Ước lượng và kiểm Định trong thống kê nhiều chiều
TRANSCRIPT
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MC LC
Trang ph baLi cm n
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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8/2/2019 c Lng v Kim nh Trong Thng K Nhiu Chiu
4/165
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
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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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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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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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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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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
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13 Chng 1
S ngibnh
Bng 1.2 : Quan h gia nng Insulin vi cn nng v lngng trong mu
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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)
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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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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
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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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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 :
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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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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 =
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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 :
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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 =
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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)
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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.
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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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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.
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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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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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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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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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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
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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
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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