Chuong 3

18

Chng 3

THIT LP M HNH D BO CHUI THI GIAN

Nh cp trn, vic thit lp m hnh d bo v kim tra tnh chnh xc

m hnh l cn thit v quan trng trong kinh doanh. V vy, chng ny ch yu

trnh by cc bc thit lp m hnh d bo trn c s thng k ton hc.

Thng thng, vic xy dng m hnh d bo tin hnh qua cc bc sau:

Bc 1: Nhn dng m hnh.

Bc 2: c lng cc tham s trong m hnh.

Bc 3: Kim nh s tn ti ca m hnh, tng tham s ng gp trong m

hnh.

Bc 4: Xc nh m hnh, p dng d liu vo kim tra tnh xc thc ca

m hnh.

3.1 Nhn dng m hnh

M hnh d liu chui thi gian c p dng trong lun ny l m hnh t hi qui

vi cc tr khc nhau c xc nh tng qut nh sau:

3.1.1 Nhn dng m hnh cho chui thi gian dng

bit m hnh t hi qui - AR d bo c ph hp khng, ngay c khi bit c

m hnh AR ph hp th vic xt m hnh t hi qui n tr th bao nhiu l va

cng khng phi l vic n gin, chng ta kho st nhng du hiu nhn dng m

hnh t hi qui ph hp l da vo th hin ca hm t tng quan ACF [1,3] v t

tng quan tng phn PACF [1,3] nh sau:

Khi ACF c dng gim nhanh dn (theo dng hnh sin hay hnh s m) v

PACF ch c duy nht mt h s tr 1 c ngha th m hnh AR(1) l m

hnh c chn tt nht (hnh 3.1a). Tuy nhin thc t do sai s trong d liu

),...,,( ,21 tptttt yyyfy

19

nn c khi ACF ca cc chui thi gian s khng tt theo dng m hon ho,

tc l n gim nhanh n 0 nhng cha tt hon ton. V PACF cng c th

c vi h s khc khng ngu nhin sau tr u tin (xem hnh 3.1b).

Hnh 3.1a Minh ha cho ACF v PACF l thuyt khi nhn dng AR(1).

Hnh 3.1b Minh ha cho ACF v PACF tnh hung thc t nhng vn cho nhn

dng AR(1).

Khi ACF gim theo hnh sin tt dn v PACF c chnh xc 2 nh nhn

tr 1 v 2 v tt ht v 0 sau tr 2 th l du hiu nhn dng ca m

hnh AR(2). Tuy nhin, s nhn dng cc qu trnh AR(2) qua cc ACF v

PACF khng phi lun lun n gin nh th, trong nhng tnh hung y

nu PACF c hai h s t tng quan ring phn u tin khc 0 r rt v cc

h s cn li khng khc 0 nhiu s lun l mt gi tt cho AR(2) (hnh

3.1c).

20

Hnh 3.1c Minh ha cho ACF v PACF l thuyt khi nhn dng AR(2.)

Nhn nh chung l hm t tng quan - ACF ca mt m hnh ph hp vi

dng AR(p) vi p>=2 s th hin mt dng suy gim theo dng hm m hay

hnh sin v PACF c nh nhn tr 1,2...p,...n v tt v 0 sau tr p.

3.1.2 Nhn dng m hnh cho chui thi gian khng dng

Nh trnh by phn trn, chui thi gian dng l chui thi gian khng bao

hm yu t xu th, hay ni khc i nhng gi tr ca n xoay quanh quanh gi tr

trung bnh ca chui. Nu chui thi gian gc c yu t xu th th chng ta chuyn

sang dng dng bng cch ly sai phn.

Sau khi ly sai phn cho chui dng, bc tip ta nhn dng m hnh da vo

hm t tng quan ACF v hm t tng quan tng phn PACF nh trn

nhn dng.

3.2 c lng cc tham s

3.2.1 c lng tham s vi phng php bnh phng cc tiu [9](Ordinary

Least Squares)

Vi phng php bnh phng cc tiu l phng php c bn c lng

cc h s hi qui ca m hnh theo chui thi gian trong lun vn ny.

T phng trnh hi qui theo chui thi gian tng qut (2.6),

n gin ta xy dng ma trn vi cc vector nh sau:

Vector b cha tt c cc tham s c lng (0, 1, , n).

Vector X l ma trn quan st vi nhng thi im trc (t-1,t-

2,). Mi dng ca ma trn cha hng s 1, lin kt vi 0, theo

sau l p quan st yt-1, yt-2.

21

Vector y l vector quan st ct hin ti.

Vector e l vector sai s c lng.

c lng cc h s hi qui bng phng php bnh phng cc tiu l mt m

hnh ton biu din gn ng i tng thc cn m phng. Gii php ny s c

nh gi sai s v c nhng kt lun mang ngha thng k,

Tng bnh phng cc phn t ca e (sai s) c vit di dng ma

trn nh sau:

Nguyn tc ca phng php bnh phng nh nht l tm c lng ca vector

thng s b l sao cho tng bnh phng sai s (3.1) nh nht. Theo phng php

ton hc, tm tr s nh nht ca tng bnh phng sai s, ta ly o hm (3.1)

theo b v cho o hm ny bng zero.

Minyy t

n

t

t

n

t

t

2^

11

2 )(

)()( XbyXbyee TT (3.1)

n

p

p

y

y

y

y .

.

.

.

.

2

1

p

b

.

.

.

.

1

0

pnnn

pp

pp

yyy

yyy

yyy

X

21

....

...........

21

11

1

1

...1

p

e

.

.

.

.

1

0

22

Nghim ca phng trnh ny chnh l c lng b ca theo phng

php bnh phng nh nht.

T (3.2) ta c:

Trong b l tr s tho mn iu kin o hm ca eTe = 0. Cc phng trnh

trung bnh trong (3.3) c gi l h phng trnh chun.

3.2.2 Cc tnh cht ca bnh phng cc tiu

Kt qu ca phng php bnh phng cc tiu l mt c lng khng chch ca

b tham s [1]. Trong m hnh hi qui tuyn tnh a bin. Tnh cht ny c th

gii thch khi tm c gi tr ca b v gii thch nh sau:

Phng trnh hi vi cc tr c lng c th vit thnh:

Trong bi l nhng c lng ca tham s m ng i, uc lng sai s m

ng c tnh cho mi thi on t

yXbXX TT )( (3.3)

ttt yy^

(3.2) 02

)(

XbXyX

ee TTT

yXXXb TT 1)( (3.4)

ktnttt yyyy ...22110^

(3.5)

23

kn

n

t

t

1

2

^2

T sai s c lng ta c th tnh c c lng phng sai ca sai s,

Sau ly cn bc 2 t sai s chun ca hi qui,

3.3 Kim nh

Phng php thng k tin hnh nh gi s ph hp ca m hnh l tnh

ton cc h s xc nh, dng thng k F [9] nh gi mc ngha tng qut ca

m hnh, tnh ton cc sai s chun ca c lng v nh gi ngha ca tng

bin c lp .

3.3.1 Kim nh ngha tng qut ca m hnh

xc nh m hnh thch hp vi d liu n mc no chng ta hnh

thnh gi thuyt kim nh:

H0 : j = 0 (1 = 2 =...= j =0)

H1 : j 0 (c t nht mt h s j 0)

Nu gi thuyt H0 trn l ng (tt c cc h s dc u ng thi bng 0)

th m hnh hi qui xy dng khng h c tc dng d on hay m t v bin

ph thuc.

Th tc kim nh cho H0: 1 = 2 = = k = 0 l ng, tnh F0 vi cng thc:

kn

n

t

t

1

2

24

(3.9)

Nu |F0| > F,k,n-k1[9]. T chi gi thuyt H0, tc l tm chp nhn H1 (m hnh c

ngha v mt thng k)

Tip theo c cc gi tr SSR, SSE, SST , chng ta thc hin tnh theo cng thc

sau:

Trong :

SSE(Sum of Square Error) : Tng bnh phng ca sai s.

SSR(Sum of Square Regression): Tng bnh phng ca cc phn d.

SST(Sum of Square Total ): Tng ca cc bnh phng.

Kim nh ny c m t tng qut trong bng phn tch phng sai [1,9] nh sau:

Ngun bin ng Tng bnh

phng

Bc t do Trung bnh bnh

phng

F0

Hi qui

Sai s / phn d

Tng

SSR

SSE

SST

k

n-k-1

n-1

MSR

MSE

MSR/MSE

Bng 3.1: phn tch phng sai

E

R

E

R

MS

MS

kn

SSk

SS

F

1

0 (3.6)

ERT

n

t

tT

n

tTT

R

n

t

tT

n

tT

T

n

t

tE

TTT

E

SSSSSS

yySSn

y

yXbSS

yySSn

y

yySS

yySSyXbyySS

2_

1

^1

2

2_

1

1

2

2^

1

)(

)(

)(

)(

)( (3.7)

(3.8)

(3.9)

25

Nhn nh nhng kt qu thng k h s xc nh chung R2 v h s xc

nh iu chnh - R2adj [1]:

Cch tnh hai h s xc nh ny u c ngha nh gi cht lng c

lng chung ca m hnh hay ni cch khc i l nh gi trnh gii thch ca

cc bin c lp cho i tng d bo. Tuy nhin, c mt s khc bit v mt

ngha ca hai h s xc nh ny. H s xc nh u tin s tng khi s bin gii

tch gia tng, do s khng c kh nng so snh cht lng hai m hnh c bin

c lp khc nhau. H s xc nh iu chnh s khng ph thuc vo bin gii

thch, do chng ta c th so snh v nh gi hai hay nhiu m hnh c bin gii

tch khc nhau.

- H s xc nh chung :

- H s xc nh iu chnh:

3.3.2 Kim nh tham s dc

Tham s dc m t s thay i ca y khi bin c lp Xj (yt--1, yt-2,) thay i,

nhng tt c cc yu t khc Xj u gi nguyn. Mi tham s dc cng chnh l

h s hi quy ring v biu hin bng du v ln c lp vi cc h s khc.

Chng ta mong i s ng gp ngha t cc bin s c lp cho m hnh. S

ng gp ca tng bin s c lp s c m t thng qua s kim nh gi thuyt

nh sau:

n

t

tt

n

t

tt

T

E

yy

yy

SS

SSR

1

_2

1

^2

2

)(

)(

11 (3.10)

1

112

n

SSkn

SS

MS

MSR

T

E

T

EAdj

(3.11)

26

H0 : j = 0 (1 = 2 =... =j =0)


Mi mt tham s dc s c kim nh mt cch c lp. Mt phng trnh hi

qui s c k ln kim nh cho k tham s h s dc.

Nu ta tm chp nhn H0 : j = 0, th ch ra rng Xj (yt-1, yt-2.) khng c ng gp

trong m hnh. Kim nh thng k cho gi thuyt ny l kim nh t [1,9]vi,

Trong :

bj : l h s c lng ca tham s m ng j

Cjj: l phn t ng cho chnh ca ma trn (XTX)1 tng ng vi bj,

Kim nh t s so snh vi gi tr ti hn tra bng vi (n-k) l bc t do. Nu gi

thuyt ban u khng b t chi (H0 ng) cho mt hay nhiu h s th c th l

gii rng cc bin ny khng c ngha trong m hnh d bo.

Gi thuyt H0 : j = 0 b bc b nu |t0| > t/2, nk1 . Tc l loi gi thit H0 chp

nhn gi thuyt H1.

3.4. M hnh tm c

M hnh t hi qui bc mt c cp ch xt tng quan gia gi tr lin

nhau (yt,yt-1) trong chui thi gian

M hnh t hi qui bc 2 xem xt nh hng ca quan h gia gi tr 2 k trc

^2

0

jj

j

C

bt

.110 ttt yy

.22110 tttt yyy

27

Cng thc t hi qui trn chui thi gian tng qut xt trn quan h ca nhiu

k trc c trnh by nh sau:

Trong :

yt Gi tr bin c lp ti thi im t,

yt-i (i = 1, 2, ..., p) Gi tr bin c lp ti thi im t-i,

o, i (i=1,..., p) H s hi qui,

p bc ca m hnh t hi qui,

t Sai s.

p

i

titit yy

1

.0

28

3.5 V d cho bi ton chui thi gian

Gi s thit lp m hnh d bo chui thi gian vi d liu ngu nhin gm 15 quan

st nh sau:

t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

yt 1.89 2.46 3.23 3.95 4.56 5.07 5.62 6.16 6.26 6.56 6.98 7.36 7.53 7.84 8.09

Cc bc thit lp m hnh d bo c thc hin ny l:

Bc 1: Nhn dng cu trc m hnh.

- Xc nh cc h s ca hm ACF v PACF

Bng 3.2 cc gi tr hm ACF v PACF cho v d m hnh chui thi gian.

- V th theo cc h s trn quan st v nhn dng m hnh:

Hnh 3.2a th quan st ACF v d m hnh chui thi gian

Lag 1 2 3 4 5 6 7 8 9 10 11 12 13

ACF .79 .58 .38 .21 .05 -.08 -.18 -.27 -.34 -.39 -.40 -.37 -.29

PACF .79 -.13 -.08 -.10 -.09 -.08 -.08 -.10 -.12 -.08 -.04 .00 .02

29

tttt yyy 22110

Hnh 3.2b th quan st PACF v d m hnh chui thi gian

Theo quan st th trn th khi ACF c dng gim nhanh dn (theo dng hnh s

m) v PACF ch c duy nht mt h s tr 1 c ngha th m hnh AR(1) l m

hnh c chn tt nht nhng chng minh gi thuyt nhn dng trn, chng ta

gi nh rng m hnh c tr l 2 nh sau:

Bc 2: c lng tham s vi phng php bnh phng cc tiu

30

Xy dng ma trn t chui d liu trn:

M hnh c vit li li vi dng ma trn y=Xb+ ta c lng c cc tham

s theo phng php OLS, tm

yXXXb TT 1)(

p

.

.

.

.

1

0

7.537.841

7.367.531

6.987.361

6.566.981

6.266.561

6.166.261

5.626.161

5.075.621

4.565.071

3.954.561

3.233.951

2.463.231

1.892.461

X

8.09

7.84

7.53

7.36

6.98

6.56

6.26

6.16

5.62

5.07

4.56

3.95

3.23

y

2

1

0

b

31

Kt qu:

T cc h s c lng trn ta c phng trnh c lng hi qui

08.0

1.1

8.0

byXXXbTT 1)(

7.53 7.36 6.98 6.56 6.26 6.16 5.62 5.07 4.56 3.95 3.23 2.46 1.89

7.84 7.53 7.36 6.98 6.56 6.26 6.16 5.62 5.07 4.56 3.95 3.23 2.46

1 1 1 1 1 1 1 1 1 1 1 1 1

XX T

7.537.841

7.367.531

6.987.361

6.566.981

6.266.561

6.166.261

5.626.161

5.075.621

4.565.071

3.954.561

3.233.951

2.463.231

1.892.461

393.4997 420.8423 67.63

420.8423 451.3932 73.58

67.63 73.58 13

XX T

4.871867 5.30788- 4.697623

5.30788- 5.811533 5.28007-

4.697623 5.28007- 5.523661

)( 1XX T

7.53 7.36 6.98 6.56 6.26 6.16 5.62 5.07 4.56 3.95 3.23 2.46 1.89

7.84 7.53 7.36 6.98 6.56 6.26 6.16 5.62 5.07 4.56 3.95 3.23 2.46

1 1 1 1 1 1 1 1 1 1 1 1 1

yX T

8.09

7.84

7.53

7.36

6.98

6.56

6.26

6.16

5.62

5.07

4.56

3.95

3.23

446.1821

479.6185

79.21

32

Tnh ton cc sai s theo phng trnh c lng hi qui nh sau:

- c lng phng sai ca sai s :

- Sai s chun ca hi qui:

Bc 3: Kim nh

* Kim nh tng qut m hnh

- Cc sai s:

- H s xc nh chung:

-H s xc nh iu chnh:

21

^

08.08.01.1 ttt yyy

11262.001268.01

2

kn

en

t

t

127.0)( 2^

1

t

n

t

tE yySS

030.28)( 2

1

^

yySSn

t

tR

.157820.127030.28 ERT SSSSSS

995.0157.28

127.01

)(

)(

11

1

_2

1

^2

2

n

t

tt

n

t

tt

T

E

yy

yy

SS

SSR

012.011

126.0

213

)( 2^

1

2

^2

tt

n

t

t yy

kn

e

33

- Kim nh s tn ti m hnh :

t gi thuyt

H0 : = 0


Cho tin cy kim nh l 95%, ta c mc ngha = 5%. Vi n=13 v k=2. Tra

bng ca phn phi F, ta tm c gi tr ti hn l : F0.05,10,1= 3.81.

Tnh ton gi tr kim nh.

|F0|> F0.05,2,10= 4.10. Do , bc b gi thuyt H0, tc l tm chp nhn H1.Nh vy

m hnh c ngha v mt thng k.

* Kim nh tham s dc:

Kim nh gi thuyt

H0 : j = 0


Vi v d ny ta c 2 bin c lp trong m hnh. Do , j=1,2. Chng ta kim nh

t kim nh mi h s hi qui vi tin cy 95%, gi tr t tnh ton s c so

snh vi gi tr t ti hn t bng phn phi Student vi (n-k-1) bc t do v mc

ngha /2=0.025. Tra bng phn phi t/2,n-k-1= t0.025,10 = 2.228.

t gi thuyt

995.0

12157.28

11127.0

1

1

112

n

SSkn

SS

MS

MSR

T

E

T

EAdj

1105

10

127.02

030.28

1

0

kn

SSk

SS

FE

R

34

H0 : 0

H1 : 0 .

Tnh

Gi tr |t02| < t0,025,10 =2.2281. Nn ta chp nhn gi thuyt H0, tc l bin ph

thuc yt-2 khng ng gp trong m hnh ny.

Tng t ta kim tra

t gi thuyt

H0 : 0

H1 : 0

Tnh

Gi tr |t01|> t0,025,10 = 2.228. Nn t chi gi thuyt H0, tc l tm chp nhn gi

thuyt H1, bin ph thuc yt-1 c ng gp trong m hnh hi qui ny.

Bc 4: M hnh tm c v d bo gi tr

Qua cc bc tnh ton v kim nh trn ta tm c m hnh hi qui bc 1

nh sau

110 tt yy

0.335^

22

2

202

C

bt

2.964^

11

2

101

C

bt

35

- Dng phng php OLS nh tnh ton trn ta c m hnh hi qui c lng cho

vic d bo:

d bo cho k th 16 ta c:

1

^

91.09.0 tt yy

262.8362.79.091.09.0 1516

^

yy

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Chuong 3