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8/3/2019 bi kinh t lng

1/18

Tiu lun Kinh t lng ti : M hnh hi quy bi

Chng I - Ni dung m hnh hi quy bi1 .Xy dng m hnh

1.1 .Gii thiuM hnh hi quy hai bin m chng ta nghin cu chng 3 thng khng kh nng gii thch hnh vi ca bin ph thuc. chng 3 chng ta ni tiu dng

ph thuc vo thu nhp kh dng, tuy nhin c nhiu yu t khc cng tc ngln tiu dng, v d tui, mc lc quan vo nn kinh t, ngh nghip Vth chng ta cn b sung thm bin gii thch(bin c lp) vo m hnh hi quy.M hnh vi mt bin ph thuc vi hai hoc nhiu bin c lp c gi l hiquy bi.Chng ta ch xem xt hi quy tuyn tnh bi vi m hnh tuyn tnh vi trongtham s, khng nht thit tuyn tnh trong bin s.M hnh hi quy bi cho tng th

ii,kki,33i,221i X...XXY +++++= (4.1)

Vi X2,i, X3,i,,Xk,i l gi tr cc bin c lp ng vi quan st i 2, 2, 3,, k l cc tham s ca hi quy i l sai s ca hi quyVi mt quan st i, chng ta xc nh gi tr k vng ca Yi

[ ] i,kki,33i,221 X...XXs'XYE ++++= (4.2)1.2. ngha ca tham sCc h s c gi l cc h s hi quy ring

m

X's

Xm

Y

=

(4.3)

ko lng tc ng ring phn ca bin Xm ln Y vi iu kin cc bin s khctrong m hnh khng i. C th hn nu cc bin khc trong m hnh khng i,gi tr k vng ca Y s tng m n v nu Xm tng 1 n v.1.3. Gi nh ca m hnhS dng cc gi nh ca m hnh hi quy hai bin, chng ta b sung thm ginh sau:

(1)Cc bin c lp ca m hnh khng c s ph thuc tuyn tnh hon ho,

ngha l khng th tm c b s thc ( 1, 2,..., k) sao cho0X...XX i,kki,33i,221 =++++ vi mi i.Gi nh ny cn c c pht biu l khng c s a cng tuyn hon hotrong m hnh.

(2) S quan st n phi ln hn s tham s cn c lng k.(3)Bin c lp Xi phi c s bin thin t quan st ny qua quan st khc hay

Var(Xi)>0.2.c lng tham s ca m hnh hi quy bi2.1.Hm hi quy mu v c lng tham s theo phng php bnh phngti thiu

Trong thc t chng ta thng ch c d liu t mu. T s liu mu chng ta clng hi quy tng th.

1 SVTH : Nguyn Th Hng KHT3


2/18


Hm hi quy mu

ii,kki,33i,221i eX...XXY +++++= (4.4)

i,kki,33i,221iiii X...XXYYYe ==

Vi cc m l c lng ca tham s m. Chng ta trng i m l c lngkhng chch ca m, hn na phi l mt c lng hiu qu. Vi mt s ginh cht ch nh mc 3.3.1 chng 3 v phn b sung 4.1, th phng phpti thiu tng bnh phng phn d cho kt qu c lng hiu qu m.Phng php bnh phng ti thiuChn 1, 2, , ksao cho

( )2n

1i

i,kki,33i,221i

n

1i

2

iX...XXYe

==

= (4.5)

t cc tiu.

iu kin cc tr ca (4.5)

( )

( )

( ) 0XX...XXY2e

...

0XX...XXY2e

0X...XXY2e

i,kn

1ii,KKi,33i,221i

k

n

1i

2

i

i,2

n

1ii,KKi,33i,221i

2

n

1i

2i

n

1ii,KKi,33i,221i

1

n

1i

2i

==

==

==

=

=

=

=

=

=

(4.6)

H phng trnh (4.6) c gi l h phng trnh chun ca hi quy mu (4.4).Cch gii h phng trnh (4.4) gn gng nht l dng ma trn. Do gii hn cachng trnh, bi ging ny khng trnh by thut ton ma trn m ch trnh bykt qu tnh ton cho hi quy bi n gin nht l hi quy ba bin vi hai bin clp. Mt s tnh cht ca hi quy ta thy c hi quy hai bin c lp c th pdng cho hi quy bi tng qut.2.2.c lng tham s cho m hnh hi quy ba binHm hi quy tng th

ii,33i,221i XXY +++= (4.7)Hm hi quy mu

ii,33i,221i eXXY +++= (4.8)Nhc li cc gi nh

(1)K vng ca sai s hi quy bng 0: ( ) 0X,XeE i,3i,2i =

(2)Khng t tng quan: 0e,ecov ji = , ij(3)Phng sai ng nht: ( ) 2ievar =

(4)Khng c tng quan gia sai s v tng Xm: ( ( 0X,ecovX,ecov i,3ii,2i ==

(5)Khng c s a cng tuyn hon ho gia X2 v X3.(6) Dng hm ca m hnh c xc nh mt cch ng n.



3/18


Vi cc gi nh ny, dng phng php bnh phng ti thiu ta nhn c clng cc h s nh sau.

33221 XXY = (4.10)

2n

1i

i,3i,2

n

1i

2n

1i

2

n

1i

i,3i,2

n

1i

i,3i

n

1i

2n

1i

i,2i

2

xxxx

xxxyxxy

i,3i,2

i,3

=

===

====(4.11)

2n

1i

i,3i,2

n

1i

2n

1i

2

n

1i

i,3i,2

n

1i

i,2i

n

1i

2n

1i

i,3i

3

xxxx

xxxyxxy

i,3i,2

i,2

=

===

====(4.12)

2.3. Phn phi ca c lng tham s

Trong phn ny chng ta ch quan tm n phn phi ca cc h s c lng 2 v 3 . Hn na v s tng t trong cng thc xc nh cc h s c lng nn

chng ta ch kho st 2 . y ch trnh by kt qu1.

2 l mt c lng khng chch : ( ) 22E = (4.13)

( ) 22n

1ii,3i,2

n

1i

2i,3

n

1i

2i,2

n

1i

2i,3

2

xxxx

xvar

=

===

=(4.14)

Nhc li h s tng quan gia X2 v X3 :

=

==

=

n

1i

2i,3

n

1i

2i,2

n

1ii,3i,2

XX

xx

xx

r32

t32XX

r = r23 bin i i s (4.14) ta c

( )( )

2

223

n

1i

2i,2

2

r1x

1var

=

=

(4.15)

T cc biu thc (4.13) v (4.15) chng ta c th rt ra mt s kt lun nh sau:(1)Nu X2 v X3 c tng quan tuyn tnh hon ho th 223r =1. H qu l

2var v cng ln hay ta khng th xc nh c h s ca m hnh

hi quy.(2)Nu X2 v X3 khng tng quan tuyn tnh hon ho nhng c tng

quan tuyn tnh cao th c lng 2 vn khng chch nhng khng hiuqu.

Nhng nhn nh trn ng cho c hi quy nhiu hn ba bin.

3. 2R v 2R1



4/18


Nhc li khi nim v 2R :TSS

RSS1

TSS

ESSR2 ==

Mt m hnh c 2R ln th tng bnh phng sai s d bo nh hay ni cch khc ph hp ca m hnh i vi d liu cng ln. Tuy nhin mt tnh cht ctrng quan trng ca l n c xu hng tng khi s bin gii thch trong m hnhtng ln. Nu ch n thun chn tiu ch l chn m hnh c 2R cao, ngi ta cxu hng a rt nhiu bin c lp vo m hnh trong khi tc ng ring phnca cc bin a vo i vi bin ph thuc khng c ngha thng k. hiu chnh pht vic a thm bin vo m hnh, ngi ra a ra tr thng k

2R hiu chnh(Adjusted 2R )2

kn1n

)R1(1R 22

=

(4.16)

Vi n l s quan st v k l s h s cn c lng trong m hnh.Qua thao tc hiu chnh ny th ch nhng bin thc s lm tng kh nng giithch ca m hnh mi xng ng c a vo m hnh.

4. Kim nh mc ngha chung ca m hnhTrong hi quy bi, m hnh c cho l khng c sc mnh gii thch khi ton bcc h s hi quy ring phn u bng khng.Gi thitH0: 2 = 3 = = k= 0H1: Khng phi tt c cc h s ng thi bng khng.

Tr thng k kim nh H0:

)kn,1k(F~k)-(n

SSR

1)-(kSSE

F =

Quy tc quyt nh Nu Ftt > F(k-1,n-k, ) th bc b H0. Nu Ftt F(k-1,n-k, ) th khng th bc b H0.

5.Quan h gia R2 v F

)kn()R1(

)1k(R

)R1)(1k(

R)kn(

E1)(1k(

E)kn(

ETSS)(1k(

E)kn(E)kn(

)kn(RSS

)1k(E

F

2

2

2

2

=

=

=

=

=

=

SS/TSS)

SS/TSS

SS)

SS

1)RSS-(k

SSSS

6. c lng khong v kim nh gi thit thng k cho h s hi quyc lng phng sai ca sai s

kn

es

n

1i

2i

2

=

=

(4.17)

2



5/18


Ngi ta chng minh c 2s l c lng khng chch ca 2, hay (22sE = .

Nu cc sai s tun theo phn phi chun th 2 )kn(22

~s)kn(

.

K hiumm m s)

(e.s == . Ta c tr thng k )kn(mmm

t~)(e.s

c lng khong cho m vi mc ngha l)(e.st)(e.st m)2/1,kn(mmm)2/1,kn(m + (4.18)

Thng thng chng ta mun kim nh gi thit H 0 l bin Xm khng c tc ngring phn ln Y.H0 : m = 0H1 : m 0Quy tc quyt nh

Nu /t-stat/ > t(n-k, /2) th ta bc b H0. Nu /t-stat/ t(n-k, /2) th ta khng th bc b H0.

7. Bin phn loi (Bin gi-Dummy variable)Trong cc m hnh hi quy m chng ta kho st t u chng 3 n y uda trn bin c lp v bin ph thuc u l bin nh lng. Thc ra m hnhhi quy cho php s dng bin c lp v c bin ph thuc l bin nh tnh.Trong gii hn chng trnh chng ta ch xt bin ph thuc l bin nh lng.Trong phn ny chng ta kho st m hnh hi quy c bin nh tnh.i vi bin nh tnh ch c th phn lp, mt quan st ch c th ri vo mt lp.Mt s bin nh tnh c hai lp nh:

Bin nh tnh Lp 1 Lp 2Gii tnh N NamVng Thnh th Nng thnTn gio C KhngTt nghip i hc Cha

Bng 4.1. Bin nh phnNgi ta thng gn gi tr 1 cho mt lp v gi tr 0 cho lp cn li. V d ta khiu S l gii tnh vi S =1 nu l n v S = 0 nu l nam.Cc bin nh tnh c gn gi tr 0 v 1 nh trn c gi l bin gi(dummy

variable), bin nh phn, bin phn loi hay bin nh tnh.7.1. Hi quy vi mt bin nh lng v mt bin phn loiV d 4.1. v d ny chng ta hi quy tiu dng cho go theo quy m h c xemxt h thnh th hay nng thn.M hnh kinh t lng nh sau:Yi = 1 + 2X i+ 3Di + i(4.19)Y: Chi tiu cho go, ngn ng/nmX : Quy m h gia nh, ngiD: Bin phn loi, D = 1 nu h thnh th, bng D = 0 nu h nng thn.Chng ta mun xem xt xem c s khc bit trong tiu dng go gia thnh th vnng thn hay khng ng vi mt quy m h gia nh Xi xc nh.

i vi h nng thn[ ] i21iii X0D,XYE +== (4.20)



6/18


i vi h thnh th[ ] i231iii X)(1D,XYE ++== (4.21)

Vy s chnh lch trong tiu dng go gia thnh th v nng thn nh sau[ ] [ ] 3iiiiii 0D,XYE1D,XYE === (4.22)

S khc bit trong tiu dng go gia thnh th v nng thn ch c ngha thngk khi 3 khc khng c ngha thng k.Chng ta c phng trnh hi quy nh sauY = 187 + 508*X - 557*D (4.23)t-stat [0,5] [6,4] [-2,2]R2 hiu chnh = 0,61H s hi quy 557 3 = khc khng vi tin cy 95%. Vy chng ta khng th

bc b c s khc bit trong tiu dng go gia thnh th v nng thn.Chng ta s thy tc ng ca lm cho tung gc ca phung trnh hi quy cathnh th v nng thn sai bit nhau mt khong 3 = -557 ngn ng/nm. C thng vi mt quy m h gia nh th h thnh th tiu dng go t hn h nngthn 557 ngn ng/nm.Chng ta s thy iu ny mt cch trc quan qua thsau:

6000Hnh 4.1. Hi quy vi mt bin nh lng v mt bin phn loi.

7.2. Hi quy vi mt bin nh lng v mt bin phn loi c nhiu hn haiphn lpV d 4.2. Gi s chng ta mun c lng tin lng c quyt nh bi s nmkinh nghim cng tc v trnh hc vn nh th no.Gi Y : Tin lngX : S nm kinh nghimD: Hc vn. Gi s chng ta phn loi hc vn nh sau : cha tt nghip i hc,i hc v sau i hc.Phung n 1:Di = 0 nu cha tt nghip i hc

Di = 1 nu tt nghip i hcDi =2 nu c trnh sau i hc



7/18


Cch t bin ny a ra gi nh qu mnh l phn ng gp ca hc vn vo tinlng ca ngi c trnh sau i hc ln gp hai ln ng gp ca hc vn ivi ngi c trnh i hc. Mc tiu ca chng ta khi a ra bin D ch l phnloi nn ta khng chn phng n ny.Phng n 2: t b bin giD1iD2i Hc vn00 Cha i hc10 i hc01 Sau i hcM hnh hi quyYi = 1 + 2X + 3D1i + 4D2i + i(4.24)Khai trin ca m hnh (4.24) nh saui vi ngi cha tt nghip i hcE(Yi )= 1 + 2X (4.25)

i vi ngi c trnh i hcE(Yi )= ( 1 + 3)+ 2X3(4.26)i vi ngi c trnh sau i hcE(Yi )= ( 1 + 3+ 4 )+ 2X (4.27)7.3. Ci by ca bin giS lp ca bin phn loiS bin giTrong v d 4.1. 21Trong v d 4.232iu g xy ra nu chng ta xy dng s bin gi ng bng s phn lp?V d 4.3. Xt li v d 4.1.

Gi s chng ta t bin gi nh sauD1iD2iVng10Thnh th01Nng thnM hnh hi quy lYi = 1 + 2X i+ 3D1i + 4D2i + i(4.28)Chng ta hy xem kt qu hi quy bng Excel

CoefficientsStandard

Error t Stat P-value

Intercept 2235,533 0 65535 #NUM!

X 508,1297 80,369801436,322396 1,08E-06

D1 -2605,52 0 65535 #NUM!D2 -2048 0 65535 #NUM!

Kt qu hi quy rt bt thng v hon ton khng c ngha kinh t.L do l c s a cng tuyn hon ho gia D1, D2 v mt bin hng X2 =-1.D1i + D2i + X2 = 0 i.Hin tng a cng tuyn hon ho ny lm cho h phng trnh chun khng cli gii. Thc t sai s chun tin n v cng ch khng phi tin n 0 nh ktqu tnh ton ca Excel. Hin tng ny c gi l ci by ca bin gi.Quy tc: Nu mt bin phn loi c k lp th ch s dng (k-1) bin gi.



8/18


7.4. Hi quy vi nhiu bin phn loiV d 4.4. Tip tc v d 4.2. Chng ta mun kho st thm c s phn bit i xtrong mc lng gia nam v n hay khng.t thm bin v t li tn binGTi: Gii tnh, 0 cho n v 1 cho nam.TL : Tin lngKN: S nm kinh nghim lm vicH: Bng 1 nu tt nghip i hc v 0 cho cha tt nghip i hcSH: Bng 1 nu c trnh sau i hc v 0 cho cha.M hnh hi quy TLi = 1 + 2KNi + 3Hi + 4SHi + 5GTi+ i(4.29)Chng ta xt tin lng ca n c trnh sau i hcE(TLi /SH=1GT=0)= ( 1 + 4)+ 2KNi7.5. Bin tng tcXt li v d 4.1. Xt quan h gia tiu dng go v quy m h gia nh. cho

n gin trong trnh by chng ta s dng hm ton nh sau.Nng thn: Y = 1 + 1XThnh th: Y = 2 + 2XD : Bin phn loi, bng 1 nu h thnh th v bng 0 nu h nng thn.C bn trng hp c th xy ra nh sau

(1) 1= 2 v 1= 2, hay khng c s khc bit trong tiu dng go giathnh th v nng thn.

M hnh : Y = a + b XTrong 1= 2 = a v 1= 2 = b.

(2) 1 2 v 1= 2, hay c s khc bit v tung gc

M hnh: Y = a + bX + cDTrong 1 = a, 2 = a + c v 1 = 2 = b.

(3) 1= 2 v 1 2, hay c s khc bit v dcM hnh: Y = a + bX + c(DX)Trong DX = X nu nu D =1 v DX = 0 nu D = 0 1 = 2 = a , 1 = b v 2 = b + c.

(4) 1 2 v 1 2, hay c s khc bit hon ton v c tung gc v dc.

M hnh: Y = a + bX + cD + d(DX)

1 = a , 2 = a + c, 1 = b v 2 = b + d.



9/18


Hnh 4.2. Cc m hnh hi quyBin DX c xy dng nh trn c gi l bin tng tc. Tng qut nu Xp lmt bin nh lng v Dq l mt bin gi th XpDq l mt bin tng tc. Mt mhnh hi quy tuyn tng qut c th c nhiu bin nh lng, nhiu bin nh tnhv mt s bin tng tc.


Quy m h, Xa. M hnh ng nht

1=

2

1

=

2

Tiu dng go, Y

Tiu dng go, Y

Quy m h, Xb. M hnh song song

1

2

1

=

2

Quy m h, Xd. M hnh phn bit

Tiu dng go, Y

Tiu dng go, Y

1=2

1

2

1

2 1

1

2

Quy m h Xc. M hnh ng quy


10/18


Chng II - Phng php lp m hnh , phn tch v d bo hin tngkinh t bng Eviews

Bng di y cho cc gi tr quan st v thu nhp (Y-USD/u ngi ), t l laong nng nghip (X1 - %) v s nm trung bnh c o to i vi nhng

ngi trn 25 tui (X2 nm )Y X1 X26 9 88 10 138 8 117 7 107 10 1212 4 169 5 108 5 10

9 6 1210 8 1410 7 1211 4 169 9 1410 5 1011 8 12

Khi ng Eviews , t ca s Eviews chn File New Workfile

Hp thoi m Workfiel nh sau :



11/18


Trong Workfile Range ta chn Undated or irregular . Trong phn Range : Startdate nhp 1 v End date nhp 15 . Sau click Ok ca s mi s xut hin lWorkfile Untitled .

nhp d liu , t ca s Eview chn Quick/Empty group, mt ca s s xuthin vi tn Group: Untitled Workfile : Untitled. Sau nhp s liu ca 3 binY,X1,X2 vo

v lu tn l Group 1 .1). Tm hm hi quy tuyn tnh mu ca Y theo X1 v X2

T ca s Eviews chn Quick ri chn tip Estimate Equation .Sau khi nhpchut chn Estimate Equation , mn hnh s xut hin ca s EquationSpecification . Trong khung Equation Specification g Y c X1 X2 . G xonglnh ny th ca s nh sau :



12/18


nhp Ok , kt qu phn tch hi quy s xut hin nh sau :

T kt qu trn ta bit c cc h s hi quy :

0 = 6.20298 , 1 = -0.376164 , 2 = 0.452514t ca s Equation : Untitled Workfile : Untiled ta chn View Representations .Mt ca s mi xut hin l



13/18


Phng Trnh :^

iY =6.202979516 - 0.3761638734*X1 + 0.4525139665*X2

chnh l phng trnh hi quy tuyn tnh mu .

2). Tm c lng phng sai ca sai s ngu nhinT bng Equation : untitled Workfile Untitled ta cS.E. of regression 1.01126491828

y chnh l ( )se =1.011265. T y ta suy ra^

2 =2(1.011265) =1.0226569

3) c lng sai s chun ca cc h s hi quyT bng Equation : untitled Workfile :Untitled ta c :

Variable Std. Error C 1.86225321921

X1 0.132723756455X2 0.119511151331Suy ra : 0( )se = 1.862253 , 1( )se = 0.132724 , 2( )se = 0.1295114)Khong tin cy i xng ca cc h s hi quy vi tin cy 95% tm cn trn v cn di ca d bo khong cc h s hi quy . Ta vo excelsau g lnh: /2 ( )t n k = t0.025(12)= 2.178813

Ta c khong c lng ca 0 l :^ ^

0 /2 0( ) ( )t n k se =6.20298 2.178813*1.862253=(2.145478,10.26048)

Tng t ta c khong c lng ca 1 l (-0.6653441, -0.0869837) v 2 l(0.1921215,0.7129064)5)Kim nh cc gi thit 0 1 0 2: 0, : 0H H = = vi mc ngha 5%



14/18


TH1 : t gi thit : 0 1: 0H = , 1 1: 0H

vi mc ngha 5% th ta bc b gi thit 0 1: 0H = v Prob = 0.01505829972 Covariances Matrix . Nhp chut ta c kt qu sau :



15/18


10) Kim nh White (c phng sai thay i trong m hnh khng )Gi thit : 20 :H l hng s 21 :H khng phi l hng s thc hin kim nh White ,sau khi c lng hm hi quy tuyn tnh

mu t ca s Equation : EQ01 ta chn View/residual tests/*WhiteHeteroskedasticity (cross terms) .Sau khi nhp chut , bng kim nhWhite xut hin nh sau

vi mc ngha 5% ta c n 2R =2.404736 V p-value l 0.790769 ln hnnhiu so vi mc ngha =5% nn ta khng c c s cng nhn gi thit H1nnta chp nhn gi thit H0 tc l 2 l mt hng s ( ngha l ko c hin tng

phng sai thay i ).11)Kim nh phn phi chunGi thit : 0H : m hnh ang xt c phn phi chun 1H : m hnh ang xt khng c phn phi chun

View -> Residual Test -> Histogram-Normality Test



16/18


T bng trn ta c JB=3.545399 v xc sut p-value = 0.169874 kh ln nn takhng c c s cng nhn gi thit 1H nn ta tm thi chp nhn gi thit 0H .

Ngha l m hnh ta ang xt c phn phi chun .12) Kim nh BG ( c hin tng t tng quan bc 1 hay khng )gi thit : 0H : m hnh ang xt c hin tng t tng quan bc 1

1H : m hnh ang xt c hin tng tng quan bc 2Sau khi c m hnh hi quy tuyn tnh ta chn View/Residual Tests/CorrelationLM Test . Nhp chut mt ca s s xut hin :

Trong khung Lag to include ca ca s Lag Sprecification ta cn chn bc ttng quan .Trong m hnh ny ta chn bng 1 .Ri nhp Ok ,bng kt qu s xuthin :



17/18


Ta c n 2R =5.972927 , v c xc sut p-value l 0.050466 xp x bng mc ngha =5% nn ta ko c c s cng nhn gi thit 1H nn ta tm thi chp nhngi thit 0H . ngha l tn ti tng quan bc 1 .



18/18


Ti liu tham kho

1 . Gio trnh Kinh t lng , trng i hc Kinh t TP.HCM .

2 . Bi tp Kinh t lng , trng i hc Kinh t TPHCM .

3. Din n sinh vin i hc Kinh t TP.HCM http://ueh.vn

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