Caohoc 17_tep1

Embed Size (px)

Citation preview

KINH T LNGCHNG TRNH NNG CAOGING VIN GIO TRNH:

: Nguyn Th Minh Kinh t lng chng trnh nng cao Nguyn Quang Dong- 2006

1

Ni dung mn hc Phn I: n phn KTL c bn: M hnh hi quy: c lng, kim nh v d bo Cc khuyt tt ca m hnh Mt s dng ca m hnh hi quy Phn II: Kinh t lng nng cao - mt s dng m hnh M hnh c gi tr tr ca bin ph thuc M hnh gm nhiu phng trnh M hnh c bin ph thuc l bin gi M hnh vi chui thi gian Phn III: Thc hnh my tnh nh gi: 40% kim tra trn my tnh/ Eviews + 60% thi vit tnh2

Phn I- M hnh kinh t lng c bn M hnh hi quy: c lng Kim nh D bo Cc khuyt tt ca m hnh Mt s dng hm hi quy

3

Gii thiu Nh kinh t: cung tin tng th lm pht tng (cc yu t khc khng i) Nh thng k: cung tin v lm pht c quan h tuyn tnh cht vi nhau( xu hng thay i rt ging nhau) Nh kinh t lng: khi cung tin tng 1% th lm pht tng 0.2% (khi cc yu t khc khng i) Tc ng ca vic tng cung tin ln lm pht? Tc ng ca vic tng chi tiu chnh ph ln tng trng kinh t? Tc ng ca vic tng gi ln doanh thu?, v.v

4

M hnh hi quy tuyn tnh Mc ch ca phn tch hi quy: Dng s liu quan st c lng nh hng ca cc bin s (bin c lp) ln gi tr trung bnh ca mt bin s no (bin ph thuc) T cc tham s c lng c: nh gi tc ng nh hng Thc hin cc d bo a ra cc khuyn ngh v chnh sch

5

M hnh hi quy tuyn tnh gii thiu V d: Q = Q ( Y, P) => hm hi quy tuyn tnh th hin quan h ny: Q = 1+ 2 Y+ 3 P + u, nu gi thit E(u) =0 => E(Q| Y, P) = 1+ 2 Y+ 3 P Nu bit chng hn 1 =10, 2 =0.6, 3 = -0.3 => Khi gi tng 1 n v => ? Khi thu nhp tng 1 n v =>? Khi Y =100, P =10 th =>? Chng ta mun bit cc j6

M hnh hi quy tuyn tnh gii thiu M hnh hi quy tng th dng tuyn tnh

Yi = 1 + 2 X 2i + 3 X 3i + .. + k X ki + u i

E (Y | X 2 ;.., X k ) = 1 + 2 X 2 + .. + k X k Cc thnh phn ca m hnh: Bin ph thuc Cc bin c lp H s chn H s gc, h s hi quy ring7

M hnh hi quy tuyn tnh gii thiu ngha ca cc h s hi quy H s chn H s gc Tuy nhin cc h s ny thng khng bit => cn c lng Hm hi quy mu: gi s c mu ngu nhin n quan st

Yi = 1 + 2 X 2i + 3 X 3i + .. + k X kic lng cho E(Y| Xj)

c lng cho cc j cha bit

8

M hnh hi quy tuyn tnh gii thiu Q: lm th no nhn c cc c lng tt Vit li hm hi quy mu:

Yi = 1 + 2 X 2i + 3 X 3i + .. + k X ki + ei => sai lch gia gi tr thc t v gi tr c lng l ei = Yi Yi

Tm ng hi quy mu m c: e12 + e22 +...en2 b nht => OLS

9

M hnh hi quy tuyn tnh c lng OLS M hnh hai bin => UL OLS l: 2 =

x y x2i 2 2i

i

; x 2i := ( X i X ); y i := (Yi Y )

2 var( 2 ) = 2 x 2i M hnh 3 bin => 2 =2

2 = ei2 /( n 2)

y i x 2 i x 3 i y i x 3 i x 2 i x 3i x22i x 3i ( x2 i x3i ) 22

2 var( 2 ) = 2 2 (1 r23 ) x 2i

Vic s dng cc c lng ny c u im g10

nh l Gauss-Markov nh l: Nu cc gi thit 1-6 c tha mn th: cc c lng nhn c t phng php OLS l: Tuyn tnh, khng chch* C phng sai nh nht trong lp cc UL KC Cc gi thit: 1. E(ui|X2i,...,Xki)=0: khng c sai s h thng 2. var(ui|X2i,...,Xki) = 2 vi mi i 3. cov(ui,uj)=0 vi mi i khc j 4. ui ~ N(0, 2) vi mi i 5. Khng c a cng tuyn hon ho gia cc bin Xj 6. Bin Xj l phi ngu nhin, nu l ngu nhin th phi c lp vi Ui 11

nh gi s b v hm hi quy c lng Vy nu cc gi thit trn tha mn th p/p OLS cho ta cc UL im tt nht cho cc tham s ca tng th Ngoi ra vi gi thit 6 v tnh chun ca u, ta bit c phn phi ca cc c lng Du ca cc h s c lng: c ph hp vi l thuyt kinh t khng? H s xc nh (h s xc nh bi): R2 , cho bit cc bin gii thch trong m hnh gii thch c bao nhiu phn trm s bin i ca bin ph thuc

12

V d minh ha Kt qu thu c t hm hi quy mc tng gi theo mc tng trong cung tin l nh sau:

p = 0.005 + 0.8m 10 gdp p,m v gdp: mc tng (%) trong gi, cung tin v GDP thc CH: con s 0.8 cho bit iu g? Khi tng cung tin 1%, liu mc tng (%) trong mc tng gi s l khong bao nhiu? => Bi ton tm khong tin cy Liu c thc s l khi tng cung tin th ga cng tng khng? => Bi ton kim nh gi thuyt thng k13

Bi ton xy dng KTC cho cc tham s Nu gi thit 6 cng c tha mn, khi cc KTC lKTC cho j

( j t / 2 , ( n k ) se( j ); j + t / 2,( n k ) se( j ))

KTC i xng KTC bn phi KTC bn tri

( ; j + t ,( n k ) se( j )) ( j t , ( n k ) se( j );+)

KTC cho 2

(n 2) 2 (n 2) 2 2 ( 2 ; 2 ); = ei2 /( n k ) / 2;n k 1 / 2;n kV d 1

14

V d (ch3bt3)Dependent Variable: Q Included observations: 20 Variable Coefficient Std. Error t-Statistic Prob. C 1373.24 171.41 8.01 0.00 P -113.42 32.03 -3.54 0.00 ADD 83.87 15.28 5.49 0.00 R-squared 0.74 Mean dependent var 460.20 Adjusted R-squared 0.71 S.D. dependent var 155.31 S.E. of regression 83.73 Akaike info criterion 11.83 Sum squared resid 119189.60 Schwarz criterion 11.98 Log likelihood -115.31 F-statistic 24.18 Durbin-Watson stat 1.94 Prob(F-statistic) 0.00

15

V d (ch3bt3) => Hm hi quy c ph hp vi l thuyt kinh t khng? Xt du ca h s c lng: ^2 = -135.70; ph hp vi ltkt ni rng.... Khi gi thay tng 1 n v th trung bnh Q thay i trong khong no? Tm KTC i xng cho 2 (- 135.7- t0.025,1732.03; -125.7+t0.025;1732.03) Khi ADD tng 1 n v th trung bnh Q tng ti a bao nhiu n v? Tm KTC bn phi cho 3: => 83.87+ t0.05;1715.28 = Cc bin ADD v P gii thch c bao nhiu % s thay i trong Q?16

Bi ton kim nh gi thuyt v tham sV d v cc gi thuyt mun kim nh: Cung tin khng nh hng n lm pht? Xu hng tiu dng cn bin 0 W = (t0.05;) = (1.66; ) 2 0 1.7 0 t qs = = = 1.1 Khng bc b H0 ) 1.5 se( 219

Bng tm tt v cp gt v min bc bLoi gi thit Hai phai

H0 i = i*

H1 i # i* i < i* i > i*

Bc b H0 khi | t |> t /2(n - k) t < - t (n - k) t > t (n - k)

Bn tri i = ( ) i* Bn phi i = ( i*)

t qs

2 * = se ( 2 )

20

Ghi ch Khi kim nh: H0: j = 0; H1: j # 0 th c 2 cch thc hin: Kim nh thng thng: dng t s t c gi tr P: nu P< th bc b H0 Khi khng ni r (mc ngha ca kim nh), hoc (1) ( tin cy dng khi tm KTC) th mc nh = 0.05

21

Kim nh F v s ph hp ca hm hi quy V s ph hp ca hm hi quy:

Y= 1+ 2TV+ 3IN +4P+ u Fqs = (R2/3) / [(1 R2) /(n -4)] Nu Fqs> f (3, n-4) => bc b H0 Cng thc chung:

n = 100; R2 = 0.68

H0: 2= 3= 4= 0; H1: c t nht 1 h s l khc 0Fqs = 68 > 3.1 Bc b H0

Nu Fqs = (R2/(k-1)) / [(1 R2) /(n -k)] >f (k-1, n-k) => bc b H0; trong k l s bin c mt trong m hnh

22

Kim nh hi quy c iu kin rng buc- kim nh F V d: Mun kim nh: c hai hnh thc qung co u khng c tc ng n li nhun H0: 2 = 0; 3 = 0 ; H1: c t nht 1 trong 2 h s ny khc 0 W = (f(m, n-k), ) = (f0.05(2,96), ) = (3.49, ) Thc hin hi quy thu hp: Y= 1+ 2P+ v, thu c R2th2 ( R 2 Rth ) / m (0.95 0.8) / 2 Fqs = = = 144 2 (1 R ) /( n k ) (1 0.95) / 96

Fqs thuc min bc b => bc b H0

23

Kim nh nh F (tip) V d ch3bt3: Hm hi quy c ph hp khng? c P v ADD u cng khng nh hng n Q? H0 : 2 = 3 = 0; H1: c t nht 1 h s khc 0 Kim nh F: c thng k F: Fqs = 24.18 > f0.05(2, 17) => bc b H0 c gi tr P ca thng k F: Gi tr P ca thng k F = 0.00< 0.05 => bc b H0

24

Bi ton d bo Tr li bi ton v mc tng gi (lm pht) Gi nh sang nm 2008: GDP tng 9%, cung tin tng 20% Khi mc tng gi (trung bnh) s l bao nhiu? Mc tng gi trung bnh s dao ng trong khong no? Mc tng gi (c bit) l bao nhiu? Mc tng gi c bit s dao ng trong khong no? Bi ton v d bo gi tr trung bnh v gi tr c bit

25

Thc hin d bo D bo bng c lng im D bo bng KTC gi tr trung bnh1 ( X 0 X ) 2 1/ 2 1 ( X 0 X ) 2 1/ 2 Yi t / 2 ( + ) < E (Y | X = X 0 < Yi + t / 2 ( + ) 2 2 n n xi xi

Gi tr c bit1 ( X 0 X ) 2 1/ 2 1 ( X 0 X ) 2 1/ 2 Yi t / 2 (1 + + ) < Y | X = X 0 < Yi + t / 2 (1 + + ) 2 2 n n xi xi

26

Tm tt ngha kinh t ca h s gc: Yi = 1 + 2 X 2i + 3 X 3i + .. + k X ki + u i Khi X2 tng 1 n v => Y tng 2 n v, khi cc bin khc khng i

ln(Yi ) = 1 + 2 ln( X 2i ) + .. + k ln( X ki ) + u iKhi X2 tng 1% th trung bnh ca Y tng 2 % n v ngha thng k ca h s gc: c khc 0 hay khng? ~ bin X tng ng c nh hng ln bin c lp khng27

V cc khuyt tt c th c ca m hnh- a

cng tuyn cao

- Phng sai ca sai s thay i - T tng quan - Dng hm sai - Tnh chun ca ssnn

28

a cng tuyn Khi nim v a cng tuyn: mi tng quan tuyn tnh gia cc bin gii thch trong m hnh CT hon ho CT khng hon ho - ch quan tm khi CT cao v d: gi du v CPI; gi tht ln v gi tht b; lao ng v vn ca doanh nghip Chng hn trong: Y= 1+ 2X2+ 3X3 + u ==> r23 cao? Y= 1+ 2X2+...+ kXk + u ==> tng quan tuyn tnh gia X2;...;Xk cao Lm sao pht hin: hi quy ph; ..v d: m Eviews29

a cng tuyn c lng OLS khi c hin tng a cng tuyn cao Vn l ULTTKC tt nht trong lp cc UL TTKC Tuy nhin n khng tt, nh sau: Xt m hnh hi quy 3 bin, khi :

2 var( 2 ) = 2 2 (1 r23 ) x 2iPhng sai ca cc UL ln => chnh xc thp KTC thng rng T s t thng nh => ? Du h s c lng c th sai .v.v30

Phng sai sai s thay i Khi nim: var(ui) = 2i Nguyn nhn: Mi quan h gia cc bin s Con ngi hc c t hnh vi trong qu kh, v.v. UL OLS khi PSSS thay i: Vn l UL tuyn tnh, khng chch, nhng khng hiu qu UL ca cc phng sai s chch Kim nh T, F mt hiu lc31

Kim nh White v PSSS thay i H0 : PSSS trong m hnh l khng i c lng m hnh gc thu c cc phn d et chy hm hi quy (trng hp c tch cho):2 e 2 = 1 + 2 X 2 + 3 X 3 + 4 X 2 + 5 X 32 + 6 X 2 X 3 + u

=> R2(1)

Nu: 2 nR 2 (1) > (k 1)

( R 2 (1)) /( k 1) F= (1 R 2 (1)) /( n k )k: s bin trong m.h 1

PSSS thay i

Tng t vi trng hp khng c tch cho V d (m eviews)32

Khc phc PSSS thay i nh dng ca phng sai thay Dng th d on dng ca phng sai Thc hin cc kim nh kim nh d on Cch khc phc: phng php bnh phng b nht tng qut (GLS): Bin i bin s m hnh mi c PPSS khng i c lng bng OLS m hnh mi ny, t suy ngc li h s cho m hnh gc V d: Y= 1+ 2TV+ 3IN +4P+ u nu PSSS c dng: var(ui) = aTV2, khi Y/TV= 1/TV+ 2+ 3IN/TV +4P/TV+ u/TV

Khi var(ui/TVi) = var(ui)/ TVi2

=a

= khng i

33

T tng quan Khi nim: cov(ui; uj) >< 0 vi i> t tng quan bc nht; AR(1) ut = 1ut-1 +..+ put-p+ vt => AR(p) v(t) l sai s ngu nhin, tha mn cc gi thit ca OLS. Hu qu khi c t tng quan: Vn l UL khng chch Phng sai c lng ca c lng

thng b chch

Cc kim nh T, F khng ng tin cy

2

cng l c lng chch =>34

Pht hin t tng quan Kim nh Durbin Watson, dng trong trng hp: AR (1) Khng c gi tr tr ca bin ph thuc l bin gii thch Khng mt quan stKhng chng c kt lun

TTQ dng 0 dL

Khng c TTQ

TTQ m 4-dL

dU

2

4-dU

435

Pht hin t tng quan Khi c gi tr tr ca bin ph thuc l bin gii thch: Durbin h kim nh B-G et = a1 + a2 Xt + 1et-1+..+ p et-p +vt => R2(1) et = a1 + a2 Xt + vvt Nu: => R2(2) hoc( R 2 (1) R 2 (2)) / p F= (1 R 2 (1)) /( n k *)

nR (1) > ( p )2 2

M hnh gc c TTQ bc p

v d Eviews36

T tng quan- khc phc Bin php khc phc: gi s TTQ c dng AR(1): ut = ut-1 +vt c lng h s t tng quan ri sau dng GLS da trn h s c lng ny, nh sau: t Y* = Y Y(-1); X* = X X(-1) Thc hin OLS hm hi quy theo bin mi: Y* = 1+ 2X* + v

37

nh dng m hnh Tha bin: => c lng OLS l khng chch, vng nhng khng hiu qu Kim nh tha bin Kim nh tha 1 bin: kim nh T Kim nh tha >= 2 bin: Kim nh F Thiu bin: => c lng OLS chch v khng vng Dng hm sai & thiu bin: Kim nh RESET Hi quy m hnh gc: Y = 1+ 2X+u , thu c c lng ca Yt v R2(1) Thc hin hi quy:

Yt = 1 + 2 X t + 3Yt 2 + .. + Yt m + ut

Thu c R2(2)38

nh dng m hnh (Tip) Nu Fqs = [(R2(2) R2(1)/(m-1)]/[ (1-R2(2))/(n-k(2)) ]> f (m, n-k(2)) Bc b H0, trong H0: hm nh dng ng Kim nh nhn t Lagrange (LM) Hi quy hm hi quy gc, thu c c lng ca Yt v R2(1) Thc hin hi quy: e = + X + Y 2 + .. + Y m + ut 1 2 t 3 t t t

Thu c R2(3). Nu nh dng sai

2 nR 2 > (m 1)

=> m hnh39

Tm tt Mc ch ca phn tch hi quy Phng php s dng UL m hnh hi quy tuyn tnh c in: OLS Cc kt qu c lng dng : Suy din v cc h s trong tng th T c cc ng dng thc t v chnh sch cc UL thu c c cc tnh cht tt, m hnh cn tha mn mt s gi thit c bn xt v 4 gi thit c bn M hnh vi bin gii thch l bin gi: xem gio trnh Tnh chun ca ssnn: xem gio trnh

40

Nhng ni dung chnh cn nh Hiu ngha kinh t/ ngha thng k ca cc h s c lng trong m hnh Hiu mt s thng k quan trng ca m hnh Nm c 4 loi khuyt tt c th c ca m hnh v hu qu Nm c cc phng php pht hin chnh/ cng thc ca cc thng k dng kim nh Hiu c cch khc phc ca tng loi khuyt tt

41

Phn II Kinh t lng nng cao

42

Chng I: M hnh t hi quy, m hnh tr phn phi v kim nh quan h nhn quYu cu: Nm c bn cht 2 loi m hnh Nm c phng php UL IV Nm c cch bin i m hnh c tr phn phi thnh m hnh t hi quy Nm c kim nh nhn qu

43

M hnh t hi quy v m hnh c tr phn phi M hnh t hi quy: L m hnh trong c t nht mt bin gii thch l gi tr tr ca bin ph thuc V d: Yt = a1 +a2Xt + a3Yt-1 + ut M hnh c tr phn phi: L m hnh trong c c gi tr hin ti v gi tr tr ca bin gii thch.Yt = a+b0 Xt+...+bk Xt-k+ ut Yt = a+b0Xt+...+bk Xt-k+..+ ut

m hnh c tr phn phi hu hn; k: chiu di ca tr

m hnh c tr phn phi v hn

b0 tc ng ngn hn, l tc ng tc th ca s ca X ln bin Y b0+...+bk+...= tc ng di hn ca X ln Y, l: ---44

u l m hnh ng: S liu theo thi gian Th hin tc ng tr gia cc bin s kinh t (chnh sch tin t v lm pht, cung-cu v gi,.v.v)

45

c lng m hnh c tr phn phi Gi s m hnh cn UL l: Yt = a+b0 Xt+...+bk Xt-k+..+ ut Phng php Alt and Tinbergen: Dng phng php OLS UL Yt theo Xt, thu c c lng ca b0 UL Yt theo Xt v Xt-1, thu c c lng ca b0 v b1;,v.v Dng qu trnh trn khi UL ca h s cui cng khng c ngha thng k, hoc du ca t nht mt h s UL thay i Nhc im ca phng php trn: Khng c nh hng ban u v chiu di ca tr Khi c lng cc tr k tip => s bc t do b gim i => cc suy din s thiu chnh xc Cc bin tr thng c tng quan cao=> vn v a cng tuyn Cn n cch tip cn khc => chuyn v dng m hnh t hi quy?46

Bin i m hnh c tr phn phi thnh m hnh t hi quy Mc ch: nhm UL cc tham s ca m hnh c tr phn phi tng: a ra cc gi nh v dng ca dy cc h s bj Dng gi nh ny chuyn m hnh v dng t hi quy Phng php Koyck: Xt m hnh c tr phn phi v hn: Yt = a+b0Xt +...+ bk Xt-k +..+ ut Gi nh: b0;b1;.. c cng du v: bk = b0 k vi 0 Cn phng php c lng mi c lng m hnh t hi quy51

Phng php bin cng c tng: Nhm gii quyt vn v s tng quan gia bin gii thch Yt-1 v sai s ngu nhin vt;bng cch thay th Yt-1 bng mt bin Zt c tnh cht: C cng tuyn cao vi bin Yt-1 Khng tng quan vi vt Bin nh vy c gi l bin cng c Thc hin: (Liviatan) chn Xt-1 lm bin cng c cho Yt-1 p dng OLS cho m hnh vi bin cng c ny52

Tr a thc Almon tng: L mt cch tip cn khc ca m hnh TPP, vi gi thit cc h s trong m hnh c th biu din c di dng a thc nh sau bi = a0 + a1i+a2i2 hoc bi = a0 + a1i+a2i2+...+arir Thc hin: dng php i bin s v sau p dng OLS V d: vi m hnh TPP c chiu di tr l 5: Yt = a+b0Xt +...+ b5 Xt-5 + utZ 0t = X t i = X t + .... + X t 50 5 0

Gi 5s r = 2. Php i bin c thc hin nh sau:Yt = a+ a0Z0t + a1Z1t + a2Z2t+ut

Z 1t = iX t i = X t + 2 X t 2 . + ... + 5 X t 5 Z 2t = i 2 X t i = X t + 4 X t 2 ... + 25 X t 50 5

53

Kim nh quan h nhn qu T phn tch hi quy ni chung khng suy ra c quan h nhn qu i vi hi quy theo chui thi gian, c th suy din c v quan h nhn qu Khi nim nhn qu Grange: X=>Y nu X gip d bo Y Y=> X nu Y gip d bo X X Y?

54

Kim nh quan h nhn qu (tip) Thc hin kim nh: H0: X khng gy ra Y; Ha: X gy ra Y Thc hin OLS: Yt = 0+ 1Yt-1+..+ mYt-m+ 1Xt-1+..+ mXt-m+ut (*) Yt = 0+ 1Yt-1+..+ mYt-m+ ut (**) Fqs = [ (R2*- R2**)/m]/[(1-R2*)/n-k] Nu Fqs> f(m, n-k) => bc b H0; X gy ra Y Tng t i vi: H0: Y khng gy ra X; Ha: Y gy ra X V d 2 (Eviews/ demo.wf1):55

Tm tt chng I Bin c lp c th c nh hng lu di n bin ph thuc => m hnh tr phn phi Yt = a+b0Xt +...+ bk Xt-k +..+ ut Mun c lng tc ng di hn v tc ng ngn hn Chuyn v m hnh t hi quy: Yt = a1 +a2Xt + a3Yt-1 + ut Khi : a2 tc ng ngn hn, a2/(1-a3) tc ng di hn ca X C 3 dng ca m hnh t hi quy Bin i Kyock: gi s v dng ca bi M hnh k vng hp l M hnh iu chnh ringvt : TTQ Vt: khng TTQ56

Tm tt chng I Nu m hnh t hi quy l m hnh hiu chnh ring th c th p dng c OLS c lng tc ng di hn v tc ng ngn hn Th no l m hnh hiu chnh ring? Y*t = a + bXt+cZt+ ut => qua qu trnh hiu chnh => Yt = a1 +a2Xt + a3Yt-1 + vt, trong vt khng t tng quan

57

Tip Nu m hnh t hi quy l m hnh k vng thch nghi hoc m hnh Kyock: OLS l khng thch hp v Yt-1 c tng quan vi ssnn=> dng phng php bin cng c Phng php BCC: tm mt bin thay th cho Yt-1 trong m hnh v UL OLS cho m hnh c thay th ny Trong gi ca Liviatan l: dng Xt-1 lm bin cng c cho Yt-1

58

M hnh nhiu phng trnh

59

Gii thiu

Y= f(X, Z,u)

Z=g(X,Y,v)

Gi thit OLS b vi phm=> Khng s dng c OLS

60

C ch lin h ngc Gii thiu: Trong m hnh c nhiu phng trnh gia cc bin s Gia cc bin ny c th c mi quan h qua li => C th tn ti tng quan gia cc bin gii thch vi sai s ngu nhin => Vi phm gi thit c bn ca OLS => ? V d1: m hnh cung - cu: quan h gia cung cu v gi ca mt loi hng ha QD = a1 +a2P + u1; QS =b1+b2P + u2 ; QD= QS (4.3)61

(4.1) (4.2)

Q: P v U2 c tng quan khng?

Gi s c c sc v cung=> ng cung dch chuyn, ko theo s thay i trong gi v sn lng P S P1 P0 D

Q1 Ban u th trng cn bng mc gi P0 v sn lng Q0

Q0

Q u2 thay i ko theo P thayi: c tng quan gia u2 v P62

V d 2: M hnh Keynes dng n gin: Ct = a1 + a2 Yt + ut Yt = Ct + It Ct v Yt c tc ng ln nhau Khi ut thay i => Ct thay i ==> Yt thay i. Ngha l ut v Yt c tng quan vi nhau: phng trnh (4.4) vi phm gi thit ca OLS. cc UL. OLS s l UL chch v khng vng63

(4.4) (4.5)

(4.4)

(4.5)

Tm tt: cc mi quan h 2 chiu gia cc bin s kinh t c th lm cho: M hnh khng xc nh c (hm cung/ hm cu?) M hnh xc nh c nhng vi phm gi thit ca phng php OLS v tnh khng tng quan gia bin gii thch v SSNN Khi cc UL thu c t OLS: Chch Khng vng Khi cn phi dng n cc phng php c lng khc => cn xem xt vn nh dng c cc phng php c lng tng ng64

nh dng Bin ni sinh: bin m gi tr ca n c xc nh t m hnh sinh Bin ngoi sinh: cc gi tr ca n c xc nh ngoi m sinh hnh (bao gm c bin tr ca bin ni sinh, ca bin ngoi sinh) Xt h M phng trnh

12Y2t + 13Y3t .. + 1M YMt + 11 X 1t + .. + 1K X Kt + u1t Y2t = 21Y1t + . 23Y3t . + 1M YMt + 21 X 1t + .. + 2 K X Kt + u 2t (4.6) YMt = M 1Y1t + M 2Y2t + M 3Y3t . + M ( M 1)Y( M 1)t + M 1 X 1t + .. + MK X Kt + u MtY1t =Cc phng trnh cu trc; cc phng trnh hnh vi, cc h s: h s cu trc65

Phng trnh rt gn: rt ra t phng trnh hnh vi, trong bin ni sinh f(bin ngoi sinh; ssnn)

Phng trnh rt gn cho v d 2: It bin ngoi sinh1 1 Yt = + It + w t 1 2 1- 2Ct = Yt = 1 + 2 It + wt 3 + 4 It + wt (4.7) (4.8)

1 2 Ct = + It + w t 1 2 1- 2

(4.7) v (4.8) l cc p.t rt gn ca (4.4) v (4.5) 2 v 4: cc nhn t ngn hn, th hin tc ng tc th cas thay i ca bin ngoi sinh It ln cc bin ni sinh (Ct v Yt) tng ng66

Nhn xt: Phng trnh cu trc (= p.t hnh vi): v phi c cha c bin ni sinh Phng trnh rt gn: v phi ch cha bin ngoi sinh OLS p dng c cho cc p.t rt gn, thu c cc j (Ti sao?) T c th suy ngc ra cc h s ca cc p.t cu trc Khi no th suy ngc ra c? Vn nh dng

67

nh dng P.t khng nh dng c: l phng trnh hnh vi m cc h s ca n khng suy ra c t cc h s ca h phng trnh rt gn V d: tr li v d v cung-cu QDt = 1 + 2 Pt + u1t QSt = 1 + 2Pt + u2t Qst = QDt Pt = 1+vt ; Qt = 2+wt; H phng trnh rt gn Bin ni sinh: Pt; Qt => 1 =(1-1)/(2-2) 2 =(12-2 1)/(2-2) t 1 v 2 khng th suy ra c 4 h s i v i68

Phng trnh nh dng c: l phng trnh hnh vi m cc h s ca n c th suy ra c t cc h s ca h phng trnh rt gn, v c chia lm 2 loi: Phng trnh nh dng ng: cc h s ca n c xc nh mt cch duy nht t cc h s ca h phng trnh rt gn Phng trnh v nh: cc h s ca n c xc nh mt cch khng duy nht t cc h s ca h phng trnh rt gn V d: QDt = 1 + 2 Pt + 3It+ u1t QSt = 1 + 2Pt + 3Pt-1+ u2t 1 + 2 Pt + 2It+ u1t = 1 + 2Pt + 3Pt-1+ u2t69

Bin ngoi sinh: It ; Pt-1 => h rt gn l: Pt = 1+ 2It+ 3Pt-1+ v1t Qt = 4+ 5It+ 6Pt-1+ v2t Trong : (*) 1= (1-1)/(2- 2); 2 = - 3/(2- 2); 3= 3/(2- 2) 4= (21-12)/(2- 2); 5 = - 3 2 /(2- 2); 6= 2 3/(2- 2) T h (*): cc h s cu trc c suy ra mt cch duy nht t cc h s rt gn => c 2 p.t cung/ cu u nh dng ng

70

Quy tc nh dng Gi M: s bin ni sinh; K: s bin ngoi sinh ca m hnh Xt phng trnh vi m bin ni sinh, k bin ngoi sinh. iu kin cn, iu kin p.t l nh dng c? iu kin cn: phng trnh ni trn l nh dng c th: K-k>=m-1 Khi K-k = m-1: phng trnh nh dng ng Khi K-k >m-1: phng trnh v nh V d: Qt= 1 + 2 Pt + u1t (1) Qt = 1 + 2Pt + u2t (2) M= 2; K = 0; m -1 = 1

P.t (1): m =2, k=0 => K-k=0< m-1=1=> khng nh dng c 71 P.t (2)?

V d: Qt = 1 + 2 Pt + 3It+ u1t QSt = 1 + 2Pt + u2t (3) (4) M = 2, K = 1

I: bin ngoi sinh; M =2; K = 1 p.t (3): k = 1; m=2, K-k = 0< m-1=1 => khng nh dng c p.t (4): k=0;m=2, K-k= m-1=1=> nu nh dng c th nh dng ng

72

iu kin cn v nh l: Trong m hnh c M phng trnh, mt p.t l nh dng c khi v ch khi tn ti t nht mt nh thc cp M-1 khc khng c xy dng t h s ca cc bin khng c trong p.t nhng c trong cc p.t khc ca m hnh Cch kim tra /k ca 1 p.t, chng hn p.t th j: Lp bng ma trn h s ca tt c M phng trnh, khng tnh h s t do Gch b cc ct m h s p.t j l khc khng Tm xem c tn ti nh thc cp (M-1) khc khng? iu kin trn gip xc nh 1 p.t l nh dng c hay khng. Vi p.t nh dng c, /k cn cho bit p.t nh dng ng hay v nh 73

V d: Y1t 10

12Y2t

Y2t 20 Y3t 30 21Y1t

13Y3t 11 X 1t . 23Y3t 21 X 1t 22 X 2t . 31 X 1t 32 X 2t

= u1t = u 2t = u 3t 41 X 3t = u 4t

Y4t 40 41Y1t 42Y2tpt 1 2 3 4 Y1 1 0 -31 -41 Y2 -12 1 0 -42 Y3 -12 -23 1 0 Y4 0 0 0 1 X1 X2 X3 0 Y4 0 0 1

p.t 1X2 -22 -32 0 X3 0 0 -43

-11 0

-21 -22 0 -31 -32 0 0 0 -43

Mi nh thc cp 3 (= M-1) bng 0 => p.t 1 khng nh dng c

74

Kim nh v s tng quan gia 1 bin gii thch v ssnn Nu khng tn ti tng quan, khi cc ULOLS s l UL vng v hiu qu. Nu c tn ti tng quan, ULOLS s chch v khng vng. Dng kim nh Hausman Kim nh Hausman c th hin nh sau: Xt m hnh Qt = 1 + 2 Pt + 3It+ 4Rt+ u1t Nghi ng Pt c tng quan vi u1t (v Qt = 1 + 2Pt + u2t Phng trnh rt gn c dng: Pt = 1+ 2It+ 3Rt+ v1t (5) c lng (5) bng OLS thu c P v v1t. c lng: Qt = 1 + 2Pt + 3v1t +u2t . Nu h s ca v1t 75 khc khng mt cch c ngha => c tng quan gia P )

c lng h phng trnh Nu kim nh Hausman cho thy c tng quan gia bin gii thch v ssnn => khng s dng c OLS Phng php thng c s dng: c lng ring l tng phng trnh (p/php thng tin khng y ) S trnh by cc p/p c lng cho 3 dng m hnh M hnh quy M hnh trong cc p/t l nh dng ng M hnh trong c cc p/t l v nh

76

M hnh quy- OLS Xt m hnh c dng quy nh sau: Y1t=10 + Y2t=20 + 21Y1t + Y3t=30 + 31Y1t+ 32Y2t + 11X1t+ 12X2t+u1t 21X1t+ 22X2t+u2t (4.9) (4.10)

31X1t+ 32X2t+u3t (4.11)

Cc sai s u1, u2 v u3 l khng tng quan vi nhau Trong : Yi: bin ni sinh; Xi bin ngoi sinh Xt (4.9): khng c bin ni sinh v phi => OLS Xt (4.10): c bin ni sinh v phi, nhng cov(Y1t, u2t) = cov(u1t; u2t) = 0 ( ga thit) => OLS Tng t cho (4.11) =>OLS Nu m hnh c dng quy, c th dng OLS UL cho77 tng phng trnh

UL phng trnh nh dng ng, p/p bnh phng b nht gin tip (ILS) Nu m hnh nh dng ng => dng ILS V d 3 (eviews) Phng php ILS gm cc bc: B1: Tm h phng trnh rt gn B2: UL tng p.t rt gn bng OLS B3: Tm UL ca h s cu trc t cc h s UL ca cc p.t rt gnKhng p dng c nu phng trnh l v nh

78

UL phng trnh v nh, p/p bnh phng b nht 2 giai on (2SLS) Nu cc phng trnh trong m hnh l v nh => dng 2SLS hoc 3SLS V d 4 Phng php 2SLS gm 2 bc sau: c lng cc phng trnh rt gn, thu c Yi c lng cc phng trnh ban u, trong cc bin Yi v phi c thay bng cc UL ca n

79

2SLS- cc u im chnh D p dng C th p dng cho tng phng trnh ring r p dng c cho c phng trnh nh dng ng, khi kt qa trng vi kt qu thu c t ILS Cho bit cc lch chun ca cc c lng Cho ngay cc UL cho cc h s Tuy nhin ch nn dng trong trng hp mu ln

80

Tm tt chng Trong m hnh nhiu phng trnh, thng thng cc bin c gii thch trong cc pt l c quan h vi nhau, khi thng gy ra hin tng cc bin v phi c tng quan vi ssnn => khi OLS l khng ph hp Khi nu phng trnh l nh dng c th c th c lng thng qua h phng trnh rt gn Nu l nh dng ng: ILS: UL OLS h p.t rt gn ri tnh ngc li cho h s ca cc phng trnh hnh vi (pt gc) Nu l v nh: dng 2SLS, UL OLS p.t rt gn ri ly kt qu UL lm bin s cho p.t hnh vi c lng tip

81

V d 3 Xt m hnh: Ct = 1 + 2 Yt + ut Yt = Ct + It ; I: bin ngoi sinh Cu hi: nh dng phng trnh (1); (2)? Xt iu kin : p.t 1pt C 1 -1 Y I (1) (2) -2 1 0 1Tn ti ma trn cp 1x1 khc khng => (1)nh dng c

Xt iu kin cn cho p.t 1: K=1, k =0 => K-k = 1; m =2; m-1 = 1=> K-k =m-1 => nh dng ng => c th thc hin c ILS82

Thc hin ILS B1: Phng trnh rt gn cho (1): Ct = 1 + 2It +vt => B2: UL p.t rt gn thu c: CONS = 258.71 + 8.04*I Ngha l: c lng ca 1 l: 258.71; ca 2 l 8.04 B3: Tnh ngc li cho c lng ca p.t hnh vi: Ct = 1 + 2 Yt + ut ; Yt = Ct + It => Ct = 1 + 2 (Ct + It )+ut ; (1- 2)Ct = 1 + 2 It +ut Ct = 1/ (1- 2) + (2 /(1- 2)) It +ut /(1- 2) 1= 1/ (1- 2) ; 2= 2/ (1- 2); 2= 2/(1+ 2); 1= 1/(1+ 2); UL ca 1 = 258.71/(1+8.04); ca 2 = 8.04/(1+8.04) (nhiu khi phi UL ton b h phng trnh rt gn th mi tnh ngc c ra cc h s ban u, y ch trnh by mt trng hp minh ha)83

V d 4 M hnh: Rt = a1 + a2 Mt + a3 Yt + a4Mt-1 + u1t Yt = b1 + b2 Rt + b3It + u2t Bin ni sinh: Rt; Yt Kim tra nh dng: p.t (4) nh dng c dng v nh => khng dng ILS c Dng 2SLS Eviews ch10bt14 (3) (4)

84