Ước lượng và kiểm Định trong thống kê nhiều chiều

8/2/2019 c Lng v Kim nh Trong Thng K Nhiu Chiu

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MC LC

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Mc lc

Li gii thiu

CHNG 1 : CC KHI NIM.................................................................................1

1. 1Vc tngu nhin nhiu chiu ...........................................................................11. 1. 1Vc tngu nhin nhiu chiu .................................................................... 1

1. 1. 1. 1nh ngha ..........................................................................................11. 1. 1. 2Hm phn phi xc sut ..................................................................... 11. 1. 1. 3Phn phi xc sut ..............................................................................1

1. 1. 2Vector trung bnh vector k vng.............................................................21. 2Ma trn hip phng sai..................................................................................... 5

1. 2. 1Ma trn hip phng sai mu ......................................................................51. 2. 2Ma trn hip phng sai tng th ................................................................61. 2. 3Ma trn tng quan ..................................................................................... 71. 2. 4Vector trung bnh - ma trn hip phng sai cho nhiu nhm con ca cc

bin ............................................................................................................101. 2. 4. 1Hai nhm ..........................................................................................101. 2. 4. 2Ba hoc nhiu cc nhm hn............................................................14

1. 3 S kt hp tuyn tnh gia cc bin .................................................................151. 3. 1Tnh cht ca mu...................................................................................... 151. 3. 2Tnh cht ca tng th ...............................................................................22

1. 4Hm mt ca mt i lng ngu nhin nhiu chiu.................................. 241. 4. 1nh ngha .................................................................................................241.4.2 Tnh cht ................................................................................................24

1. 5 Phn phi i lng ngu nhin nhiu chiu ................................................... 241. 5. 1nh ngha .................................................................................................241. 5. 2Tnh cht .................................................................................................... 25
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1. 6 Phn phi chun nhiu chiu............................................................................261. 6. 1Hm mt phn phi chun mt bin.....................................................271. 6. 2Hm mt ca phn phi chun nhiu chiu..........................................281. 6. 3Tng qut ha phng sai tng th ...........................................................281. 6. 4Tnh cht phn phi chun ca bin ngu nhin nhiu chiu....................301. 6. 5c lng trong phn b chun nhiu chiu ............................................36

1. 6. 5. 1c lng hp l ti a (MLE) .......................................................361. 6. 5. 2Phn phi ca y v S ......................................................................38

CHNG 2 : C LNG KHNG CHCH TUYN TNH............................402. 1M hnh thng k tuyn tnh tng qut hng y ........................................402. 2c lng khng chch cho m hnh thng k tuyn tnh tng qut hng y

...................................................................................................................... 422. 2. 1nh l 2.1 (Gauss Markov) ....................................................................422. 2. 2B 2.1....................................................................................................432. 2. 3H qu 2.1.................................................................................................. 44

2. 3M hnh thng k tuyn tnh vi hng khng y ......................................462. 4c lng khng chch cho m hnh thng k tuyn tnh hng khng y ..

..........................................................................................................................462. 4. 1nh l 2.2 ..................................................................................................462. 4. 2B 2.2................................................................................................... 472. 4. 3nh l 2.3 ( Gauss Markov ) .................................................................. 482. 4. 4c lng bnh phng b nht mrng ................................................. 492. 4. 5T hp tuyn tnh tt nht ca thng k th t .........................................52

2. 5ng dng trong m hnh c lng tham s hi quy nhiu chiu ..................592. 5. 1Hm hi quy tng th (PRF)......................................................................592. 5. 2Dng ma trn ca hm hi quy.................................................................. 60

2. 5. 2. 1Hm hi quy tng th PRF ...............................................................602. 5. 2. 2Hm hi quy mu SRF .....................................................................60

2. 5. 3c lng bnh phng b nht thng thng (OLS)..............................612. 5. 3. 1

Gii thiu..........................................................................................61
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2. 5. 3. 2iu kin cn.................................................................................... 622. 5. 3. 3Nghim h phng trnh chun ........................................................672. 5. 3. 4iu kin .....................................................................................69

2. 6Xy dng thut ton hi quy cho lp trnh trn my tnh.................................722. 6. 1Bi ton xy dng phng trnh siu phng hi qui. ................................722. 6. 2Bi ton tnh h s tng quan ring ........................................................722. 6. 3Bi ton hi quy tng bc .......................................................................732. 6. 4M t phng php tnh ton.....................................................................74

2. 6. 4. 1Cc k hiu s dng .........................................................................742. 6. 4. 2Phng php tnh ton...................................................................... 74

2. 6. 5Xy dng hm tnh nh thc ca ma trn (sau s dng hm ny tnhnh thc ca ma trn covarian L_Da)......................................................75

2. 6. 5. 1Phn 1 ...............................................................................................752. 6. 5. 2Phn 2 ...............................................................................................762. 6. 5. 3Xy dng hm tnh nh thc ca ma trn khi bi 1 hng 1 ct .......

..........................................................................................................772. 6. 6Bi ton v tng quan ring.....................................................................782. 6. 7Bi ton v hi quy tng bc................................................................... 782. 6. 8Lu thut ton ca ba bi ton nu trn................................................79

CHNG 3 : KIM NH GI THIT TRN VECTK VNG...................823. 1Mu thun gia kim nh nhiu chiu v mt chiu ......................................823. 2Kim nh trn vi bit ......................................................................83

3. 2. 1Nhc li kim nh n bin gi thit0

:H0

= vi bit...........833. 2. 2Kim nh nhiu chiu cho gi thit :

0 0: H = vi bit ........... 843. 3Kim nh gi thit trn khi cha bit ....................................................89

3. 3. 1Nhc li kim nh n bin cho gi thit 0 :H 0 = khi cha bit ......................................................................................................................89

3. 3. 2 ca Hotelling kim nh gi thit2T 0 0: H = vi cha bit ....... 903. 4 So snh hai vetor trung bnh ............................................................................95
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3. 4. 1Nhc li hai mu mt chiu vi kim nh t Test ................................ 953. 4. 2Kim nh vi hai mu nhiu chiu ......................................962 TestT

3. 5Kim nh trn tng bin ring l vi iu kin bc b gi thit 0H bi........................................................................................................1002T tes t

3. 6Thao tc tnh ton ca - Thu c t hi quy nhiu chiu...............1062T 2T3. 7Kim nh cc cp quan st ............................................................................108

3. 7. 1Trng hp mt chiu .............................................................................1083. 7. 2Trng hp nhiu chiu ..........................................................................110

3. 8Kim nh thm thng tin...............................................................................1133. 9 Phn tch hnh th ...........................................................................................118

3. 9. 1Phn tch hnh th mt mu .....................................................................1183. 9. 2Phn tch hnh th hai mu ......................................................................121

CHNG 4: KIM NH GI THIT TRN MA TRN HIP PHNG SAI

...................................................................................................................................... 1304. 1Gii thiu........................................................................................................1304. 2Kim nh m hnh d kin cho ...............................................................130

4. 2. 1Kim nh gi thit H0: 0 = .............................................................1304. 2. 2Kim nh tnh cu ..................................................................................1324. 2. 3Kim nh ( )20 1:H I+ J = ...............................................135

4. 3 So snh cc kim nh ma trn phng sai ....................................................1384. 3. 1Kim inh phng sai bng nhau ............................................................1394. 3. 2Kim nh bng nhau cc ma trn hip phng sai nhiu chiu.............140

4. 4Kim nh tnh c lp ................................................................................... 1454. 4. 1c lp ca hai vector con ......................................................................1454. 4. 2Sc lp ca nhiu vectors con ............................................................1474. 4. 3Kim nh c lp cho tt c cc bin .....................................................151

Ti liu tham kho

Ph lc
http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http://0.0.0.0/http: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Li Gii Thiu

c lng v kim nh l cc bi ton c ngha ln trong thng k. T mungu nhin kho st c ta c tha ra nhng nhn nh st vi tng th c c

nhng don tng i chnh xc v mt hin tng ca x hi hay cc bin ng

trong tng lai nc ta hin nay, phn tch thng k nhiu chiu cha c quan

tm mt cch ng k trong cc trng i hc v cao ng. c lng v kim nh

li l bi ton c ngha quan trng trong vic phn tch hi quy v phng sai nhiu

chiu. Bt ngun t nhng suy nghtrn, vi s hng dn ca Thy v s nghin cu

ca bn thn, tc gi xin c gii thiu lun vn thc sca mnh vi ti :c Lng v Kim nh Trong Thng K Nhiu Chiu.

Ni dung ch yu ca lun vn ny nhm gii thiu :

Hm ( )g F no i vi n c c lng tuyn tnh khng chch. Tm trong lptt c cc c lng tuyn tnh khng chch ca ( )g F c lng c phng sai

b u nht. T ng dng trong m hnh c lng tham s hi quy v xy

dng cc thut ton cho bi ton tm siu phng hi quy. Cc kim nh gi thit ch yu l trn vector k vng v ma trn hip phng sai.

Phn tch lm ni r u im ca vic s dng kim nh nhiu bin trong

thng k nhiu chiu thay v s dng kim nh mt bin thng thng.

Da vo ni dung cbn trn lun vn gm bn chng vi b cc nh sau

Chng 1 : CC KHI NIM

Chng ny nhm gii thiu slc v cc khi nim, thuc tnh ca bin ngu

nhin nhiu chiu. Gii thiu r v cc tnh cht ca phn phi chun nhiu chiu. yl phn phi quan trng trong bi ton c lng v kim nh.

Chng 2 : C LNG KHNG CHCH TUYN TNH

Gii thiu cc nh l (Gauss Markov) v b dng gii quyt bi ton

c lng cho m hnh thng k tuyn tnh vi hng y v hng khng y .

T l thuyt c c ta xy dng m hnh ng dng c lng tham s hi

quy bng phng php bnh phng b nht. Sau l ng dng xy dng thut


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ton tm phng trnh siu phng hi quy. Cui chng l s thut ton tm phung

trnh siu phng hi quy v hi quy tng bc.

Chng 3 : KIM NH GI THIT TRN VECTK VNGTa tm thy t chng ny cc kim nh gi thit

0 0: H = cho trng hp ma

trn hi p phng sai bit hoc cha bit. Bi ton kim nh gi thit

vi i thit

1: oH = 2 21 1: H tc l so snh hai vector trung bnh ca hai

mu ngu nhin nhiu chiu cng c trnh by chng ny. Hoc l bi ton kim

nh cc c p quan st t ghp ni hai mu nhiu chiu cng c tho lun kh k

trong chng nyu im ca chng ny l cc phn u c xy dng t m hnh n chiu v

pht trin thnh m hnh a chiu, gip ngi c c th so snh u im ca kim

nh nhiu chiu so vi dng kim nh mt bin cho bi ton kim nh nhiu chiu.

ng thi cc v dc trnh by c th vi kt qu r rng lm sng t hn phn l

thuyt c trnh by.

Chng 4 : KIM NH GI THIT TRN MA TRN HIP PHNG SAI

Trong chng ny , bao gm ba loi hnh cbn ca kim nh gi thit : (1) mhnh d kin ca ma trn hip phng sai, (2) hai hoc nhiu ma trn phng sai bng

nhau, v (3) chc chn thnh phn ca ma trn phng sai l 0, ko theo tnh c lp

tng ng ca cc bin ngu nhin (chun nhiu chiu). Trong hu ht trng hp,

chng ta s dng xp x t s hp l. Kt qu thng k kim nh thng lin quan n

t s xc nh ca cc ma trn hip phng sai mu vi gi thit khng v vi i

thit khc khng.

u im ca chng ny l bn cnh phn trnh by l thuyt u c km theo ccv d rt c th vi cc kt qu rt r rng v c lin thng vi cc kt qu ca chng

1 v 3. iu ny gip chng ta c ci nhn r lin h gia cc bi ton kim nh vi

phng sai v cu trc ca ma trn hip phng sai.

Nhm gip lun vn cht ch hn v l lun , cui lun vn l ph lc cc bng tra

ca phn phi nh : phn phi chun, phn phi chi bnh phng, phn phi Student,

phn phi Fisher


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1 Chng 1

CHNG 1 : CC KHI NIM

1. 1

Vc tngu nhin nhiu chiu :1. 1. 1 Vc tngu nhin nhiu chiu

n

1 2 nX (X , X ,...,X ) : ( , ,P)= F R

X l hm o c, tc l nghch nh ca mi hnh hp u l tp thuc .F

1. 1. 1. 1 nh nghaTh t trong : vinR n1 2 n 1 2 na (a ,a ,...a ), b (b ,b ,...b )= = R , ta ni a b

nu i ii 1,2,..., n : a b =

Hnh h p trong :nR

1 2 n[a, b] {x (x , x ,..., x ) : a x b}= =

1 2 n k k k (a, b] {x (x , x ,..., x ) : a x b k 1,2,...,n}= = < =

1. 1. 1. 2 Hm phn phi xc sut( ) ( ){ } nXF x : P : X x x= <

- L hm n iu khng gim : X Xx y F (x) F (y ) - Lin tc phi, c gii hn tri :

k 0 X k X 0 k 0 X k X 0x x F (x ) F (x ) ; x x F (x ) c F (x )

- Tin ti 0 khi vi mt ch s j no jx - Tin ti 1 khi x +

1. 1. 1. 3 Phn phi xc sutj

j n 1 1 n n

n

n m

m 1 m m 1 m

X X 1 1 n n

m {0,1}

P ((a, b]) : ( 1) F (b a ,..., b a ) 0 a, b : a b

=

Tnh ngha trn ta c th ni rng ra mt o XS trn nR

- nX XP ( ) 0, F ( ) 1 = =R-

n n

X XP ( A) 1 P (A) A ( ) = R RB

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2 Chng 1

- nX X XP (A B) P (A) P (B) A,B ( ),A B = + = RB

- n

X k X k 1 2 i k k 1 k 1

P ( A ) P (A ) A ,A ,... ( ),A A i k

= == = R

B

* H qu :

Trong nghin cu cc i lng ngu nhin nhiu chiu, c thdng cc o xc sut trn (phn phi XS ca LNNNC) thay cho o xc sut P trn

.

nR

1. 1. 2 Vector trung bnh vector k vng :Cho y l biu din ca mt vector ngu nhin p bin o c trn n v mu.

Nu n vectors ring lc quan st trong mu : , th :1 2y , y ,..., yn

1

2yi

p

y

y

y

=

Vector trung bnh mu y c th c thc tm tng t nh n vectorc

quan st hoc c tnh bi trung bnh ca mi mt p bin ring l :

1

2

1

1y y

n

i

i

p

y

y

n

y

=

= =

(1.1)

y c th nh : 22 1n

iiy y

== n . Do 1y l trung bnh ca n quan st trn

bin u tin, 2y l trung bnh ca bin th hai v c nh th.

Tt c n vectorc quan st c thc chuyn vn vector

dng v c lit k trong ma trn Y nh sau

1 2y , y ,..., yn

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3 Chng 1

nv

dng

Cc bin

nv

dng

Cc bin

y n thng l ln hn p. D liu c sp xp theo dng bng bng vic truy

nhp vo cc vector quan st theo hng ch khng phi l theo ct. Ch rng ch

s di u tin i tng ng vi cc n v dng v ch s th hai j chn cc

bin. Quy c ny sc mc nh cho bt k cc trnh by tng t .

C th b sung mt cch th 2 tnh y , ta c th c c y tY . Ta tnh

tng n d liu vo t mi ct ca Y v chia cho n. iu ny c thc biu din

t hng dn sau :

1 2

'j A , ,...,i i ii i i

a a a

= p ,

1

2Aj

jj

jj

n jj

a

a

a

=

Vy nn ta c :

1''y j Yn= (1.3)

y ' l vector dng ca :

1

1

1

j

=

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4 Chng 1

Mt minh ha th hai ca 'Y l :

( )

12

22

2

1

2

1 1 1, , ...,n

i

i

n

y

yy

y

=

=

Ta c th bin i'

y thu c :

1 'y Y jn

= (1.4)

By gita cp n tng th. Trung bnh ca y trn tt c cc gi tr c th

c trong tng thc gi l vector k vng l thuythoc lgi trk vngca y.

N c nh ngha nh l vector ca gi tr k vng ca mi mt bin.

( )

( )

( )

( )

11 1

22 2y

p pp

E yy

E yyE E

y E y

= = =

= (1.5)

y j l k vng l thuyt ca bin th j.

iu ny cho thy rng gi tr k vng ca mi mt trongjy y l j ,

chnh l ( ) jjE y = , do gi tr k vng ca y (trn tt c cc gi tr ca mu) l

( )( )( )

( )

11 1

222y

ppp

E yy

E yyE E

y E y

= = =

= (1.6)

Thnh ra y l mt c lng khng chch ca . Ta nhn mnh li, y s

khng bao gibng ti

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5 Chng 1

1. 2 Ma trn hip phng sai :1. 2. 1 Ma trn hip phng sai mu :Ma trn hip phng sai mu ( )S jks= l ma trn ca cc phng sai v hip

phng sai mu vi p bin

( )

11 12 1

21 22 2

1 2

S

p

p

jk

p p pp

s s s

s s ss

s s s

= =

(1.7)

Trong S, phng sai mu ca p bin nm trn ng cho ma trn. Tt c cc

hi p phng sai xut hin ngoi ng cho chnh ca ma trn. Vij dng (ct)

bao gm cc hip phng sai ca yj vi p - 1 bin khc.

Ta a ra hai cch tip cn thu c S. u tin l cc php tnh ring l

ca jks . Phng sai mu ca phng sai ca bin j,2

jj js s= c tnh bi cng

thc :

( )2

12

1

n

ii

jj

y ys s

n

=

= =

(1.8)

Hoc l :

22

2 1

1

n

iiy ny

sn

=

=

(1.9)

Nu dng ctj ca Y th :

( )2

2

1

1

1

n

jj j ij j

i

s s yn =

= =

y (1.10)

221

1ij j

i

y nyn

=

(1.11)

y jy l trung bnh ca j cc bin.

Hip phng sai caj x kcc bin, jks c tnh bi :

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6 Chng 1

( )( )11

n

i ii

xy

x y ys

n

=

=

(1.12)

Hoc

1

1

n

i iixy

y nx ys

n=

=

(1.13)

Nu dngj v kct ca Y th :

( )(1

1

1

n

jk ij ik )j ki

s y y y yn =

=

(1.14)

1

1ij ik j k

i

y y ny yn

=

(1.15)

1. 2. 2 Ma trn hip phng sai tng thNu yl mt vector ngu nhin c ly t bt k mt gi tr no ca tng th

nhiu chiu , ma trn hip phng sai ca tng thc nh ngha l

(1.16)( )

11 12 1

21 22 2

1 2

cov y

p

p

p p pp

= =

Ma trn hip phng sai tng thtrn cng c thc tm nh sau :

( )( )'

y y E = (1.17)

Ma trn ( p x p) l ma trn ngu nhin. Gi trc k vng ca mt ma

trn ngu nhin c xc nh nh l mt ma trn nhng gi trc k vng ca

s tng ng cc phn t. Ta s thy c s xy dng ma trn phng sai hip

phng sai mu ca p chiu nh sau :

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7 Chng 1

( )( ) ( )

( ) ( )( ) ( )( )

( )( ) ( ) ( )( )

( )( ) ( )( ) ( )

( ) ( )( ) ( )( )

1 1

2 2

1 1 2 2

2

1 1 1 1 2 2 1 1

2

2 2 1 1 2 2 1 1

2

1 1 2 2

2

1 1 1 1 2 2 1 1

'y y , , ...,

p p

p p

p p

p p

p p p p p p

p p

y

yE E y y y

y

y y y y y

y y y y yE

y y y y y

E y E y y E y y

= =

=

=

( )( ) ( ) ( )( )

( )( ) ( )( ) ( )

2

2 2 1 1 2 2 1 1

2

1 1 2 2

11 12 1

21 22 2

1 2

p p

p p p p p p

p

p

p p pp

E y y E y E y y

E y y E y y E y

=

V ( )jk jk E s = = vi mij v k nn ma trn hip phng sai mu S l mt

c lng khng chch ca :

( )SE = (1.18)

1. 2. 3 Ma trn tng quan :Tng quan mu gia (j x k) cc bin c nh ngha bi :

jk jk

jk

j kjj kk

s sr

s ss s= = (1.19)

Ma trn tng quan mu l tng t ma trn hip phng sai vi s tng

quan trong khng gian ca cc phng sai.

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8 Chng 1

(1.20)( )

12 1

21 2

1 2

1

1

1

R

p

p

jk

p p

r

= =

V ddng th 2, bao gm tng quan ca vi mi thnh phn ca y

(bao gm c tng quan ca vi chnh n, l 1). Dnhin ma trn tng quan l

ma trn i xng, v . Ma trn tng quan c th thu c t ma trn hip

phng sai, v ngc li,

2y

2y

jk kjr r=

( )( )

11 22

1 2

D diag , ,...,

diag , ,...,

s pp

p

s s s

s s s

=

=

0 0

0 0

0 0

s

s

s

=

(1.21)

Ta c :

1 1R D SDs s = (1.22)

S D RDs s= (1.23)

Tng t ma trn tng quan tng th v c nh ngha l :

(1.24)( )

12 1

21 2

1 2

11

1

P

p

p

jk

p p

= =

y :

jk

jk

j k

=

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15/165

9 Chng 1

V d 1.1 : Cho bng d liu sau, vi ba bin c o (mu th l 100g )ti mi

a im khc nhau y1=calcium trong t, y2=lng calcium c chuyn i,

y3=calcium c trong cy ci xanh.

a im

Bng 1.1 : Lng calcium trongt v trong cy ci xanh

tnh c ma trn phng sai mu cho ct d liu calcium ca bng. Ta

tnh tng bnh phng ca mi mt ct v tng cc kt qu mi cp ca ct, ta minh

ha php tnh ca 13s :

T cc cng thc trn ta d dng tnh c :

1 28 1.y = v 3 3 089.y =

T cng thc (1.14) v (1.15), ta c :

Tip tc vi s tng t ta c c ma trn hip phng sai l :

c c ma trn tng quan vi d liu trn ta c th tnh ton ring l

bng cch dng cng thc ( 1.19) . Hoc c th dng trc tip t th thut ma trn :

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16/165

10 Chng 1

1 1R D SDs s =

Ma trn ng cho chnh Ds c th tm bng cch ly cn bc hai trn cc

ng cho chnh ca ma trn S :

T ta c :

1. 2. 4 Vector trung bnh - ma trn hip phng sai cho nhiu nhm conca cc bin :

1. 2. 4. 1 Hai nhm :Nhiu khi mt kho st no quan tm n hai dng khc nhau ca bin, c

hai cng c o trn mt n v mu. Mt s hnh vi c quan st trong lp hcdnh cho sinh vin, v trong cng mt khong thi gian nht nh (cc n v c

bn thc nghim) mt s hnh vi ca gio vin cng c quan st. Kho st mun

nghin cu mi lin h cc bin ca hc sinh v cc bin ca gio vin.

Ta s biu din hai nhm vector bi y v x vi p bin trong y v q bin trong

x. V vy, mi mt vector quan st trong mu l c phn chia l :

(1.25)

1

1

1 2y

, , , ...,x

i

ipi

i i

y

yi n

x

xp

= =

S S

S

S S

yy yx

xy xx

=

(1.26)

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17/165

11 Chng 1

y l (p x p), S l ( p x q ) ,Syy yx Sxy l ( q x p) v Sxx l (q x q). Cng cn

ch rng v tnh cht i xng ca S nn

'S Sxy y= x

(1.27)

Vy nn, ta c th vit :

(1.28)'

S SS

S S

yy yx

yx xx

=

minh ha ta cho p = 2 v q = 3, ta c :

Cc mu trong mi v SS ,S ,Syy yx xy xx c biu din r rng trong minh

ha ny. V d dng u ca S l hip phng sai ca vi miyx 1y 1 2 3, ,x x x . Dng

th hai l biu din hip phng sai ca vi ba bin cax. Mt khc ta cng c

dng u ca S

2y

xy l cc hip phng sai ca 1vi v v c th Nh

vy :

1y 2y

'S Sxy y= x

Tng t, i vi tng th kt qu ca vic phn chia cc vector ngu nhin l :

( )

( )

y y

x x

y

x

EE

E

= =

(1.29)

ycov

x

yy yx

xy xx

= =

(1.30)

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18/165

12 Chng 1

y 'xy = yx . Ma trn con yy l ma trn hi p phng sai ( p x p) cha

phng sai ca trn ng cho chnh v hip phng sai ca mi

v nm ngoi ng cho. Tng t nh vy,1 2

, , ...,p

y y yj

y

ky xx l ma trn hip phng sai

( q x q ) ca 1 2, , ..., kx x x . Ma trn yx l ( p x q ) v bao gm hip phng sai ca

mi vi mijy kx . Ma trn hi p phng sai yx th cng c biu din bi

( )cov y,x tc l :

( )cov y,x yx= (1.31)

Cn ch s khc nhau trong ngha gia trong cng thc

(1.30) v

ycov

x

=

( )cov y,x yx= trong cng thc (1.31).y

covx

bao gm mt vector

n cha p+q bin, v ( )cov y,x th bao gm hai vector.

Nu x v y l c lp th 0yx = . iu ny c ngha l mi mt bin

u khng tng quan vi mi

jy

k v th nn 0j ky x = cho

1 2 1 2, , ..., ; , , ...,j p k q= = .

V d 1.2: Reaven v Miller (1979; Andrews v Herzberg 1985,

pp. 215-219) o lng nm bin so snh gia ngi bnh thng v ca bnh nhn

i tho ng . Trong Bng 1.2 ta ch xt mt phn d liu cho ngi bnh

thng. Ba bin chnh c quan tm l :

1x = lng ng khng c dung np

2x = isulin dng cn bng lng ng c ung

3x = khng isulin

Thm hai b sung cc bin nh cng c quan tm l :

1y = quan h trng lng

2y =Lu lng ng huyt

__________________________________________________________________


19/165

13 Chng 1

S ngibnh

Bng 1.2 : Quan h gia nng Insulin vi cn nng v lngng trong mu

__________________________________________________________________


20/165

14 Chng 1

Vector trung bnh c phn chia theo cng thc l :

Ma trn hip phng sai c phn chia nh trong phn tch trn s l :

Lu l ma trn vSyy Sxx l cc ma trn vung, v Sxy l ma trn chuyn

v ca Syx

1. 2. 4. 2 Ba hoc nhiu cc nhm hn :Trong mt s tnh hung ba hay nhiu hn cc nhm rt c quan tm. Nu

vectory quan st c phn chia nh sau :

1

2

y

yy

yk

=

,

y c1y 1p cc bin, c2y 2p ,,v cyk kp cc bin vi :

1 2 ... kp p p p= + + + Vector trung bnh mu v ma trn hip phng sai c cho bi :

1

2

y

yy

yk

=

(1.32)

__________________________________________________________________


21/165

15 Chng 1

(1.33)

11 12 1

21 22 2

1 2

S S S

S S S

S

S S S

p

p

p p pp

==

V d nh ma trn con ( )2 2S k kp x p bao gm phng sai v hip phng

sai ca cc bin trong vi cc bin trong .2y yk

Tng ng vi tng th ta c kt qu nh sau :

1

2

k

=

, (1.34)

(1.35)

11 12 1

21 22 2

1 2

k

k

k k kk

=

1. 3 Skt hp tuyn tnh gia cc bin1. 3. 1 Tnh cht ca mu :

Ta thng quan tm n s kt hp tuyn tnh gia cc bin .Trong

phn ny chng ta s kho st trung bnh, phng sai v hip phng sai ca s kt

hp tuyn tnh.

1 2, , ..., py y y

Cho l cc l h s v c xem nh l s kt hp tuyn tnh ca

cc yu t ca vector y,

1 2, , ..., pa a a

1 1 2 2'... a yp pz a y a y a y= + + + = (1.36)

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22/165

16 Chng 1

y . Nu cng mt h s ca vectorac p dng cho

mi trong mu , ta c :

( 1 2'a , ,..., pa a a= )

yi

1 1 2 2 1 2'z ... a y , , ,....,i i i p ip ia y a y a y i p= + + + = = (1.37)

Trung bnh mu ca z c thc tm thy bi trung bnh cng ca n gi tr

, ,, hoc l nh mt kt hp tuyn tnh ca1 1'a yz = 2 2

'a yz = 'a ynz = n y cc

vector trung bnh mu ca .1 2y , y ,..., yn

1

1 'a yn

ii

z zn =

= =

(1.38)

Kt qu trn y l tng t nh kt qutrng hp n bin z a y= y

1, , ...,i iz ay i= = n

nTng t nh vy, phng sai mu ca c thc tm

nh l phng sai mu ca hoc trc tip ta v S , y S l ma trn

hip phng sai ca

1'a , ,...,i iz y i= =

1 2, , ..., nz z z

1 2

y , y ,..., yn

( )

2

12

1

'a Sa

n

ii

z

z zs

n

=

=

=

2

(1.39)

Ch rng l m hnh nhiu bin t kt qun bin2 'a Sazs =2 2

zs a s=

y v1' , , ...,i iz a y i= = n2s l phng sai ca 1 2, , ..., ny y y

V phng sai l lun khng m, ta c v thnh ra cho mi a.

Do t nht S l na xc nh dng. Nu cc bin l lin tc v khng quan h

tuyn tnh, v nu n-1> p (do S hng y ) th S c xc nh dng ( vi xc

xut l 1 )

2 0zs > 0'a Sa >

Nu ta xc nh mt kt hp tuyn tnh khc

y l vector h s ( hng s) khc . V th hip phng sai

ca z v w c cho bi :

1 1 2 2

'b y ... p pw b y b y b= = + + + y

)( 1 2'b , ,..., pb b b= 'a

__________________________________________________________________


23/165

17 Chng 1

( )( )11

'a Sb

n

i ii

zw

z z w ws

n

=

=

= (1.40)

Tng quan mu gia w v z sn sng nhn c l :

( )( )2 2'

' '

a Sb

a Sa b Sb

zwzw

z w

sr

s s= = (1.41)

Gita s biu din lun hai vector h s ( hng s ) a v b l v to

iu kin thun li v sau khi mrng nhiu hn hai vectors nh vy. Cho :

1a 2a

1

2

,

,aA a

=

v nh ngha:

1 1

22

'

'

a yz

a y

z

z

= =

V sau ta c th c nhn ty t biu din ca biu thc :

1

2

'

,

az y Aya

= =

Nu ta c lng hai chiu t cho mi p bin trong mu. Chng ta

nhn c v gi tr trung bnh ca z trn mu c thc

tm thy t

zi yi

1 2z Ay , , ,...,i i i= = n

y :

1 1

2 2

'

' yz yz az a

= =

(1.42)

1

2

'

'z y Ay

a

a

== =

(1.43)

Ta c th dng (1.39) v (1.40) xy dng ma trn hip phng sai mu cho z :

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24/165

18 Chng 1

(1.44)1 1 2

2 1 2

2

1 1 1 2

2

2 1 2 2

' '

' '

a a a aS

a a a a

z z z

z

z z z

s s S S

s s S S

= =

Bi v :

1 1 1 2

2 1 2 2

' ''

' '

a a a aASA

a a a a

S S

S S

=

Yu t ny a n :

(1.45)( )1 1 22

''

'

aS S a , a A

az

= =

SA

Kt qu hai bin trn c th sn sng m rng nhiu hn hai kt hp

tuyn tnh. Nu chng ta c k php bin i tuyn tnh, chng ta c th biu din

nh sau :

1 11 1 12 2 1 1

2 21 1 22 2 2 2

1 1 2 2

'

'

'

... a y

... a y

... a y

p p

p p

k k k kp p k

z a y a y a y

z a y a y a y

z a y a y a y

= + + + =

= + + + =

= + + + =

Hoc bng k hiu ma trn l :

1 11

2 2 2

' '

' '

' '

a y a

a y az y Ay

a y ak k k

z

z

z

= = = =

y z l ( k x 1 ) chiu, A l ( k x p ) chiu, v y l ( p x 1 ) chiu ( chng ta quy

c l k ). Nu l nh tr cho tt c ccp z Ayi = i n1y , ,...,i i = iu ny cho

bi (1.38) Vectror trung bnh mu ca z l :

__________________________________________________________________


25/165

19 Chng 1

1 1

22

' '

''

''

a y a

aa y

z y Ay

aa y kk

= = =

(1.46)

Mrng t biu din (1.44) ta c ma trn hip phng sai mu ca trthnh :z

(1.47)

1 1 1 2 1

2 1 2 2 2

1 1

' ' '

' ' '

' ' '

a Sa a Sa a Sa

a Sa a Sa a SaS

a Sa a Sa a Sa

k

kz

k k k

=

k

( )

( )

( )

1 1 2

2 1 2

1 1

'

'

'

a Sa , Sa , Sa

a Sa , Sa , Sa

a Sa , Sa , Sa

k

k

k k

=

( )

1

21 2

'

'

'

a

aSa , Sa , ,Sa

a

k

k

=

(1.48)( )

1

21 2

'

''

'

aa

S a , a , ,a ASA

a

k

k

= =

Ch rng t (1.47) v (1.48) ta c :

(1.49)( )1

' 'tr ASA a Sak

i i

i=

=

bin i tuyn tnh yu hn l :

1 2z Ay b, , , ...,i i i n= + = (1.50)

Vector trung bnh v hip phng sai mu c cho bi :

z Ay b= + (1.51)

(1.52)'S ASAz =

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26/165

20 Chng 1

V d 1.3: Timn (1975, p. 233; 1980, p. 47) bo co v kt qu ca mt th

nghim y ch l tm kim tr li p t nm v tr trong mt ch. Cc

bin l thi gian p ng cho jp t 1 2 5, , ...,jy j = . D liu c cho trong

Bng 1.3 sau:

S ch

Bn 1.3: Thi ian n cho nm tca m t ch

Nhng bin ny c quy c ( o trn cng mt n v tng t nh trung

bnh v phng sai ) v cc nh nghin cu c th mun kim tra mt s kt hp

tuyn tnh n gin. Xem xt cc kt hp tuyn tnh sau y l mc ch minh

ha :

Nu z l c tnh cho mi ca 11 quan st, chng ta c c

vi trung bnh1 2 11288 155 146, , ....,z z z= = = 197 0.z = v phng sai

. Ta cng c cng mt kt qu nh vy nu dng cng thc (1.38) v

(1.39) th vector trung bnh mu v ma trn hip phng sai cho d liu trn l :

2 2084 0.zs =

__________________________________________________________________


27/165

21 Chng 1

K , v (1.38) nn ta c :

v t (1.39) :

By gita s nu ra mt t hp tuyn tnh th hai :

Trung bnh mu v phng sai ca w l :44 45

'w b y .= = v 2 605 67'b Sb .ws = =

Ma trn hip phng sai mu ca z v w c tnh bi (1.40) l :

40 2'a Sb .z ws = =

Tip tc dng cng thc (1.41)ta tm c tng quan ca z v w l :

By gichng ta xt ba hm tuyn tnh :

Ta cng c th biu diu di dng ma trn nh sau :

__________________________________________________________________


28/165

22 Chng 1

hoc :

Vector trung bnh mu cho bi (1.46) l :

Ma trn hip phng sai mu c cho bi : l :'S ASAz =

Ma trn hip phng sai c th bin i n mt ma trn tng quan bi

cng thc (1.22):

Sz

y :

L nhn c t cn bc hai cc phn t trn ng cho chnh ca S .z

1. 3. 2 Tnh cht ca tng th :Cc kt qu v s kt h p tuyn tnh trn c bn sao trong tng th. Cho

y a l mt vector h s (hng s ). Trung bnh mu caz s l:'a yz =

__________________________________________________________________


29/165

23 Chng 1

( ) ( ) ( )' 'a y a y a'E z E E = = = (1.53)

V phng sai ca tng th l :

( )2 ' 'var a y a az = = (1.54)

Cho y b l vector h s (hng s) khc a. Hip phng sai tng

th ca v l :

'b yw =

'a yz = 'b yw =

( ) 'cov , a bz wz w = = (1.55)

T cng thc tng quan cax vy:

( )( )( )

( ) ( )22

,x yxy

xy

x yx y

E x ycorr x y

E x E y

= = =

Ta c tng quan tng th caz v w l :

( )

( )( )

'' '

' '

a ba y , b y

a a b b

zwzw

z w

corr

= = =

(1.56)

Nu Ay l biu din cho nhiu kt hp tuyn tnh, vector trung bnh mu v

ma trn hip phng sai cho bi :

( ) ( )Ay A y AE E= = (1.57)

( )'

cov Ay A A= (1.58)

Php bin i tuyn tnh tng qut hn z Ay b= + c vector trung bnh mu

v ma trn hip phng sai l :

( ) ( )Ay b A y b A bE E+ = + = + (1.59)

( ) 'cov Ay b A A+ = (1.60)

__________________________________________________________________


30/165

24 Chng 1

1. 4 Hm mt ca mt i lng ngu nhin nhiu chiu1. 4. 1 nh nghaTa ni rng i lng ngu nhin nhiu chiu X l lin tc nu tn ti mt

hm s sao chon nf : ( , ) : [0, )+= = R R

n 2 1x x xx

n

X 1 2 n 1 2 n 1F (x) f ( t )d t ... f (t , t ,..., t )dt dt ...dt x (x , x ,..., x )

= = = R2 n

Lc hm f c gi l hm mt (XS) ca vc tNN X .

1. 4. 2 Tnh chti) n 2 1

n 2 1

b b bb

n

X 1 2 n 1 2

a a a a

P ((a, b]) f ( t )d t = ... f (t , t ,..., t )dt dt ...dt a,b= Rn

ii) E(X) t .f ( t )d t

=

iii) TVar(X) ( t E(X)).( t E(X)) .f ( t )d t

=

1. 5 Phn phi i lng ngu nhin nhiu chiu :1. 5. 1 nh nghaCho l mt i lng ngu nhin nhiu chiu. V hm

mt xc sut kt hp ca chng l f

1 2 nX (X ,X ,...,X )=

X(X1, X2, , Xn). Tng t nh trc y,

chng l c lp nu hm mt xc sut chung l tch ca mi hm mt xc

sut ring l. V vy, chng ta c

fX(X1, X2, , Xn) = fX1(X1) . fX2(X2) . . . fXn(Xn)

Trong trng hp c bit khi mi gi tr x c phn phi ging nhau v c lp

ln nhau, chng ta c

fX(X1, X2, , Xn) = fX(X1) . fX(X2) . . . fX(Xn)

Trong fX(x) l hm phn phi chung ca mi gi tr x.

__________________________________________________________________


31/165

25 Chng 1

1. 5. 2 Tnh cht :a.

Nu a1, a2, , an l hng s hoc khng ngu nhin th

E[a1X1 + a2X2 + . . . + anXn] = a1E(X1) + a2E(X2) + . . . + anE(Xn).

Gi tr k vng ca mt t hp tuyn tnh cc s hng bng t h p tuyn

tnh ca mi gi tr k vng ring l.

b. Nu mi Xiu c gi tr trung bnh bng nhau th E(Xi) = , chng ta c( )i iE a .X a = i

c bit, nu tt c h s aiu bng nhau v bng (1/n) th chng ta c:

( )ix

E E Xn

= =

Gi tr k vng ca gi tr trung bnh ca cc bin ngu nhin c phn

phi ging nhau s bng gi tr trung bnh chung ca chng.

c. ( ) ( ) ( )2

i i i i i i j i j

i jVar a .X a .Var x a .a cov X .X

= + trong cc h s ai c gi thit l hng s hoc khng ngu nhin.

d. Nu tt c cc bin X1, X2, , Xnu c lp th mi cp tng quanv hip phng sai s bng 0 hay Cov(x

ij

i, xj) = 0 = vi mi i j.ij

e. T (c) v (d) ta c th rt ra kt lun rng khi bin x c lp th( ) (2i i i i iVar a .X a .Var x ) =

v s hng hip phng sai s khng tn ti na.

Do , phng sai ca tng cc bin ngu nhin c lp s bng tng cc

phng sai. c bit, nu tt c cc gi tr phng sai u bng nhau,

ngha l vi mi i th( ) 2iVar X =

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32/165

26 Chng 1

( ) 2 2i i iVar a .X a =

f.Nu tt c cc X1, X2, . . ., Xnu l bin ngu nhin c lp ngha l tpbin X i c phn phi chun vi gi tr trung bnh i v phng sai hay

c th hin bng k hiu X

2

i

i ~ N(i, ) th t h p tuyn tnh ca tp

bin x cho trc c dng

2

i

a1X1 + a2X2 + . . . + anXn

cng s c dng phn phi chun vi gi tr trung bnh l

a1 1 + a2 2 + . . . + an n

v gi tr phng sai l .2 2 2 2 2 21 1 2 2 n na a .... a + + +

Trong k hiu php ly tng, chng ta c th vit nh sau

( ) ( ) ( )2 2i i i i 1 iU a X ~ N a , a =

g. Nu tt c cc X1, X2, . . ., Xnu c lp v c phn phi ging nhau tuntheo phn phi chun N(, ) th gi tr trung bnh ca chng l2

i

1X

n= X s c dng phn phi chun vi gi tr trung bnh bng v

phng sai bng2

n

, ngha l X ~ N

2

,n

. Tng t, chng ta c z =

( )n X

~ N(0, 1).

1. 6 Phn phi chun nhiu chiu :a s kim nh ca bin ngu nhin mt chiu v cc khong tin cy da trn

phn phi chun n chiu. Tng t nh vy, phn ln cc phng php ngu

nhin a chiu c phn phi chun nhiu chiu nh chnh nn tng csca n.

C nhiu ng dng hu ch cho phn phi chun nhiu chiu. Phn phi c

thc m t bng cch s dng : trung bnh, phng sai v hip phng sai.

__________________________________________________________________


33/165

27 Chng 1

th ca bin ngu nhin hai chiu c xu hng biu th tuyn tnh. Hm tuyn tnh

ca bin ngu nhin nhiu chiu c phn b chun thng l chun tc. Nh trong

trng hp mt chiu biu hin thun li ca hm mt l mn chnh n lm

ngun gc cho nhiu tnh cht v cc kim nh thng k. Thm ch khi d liu

khng phi l chun tc nhiu chiu th chun tc nhiu chiu c th x l bng cc

xp x c li. c bit l trong nhng kt lun lin quan n vector trung bnh mu,

thng c xp x chun tc nhiu chiu bi nh l gii hn trung tm.

Khi hm mt chun nhiu bin c m rng t hm mt chun mt

chiu n c tha hng nhiu tnh cht c trng. Ta s nhc li hm mt ca

phn phi chun mt bin trong mc 1.6.1 v sau s mrng m t hm mt

phn phi chun ca bin ngu nhin nhiu chiu trong mc 1.6.2. Cc mc cn

li ca chng dnh cho vic nghin cu m rng cc tnh cht ca phn phi

chun nhiu chiu.

1. 6. 1 Hm mt phn phi chun mt bin :Nu mt bin ngu nhin y, vi trung bnh v phng sai 2 , c phn phi

chun th hm mt ca n c cho bi cng thc :

( ) ( )2 22

2

1

2,yf y e y

= < < (1.61)

Khi biny c hm mt (1.61), ta ni rngy c phn phi chun ( )2,N .

Hm s ny thng c biu din minh ha bi th hnh qu chung trong hnh

1.1 khi cho 10 = v 2.5 =

Hnh 1.1 : th hm mt phn phi chun

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28 Chng 1

1. 6. 2 Hm mt ca phn phi chun nhiu chiuNu bin y c phn phi chun nhiu chiu vi vector trung bnh v ma

trn hip phng sai , hm mt c cho bi :

( )( )

( ) ( )1 2

1 2

1

2

'y y y

pf e

=

(1.62)

y p l s lng cc bin trong y c mt (1.62) ta ni y c phn phi

hoc n gin y l( ,pN ) ( ),pN .

S hng ( ) ( )( ) (12 2 2y y )y

= trong s m ca hm mt phn

phi chun mt chiu l bnh phng khong cch t y n , n v lch

chun . Tng t nh vy s hng ( ) ( )1y y trong s m ca hm mt

ca phn phi chun nhiu chiu l bnh phng khong cch suy rng tyn

hoc khong cch Manhalanobis

( ) ( )2 1'

y y = (1.63)

Khong cch ng bin vi s lng ca p bin.

H s ca hm m(1.62) ,1

2 xut hin nh s tng t ca 2 trong (1.61).

1. 6. 3

Tng qut ha phng sai tng th :Ta bit rng S nh l mt tng qut ha phng sai mu. Tng t l

tng qut ha phng sai tng th. Nu 2 l b trong phn phi chun mt bin,

th gi tr y tp trung gn trung bnh . Tng t gi tr nh ca trong trng hp

nhiu chiu chng t rng ,y s tp trung gn trong khng gian p chiu hoc l cc

a cng tuyn tnh gia cc bin.

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29 Chng 1

S hng a cng tuyn tnh ch ra tng quan cao gia cc bin. Trong

trng hp c li s bc t hn p.

(a) nh (b) ln

Hnh 1.2 : M t ca hn hi chun hai chiu

Trong s hin din ca a cng tuyn tnh mt hoc nhiu hn gi tr ring ca

ma trn s gn 0 v s nh nh vy l kt qu ca cc gi tr ring.

Hnh 4.2 cho thy rng, trong trng hp hai chiu, mt php so snh ca mt phn

b vi nh v mt phn b vi ln hn. Mt cch khc biu din mt

tp trung cc im trong phn phi chun hai chiu l biu ng vin. Hnh 4.3

biu din biu ng vin cho phn phi hai chiu hnh 4.2 . Mi mt ellipse

bao hm mt t l khc nhau gia cc vectoryc quan st.. Mt ct ca mt

chun hai bin mt vng ca ellipse l ni bao gm cc t l cc quan st.

(a) nh (b) ln

Hnh 1.3: Biu ng vin cho cc phn phi trong hnh 1.2

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30 Chng 1

c Hnh 1.2 v 1.3, nh xut hin hnh bn tri v ln hn xut hin

hnh bn phi. Trong Hnh 1.3a c s tng quan cht hn gia y1 v y2. Trong

Hnh 1.3b phng sai ln hn. Trong thc t cho mt p bin bt k, Nu gim s

tng quan gia cc bin hoc l tng phng sai th dn ti mt ln hn.

Hnh 1.4 : th hm mt phn phi chun hai chiu

1. 6. 4 Tnh cht phn phi chun ca bin ngu nhin nhiu chiu :Di y l mt s tnh cht ca vector ngu nhin y (p x 1) c phn phi

chun ( ),pN :

1 Tnh chun tc ca kt hp tuyn tnh cc bin trong y :

(a).Nu a l mt vector h s ( hng s ),th hm tuyn tnhl chun n bin :1 1 2 2

'a y ... p pa y a y a y= + + +

Nu y l th l( ),pN 'a y ( )' 'a ,a apN

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31 Chng 1

Trung bnh v phng sai ca c cho bi cng thc (1.53) v (1.54).

Nh vy th

'a y

( )' '

a y aE = v

( )

'

cov Ay A A= cho bt k mt vector ngu

nhin y. By gichng ta c thm thuc tnh c phn phi chun nu y c

phn phi

'a y

( ),pN

(b). Nu A l ma trn h s (q x p) c hng l q. y , q dng kt hptuyn tnh trong A

q p

y c phn phi chun nhiu chiu :

Nu y c phn phi

( ),

p

N th Ay c phn phi

( )'A,A A

p

N

y mt ln na nhc li ( )Ay AE = v ( ) 'cov Ay A A= nhng by

gic thm cc tnh nng ca q bin trong Ay vi phn phi chun nhiu

chiu.

2 Bin c chun ha :

Chun ha vectorz c tht c bng hai cch sau :

( ) ( )1

z T y

= (1.64)

y l ch sc dng bi phng php Cholesky hoc :'T T =

( ) ( )1

1 2z y

= (1.65)

y1 2

l cn bc hai ca ma trn i xng ca c xc nh bi1 2 1 2 'A CD C= . Nh vy m 1 2 1 2 = . Trong c hai cng thc (1.64) v

(1.65) vector ca bin ngu nhin c chun ha c tt c trung bnh bng 0

v phng sai bng 1 v tt c cc h s tng quan bng 0. Trong c hai trng

hp t (1b) ta thy z c phn phi chun nhiu chiu :

Vy nu y l ( ,pN ) th z l ( ),0 IpN

3 Phn phi chi ( khi )- bnh phng:

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32 Chng 1

Mt bin ngu nhin chi-bnh phng vi p bc t do c xc nh nh l

tng bnh phng p bin ngu nhin chun c lp. V vy, nu z l vector

c chun ha xc nh nh trong (1.64) v (1.65) th c phn

phi

2

1

'z zp jj z= =2 vi p bc t do, k hiu l 2p hoc ( )

2 p . T mt trong hai cng

thc (1.64) v (1.65) ta c c ( ) (1''z z y ) y = . Do ,

Nu y c phn phi ( ),pN th ( ) (1'

y ) y c phn phi 2p . (1.66)

4 Tnh chun tc ca phn phi bin duyn :

(a).Bt k mt nhm con no ca yu c phn phi chun nhiu chiu, vi vectortrung bnh tng ng vi vector con ca v ma trn hip phng sai tng

ng vi ma trn con ca . minh ha iu ny, cho vector

( )'

1 1 2, ,...,y ry y y= l vector con ny cha r phn t u ca y v

( )'

2 1,...,y ry y+= p bao gm p r phn t cn li. Nh vy y, v c

phn chia nh sau :

1 1 11

2 2 21

, ,y

y y

= = =

12

22

y v l ( r x 1) v1y 1 11 l ( ). Nh vy l c phn b chun

nhiu chiu.

r xr 1y

Vy nu y l th l( ),pN 1y ( )1 11,rN

Cng cn nhc li l ta c ( )1 1y E = v ( )1cov y 11= cnh cho bt k

mt vector ngu nhin no c phn chia theo cch ny. Nu y l ngu nhin

p bin phn b chun th l r bin phn b chun.1y

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33 Chng 1

(b).Mt trng hp c bit ca cc kt qu trc, vi mi yj trong y c phnphi chun n :

Nu y l th l( ,pN ) jy ( ), , 1,2,...,j jjN j = p

Cch o vn ny khng thc sng. Nu mt ca mi mt trong y

l phn b chun th khng nht thit yphi l phn b chun nhiu chiu theo

nhtrn.

jy

Trong ba tnh cht ti p theo, ta cho mt vectorc quan st v phn chia

thnh hai vector phc k hiu bi y v x , y y l (p x 1)v x l (q x 1).

Hoc, ngoi ra, cho php x l i din cho mt s b sung cng xt vi cc

bin trong y. T kt qu trc ta c :

, covy y

x x

yy yxy

xy xxx

E

= =

Trong tnh cht 5, 6, 7, chng ta gi thit rng :

y

x

l ,

yy yxy

p q

xy xxx

N +

5 Tnh c lp :

(a).Nu Vector con y v x l c lp th 0yx = (b).Hai bin ring l v c lp nujy ky 0jk = . Ch rng iu ny khng

thc sng cho cc bin ngu nhin khng c phn b chun.

6 Phn b c iu kin :

Nu y v x khng c lp, tc l 0yx phn phi c iu kin ca yi vi

x, ( )y xf l phn b chun nhiu chiu vi :

( ) ( )1y x x y yx xxE= + x (1.67)

( ) 1cov y x yy yx xx xy= (1.68)

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34 Chng 1

Ch rng ( )y xE l mt vector ca cc hm tuyn tnh ca x, trong khi

( )cov y x l mt ma trn khng ph thuc vo x. Xu hng tuyn tnh trong

cng thc (1.67) cnh cho bt k cc cp bin. Nh vy ta c th dng cng

thc trn kim tra tnh chun tc, kim tra th phn tn hai chiu ca tt c

cc cp bin v cng xem xt bt k mt xu hng phi tuyn no. Trong cng

thc (1.67) ta c c siu chnh trong vic dng phng sai v tng quan

o s lin h gia hai bin ngu nhin c phn b chun. Nh trnh by

phn trc hip phng sai v tng quan ch tt cho vic o mi lin h ca

cc bin c xu hng tuyn tnh v thc t s khng ph h p cho bin ngu

nhin phi chun vi mi lin h phi tuyn. Ma trn 1yx xx trong cng thc

(1.67) c gi l ma trn ca cc h s hi quy (matrix of regression

coefficients) bi v n lin h t ( )y xE n x.

7 Phn phi ca tng ca hai vector ph :

Nu y v x c cng c(p x 1) v c lp th :

ly x+ ( ), p d d yy xxN + + (1.69)

y x l ( ), p d d yy xxN + (1.70)

Trong phn cn li ca mc ny, ta s minh ha trng hp c bit ca tnh

cht 6 ca phn b chun hai bin. Cho

u

y =

C phn phi chun hai bin vi :

( )uy

x

E

=

, ( )

2

2cov u

y yx

yx x

= =

Tnh ngha ( ) ( ) ( ),f y x g y x h x= y ( )h x l hm mt ca x v

l hm mt kt hp cay vx. Do :( ,g y x)

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35 Chng 1

( ) ( ) ( ),g y x f y x h x=

V bi v bn phi l mt tch s. Chng ta c gng tm kim mt hm cay v

x, hm ny th c lp vi x v mt ca n c chc nng nh l ( )f y x .

T hm tuyn tnh ca y v x l chun tc bi tnh cht (1a) , ta c th xem

y x nh th. Ta s c gng tm gi tr ca sao cho y x v x l c

lp vi nhau.

Khi m z y x= v x l chun tc v c l p th tm( )cov , 0x z =

(cov , )x z ta biu dinx vz nh l hm s ca u,

( ) ( )0, 1 0, 1 'u a uy

xx

= =

= ,

( )1, 'u b uz y x = = = By gi:

( ) ( )cov , cov ,' ' 'a u b u a bx z = = [t cng thc (1.55)]

( ) ( )2

2 2

2

1 10, 1 ,=

y yx

xy x yx x

yx x

= =

Khi ( )cov , 0x z = ta thu c 2= yx x v z y x= trthnh :

2

yx

x

z y x

=

Bi v tnh cht (1a) , mt ca ( )2

yx xy x l chun tc vi :

2 2

yx yx

y x

x x

E y x

=

( )2var var ' 'b u b byx

x

y x

= =

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42/165

36 Chng 1

2

2

2 22

2

1

1,y yxyx yx

yx y

x xyx xx

= =

i vi mt gi tr ca x, chng ta c th biu din y nh sau

( )y x y x = + y x l i lng cnh tng ng vi gi tr cax

v y x l phn tn ngu nhin. Nn ( )f y x l chun tc, vi :

( ) ( ) ( )y x yE y x x E y x x x x = + = + = +

( ) 2 2varyx

y

x

y x

=

1. 6. 5 c lng trong phn b chun nhiu chiu :1. 6. 5. 1 c lng hp l ti a (MLE) :

Khi mt phn phi l phn b chun nhiu chiu c ginh cnh cho

mt tng th, c lng cc tham s thng c tm bi phng php hp l cc

i. K thut ny da trn tng n gin, cc vectorc quan st

c xem nh l bit trc v gi tr ca

1 2, ,...,y y yn

v c tm nh mt ti a ha

mt ng thi ca y c gi l hm h p l. Cho mt phn b chun nhiu

chiu th c lng hp l ti a ca v l :

y= (1.71)

( )( )'

1

1 1 1y y y y W S

n

i i

i

n

n n= n

= = = (1.72)

y ( )( )'

W y y y yi i= v S l ma trn hip phng sai mu c nh ngha

bi cng thc (1.7) Khi c s chia l n thay v n 1, n l mt c lng chch

v ta thng dng thay th

S

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37 Chng 1

By gi ta cho mt vector y c hiu chnh v xem nh l mt c lng

hp l ti a ca V t cu to ca mu ngu nhin, chng c lp, v mt

ng thi l tch ca cc mt ca

yi

y . Hm hp l s l :

( ) ( )( )

( ) ( )' 1 2

1 2 1 21 1

1, ,..., , , , ,

2

y y y y y y n n

n i pi i

L f e

= =

= =

( )( ) ( )

' 1

12

2

1

2

y y n

i

np ne

= =

(1.73)

thy rng y= ti u l hm hp l. Chng ta bt u bi vic cng tr

y trong cng thc m(1.73) :

( ) ( ),

1

1

1

2y y y y y y

n

i i

i

=

+ +

Khi iu ny c khuyt i trong cc s hng ca yi y v y hai trong

bn kt qu ca cc s hng b trit tiu bi v ( )y yii v (4.13) trthnh :

( )( ) ( ) ( ) ( )

' '1 1

12 2

2

1

2

y y y y y y n

i iin

np nL e

= =

(1.74)

V l xc nh dng, ta c1 ( ) ( )'

1 2 0y y n v

( ) ( )'

1 20

y y ne

< 1 ti u ha xy ra khi s m l 0, lc ny L c ti u

ha khi y=

c lng hp l ti a ca ma trn tng quan tng th P [ xem cng thc

(1.24)] l :

P R=

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38 Chng 1

Mi quan h gia cc bin chun nhiu chiu l tuyn tnh, iu ny c

cp phn trc. Nh vy v ch phc v tt cho phn b chun nhiu

chiu. Bi v chng cho c trong mi quan h tuyn tnh [ xem mc 1.6.4].

Nhng c lng ny khng hu ch cho cc phn b phi chun.

S R

1. 6. 5. 2 Phn phi ca y v S :Xy dng phn phi ca

1y y

n

iin

== ta c th chia thnh hai trng hp :

1- Khi y da trn csmu ngu nhin t phn phi chun nhiuchiu , th

1 2, ,...,y y yn

( ,pN ) y c phn phi ( ),pN n .

2- Khi y da trn c smt mu ngu nhin t phn phi phichun nhiu chiu tng th vi vector trung bnh

1 2, ,...,y y yn

v ma trn hip phng

sai , vi rng n, y c xp x ( ),pN n . R hn na, kt qu ny

c bit nh l nh l gii hn trung tm nhiu chiu : Nu y l vector

trung bnh ca mt mu ngu nhin t mt tng th vi vector

trung bnh

1 2, ,...,y y yn

v ma trn hip phng sai , th khi , phn phi can

( )y n xp x ( ),0pN

C p bin trong S v2

p

hip phng sai, cho tng cng c :

( ) ( )1 12 2 2p p p p pp p + + = + =

u vo khc bit nhau. Phn phi ng thi ca ( )1 2p p + cc bin khc nhau

ny trong ( ) ( )( )'

1W S y y y yi iin= = l phn phi Wishart, k hiu l

( )1,pW n y l s bc t do.1n

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39 Chng 1

Phn phi Wishart c s chiu tng t nh phn phi 2 v n c s

dng mt cch tng t. Trong tnh cht 3 mc 1.6.4, phn phi 2 ca mt bin

ngu nhin c nh ngha l tng bnh phng ca cc bin ngu nhin chun c

lp :

( )2

2

21 1

n ni

i

i i

yz

= =

= l phn phi ( )2 n

Nu y c thay th cho th ( ) ( )2

2 21ii y y n s2 = c phn

phi ( )2 1n . Tng t cng thc xc nh ca bin ngu nhin c phn b

Wishart :

( )( )'

1

y y n

i i

i=

l ( ),pW n

y l c lp v c phn phi nh1 2, ,...,y y yn ( ),pN Khi y c thay th

cho phn phi phn cn li Wishart vi t hn mt bc t do :

( ) ( )( )'

1

1 S y y y yn

i i

i

n=

= l ( )1,pW n

Cui cng, lu l khi ly mu t phn phi chun nhiu chiu, y v S l c

lp.

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46/165

40 Chng 2

CHNG 2 : C LNG KHNG CHCH TUYN TNH

Lp tt c cc i lng khng chch tuyn tnh ca hm ( )g F no y l htt c cc c lng khng chch cc hm tuyn tnh ca i lng ngu nhin

c quan st.

Trong chng ny ta s nghin cu hai vn :

1) Hm ( )g F no i vi n c c lng tuyn tnh khng chch2) Tm trong lp tt c cc c lng tuyn tnh khng chch ca ( )g F c

lng c phng sai b u nht.

2. 1 M hnh thng k tuyn tnh tng qut hng y :M hnh thng k tuyn tnh tng qut vi hng y bao gm vector ngu

nhin n chiu quan st c Yc biu din di dng :

Y X = + (2.1)

Trong Xl ma trn cp n x p bit, vector l vector ct p chiu v l

vector tham s cha bit, cn l vector sai s ngu nhin n chiu vi

(2.2)2 0 ; ITn

E E = =

Vi 20 < < , 2 ni chung l cha bit, l ma trn n v cp n. M

hnh c gi l m hnh tuyn tnh hng y nu hng

In

( )r X p= .

M hnh tuyn tnh l trng hp c bit vi m hnh tuyn tnh cng dng nhng

vi :2 STE = (2.3)

vi S l ma trn cp n bit, c hng bng n.

Tuy nhin ta c tha m hnh (2.1), (2.3) v (2.1),(2.2). Tht vy, v S l

ma trn hip phng sai xc nh dng nn c tn ti ma trn khng suy bin D

c p n x n sao cho .V DDT=

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47/165

41 Chng 2

Nu t ta c1*Y D= Y ** *Y X = + trong , do

1 1* *X D X, D = =

*

0E = v ( ) ( )2 1 1 2* *

D S D I

T T

nE

= = . Nh vy m hnh tuyn tnh ivi tha mn (2.2). Do gi thit (2.2) khng lm gim tnh tng qut ca m

hnh.

*Y

Nu ma trn S suy bin th khng th p dng c phng php trn v cn

phi xt mt l thuyt tng qut hn.

Ch rng p vector ct ca X l vector n chiu nm trong mt a tp tuyn

tnh (khng gian con) ca khng gian n chiu. V vy vi bt k vector p chiu ,

vector c chiu l n v nm trong a tp tuyn tnh p chiu cm sinh bi p

ct ca X. K hiu a tp l . Gi s

X=

pD

( ) ( ) ( ){ }1 2 , ..., p l cstrc chun ca

. Khi vector c th biu din di dngpD X=( )

1

p i

iic

== . Khong

cch gia vector pD v Yt cc tiu khi l hnh chiu trc giao ca Yln

.

pD

Gi s l vector sao cho :

X=

l c lng bnh phng b nht ca . tm ta rng X l hnh

chiu trc giao ca Y ln , do pD

( )( )

( ) 0 1 Y X , , ..., .T

i i p = =

V vector ct bt k ca X l t hp tuyn tnh ca ( ) ( ) ( )1 2 , ..., p nn tha mn

phng trnh :

( )X Y X 0T =

Hoc phng trnh chun Gauss Markov :

X X X YT = T (2.4)

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48/165

42 Chng 2

V l ma trn khng suy bin nnX XT

( )

1 X X X YT T

= (2.5)

R rng l c lng bnh phng b nht l duy nht. Ta s chng minh rng

c lng khng chch tuyn tnh tt nht ca hm tuyn tnh lT T khi m

hnh l m hnh hng y .

2. 2 c lng khng chch cho m hnh thng k tuyn tnh tng qut hngy :

2. 2. 1 nh l 2.1 (Gauss Markov) :Gi s Y X+= l m hnh tuyn tnh hngy , l hm tuyn tnh

ca . Khi c lng

T

T l c lng khng chch tt nht ca , trong

c xc nh bi

T

( )1

X X X YT T

= .

Chng minh : Trc tin ta chng minh rng T l c lng khng chch ca

. Tht vy , T

( ) ( )1 -1T T T T T X X X EY= X X X X= T TE

= T

Gi s l mt c lng khng chch ca bt k ca , tron l ma trn

cp p x n no y. cho l c lng khng chch ta phi c :

AY A

AY

AY A Y AX E E= = =

tc l AX Ip= .

Ta s chng minh rng :

( ) ( )T T AYD D

vi bt k v bt k trong D l k hiu ca ton t hip phng sai ng vi

, hoc tng ng

__________________________________________________________________


49/165

43 Chng 2

( ) ( )T AY TD D

Tc l ta phi chng minh c rng ( )

( ) AY D D l ma trn xc nh khngm.

t vi-1 TQ=A - V X TV X X= , khi :

( ) ( ) ( ) 2 12 AY QY QXVD D D = + +

V nnAX Ip= ( ) ( )1 1QX V AX I V 0p = = , v v vy

( ) ( ) ( ) AY QYD D D =

l ma trn xc nh khng m.

D dng thy rng

( ) 2T - V TD = 1 .

By gita hy xc nh c lng khng chch ca2

2. 2. 2 B 2.1 :c lng khng chch ca 2 trong m hnh tuyn tnh c hngy c

dng :

( )( )2 11

1 TY X X X XT Tn p

=

Y . (2.6)

Chng minh :Ch rng ma trn l ma trn ly ng, tc l :-1 TI XV X

( )2

1 - T -1XV X I XS XT = .

Hn na ( )-1 TX I XV X X 0 = , do

__________________________________________________________________


50/165

44 Chng 2

( )( ) ( )( )( )

( ) ( )2 2

T -1 T T -1 T

-1 T

Y I XV X Y Y I XV X

I XV X

E tr D

tr n p

=

= =

trong ( ).tr l k hiu vt ca ma trn vung ( tc l tng cc phn t trn ng

cho chnh ) v ta cng cn ch n tnh cht ca ( ).tr l ( ) ( )tr AB tr BA= . T

suy ra rng nu 2

cho bi (3.6) th 2

2E = .

2. 2. 3 H qu 2.1 :Xt m hnh tuyn tnh tng qut (2.1), (2.3). Khi c lng khng chch

bnh phng b nht ca l :

( )-1-1 T T -1 XS X X S Y= , (2.7)

Cn c lng khng chch cho 2 l :

( )( )2 1 T -1 -1 T -1 T -1Y S S X X S X X S Y

n p =

. (2.8)

Tht vy, nu p dng (2.3) v b 1.1 cho m hnh (2.1), (2.2) trong thay

bi v bi ta s nhn c (2.7), (2.8).

Y-1D Y X -1D X

V d 2.1 : Gi s l vector quan st n chiu cY ( )Y , 1,...,11 1T

E = = ,

< < v ma trn hip phng sai l . Hy tm c lng khng

chch tuyn tnh vi phng sai b nht ca

2 2 0,S >

.

Ta c m hnh

Y 1 = +

trong 2 STE = .

Do p dng h qu trn ta c :

-1

-1

1 S Y

1 S 1

T

T = , (2.9)

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45 Chng 2

( )( )2 1 T -1 -1 T -1 T -1Y S S 1 1 S 1 1 S Y

n p =

.

c bit nu ( )21S ,...,diag 2n = , trong ( ).diag l k hiu ca ma trn ng

cho, ta c :

2

1

2

1

Y

1

n

i ii

n

ii

=

=

=

1 1 1

22 2 2 2 2i i j

,

1Y Y Y

n n n

i i j

i i j in p

= = =

=

i

cn

1

2

1

n

i

i

D

=

=

Khi 2 bit (cho 2 1 = ) th c lng khng chch tuyn tnh vi

phng sai b nht ca cho bi (3.9) trng vi c lng khng chch v

phng sai b nht khi Y c phn b chun ( )1,SN .Tuy nhin c trng hp c lng tuyn tnh khng chch bng phng

php bnh phng b nht rt l km hiu qu , chng hn ly mu ngu

nhin t phn bu trn (

1,..., nX X

)0, th c c lng khng chch vi phng sai b

nht l :

( )

1

n

nX

n

+=

Cn c lng tuyn tnh bnh phng b nht, nh ta d dng thy l :

1 2

X

=

1 l c lng c phng sai ln hn ng k so vi phng sai ca

khi n ln.

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46 Chng 2

2. 3 M hnh thng k tuyn tnh vi hng khng y

Xt m hnh tuyn tnh tng t nh(2.1) v (2.2) dng :

Y X+= , (2.10)

vi :

2 ITE = , (2.11)

cn

( )Xr p<

M hnh nh vy gi l m hnh tuyn tnh hng khngy .

i vi m hnh tuyn tnh hng khng y

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