Qua Trinh Ngau Nhien - To Van Ban

8/2/2019 Qua Trinh Ngau Nhien - To Van Ban

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Ts t vn ban

Bi gingXc sut thng k

VQu trnh ngu nhin

(Dnh cho cc lp cao hc k thut - HVKTQS)

PHIn BN 09/05 - 12/05 - 08/06 - 11/06 - 20/03/07 - 15/05/07 - 10/7/2007 - 05/09/07(Cha hon thin)

H ni - 2005 - 2006 - 2007



M C L C

Phn Chng Ni dung trang

Mc lc 2

Li ni u 5

Cc k hiu hay s dng 7Phn I Xc sut Thng k 9

Chng I Kin thc b sung v xc sut 9

1.1. Cc bin ngu nhin quan trng 9

1.1. Bin nhu nhin chun 8

1.2. Vc t ngu nhin chun 11

1.3. M rng khi nim mt i vi BNN ri rc 17Cu hi v bi tp Chng I 20

Chng II

Chng III 3.5.S hi t ca dy cc BNN

3.5.1. Cc dng hi t

3.5.2. Cc nh l gii hn

23

23

25

Chng IV L thuyt c lng

Phn II Qu trnh ngu nhin 32Chng V Nhng khi nim tng qut 32

5.1. M u

5.1.1. Cc nh ngha

5.1.2. Phn loi s b

5.1.3. V d v QTNN

5.1.4. H cc phn b hu hn chiu

32

32

33

34

35

5.2. Mt s lp cc qu trnh ngu nhin

5.2.1. Qu trnh cp II

5.2.2. Qu trnh s gia c lp

5.2.3. Qu trnh dng (QT dng theo ngha hp, dng

theo ngha rng, dng ng thi)

5.2.4. Qu trnh Gauss

36

36

38

39

45

5.3.Tnh cht ergodic v trung bnh thi gian 46

2



5.3.1. Gii thiu

5.3.2. Ergodic k vng

5.3.3. Ergodic phng sai, t hip phngsai, PS cho

5.3.4. Cc loi ergodic khc

5.3.5. o hm tng quan

46

47

50

54

55

5.4.Lin tc, o hm, tch phn

5.4.1. Lin tc (theo xc sut, theo trung bnh)

5.4.2. o hm (theo bnh phng trung bnh)

5.4.3. Tch phn (theo bnh phng trung bnh)

57

57

59

61

5.5.Hai QTNN quan trng

5.5.1. QT Poisson (nh ngha, xc sut ng thi n

chiu, hm t tng quan, dy thi im n, xc

nh cng dng n, cc bin th, nhiu bn,sinh cc qu o)

5.5.2. QT Wiener (. ngha, cc tnh cht, sinh qu o)

5.5.3. Gii thiu v cc QTNN khc

65

65

75

74

77

5.6. Qu trnh ngu nhin phc

Cu hi l thuyt v bi tp chng V

77

79

Chng VI X l cc QTNN 86

6.1.Mt ph cng sut

6.1.1. Vn nghin cu QTNN trong min tn s

6.1.2. Mt ph cng sut

6.1.3. Mt ph cng sut cho

6.1.4. Mt ph cng sut cho QT thc khng dng

6.1.5. Mt ph cng sut cho dy ngu nhin

6.1.6. Mt s m hnh nhiu (nhiu trng, nhiu nhit,

nhiu trng thng di, nhiu mu, nhiu bn)

6.1.7. Ph cng sut ca QTNN phc(V d: Ph vch, hiu ng Doppler)

86

86

89

93

95

97

99

103

6.2.Cn bn v h tuyn tnh

6.2.1. H tuyn tnh tng qut

6.2.2. H tuyn tnh bt bin theo thi gian

6.2.3. H nhn qu v h n nh

107

107

109

112

3



6.2.4. Trng hp h ri rc 113

6.3. H tuyn tnh vi u vo ngu nhin

6.3.1. Vn u ra

6.3.2. Cc c trng xc sut ca QT u ra

6.3.3. p ng h LTI ri rc vi u vo ngu nhin

6.3.4. Cc v d (H l tng, Lc bc nht, Trung bnh

trt, Ph ca QT o hm)

115

115

117

120

122

6.4. Qutrnht hi quy trung bnh ng

6.4.1. Qu trnh t hi quy AR

4.4.2. Qu trnh trung bnh ng MA

6.4.3. Qu trnh ARMA

124

124

128

130

6.5. Qu trnh thng di v iu ch

6.5.1. Qu trnh thng di

6.5.2. Nhiu trong h thng tin iu bin AM

6.5.3. Nhiu trong h thng tin iu tn FM

133

133

138

142

6.6. Lc phi hp

6.6.1. Trng hp tng qut

6.6.2. Lc phi hp cho nhiu mu

6.6.3. Lc phi hp cho nhiu trng

6.7. c lng tuyn tnh ti u

6.7.1. t bi ton

6.7.2. Bi ton l trn Lc Wiener bt kh thi

6.7.3. Lc Wiener kh thi

Cu hi l thuyt v bi tp Chng VI

147

147

148

149

151

151

153

155

159

Chng VII

(d tr)

Qu trnh Markov

Xch Markov

Qu trnh Markov vi thi gian lin tc

Phn III Ph lc A - Cc bng thng k

Ph lc B - Php bin i Fourier

Bng B-1 Tnh cht ca php bin i Fourier

Bng B-2. Cp php bin i Fourier

171

171

172

Ti liu tham kho 173

4



Chng 1. kin thc b Sung v xc sut1.1.Cc bin ngu nhin quan trng1.1.1.Bin ngu nhin ri rc

Tn K hiu Xc sut P{ }X k= K vng Phng sai

Nh thc B(n,p) k k n k nC p (1 p) ;k 0,1,..., n = np np(1-p)

Poisson P( )ke

; k 0,1,...k!

=

Hnh hc G(p) p(1-p) k=0,1,2,...k;

1 p

p

2

1 p

p

Siu hnhhc

H(N,n,p)

k n kNp N Np

nN

C C; k 0,1,..., n

C

= ........ .........

Cc lut phn b ri rc khc: u ri rc, nh thc m,...

1.1.2Bin ngu nhin lin tcTn K hiu Mt K vng Phng sai

u U([a;b])1

; a x bb a

a b

2

+

2(b a)

12

M E( ) xe ; ,x>0 1/ 21/

Cauchy C ( , ) 2 2/ [ ( (x ) )] + Khng tn tiKhng tn

ti

Chun N(m, 2 )2

22

1 (x m)exp ( 0)

22

>

m 2

Gamma (r, ) r 1 x( x) e ; , r, x 0(r)

>

r

2

r

Khi bnhphng

2 (n) n x n

12 2 2x e /(2 (n / 2); x 0,n 1,2,...

> =

n 2n

Student T(n)((n 1) / 2))

n (n / 2)

+

2(n 1) / 2x(1 )

n

++ 0n

n 2

Fisher-Snecdecor

F(n,m)n 2 n m

2Bx (m nx) ;

+

+ 2 m, n, x > 0.......... ..........

Weibul W( , ) 1 xx e ; , , x 0 >

1

(1 1/ )

+

..........

Lga chun2LN(m, )

21

22

1 (ln x m)x exp ; , x 0

22

>

exp2

m2

+

Rayleigh2(x a) / b2

(x a)e , x ab

ba4

+

4b

4



Lu : vi u>0 hm Gamma.u 1 to

(u) t e dt =

Tnh cht: ;(u 1) u (u) + = (n) (n 1)! ; (1/ 2) = = .Cc lut phn b lin tc khc: B ta, tam gic,...Bin ngu nhin chun rt quan trng ta dnh ra 1 phn ring.

.1.2. Bin ngu nhin chun

1.2.1.Tnh cht hm mt . f(x) =2

22

1 (x m)exp ( 0)

22

>

+Hm mt xc nh trn ;+f(x) > 0: th nm trn trc honh;+Trc Ox l tim cn ngang;

+Gi tr cc i2

1

2

, t c ti x = m;

+ th i xng qua ng thng x=m, c dng hnh chung (Hnh 1.1).

Hnh 1.1. th hm mt ca phn b chun.

2

1

2

m xO

1.2.2.Cc tham s c trng2

E[X] m;

D[X] .

=

= (1.1)

Nh vy nhn thy rng, ch cn bit k vng v phng sai l c th bitmt f(x) v do hon ton bit v phn b chun. Cn c th tnh c

+ chch Skew(X) =3

3

E[(X EX) ]

= 0;

+ nhn Kurt(X) =4

4

E[(X EX) ]

- 3 = 0. (1.2)

1.2. 3.Bnn chun ho (chun tc).X c gi l bin nn chun tc nu X N(0,1).Hm mt ca n cho bi

8



2x

21

(x) e2

=

. (1.3)

c im : -Gi tr ca c lp bng vi x(x) {0;4];

- th i xng qua trc tung;

-Hm phn b tng ng2

tx2

1F(x) e dt

2

=

(1,4)

cng c lp bng. Tuy nhin, tit kim bng, thay cho F(x), ngi ta lpbng gi tr ca hm Laplace:

2tx

20

1

(x) e dt,2

= x[0; 3]. (1.5)

Vi x > 3, coi (x) 1

2.

Hnh 1.2. th hm mt chun ho (a) v th hm Laplace (b).

Khi cn tnh F(x) qua (x) hay ngc li, dng cng thc :

F(x) =1

2+ (x). (1,6)

Cng thc sau rt c ch tnh xc sut X nm trn on no :

[ ]{ }P X a;b (b) (a). = (1,7)

1.2.4.Bin i tuyn tnh bnn chun.

+Cho X N(m, ) Y= a X+b c phn b chun.2 a,b ,

T d thy aX+b N(am+b, 2 2a ).

+H qu. X 2X m

N(m, ) U

=

N(0,1). (1.8)

9



H qu ny cho ta phng php thun li tnh { }P X [a;b] :

[ ]{ }P X a;b =Pa m X m b m

b m a m( ) ( ).

=

(1.9)

1.2.5.Phn v .Phn v chun mc , k hiu U , l gi tr xc nh bi

{ }P U U> = , vi U N(0,1)

2t

2

U

1e dt

2

+

=

. (1.10)

Hnh 1.3. Phn v chun mc .Tnh cht: 1U U = . (1.11)

Mt s gi tr c bit: (1.12)0,10 0,025

0,05 0,01

U 1,280; U 1,960;

U 1,645; U 2,326.

= =

= =

Lu : Nhiu ti liu khng lp bng ca U m lp bng ca hocp u

vi

{ }P U p< = ; { }P U u< = .1.2.6. Sai s trung gian, dng mt chun dng trong pho binh.

Cho X , U( 2N m, ) l phn v chun mc , t== 6745,0UL 25,0 ;

.4769,02/U 25,0 == (1.13)Chng ta c th vit li hm mt ca X di dng

( )2 2(x m) / Lf x e

L

2=

. (1.14)

R rng l , nu m = 0 th

{ } 5,0LXLP =


9/187



Nu X v Y c lp th chng khng tng quan. Ngc li khng ng:Tn ti nhng BNN X v Y khng tng quan, song chng khng c lp.

i vi 2 BNN chun X, Yth X v Y c lp X v Y khng tng quan.b) Trng hp tng qut.

Cho l VTNN vi cc thnh phn l nhng BNN

bnh phng kh tch. t

1 T1 n

n

X

X ... (X ,...,X )

X

= =

1 1

n n

E[X ] m

m E[X] ... ...

E[X ] m

= = =

- vc t k vng;

Ma trn tng quan ca X cho bi

( )ij i jR (R ) E[X X ]= = .R rng .2ii iR E[X= ]Ma trn hip phng sai ca X cho bi

ij( ) = = Cov(X) = E[(X-m) . (1.18)T

(X m) ]

Lu :2 2i i i i iiD[X ] E[(X m ) ] = = =

j

- phng sai ca .iX

ij

= - hip phng sai ca .i i j j i

E[X m )(X m )] Cov(X , X ) =i j

X ,X

i j i i j jij

i j i j

Cov(X ,X ) E[(X m )(X m )]

D[X ]D[X ] D[X ]D[X ]

= = - h s tng quan ca .i jX ,X

R -ma trn cc h s tng quan.ij( )=

c)Tnh cht 1) ij 1, i, j. (1.19)

2) Nu cc thnh phn X1 c lp th khng tng

quan v R= ma trn cho,n j,...,X iX ,X

ij(R ) ij( ) -ma trn n v . Ngc li khng ng.

3) v R i xng , xc nh khng m.1.3.2. VTNN chun, cc tnh cht quan trng.

VTNN X= c gi l VTNN chun ( X gi l c phn b

chun trong

T1 n(X ,...,X )

n ) nu t hp tuyn tnh bt k cc thnh phn ca n c phn bchun.

12



Ni cch khc, u1,...,un, BNNY= 1 1 n nu X ... u X+ + c phn b chun.

H qu. Tng thnh phn ca VTNN chun lBNN chun.Lu : iu ngc li ni chung khng ng: Tng thnh phn ca VTNN

l chun chun.T1 nX (X ,...,X )= T1 nX (X ,...,X )=By gi gi m = E[X] l vc t k vng v = Cov(X) l ma trn hip

phng sai ca X (d thy tn ti ), phn b chun c k hiu bi N(m, ).VTNN chun X c vc t k vng m v ma trn hip phng sai c k hiubi

X N(m, ).+ Nu nh thc ca bng 0 th VTNN chun X c gi l suy bin. t

(hnh ca ), tn ti khng gian con k chiu cak Rang( )= n chiu caX trn khng gian ny l VTNN chun khng suy bin.

Mnh - nh ngha. Gi s X l VTNN chun vi ma trn tng quan .Nu det( ) 0 th X c gi l VTNN chun khng suy bin v mt ca ncho bi

f(x)= T 1n / 2 1/ 2

1 1exp (x m) (x m)

2(2 ) (det )

, nx . (1.20)

Nh vy, vc t gi tr trung bnh m v ma trn hip phng sai hon tonxc nh phn b chun; cc thng tin v m men cp cao hn l khng cn thit.

t1

n

G .

=

D thy 1G =1

n

1/

.

1/

, vi i iD[X ] =

Li t ;1 1R G G =

D thy = GRG ; 1 1 1G R G 1 = ;

11 1n1

n1 nn

D ....D1

R ..............det(R)

D ....D

=

trong l phn ph i s ca trong ma trn R. Thay vo (1.20) ta cijD ijR

13



f(x) =n

j ji iijn / 2 1/ 2

i ji, j 11 n

x m1 1 x mexp D .

2... (2 ) (det R) =

(1.21)

Mnh . Cho X = T1 n(X ,...,X ) N(m, ): . Khi l cc BNN

c lp khi v ch khi khng tng quan

1X ,...,Xn

n1X ,...,X

( l ma trn cho:

21

2n

.

=

)

1.3.3.Bin i tuyn tnh VTNN chun.

Mnh . Cho X N(m, ), A- ma trn cp kn tu cn kb bt k.

Khi VTNN Y=AX+b c phn b chun trn k vi T

E[Y] Am b;

Cov(Y) A A .

= +

=

H qu. Gi s XN(m, ) l VTNN chun trong n . Khi tn ti matrn trc giao A sao cho

U = A(X-m) N(0, D):trong D l ma trn cho, cc phn t trn ng cho chnh ca n khng m.

Nu X khng suy bin (det 0 ) th cc phn t trn ng cho chnh caD dng.

Chng minh. Ta chng minh cho trng hp det 0 . Khi , i xng,

xc nh dng, vy tn ti ma trn trc giao F c cc vc t ct ei l cc vc tring ca vi cc gi tr ring i tng ng sao cho

D =1

1 T

n

F F .

=

(1.22)

l ma trn cho. V xc nh dng nn cc gi tr ring i 0 > . t A =1

F th

E[U] = 0 ; Cov(U) = E T T T[F (X m)(X m) F] F F D = = . (1.23)

Khi U l VTNN chun, quy tm, cc thnh phn c lp. Bi v mi php bin i trc giao chnh l mt php quay trong n nn ta cth pht biu h qu trn bng li nh sau:

i vi mi VTNN chun, ta c th dng mt php quay thch hp binn thnh VTNN chun vi cc thnh phn c lp.

14



Hnh 1.4.ng ng mc ca mt chun 2 chiu.

O x

1.3.4. Mt s BNN lin quan n VTNN chun.

Mnh . c lp cng phn b chun N(0,1) th1X ,...,Xn

2 2 21 nY X ... X (n)= + + : . (1.24)Mnh . U N(0,1) , V , U, V c lp th: 2 (n):

U

TV / n

= T(n): . (1.25)

Mnh (Fisher). Nu X = l VTNN n chiu sao cho cc

thnh phn l nhng BNN c lp, cng phn b chun N(m,

T1 n(X ,..., X )

2 ) th :

a)n

i

i 1

1X X

n =

= v ( )n

2 2i

i 1

1S X

n

X

=

= l hai BNN c lp;

b)

2

22 n2i

2i 1

X N(m, );n

X XnS(n 1).

=

=

:

:

(1.26)

H qu. c lp cng phn b chun N(m,1X ,...,Xn2 ) th

n 2i

i 1

X mn

1(X X)

n 1 =

:T = T(n-1). (1.27)

1.3.5.Mt s phn v khc.

a) . Phn v mc ca phn b Khi bnh phng vi n bc t do, k

hiu l , l gi tr xc nh t biu thc:

2 (n)2

(n)

15



{ }2P X (n)> = , 0 1< < trong .2X (: n)

b) . Phn v Student mct (n) vi n bc t do, k hiu l , l gi tr

xc nh t biu thc:

t (n)

{ }P T t (n)> = , 0 1< 30.

Ngi ta lp bng gi tr ca 2 (n) v t (n) vi nhng gi tr khc nhauca v n.

2 t (n)(n)

Hnh 1.5. Phn v ca phn b Khi bnh phng(a) v ca phn b Student (b).

1.3.5.Vc t ngu nhin chun 2 chiu.Cho Z = (X,Y) l VTNN chun 2 chiu (khng suy bin) vi vc t k vng

m = ( v ma trn h s tng quan) .T

1, 2m m1

R1

=

Theo cng thc (1.21),

mt ng thi ca Z cho bif(x,y) =

2 2

1 1 222 1 1 2 21 2

1 1 x m x m x m x mexp 2

2(1 )2 1

2

+

.(1.28)

D dng tnh cE[X1] = m; D[X] =

21 ;

E[X2 ]= m; D[X] =22 ; XY . = (1.29)

c bit, nu X v Y c lp = 0 ( X v Y khng tng quan), mt ng thi cho bi

16



f(x,y) =

2 2

2

1 2 1 2

1 1 x y mexp

2 2

+

. (1.30)

i vi m men bc cao chng ta c kt qu quan trng sau y:

Nu (X, Y) l VTNN chun quy tm th2 2 2 2 2E[X Y ] E[X ]E[Y ] 2E [XY]= + . (1.31)

By gi chn X = Y N(0,: 2) th 4E[X ] 3 4= v chng ta nhn c cngthc tnh nhn (1.2).

1.3.6. Mt chun 2 chiu dng trong pho binh - Elp tn mt. nghin cu mc tn mt ca n ri trn mt phng nm ngang, ngi

ta lp h trc Oxy vi gc O trng vi mc tiu (im ngm bn), trc Ox lhng bn. Tng t nh (1.13) t

(1.32)

D 1 0,25 1

H 1 0,25 2

L U 0,6745 ;

L U 0,6745 .

= =

= =

nh lut tn mt khng nh rng, to im n ri (X, Y) tun theo lut

chun vi hm mt (1.30), m1 = m2 = 0. C th vit li mt ny di dng

2 22

2 2D H D H

x yf (x, y) exp

L L L L

2 = +

(1.33)

trong LD - sai s trung gian v tm, LH - sai s trung gian v hng .

i vi hu ht cc pho thng dng, LD ln gp 10 15 ln LH.

Elip tn mt (E) l elp c cc bn trc 4LD, 4LH (c ti liu ghi l LD, LH).

Xc sut im n ri (X,Y) nm ngoi elip tn mt rt nh, c th b qua:

( ) ( ){ } ( )2 2

2

0,25X Y

X YP X,Y E P 4U 0,025

= +

. (1.34)

Ngi ta chia (E) thnh cc vng vi t l % xp x n ri vo (Hnh 1.5);

nh c th tnh d dng xc sut n ri vo min G cho trc no .y

22LD

71625LH

25167

x

17



Hnh 1.6 . Elip tn mt vi thang chia .

1.4.M rng khi nim mt i vi BNN ri rc.+Chng ta bit rng, nu X l BNN lin tc vi hm phn b F(x) v hm

mt f(x) th:

*dF(x)

f (x) , xdx

=

=

;

0.

; (1.35)

* ; (1.36)f (x) 0; f (x)dx 1

* . (1.37){ }b

a

P a X b f (x)dx < =

+ m rng khi nim hm mt cho BNN ri rc tr

c ht ta

a rahm bc nhy n v, l hm:

1 khi x 0u(x)

0 khi x

=



j x1(x) e d

2

=

. (1.41)

Hm delta c th hin bng vc t n v //Oy (Hnh 1.7.). N c th coi

l o hm ca hm bc nhy n v:

du(x)(x)

dx =

k 0 h;h,k 0

u(h) u(k)lim

h k< <

=

(1.42)

Khi , nu X l BNN ri rc tp trung ti { }ix ,i 1,2,...= vi

{ }i ip P X x ,= = ii 1

p 1

=

i

, th c th coi X c mt

. (1.43)ii 1

f (x) p (x x )

=

Mt ny tho mn cc tnh cht (1.36) - (1.37) ca hm mt thng

thng. Ngoi ra, c th coi n l o hm ca hm phn b:dF(x)

f(x)dx

= .

c bit, hm delta ti a chnh l hm mt ca BNN hng s X = a; s d

nhvy l v :

du(x a)(x a)

dx

= ; (x a) 0; (x a)dx 1

= . (1.44)

Ngi ta cng hay xt hm khi lng xc sut ca BNN X

{ }p(x) P X x , x .= = V d. Cho X l BNN vi bng xc sut

Hm mt (suy rng) v hm khi lng xc sut th hin Hnh 1.8.

Hnh 1.8. Hm mt (a) v hm khi lng xc sut (b) ca BNN ri rc.

X 1 2 4

P 0,5 0,3 0,2

0,5

O x1 2 4

y

0,5

O x1 2 4

y

0,5 khi x 1

0,3 khi x 2 =

p(x)0,2 khi x 4

0 trai lai

=

= =

19



Cu hi Chng I

1.1 Nu mt s hiu bit v bin ngu nhin ri rc.1.2 Nu mt s hiu bit v BNN lin tc, 4 lut phn b lin tc.1.3 BNN chun: nh ngha, tnh cht hm mt , cc tham sc trng, BNN

chun tc, bin i tuyn tnh, phn v U .1.4 BNN chun: nh ngha, sai s trung gian, dng mt BNN chun dng

cho pho binh, qui tc 2 ,3 .1.5 Vc t k vng, ma trn tng quan, ma trn hi p phng sai ca vc t

ngu nhin n chiu; vi tnh cht.1.6 Vc tngu nhin chun: nh ngha, tnh cht.1.7 Bin i tuyn tnh VTNN chun.1.8 Mt s bin ngu nhin lin quan n VTNN chun.1.9 Phn v , vc tngu nhin chun 2 chiu.2 (n); t (n)

1.10 Mt chun 2 chiu dng trong pho binh, elip tn mt.1.11 Khi nim mt vi BNN ri rc.Bi tp chng I.

1.1. Chng t rng nua) X th E[X] = np; D[X] =np(1-p).B(n,p):b) X thP( ): E[X] ; D[X]= = ;c) th E[X] =(1-p)/p .X G(P):

1.2. Chng t rng nu X [ ]U a;b: th E[X] =2a b (b a)

; D[X] .2 1

+ =

2

1.3. Cho X ; chng t rng E[X] =m.2N(m, ):1.4. Cho X . Vit ra hm mt ca X v tnh cc sc sutN(2,9):

a) { } { }P 0 X 1 ; b)P 1 X 4 .<

1.5. Cho X . Tm mt ca Y = 2X 3. Tnh E[Y], D[Y].N(0,1):

1.6. Cho X .TnhN(0,1): { } { } { }P X 1,645 ; P X 1,960 ; P X 1,960 .> > >

1.7. Vit mt ca phn b chun, bit rng n c k vng 0 v sai s trunggian 2. Tnh { } { }P 2 X 2 ; P 0 X 2 .

1.8. ng knh ca vin bi c phn b chun vi trung bnh 20 v lch

chun 0,5. Quy tc khng nh cho ta iu g?2 ; 3

1.9. Cho (X,Y) : .N(0, )

a) Gi s , kt lun g v tnh c lp gia X v Y?1 0

0 3

=

b)Gi s , tm h s tng quan gia X v Y.4 1

1 9

=

20



1.10. Ma trn no sau y l ma trn hip phng sai?

a) ; b) ; c) ; d) .2 0

0 1

2 1

1 1

2 1

0 1

1 1

1 1

1.11. Cho U=(X,Y,Z) N(m, ): ,trong Tm (0,1, 2)= ;

1 0,5 0,5

0,5 1 0

0,5 0 4

=

, )

.

Tm phn b ca T = X 2Y + 3Z .

1.12. Cho , trong X

U N(0Y

=

:

1 0,5

0,5 1

=

10

210

.

Tm VTNN dng sao cho 2 thnh phn ca V, tc l

aX+bY v cX+dY l 2 BNN c lp.

aX bYV

cX dY

+= +

1.13. Cho l nhng BNN c lp cng phn b chun N(0,1).t

. Tm a, b 1X ,...,X

2 2 21 5 1X X ... X ; Y X ... X= + + = + +

{ } { }P X a 0,05; P Y b 0,05> = < = .

1.14. Gi s im n ri (X,Y) c phn b chun, trong lch hngX v lch tm Y khng c sai s h thng (tc l k vng 0), c lp, v cng lch chun 4 mt. Tnh xc sut n ri vo vng trn tm O bn knh 3mt.

1.15. Gi s im n ri (X,Y) c phn b chun, trong lch h

ngX : , lch tm Y , X v Y c lp. Tnh xc sut n ri vovng trn bn knh 3 mt, tm ti im ngm bn..

N(0, 4) N(0,5):

1.16.c lng xc sut n trng vo xe tng, bit rng ta ngm bn voim gia ca phn di ca xch v sau khi v xe ln h trc vi elp tn mt ththu c hnh v sau y.

y

22LD

71625LH25167

x

Hnh 1.9 . Xe tng trong h thng elip tn mt .

21



1.17. Cho , tnhX U[a;b: ] P{ X EX 2 } < , P{ X EX 3 } < vi

DX = ; so snh vi cng thc (1.17).1.18. Gi s . Chng t rng X l BNN khng c tr nh theo

nghaX E( ):

P{X s t | X t} P{X s}, t,s 0> + > = > .1.19. Cho , X v Y c l p. Chng minh rngX E( ), Y E(: : ) X Y+

. Mrng kt qu sang trng hp c nhiu bin ngu nhin.E( ) + :1.20. Bit rng mt ca BNN X c dng

xkxe khi x 0

f(x)0 x

=



Chng 3VC TNGU NHIN

3.5. SHI T CA DY CC BIN NGU NHIN

Mc ny trnh by bn dng hi t thng dng nht ca dy cc BNN cngnh nhng nh l ht nhn ca l thuyt xc sut v s hi t ca dy cc BNN

c lp: lut s ln, nh l gii hn trung tm. S hi t ca dy cc BNN dng

khc nh xch Markov, martilgal c trnh by chuyn kho khc.

3.5.1. Cc dng hi ta)nh ngha. Gi s X v , n = 1, 2, l cc BNN cng xc nh trn

khng gian xc sut (nX

), ,P .

(i) Ta ni dy cc BNN { }nX hi t chc chn ti BNN X v vit

(hay (cc)) nuccnX X nX Xn

nlim X ( ) X( ),

= .

(i) Ta ni dy cc BNN { }nX hi t hu chc chn ti BNN X v vit

(hay (hcc)) nu tn ti bin c vi P(A) = 1 saocho

hccnX X nX X A

nnlim X ( ) X( ), A

= .

(ii) Ta ni dy cc BNN { }nX hi t theo xc sut ti BNN X v vit

nuP

nX X{ }n

nlim P X X 0, 0.

= >

(iii) Ta ni dy cc BNN { }nX hi t trung bnh cp p (0 < p < ) ti BNN

X , v vit (hay theo trung bnh c p p), nuLPnX X nX Xp

nE X , n< vp

nnlim E X X 0

= .

(4i) Ta ni dy cc BNN { }nX hi t theo lut n BNN X v ta vit

hay nuLnX X XnF XF

X Xnnlim F (x) F (x) =

ti mi im lin tc ca hm phn b XF (x).T bt ng thc Liapunov (xem 5.2.1), hi t trung bnh cp p s suy ra hi

t trung bnh cp q vi 0 < q < p. Tuy nhin trong thc tng dng th hi t trungbnh cp hai l quan trng nht; hi t trung bnh cp hai cn gi l hi t bnhphng trung bnh hay MS - hi t (mean square convergence), k hiu

MSnX X, (n ) hay nl.i.m. X X=

23



(l.i.m. l vit tt ca ch limit in mean).Ring vi hi t theo lut, cc BNN v X c th xc nh trn nhng

khng gian xc sut khc nhau.nX

b. Tiu chun Cauchy vshi tp dng tiu chun Cauchy v s hi t ca dy s thng thng cng nh

mt vi k thut khc, chng ta pht biu cc tiu chun Cauchy sau y v s hit ca dy cc BNN. u im ca cc tiu chun Cauchy l khng cn s c mtca BNN gii hn X.

nh ngha: Ta ni dy cc BNN { }nX l dy Cauchy (hay dy cbn) huchc chn, theo xc sut hay theo bnh phng trung bnh nu ln lt tho mncc tnh cht sau:

+ Dy { }nX ( ),n 1,2,... = l dy Cauchy vi hu ht ;

+ ,0 > { }n mP X X 0 khi n, m ;

+2

n mE X X 0 khi n,m .

nh l (tiu chun Cauchy). Dy cc BNN { }nX hi tn BNN X no :(i) hu chc chn khi v ch khi n l dy Cauchy hu chc chn.(ii) theo xc sut khi v ch khi n l dy Cauchy theo xc sut;(iii) theo bnh phng trung bnh khi v ch khi n l dy Cauchy theo bnh

phng trung bnh.c gi c th tham kho chng minh trong cc cun sch chuyn bit v

xc sut nh [4], [11].c. Mi quan h gia cc dng hi tChng ta s pht biu nh l sau y nu ln mi quan h gia cc dng hi

t va nu.

nh l. (i) Dy { }nX hi t hu chc chn s hi t theo xc sut:hcc

nX X P

nX X .

(ii) Dy { }nX hi t bnh phng trung bnh th cng hi t theo xc sut:MS P

n nX X X X.

(iii) Dy { }nX hi t theo xc sut th cng hi t theo lut:

X XP Ln nX X. Mi quan h gia cc dng hi t th hin gin sau

hu chc chn Theo xc sut

Theo trung bnh

Theo lutChc chn

Ngoi ra, quan h sau y cng hay c s dng:nh l. Nu dy cc BNN {X hi t theo xc sut n X th c th tch ra

mt dy con { hi t hu chc chn n X.n}

nkX ,k 1,2,...}=

24



3.5.2. Cc nh l gii hn.Cc nh l gii hn mc nh ny kho st dng iu ca tng cc BNN

c lp cng phn b khi s cc s hng tng ln v hn, bao gm lut yu s ln,lut mnh s ln v nh l gii hn trung tm. Trc ht, chng ta tm hiu btng thc Chebyshev. Ngoi vic dng chng minh lut yu s ln, bt ngthc ny cn c ng dng vo nhiu mc ch khc.

a. Btng thc ChebyshevGi s X l BNN vi k vng EX v phng sai DX hu hn. Khi ,

xy ra bt ng thc:0 >

{ } 2D[X]

P X E[X] .

(3.5.1)

Chng minh. Chng ta chng minh cho trng hp X c hm mt , ta c:E[X]

2 2D[X] (x E[X]) f (x)dx (x E[X]) f (x)dx+

=

{ }2 2

E[X] (x E[X]) f (x)dx P X E[X] .

+

++ Nhn c pcm.

i khi dng sau y ca (3.5.1) cng rt tin li:

{ } 2D[X]

P X E[X] 1 . <

c bit, khi l mt s nguyn ln lch chun chng ta thu c

{ } 21

P X E[X] n 1 .n

<

Nu chn n = 3 th

{ }8

P X E[X] 3 .9

< (3.5.2)

Bt ng thc (3.5.2) cng c pht biu di dng quy tc 3 :Mi BNN khng lch khi gi tr trung bnh ca n mt lng 3 vi xc

xut kh ln.

Chng ta thy xc sut kh ln y ch l 8/9, thp hn rt nhiu so vi0.9973 trng hp X c phn b chun theo cng thc (1.17). Nh vy, nu bitthm thng tin v tnh chun ca BNN X, chng ta c nhng khng nh mnh

hn v kh nng xut hin bin c{ }X E[X] 3 . < b. Lut yu slnCho dy BNN { }nX c l p, cng phn b vi k vng iE[X ] m= v

phng sai hu hn. Khi , vi mi2iD[X ] = 0 > cnh,

1 n

n

X ... Xlim P 1.

n + +

< =

(3.5.3)

25Chng minh. T gi thit suy ra



21 n 1 nX ... X X ... XE ; D

n n

+ + + + = = .

n

p dng bt ng thc Chebyshev chng ta c2

1 n2

X ... XP 1

n n

.

+ + <

Chuyn qua gii hn khi nhn c pcm. n Theo cc dng hi t xt n mc 3.5.1, lut yu s ln chnh l:i vi dy BNN c lp, cng phn b vi phng sai hu hn, dy trung

bnh cng hi t theo xc sut n k vng chung ca dy.nh l ny c nu ra bi Bernoulli cui th k 17 nh l thnh cng

u tin ca l thuyt xc sut non tr. Thc ra, vi cng gi thit, chng ta cnthu c s hi t hu chc chn, dng hi t mnh hn hi t theo xc sut. lni dung ca lut mnh s ln, cng trnh thuc v Kolmogorov.

c.Lut mnh slnCho dy BNN { }nX c l p, cng phn b vi k vng iE[X ] = v

phng sai D[ hu hn. Khi 2iX ] =

1 n

n

X ... XP lim 1

n+ + = =

. (3.5.4)

Chng minh y nh l ny kh su sc v chng ta b qua.Nh vy, vi iu kin nu ra, dy trung bnh cng ( )1 nX ... X / n+ + hi

t hu chc chn n k vng .V d 5. Xt dy cc php th Becnoulli.

26

i1

X

0

=

Trung bnh cng ( ) 1 nnX ... X

Xn

+ += bng - tn sut xut hin bin c

A trong n php thu tin. Vi P = P(A) ta c

nf

i iE[X ] p; D[X ] p(1 p).= =

nu bin c A xy ra php th th i

nu tri li

Theo lut mnh s ln, tn sut hi t hu chc chn n E[ iX ] p P(A).= =Nh vy, lut mnh s ln l cston hc ca nh ngha thng k v xc

sut, a ra giai on u ca l thuyt ny.V d 5.. Hnh 3...(a) trnh by kt qu m phng vi BNN m X vi k

vng E[X] = 1. Theo cc gi tr , chng ta tnh ton c trung bnh cngiX

( ) 1 nn X ... XX .n+ +=

Sau khong 200 php th chng ta dng nh nhn c sn nh.Hnh 3....(b) ch ra hnh nh ca n{(X) } vi 50 gi tru, s bin ng

dng nh cn ln.



( )nX

10 30 50400 1000200

n

(a) b

Hnh 3. . Trung bnh cng ca dy BNN m k vng 1.

Nhn xt: S hi t ca dy n{(X) } thng l chm hn rt nhiu so vi shi t ca dy tt nh hay gp thng thng.

V d, nu chng ta cn mt ngng xc sut ( tin cy) 95%, i viBNN c , theo bt ng thc Chebychev chng ta c tha ra bng sau yv s php th cn thit trung bnh cng

1 =

n(X) lch khi k vng E[X] mt

lng b hn .Sai s tuyt i 0,1 0,01 0.001

S php th n 2 000 200 000 2 000 000

Gn y ngi ta a ra nhng bt ng thc tinh vi hn, s php th cnthit gim c5 ln.

d)nh l gii hn trung tm.

n(X) nghin cu t m hn v , hy chun ho n bng cch t

n

n

(X)

T n

= .Z ] 1.

(3.5.5)R rng v D[nE[T ] 0= n = Nh vy, k vng v phng sai ca

BNN gii hn (nu c) vn l 0 v 1 tng ng. Cu hi t ra l: Hm phn bhay hm mt (nu c) ca BNN gii hn s ra sao?

nh l sau y tr li cho cu hi ny.nh l (nh l gii hn trung tm). Cho dy BNN { }nX c lp, cng

phn b vi k vng iE[X ] = v phng sai2

iD[X ] = hu hn. Khi i

vi dy { }iT xc nh theo (3.5.5) xy ra ng thc:

{ }

2tx 2

nx

1lim P T x e dt2

< = (3.5.6)

V phi ca (3.5.6) chnh l hm phn b chun tc F(x). Nh vy, nh lgii hn trung tm khng nh rng: i vi dy BNN c lp cng phn b v

phng sai hu hn, dy chun ho ca trung bnh cng hi t theo lut n phnb chun tc.

nh l c cng bu tin bi Laplace cho dy { } vida vo cng thc Stirling. nh l c chng minh theo phng php hm c

nX nX ~ B(1, p)

27



trng. c gi cng c th tham kho chng minh t m khc [ ]. V tm quantrng c bit, ngi ta pht trin nh l ny theo rt nhiu hng khc nhau.

Nhn xt. Ngi ta nhn thy tc hi tnh l gii hn trung tm khtt, th hin bt ng thc

3i i

x 3x

E X EXSup F (x) F(x) C

n<



2tx2

0

1(x) e dt

2

=

- hm Laplace.

Ngi ta thy rng xp x l tt nu np > 5; nq > 5 hoc npq > 20.Cng thc hiu chnh. Vi cc iu kin va nu cho n, p, q ngi ta thy

c th lm tt hn xp x (3.5.9) bng cch tng ln na n v, v gim ina n v. C th, nn s dng xp x sau:

2k 1k

n 1 2 2 1P (k k ) (x ) (x ), (3.5.11)

trong

1 2

1 2

1 1k np k

2 2x ; xnpq npq

np +

= =

>

vi iu kin hoc npq > 20.np 5; nq 5>V d: Xc sut trng ch ca mt x th khi bn mt vin n vo bia l

0,75. Tnh xc sut x th bn 100 vin c 81 pht trng ch trln.Gii. np, nq > 10, chng ta c th p dng cc kt qu nu trn.

100 2 1P P (81 100) (x ) (x )= ;

181 100.75

x 1,38100.0,75.0.25

= ;

29

2100 100.75

x 5,77.100.0,75.0.25

=

(1,38) 0,4162; (5,77) (3,0) 0,5000 = =

Tra bng ta c ;P 0,5000 0,4162 0,0838 8% = = .

Nu ta dng cng thc hiu chnh th

181 0,5 100.75

x 1,27

100.0,75.0.25

= ; 2100 0,5 100.75

x 5,889.

100.0,75.0.25

+ =

(1,27) 0,398; (5,77) (3,0) 0,5000 = =

;

P 0,500 0,398 0,102 10% = = .Xp x (3.5.9) v do (3.5.10) l tt vi cc diu kin a ra. Cc diu

kin ny tho mn, chng hn khi n ln hoc khi p gn vi 1/2. Khi n nh, ngi ta lp bng gi tr cho cc xc sut . Khi n khng ln

lm, hoc khi

(n 20) nP (k;p)p 0 haykhi p 1, cc cng thc trn khng cn chnh cc na,

ngi ta s dng nh l gii hn Poisson sau y.nh l (nh l gii hn Poisson). Gi s trong lc Becnoulli

p P(A),= v khi m npn const= = thk

nnlim P (k) e

k!

= .

Chng minh. T cng thc Becnoulli ta c

k k k k (n k)n n

n(n 1)...(n k 1)P (k) C p (1 p) p (1 p)

k! += = .

T ch p / n= suy ra



k n

nn(n 1)...(n k 1)

P (k) 1k! n n

+ =

k

n kk 1 k 11.(1 )...(1 ) 1

k! n n n

=

.

Chuyn qua gii hn khi (k cnh) ta cn

nnlim P (k)

=

n( )( ) k k k

nlim 1 e

k! n k!

=

.

ngha. Khi n ln cn np = khng ln lm ( 5 chng hn), xy ra cngthc xp x

k

nP (k) e k!np

=

(3.5.12)

Xp x l tt khi n > 50; p < 0,1.V xc sut p l b nn bin c A rt t khi xy ra khi thc hin mt php th.

Chnh v th, nh l gii hn Poisson cn c gi l nh lut v cc s kinhim hoi.

Ngi ta lp bng cc gi tr cak

ek!

.

C ti liu li lp bng gi tr

chokm

k 0

e

k!

=

. Cc xc sut n 1 2P (k k ) c th d dng tnh c t (3.5.12) v

cc bng ny.V d. Xc sut mt loi my bay trn mt tuyn ng nht nh b tai

nn l 4p 10= . Tm xc sut trong 1000 ln bay c :a) mt ln b tai nn; b) c t nht 1 ln b tai nn.

Gii. Ta c th p dng nh l gii hn Poisson vi n = 1000, p = 0,0001;.np 0,1 = =

10,1

a 1000(0,1)

P P (1) e 0,0905.1!

= =

00,1

b 1000 1000(0,1)

P P (1 1000) 1 P (0) 1 e 0,0952.0!

= = =

Nu by gip = 0,001, tnh ton tng t ta c a bP 0,3679; P 0,6321. y l nhng xc sut kh ln.

Cu hi n tp chng III3.1. Dy BNN hi t hcc, theo xc sut, bnh phng trung bnh v theo lut.

Mi quan h gia cc dng hi t ny.3.2. Pht biu v chng minh bt ng thc Chebyshev, lut yu s ln. Pht

biu lut mnh s ln.3.3. nh l gii hn trung tm.

30


29/187

Phn IIQU TRNH NGU NHIN

Chng 5NHNG KHI NIM TNG QUT

5.1. MU

Trong chng ny chng ta s nghin cu nhng khi nim cn bn nht v

qu trnh ngu nhin (QTNN). Khi u, l thuyt QTNN c pht trin gn vi

dao ng v nhiu trong nhng h vt l. QTNN a ra nhng m hnh hu hiu

nghin cu cc lnh vc khc nhau nh vt l thng k, thng tin lin lc, phn

tch chui thi gian, phn tch hot ng mng my tnh v khoa hc qun l.5.1.1. Cc nh ngha

*Gi s I l tp v hn no cn ( )S, P, l khng gian xc sut cbn.

H cc bin ngu nhin (BNN) { }tX , t I cng xc nh trn ( )S, P, c gi l

hm ngu nhin.

Tp ch s I c gi l tp xc nh; tp gi tr E ca cc BNN c

gi l tp gi tr ca hm ngu nhin.

tX

Khi I , tham bin t thng ng vai tr thi gian (cng c th c ngha khc) v chng ta gi { }tX ,t I l QTNN. Hn na, nu I l tp m c

th ta gi { }tX , t I l QTNN vi thi gian ri rc hay dy cc BNN. c bit, ta

gi { }nX , nN hoc { }n 0 0X , n n ,n 1,...= + l dy cc BNN mt pha. Nu I =Z

th ta gi { }nX , nZ l dy cc BNN hai pha.

Khi I l mt khong suy rng ca , chng ta giR { }tX ,t I l QTNN thi

gian lin tc, n gin l QTNN.

i vi cc trng h p khc, v d kI = R , { }tX , t I c gi l trng

ngu nhin. Chng ta s nghin cu ch yu QTNN thi gian ri rc hoc thi

gian lin tc. Trong quyn sch ny nu khng ni g thm, QTNN xem xt l

QTNN thc, tp gi tr l tp con ca . Khi tp gi tr l tp con ca ,

chng ta c QTNN phc, chng ta s ni r QT l phc.

R C

*Thc cht, chng ta ang cp n hm 2 bin X {X(t, ), t I, S}=

sao cho:

Khi cnh, l mt BNN;t I X(t, )

32


30/187

Khi cnh, l mt hm s thng thng trn I, c gi l

mt quo (tn khc: th hin, hm chn) ca QT ng vi kt cc ca th

nghim ngu nhin.

S X(t, )

*Cng l chnh ng, v i khi li l thong nht khi coi QTNN nh l

mt nh xng mi vi quoS X(., ) - mt hm s thng thng trn I.Thut ng hm ngu nhin c l xut pht t quan im ny.

Nh vy, chng ta li c th coi QTNN nh l h ca cc quo hay l

mt tng th (ensemble) ca cc th hin (hay cc hm chn) ca n.

S khc bit gia cc quan nim nu trn v QTNN nm mc ch v

phng php nghin cu.

tin li chng ta hay vit X(t) thay cho , cng nh hay vittX ( )X t,

thay cho ( )tX l gi tr ca qu trnh (QT) ti thi im t khi kt qu ca th

nghim ngu nhin l (khi xy ra bin c scp S ). Chng ta cng hay k

hiu QT bi {{X(t, ), t I, S} X(t), t I}, {X(t)} hay n gin l X.

5.1.2. Phn loi sb.

Ty theo tp ch s I v khng gian gi tr E (trong Vt l, E c gi l

khng gian trng thi), ngi ta phn QTNN lm bn loi sau y:

i) I lin tc, E lin tc: QTNN vi thi gian lin tc v trng thi lin tc

(tn khc: QTNN lin tc);

ii) I lin tc, E ri rc: QTNN vi thi gian lin tc v trng thi ri rc (tnkhc: QTNN ri rc);

iii) I ri rc, E lin tc: QTNN vi thi gian ri rc v trng thi lin tc (tn

khc: dy ngu nhin lin tc ).

iv) I ri rc, E ri rc: QTNN vi thi gian ri rc v trng thi ri rc (tn

khc: dy ngu nhin ri rc).

Theo cc tnh cht ca quo, ngi ta c th phn loi QTNN mt cch t

m hn. Chng hn, khi I l khong suy rng no ca R, ta ni { }tX , t I l:

* QT lin tc theo quo nu hu ht cc quo ca n l hm lin tc;* QT bc nhy nu hu ht cc quo ca n l hm bc thang

Lu rng trong cun sch ny, thut ng hu ht hay hu chc chn

c s dng vi ngha rng, cc tnh cht kn xy ra vi xc sut 1.

Hnh 5.1(a) m t mt quo in hnh ca QTNN lin tc {X(t)}. Hnh

5.1(b) m t dy ngu nhin c c bng cch ly mu QT {X(t)} theo chu k T0

chn trc: ( )n 0X X n T , n 0,1,2,...= = Hnh 5.1(c) m t quo ca QT du ca

QT ban u: Khi X(t) dng, Y(t) nhn gi tr 1; gi tr -1 c nhn ti nhng

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thi im cn li. Dy ri rc Hnh 5.1(d) m t dy mu vi chu k ly mu T0

ca QT du ( ){ }Y t .

a

c

(d)

(b)

Hnh 5.1. Cc dng quo ca qu trnh ngu nhin

5.1.3. V d v QTNN

V d 5.1.in tm in tm l bc tranh ca tim ngi thc hin bng sng in trng

hay siu m). C nhiu nhiu trn nhng bc tranh ny bi v sng in trng

(hay siu m) c phn x t nhiu ni bn trong lng ngc.

V d 5.2.u ra ca knh thng tin thng b mo do nhiu in t. Lu

rng c tn hiu u vo cng nh tn hiu iu chu c khng gian trng thi

gin on, tn hiu u ra li c khng gian trng thi lin tc. C ba tn hiu c

thi gian lin tc.

V d 5.3.Tn hiu m thanh. Tn hiu m thanh c thc xem l ngunhin t ch dy cc m lng to nn tn hiu l bt nh. C khng gian trng

thi v thi gian u lin tc.

V d 5.4.Tn hiu FM vi nhiu. Tn hiu FM (tn hiu iu ch tn s)

in hnh c dng:

( ) ( )t

c f0

X t cos 2 f t 2 k a s ds

= +

,

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trong fc l tn s mang, thng nm trong di tn 88 108MHz , kf l h s

iu ch my pht (transmitter`s modulation constant), cn a(s) l tn hiu m

thanh cn truyn i. Thm ch khng c nhiu, X(t) l tn hiu ngu nhin t ch

a(s) l ngu nhin.

V d 5.5.Sng sin ngu nhin. Cho [ ]U U 0;1: l bin ngu nhin c phnbu trn [0;1]. Xt qu trnh

( ) ( ) ( )X t, U sin 2 t , t = R.

Mi hm mu ca n l mt hm hnh sin theo thi gian vi bin ngu

nhin.

V d 5.6.Dy nhiu trng. Dy cc BNN {X(n)} c gi l dy nhiutrng nu n quy tm (k vng khng), cng phng sai v khng tng quan.Mt quo in hnh ca dy nhiu trng vi 2 1 = th hin Hnh 5.2. Dy

nhiu trng v QTNN nhiu trng c vai tr quan trng trong nghin cu QTNN.

Hnh 5.2. Mt thhin ca dy nhiu trng Gauss vi phng sai 1.

5.1.4. H cc phn b hu hn chiuGi s ( ){ }X t , t I l QTNN. i vi thi im 1t I c nh, X(t1) l

bin ngu nhin vi hm phn b FX(x1,t1) xc nh bi

( ) ( ){ }X 1 1 1 1F x , t P X t x= < . (5.1.1)

By gigi s { }1 nJ t ,..., t= l mt t p con hu hn ca I. Hm phn b

ng thi ca ( ) ( )1 nX t ,...,X t :

( ) ( ) ( ){ }X 1 n 1 n 1 1 n nF x ,..., x ; t ,..., t P X t x ,...,X t x= < < (5.1.2)

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c gi l hm phn b hu hn chiu ( y c n chiu) ca qu trnh

( ){ }X t ng vi tp ch s J. Tp cc hm phn b hu hn chiu c gi l h

cc hm phn b hu hn chiu.

R rng, h cc hm phn b hu hn chiu c hai tnh cht sau y:

i) Tnh cht i xng: Hm phn b hu hn chiu khng thay i nu tahon v b ch s (1, 2,,n). V d, khi hon v hai ch su chng ta c:

( )X 1 2 n 1 2 nF x , x ,..., x ; t , t ,..., t = ( )X 2 1 3 n 2 1 3 nF x , x , x ,..., x ; t , t , t ,..., t . (5.1.3)

ii) Tnh nht qun theo ngha

( ) ( )X 1 n 1 n X 1 n 1 1 n 1xnlim F x ,..., x ; t ,..., t F x ,..., x ; t ,..., t

= . (5.1.4)

Ngc li, cho h hm phn b hu hn chiu tho mn hai tnh cht nu

trn th tn ti QTNN ( ){ }X t , t T vi h hm phn b hu hn chiu cho.

chnh l ni dung ca nh l tn ti ca Kolmogorov (ngi Nga).Rt nhiu tnh cht quan trng ca QTNN c quy nh bi tnh cht ca

cc hm phn b hu hn chiu ca n, trong quan trng nht l hm phn b

mt chiu F(x; t) v hm phn b hai chiu F(x1, x2; t1, t2).

Thng thng, chng ta phi nghin cu ng thi mt s QT. T, m

rng (5.1.2), chng ta a vo khi nim hm phn bng thi ca hai qu trnh

( ){ } ( ){ }X X t ,Y Y t= = :

( )XY 1 n 1 m 1 n 1 mF x ,..., x , y ,..., y ; t ,..., t ,s ,...,s = ( ) ( ) ( ) ( ){ }1 1 n n 1 1 m mP X t x ;...;X t x ;Y s y ;Y s s= < < < < . (5.1.5)

c gi cng c th d dng mrng sang trng hp c hu hn QT.

5.2. MT S LP QU TRNH NGU NHIN

5.2.1. Qu trnh cp II

QTNN ( ){ }X X t , t I= c gi l QT c p p (p > 0) nu vi mi

( )t I, X t l bin ngu nhin kh tch cp p, tc l ( )p

E X t < .T bt ng thc Liapunov (xem [4 ], Phn III tr 127)

( )( ) ( )( )1/ q 1/ pq p

E X t E X t , 0 q p < < (5.2.1)

suy ra nu QT l cp p th n cng l cp q vi 0 < q < p.

Trng hp quan trng nht khi p = 2, lc ta c QT cp hai.

i vi QT cp hai ( ){ }X X t , t I= t

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( )

( ) ( ) ( )( )X

X

t E (X(t));

R t,s E X t X s , t,s I

=

= ;

( ) ( ) ( )( )XC t,s Cov X t ,X s , t,s I= . (5.2.2)

v gi ( )X t l hm k vng, RX (t,s) l hm t tng quan v CX(t,s) l hm t

hip phng sai ca qu trnh X cho.

Khi bit hm k vng , hm t hi p phng sai v hm t tng quan c

th biu din qua nhau theo cng thc

( ) ( ) ( ) ( )X X X XC t,s R t,s t s= . (5.2.3)

Cng ty, nu QT l quy tm, tc l hm trung bnh bng 0, th hm t

tng quan v hm t hip phng sai trng nhau.

Nhiu khi chng ta cn tm ( ) ( )2

E X t X s l k vng ca bnh phng

s gia X(t) X(s) ca qu trnh ti hai im t, s. i lng ny c th tnh thng

qua hm t tng quan:

( ) ( ) ( ) ( ) ( )2

X X XE X t X s R t, t 2R t,s R s,s = + . (5.2.4)

Trong k thut, thng th ( ){ }X t l dng sng theo thi gian ca in p

hay dng trn khng 1 m. Cng sut ca QT ti thi im t l X2(t) v cng sut

trung bnh ti thi im ny l E(X2(t)), khiu bi PX(t). Nh vy, trong biu

thc (5.2.2) cho t = s ta nhn c cng sut trung bnh( ) ( ) ( )2X XP t E[X t ] R t, t= = (5.2.5)

v hm phng sai

( ) ( ) ( )( ) ( ) (22 2

X X X X Xt C (t, t) E X t t R t, t t = = = ) . (5.2.6)

chun ho hm t hip phng sai, ngi ta dng h s t tng quan

( )( )

( ) ( )X

XX X

C t,st, s

C t, t C s,s = . (5.2.7)

Hm trung bnh v hm t tng quan l hai thng k quan

trng nht ca QT. Tuy nhin, tnh chng phi thng qua trung bnh tng th,

tc l phi bit mt xc sut hai chiu ca QT - iu rt kh thc hin trong

thc t. bi 3.5 tip theo chng ta s gii thiu cch tnh xp x cc hm ny

trong tnh hung khi vic ly trung bnh tng th khng th lm c.

X (t ) XR (t,s)

i vi hai QTNN cng xc nh trn I v khng gian xc sut ( )S, P, , t

( ) ( ) ( )XYR t,s E[X t Y s= ], (5.2.8)

( )XY X YC t,s E[(X(t) (t)) (Y(s) (s))]; t,s I= , (5.2.9)

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XYXY

XY XY

C (t,s)(t,s)

C (t, t) C (s,s) = (5.2.10)

v c gi ln lt l hm tng quan cho, hm hip phng sai cho v h s

tng quan cho ca hai qu trnh X v Y.

D thy quan h (5.2.3) c th mrng cho hm hip phng sai cho:( ) ( ) ( ) ( )XY XY X YC t,s R t,s t s= . (5.2.11)

Hai qu trnh X v Y c gi l khng tng quan nu:

( )XYC t,s 0, t,s I= . (5.2.12)

iu ny tng ng vi

( )XYR t,s E[X(t) ].E[Y(t)], t ,s I= .

Hai qu trnh X v Y c gi l trc giao nu:

( )XYR t,s E[X(t)Y(s)] 0, t,s I= = . (5.2.13)R rng, nu mt trong hai QT l quy tm th hai khi nin khng tng

quan v trc giao l tng ng:

X quy tm: X,Y khng tng quan X, Y trc giao.V d 5.7. V d tm thng v QTNN l tn hiu tt nh X(t) = f(t), trong

f(t) l hm s cho trc. i vi trng hp ny,

( ) ( ) ( ) ( ) ( )X Xt f t ; R t,s f t f s = = .

V d 5.8. Xt qu trnh ( ){ }X t vi

( ) ( ) 0,2 t sX Xt 3; R t,s 9 4e = = + .

Hy tm k vng, phng sai, hi p phng sai ca cc bin ngu nhin

Z = X(5); W = X(8).

Ta c E[Z] = E[W] = 3

( ) ( )

( )

2 2X X

X

E[Z ] R 5,5 13; E[W ] R 8,8 13;

E[Z W] R 5,8 11,195.

= = =

= =

=

Vy D[Z] = D[W] = 13 - = 42

3Cov(Z, W) = E[Z W] E[Z]E[W] = 2,195.

5.2.2. Qu trnh s gia c lp

nh ngha. QTNN ( ){ }X X t , t I= c gi l QT s gia c lp nu

cc s gia ca n trn nhng khong thi gian ri nhau l nhng BNN c lp:

Vi mi t0, t1, , tn trn I: t0 < t1 < < tn, cc s gia

( ) ( ) ( ) ( ) ( )0 1 0 n nX t ,X t X t ,...,X t X t 1

l nhng BNN c lp.

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Thm vo , nu lut phn b ca X(t) X(s) ch ph thuc vo hiu t s

th ta gi X l qu trnh s gia c lp thun nht.

V d 5.9. Cho { }n , n 0,1,... = l dy BNN c lp. Dy tng ring {Xn}

0 0 1 0 1 2 0 1 2X ; X ; X ,...= = + = + +

lp thnh dy s gia c lp (hy chng minh !).nh ngha. Cho { }tX X , t I= l QT c p hai: ( )

2E X t t I.< Ta

ni X l QT s gia khng tng quan nu hai s gia ca n trn nhng khong thi

gian ri nhau l nhng BNN khng tng quan.

C th l, vi t0, t1, t2, t3 bt k trn I sao cho t0 < t1 < t2 < t3 ta c:

1 0 3 2Cov(X(t ) X(t ), X(t ) X(t )) 0 = .

R rng l, nu X l QT s gia c lp v l QT cp hai th X l QT s gia

khng tng quan. iu ngc li ni chung khng ng (c nhng phn v dchng tiu ny).

5.2.3. Qu trnh dng

QT dng c vai tr c bit quan trng trong v tuyn in cng nh nhiu

ngnh khc. Vi th, qu trnh dng c dng iu bt bin theo thi gian.

Chng ta phn QT dng lm hai loi: theo ngha hp v theo ngha rng.

a) Qu trnh dng theo ngha hp

nh ngha. Ta ni QTNN { }tX X , t I= l QT dng theo ngha hp (hay

QT dng mnh) nu vi s t nhin n bt k, vi J = {t1,, tn} l tp con ty ca I v vi s thc h bt k sao cho { }1 nK t h,..., t h I= + + , cc VTNN

1 n(X(t ), ...,X(t )) v 1 n(X(t h),...,X(t h))+ +

c cng lut phn b.

Ni ngn gn, l qu trnh c h phn b hu hn chiu bt bin vi php

dch chuyn thi gian.

Thng ngi ta xt trng hp I = R; cng c th xt cc trng hp

khc, v d [ )I 0;= + hay [ ]I a; b= .Thc t, vic kim tra iu kin dng ni trn rt kh khn. Tuy nhin, li

c th pht biu mt lot iu kin cn (nhng khng ) mt qu trnh (QT) l

dng theo ngha hp. Nu vi phm d 1 trong cc iu kin ny th khng nh

rng QT khng l dng theo ngha hp.

* Cho n = 1. Chng ta nhn c

( ) ( )X XF x, t F x, t h , x , t,h : t, t h I= + +R .

Nh vy, i vi QT dng theo ngha h p hm phn b mt chiu khng ph

thuc vo thi gian.

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* Nu QT l cp k (k nguyn khng m) thk

kE(X (t)) , t I= .

c bit, nu QT l cp hai th hm trung bnh v phng sai l hng s:

( ) ( )

( ) ( )

X

2 2X

t E[X t ] ;

t D[X t ] , t

= =

= = I.

* By gichn n = 2, t0 cnh ty trn I nhn c:

( ) ( )( )X 1 2 1 2 X 1 2 0 0 2 1F x , x ; t , t F x , x ; t , t t t .= +

Nh vy, mt iu kin l: Hm phn b hai chiu ch ph thuc hiu

thi gian:

( ) ( )X 1 2 1 2 X 1 2F x , x ;t , t F x , x ;= vi 2 1t t = .

Nu cc hm phn b c mt th iu kin ny tng ng vi

( ) ( )X 1 2 1 2 X 1 2 2 1f x , x ; t , t f x , x ; t t= .

* Tng t, hm t tng quan RX(t,s) v hm t hip phng sai CX(t,s)

ch ph thuc vo hiu thi gian:

( ) ( ) ( ) ( )X XR t,s E[X t X s ] R t s= = ;

( ) ( ) ( )( ) ( )X XC t,s Cov X t ,X s C t s= = .

Thc vy,

( ) ( ) ( )X2 2

R t,s xydF x, y; t,s xydF x, y; t s= = R R

ch ph thuc vo hiu t - s. Tng t cho hm ( )XC t,s .

Nhng trnh by v dng theo ngha hp nu trn c thc mrng thnh

dng theo ngha hp ng thi ca hai QT:

Hai QT ( ){ }X t v ( ){ }Y t c gi l dng theo ngha hp ng thi nu

cc hm phn bng thi ca chng (xc nh theo (5.1.5)) bt bin vi php

dch chuyn thi gian.

b) Qu trnh dng theo ngha rngnh ngha. Gi s { }tX X , t I= l QT c p hai. Ta ni rng X l QT

dng theo ngha rng nu:

i) Hm k vng l hng s

( ) ( )X Xt E[X t ] , t = = I ;

ii) Hm t tng quan ch ph thuc vo hiu thi gian

( ) ( ) ( ) ( )X XR t,s E[X t X s ] R , t s= = =

trong ( )XR l hm no ca bin .

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tin li, QT dng theo ngha rng c gi tt l QT dng.

Lu rng nu xy ra (i) th iu kin (ii) tng ng vi

ii) ( ) ( ) ( ) ( ) ( ) ( )X X X XC t,s E[(X t t )(X s s ) C , t s= = =

l hm ch ph thuc vo hiu t s.

Ngi ta cng gi hm ( )XR , ( )XC ln lt l hm t tng quan v thip phng sai ca qu trnh dng X.

Cc iu kin (ii) v (ii) c vit di dng tin li sau y:

( ) ( )

( ) (X X

X X

R t , t R

C t , t C .

+ =

+ = )

;

Phng sai ca QT dng l hng s:

( ) ( ) ( )2 2X XD[X t ] t C 0= = = , t I .

Ngi ta gi l k vng chung,X 2X l phng sai chung ca QT X.R rng, mi QT dng theo ngha hp v l cp hai s l mt QT dng theo

ngha rng. Ngc li ni chung khng ng (c nhng v d minh hoiu ny).

Sau y chng ta nu ra mt s tnh cht ca hm t tng quan.

nh l 5.1. Cho ( )XR l hm t tng quan ca QT dng nhn gi tr

thc ( ){ }X t , t R . Khi :

i) ( )XR l hm chn, tc l ( ) ( )X XR R , = R.

ii) Hm ( )XR t cc i ti 0 = :

( ) ( )X XR R 0 , R.

iii) ( )XR l hm xc nh khng m theo ngha:

Vi mi b 2n s thc t1,, tn, b1,,bn bt k lun xy ra bt ng thc:

( )n

i j X i ji, j 1

b b R t t 0=

. (5.2.14)

Chng minh.i) ( ) ( ) ( ) ( ) ( )X X X X XR R 0 R 0, R ,0 R = = = = .

ii) p dng bt Cauchy-Schwarz, ta c

( ) ( ) ( ) ( )

( ) ( )( ) ( )X X

1/ 22 2X

R R ,0 E X X 0

E X E X 0 R 0 .

= =

=

iii) ( ) ( ) ( )2n n

i i i j i j

i 1 i, j 1

0 E b X t b b E[X t X t ]

= =

=

=

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42

) (n

i j X i ji, j 1

b b R t t=

= .

Cu hi t nhin t ra l, vn ngc li phi chng cng ng? nh

l khng km chng minh sau y s l cu tr li cho vn ny.

nh l 5.2. Hm s ( )R , R l hm t tng quan ca mt QTNN thc,dng khi v ch khi ( )R l hm xc nh khng m.

Kim tra tnh xc nh khng m ca mt hm s cho trc l iu kh kh

khn. nh l sau y (xem [12], tr 404) nu ln mt iu kin mt hm cho

trc l hm t tng quan.

nh l 5.3.(Polya). Hm chn ( )R , R l hm t tng quan ca mt

QTNN thc, dng no nu tho mn hai iu kin sau y:

i) ( )R l hm li trn ( )0;+ ;ii) ( )lim R c ;

= c - hu hn.

V d 5.10. Xt dao ng iu ho vi bin v tn s hng s, pha ngu nhin:

( ) ( )X t Asin t + , t= R

trong A, l cc hng s, c phn bu trn [0;2]. Chng ta c:

( )2

0

AE[X(t)] sin t + d 0

2

= =

;

( ) ( )( ) ( )

( )

X

2

2

R t , t E Asin t + .Asin t +

AE cos cos 2 t + 2

2

Acos ,

2

+ = +

= +

=

l hm s ch ph thuc vo .

Vy X dng. Hn na chng minh c (xem [15] tr 90 -91):

Cho l hng s, A v l hai BNN c lp; { }X(t),tR xc nh Vd 5.11 l QT dng mnh khi v ch khi c phn bu trn [0;2].

V d 5.11. Xt sng sin ngu nhin

( ) ( )X t Asin 2 t , t= R

trong A l bin ngu nhin c phn b u trn [0;1]. D thy

( )X1

t sin 2 t con2

= st , vy ( ){ }X t khng l QT dng (theo ngha rng).

Chng ta cng c th tnh c


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( )( ) ( )

( )( ) ( )

X Xsin 2 t sin 2 s sin 2 t sin 2 s

R t,s ; C t,s3 1

= =

2.

nh ngha. QTNN ( ){ }X t , tR c gi l tun hon theo bnh phng

trung bnh (hay MS - tun hon) nu tn ti s thc t0 sao cho

.20E[X(t t ) X(t)] 0, t+ = RTa gi t0 l MS - chu k ca qu trnh.

Tnh ngha suy ra ngay rng, vi xc sut 1,vi mi t

0X(t t ) X(t)+ = .

Lu : Khng suy ra ( ) ( ){ }0P :X t t , X t, , t 1 + = = .

nh l 5.4. Nu i vi dng ( ){ }X t xy ra ng thc ( ) ( )X XR 0 R t= 0

th ( ){ }X t l MS - tun hon vi MS - chu k l t0.

Lu rng hm t tng quan ca QT dng c th c thnh phn hng s

khc khng, c th c thnh phn tun hon, c th c thnh phn tt dn nhanh

hoc chm. Hnh 5.3 a ra cc dng in hnh ca hm t tng quan.

XR ( )

XR ( )

XR ( )

XR ( )

Hnh 5.3.Cc dnh in hnh ca hm ttng quan ca QT dngi vi QT dng ( ){ }X t , h s tng quan (5.2.7) trthnh

( )( )( )

XX

X

C,

C 0

= R. (5.2.15)

c)Dngng thi.nh ngha. Ta ni hai QT { }X(t)},{Y(t) l dng ng thi nu mi QT

{ }X(t) ,{ }Y(t) l dng, hn na hm tng quan cho ca chng ch ph thucvo hiu thi gian:

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XY XYR (t,s) E[X(t)Y(s)] R (t s).= = i hi ny c vit li di dng tin li sau y:

XY XYR (t , t) E[X(t )Y(t)] R ( ).+ = + = (5.2.16)

Hm c gi l hm tng quan cho ca hai QT {XYR () }X(t)},{Y(t) .Khi mi QT X, Y l dng, rng buc (5.2.16) tng ng vi hm hip

phng sai cho ch ph thuc vo hiu thi gian:C (XY XY XY X Yt , t) C ( ) R ( ) .+ = = (5.2.17)

Khc vi hm t tng quan, hm tng quan cho ni chung khng chn.Sau y l mt s tnh cht ca hm tng quan cho hai QT dng ng thi.

nh l 5.5.i vi hai QT thc dng ng thi { }X(t)},{Y(t) ta c:

XY YX(i) R ( ) R ( ); =

XY X Y(ii) R ( ) R (0) R (0) ;

(iii) X YXYR (0) R (0)

R ( )

2

.+

Chng minh. Tnh cht (i) trc ti p suy t nh ngha. T bt ng thcCauchy-Schwarz v t tnh dng suy ra

( 2 2 2E[X(t )Y(t)]) E[X (t )]E[Y (t)]+ +

= = 2 2 X YE[X (0)].E[Y (0)] R (0) R (0),chng ta nhn c (ii). Tnh cht (iii) suy ra t (ii) v bt ng thc Cauchy:

X YX Y

R (0) R (0)R (0)R (0) .

2

+

Mt s dng c th ca hm t tng quan cho th hin Hnh 5.4.

Hnh 5.4.Hm tng quan cho ca hai QT thc, dngng thi

Mt li ch ca hm tng quan cho l nh ta c th tnh c hm ttng quan ca tng hai QT.

nh l 5.6.i vi hai QT dng ng thi { }X(t)},{Y(t) ta c

(5.2.18)X Y X Y XY YXR ( ) R ( ) R ( ) R ( ) R ( ).+ = + + +

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Chng minh. Ta cE[X(t ) Y(t )][X(t) Y(t)]+ + + +

= E[X(t )X(t)] E[Y(t )Y(t)] E[X(t )Y(t)] E[Y(t )X(t)]+ + + + + + +

= X Y XY YXR ( ) R ( ) R ( ) R ( ). + + +

5.2.4. Qu trnh Gauss

QTNN ( ){ }X t , t I c gi l qu trnh Gauss (hay QT chun) nu cc

phn b hu hn chiu cu n l chun. Ni cch khc, i vi mi tp con hu

hn J = {t1,...,tn} I, vc tngu nhin c phn b chun.1(X(t ), ...,X(t ))n

Theo nh ngha ca VTNN chun, iu kin ny chnh l:

1 na ,...,a R, BNN ( ) ( )1 1 n na X t ... a X t+ + c phn b chun.

n gin cch vit, gi s ( )iX X ti= , t

;1 1

n n

X E[X ]

[X ]

X ; m E(X)

X E

= = =

g g1

n

xx ;

x

=

g ( )1 nC Cov X ,...,X .=

Hm mt ng thi ca cc bin ngu nhin cho bi1X ,...Xn

( )( )

( ) (T 1

1 n

1x m C x m

2X ,...,X 1 n n / 2n / 2

1f x ,..., x e

2 det C

) =

. (5.2.19)

c bit, phn b mt chiu v phn b hai chiu l cc phn b chun.

Mt ng thi (5.2.19) hon ton quyt nh bi cc phn b mt chiu(v ) v cc phn b hai chiu (vi i iim E[X ], C D[X ]= = i ( )ij i jC Cov X ,X= vi

i j). V vy, nu QT cho l dng (theo ngha rng) th cc gi tr s bt

bin i vi php dch chuyn thi gian, v do , mt ng thi (5.2.19)

cng s bt bin. T chng ta c:

im , Cij

nh l5.7.Nu qu trnh Gauss l dng (theo ngha rng) th n cng l QT

dng theo ngha hp.

Hai lp QTNN c bit quan trng l QT Poisson v QT Wiener scgii thiu 5.5.

45


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5.3. TNH CHT ERGODIC V TRUNG BNH THI GIAN5.3.1. Gii thiu

Ergodic l mt tnh cht tinh vi, thot u kh c th chp nhn c n, th

nhng li rt hu ch v c s dng rng ri. Khi nim ny c nh ngha

theo nhiu dng khc nhau: theo ngha bnh phng trung bnh hay theo ngha hu

chc chn; theo k vng, theo hip phng sai hay theo m men cp p no ; p

dng cho qu trnh dng hay cho qu trnh bt k. Nhng nh l ergodic c

pht hin u tin vo na u th k XX bi J.Von.Neumann, B.Birkoff (M),

A.Ia. Khinchin (Nga).

Ni mt cch ngn gn, tnh cht ergodic m bo rng:

Nu mt tham s thng k no ca QT c tnh bng k vng, tc l

trung bnh tng th th tham s cng c thc tnh theo trung bnh thi gian

i vi mt quo n l.

hnh dung ra tnh ergodic, chng ta xt v d sau y. Sp xp thi gian

trong ngy theo pht ta c th coi { }t I 0, 00; 0, 01;...; 24, 00 = . Gi spht th

t, s gi tin chuyn qua mt nt no ca mt mng my tnh l N(t). S gi tin

trung bnh chuyn qua nt lc 10 gil E[N(10,00)].

tnh gi tr ny, chng ta ghi li kt qu quan st ca N ngy, v nh N= 100. Theo lut mnh s ln ta c th xp x

( ) ( )( )1 1001

E[N(10,00)] N 10,00 ... N 10,00100

+ +

trong Ni(10,00) l s gi tin chuyn qua nt trong 1 pht vo lc 10 gingy

th i.

Nu qu trnh ( ){ }N t c tnh ergodic, chng ta ch cn quan st trong mt

ngy no , vi thi on kh ln no - chng hn N = 100 pht quanh 10

gi, v nh t 9 gi00 n 10 gi40. Chng ta ch vic ly s gi tin chuyn qua

nt trong 100 pht chia cho 100 sc xp x cho gi tr trung bnh cn tm.

n gin, ngi ta c th tin hnh hai th nghim trn theo phng php

m phng thng k. K c khi y, chng ta thy li ch ln lao ca vic ly trung

bnh theo thi gian. Tin ch ca vic ly trung bnh thi gian ln n mc ngi

ta c tin hnh phng php ny, d rng qu trnh c th khng l ergodic.

Chng ta a ra sau y hai dng ca tnh ergodic: ergodic k vng v

ergodic hip phng sai, v hu nh ch p dng cho QT dng.


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5.3.2. Ergodic k vng

nh ngha. i vi hm s f(t),t cho tr c, trung bnh thi gian ca

f(t) xc nh biT

TT

1lim f (t)dt

2T +

, (5.3.1)

v c k hiu l A[ f(t)]. Ton t A[ . ] c gi l ton t trung bnh thi gian.

Trng hp {f(t)} l QTNN, gii hn a ra c hiu theo bnh phngtrung bnh.

nh ngha. QTNN dng nhn gi tr thc ( ){ }X t c gi l ergodic k

vng nu k vng ca QT bng trung bnh thi gian ca mt quo bt k:

( ) ( )T

TT

1E[X t ] lim X t dt,

2T+ = (5.3.2)

gii hn theo bnh phng trung bnh.

t ( )T

TT

1X X

2T = t dt . (5.3.3)

TX l gi tr trung bnh thi gian trn [- T; T] ca { }X(t) . Khi , gii

hn l gi tr trung bnh ca quo trn ton trc s. i vi QTNN

tng qut, gii hn nu l mt BNN, tc l ph thuc vo

T

TtlimX

S . Tuy nhin:

i vi QT ergodic, chng ta c th ly trung bnh thi gian ca mt quo bt k lm trung bnh tng th ca qu trnh :

E[X(t)] A[X(t)]= .

nh l 5.8. Gi s ( ){ }X t l QTNN dng, nhn gi tr thc vi hm trung

bnh v hm t hip phng sai ( )XC . iu kin cn v ( ){ }X t l QT

ergodic k vng l

( )T

X

T 0

1lim 1 C d 0

T T+

=

. (5.3.4)

Chng minh. Theo tnh cht ca tch phn (xem mc 5.4.3) chng ta c:T

TT

1E[X ] E[X(t)]dt .

2T = =

T , ( )TX T theo bnh phng trung bnh khi v ch khi

( )2T TD[X ] 0 T = . Li p dng tnh cht ca tch phn v tnh dng ca

( ){ }X t chng ta i ti


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( )( ) ( )( )T T

2T T

T T

1 1D[X ] E X t dt X s ds

2T 2T

= =

( )( ) ( )( )T T

2T T

1E X t X s dtds

4T =

( )T T

X2T T

1C t s dtd

4T = s

s

.

i vi tch phn kp cui cng, bng cch i bin u t s , v t= = + ta i ti:

( )2T

X2T

11 C

2T 2T

d

.

(Nu cc bn kh khn trong vic xc nh cn ca bin tch phn u, v, trc

By gis dng tnh chn ca hm C() ta c

u 2 = . Cng c th dng lc

vi phn tnh tch phn ny).

( ) ( )u t s 2 ; v t s 2= = + ,

trong tch phn n thu c t

ht hy dng php quay s

t

T

T

( )2T

2T T X

0

1D[X ] 1 C d

T 2T

= =

. (5.3.5)

T nhn c kt lun ca nh l.

Nhn xt. Theo quan im ca thng k ton, XT chnh l mt c lng

khng chch v vng ca k vng . Hn na, theo bt ng thc Chebychev

(3.5.1), phng sai tnh theo (5.3.5) cn cho php chng ta tm khong tin cy

cho c lng ny. Chng hn, tin cy

2T

( )T T T TX 4,47 ; X 4,47 + ( )T T T T3 ; X 3 )( X +

T

l ln hn 95% (ln hn 8/9).

Nh vy khi l mt c lng thonh ca k vng .T , X


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trong nhiu trng hp.

H qu.

a) Nu tch phn hi t th qu trnh( )X0

C

d ( ){ }X t l ergodic k vng.

b) Nu ( ) ( )2X XR hay tng ng

( ) ( )XC 0

th ( ){ }X t l QT ergodic k vng.

Chng minh.

a) l hin nhin. chng minh b) gi s 0 > cho trc. Khi tm c

T0 > 0 ( )C < / 2 vi . T0T>

( ) ( )TT 0

X X0 0

1 1C d C d

T T = ( )

( )T 0 X 0X

T0

T C 0 T T1C d

T T 2

T

+ +

vi T ln.

Khng phi mi QT dng u l ergodic k vng, xt v d sau y.

V d 5.12. Xt ( ){ }X t vi ( )X t U= , trong U l bin ngu nhin vi

E(U) = m; 0 < D[U] < + .

D thy ( ){ }X t l QT dng, cc quo u l nhng ng thng nmngang, ( ) ( )tX U = vi mi S .

Bi v D[U] > 0 nn bin c: ( ){ }:U m c xc sut dng, v do

( ) ( )TTlim X U( ) m E X t

= = .

Vy ( ){ }X t khng l QT ergodic k vng.

V d 5.13.i vi QT dng ( ){ }X t vi ( ) cXC qe = chng ta c

( ) ( ) ( )T T c cTX0 0

1 1 qC d qe d 1 e 0 TT T cT

= = .

Theo nh l Slutsky, ( ){ }X t l ergodie k vng. Ngoi ra, theo (5.3.5)

chng ta tnh c phng sai 2T nh sau:

2T2 cT T

0

1D[X ] 1 qe d

T 2T = =

( )

2cTq 1 e1 0

cT 2cT

T=

.

Cc kt qu k trn cho QT dng, thc, thi gian lin tc c mrng cho

dy dng, thc.


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nh ngha. Dy ngu nhin dng nhn gi tr thc ( ){ }X n c gi l

ergodic k vng nu:

( ) ( ) ( )N

Nn N

1X X n E[X n ],

2N 1 =N= =

+ ,

gii hn theo bnh phng trung bnh.nh l 5.10. Cho dy ngu nhin dng nhn gi tr thc ( ){ }X n vi hm t

tng quan CX(n). Dy ( ){ }X n l ergodic k vng khi v ch khi phng sai 2N

ca XN

( )N

2N N X

n N

n1D[X ] 1 C n

2N 1 2N 1=

= = + +

(5.3.7)

dn n khng khi N .

inh l 5.11 (nh l Slutsky). Dy dng nhn gi tr thc ( ){ }X n l

ergodic k vng khi v ch khi

( )N

XN n 0

1lim C n 0.

N == (5.3.8)

Nhn xt. Nu dy dng ( ){ }X n l ergodic k vng th

( )N

Nn N

1X X

2N 1 ==

+ n

s l c lng khng chch v vng ca k vng vi phng sai c tnh

theo (5.3.7).

V d 5.14. Xt dy dng ( ){ }X n vi ( ) nXC n Pa= , (0 < a


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o bt k, c th l:

( )( ) ( )( )T

2 2

TT

1V E[ X t ] lim X t dt

2T+ = = , (5.3.9)


Ni mt cch ngn gn, i vi QT ergodic phng sai, trung bnh tng thv trung bnh theo thi gian ca bnh phng cc lch (X(t) - )2 l nh nhau:

.( )( ) ( )( )2 2

E[ X t ] A[ X t ] =

Lu rng trong nhiu trng h p, ta c th coi qu trnh quy tm ho

( )({ ) }2X t - k hiu l ( )2X - l QT dng. Nh vy, iu kin ergodicphng sai ca QT ( ){ }X t cng chnh l iu kin ergodic k vng ca QT

( )({ ) }2X t , l ( ) ( )T 2XT0

1lim C d 0T +

= . Bng cch khai trin d thy

22 2

(X )C ( ) E[(X(t+ )- ) (X(t) - ) ] -

= 4 ,

iu kin trnn n gin hn mt cht:

iu kin cn v ( ){ }X t l ergodic phng sai lT

2 2 4 2X X

T0

1lim E[(X(t+ ) ) (X(t) ) ]d = C (0)

T = . (5.3.10)

Trong h thc (3.5.10) chng ta cn n nhng m men cp 4. Tuy nhin,

vi qu trnh Gauss, vn trnn n gin hn.

H qu. Cho ( ){ }X t l QT dng Gauss vi k vng bit. QT ( ){ }X t

l ergodic phng sai khi v ch khi

( )T

2X

T0

1lim C d 0.

T+ = (5.3.11)

Khi , ( ){ }X t cng l qu trnh ergodic k vng.

Chng minh.i vi qu trnh Gauss, p dng h thc (1.31) ta c

22 2

X(X )C ( ) E[(X(t+ ) - ) (X(t) - ) ] -

= 4

X

4 2 4 2X X( 2E [(X(t+ ) - )(X(t) - )]) - 2C ( )= + =

By gich vic p dng nh l Slutsky. Phn cn li suy t khng nh

( ) ( ) ( )2T T

2X X

0 0

1 10 C d C d 0 , T

T T .

Nhn xt. Nu ( ){ }X t l ergodic phng sai th trung bnh thi gian trn


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] )[ T;T ca bnh phng lch ( )(2

X t xc nh bi

( )( )T

2T

T

1V X t

2T = dt

s l c lng khng chch cho phng sai V = D[X(t)] = CX(0). Phng sai ca

c lng ny cho bi (5.3.5), trong CX() cn phi thay th bi( )

( )2XC

.

b) Trng hp k vng cha bit.

By gichng ta khng quan tm n ergodic phng sai na, vn l lm

th no c lng V. Bi v cha bit, chng ta c th dng trung bnh thi

gian XT theo (5.3.2) c lng, ri sau tnh c lng

( )( ) ( )T T

2 2T T

T T

1 1V X t X dt X t dt

2T 2T = = 2TX . (5.3.12)

Xc nh cc tnh cht thng k ca kh kh khn. khc phc, chng

ta da vo nhn xt sau y. Ni chng, l mt c lng chch ca phng

sai V. Tuy nhin, khi T ln, chch c th b qua. Hn th, phng sai c

thc xp x bi phng sai ca V

TV

TV

TV

T l c lng ca V khi bit. Trong

nhiu trng h p, sai s bnh phng trung bnh l nh hn

vi gi tr T ln. T, dng c lng V c th s tt hn

dng V

2T

E[(V V) ]

2TE[(V V) ] TV

Tc lng V k c khi bit.

c) Ergodic thip phng sai

Nu ( ){ }X t c k vng bit, chng ta c th xt ( ){ }X t . Nu

( ){ }X t c k vng cha bit, chng ta c th dng X T theo (5.3.2) c

lng v xt ( ){ }TX t X vi lu rng kt qu l kh chnh xc vi T ln.

Nh vy, chng ta c th gi s rng QT l quy tm, tc l E(X ) = 0.t

i vi cnh, qu trnh tchR ( ) ( ){ }X t X t+ c th coi l dng vik vng ( )XC . p dng cc kt qumc 5.3.2 c lng ( )XC , chng ta

c th dng trung bnh thi gian

( )( ) ( )T

X TT

1C Z

2T = t dt, (5.3.13)

vi ( ) ( ) ( )Z t X t X t= + .

y l c lng khng chch ca ( )XC , phng sai ca n cho bi

(5.3.5), trong ( )XC cn phi c thay th bi hm t hip phng sai ca


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qu trnh ( ){ }Z t :

( ) ( ) ( ) ( ) ( ) ( )2Z XC E X t+ + X t+ X t+ X t C = .

p dng nh l Slutsky ta i n kt lun:

Qu trnh ( ){ }X t l ergodic t hip phng sai nu v ch nu

( )T

ZT

0

1lim C d 0.

T+ = (5.3.14)

Nu by gigi thit thm ( ){ }X t l qu trinh Gauss th

( ) ( ) ( ) ( )2Z X X XC C C C = + + .

Trong trng hp ny (5.3.5) cho ta

( )( )( )

( ) ( ) ( )T

2X X X X

T 0

2D C C C C d

T

= + +

.

Nu ( )XC 0 th ( ) ( )ZC 0 , t ta nhn c kt qu sau y:

H qu. Nu qu trnh dng Gauss nhn gi tr thc c ( )C 0 ( )

th n l QT ergodic t hip phng sai.

Tnh cht ergodic k vng v ergodic t hip phng sai hay c s dng

hn c. Chnh v th, qu trnh c hai tnh cht ny cn c gi l ergodic suy

rng (xem [13] tr 93). Chng ta xt thm mt loi ergodic na lin quan n hai

qu trnh.d) Ergodic hip phng sai cho.

Hai QT nhn gi tr thc, dng ng thi ( ){ }X t v ( ){ }Y t c gi l

ergodic hip phng sai cho nu tng QT l ergodic t hip phng sai, hn na

hip phng sai ca chng

( ) ( )( ) ( )( )XY X YC E[ X t+ Y t = ]

c thc tnh thng qua trung bnh thi gian

( ) ( )( ) ( )(T

XY X YT

T1C lim X t Y t2T+

= + )dt , (5.3.15)


Ging nh tin hnh mc c), bng cch quy tm ho, chng ta c th

coi ( ){ }X t v ( ){ }Y t l quy tm. Trung bnh thi gian

( ) ( ) ( )T

XY TT

1C X t+

2T = Y t dt , (5.3.16)

l mt c lng khng chch ca ( )XYC v phng sai ca c lng ny


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c tnh theo (5.3.5), ( )XC phi c thay bi ( )XYC .

Nu c ba hm ( )XC , ( ) ( )Y XYC , C u dn n 0 khi th

( ){ }X t v ( ){ }Y t l ergodic hip phng sai cho. ()

5.3.4. Cc loi ergodic khc

Xt mt quo { }X(t, ), t R ca QT X. Vi x cnh, hm s

1 X(t, )u(x X(t, ))

0 X(t, )

x

x


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th { }X(t) c gi l ergodic hm mt .

* Nu vi k > 0,k kE(X (t)) A[X (t, )],=

th { }X(t) c gi l QT ergodic m men cp k.

Vn trung bnh thi gian v ergodic

Mt cch y nht:

Nu tt c cc c trng xc sut ca QT tnh thng qua trung bnh tng th

u c thc tnh thng qua trung bnh thi gian ca cc c trung tng ng

th QT c gi l QT ergodic.

Ngi ta cng tm c cc iu kin (gi l iu kin egrodic), ch yu

vi qu trnh dng c c tnh cht .

Tnh ergodic l mt dng rt hn ch ca tnh dng v tht l kh khn

kim tra xem trong tnh hung vt l c th no , gi thit ergodic thong hay

khng. D sao, chng ta vn thng gi thit QT l ergodic n gin ho bi

ton. Trong th gii thc, chng ta vn buc lng phi lm vic vi ch mt hm

mu ca qu trnh. Khi y, d mun hay khng, chng ta vn phi tm gi tr trung

bnh, hm t tng quan ch t mt hm dng sng theo thi gian. T gi thit

ergodic, chng ta c th coi nhng gi tr tnh c l nhng tham s thng k ca

qu trnh. Nhiu ngi cm thy kh chp nhn nhng li bn lun ny. Tuy nhin

cn phi nhrng, l thuyt ca chng ta ch phc v m hnh ho nhng iu

xy ra trong th gii thc.

5.3.5. o hm tng quan.

Thc t khng bao gita c tho c hm tng quan ca hai QT bi v

ta khng th no bit ht cc th hin ca chng. Ni chung, chng ta ch bit mt

on ca mt quo ca mi qu trnh. Nh vy, chng ta cn phi gi s rng,

cc QT ca chng ta l ergodic ng thi (hay t ra l ergodic tng quan ng

thi). Thc ra cc gii thit ny- gi l gi thit ergodic- t quan trng, ta vn tin

hnh cng vic ca ta cho d cc QT khng phi l ergodic.

Trn tinh thn , ngi ta to ra nhng thit b vt l o hm tng

quan ca cc QTNN, gi l tng quan k (correlometer) o hm tng quan

cho ca hai QTNN { }X(t),Y(t) .

S khi ca tng quan k th hin Hnh 5.6. N bao gm hai b gi

chm, b nhn v b tch phn. Th hin x(t) v y(t) ca hai QT { }X(t) v { }Y(t)

c gi chm ln lt l T v T n v thi gian; tip theo, tch ca cc hm

sng gi chm c to thnh. Sau na, tch ny c tch lu li qua b tch

phn u ra l tch phn trn on

[ ]1 1t ; t 2T+ vi di 2T.


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y(t)

x(t)

A

0 1R (t 2T)+

B Gi chm

B nhn ( )t 2T1t1

1dt

2T

+ g

Gi chm

Hnh 5.6. S khi tng quan k

Gi s tn hiu tn ti t ra t thi im T, cn dng tu . D thy u

ra nhn c

1t

t T1

0 1t T1

1R (t 2T) x(t )y(t)dt

2T

+

+ = + .

V d, nu ta chn , t tnh ergodic ta c1t = 0

T

0 XT

1Y XY(2T) X(t )y(t)dt A ( ) R ( )2T

= + = R .

Cho thay i (thng qua cc b gi chm) ta c tho xp x hm tng

quan cho ca X v Y.

Khi thay i cc im ni A, B v p dng c hai u vo l x(t) hoc c hai

u vo l y(t), ta c tho hm t tng quan XR ( ) , YR ( ) .

Cn lu di on ly tch phn 2T phi ln. Vi mt s QT c th,

ta c th tnh c di 2T ti thiu sai s tng i nh, v d .5%Rt nhiu s khc c thit ko hm t tng quan. V d, ngoi

vic gi nguyn hai b gi chm, ngi ta b tr b cng thay cho b nhn. Sau

b cng l b lc bnh phng (dng it) ri mi n b lc m tn.

ng lu tm nht phi kn cc tng quan k quang hc ca Michelson.

Chi tit xem [8 ] tr 439 441.

5.3.6. c lng dy tng quan.

i vi dy dng (v ergodic!) {X(n)}, ngi ta dng 2 c lng sau ycho dy t tng quan:

a) c lng ttng quan khng chch

N k

X Xi 1

1R (k) R ( k) X(i k)X(i) (0 k N)

N k

== = +

. (5.3.17)

V nn y l c lng khng chch. Nu thm githit {X(n)} l Gauss, quy tm th y l c lng vng. Tuy nhin, c th

X XE[R (k)] R (k)=

{ }XR (k) khng t cc i ti k = 0, t ma trn tng quan c th khng xc

nh khng m.


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b)c lng ttng quan chch

N k

X Xi 1

1R (k) R ( k) X(i k)X(i) (0 k N)

N

== = + (5.3.18)

y l c lng chch, song l c lng tim cn khng chch. N cngl c lng vng. u im ni bt ca c lng ny l n t cc i ti k = 0

v ma trn tng quan lun lun xc nh khng m.Ngi ta thy rng, ch nn c lng vi k N /10 .Khi N ln, sai s bnh phng trung bnh ca c lng khng chch l ln

hn so vi c lng chch. Vy khi c th, hy dng c lng chch, s tthn c lng khng chch.

5.4. LIN TC, O HM, TCH PHN

Lin tc, o hm, tch phn l nhng khi nim trung tm nghin cu,

dng iu ca h tt nh, tn hiu u vo l hm xc nh ca thi gian.

Chng ta cn hon thin v mt xc sut, tin hnh nhng nghin cu su sc hn,sang h vi u vo ngu nhin. Mi khi nim nu u lin kt cht ch vi

mt trong ba dng gii hn: hu chc chn, theo xc sut v theo trung bnh cp p

(xem 3.5). Tuy nhin trong thc t, hi t bnh phng trung bnh (MS - hi t)

l quan trng nht. V th, chng ta s hng s ch vo dng hi t ny. Trong

mc ny, chng ta ch xt nhng QT vi thi gian lin tc.

5.4.1. Lin tc

a) Lin tc theo xc sut

nh ngha. Cho QTNN ( ){ }X X t , t I= . Qu trnh X c gi l lin tctheo xc sut ti nu:0t I

( ) ( ){ }0t t0

0, lim P X t X t 0

> > = . (5.4.1)

Nu X lin tc theo xc sut ti mi im 0t J I th X c gi l lin tc

theo xc sut trn J.

b) Lin tc theo trung bnh

Gi s( ){ }

X X t , t I= l QT cp p (p > 0), tc l (xem mc 5.2.1):

( )p

E X t , t I< .

nh ngha. Qu trnh cp p ( ){ }X t , t I c gi l lin tc trung bnh cp

p ti nu0t I

( )p

t0t t0lim E X t X 0

= . (5.4.2)

Nu ( ){ }X t lin tc trung bnh cp p ti mi im 0t J I th n c gi

l lin tc trung bnh cp p trn J.


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T bt ng thc Chebyshev v bt ng thc (5.2.1) suy ra:

+ Lin tc trung bnh cp p (p > 0) th lin tc theo xc sut;

+ Lin tc trung bnh cp p th lin tc trung bnh cp q vi 0 < q < p.

Trong ng dng, ngi ta xt ch yu qu trnh cp hai. Chng ta tch lin

tc trung bnh cp hai (gi l lin tc bnh phng trung bnh) ra mc ring.

c) Lin tc bnh phng trung bnh

nh ngha. QTNN ( ){ }X X t , t I=

I

c gi l lin tc bnh phng trung

bnh (hay MS - lin tc) ti nu0t

( ) ( )2

0t t0lim E X t X t 0

= . (5.4.3)

Nu ng thc xy ra ti mi im 0t I , qu trnh ( ){ }X t c gi l lin

tc bnh phng trung bnh (hay MS lin tc) trn I.

nh l sau a ra mt iu kin thng qua hm t tng quan, d dng

kim tra trong thc hnh.

nh l5.12. Gi s ( ){ }X X t , t I= l QT cp hai vi hm trung bnh, hm

t tng quan v hm t hi p phng sai ln lt l ( ) ( ) ( )X X Xt ,R t,s ,C t,s .

Khi ,

a) Nu X l MS lin tc th ( )X t l hm s lin tc.

b) X l MS- lin tc khi v ch khi hm hai bin ( )XR t,s lin tc.

c) X l MS lin tc khi v ch khi cc hm s ( )X t v ( )XC t,s l nhng

hm lin tc.

Chng minh.

a) Theo bt ng thc Schwarz chng ta c:

( ) ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

X X 0 0

20

0 t t E X t X t E X t X t

E[X t X t ] 0 t 0 .

=

0

Vy ( )X t l hm lin tc.b)iu kin cn. Khi th0 0t t , s s ( ) ( ) ( ) ( )0 0X t X t , X s X s theo

bnh phng trung bnh. T tnh cht ca k vng chng ta nhn c

( ) ( ) ( ) ( ) ( ) ( )X 0 0 X 0 0R t,s E[X t X s ] E[X t X s ] R t ,s= = ( )0 0t t , s s .

(Bn no kh khn khi dn ra gii hn va nu, hy dng ng nht thc)

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )0 0 0 0X t X s X t X s X t X t X s X s = +

( ) ( ) ( )( )

( ) ( ) ( )( )

0 0 0X t X s X s X s X t X t+ + 0 (*)

ri s dng cc bt ng thc quen bit).


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iu kin . Dng cng thc (5.2.4) v tnh lin tc ca RX ta c

( ) ( ) ( ) ( ) ( ) ( )2

0 X X 0 X 0 X 0 0E X t X t R t, t R t, t R t , t R t , t 0 = + 0t )(t

c) y l h qu ca (a), (b) v ng nht thc (5.2.3).

Lu . Thc ra, da vo (*) chng ta cn c th pht biu :

QT {X(t)} l MS - lin tc ti khi v ch khi lin tc ti ;{X(t)} l MS - lin tc khi v ch khi lin tc ti vi mi

ot XR (t,s) o o(t , t )

XR (t,s) o o(t , t ) ot I .

V d 5.15.i vi sng sin ngu nhin V d 5.11 chng ta tnh c

( )( ) ( )

Xsin 2 t sin 2 s

R t,s3

= l hm lin tc theo hai bin t, s. Vy y l qu

trnh MS - lin tc.

Lu . i vi QT Poisson s xt mc 5.5.1 chng ta s tnh c

( ) ( )2

XR t,s min t,s ts= + l hm lin tc. T, QT Poisson l MS lin tc. Tuy nhin, mi quo ca

QT ny u lm hm bc thang v do khng lin tc.

Trong thc t, lm th no kim tra tnh lin tc bnh phng trung bnh?

p dng nh l trn, chng ta mong mun bit dng ca hm t tng quan

( )XR t,s . Trong hu ht cc tnh hung thc t, chng ta ch c th tnh ( )XR t,s

mt cch xp x thng qua t chc ca n, m t chc li l mt hm bc

thang! V vy, ch tr khi t chc qu r rng vi phm tnh lin tc, chng ta

thng phi cng nhn rng, QT l MS lin tc.

5.4.2. o hm

nh ngha. QTNN ( ){ }X X t , t I=

I

c gi l kh vi theo bnh phng

trung bnh (hay MS - kh vi) ti 0t nu tn ti BNN ( )0X t sao cho

( ) ( )( ) (0

X t X tX t 0

+

) (5.4.4)

gii hn theo bnh phng trung bnh, chnh l

( ) ( )( )

2

00

X t X tlim E X t 0

+ =

.

Khi , ( )0X t c gi l o hm ca qu trnh ( ){ }X t , t I ti im t0.

Nu (5.4.4) xy ra ti mi im 0t I th qu trnh ( ){ }X t c gi l MS -

kh vi, qu trnh ( ){ }X t c gi l o hm ca qu trnh ( ){ }X t .

nh l 5.13. Gi s ( ){ }X t l QTNN c p hai vi hm t tng quan

( )XR t,s . Nu ( )XR t,s l hm kh vi cp hai ti im (t0, t0) th qu trnh ( ){ }X t


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l MS - kh vi ti im t0.

Chng minh. Theo (5.2.4) chng ta c:2

0 0 0X(t ) X(t ) X(t ) X(t )h( , ) E+ + 0 =

( ) ( ) ( ) ( )2 20 0 02 21 1E X t X t E X t X t = + + + 0

( ) ( )( ) ( ) (( )0 0 02

E X t X t X t X t + +

)0

( ) ( ) (X 0 X 0 0 X 0 021

R t , t 2R t , t R t , t ) = + + + +

( ) ( ) (X 0 X 0 0 X 0 021

R t , t 2R t , t R t , t ) + + + + +

( ) ( ) ( ) (X 0 0 X 0 0 X 0 0 X 0 02 R t , t R t , t R t , t R t , t + + + + +

) .

Khai trin Taylorn cp hai hm ( )XR t,s ti (t0, t0) chng ta nhn c

( ) ( )h , 0 , 0 . Theo tiu chun Cauchy v s hi t, tn ti BNN bnh

phng kh tch ( )0X t sao cho( ) ( )X t X t+

( )0X t (MS - hi t).

Nh vy, nu QT cp hai l MS - kh vi th QT o hm li l qu trnh cp

hai. Tnh cht sau y cho php chng ta tnh hm trung bnh v hm t tngquan ca QT o hm cng nh hm t tng quan cho ca n vi qu trnh xut

pht (xem chng minh trong [6],tr 221).

nh l 5.14. Gi s ( ){ }X t , t I l QTNN MS- kh vi vi hm trung bnh

( )X t , hm t tng quan . Khi XR (t,s) ( )X t l hm kh vi; qu trnh

( ){ }X t l QT cp hai vi cc tnh cht sau:

( ) ( ) ( )X Xt E(X t ) t ; = =

( ) ( ) ( )( )2 X

XR t,s

R t,s E (X t X s )t s

= =

;

( ) ( ) ( )( )X

X XR t,s

R t,s E(X t X s )t

= =

. (5.4.5)

V d 5.16. Cho QT dng ( ){ }X t vi hm t tng quan .

Chng t rng y l qu trnh MS - kh vi. Tnh cc hm trung bnh, hm t

tng quan v tng quan cho.

( )23

XR 10e =


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Gii. l hm kh vi c p hai theo hai bin t, s, vy( ) ( )23 t s

XR t,s 10e =

( ){ }X t l qu trnh MS - kh vi.

Gi s ( )EX t = , ta c

( ) ( )( ) ( )X dt E X t 0dt = = = ;

( ) ( ) ( )( ) ( )

( ) ( ) ( )

2 23 t sX

2 23 t s

R t,s E X t X s 10et s

60e 6 t s t s 1 ;

= =

= + +

( ) ( ) ( )( ) ( ) ( ) ( )2 23 t s 3 t s

X XR t,s E X t X s 10e 60e t st

= = =

.

V ( )X t khng i cn hm tng quan XR ch ph thuc vo t s nn

( ){ }X t l QT dng. Hn na, hm tng quan cho X XR cng ch ph thuc

vo t s nn X v l dng ng thi.X

Tnh cht va nu xy ra khng chv d ny m cn QT dng bt k:

H qu. Nu hm tng quan ( )XR ca QT dng X l kh vi lin tc cp

hai th X l qu trnh MS - kh vi, QT o hm X l dng, hn na X v X ldng ng thi vi

2XXX X 2

dR ( ) d R ( )R ( ) ; R ( )

d d

X = =

. (5.4.6)

5.4.3. Tch phna)Tch phn trn on hu hn. Ging nh tch phn Riemann vi hm tt

nh, chng ta c th xy dng tch phn ca mt QTNN trn on [a;b] nh sau:

nh ngha. Gi s ( ) [ ]{ }X t , t a;b l QTNN cp hai. ng vi php phn

hoch on [a;b] bi cc im chia t0, t1, , tn vi a = t0 < t1 < < tn = b, ta lp

tng tch phn

( )n

n ii 1

S X s=

= i

trong s i l im ty trn on [ ]i 1 i i i it ; t ; t t 1 = . Nu khi sao chon

{ }imax ,i 1,...,n 0 = , tn ti gii hn (theo bnh phng trung bnh) ca dy

{ }nS n mt BNN Y xc nh, khng ph thuc vo cch phn hoch on [a;b]

cng nh cc im trung gian si, th qu trnh ( ){ }X t c gi l MS - kh tch


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trn on [a;b] cn Y c gi l tch phn (theo ngha bnh phng trung bnh

hay MS tch phn) ca QTNN ( ){ }X t trn [a;b]:

( )b

a

Y X t d= t .

nh l sau cho php chng ta d dng kim tra iu kin kh tch thng quahm t tng quan.

nh l 5.15. Gi s ( ) [ ]{ }X t , t a;b l QTNN vi hm t tng quan

( )XR t,s . Nu tn ti tch phn kp

( )b b

Xa a

R t,s dtds (5.4.7)

th ( ){ }X t l MS - kh tch trn [a;b].

Chng minh. Theo tiu chun Cauchy, chng ta bit rng tch phn ca

( ){ }X t tn ti khi

( ) ( ) ( )( )

2n m

i i i 1 i i i 1i 1 j 1

E X s t t X v u u = =

0

)

,

khi . (5.4.8)n,m 0

Khai trin ngoc vung chng ta c ba s hng. Bin i cc s hng ny,

v di vi s hng th hai ta nhn c:

( ) ( )( )( )n m i j i i 1 j j 1i 1 j 1

2E X s X v t t u u = =

= ( )( )( )n m

X i j i i 1 j j 1i 1 j 1

2 R s ,u t t u u = =

( ) (b b

Xa a

2 R t,s dt ds n,m .

Tng t vi mi s hng cn li, chng u dn n suy

ra (5.4.8) v nhn c pcm.

( )

b b

Xa a R t,s dtds

H qu.

1) Nu ( )XR t,s lin tc trn hnh vung [a; b] [a; b] th qu trnh ( ){ }X t l

MS - kh tch trn [a; b].

2) Mi qu trnh MS - lin tc l mt qu trnh MS - kh tch (trn on [a; b]).

Tch phn va xy dng (cn c gi l MS tch phn) c cc tnh cht ca

tch phn xc nh thng thng: tnh cht tuyn tnh, tnh cht xc nh khng m, h


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thc Sacl. N cng tho mn cng thc Newton Leibnitz, ta cng c th tnh c

cc c trng xc sut ca n, c th l:

i) Nu qu trnh ( ) [ ]{ }X t , t a;b l MS - kh vi lin tc (ngha l ( ){ }X t

l MS - lin tc) trn [a;b] th

( ) ( ) ( )b

a

X t dt X b X a = . (5.4.9)

ii) Gi s QT ( ) [ ]{ }X t , t a;b l MS - lin tc. t . Khi ( ) ( )t

a

Y t X s ds=

* ( ){ }Y t , t [a;b] l qu trnh MS - kh vi v ( ) ( )Y t X t = .

* Hm k vng v hm t tng quan ca { }Y(t) cho bi:

t t t

Y Xa a a(t) E[ X(s)ds] E[X(s)]ds (s)ds; = = =

t t

Ya a

R (t,s) E X(u)du X(v)dv

=

. (5.4.10)t s t s

Xa a a a

E[X(u)X(v)]dudv R (u, v)dudv= =

*Nu {X(t)} l QT Gauss th {Y(t)} cng l QT Gauss.

Nhn xt. (i) Nu( ){ }

X t l MS - kh tch th khng nht thit cc quo

( )X , l kh tch Riemann vi xc sut 1. Tuy nhin, nu gi thit thm rng cc

quo ( )X t, kh tch Riemann vi xc sut 1 (v d, vi QT lin tc theo qu

o hay QT bc nhy b chn) th c th thy rng, vi xc sut 1

( ) ( )b

a

Y X t, = dt

i

.

Ni cch khc, tch phn c th tnh thng qua tch phn ca mi

quo nu tch phn cng nh tch phn ca mi quo tn ti.

( )b

a

X t dt

Sdnh vy v tng tch phn ( )n

ii 1

X s=

ti cng l tng tch phn

ca hm mu tng ng, v t ch, t s hi t theo trung bnh c th trch ra mt

dy hi t hu chc chn.

(ii) i vi QTNN phc, cc kt qutrn vn cn ng vi lu rng, k

hiu tr tuyt i . by giphi hiu l m un ca s phc tng ng; v d,

2X(t) by giphi hiu l .*X(t)X (t )


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b)Tch phn suy rng

Gi s {X(t) } l QT cp hai, o c (*) vi hm t tng quant

XR (t,s) sao cho vX XP (t) R (t, t)=2

h(t,s) l nhng hm kh tch trn mi on

hu hn. Khi , theo nh 5.15, tn ti

[ ]b

a; ba

Y (t) h(t,s)X(s)d= s

cn v l tn ti tch phn hai lp

[a;b]

b b b b

Y Xa a a a

R (u, v)dudv h(t, u)h(t, v)R (u, v)dudv= .

By gitb

aab

h(t,s)X(s)ds l.i .m. h(t,s)X(s)ds+

+

= .

Gii hn ny tn ti khi v ch khi tch phn hai lp

Xh(t,u)R (u, v)h(t, v)dudv+ +

(5.4.11)

hi t.Nu iu kin (5.4.11) tho mn vi mi t th QT { }Y(t),t thu

c l QT cp hai vi hm t tng quan

Y XR (t,s) h(t, u)R (u, v)h(s, v)dudv+ +

= . (5.4.12)

Nu hm h(t,s) c tnh cht h(t,s) h(t s), t, s= (xy ra, v d khi h(t,s)l hm p ng xung ca h LTI bt bin theo thi gian (xem mc 6.2.2)), iukin nu trn trthnh i hi hi t ca tch phn hai lp

Xh(t u)R (u, v)h(t v)dudv+ +

, (5.4.13)

trong 2h(t) v kh tch trn on hu hn bt k. Hm t tng

quan ca QT Y trthnhX XP (t) R (t, t)=

Y XR (t,s) h(t u)R (u, v)h(s v)dudv+ +

= . (5.4.14)

---------------

(*) Vi nhng ngi quan tm n ng dng c th b qua khi nim o c ca QTNN.

QTNN {X(t), } c gi l o c nu nh xt I

( )( )[ I ]X : I S; ( , ) % l nh x o c, trong l

- i s Borel trn l - i s Bo

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