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LA CHN MHNH HI QUY
Chng 8
1. Cc tiu chun ca m hnh
Tnh tit kim :m hnh cng n gin cng tt
Tnh ng nht : cc tham s c lng l duy nht chocng mt tp hp s liu
Tnh thch hp :R2 v R2hiu chnh cng gn 1 cng tt
Tnh bn vng : m hnh phi da trn mt c s lthuyt no
C kh nng d bo tt : m hnh cho kt qu d bost vi thc t
2. Cch tip cn la chn m hnh
C hai hng tip cn
a.Xc nh s bin c lp
T n gin n tng qut :B sung bin c lpt t vo m hnh
T tng qut n n gin : u tin, xt m hnhy cc bin c lp c xc nh . Sau
tin hnh loi tr nhng bin khng quan trng rakhi m hnh
2. Cch tip cn la chn m hnh
Kim tra cc bnh ca m hnh
b.Kim tra m hnh c vi phm gi thit haykhng
a cng tuyn
Phng sai thay i
T tng quan
2. Cch tip cn la chn m hnh
Cn da vo
c.Chn dng hm
Cc l thuyt kinh t
th biu din
Cc kt qu thc nghim
2. Cch tip cn la chn m hnh
Gi tr ca hm hp l log-likelihood(L)
d.Mt s tiu chun khc
Gi tr ca L cng ln chng t m hnh cng phhp
22
2
1)2ln(
2ln
2i
Unn
L
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2. Cch tip cn la chn m hnh
Tiu chun AIC (Akaike info criterion)
d.Mt s tiu chun khc
Gi tr ca AIC cng nh chng t m hnh cngph hp
nk
en
RSSAIC
2
2. Cch tip cn la chn m hnh
Tiu chun Schwarz (Schwarz criterion)
d.Mt s tiu chun khc
Gi tr ca SC cng nh chng t m hnh cngph hp
nk
nn
RSSSC
2
2. Cch tip cn la chn m hnh
Nu ch n phc tp ca m hnh th thng ch n tiu chun SC
d.Mt s tiu chun khc
Nu xt s liu theo thi gian th thng dng tiu chunAIC
Lu l bin ph thuc xut hin trong m hnh phi
cng dng
Kt qu hi quy bng Eviews nh sau :
3. Cc sai lm thng gp khi chn m hnh
Gi s m hnh ng l :Yi = 1 + 2X2i+3X3i + Ui (a)
Nhng ta li chn m hnh :
Yi = 1 + 2X2i + Vi ( b)
hu qu :
a. B st bin thch hp
Gi s m hnh ng l :Yi = 1 + 2X2i + Ui (a)
Nhng ta li chn m hnh (c thm X3):
Yi = 1 + 2X2i + 2X3i + Vi (b)
hu qu :
3. Cc sai lm thng gp khi chn m hnh
b. Tha bin
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4. Pht hin nhng sai lm
Xt hm hi qui : Yi = 1+ 2X2i+ 3X3i+ 4X4i+ 5X5i + Uia. Pht hin tha
bin
- Trng hp nghi ngX5l bin thakim nhH0 : 5 = 0 (Kim nh bng cch
no?)
Nu chp nhn H0 X5khng cn thit. (C th sdng redundant test ca
Eviews)
Trng hp nghi ngX3 v X5l cc bin khng cnthitkim nh gi thit ng
thi
H0 : 3= 5 = 0
(S dng kim nh Wald)
Kim nh Wald cho m hnh sau .
P_value = 0,9688>0,05
chp nhn H0
Redundant variables Test
P_value = 0.0022 < 0,05
bc b H0, => Khngtha bin X2
Xt m hnh : Yi = 1 + 2Xi + Ui (*)
Gi s nghi ng m hnh b st bin Z kim tra bng cch :
- Nu c s liu ca Z :+ Hi qui m hnh Yi = 1+2Xi+3Zi +Ui+ Kim nh H0
: 3= 0. Nu bc b H0 th mhnh ban u b st bin Z.
- Hoc dng Omitted variable test
4. Pht hin nhng sai lmb. Kim nh cc bin b b st
Omitted variables Test
P_value = 0.8200>0,05
chp nhn H0, =>Khng b st bin X4
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Nu khng c s liu ca Z : dng kim nhRESET ca Ramsey.
Ramsey xut s dng lm xp x cho Zi.Bc 1: Hi qui m hnh (*), thu lyBc
2: Hi qui Yitheo cc bin c
lp trong (*) v (m hnhny gi l m hnh (new)) .
Bc 3: Kim nh H0: cc h s cang thi bng 0.
Nu bc b H0m hnh (*) b st bin.
Kim nh RESET ca Ramsey :
32
,
ii YY
iY
3
i
2
i Y,Y
3
i
2
i Y,Y
3
i
2
i Y,Y
Kim nh RESET ca Ramsey :
P_value = 0.0000
