05 tvu sta301_bai3_v1.00131012140

Bi 3: M hnh hi quy tuyn tnh n

STA301_Bi 3_v1.0013101214 23

BI 3. M HNH HI QUY TUYN TNH N

Mc tiu

Sau khi kt thc bi, hc vin s hiu c nhng vn sau y: tng ca phng php bnh phng

ti thiu (OLS) v cch s dng OLS c lng cc h s hi quy.

ngha ca cc h s hi quy c lng. Cc gi thit c bn ca phng php OLS. H s xc nh r2 o ph hp ca

hm hi quy. Khong tin cy v kim nh gi

thuyt cho cc h s hi quy. Phn tch phng sai kim nh v

s ph hp ca m hnh. D bo.

Ni dung Hng dn hc

Phng php OLS. Cc gi thit c bn ca phng php bnh

phng ti thiu. H s xc nh r2 o ph hp ca hm

hi quy mu. c lng khong cho h s hi quy. Kim nh gi thuyt v cc h s hi quy. Phn tch phng sai trong m hnh hi quy. D bo.

ngh hc vin n li phn c lng v kim nh gi thit trong mn l thit xc sut v thng k ton.

Theo di k bi ging. Xem cc v d cho mi phn bi ging. Lm cc v d v tr li cu hi trc nghim.


24 STA301_Bi 3_v1.0013101214

TNH HUNG DN NHP

Tnh hung

Cng ty du n Tng An ang xem xt vic gim gi bn sn phm (loi bnh 5 lt) tng lng hng bn ra, ng thi qung b sn phm ca mnh n khch hng. Ngi qun l ca cng ty mun tnh ton xem nu sn phm ny c gim gi i 1000 ng/lt th lng hng trung bnh bn ra s thay i th no. ng thi, nu nh gim gi 1000 ng cho 1 lt m lng hng bn thm c l nhiu hn 50000 sn phm th cng ty s tin hnh 1 chin dch khuyn mi trong 1 thng vi gi gim i l 10000/lt. tin hnh nghin cu ny, phng marketing ca cng ty da vo cc s liu bn hng ca cng ty trong vng 15 thng qua (n =15 quan st) thu thp s liu v gi bn (P) v lng bn (Q) cho loi du n ny. Nghin cu vin sau khi tin hnh cc thng k m t quyt nh dng hm cu dng tuyn tnh xem xt nh hng ca gi n lng bn: i 1 2 i iQ P u .

Dng s liu ca mu, c lng c hm hi quy mu c dng

i iQ 6227 30.43P .

Cu hi

Theo kt qu ca m hnh, khi gi gim 1 n v, lng hng bn ra thay i th no? Liu khi gi gim i 1000 ng 1 lt th lng hng bn thm ln hn c 50000 sn phm

nh cc nh nghin cu mun kim tra khng? Gi bn quyt nh bao nhiu % trong s thay i ca lng bn? Nu gi bn l 150000 ng 1 bnh th lng bn d bo l bao nhiu?


25

Ni dung bi ny gii thiu mt m hnh hi quy n gin nht v a ra cc phng php c lng, kim nh gi thit v d bo. l m hnh hi quy tuyn tnh n hay cn c gi l m hnh hi quy 2 bin, m hnh cp n mt bin c lp X v mt bin ph thuc Y. Trong bi ny chng ta s c lng hm hi quy tng th PRF da trn thng tin mu. Mc d c rt nhiu phng php c lng hm hi quy tng th nhng chng ta s s dng phng php thng dng l phng php bnh phng ti thiu (OLS) (Ordinary Least Square).

3.1. c lng tham s hi quy bng phng php bnh phng ti thiu

BI TON

Cho bin c lp X v bin ph thuc Y, gi s ta c hm hi quy tng th (PRF) c dng tuyn tnh:

i i i 1 2 i iY E(Y | X ) u X u (3.1)

Vi mt mu quan st 1 1 2 2 n n(X ,Y ),(X ,Y ),..., (X ,Y )

Ta c: hm hi quy mu (SRF)

i 1 2 i Y X (3.2)

v: i 1 2 i i i i Y X u Y u (3.3)

1 2 , l cc c lng ca i i

i i

x X xy Y y

, iu l c lng

ca iu , iu c coi l phn d.

T (3.3) ta c: i i iu Y Y .

Vn t ra l s dng cc d liu ca X v Y tm c lng tt nht cho 1 2, tha mn tng bnh phng cc phn d t gi tr nh nht.

Tc l ta cn phi xc nh 1 2 , sao cho: n n

2 21 2 i i 1 2 i

i 1 i 1

f ( , ) u (Y X )

t min.

Trong cc bi ging v gii tch nhiu bin ta c trang b phng php tm gi tr cc tiu, cc i ca hm f (X,Y) . Vy hm 1 2 f ( , ) t gi tr nh nht

th 1 2 , phi l nghim ca h phng trnh

n1 2

i 1 2 ii 11

n1 2

i i 1 2 ii 12

f ( , ) 2(Y X ) 0

f ( , ) 2X (Y X ) 0

(3.4)

Suy ra:

n n

1 2 i ii 1 i 1

n n n2

1 i 2 i i ii 1 i 1 i 1

n X Y

X X X Y

(3.5)


26 STA301_Bi 3_v1.0013101214

Ta c: n n n

i i i ii 1 i 1 i 1

1 1 1X X ; Y Y ; XY X Yn n n

n n

2 2 2 2i i

i 1 i 1

1 1X X ; Y Y .n n

Phng trnh (3.5) dn n:

1 2

21 2

X Y X X XY

(3.6)

Gii h phng trnh (3.6) ta thu c nghim

2 2 2

1 2

XY (X)(Y)X (X)

Y X

(3.7)

Ta t n n

2 2 2 2 2YY i i

i 1 i 1S (Y Y) Y n(Y) nY n(Y)

n n2 2 2 2 2

XX i ii 1 i 1

S (X X) X n(X) nX n(X)

n n

XY i i i ii 1 i 1

S (X X)(Y Y) X Y n(X)(Y) nXY n(X)(Y)

Khi (3.7) c th vit li l

XY2

XX

1 2

SS

Y X

Phng php tm cc c lng 1 2 , nh trn c gi l phng php bnh phng ti thiu.

3.1.1. Tnh cht ca tham s hi quy mu c lng bng phng php bnh phng ti thiu. Phng php bnh phng ti thiu em li cc c lng vi cc tnh cht nh sau: ng vi mt mu 1 1 2 2 n n((X ,Y ), (X ,Y ),...(X ,Y )) cho trc, h s 1 2 , c xc

nh duy nht. ng thng ca phng trnh hi quy mu (SRF) i 1 2 i Y X i qua im c

to gi tr trung bnh (X,Y).

Gi tr trung bnh ca cc c lng ca iY bng gi tr trung bnh ca cc quan st

iY tc l: iY Y hay n n

i ii 1 i 1

1 1Y Y .n n


27

Gi tr trung bnh cc phn d iu bng 0

n

ii 1

u 0.

Cc phn d iu v iY khng tng quan, tc l:

n

i ii 1

u Y 0.

Cc phn d iu v iX khng tng quan, tc l:

n

i ii 1

u X 0.

By gi ta s chng minh mt s tnh cht trn: o Hin nhin v h phng trnh (3.6) c nghim duy nht.

o Hin nhin v gi tr ca 1 2 , l mt hm ca mu.

o Thay im (X,Y) vo phng trnh hi quy mu, ta c:

1 2 Y X

1 2 Y X .

o Ta c: n n

i 1 2 ii 1 i 1

1 1 Y Y Xn n

1 2 XY.

o Ta c: i i iu Y Y . Suy ra ngay

n n n n

i i i i ii 1 i 1 i 1 i 1

u (Y Y ) Y Y nY nY 0.

o R rng t: n n n n

2i i i i i i i i

i 1 i 1 i 1 i 1

u Y (Y Y )Y Y Y Y

n n2

i 1 2 i 1 2 ii 1 i 1

Y ( X ) ( X )

2 2 21 2 1 1 2 2

n Y n XY n( 2 X X )

n2 2 2 2

i i 1 1 2 2 1 2 1 1 2 2i 1

1 u Y ( X) ( X X ) ( 2 X X ) 0.n

Vy n

i ii 1

u Y 0.

(3.8)


28 STA301_Bi 3_v1.0013101214

o D dng thy n n

i i i 1 2 ii 1 i 1

u Y u ( X )

n n

1 i 2 i ii 1 i 1

u u X

.

T tnh cht 4 v 5 ta c

n n

i i ii 1 i 1

u u Y 0

.

Vy ta c: n

i ii 1

u X 0.

V D 3.1

Thu thp s liu v im hc tp ca hc sinh v mc thu nhp hng nm ca b m ta c bng s liu sau:

Thu nhp (x) (triu/nm) 45 60 30 90 75 45 105 60

im trung bnh (y) 8.75 7.5 6.25 8.75 7.5 5.0 9.5 6.5

Hy tm hm hi quy mu v tnh cc c trng ca n

3.1.2. Cc gi thit c bn ca phng php bnh phng ti thiu Khi phn tch hi quy, mc ch ca chng ta l tm phng trnh hi quy mu thng qua vic c lng cc h s 1 2, . Da vo d liu mu ta thu c cc c lng

tng ng l 1 2 , . Nhng 1 2 , l cc c lng im ca 1 2, . V th ta cha bit c cht lng ca cc c lng ny th no. Ta cn a ra mt s cc gi thit ca phng trnh bnh phng ti thiu thu c cc c lng tt nht cho 1 2, . T ta cng s thu

c gi tr iY l c lng tt nht cho iE(Y | X ) . Cht lng ca cc c lng s ph thuc vo cc yu t sau: Dng hm ca m hnh c chn. Ph thuc vo cc iX v iu .

Ph thuc vo c ca mu. Vn v dng hm ca m hnh c la chn chng ta s xem xt bi 7. Ta s a ra cc gi thit cho iX v iu cc c lng thu c khng chch v c phng sai nh nht. Gi thit 1: Bin gii thch X c gi tr quan st iX khc vi t nht 1 gi tr cn

li, tc l phng sai mu hiu chnh khng suy bin:

n'2 2X i

i 1

1S (X X) 0.n 1


29

Gi thit 2: Gi tr trung bnh ca sai s c th mang du m hoc dng i vi mi gi tr quan st nhng v mt trung bnh th bng 0.

Gi thit 3: Cc gi tr ca X c cho trc v khng ngu nhin, tc l mi iX

c cho trc v khng phi l bin ngu nhin. iu c ngha l iX v iu l khng tng quan vi nhau.

i i i i i i

i i i i

CoV(X ,u ) E(X u ) E(X ) E(u )X E(u ) X E(u ) 0.

Gi thit ny c mt ngha rt quan trng l nu X v u c c tng quan th khi X thay i, u cng s thay i. V th gi tr k vng ca Y s khc 1 2X.

Gi thit 4: Phng sai sai s thun nht (khng i)

2i jVar(u ) Var(u ) i j .

Gi thit 5: Khng c tng quan gia cc iu , tc l:

i jCoV(u ,u ) 0 i j .

Vi cc gi thit nu, khi ta c tnh cht ca cc c lng theo phng php bnh phng ti thiu nh sau: nh l Gauss-Markov Gi s ta c m hnh hi quy tuyn tnh, khi vi cc gi thit 1-5 ta c c lng bnh phng ti thiu l cc c lng tuyn tnh khng chch v c phng sai nh nht trong lp cc c lng tuyn tnh khng chch. nh l Gauss-Markov cho mt khng nh l cc c lng 1 2 , ca 1 2, c c bng phng php bnh phng ti thiu l cc

c lng khng chch v c phng sai ti thiu trong cc c lng khng chch ca 1 2, .

3.1.3. Sai s ca phng php bnh phng ti thiu

Trong phn 3.1 ta c cc c lng 1 2 , ca 1 2, theo phng php bnh phng ti

thiu l

2 2 2

1 2

XY (X)(Y)X (X)

Y X .

t: i ii i

x X Xy Y Y


30 STA301_Bi 3_v1.0013101214

Khi ta c:

1 2 Y X

n n2

2 i i ii 1 i 1

x y x .

Vi cc gi thit 1-5 ca phng php bnh phng nh nht, ta c phng sai v lch chun ca cc c lng l

2

2 n2i

i 1

Var( )x

; 2 n

2i

i 1

se( ) ;x

n2i

2i 11 n

2i

i 1

XVar( )

n x

;

n2i

i 11 n

2i

i 1

Xse( ) ,

n x

vi iVar(u ) , se: sai s tiu chun (standard error).

Do 2 cha bit nn da vo d liu mu cho ta thu c c lng ca 2 l 2 c xc nh bng cng thc sau:

n n2 2i i

2 i 1 i 1

u u

n 2 n 2

l sai s tiu chun ca c lng (standard error of the estimate).

3.2. H s xc nh 2r o ph hp ca hm hi quy mu: Cho hai bin X v Y, xc nh mi quan h ca X v Y c dng tuyn tnh hay khng ta a ra mt i lng o mc ph thuc tuyn tnh gia X v Y.

Ta c: i i i Y Y u

i i i i i Y Y Y Y u Y Y u

i i i y y u (3.9)

Bnh phng hai v ca (3.9) ta c: n n n n

2 2 2i i i i i

i 1 i 1 i 1 i 1

y y u 2 y u

n n2 2i i

i 1 i 1

y u

n n2 2 22 i i

i 1 i 1

x u

(3.10)


31

t: n n

2 2i i

i 1 i 1TSS y (Y Y)

(3.11)

TSS (Total sum of squares) gi l tng bnh phng cc sai lch gia iY vi gi tr

trung bnh Y . n n n

2 2 2 2i i i 2 i

i 1 i 1 i 1

ESS (Y Y ) y x

(3.12)

ESS (Explained sum of squares) l tng bnh phng cc sai lch gia gi tr iY v trung bnh ca n.

n2i

i 1

RSS u .

(3.13) (3.12)

RSS (Residual sum of squares) l tng tt c cc bnh phng sai lch gia gi tr quan st iY v gi tr iY nhn c t hm hi quy hay gi l tng cc phn d. T (3.10), (3.11), ( 3.12), (3.13) ta c:

TSS ESS RSS (3.14)

Chia hai v cho TSS ta c:

ESS RSS1TSS TSS

n n2 2

i ii 1 i 1n n

2 2i i

i 1 i 1

(Y Y) u

(Y Y) (Y Y)

(3.15)

t:

n2

i2 i 1

n2

ii 1

(Y Y)ESSr .TSS (Y Y)

T (3.14) v (3.15) ta c: 2 RSSr 1TSS

(3.16)

Ta c:

n n n2 2 2 2 2

2i 2 i 2 i2 i 1 i 1 i 1 X

2n n n 22 2 2 Yi i i

i 1 i 1 i 1

y x (X X)SrSy y (Y Y)

(3.17)

trong : n

2 2X i

i 1

1S (X X)n 1

;

n2 2Y i

i 1

1S (Y Y)n 1


32 STA301_Bi 3_v1.0013101214

l phng sai mu ca X v Y. Ngoi ra v

n

i ii 1

2 n2i

i 1

x y

x

nn (3.17) c th c vit

li nh sau:

2n

i ii 12

n n2 2i i

i 1 i 1

x yr

x y

(3.18)

T (3.18) ta c: n n n n

i i i i i ii 1 i 1 i 1 i 1

n n n n2 2 2 2i i i i

i 1 i 1 i 1 i 1

1x y X Y ( X )( Y )nr

x y (X X) (Y Y)

n n n

i i i ii 1 i 1 i 1

n n n n2 2 2 2i i i i

i 1 i 1 i 1 i 1

n X Y ( X )( Y )

n X ( X ) n Y ( Y )

Ta thy rng r chnh l h s tng quan mu ca X v Y. Cc tnh cht ca h s tng quan: r c th m hoc dng.

1 r 1.

r c tnh cht i xng r(X,Y) r(Y,X).

Nu X aX c v Y bY d, a, b > 0, c, d l hng s ta c r(X ,Y ) r(Y,X) .

Nu X,Y c lp th r = 0. r o ph thuc tuyn tnh gia X v Y.

3.3. Phn b xc sut ca cc tham s hi quy mu

Trong phn trc ta thu c cc c lng im ca 1 v 2 theo phng php bnh phng nh nht (OLS) da trn cc gi thit c bn v sai s ngu nhin iu l:

iE(u ) 0.

2iVar(u ) .

i jCov(u ,u ) 0 , i j .


33

Khi cc c lng im thu c tng ng l 1 2 , c tnh cht khng chch v c phng sai nh nht. Tuy nhin, cc c lng im khng cho ta bit c sai lch ca chng so vi gi tr thc, v vy c lng khong cho ta nhiu thng tin hn so vi c lng im. c th tm c c lng khong cho cc tham s 1 2,

chng ta cn xc nh c phn phi xc sut ca 1 v 2 . Cc phn phi xc sut

ny ph thuc vo phn phi xc sut ca iu . Vy ta a thm gi thit v phn phi

xc sut ca iu nh sau:

Gi thit: iu c phn phi chun 2N(0; ) ,

Vi gi thit thm vo , 1 2 , cn c cc tnh cht sau:

1 2 , l cc c lng vng, tc l khi c mu ln th chng hi t n gi tr

1 2, .

1 c phn phi chun vi

1 1E( ) ,

n2i

2 2i 11 1 n

2i

i 1

XVar( )

n x

(3.19)

tc l 21 1 1 N( ; ) . T bin ngu nhin

1 1

1

Z

c phn phi chun tc N(0;1).

2 c phn phi chun vi:

2 2E( ) ,

22

2 2 n2i

i 1

Var( )x

(3.20)

tc l 22 2 2 N( ; ) . Do bin ngu nhin 2 22

Z

c phn phi chun tc

N(0;1).

Thng k 2

22

(n 2)

c phn phi khi-bnh phng vi n 2 bc t do.

Cc c lng 1 2 , c phng sai nh nht trong s cc c lng khng chch

ca 1 2, .

Ta c i 1 2 i iY X u . T gi thit ca iu ta thu c cc thng k Z v 2 c

quy lut phn phi chun tc v khi bnh phng vi (n 2) bc t do. Vy ta c

th tm c khong c lng cho cc tham s 1 2, v 2 .


34 STA301_Bi 3_v1.0013101214

3.4. c lng khong cho h s hi quy

Trong mc 3.3 vi gi thit v phn phi chun 2N(0; ) ca iu ta c:

21 1 1

N( ; )

22 2 2

N( ; )

vi cc phng sai 2 21 2, c xc nh trong

(3.19) v (3.20). Tuy nhin v phng sai 2 cha bit, nn cc phng sai 2 21 2, cng

cha bit, v vy ta dng c lng khng chch ca 2 l: n

2i

2 i 1u

RSS .n 2 n 2

Khi cc thng k:

1 11

1

T Se( )

v 2 22

2

T Se( )

vi: 1 1 Se( ) Var( ) ; 2 2 se( ) Var( ) .

Cc thng k ny c phn phi student vi (n 2) bc t do. ng thi, thng k 2

22

(n 2)

c phn phi khi bnh phng vi (n 2) bc t do.

3.4.1. Khong c lng cho 1

Vi tin cy 1 cho trc, ta c:

(n 2) (n 2)12 2

P t T t 1 ,

vi 2

(n 2)t l phn v mc 2 ca phn phi Student

1T , tc l:

2 2

(n 2) (n 2)1 1

1

P{ t t } 1se( )

.

T dn n

2 2

(n 2) (n 2)1 1 1 1 1

P{ t se( ) t se( )} 1 .

Vy vi mu c th ta c khong c lng cho 1 l:

2 2

(n 2) (n 2)1 1 1 1 1

( t se( ); t se( )) .


35

3.4.2. Khong c lng cho 2

Tng t nh trn ta c, vi tin cy 1 cho trc th:

(n 2) (n 2)2 22

2 22

P t T t 1Se( )

.

T ,

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

P t Se( ) t Se( ) 1 .

Vy vi mi mu c th ta c khong c lng cho 2 l:

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

t Se( ); t Se( )

3.4.3. Khong c lng cho 2 Ta thy thng k

22

2

(n 2)

c phn phi khi-bnh phng vi (n-2) bc t do. Do :

22 2 21 / 2;n 2 / 2;n 22

(n 2)P{ } 1

vi 21 / 2;n 2 v 2

/ 2;n 2 l cc gi tr phn v mc 1 / 2 v / 2 ca phn phi 2 (n 2) .

T ta c: 2 2

22 2

/ 2;n 2 1 / 2;n 2

(n 2) (n 2)P 1

.

Vy vi mu c th v tin cy 1 , ta c khong c lng cho 2 l: 2 2

22 2

/ 2;n 2 1 / 2;n 2

(n 2) (n 2)( ; )

.

3.5. Kim nh gi thuyt v cc h s hi quy Kim nh gi thuyt thng k l mt trong nhng nhim v quan trng ca nh kinh t lng. Chng hn, trong m hnh hi quy (3.1) ta thy nu 2 0 th Y s c lp vi X, tc l X khng nh hng ti s thay i ca Y . Tuy nhin, ta li cha bit 2 c bng 0 hay khng v vy ta cn kim nh gi thuyt ny. Trong cc mc trc, chng ta a ra cc c lng im v c lng khong cho h s hi quy 1 2, . Cc c lng khong ny s gip ta gii quyt bi ton

kim nh gi thuyt v 1 2, .


36 STA301_Bi 3_v1.0013101214

Ta bit bi ton kim nh gi thuyt gm cc bc c bn sau: Bc 1: Thit lp gi thuyt 0H v i thuyt 1H .

Bc 2: Xy dng tiu chun thng k kim nh, xc nh quy lut phn phi xc sut ca tiu chun thng k khi gi thuyt 0H c cho l ng.

Bc 3: Xy dng min bc b gi thit W ng vi mc ngha cho trc. Bc 4: So snh gi tr mu (quan st c) ca tiu chun thng k bc th 2

vi min bc b gi thuyt W bc 3 a ra kt lun bc b hay chp nhn gi thuyt 0H .

3.5.1. Kim nh gi thuyt cho 1

Ta a gi thuyt 0H : *

1 1 v i thuyt 1H : *

1 1 hoc 1H : *

1 1 hoc 1H : *

1 1 .

Ch rng nu gi thit H0 l ng th: thng k

1 11

1

T Se( )

c phn phi Student vi n 2 bc

t do. Ta s da vo thng k ny tin hnh kim nh gi thuyt cho 1 . Ta c cc bi ton kim nh gi thuyt sau: Bi ton 1: Kim nh hai pha

*0 1 1

*1 1 1

H :

H :

Min bc b: (n-2) (n-2)/ 2 / 2W ( ; t ) (t ; ) vi (n-2)pt l phn v mc p (p = /2) ca

phn phi Student 1T .

Bi ton 2: Kim nh mt pha (phi) *

0 1 1*

1 1 1

H :

H :

Min bc b: (n-2)W= t ; , vi (n-2)t l phn v mc ca phn phi Student 1T .

Bi ton 3: Kim nh mt pha (tri) *

0 1 1*

1 1 1

H :

H :

Min bc b: (n-2)W ( ; t ) .

3.5.2. Kim nh gi thuyt cho 2

Ta c gi thuyt *0 2 2H : vi i thuyt *

1 2 2H : hoc *

1 2 2H : hoc *

1 2 2H : .


37

Trong mc 3.4 ta cng thy nu gi thuyt 0H ng th thng k

2 22

2

T Se( )

c phn phi Student vi n 2 bc t do. Do , ta c th tin hnh cc bi ton kim nh gi thuyt sau cho 2 :

Bi ton 1: Kim nh hai pha *

0 2 2*

1 2 2

H :

H :

Min bc b: (n-2) (n-2)/ 2 / 2W ( ; t ) (t ; ) (n-2)pt l phn v mc p ca phn phi Student 2T .

Bi ton 2: Kim nh mt pha (phi)

*

0 2 2*

1 2 2

H :

H :

Min bc b: (n-2)W (t ; ) , vi (n-2)t l phn v mc ca phn phi Student 2T .

Bi ton 3: Kim nh mt pha (tri)

*

0 2 2*

1 2 2

H :

H :

Min bc b: (n-2)W ( ; t ).

3.5.3. Kim nh gi thuyt cho phng sai 2

Gi thuyt 2 20 0H : , vi mt trong cc i thuyt

2 21 0H : ,

2 21 0H : ,

2 21 0H : .

Ta c nu 0H ng th thng k

22

2

(n 2)

c phn phi khi bnh phng vi n 2 bc t do. p dng kt qu , ta c th gii quyt cc bi ton kim nh i vi 2 nh sau: Bi ton 1: Kim nh hai pha

2 20 0

2 21 0

H :

H :


38 STA301_Bi 3_v1.0013101214

Min bc b: 2 21- / 2;n 2 / 2;n 2W (0; ) ( ; )

trong 2p;n 2 l phn v mc p ca phn phi2 .

Bi ton 2: Kim nh mt pha (phi)

2 2

0 02 2

1 0

H :

H :

Min bc b 2 ;n 2W= ;+ . Bi ton 3: Kim nh mt pha (tri)

2 20 0

2 21 0

H :

H :

Min bc b: 21- ;n 2W= 0; .

3.5.4. Phng php xc sut ngha (p-value)

Vi mt mu c th ta c gi tr quan st ca thng k iT (i 1,2) l: *

i iiqs

i

t Se( )

Ta c: p-value i iqsP T t i 1, 2 Xc sut ny gi l xc sut ngha, y chnh l xc sut mc sai lm loi 1 (tc l xc sut bc b

0H khi 0H ng).

Ta thy rng nu xc sut ngha cng cao th hu qu vic bc b 0H khi 0H ng cng nghim trng, nu xc sut ngha cng nh th hu qu ca vic bc b sai 0H cng t nghim trng. Vy khi cho trc mc ngha (y l xc sut gii hn c bc b 0H ), nu xc sut ngha khng vt qu th ta c th bc b 0H m khng s phm sai lm nghim trng, cn nu

xc sut ngha ln hn th cha c c s bc b 0H .

By gi ta c th s dng xc sut ngha tin hnh cc bi ton kim nh i vi cc tham s 1 2, .

Kim nh hai pha

*

0 i i*

1 i i

H :

H :

i = 1, 2

CH

Phng php kim nh trn c gi l phng php kim nh theo min tiu chun m ta bit trong gio trnh xc sut thng k. Ngoi phng php trn ta cn c phng php kim nh gi thuyt theo p-value xc sut ngha, phng php ny cng c gii thiu trong gio trnh xc sut-thng k.


39

Bc 1: Tnh *

i iiqs

i

t Se( )

;

Bc 2: Tnh p-value

p-value = i iqsP T t hoc i iqsT t

i iqs2P T t . Bc 3: So snh xc sut ngha vi mc ngha xc nh t trc, nu p-value th bc b 0H , cn nu p-value th chp nhn gi thuyt 0H .

Kim nh mt pha (phi) *

0 i i*

1 i i

H :

H :

i =1, 2

Bc 1: T mu s liu c c, thnh lp thng k

*i i

iqsi

t Se( )

;

Bc 2: T thng k , tnh xc sut ngha p-value = i iqsP T t . Bc 3: So snh xc sut ngha vi mc ngha xc nh t trc, nu p-value th bc b gi thuyt 0H , cn nu p-value th chp nhn gi

thuyt 0H .

Kim nh mt pha (tri)

*0 i i

*1 i i

H :

H :

i = 1, 2

Bc 1: Tnh*

i iiqs *

i

t

Se( )

;

Bc 2: Tnh p-value = iqs1 P T t . Bc 3: So snh xc sut ngha vi mc ngha xc nh t trc, nu p-value th bc b gi thuyt 0H , cn nu p-value th chp nhn gi

thuyt 0H .

V D 3.2

T v d 3.1 hy: a) Tm khong c lng cho cc h s hi quy vi tin cy 95%. b) Vi mc ngha 5% c th kt lun thu nhp ca b, m c nh hng ti kt qu hc

tp ca con ci hay khng? c) Tnh ESS, TSS.


40 STA301_Bi 3_v1.0013101214

Gii: Theo bo co ca Eviews cho v d 3.1 ta c:

a) Ta c cc gi tr c lng ca 1 2, l 1 2 4.785256, 0.042094 v sai s

chun l: 1 2 Se( ) 1.195385, Se( ) 0.017601. V c mu n = 8, vi mc tin

cy 0.05 , tra bng phn phi student ta c: (7)0.025t 2.364624 . Vy ta c cc

khong c lng cho 1 2, l:

1

1

4.785265 2.364624x1.195385; 4.786265 2.36462x1.195385

1.958629; 7.611901 .

Tng t ta c: 2 2.78634; 2.86693 .

b) Ta cn kim nh bi ton sau:

0 2

1 2

H : 0H : 0

Cch 1: Ta c gi tr tiu chun thng k ca bi ton trn l:

22

2

0.042094 t 0.0539 0.017601Se( )

.

Vi mc ngha 5%, tra bng phn phi student ta c: (7)0.025t 2.364624 .

Vy min bc b ca bi ton l: W = ; 2.364624 2.364624; . Ta thy gi tr tiu chun thng k 2t W , do cha bc b c H0. Nh vy c th kt lun thu nhp ca b m khng nh hng n kt qu hc tp ca con ci mt cch c ngha. Cch 2: Ta thy gi tr p- value = 0.0539 > 0.05 v vy cha th bc b c H0.


41

c) T kt qu trong bng ta c r2 = 0.488035, RSS = 8.155499, do theo cng thc

2 RSSr 1TSS

ta c : TSS = RSS/(1 r2) = 8.155499/ (1 0.488035) = 15.9288. ng thi ta li c cng thc: TSS = ESS + RSS, do ta c: ESS = TSS RSS = 15.9288 8.155499 = 7.774301.

3.6. Phn tch phng sai trong phng trnh hi quy

Trong phn ny chng ta xt bi ton kim nh gi thuyt v h s hi quy 2 theo mt phng php

khc, l phng php phn tch phng sai.

Ta xt bi ton kim nh 0 21 2

H : 0H : 0

(*)

Gi thuyt 0H ni ln rng bin X khng nh

hng ti Y, khi ta bc b gi thuyt 0H cng c ngha l ta bc b gi thuyt cho

rng bin X khng c nh hng ti bin Y.

Trong cc phn trc ta thy nu nh gi thuyt 0H l ng, tc l: 2 0 , th thng k

2

2 2

(n 2) RSS

c phn phi khi - bnh phng vi n 2 bc t do, cn thng k 2ESS

cng c c phn phi khi-bnh phng vi 1 bc t do. Mt khc hai thng k c lp vi nhau, vy thng k

2 2

22

ESS TSSr r n 21F RSS TSS 1 r 1(1 r )n 2 n 2

c phn phi Fisher vi s bc t do l: 1;n 2 . T , vi mc ngha cho trc, min bc b cho bi ton kim nh ang xt l W= f 1;n 2 ; . ngha: Cch tip cn theo hng phn tch phng sai nh trn cho php ta a ra cc phn on v ph hp ca m hnh hi quy ang xt. C th, nu thng k F c gi tr rt ln (ng vi xc sut ngha rt nh) th ta c th kt lun m hnh c lp ph hp vi s liu quan st. Cn nu thng k F c gi tr nh n mc xc sut ngha tng ng ca n ln hn mc ngha nh (bng 5% chng hn) th r rng m hnh l khng ph hp vi s liu, lc cn tm m hnh khc. Ta c bng phn tch phng sai ngn gn nh sau:


42 STA301_Bi 3_v1.0013101214

Ngun bin thin Tng bnh phng Bc t do Phng sai

X

n

2 2i

i 1

ESS x 1 ESS1

Phn d

n

2i

i 1

RSS u n 2

RSS

n 2

Tng TSS n 1

3.7. ng dng ca phn tch hi quy, bi ton d bo

Mt trong cc ng dng ca phn tch hi quy l d bo cho bit gi tr ca X l 0X , ta cn d bo gi tr ca Y l

0Y , khi thay gi tr 0X vo phng trnh hi quy mu

ta nhn c gi tr c lng ca Y l 0Y tha mn

phng trnh: 0 1 2 0 Y X .

Gi tr thc 0Y tha mn phng trnh 0 1 2 0 0Y X u , vi 0u l sai s.

Ta c : 0 0 1 1 2 2 0 0 Y Y ( ) ( )X u .

ng thi

1 1 2 2 E( ) ;E( ) v 0E(u ) 0.

Do : 0 0 0 0 E(Y Y ) 0 E(Y ) Y .

Vy c lng 0Y l mt c lng khng chch ca 0Y .

Ngoi ra, phng sai ca 0 0Y Y c tnh theo

0 0 1 1 2 2 0 0 Var(Y Y ) Var[( ) ( )X u ]

21 1 0 2 2 0 1 1 2 2 0

Var( ) (X ) Var( ) 2X Cov( ; ) Var(u )

222 2 2 20

0xx xx xx

x1 X X2xn S S S

22 20

xx

(X X)11 Xn S

trong : n n n

2 2 2 2xx i i i

i 1 i 1 i 1

S X (X X) X n(X)

.

Do phng sai 2 cha bit, ta thay 2 bng c lng khng chch 2 .

Khi ta c thng k 0 00 0

Y Yt Se(Y Y )

c phn phi Student vi n 2 bc t do.

Vy vi mc ngha cho trc ta c khong c lng 0Y l:


43

NG DNG n 2 n 2

0 0 0 0 0 0 02 2

Y t Se(Y Y ) Y Y t Se(Y Y ) (3.21)

Cng thc (3.21) cho ta khong c lng v gi tr 0Y ca Y khi cho bit trc gi

tr 0X ca X.

Bi ton trn c th pht biu di mt dng tng ng khc nh sau (Bi ton d bo gi tr trung bnh): Cho trc gi tr 0X ca X, cn c lng gi tr trung bnh

ca Y khi 0X X , tc l c lng gi tr 0E(Y | X X ) .

Ta c:

0 1 2 0E(Y | X ) X ,

0 1 2 0 Y X .

T , kt hp vi (3.19) v (3.20), ta thy

0 0 1 1 2 2 0 Y E(Y | X ) ( ) ( )X

2

2 00 0

xx

(X X)1Var(Y E(Y | X ))n S

.

Do 2 cha bit, ta dng c lng 2 , dn n: 22

2 00 0

xx

(X X)1 Var(Y E(Y | X ))n S

.

K hiu: 0

2 0 0Y

S Var(Y E(Y | X )) ,

khi y thng k

0

0 0

Y

Y E(Y | X )tS

.

c phn phi Student vi n 2 bc t do. p dng kt qu trn, ta c th c lng gi tr trung bnh c iu kin

0E(Y | X ) bng biu thc sau:

NG DNG

0 0

n 2 n 2 0 0 0Y Y2 2

Y t S E(Y | X ) Y t S (3.22)


44 STA301_Bi 3_v1.0013101214

TM LC CUI BI

Phng php OLS

Gi s c 1 mu v 2 bin X v Y.

Ta cn c lng cc tham s trong m hnh PRF: i i i 1 2 i iY E Y | X u X u

tc l i tm cc h s trong m hnh: i i i i i i i Y X u Y u .

tng ca phng php OLS l tm 1 ng SRF sao cho cc gi tr c lng iY cng gn vi cc gi tr quan st Yi cng tt. V vy, ta i tm min cho hm sau:

n n

2 21 2 i i 1 2 i

i 1 i 1

f , u (Y X ) .

Nh vy phng php OLS s ti thiu ha tng bnh phng cc phn d: n

2i

i 1

RSS u min.

Ta c cng thc cho cc h s c lng l: 1 2 Y X ;

n

i ii 1

2 n2i

i 1

x y

x

vi i i i ix X X, y Y Y.

Cc h s c lng trong m hnh

H s 1 2 , c xc nh duy nht ng vi mt mu i iX , Y

1 2 , l cc c lng im ca 1 2, .

Cc gi thit c bn ca phng php OLS v cc khuyt tt tng ng ca m hnh Di y l cc gi thit cn lu : Gi thit 1: M hnh hi quy phi c dng tuyn tnh. Gi thit 2: Cc gi tr ca X c gi thit l phi ngu nhin v khng tng quan vi cc sai s ngu nhin, tc l :

i i i i i i

i i i i

CoV X ,u E X u E X E u

X E u X E u 0.

Gi thit 3: Trung bnh ca cc nhiu ngu nhin bng 0: E( iu /Xi) = 0.

Gi thit 4: Phng sai ca cc nhiu ngu nhin l khng i: 2i jVar u Var u . Ch : Gi thit 4 khng tho mn, ta ni c hin tng phng sai ca sai s thay i.

Gi thit 5: Khng c tng quan gia cc nhiu ngu nhin: i jCoV u ,u 0 . Ch : Gi thit 5 khng tho mn, ta ni c hin tng t tng quan. Gi thit 6: S quan st n phi ln hn tng s tham s trong m hnh.


45

nh l Gaus-Markov: Vi cc gi thit cho ca phng php bnh phng ti thiu tho mn, c lng bnh phng ti thiu l cc c lng tuyn tnh khng chch v c phng sai nh nht trong lp cc c lng tuyn tnh khng chch.

r2 o ph hp ca hm hi quy, gi tr ca r2 cho bit bao nhiu phn trm s bin thin ca bin Y c gii thch bi bin X hoc bi hm hi quy mu.

ngha khong tin cy:

KTC cho 1: n 2 n 21 1 a 1 1 a 12 2

t Se ; t Se

KTC cho 1 cho bit trung bnh ca Y thay i th no khi X = 0. KTC cho 2:

a a2 2n 2 n 22 2 2 2 2 t Se ; t Se KTC cho 2 cho bit trung bnh ca Y thay i th no khi bin X thay i 1 n v.

Kim nh gi thit: Trong m hnh E(Y/Xi) = 1 + 2Xi: Ta mun kim tra H0: j = j* (j = 1,2).

Kim nh Gt cho 1 = 1* cho bit trung bnh ca Y c bng 1* khi X = 0 hay khng. Kim nh Gt cho 2 = 2* cho bit tc thay i ca trung bnh ca Y khi bin X thay i 1 n v c bng 2* hay khng.

Phn tch phng sai kim nh v s ph hp ca m hnh. kim nh s ph hp ca m hnh hi quy tuyn tnh so vi s liu, ta c th tnh cc tng bnh phng sai s ESS, RSS v TSS, t xc nh thng k F c phn phi Fisher ri tin hnh kim nh gi thuyt i vi thng k .

D bo. T s liu mu, ta c lng c m hnh hi quy thc nghim, t c th d bo c gi tr ca bin ph thuc mi khi c mt gi tr mi ca bin c lp.


46 STA301_Bi 3_v1.0013101214

CU HI THNG GP

1. Ngoi phng php OLS th c phng php no khc c lng m hnh hi quy mu khng?

2. Trong phng php OLS, trong mi trng hp, ta u phi gii h phng trnh tm cc c lng ng khng?

3. Nu mt m hnh hi quy bi vi nhiu bin th vic dng phng php OLS c thun tin khng?

4. Khi c lng cc h s bng OLS, lm th no nh gi c cht lng ca chng?

5. Ti sao phi xem xt cc gi thit ca phng php OLS?

6. nh gi ph hp ca m hnh hi quy vi cc s liu ca mu, ta dng tiu ch no?

7. C nht thit phi xy dng c m hnh hi quy mu vi r2 phi ln?

8. Trong kim nh gi thit, vic dng phng php xc sut ngha (p-value) c th thay cho phng php kim nh thng thng hay khng?

CU HI TRC NGHIM

1. Cng thc no sau y th hin phng php bnh phng ti thiu (OLS)?

A. n n

i i ii 1 i 1

u Y Y min

B. n n

i i ii 1 i 1

u Y Y min

C. 2n n

2i i i

i 1 i 1

u Y Y min

D. 2n n

2i i i

i 1 i 1

u Y Y max

2. Cho m hnh hi quy: Y = 20 + 0.75X. Tnh gi tr phn d ti im X = 100, Y = 90 A. 5 B 5

C. 0 D. 15.

3. Bc t do trong kim nh t vi m hnh 2 bin v c 20 quan st l: A. 20 B. 22

C. 18 D. 2

4. R2 cho bit: A. Tng quan gia X v Y. B. S bin thin ca Y.

C. Hip phng sai gia X v Y. D. Phn bin thin ca Y c gi thch bi X

5. Cho m hnh vi TSS = 0.9243, RSS = 0.2137. Tm r2 A. 0.7688 B. 0.2312

C. 0.3007 D. 0

Economy & Finance

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