maxXi , b (N

i=1

EXi2)1/2. p dng nh l 1.7 chng 1 suyra

P{|SN ESN| > t} 2exp[ t2

2b2(2 exp(2at

b2))]. (2.11)

Hn na t nh l 1.8 v tnh cht Ni=1

di = SN ESN cng vi b 2.28 ta suy ra rng: nu 1 0. Th

P{|SN ESN| > t} 2exp( tq

Cqa2) (2.12)

vi Cq > 0 ch ph thuc vo q.

45


49/83

By gi, chng ta quay li vn chnh ca chng ny vi nhngp dng ca bt ng thc ng chu nh l 1.11 cho tng cc binngu nhin c lp. Bt ng thc ny l cng c rt mnh v hiu qu

cho vic nghin cu su hn tnh cht kh tch ca tng cc bin ngunhin c lp c bt u phn trc, v nghin cu v cc nh lv gii hn hu chc chn chng sau.

u tin, ta s pht biu v chng minh mt c lng ui catng cc bin ngu nhin c lp da vo bt ng thc ng chu, vik hiu (Xi)iN l dy khng tng, c sp sp t (Xi)iN.nh l 2.29. Cho cc bin ngu nhin(Xi)iN l c lp, i xng vnhn gi tr trong khng gian Banach B. Vi mi s nguyn k

q v

cc s thc s, t > 0

P{N

i=1

Xi > 8qM+2s+t} (Koq

)k+P{k

i=1

Xi > s}+exp( t2

128qm2).

(2.13)

Vi M = EN

i=1

ui , m = E(supfD

(N

i=1

f2(ui))1

2 ) , ui = XiI{Xi sk} (i N)

Chng minh. Ta s chia chng minh thnh 3 bc.

Bc 1: Vi Xi l bin ngu nhin tu v nu s k

i=1

Xi

th N

i=1

Xi s + N

i=1

ui vi ui = XiI{Xi sk} (i N).Thc vy, nu k hiu J l tp cc s t nhin i N sao cho Xi > s

k

th CardJ k (v nu ph nhn th s ki=1 Xi) v khi :

N

i=1

Xi iJ

Xi + i/J

Xi k

i=1

Xi + N

i=1

ui s + N

i=1

ui.

Bc 2: Bi tnh i xng, c lp ca (Xi), nn X = (Xi)iN ccng phn phi vi (iXi)iN (vi (i) l dy Redemache c lp ca X)

gi s rng chng ta cho A BN sao cho P{X A} 12

.

Nu khi X

H = H(A,q,k) th tn ti (theo nh ngha ca H )

46


50/83

j k v x1,...,xq A sao cho

{1,...,N

}=

{i1,...,ij

} I vi I =

q

l=1{i N; Xi = xli}cng vi bc 1 ta c th vit : nu s

ki=1

Xi th

N

i=1

iXi s + N

i=1

iui s +j

l=1

iuil + iI

iui

s +j

l=1

Xl + iI

iui 2s + iI

iui

iu c ngha l, vi mi r q; s , t > 0 khi s k

i=1

Xi th

N

i=1

iXi > 2s + t suy ra iI

iui > t, vy ta c

{N

i=1

iXi > 2s + t} {k

i=1

Xi > s}{iI

iui > t}.

T y ta c

P{N

i=1

iXi > 2s + t} = P{k

i=1

Xi > s} + P{iI

iui > t}

P{k

i=1

Xi > s} + PXP{iI

iui > t} = P{k

i=1

Xi > s}

+

{X /H}

P{iI

iui > t}dPX +

{XH}

P{iI

iui > t}dPX.

p dng cng thc ng chu nh l 1.11 (cng thc (1.7)) th

{X /H} P{iI iui > t}dPX P{X / H} (Ko

q)k.

47


51/83

Vy ta c

P{N

i=1iXi > 2s + t}.

(Koq

)k + P{k

i=1

Xi > s} +

{XH}

P{iI

iui > t}dPX. (2.14)

Bc 3: Ta chn A v c lng cho P{N

i=1

iui > t} vi vic chn A

. p dng bt ng thc (1.5) vi ch M(X) XP

{i ixi > 2Ei ixi + t} P{

i

ixi > M(X) + t} 2exp(t2/82) (2.15)

by gi ta xt A = A1 A2 vi

A1 = {x = (xi)iN : EN

i=1

ixiI{xis/k} 4M}

A2 = {x = (xi)iN : supfD

Ni=1

f2(xi)I{xis/k}1/2 4m}.

Khi ta c P{X A} 1/2 v vy ta c th p dng (2.14), tuy nhinta thy:

V E(iI

ixi) = 0 vi I N nn p dng bt ng thc (1.1) th

EiIJixi EiJ ixi.Vy ta c nhn xt trung bnh Rademacher l n iu tng , tc lE

iJixi n iu tng ca J N.

Theo nh ngha ca I, th vi mi i I, ta c th c nh l(i) saocho 1 l(i) q vi Xi = xl(i)i . Ly Il = {i : l(i) = l}, 1 l q, khi ql=1Il = I v Il1 Il2 = vy ta c

E

iI iui =E

q

l=1iI(l)

iui q

l=1

EiI(l)

ixl

iI{xlis/k}.

48


52/83

Li do tnh n iu ca trung bnh Rademacher, v nh ngha caA , suy ra:

EiI(l)

ixliI{xlis/k} EN

i=1

ixliI{xlis/k} 4M.

Vy chng ta cEiI

iui 4qM. ()

Tng t, do tng supfD

iI

f2(ui) cng l n iu theo I N nn

supfDiI

f2(ui) = supfD

ql=1

iI(l)

f2(ui) q

l=1

supfD

(iI(l)

f2(xi)I{xis/k})

q

l=1

supfD

(N

i=1

f2(xi)I{xis/k}) q(4m)2.

Vy ta c2 = sup

f

DiIf2(ui) 16qm2. ()

T (2.15) v (*)(**) ta c

P{

i

iui > 8qM + t} P{

i

iui > 2E

i

iui + t}

2exp(t2/82) 2exp(t2/128qm2).T iu ny cng cng thc (2.14) vi ch thay t bi t + 8qM, ta ciu phi chng minh.

Nhn xt 2.30. Nhn xt u tin M,m c nh ngha thng qua binngu nhin cht ct ui nu ta khng cn quan tm n tnh cht ct,chng hn bin ngu nhin b chn th (2.13) v tri c th thay th2s bi sNhn xt tip theo v mt vi c lng cho M v m.

Th nht:

m2

supfDN

i=1f2

(ui) + 8

Ms

k . (2.16)

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53/83

Tht vy, p dng ln lt cc bt ng thc Holder, bt ng thc(2.4) v nguyn l co nh l 2.8 ta c:

m2 = (E supfD

(N

i=1

f2(ui))12 )2 E(sup

fD

Ni=1

f2(ui))

supfD

Ni=1

Ef2(ui) + 8E

i

uiui supfD

Ni=1

Ef2(ui) + 8E

i

uisk

( v ui sk ).Th hai:

m2

2M2. (2.17)

Tht vy: theo nh l 1.3 (TH p = 1) ta c

m E(supfD

Ni=1

2f(ui)) = E(sup

fD

2f(

Ni=1

ui))

2M.

Th ba:

m2 N

i=1 Eui2. (2.18)

(iu ny c suy ra t bt ng thc Holder v tnh cht: vi mif D th f2(u) u2 ).

Mt nhn xt quan trng bc ba ca chng minh trn l tnhn iu ca trung bnh Rademacher, n c s dng khi nh gi k

vng EN

i=1

iXi. Di y l mt p dng na cho nhn xt ny.

H qu 2.31. Cho (Xi)iN l bin ngu nhin i xng v c lp trongL1(B) v t M = E

Ni=1

Xi. Th vi mik q v s > 0.

P{EN

i=1

iXi 2qM + s} (Koq

)k + P{k

i=1

Xi > s}. (2.19)

(Chng minh h qu c trnh by trong [5])Tip theo ta quay li vi mt s ng dng ca nh l 2.29. u tin,ta tr li cu hi v tnh kh tch ca chui hi t hu chc chn ca

50

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cc bin ngu nhin c lp quy tm b chn. Nh nh l 2.23 chng minh, dng hm m ca chui l kh tch. Nhng ta bit trongtrng hp thc th chui nay l kh tch cho c hm exp(x ln+ x), v

n l hm tt nht c th cho tnh kh tch.Tht vy, ly (Xi)i1 l dy cc bin ngu nhin c lp xc nhbi

P{Xi = 1} = (2i2)1 v P{Xi = 0} = 1 i2.Khi th

i

EX2i < . Tuy nhin, nu SN =i

Xi v vi , > 0, vi

mi N, ta c

E exp(|SN|(ln+ |SN|)1+) =N

k=0exp(k(ln+ k)1+)P{|SN| = k}

exp(N(ln N)1+)P{|SN| = N} exp(N(ln N)1+)N

i=1

P{Xi = 1}

= exp(N(ln N)1+)N

i=1

2i2 = exp(N(ln N)1+ 2N

i=1

ln

2i)

exp(N(ln N)1+2N(ln 2N)) = exp N((ln N)1+2 ln Nln2)).tin ti v cng khi N tin ti v cng.

Di y l nh l m rng tnh cht kh tch ny cho trng hpvc t.

nh l 2.32. Cho (Xi) l bin ngu nhin c lp nhn gi tr trongkhng gian Banach B sao cho S =

i

Xi hi t hu chc chn v

Xi < a vi mii th vi mi < 1/aE exp(S(ln+ S) < .

NuXi ch quy tm th vi mi < 1/2a iu trn ng.

Chng minh. Vi mi N ta t SN =N

i=0

Xi , Bi v khi p dng b

Fatou ta c

E exp(S(ln+ S) infNE exp(SN(ln+ SN)

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supNE exp(SN(ln+ SN)

do ta ch cn chng minh

supNE exp(SN(ln+ SN) <

l .Ta thy S =

i

Xi hi t hu chc chn nn supn

Sn hu chcchn v vy theo nh l 2.17 th E sup

iSip < li p dng b

FatouM = sup

iE

S

ip

E sup

i S

ip 0

P{N

i=1

Xi > 8qM+s+t} (Koq

)k+P{k

i=1

Xi > s}+exp( t2

128qm2)

li theo (2.17) ta c

P{N

i=1

Xi > 8qM+s+t} (Koq

)k+P{k

i=1

Xi > s}+exp( t2

256qM2)

Theo gi thit ta cng suy rak

i=1

Xi ka h.c.c, v vy m

P

{

N

i=1 Xi > 8qM + ka + t} (Ko

q )

k

+ exp(t2

256qM2)

khi vi mi u > 0 ln ly > 0 , v t t = u, k = [(1 2)a1u]v q = [

2a

256M2u

lnu] khi

(Koq

)k = exp(k ln(q/Ko)) exp((1 2)a1u ln(q/Ko))

exp((1 3)a1u ln u)

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(chn > 0 nh c iu trn v k q) V hn na theo cch xcnh q th

q 2

a256M2 ulnu 256qM2 at2

u ln u t2

256qM2 ua1 ln u

V vy, khi u uo(M,a,) = 8qM + ka + t ln th

P{SN > u} P{SN > 8qM + ka + t} (Koq

)k + exp( t2

256qM2)

exp((1 3)a1u ln u) + 2 exp(ua1 ln u) Cexp(( + )u ln u).

Khi > 0 nh, hayexp(u ln u)P{SN > u} Cexp(u ln u)

vyE exp(SN(ln+ SN) C.

N l ng vi mi N, nn ta c iu phi chng minh.Trng hp bin ngu nhin ch quy tm, ta dng phng php i

xng ha v p dng b 2.7 ta c pcm.

Khi p dng bt ng thc J.Hoffmann- Jorgensen, nh l 2.16 chng minh s tng ng gia cc momen. y, vi bt ngchu ta s tin xa hn mt bc. Ch ra s ph thuc ca hng s nhntrong bt ng thc momen Lp vo p.

nh l 2.33. C mt hng s K sao cho vi mi dy hu hn(Xi) ccbin ngu nhin k vong 0, c lp trong Lp(B)

i

Xip K plnp

(

i

Xi1 + maxi

Xip).

Chng minh. Ta c th gi thit Xi (i N) l i xng (v nu khngta dng phng php i xng ho v p dng b 2.7 s c iuphi chng minh).

Vi mi s nguyn r, ta t X(r)N = Xj vi Xj l maximum th

r ca mu (Xi)iN (v X(r)N = 0 vi r > N) nu t M = N

i=1Xi1

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th theo nh l 2.29 v (2.17) ch ra rng vi mi k q v s, t > 0

P

{

N

i=1 Xi > 8qM+2s+t} Ko

q k

+P{

k

i=1 Xi > s}+exp(t2

256qM2). (

)

chng minh nh l ta c th gi s M 1 v X(1)N p 1 ( ta chiahai v cho max{X(1)N p, M} s c khng nh trn)

c bit, vi mi u > 0 ta c

P{X(1)N > u} EX(1)N p

up up.

Hn na, vi X(r)N () > u suy ra X(r1)N () > u v X

(1)N () > u v vy

P{X(r)N > u} P{maxiN

X > u}P{X(r1)N > u}

tip tc p dng vi r thay bi r-1 v bng phng php quy np theor ta c

P{X(r)N > u} (P{X(1)N > u})r urp.

Cho u 1 l c nh chng ta p dng cng thc trn vi r = 2 vu bi u2/3 th P{X(2)N } u4p/3. V vy m phn b ca tp

{X(1)N u, X(2)N u2/3, X(l)N 2}

c xc sut nh hn hoc bng

P{X(1)N > u} + 2u4p/3.

Chng ta p dng (*) vi k l s nguyn nh nht ln hn u. Khi ,trn tp{X(1)N u, X(2)N u2/3, X(l)N 2}

thk

r=1

X(1)N + lX(2)N + (k 1)X(l)N u + lu2/3 + 2(k 1)

u + (ln u + 1)u2/3 + 2u 4u + 3u1/3u2/3 = 7u = Cu

54


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cng vi q l s nguyn nh nht ln hn hoc bng u1/2, s = Cu, t = uta c:

P{N

i=1

Xi > 2(C+ 10)u} P{N

i=1

Xi > 8qM + 2s + t}

Koq

k+ P{

ki=1

Xi > s} + 2 exp( t2

256qM2)

= exp

k ln

q

Ko

+ P{

k

i=1Xi > Cu} + 2 exp( t

2

256qM2)

expu lnu1/2Ko

+ P{X(1)N > u} + 5u4p/3 + 2 exp(

u3/2

256)

nhn hai v vi up v ly tch phn hai v ta c iu phi chng minh.

55


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Chng 3

Lut mnh s ln

Chng ny, chng ta nghin cu lut mnh ca s ln cho tng ca binngu nhin c lp trong khng gian Banach. y, nh l 2.29 caphn 2.3 c dng rt hiu qu. Chng ta ch nghin cu lut mnhs ln dng Kolmogorov. Chng c chia lm 2 phn.

3.1 Pht biu chung cho nh l gii hnCho (Xi)iN l dy cc bin ngu nhin c lp nhn gi tr trong B,gi (ai) l dy v hn cc s dng, tng. Chng ta s nghin cu tnh

cht hu chc chn ca dySnan

. u tin ta a ra mt mnh n

gin sau:

B 3.1. Cho (Yn) v (Yn) l dy c lp cc bin ngu nhin sao

cho dy(Yn Yn) l b chn hcc (tng ng hi t hcc v 0), v (Yn) lb chn (tng ng hi t v 0) theo xc sut. Th (Yn) l b chn hcc(tng ng hi t hcc v 0).

(Chng minh nh phn cui ca nh l Levy-Ito-Nisio, nh l 2.6).Cho cc bin ngu nhin c lp (Xi). Ly (Xi) l bn sao c lp

ca (Xi) v t

Xi = Xi Xi l cc bin ngu nhin c lp v i

xng. Theo b 3.1, nghin cu tnh h.c.c caSn

an

khi bit tnh cht

56

60/83

theo xc sut ca n, th ch cn nghin cu (

ni=1

Xian

) l . Sau y ta

s nu mt b hu ch v cc tnh cht theo xc sut ca Snan lin hvi tnh cht tng ng trong Lp.

B 3.2. Cho (Xi) l dy cc bin ngu nhin c lp, i xng,

nhn gi tr trong B. Nu dySnan

l b chn (tng ng hi t v 0)

theo xc sut, cho p > 0 v mt dy b chn(cn) cc s dng th dy(E

ni=1 Xi1{Xicnan}/anp) l b chn (tng ng hi t v 0).

Chng minh. Ta chng minh cho pht biu hi t.Xt dy (cn) b chn bi C. Theo bt ng thc (2.8) cho mi n ta

c:

En

i=1

Xi1{Xicnan}p 2.3pE(maxiN

Xip1{Xicnan}) + 2(3t0(n))p

vi

t0(n) = inf{t > 0 : P{n

i=1

Xi1{Xicnan} > t} < (8.3p)1}.

VSnan

hi t theo xc sut v 0; s dng nguyn l co (nh l 2.8) ta

c:

P

{

n

i=1 Xi1{Xicnan} > t} < 2P{Sn > t} = 2P{Sn

an>

t

an}

li doSnan

P 0 > 0; n0 : n > n0 :

P{Snan

> } 12

(8.3)p1.

Vy

n > n0 tht0(n)

an<

t0(n) < an. (1).

57

61/83

Hn na

Emaxi

n

Xip1{Xicnan} =

0

P(maxi

n

Xi1{Xicnan} > t)dtp

can0

P{maxin

Xi > t}dtp = apnc0

P{maxin

Xi > ant}dtp

2apnc0

P{Sn > ant}dtp (2).

(c c l do bt ng thc Levy).T (1) v (2) ta c, vi n > n0 th

E ni=1

Xi1{Xicnan}/anp

2.3p(p +c0

P{Snan

> t}dtp) 2.3p(p + p +c

P{Snan

> t}dtp).

Do t [, c] th

P

{Sn

an> t

} P

{Sn

an>

} p.

Khi n ln. V vy m khi n ln:

En

i=1

Xi1{Xicnan}/anp 2.3p(2p + p(c )p) 2.3p(2 + cp)p

hay

En

i=1

Xi1{Xicnan}/an

p

0.Trng hp b chn, chng minh tng t.

Tip tc, khi ta xt dy (an) c tnh cht sau: tn ti dy con (amn)sao cho vi mi n: camn amn+1 Camn+1 ( y 1 < c < C < +).Trong phn tip theo ta lun gi thit iu ny ng; v vi mi n tat I(n) = {mn + 1; ...; mn}. B tip theo m t s rt gn vo trongcc tp I(n) khi nghin cu

Sn

an.

58

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B 3.3. Cho (Xi) l bin ngu nhin c lp; i xng. DySnan

l

dy b chn h.c.c (hoc hi t h.c.c v 0 ) nu v ch nu iu tng ng

ng cho (iI(n) Xi/amn).Chng minh. Ta chng minh cho pht biu hi t:

iu kin cn:Snan

hcc 0 suy ra Smnamn

hcc 0.

Nhng do (an) l dy tng, nnSmn1amn

hcc 0 v vy m

Smnamn

Smn1amn = iI(n)Xi

amnhcc 0.

iu kin :

iI(n)

Xi

amn

hcc 0, cng vi tnh c lp ca dy ( iI(n)

Xi)n,

nn theo b Borel - Catelli suy ra > 0,

n=1

P

iI(n)Xi

an > < .Theo bt ng thc Levy suy ra:

n=1

P

supkI(n)

k

i=mn1+1Xi

amn>

.

Theo b Borel-Cantelli suy ra

supkI(n)

ki=mn1+1

Xi

amn

hi t h.c.c ti0. T y, ta suy ra vi hu ht w, > 0, l0 sao cho l > l0

supkI(l) k

i=ml1+1

Xi(w) < aml.

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Vi j l0 khi vi n sao cho mj1 < n mj th theo bt ng thctam gic ta c:

Sn(w) Sml01(w) +j

1

l=l0

iI(l)

Xi(w) + n

i=mj1+1

Xi(w)

Sml01(w) +j1l=l0

aml + amj

Sml01(w) +i

l=1aml

Sml01(w) +j1l=0

Cancl

(V t gi thit ca (ai) nn: aml 1

caml+1 ...

1

cjlamjl 0 tu , v gi thit v ai tho mn camn amn+1 amn+1 nnamn than vy ta c iu phi chng minh.

T b 3.1 v b 3.3 ta c h qu sau:

H qu 3.4. Cho (Xi) l bin ngu nhin c lp nhn gi tr trong B

thSnan

h.c.c 0 khi v ch khi Snan

P 0 v Smnamn

h.c.c 0. Tng t cho tnh bchn.

Chng minh. Tht vy, iu kin cn l hin nhin. Ta ch cn chngminh iu kin .

Xt Xi l bn sao c lp ca Xi khi Xi = Xi Xi l i xng

ha ca Xi.

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V

Snm

anm

hcc

0

Smn

amn

hcc

0

iI(n)

Xi

amn

P

0.

Theo b 3.3 suy ra Snan

hcc 0.

Li doSnan

P 0, v b 3.1, ta c Snan

hcc 0.

T b 3.3v b Borel-Cantelli, ta c nhn xt: Vi bin ngunhin i xng th, s hi t ca chui dng n P{ iI(n) Xi > an} ssuy ra s hi t h.c.c ca

Snan

.

V vy phn tip theo di y ta s xem xt iu kin cn v chui

nP{

iI(n)Xi > an} hi t.

nh l 3.5. Cho (Xi) l mt dy bin ngu nhin i xng, c lpnhn gi tr trong B. Gi s, tn ti mt s nguynq 2K0 v mt dy(kn) cc s nguyn dng, sao cho iu sau y ng:

n

(K0q

)kn < (3.1)

n

P{kn

r=1

X(r)I(n) > amn} < . (3.2)

Vi > 0, t :

Mn = E iI(n)

Xi1{Xiamn/Kn}

n = supfD

(

iI(n)E(f2(Xi)1{Xiamn/Kn}))

1/2.

NuL = limn

supMnamn

< v vi > 0

n

exp(2

a2mn/

2n) < . (3.3)

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65/83

Th ta c:

nP{iI(n)

Xi > 102(,,q,L)amn} < . (3.4)

Vi(,,q,L)amn = + qL + (L +

2)1/2(q logq

K0)1/2

+ qL + (L + 2)1/2.Ngc li, nu (3.4) ng vi mt s (tng ng vi tt c) > 0 v

(3.1), (3.2) tha mn th L < (tng ng L = 0) v (3.3) ng chomt vi (tng ng, tt c) > 0.

( y: X(r)I(n) = Xj viXj l gi tr ln th r trong tp (Xi)iI(n)cn vir > CardI(n) thX(r)I(n) = 0).

Chng minh. iu kin cn: V L = limn

supMnamn

< n0 : n > n0

Mnamn

< L +

q 8qMn < 8(qL + )amn.

Vy

102(,,q,L)amn 8qMn + 2amn + 102(L + 2)1/2(q logq

K0)1/2amn.

V vy m p dng nh l 2.29 cho (Xi) ; K = Kn q vi n ln;s = amn v

t = 102(L + 2)1/2(q logq

K0)1/2amn.

Ta c:

P{

iI(n)Xi > 102(,,q,L)amn}

P{

iI(n)Xi > 8qMn + 2amn + 102(L + 2)1/2(q log

q

K0)1/2amn}

K0q kn

+P

{

kn

r=1 X(r)I(n) > amn} + 2exp t2

128m2.62

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Theo (2.16) th m2 2n + 8amn

knkhi :

exp2a2mn

2n > exp t2

128m2.Vy

n

P{

iI(n)Xi > 102(,,q,L)amn} <

(Do gi thit, ba chui 3.1, 3.2, 3.3 hi t).iu kin .

Ta chng minh cho trng hp > 0 (Trng hp > 0 lmtng t)

P{ iI(n)

Xi > amn} < . ()

t Xni = Xi1{Xiamn/kn}, i I(n). Khi theo nh ngha ca can ta c th chn c fn D sao cho:

2n 2

iI(n)E(f2n(X

ni ) 22n.

p dng nguyn l co (nh l 2.8) ta c:n

P{

iI(n)fn(X

ni ) > amn}

n

P{

iI(n)Xni > amn}

2

n

P{

iI(n)Xi > amn}.

Ly > 0, nu > 0 sao cho 2 > ln q

K0

(hay

K0q

> e2) khi

nu a2

mn > kn

2

n ta c:K0q

kn> e

2kn > e(2a2mn/

2n).

Theo 3.1 3.3 ng.Vy ta ch cn chng minh vi a2mn kn2n. Ta nh li cc tham s

r, (r), k(r) trong nh l 1.9 vi r l s c nh c chn. Ly > 0 nh cho (1 + r)2 2 v 2 (r) sao cho

(amn)amn(kn) 2

n (r) iI(n)

Ef2

n(Xn

i )

63

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Li do: (*) nu

iI(n)

Xi

amn

hi t v 0 theo xc sut nn theo b

3.2, ta cL = lim sup

Mnamn

= 0.

Li do tnh trc giao

iI(n)Ef2n(X

2i )/a

2mn

0 nn vi n ln

amn K(r)iI(n)Ef

2n(X

2i )

1

2

.

By gi p dng nh l 1.9 ta c

P{

iI(n)fn(X

(n)i ) > amn} exp((1 + r)2a2mn/2n).

Li v (1 + r)2 2 nn ta c

P{iI(n) f

n(X(n)

i ) > amn} exp(2

a2mn/

2n).

Ly tng theo n, kt hp vi iu kin (3.4), ta suy ra (3.3).

B tip theo ta hy a ra cc iu kin d kim tra hn cc iukin (3.1), (3.2).

B 3.6. Vi cc iu kin nh trong nh l3.5, gi s rng viu > 0n

P{maxiI(n)

Xi > uamn} < (3.5)

v viv > 0, vi min v t, 0 < t 1

P{maxi

I(n)

Xi > tvamn} n exp1

t(3.6)

64

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vin

sn < vis nguyn no . Khi miq > K0, tn ti dy(kn)v cc s nguyn sao cho

n(Ko

q)kn < v tha mn

n

P

knr=1

X(r)I(n) > 2s(u + v(ln

q

K0)1)amn

< .

Chng minh. Ta c

{X(2s)I(n) > tvamn} 2s

i=1

{X(i)I(n) > tvamn}.

Suy ra

P{X(2s)I(n) > tvamn} 2s

i=1

P{X(i)I(n) > tvamn}

iI(n)P{Xi > tvamn}

2s

p dng nh l 1.10 cng vi gi thit cho ca b ta c

iI(n)

P{Xi > tvamn} 2P{max Xi > tvamn}

2n exp

1

t

.

Vy

P{X(2s)I(n) > tvamn}

2n exp

1

t

2s.

Chn t = t(n) = ln 1n1

v chn kn = 2sln qK01 ln 1n. Khi

K0q

kn

K0q

2sln qK01 ln 1n

K0q

2s ln 1n

q

K0

= (n)2s = sn.

65

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Vn

sn < nnn

K0q

kn< . By gi ta c

knr=1

X(r)I(n) =2s

r=1

X(r)I(n) +

knr=2s+1

X(r)I(n) 2sX(1)I(n) + knX(2s)I(n).

V vy

P

kn

r=1

X(r)I(n) > 2s(u + v

ln

q

K0

1)

amn

P{2sX(1)I(n) > 2suamn} {knX(2s)I(n) > 2svln qK01

amn}P{X(1)I(n) > uamn} + P{X(2s)I(n) > t(n)vamn},

vi ch rng, khi ly tng theo n thn

P{X(1)I(n) > uamn} =

n

P{maxiI(n)

Xi > uamn} <

vn

P{X(2s)I(n) > t(n)vamn}

n

2n exp

1

t(n)

s 2s

n

sn < .

Khi ta c iu phi chng minh.

Khi cc bin ngu nhin Xi l c lp cng phn phi v dy (ai) lchnh quy, tc l k n, ap2n 2nkap2k (vi p > 0 no ) th ta c b sau:

B 3.7. Cho (an) l chnh quy (tc l tn ti p k n, ap2n 2nkap

2k). Gi s Xi l mt dy cc bin ngu nhin c lp cng phn

phi viX. Nu tn tiu > 0 n

2nP{X > ua2n} < th vi min v0 < t 1 ta c

P{max Xi > tua2n} 2nP{X > tua2n} 1t2p

n

vin

n < .66

70/83

Chng minh. Vi mi n, t

n = 2nP{X > ua2n}.

Khi tn ti dy (n) sao cho n n, n 2n+1 vi mi n, vn

n ua2n} = 1, i = max{i; 12i1}...).Ly 0 < t < 1 v k 1 sao cho 2k tp 2k+1.khi :

Nu k < n ta ctpap2n 2kap2n ap2nk

suy ra ta2n a2nk, v vy m

P{X > tua2n} P{X > ua2nk} 2nP{X > tua2n} 2knk 2knk 4t2pn

(do 2tp 2k).Nu k > n th

2nP{X > tua2n} 2n 4t2p2n.

Vy khi ta chn n = 4 max{n, 2n} thn

n 4

n

n +

n

2n

<

v2nP{X > tua2n} 1

t2pn.

ng thi ta c

{maxiI(n)

Xi > tua2n} =

iI(n){Xi > tua2n}

suy ra

P{max Xi > tua2n}

iI(n)P{Xi > tua2n} = 2nP{X > tua2n}.

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3.2 Cc lut s ln

Phn ny ta p dng nh l 3.5 tng qut phn trn a ra cc

lut s ln c th hn trong khng gian Banach.Ta p dng vi an = n, mn = 2n v chng ta lun ni: (Xn) tun

theo lut mnh s ln nuSnn

hcc 0 v (Xn) tun theo lut yu s lnnu

Snn

P 0.nh l 3.8. Cho 0 < p < 2, ly (Xi) l dy bin ngu nhin c lpcng phn phi (ging nh X), nhn gi tr trong khng gian Banach.

Khi

Sn

n1p

hcc

0 nu v ch nuEXp

< vSn

n1p

P

0.Chng minh. iu kin cn. Ta c

Sn

n1

p

hcc 0 th hin nhin Snn

1

p

P 0

Hn naSn

n1

p

hcc 0 th hu ht ch c hu hn n Sn()n

1

p

> 1.

Theo b Banach-Catelli th

nP{X > n 1p} <

nP{Xp < n} < EXp < .

iu kin . Nu ta xt X l bn sao c lp ca X, tX = X X. Khi E Xp Cp(EXp + EXp) <

vSnn

1

p

P 0. V vy nn nu ta chng minh c kt lun ny i vi(

Xi) th p dng b 3.1 suy ra kt qu ng i vi (Xi).

Bi vy ta c th gi s (khng mt tnh tng qut) bn thn X li xng. Khi theo b 3.3 , ta suy raiI(n)

Xi

2np

hcc 0.

Li theo b Borel-Catelli ta suy ra

n P

iI(n)Xi

2n

2

> =n P iI(n) Xi > 2np <

68

72/83

vi I(n) = {2n1 + 1; ; 2n}.Phn tip theo ta s chng minh bng 2 cch.Cch 1. Vi mi n, t ui = ui(n) = Xi1

{Xi

2np

}, i I(n). Ta c

{i I(n) : ui = Xi} 2n

i=2n1+1

{Xi > 2np}.

Suy ra

P{i I(n) : ui = Xi} 2n

i=2n1+1

P{Xi > 2np} = 2nP{X > 2np}.

Do n

P{i I(n) : ui = Xi} n

2nP{X > 2np}

=

n

2nP{Xp > 2n} < ,

(v EXp < ). Li do

iI(n) Xi > 2np iI(n) ui > 2

np {i I(n) : ui = Xi} .

Vy chng minh

n

P

iI(n)Xi

> 2np <

Ta ch cn chng minh

n

PiI(n) ui > 2np < .

Tuy nhin, vSn

n1

p

P 0 nn theo b 3.2 ta c lim E

n

i=1

ui

2np

= 0hay

lim

1

2npE

iI(n) ui = 0.

69

73/83

Vy ta ch cn chng minh

n PiI(n) ui EiI(n) ui > 2np < .

(suy ra t tnh cht: Nu an 0 v > 0 thn

P{Xn an > } <

suy ra > 0, nP{Xn > } < . Tht vy, v > 0 th

nP{Xn an >

2} <

nhng n0 : n > n0 th an < 2

, do n > n0 ta c

P{Xn an > 2} P{Xn > },

l iu phi chng minh.)Bi bt ng thc (2.10) v tnh cht cng phn phi ta suy ra

n

P{

iI(n)ui E

iI(n)

ui > 2np} 12

n

1

22np

iI(n)

Eui2

12

n

1

22np

2nEX21{X2np } =1

2

n

1

2n(2

p1)EX21{X2np }

=1

2E

X2

n1

2n(2

p1) 1{2nXp}

(v EXp < ). Ta c iu phi chng minh.Cch 2. V EXp < nn vi mi > 0 th

n

2nP{X > 2np} < .

Vi c nh, theo b 3.7 (p dng vi u = , an = n1

p ) ta cn; v 0 < t 1 th

P{maxiI(n) Xi > t2

np} 1t2p n

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74/83

vin

n < . T , vi t = 1 th

n P{maxiI(n) > 2np

} n n < .Vi 0 < t 1 th 1t2p < exp(1t ) nn

n

P{maxiI(n)

Xi > t2np} n exp(1t

).

V vy, theo b 3.6, tn ti (kn) sao cho

nK0q

kn

=n

2kn =

n12

kn

<

v n

P

kn

r=1

X(r)I(n) > 52

np

nP

Kn

r=1X

(r)I(n) > 2 1 ( + (log 2)1)2

np

< .

Tip tc, ta p dng nh l 3.5 vi q = 2K0 v vi ch t b 5 thL = 0; hn na v

2n supfD

iI(n)

E

f2(Xi)1{Xiamn/Kn} .

iI(n)EXi21Xi2n/p 2nEX21X2n/p.

V vy mn

22n/p2n

n

1

2n(2p 1)EXp1X2n/p

<

(do EXp < , lp lun nh phn trn).Vy iu kin (3.3) ca nh l 3.5 c tha mn vi mi > 0 nn ta c

n

P{ iI(n) X

i > 102

(5 + 2K0)2n/p

} <

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75/83

V vy ta hon thnh chng minh.

Tip theo ta nh li tnh cht trong trng hp khng gian hu hnchiu nh sau:Khi EXp < (0 < p < 2) (v EX = 0 vi 1 p < 2) th Sn

n1/p 0

(nh l MarcinKiewicz- Zygmund). y ta s dng phng php lp lun xp x chng minh mt phntnh cht ny (vi 0 0 ta c thcho mt bin ngu nhin hu hn chiu Y (c k vng 0 khi p = 1 vEX = 0) sao cho EX Yp < .Gi Tn l tng ring ca bn sao c lp (Yi) ca Y. Khi theo btng thc tam gic (vi p 1).

ESn Tnp n

i=1

EXi Yip = nEX Yp n

E Snn1/p

Tnn1/p

p

nhng v Y c s chiu hu hn nn theo phn trn suy raTn

n1/pP 0.

Cng vi bt ng thc sau:

P{ Snn1/p

> 2} P{ Snn1/p

Tnn1/p

} + P{ Tnn1/p

}

E Sn

n1/p Tn

n1/pp

p+ P{ Tn

n1/p }

Suy raSn

n1/pP 0.

V vy, p dng nh l 3.8 ta c pht biu sau:

Vi 0 < p 1 th Snn1/ph.c.c 0 khi v ch khi EXp < (v EX = 0

72

76/83

vi p = 1).c bit, ta c cc nh l v lut mnh s ln ca Kolmogorov chobin ngu nhin c lp cng phn phi, c m rng cho khng gian

Banach tch c nh sau:H qu 3.9. Cho X l bin ngu nhin Borel vi gi tr trong khng

gian Banach tch c B thSnn

h.c.c 0 khi v ch khi EX < vEX = 0.

Phng php chng minh xp x trn khng m rng c chotrng hp 1 < p < 2 c. Tnh cht ng vi trng hp 1 < p < 2khng ng cho khng gian Banach tng qut; v d tip theo s chng

minh iu .V d 1. Trong C0- khng gian Banach tch c gm cc dy s thctin n 0, c trang b chun sup; Khi vi mi dy gim (n) ccs thc hi t v 0 th tn ti mt bin ngu nhin b chn hu chc

chn v i xng, X sao choSn

nnkhng hi t v 0 theo xc sut.

Chng minh. Ly (k)k1 l i lng ngu nhin vi phn phi

P{k = 1} = P{k = 1} =1

2(1 P{k = 0}) =1

ln(k + 1)

t k =

n vi 2n2 k < 2n v ly X l bin ngu nhin radon,nhn gi tr trong C0 vi ta (kk)k1 th X r rng l bin ngunhin i xng (do tng thnh phn kk l bin ngu nhin thc, ixng).

Tuy nhinSn

nnkhng hi t v khng theo xc sut.

Tht vy, vi (ki)i l bn sao c lp ca (k).

Vi mi k, n 1 t Enk = ni=1

{ki = 1}; An = k2n

Enk.

Khi ta c:

P(Enk) = P(n

i=1

(ki = 1)) =n

i=1

P{ki = 1} = (ln(k + 1))n

V vy m

P(An) = 1 P(k2n E

c

nk) = 1 k2n (1 P(Enk))

73

77/83

= 1 k2n

1 1

[ln(k + 1)]n

Ta thy P(An)

0.

Nu ta k hiu (ei) l c s chnh tc ca C0 th Xi =

k=1

kkiek v vy

m

Snnn

=

ni=1

k=1

kkiek

nn=

k=1

1

nn

ni=1

ki

kek

Khi trn An th

Snnn maxk2n 1nn k|i=1 nki| 1nnnn = 1nDo n 0 nn > 0 th

lim infP{Snnn

> } liminfP(An) = 1.

Sau khi c s m rng lut mnh s ln ca bin ngu nhin cng

phn phi vi gi tr trn khng gian Banach. Ta s nghin cu lutmnh s ln cho cc bin ngu nhin c lp, nhng khng cng phnphi.

nh l 3.10. Cho (Xi) l mt dy cc bin ngu nhin radon c lp,nhn gi tr trong khng gian Banach B. Gi s rng

Xii

h.c.c 0. (3.7)

Snn

P 0. (3.8)Gi s rng tn ti v > 0; n v t : 0 < t 1.

P{max Xi > t.v.2n} n exp(1t

). (3.9)

Vi

n sn < vis > 0 v rng vi mi > 0

n exp(22n

supfD iI(n) E(f2(Xi)1{Xi2n})) < (3.10)74

78/83

yI(n) = {2n1+1; ; 2n} th lut mnh s ln ng, tc Snn

h.c.c 0.

Chng minh. Ta nh ngha Yi = Xi1{Xi i} bi (3.7):Xi

i

h.c.c

0;nn hu ht th i() i > i() ta c Xi

i 1 Xi i hay

Yi = Xi hu chc chn vi i ln. V vy n l cho kt qu vi dy(Yi) thay cho Xi hay chng ta c th gi s rng Xi i h.c.c.

Nu ta gi (Xi) l bn sao c lp ca (Xi); xt dy bin ngu nhini xng (Xi Xi), khi cc gi thit ca (Xi) u tha mn vi(Xi Xi) chng hn iu kin (3.9):

P{maxiI(n) Xi Xi > t.(2v).2n}

P{maxiI(n)

Xi > tv2n} + P{max Xi > tv2n} 2n exp(1

t)

vi

n(2n)s < bi vy nu ta chng minh c Sn

nh.c.c 0 th theo

b 3.1 cng vi iu kin (3.8) suy ra ngaySnn

h.c.c 0. Vy ta c th

gi thit bn thn (Xi) l i xng.Tip theo ta s kim tra cc iu kin p dng nh l 3.5. V

gi thit (3.8) suy raS2n

2n

vS2n1

2n

hi t theo xc sut v 0 nn

iI(n)

Xi

2n

hi t theo xc sut v 0; p dng b 3.2 suy ra

L = lim sup

Mnamn = lim sup

Mn2n = 0

Li doXii

h.c.c 0 nn > 0 thn

P{maxiI(n)

Xi > u.2n} <

Vy theo b 3.6 (vi ch cng iu kin (3.9) ta suy ra:

q 2k0 tn ti dy (kn) cc s nguyn sao cho nk0q kn < v tha

75

79/83

mn

nP

kn

r=1X

(r)I(n) > 2s

u + v(ln

q

k0)1

amn

<

Li cng vi gi thit (3.10) suy ra cc gi thit ca nh l 3.5 ctha mn. Vy u > 0, > 0 ta c

n

P

iI(n)Xi 102

2s(u + v(ln

q

k0)1) + q

2n

<

V vy > 0, chn u, > 0 sao cho

= 1022s(u + v(ln qk0)1) + qta c:

n

P{

iI(n)Xi > 2n} <

li p dng b 3.3 suy ra:Snn

h.c.c 0H qu 3.11. Vi gi thit ca nh l3.10nhng iu kin (3.9) thay

bi iu kin (3.9):n

12np

iI(n)EXip

s< (vi p > 0 v s > 0 no ) th lut mnh

s ln tha mn.

Tht vy, ta p dng bt ng thc Markov, vi mi n:

P{Xi > tv2n} EXip

(tv2n)p

Suy ra iI(n)

P{Xi > tv2n} 1(tv)p

1

2np

iI(n)

EXip

Tuy nhin ta bit vi 0 < t 1 th 1t > 1 v khi tn ti C(p) 1

(tv)p C(p)exp 1

t

76

80/83

v vy m

iI(n)P{Xi > tv2n

} C(p)exp

1

p1

v

p

1

2

np iI(n)EXip

vy ta c gi thit (3.9) ca nh l 3.10 vi

n = C(p).1

vp1

2np

iI(n)

EXip

v

n sn = Cs(p) 1vpsn 12np iI(n) EXips

0

iI(n)

P{Xi > t}

Ta xt iu kini

EXipip

< , ta thy iu kin ny suy ra t iukin (3.7), (3.8) v (3.10) (vi p 2). V vy ta c h qu sau:H qu 3.12. Cho (Xi) l dy bnn c lp vi gi tr trongB. Nu tn

ti1 p 2 :i

EXipip

< th lut mnh s ln ng khi v ch khilut yu s ln ng.

77


81/83

Kt lun

Lun vn cp n mt vn quan trng ca l thuyt xcsut. l vic nghin cu tng cc bin ngu nhin c lp vi gi tr

trong khng gian Banach. N l s m rng cho trng hp s chiu vhn ca tng cc bin ngu nhin c lp. Tuy nhin, trong trng hpny thiu i tnh cht trc giao, v vy vic m rng ny cn phi ccc cng c khc c trnh by r rng trong lun vn nh: phngphp i xng ho, phng php dng dy bin Rademacher, phngphp dng cc bt ng thc martingale thc v c bit l phngphp ng chu. Ti liu tham

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