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KINH T LNG
Gio trnh kinh t lng -Trng i hc Kinh t TP.H Ch Minh
Bi ging Kinh t lng -Trng i hc Kinh t quc dn -Nh xut bn thng k
Phn mm s dng: Eviews, Mfit, SPSS, STATA
Phng tin hc tp: my vi tnh, my tnh b ti
Bi ging in t
Ti liu tham kho
Cu trc hc phn
Chng 1: M u
Chng 2: M hnh hi quy hai bin
Chng 4: M hnh hi quy vi bin gi
Chng 6: T tng quan
Chng 5: Phng sai ca sai s thay i
Chng 7: a cng tuyn
Chng 8: Kim nh v la chn m hnh
Chng 3: M hnh hi quy nhiu bin
Hot ng ca sinh vin
* Ln lp y theo quy ch
* Thc hin bi kim tra iu kin
* Lm ti tho lun theo nhm
* T nghin cu mt s vn trong mn hc
* Thc hnh phn mm Eviews theo hng dn
Knh lin lc gia ging vin v sinh vin
a ch: [email protected]
KIN THC B TR
1. TON CAO CP
2. L THUYT XC SUT V THNG K TON
TON CAO CP
1.1 MA TRN
Cc khi nim v ma trn: vung, i xng, n
v, chuyn v, phn b i s, nghch o, hng,
nh thc
Cc php ton v ma trn: cng, nhn 2 ma trn
Cch xc nh ma trn nghch o.
TON CAO CP
TON CAO CP
1.2 CC TR CA HM NHIU BIN
Bi ton: Cho hm s
y= f(x1, x2,,xn)
Tm cc tr ca hm s trn nu c
TON CAO CP
1.2 CC TR CA HM NHIU BIN
Phng php:
Tm tt c cc o hm ring cp 1 fxi
Gii h {fxi =0} Tm im dng M(x1, x2,,xn)
Xt dng ton phng xc nh m (dng)
cc i ( cc tiu)
TON CAO CP
L THUYT XC SUT V
THNG K TON
2.1 I LNG NGU NHIN
K hiu: X, Y, Z,
Cc tham s c trng: E(X), Var(X), Se(X)
Cc quy lut phn phi xc sut quan trng:
N(,2) , 2(n), T(n), F(n1,n2)
2.1 I LNG NGU NHIN
Cc cng thc xc sut quan trng:
Nu T~T(n), th
(n)
P(T > t )
P(|T|< t(n)/2) = 1-
P(T t(n)/2) =
Cc cng thc xc sut quan trng:
Nu F~ F(n1,n2), th
1 2(n ,n )
P(F > f )
2.1 I LNG NGU NHIN
TNG QUAN GIA CC LNN
Hip phng sai ( Covariance)
Cov(X,Y)= 0: X,Y khng tng quan
Cov(X,Y)=Cov(Y,X)
Cov(X,X)=VAR(X)
Cov X,Y E X E X Y E Y
2.1 I LNG NGU NHIN
TNG QUAN GIA CC LNN
H s tng quan
Ch : ||1
-1 0 1
TQ m Khng TQ TQ dng
xy
Cov X,Y
Se(X).Se(Y)
2.1 I LNG NGU NHIN
TNG QUAN GIA CC LNN
Ma trn hip phng sai ca cc LNN
2.1 I LNG NGU NHIN
2.1 I LNG NGU NHIN
2.2 C LNG THAM S BNG KHONG T.C
Xy dng thng k
Tm khong tin cy ngu nhin
Trn mu, xc nh khong tin cy c th
Kt lun
2.1 I LNG NGU NHIN
2.3 KIM NH GI THUYT THNG K
PP truyn thng
Xy dng tiu chun kim nh
Tm min bc b H0
Trn mu, xc nh gi tr thc nghim
So snh gttn vi min bc b v kt lun
Kt lun
Phng php P-value
Tnh P-value
So snh P-value theo 2 trng hp sau:
2.3 KIM NH GI THUYT THNG K
0.050.01
Bc b H0
Cha c
s bc b H0
Bc b H0 (cha
vng chc)
Phng php P-value
a) Cha bit mc ngha
2.3 KIM NH GI THUYT THNG K
Phng php P-value
b) bit mc ngha
Bc b H0
Cha c
s bc b H0
2.3 KIM NH GI THUYT THNG K
CHNG 1: M U
1.1 TNG QUAN V KINH T LNG
1.2 CC KHI NIM C BN
1.1 TNG QUAN V KINH T LNG
1.1.1 Khi nim
1.1.2 Ni dung nghin cu
1.1.3 Qu trnh phn tch kinh t lng
1.1.4 ngha ca kinh t lng
1.1.1 Khi nim
Da trn c s ca cc mn khoa hc: Kinh t hc,
Thng k, Thng k ton v Ton hc v Tin hc.
Nhm: nh lng cc mi quan h kinh t;
d bo cc bin s kinh t; phn tch cc
chnh sch kinh t.
Econometrics- o lng kinh t
1.1.2 Ni dung nghin cu
Thit lp cc m hnh ton hc m t mi quan h
gia cc bin kinh t
o lng mc nh hng ca cc bin kinh t
ny n cc bin kinh t khc
Da vo cc m hnh ton hc d bo cc hin
tng kinh t
1.1.3 Qu trnh phn tch kinh t lng
Nu gi
thit
D bo v
xut
chnh sch
KGT v
MH
L cc
tham s
ca MH
Thit lp
MH ton
Thu thp
s liu
1.1.4 ngha
Kim chng c l thuyt kinh t c ph hp
hay khng ra quyt nh ng n trong hot
ng kinh doanh.
Trang b mt phng php lng ha cc mi
quan h kinh t.
1.2 CC KHI NIM C BN
1.2.1 Phn tch hi quy
1.2.2 M hnh hi quy tng th v MHHQ mu
1.2.3 Sai s ngu nhin
1.2.1 Phn tch hi quy
Phn tch hi quy l nghin cu s ph thuc ca
mt bin Y( bin ph thuc) vo mt hay
nhiu bin Xj khc ( bin c lp)
Gi thit: Y l bin ngu nhin, Xj l cc bin
phi ngu nhin.
Phn tch hi quy nhm:
c lng gi tr ca bin ph thuc khi bit
gi tr ca cc bin c lp
Kim nh gi thuyt v s ph thuc
D bo gi tr trung bnh v c bit ca bin
ph thuc.
1.2.1 Phn tch hi quy
1.2.2 M hnh hi quy tng th v
m hnh hi quy mu
( 1 )
M hnh hi quy tng th (hm tng th - PRF)
l hm c dng tng qut
ji 2i 3i ki2E(Y / X = X ) = f(X ,X ,...X )k
j j
Y l bin ph thuc
X2 , X3,.., Xk l cc bin c lp
k = 2: (1) l MHHQ n
k > 2: (1) l MHHQ bi
1.2.2 M hnh hi quy tng th v
m hnh hi quy mu
( 2 )
M hnh hi quy mu (hm mu - SRF) l hm
c dng tng qut
2i 3i kiY = f(X ,X ,...X )
ji 2E(Y / X = X )k
j jY l c lng im ca Yi v
f l hm c lng ca hm f
1.2.2 M hnh hi quy tng th v
m hnh hi quy mu
1.2.3 Sai s ngu nhin v phn d
( 3 ) ji 2U = Y -E(Y / X = X )k
i i j j
Sai s ngu nhin ( nhiu ngu nhin) ca hm hi
quy tng th:
1.2.3 Sai s ngu nhin v phn d
Phn d: ( 3 )
1.2.3 Sai s ngu nhin
Khi hm hi quy tng th (1) c th biu din
di dng
2i 3i ki= f(X ,X ,...X ) + Ui iY
Chng 3: M HNH HI QUY
TUYN TNH NHIU BIN
3.1. Cc khi nim c bn v MHHQ TT nhiu bin
3.2. Xy dng MHHQ mu bng PPBPNN
3.3 . c lng v KGT v cc h s hi quy tng th
3.4. H s xc nh bi v KGT ng thi
3.5. Phn tch hi quy v d bo
3.1 Cc khi nim c bn v MHHQ
tuyn tnh nhiu bin
3.1.1. MHHQ TT tng th v mu
3.1.2. Phn d
3.1.3. Dng ma trn ca MHHQ
3.1.1 MHHQ tng th v MHHQ mu
a) MHHQ tng th
i 1 2 2i k ki iY X ... X U
(3.1)
Hay
kj ji 1 2 2i k kij 2E Y X X X ... X
(3.1*)
Xt k bin Y , X2 , X3, , Xk
* ngha ca cc h s hi quy
Khi tt c cc bin c lp u nhn gi tr bng
0, th gi tr trung bnh ca bin ph thuc Y l
Khi bin Xj tng ln 1 v, cc bin c lp cn
li khng i, th gi tr trung bnh ca bin ph
thuc Y tng ln (gim xung) v
1 :
:j
j
3.1.1 MHHQ tuyn tnh tng th v mu
b) MHHQ mu
1 2 ki 2i kiY X ... X (3.2)
3.1.1 MHHQ tuyn tnh tng th v mu
3.1.2 Phn d
i i ie Y Y (3.3)
3.1.3 Dng ma trn ca MHHQ
Y
x
u
3.1.3 Dng ma trn ca MHHQ
Dng ma trn ca (3.1*) l
Dng ma trn ca (3.2) l
(3.4)
(3.5)
(3.6)
3.2 Xy dng MHHQ mu bng
phng php bnh phng nh nht
3.2.1. Cc gi thit c bn ca MHHQ
3.2.2. Phng php bnh phng nh nht (OLS)
3.2.3. Cc tnh cht ca cc c lng bnh phng
nh nht(LBPNN)
3.2.1Cc gi thit c bn ca MHHQ
jX j 1,k
iE U 0 i 1,n
Gi thit 1:
~ Ma trn X hon ton xc nh
Gi thit 2:
l cc bin s
3.2.1Cc gi thit c bn ca MHHQ
2iVar U i 1,2,...,n
Gi thit 6:
Gi thit 5:
Gi thit 3:
1TX X
rg(X) = k
2iU ~ N 0;
Gi thit 4: i jCov U ,U 0 i 1 j
3.2.2 Phng php bnh phng nh nht
a) Khi nim
L phng php xy dng MHHQ mu sao cho tng
bnh phng cc phn d t gi tr nh nht
2
1 2 k i , ,..., : e min
3.2.2 Phng php bnh phng nh nht
b) Cng thc xc nh cc h s hi quy mu
(3.7)
T TX X, X Y l hai ma trn c s
2i 3i ki
2
2i 2i 2i 3i 2i ki
T 2
3i 3i 2i 3i 3i ki
2
ki ki 2i ki 3i ki k k
n X X ... X
X X X X ... X X
X X X X X X ... X X
... ... ... ... ...
X X X X X ... X
i
i 2iT
i ki k 1
Y
YXX Y
...
YX
3.2.2 Phng php bnh phng nh nht
Cc h s hi quy mu c xc nh trong cng
thc (3.7) gi l cc c lng bnh phng nh
nht ca
3.2.2 Phng php bnh phng nh nht
Ch :
Trong trng hp MH c 3 bin Y, X, Z, xc
nh hai ma trn c s, ta cn tm 9 tng sau:
2
i i i i
2
i i i i
2
i i i i
Y Y X Z
X X YX
Z Z YZ
3.2.2 Phng php bnh phng nh nht
V D 3.1
Yi 84 90 92 96 100 108 120 126 130 136
Xi 8 9 10 9 10 12 13 14 14 15
Zi 9 8 8 7 7 8 7 7 6 6
V D 3.1
Y: doanh s bn ra trong mt thng (triu ng)
X: chi ph dnh cho qung co trong mt thng (triu
ng)
Z: gi bn 1 sn phm (ngn ng)
V D 3.1
1 2 3
i i i Y X Z
BI TP 3.1 Bng PPBPNN, xy dng hm hi
quy mu di dng sau:
Nu ngha ca cc h s hi quy mu va tm
c
3.2.3 Tnh cht ca cc LBPNN
Tnh cht 1:
Tnh cht 2: i1 Y = Y = Yn
i
1Y = Y
n j ji
1X = X (j = 2,k)
n
Tnh cht 3:
Tnh cht 5:
ie 0
i ie Y 0
n
i jii 1
e X 0
Tnh cht 4:
3.2.3 Tnh cht ca cc LBPNN
Tnh cht 6 (nh l Gauss Markov):
Tuyn tnh:
Khng chch
Phng sai nh nht
n
j ji i
i=1
= t Y
j jE( ) =
' 'j j j
'
j j :E( )=
Var( ) = min Var( )
3.2.3 Tnh cht ca cc LBPNN
3.3 c lng v KGT v
cc h s hi quy tng th
3.3.1. Ma trn hip phng sai ca cc h s hi quy
mu
3.3.2. c lng cc h s hi quy tng th
3.3.3. Kim nh gi thuyt v cc h s hi quy tng
th
3.3.1. Ma trn hip phng sai ca
cc h s hi quy mu
a) nh ngha:
1 1 2 1 k
2 1 2 2 k
k 1 k 2 k
Var( ) cov( , ) ... cov( , )
cov( , ) Var( ) ... cov( , )cov() =... ... ... ...
cov( , ) cov( , ) ... Var( )
b) nh l:
(3.8)
H qu:
cjj l phn t th j nm trn ng cho
chnh ca ma trn (XTX)-1
(3.9)
3.3.1. Ma trn hip phng sai ca
cc h s hi quy mu
Ch :
1)2
2i
e =
n - k
2
ie2) Cng thc tnh
2 T T T
ie = Y Y- X Y
(3.10)
(3.11)
3.3.1. Ma trn hip phng sai ca
cc h s hi quy mu
3.3.1. Ma trn hip phng sai ca
cc h s hi quy mu
BI TP 3.2 Lp cng thc tnh trong
trng hp m hnh c 3 bin Y, X, Z
2
ie
3.3.2. c lng cc h s hi quy tng th
Bi ton: Vi tin cy =1- , hy c lng j
Gii
Chn thng k j j
j
-T = ~ T(n -k) (j =1,k)
se( )
Vi tin cy =1-, xc nh phn v n k
2
t
n k
2
P T t
3.3.2. c lng cc h s hi quy tng th
Khong tin cy ca j
(3.12)
3.3.2. c lng cc h s hi quy tng th
Trn mu, tnh cc gi tr ca
Kt lun
V D 3.2
BI TP 3.3 S dng s liu trong v d 3.1, hy xc
nh khong tin cy 95% ca 2
3.3.3. Kim nh gi thuyt v
cc h s hi quy tng th
Bi ton: Vi mc ngha .Kim nh gi thuyt
v j theo mt trong 3 bi ton sau:
0
0 j j
0
1 j j
H :
H :
0
0 j j
0
1 j j
H :
H :
0
0 j j
0
1 j j
H :
H :
Bi ton 2:
Bi ton 3:
Bi ton 1:
Chn tiu chun kim nh:
T~ T n kNu H0 ng th
Vi mc ngha
3.3.3. Kim nh gi thuyt v
cc h s hi quy tng th
Bi ton 1: xc nh phn v n k
2
t
Ta c: n k
2
P T t
n k
2
W t : t t
3.3.3. Kim nh gi thuyt v
cc h s hi quy tng th
Min bc b H0:
Bi ton 2: xc nh phn v n kt
Ta c: n kP T t
n k W t :t t
3.3.3. Kim nh gi thuyt v
cc h s hi quy tng th
Min bc b H0:
Bi ton 3: xc nh phn v n kt
Ta c: n kP T t
n k W t :t t
3.3.3. Kim nh gi thuyt v
cc h s hi quy tng th
Min bc b H0:
Tnh gi tr thc nghim:
So snh t vi W kt lun theo quy tc kim nh
Kt lun chung
3.3.3. Kim nh gi thuyt v
cc h s hi quy tng th
V D 3.3
BI TP 3.4 S dng s liu trong v d 3.1,
a) Vi mc ngha 0,01 hy kim nh gi thuyt:
gi bn khng nh hng ti doanh s bn ra.
b) Vi mc ngha 0,05, liu c th ni rng khi
gi bn khng i, chi ph dnh cho qung co tng
ln 1 triu ng/1 thng, th doanh thu trung bnh
tng cao hn 6 triu hay khng?
3.4. H s xc nh bi v kim nh
gi thuyt ng thi
3.4.1 Cc k hiu
3.4.2 H s xc nh bi v h s xc nh bi
hiu chnh
3.4.2 Kim nh gi thuyt ng thi
1. TSS
2. ESS
3. RSS
2( )iY Y
2( )iY Y
2 2( )i i iY Y e
3.4.1 Cc k hiu
TSS= ESS + RSS (3.12)
22
iY nY
3.4.2 H s xc nh bi v h s xc
nh bi hiu chnh
1ESS RSS
TSS TSS R2 =
nh ngha :
(3.13)
a) H s xc nh bi
2
i2
22
i
eR =1-
Y - nY
Hay
ngha : o mc ph hp ca hm hi quy
3.4.2 H s xc nh bi v
h s xc nh bi hiu chnh
a) H s xc nh bi
Tnh cht
1. 0 R2 1
2. Nu k1< k2 th R12 R2
2
nh ngha : 2
2 n 1R 1 1 Rn k
(3.14)
3.4.2 H s xc nh bi v h s xc
nh bi hiu chnh
b) H s xc nh bi hiu chnh
ngha :
Dng quyt nh vic c nn a thm bin mi
vo m hnh hay khng
Vic a thm bin mi vo m hnh l cn thit
chng no cn tng, v h s gc ng vi bin
c ngha thng k
2
R
3.4.2 H s xc nh bi v
h s xc nh bi hiu chnh
b) H s xc nh bi hiu chnh
3.4.3. Kim nh gi thuyt ng thi
Bi ton: Vi mc ngha , kim nh gi thuyt
tt c cc bin c lp X2 ,X3,..,Xk u khng nh
hng ti bin ph thuc Y
),(:
...:
kjshmtnhttH
H
j
k
20
0
1
320
Bi ton:
2
0
2
1
: 0
: 0
H R
H R
3.4.3. Kim nh gi thuyt ng thi
- Tiu chun kim nh
2
2
R n -kF = .
1-R k -1
Nu gi thit H0 ng th F ~ F(k-1,n-k)
3.4.3. Kim nh gi thuyt ng thi
(3.15)
Tm min bc b H0 , vi mc ngha , xc nh
k 1,n kP F f k 1,n kf
W = f :f > f (k -1,n -k)
3.4.3. Kim nh gi thuyt ng thi
Tnh gi tr thc nghim:
So snh gi tr thc nghim f vi W
Kt lun chung
2
2
R n -kf = .
1-R k -1
3.4.3. Kim nh gi thuyt ng thi
V D 3.4
BI TP 3.5 S dng s liu trong v d 3.1, vi
mc ngha 0.01, hy kim nh gi thuyt: c hai
yu t chi ph dnh cho qung co v gi bn u
khng nh hng ti doanh s bn ra.
3.5. Phn tch hi quy v d bo
3.5.1 D bo gi tr trung bnh
3.5.1 D bo gi tr trung bnh
Gi s gi tr ca cc bin c lp X2 ,X3 ,,Xk
tng ng l X20 ,X30 ,, Xk0 . K hiu:
X0
20
30
k0
1
X
X
X
3.5. Phn tch hi quy v d bo
-12 T T0 0 0 Var Y X X X X (3.17)
(3.16)
-12 2 T T0 0 0 0 0 Var Y - Y = Var Y + 1+X X X X
(3.18)
3.5. Phn tch hi quy v d bo
Vi tin cy cho trc, hy d bo gi tr trung
bnh ca Y l
Bi ton
3.5.1 D bo gi tr trung bnh
2 20 3 30 k k0E Y X X ,X X ,...,X X
Hay E(Y/X0)
Vi tin cy xc nh phn v
0 00
0
Y - E Y XT = ~ T n - k
Se Y
1 n k
2
t.
Chn thng k
Ta c: n k
0
2
P T t
Gii
3.5.1 D bo gi tr trung bnh
0E Y X
n-k n-k0 0 0 02 2
Y - t Se Y ;Y + t Se Y
Vy khong tin cy ca l :
n-k n-k0 0 0 0 02 2
P Y - t Se Y < E Y X < Y + t Se Y =
2.5.1 D bo gi tr trung bnh
V D 3.5
BI TP 3.6 S dng s liu trong v d 3.1, vi
tin cy 98%, hy d bo doanh s bn ra
trung bnh trong mt thng ca cc ca hng c
chi ph dnh cho qung co l 10 triu ng/ 1
thng v gi bn l 8 ngn ng/ 1 sn phm.
Vi tin cy cho trc, hy d bo gi tr c
bit Y0 ca Y khi cc bin c lp X2 ,X3 ,,Xk
nhn cc gi tr tng ng l X20 ,X30 ,, Xk0
Bi ton
3.5.2 D bo gi tr c bit
Vi tin cy xc nh phn v
0 00
0 0
Y YT = ~ T n - k
Se Y Y
1 n k
2
t.
Chn thng k
Ta c: n k
0
2
P T t
Gi i
3.5.2 D bo gi tr c bit
n-k n-k0 0 0 0 0 02 2
Y - t Se Y Y ;Y + t Se Y Y
Vy khong tin cy ca Y0 l :
n-k n-k0 0 0 0 0 0 02 2
P Y - t Se Y Y < Y < Y + t Se Y Y =
3.5.2 D bo gi tr c bit
V D 3.6
BI TP 3.7 S dng s liu trong v d 3.1,
vi tin cy 98%, hy d bo doanh s bn ra
trong mt thng ca cc ca hng c chi ph
dnh cho qung co l 10 triu ng/ 1 thng v
gi bn l 8 ngn ng/ 1 sn phm.
BI TP CHNG 3
Y 2 2.1 2.4 2.6 2.9 3 3.2 3.5 3.6 3.8 4 4
X 1.2 1.5 1.6 1.7 1.8 2 2.2 2.3 2.5 2.3 2.5 2.8
Z 0.4 0.4 0.5 0.5 0.6 0.6 0.7 0.7 0.8 0.9 0.8 0.7
iY
iX
iZ
Yi : li nhun ca doanh nghip i (t VND/1 nm)
Zi :chi ph hot ng ca doanh nghip i (t VND/1
thng)
Xi : ngun vn huy ng ca doanh nghip i (t VND/1
nm)
BI TP CHNG 3
1 2 3
i i i Y X Z
Nu ngha ca cc h s hi quy mu va tm
c
BI TP CHNG 3
1. Bng PPBPNN, hy xy dng m hnh hi quy
mu c dng sau:
BI TP CHNG 3
2. Tm khong tin cy 98% ca 3
3. Vi mc ngha 1%, KGT ngun vn khng
nh hng ti li nhun ca DN
4. Vi mc ngha 5%, KGT c hai yu t
ngun vn v chi ph khng nh hng ti li
nhun ca DN
BI TP CHNG 3
5. Vi tin cy 95%, d bo li nhun trung
bnh ca DN khi ngun vn huy ng l 3 t
VN/ 1 nm, v chi ph hot ng l1.2 t VN/1
thng.
6. Kim tra kt qu bng Eviews
Chng 4:
M HNH HI QUY CHA BIN GI
4.1. Khi nim v bin gi
4.2 . Cc m hnh c cha bin gi
4.3. ng dng ca bin gi
4.1. Khi nim v bin gi
4.1.1 Bin s lng (nh lng):
4.1.2 Bin cht lng (nh tnh):
4.1.3 Bin gi:
Bin gi ch nhn hai gi tr : 0, 1
4.1. Khi nim v bin gi
4.1.4 K thut bin gi:
1/ Nu bin cht lng c m thuc tnh, lng ho
bin ny ta cn dng m-1 bin gi
3/ Thuc tnh m tt c cc bin gi u nhn gi tr
bng 0 gi l thuc tnh c s
2/ Mi bin gi, gn bng 1 vi mt thuc tnh v bng
0 nu khng c thuc tnh .
4.2 . Cc m hnh c cha bin gi
4.2.1 M hnh c mt bin c lp l bin cht
lng c hai thuc tnh
4.2.2 M hnh c mt bin c lp l bin cht
lng c nhiu hn hai thuc tnh
4.2.3 M hnh cha nhiu bin c lp u l
bin cht lng
4.2.4 M hnh hn hp
4.2.1 M hnh c mt bin c lp l
bin cht lng c hai thuc tnh
Z=1 nu l A
Z=0 nu l B
Xt mi quan h Y v bin cht lng c hai thuc
tnh l A v B.
Gi s MHHQ c dng:
(4.1) 1 2 iE Y / = + ZiZ Z
V d:
Y l nng sut ca x nghip (vsp/ 1 gi)
Z=1 nu s dng cng ngh sn xut l A
Z=0 nu s dng cng ngh sn xut l B
18 3,2i iY Z
4.2.1 M hnh c mt bin c lp l
bin cht lng c hai thuc tnh
4.2.2 M hnh c mt bin c lp l bin
cht lng c nhiu hn hai thuc tnh
Xt mi quan h: Y vi bin cht lng c m thuc
tnh (m>2)
lng ho bin cht lng trn cn dng m-1
bin gi: Z2, Z3,, Zm
Zj =1 nu bin cht lng c thuc tnh j
Zj =0 nu bin cht lng khng c thuc tnh j
Gi s MHHQ c dng:
1 2 2i 3 3i mi2E Y / = + Z + Z ...+ Zm
j ji mjZ Z
(4.2)
4.2.2 M hnh c mt bin c lp l bin
cht lng c nhiu hn hai thuc tnh
V d: iiii ZZZY 432 5.18.46.525
Trong Y l doanh s bn hng trong 1 ngy (triu
ng)
4.2.2 M hnh c mt bin c lp l bin
cht lng c nhiu hn hai thuc tnh
+) Z2 = 1 nu l ma xun
Z2 = 0 nu khng l ma xun
+) Z3 = 1 nu l ma h
Z3 = 0 nu khng l ma h
+) Z4 = 1 nu l ma thu
Z4 = 0 nu khng l ma thu
iiii ZZZY 432 5.18.46.525
4.2.3 M hnh c nhiu bin c lp u
l cc bin cht lng
Xt m hnh: Y, v k-1 bin c lp u l cc bin
cht lng.
Khi s bin gi cn s dng l:
Bin c lp th j c mj thuc tnh.
( 1)jm
4.2.3 M hnh c nhiu bin c lp u
l cc bin cht lng
V d:
iiiiii ZZZZZY 65432 5.32.25.18.16.55
Y : thu nhp 1 thng (triu ng)
Z2 = 1 nu l bc s
= 0 nu khng l bc s
Z3 = 1 nu l cng nhn
= 0 nu khng l CN
Z2 = Z3 = 0 nu l gio vin
iiiiii ZZZZZY 65432 5.32.25.18.16.55
Z4 = 1 nu c bng TC
= 0 nu khng c bng TC
Z5 = 1 nu c bng H
= 0 nu khng c bng H
Z4 = Z5 = Z6 = 0 nu c bng C
Z6 = 1 nu c bng sau H
= 0 nu khng c bng sau H
4.2.4 M hnh hn hp
cc bin c lp c c bin s lng v bin
cht lng
V d:iiii ZZXY 21 8.31.23.05.6
Y : mc chi tiu ca mi h gia nh 1 thng
(triu ng)
X: thu nhp ca mi h gia nh trong 1 thng
(triu ng)
iiii ZZXY 21 8.31.23.05.6
Z1 = 1 nu nng thn
= 0 nu khng nng thn
Z2 = 1 nu min ni
= 0 nu khng min ni
Z1 = Z2 = 0 nu thnh ph
4.3. ng dng ca bin gi
4.3.1. Phn tch ma
4.3.2. So snh 2 hi quy
4.3.3. Hi quy tuyn tnh tng khc
4.3.1. Phn tch ma
Dng bin gi a yu t ma (thi gian) vo
m hnh hi quy
4.3.2. So snh 2 hi quy
Bi ton:
Trn mu s liu 1:
Trn mu s liu 2:
i 1 2 i iY = + X + U (1)
i 1 2 i iY = + X + U (2)
? So snh (1) v (2)
Z=1 nu s liu thuc mu 1
=0 nu s liu thuc mu 2
i 1 2 i 3 i 4 i i iY = + X + Z + X Z +U (4.3)
Xt m hnh:
4.3.2. So snh 2 hi quy
i 1 2 i 3 i 4 i i iY = + X + Z + X Z +U
Nu quan st i thuc mu s liu 1, th (4.3):
1 2 i= + Xi iY U
1 3 2 4 iY = ( + )+( + )Xi iU (1)
(2)
Nu quan st i thuc mu s liu 2, th (4.3):
(4.3)
4.3.2. So snh 2 hi quy
(1) (2) 3 = 4 =0
Trong m hnh (4.1), KGT:
H0 :3 = 4 =0
Nu bc b H0 th hai hi quy khc nhau
(Dng kim nh Wald)
4.3.2. So snh 2 hi quy
4.3.3. Hi quy tuyn tnh tng khc
Bi ton:
Gi s c chui thi gian (s liu theo thi gian)
v hai bin Y v X
? ng hi quy c b gp khc khi i qua t0 hay khng
t0 : thi im chuyn i
4.3.3. Hi quy tuyn tnh tng khc
O Xt
E(Y/Xt)
Xt0
4.3.3. Hi quy tuyn tnh tng khc
Xt bin gi Z
0
t
0
0 t tZ =
1 t > t
0t 1 2 t 3 t t t tY = + X + (X -X )Z +U
Xt MHHQ
(4.4)
4.3.3. Hi quy tuyn tnh tng khc
1 2 tY = + Xt tU
01 3 t 2 3 t
- X + + Xt tY U
0t 1 2 t 3 t t t tY = + X + (X -X )Z +U (4.4)
Trc thi im t0 , (4.4):
Sau thi im t0 , (4.4):
ng hi quy (4.4) khng b gp khc khi qua t0:
3 =0
Trong m hnh (4.4), KGT:
H0 :3 =0
Nu bc b H0 th ng hi quy b gp khc
4.3.3. Hi quy tuyn tnh tng khc
Yi 9 8 9 12 11 10 13 11 13 14 16 18
Xi 16 15 15 14 14 13 13 12 12 12 11 10
Zi 1 0 0 1 1 0 1 0 0 1 0 1
Y: Doanh s bn ra trong mt ngy (triu ng)
X: Gi bn (ngn ng/ . v)
Z= 0: Ca hng thnh ph,
Z= 1: Ca hng nng thn
BI TP CHNG 4
1 2 3
i i i Y X Z
Nu ngha ca cc h s hi quy mu va tm
c
BI TP CHNG 4
1. Bng PPBPNN, hy xy dng m hnh hi quy
mu c dng sau:
2. Tm khong tin cy 98% ca 2
3. Vi mc ngha 1%, KGT a im khng
nh hng ti doanh s bn ra
4. Vi mc ngha 5%, KGT c hai yu t gi
bn v a im u khng nh hng ti doanh
s bn ra.
BI TP CHNG 4
BI TP CHNG 4
5. Vi tin cy 95%, d bo doanh s bn ra ca
nhng ca hng thnh ph c gi bn l 14 ngn
ng/ 1 sn phm.
6. Kim tra kt qu bng Eviews
CC KHUYT TT CA MHHQ
Chng 5. Phng sai ca sai s (PSSS) thay i
Chng 6. T tng quan (TTQ)
Chng 7. a cng tuyn
5.1. Bn cht, nguyn nhn, hu qu ca hin tng
5.2. Pht hin hin tng
5.3. Khc phc hin tng ( T c sch)
Chng 5
Phng sai ca sai s (PSSS) thay i
a) Bn cht:
Vi phm gi thit Var(Ui)=2 (i)
Tc l: Var(Ui) = i2
5.1. Bn cht, nguyn nhn, hu qu
ca hin tng
b) Nguyn nhn:
-Do bn cht ca s liu
- Do m hnh thiu bin quan trng hoc dng hm sai
5.1. Bn cht, nguyn nhn, hu qu
ca hin tng
c) Hu qu:
-Cc c lng OLS vn l l khng chch, nhng
phng sai ca chng l l chch
- Kt qu ca bi ton c lng v kim nh gi
thuyt v cc h s hi quy khng cn ng tin cy.
5.1. Bn cht, nguyn nhn, hu qu
ca hin tng
1.Phng php th:
- V th phn d ei hoc ei2 theo chiu tng ca
mt bin Xj no .
- Nu vi cc gi tr khc nhau ca Xj, rng ca di
th thay i th c th ni m hnh xy ra hin tng
PSSS thay i
5.2. Pht hin hin tng
2. Kim nh Park:
gi s
5.2. Pht hin hin tng
2 2
2ln ln lni i iX v
2 2 2ln ln lni i ie X v
2. Kim nh Park:
- Hi quy MH gc, thu c ei v
- Hi quy m hnh
nu MH gc c 2 bin
(5.1)
nu MH gc c nhiu bin
5.2. Pht hin hin tng
2 2 2ln ln lni i ie X v
2 2 2ln ln lnYii ie v
2. Kim nh Park:
- Kim nh gi thuyt H0: 2= 0 Nu H0 b bc b th
MH khng xy ra hin tng PSSS thay i.
5.2. Pht hin hin tng
3. Kim nh Glejser:
Tng t kim nh Park, ch khc MH bc 2 l
mt trong cc MH sau:
5.2. Pht hin hin tng
1 2i i ie X v
1 2
1i i
i
e vX
1 2i i ie X v
1 2
1i i
i
e vX
(5.2)
4. Kim nh White:
Ngi ta chng minh c rng:
Nu U khng tng quan vi cc bin c lp, bnh
phng ca cc bin c lp v tch cho gia cc bin
c lp. Th phng sai ca cc l OLS tim cn vi
phng sai ng, khi n ln
5.2. Pht hin hin tng
5.2. Pht hin hin tng
Dependent Variable: LOG(E^2)
Method: Least Squares
Date: 10/02/13 Time: 05:21
Sample (adjusted): 2 11
Included observations: 10 after adjustments Variable Coefficient Std. Error t-Statistic Prob. C 30.43261 10.58777 2.874316 0.0207
LOG(YMU) -6.499673 2.265440 -2.869056 0.0209 R-squared 0.507131 Mean dependent var 0.074239
Adjusted R-squared 0.445522 S.D. dependent var 1.570787
S.E. of regression 1.169660 Akaike info criterion 3.328159
Sum squared resid 10.94483 Schwarz criterion 3.388676
Log likelihood -14.64079 Hannan-Quinn criter. 3.261772
F-statistic 8.231482 Durbin-Watson stat 3.239316
Prob(F-statistic) 0.020859
= 5%, pht hin hin tng phng sai ca
sai s thay i
Dependent Variable: ABS(E)
Method: Least Squares
Date: 10/02/13 Time: 05:24
Sample (adjusted): 2 11
Included observations: 10 after adjustments Variable Coefficient Std. Error t-Statistic Prob. C -5.758154 3.963283 -1.452875 0.1843
1/SQR(X) 23.93605 13.08790 1.828869 0.1048 R-squared 0.294829 Mean dependent var 1.450000
Adjusted R-squared 0.206682 S.D. dependent var 1.479385
S.E. of regression 1.317664 Akaike info criterion 3.566455
Sum squared resid 13.88992 Schwarz criterion 3.626972
Log likelihood -15.83228 Hannan-Quinn criter. 3.500068
F-statistic 3.344762 Durbin-Watson stat 3.096740
Prob(F-statistic) 0.104819
= 5%, pht hin hin tng phng sai ca
sai s thay i
Heteroskedasticity Test: White F-statistic 2.559931 Prob. F(3,5) 0.1683
Obs*R-squared 5.451046 Prob. Chi-Square(3) 0.1416
Scaled explained SS 1.399115 Prob. Chi-Square(3) 0.7057
Test Equation:
Dependent Variable: RESID^2
Method: Least Squares
Date: 10/02/13 Time: 05:27
Sample (adjusted): 2 10
Included observations: 9 after adjustments Variable Coefficient Std. Error t-Statistic Prob. C 621948.9 891962.7 0.697281 0.5167
(SQR(X))^2 -1991.700 642003.9 -0.003102 0.9976
(ABS(X-Z))^2 240.5713 6369.040 0.037772 0.9713
(LOG(X^2))^2 -22216.80 340723.5 -0.065205 0.9505
= 5%, pht hin hin tng phng sai ca
sai s thay i
5.3. Khc phc hin tng
Bi ton: Gi s MH gc Yi = 1+ 2Xi+Ui
C xy ra hin tng PSSS thay i. Khc phc
hin tng trn
5.3. Khc phc hin tng
Trng hp 1: 2i bit
? Bin i MH gc v MH no? Ti sao
? Dng phng php no c lng MH sau
khi bin i
? Trng s trong phng php trn l g? Trong
thc t c th thay trng s trn bng gi tr ca
bin no
5.3. Khc phc hin tng
Trng hp 2: 2i cha bit
? C th p dng nhng gi thit no khc phc
? Nu cch khc phc tng ng vi mi gi thit?
Gii thch ti sao
6.1. Bn cht, nguyn nhn, hu qu ca hin tng
6.2. Pht hin hin tng
6.3. Khc phc hin tng (T c sch)
Chng 6. T tng quan
6.1. Bn cht, nguyn nhn, hu qu
ca hin tng
a) Bn cht:
Vi phm gi thit :
Cov(Ui,Uj)=E(UiUj)=0 (ij)
Tc l: Cov(Ui,Uj) 0
a) Bn cht:
T tng quan bc 1:
(6.1) AR(1)
T tng quan bc p:
(6.2) AR(p)
ttt UU 1
tptpttt UUUU ...2211
6.1. Bn cht, nguyn nhn, hu qu
ca hin tng
b) Nguyn nhn:
- Tnh qun tnh ca cc L kinh t theo thi gian
- Hin tng mng nhn
- Tnh tr ca cc bin kinh t
- Phng php thu thp v x l s liu
- Sai lm khi chn m hnh
6.1. Bn cht, nguyn nhn, hu qu
ca hin tng
6.1. Bn cht, nguyn nhn, hu qu
ca hin tng
6.2. Pht hin hin tng
1. th phn d:
- V th phn d theo quan st ( theo thi gian)
- Nu th tun theo mt quy lut no th kt
lun c t tng quan
2. Kim nh d Durbin- Watson:
Bi ton: Pht hin t tng quan bc 1 trong MH
Phng php:
- Hi quy MH gc, thu c
(5.5)
n
t
t
n
t
tt
e
ee
d
1
2
2
2
1
6.2. Pht hin hin tng
2. Kim nh d Durbin- Watson:
Phng php:
- Da vo 3 thng s: n, k=k-1, , tra bng xc
nh dU v dL v biu din trn trc s
- Xc nh khong cha d, v kt lun theo quy tc
kim nh
6.2. Pht hin hin tng
2. Kim nh d Durbin- Watson:
Phng php:
0 dl du 2 4-du 4-dl 4
TTQ dng Khng x Khng c TTQ Khng x TTQ m
6.2. Pht hin hin tng
3. Kim nh B-G (Breush- Godfrey):
Bi ton: Pht hin t tng quan bc p trong MH
(*)
Phng php:
- Hi quy MH gc (*), thu c ei
- Hi quy m hnh
(6.3)
i 1 2 2i k ki iY X ... X U
i 1 2 2i k ki 1 i 1 p i p ie X ... X e ... e V
6.2. Pht hin hin tng
3. Kim nh B-G (Breush- Godfrey):
Phng php:
- Kim nh gi thuyt H0: 1= 2== p=0
Tiu chun kim nh: 2=n.R2
nu H ng th 2~2(p)
Min bc b H0: )}(;{222 pW tntn
6.2. Pht hin hin tng
Dependent Variable: ABS(Y-Y(-1))
Method: Least Squares
Date: 09/25/13 Time: 15:11
Sample (adjusted): 2 10
Included observations: 9 after adjustments Variable Coefficient Std. Error t-Statistic Prob. SQR(X) 2.876950 3.710844 0.775282 0.4676
Z^3 -0.004958 0.018560 -0.267137 0.7983
EXP(X/Z) -0.350296 0.983669 -0.356111 0.7339 R-squared 0.128945 Mean dependent var 5.777778
Adjusted R-squared -0.161407 S.D. dependent var 2.905933
S.E. of regression 3.131682 Akaike info criterion 5.382219
Sum squared resid 58.84461 Schwarz criterion 5.447961
Log likelihood -21.21999 Hannan-Quinn criter. 5.240349
Durbin-Watson stat 1.602216
= 5%, pht hin hin tng t tng quan
Breusch-Godfrey Serial Correlation LM Test: F-statistic 0.557425 Prob. F(3,3) 0.6784
Obs*R-squared 3.191213 Prob. Chi-Square(3) 0.3631 Variable Coefficient Std. Error t-Statistic Prob. C -22.56983 26.01354 -0.867618 0.4494
X 0.682649 0.900053 0.758454 0.5033
Z 2.016505 2.310685 0.872687 0.4471
RESID(-1) -0.154869 0.527005 -0.293865 0.7880
RESID(-2) -0.608188 0.546628 -1.112617 0.3470
RESID(-3) 0.468176 0.699539 0.669264 0.5512
= 5%, pht hin hin tng t tng quan
6.3. Khc phc hin tng
Bi ton: Gi s MH gc Yi = 1+ 2Xi+Ui
C xy ra hin tng TTQ bc 1. Khc phc hin
tng trn
6.3. Khc phc hin tng
? Phng php no dng khc phc TTQ bc 1?
Ti sao
? Nu h s t hi quy bc 1 cha bit th c th
dng nhng phng php no c lng
? Xc nh gi tr c lng ca trong mi
phng php
7.1. Bn cht, nguyn nhn, hu qu ca hin tng
7.2 . Pht hin hin tng
7.3. Khc phc hin tng (T c sch)
Chng 7. a cng tuyn
7.1. Bn cht, nguyn nhn, hu qu
ca hin tng
a) Bn cht:
Vi phm gi thit : cc bin c lp khng
c quan h ph thuc tuyn tnh
1TX X
rg(X) = k
a) Bn cht:
a cng tuyn hon ho
a cng tuyn khng hon ho
2
2 2 3 3 ... 0 ( 0)i i k ki iX X X i
2
2 2 3 3 ... 0 ( 0)i i k ki i iX X X V i
7.1. Bn cht, nguyn nhn, hu qu
ca hin tng
b) Nguyn nhn:
- Do bn cht ca cc mi quan h gia cc bin
c lp
- M hnh dng a thc
- Mu khng mang tnh i din
7.1. Bn cht, nguyn nhn, hu qu
ca hin tng
7.1. Bn cht, nguyn nhn, hu qu
ca hin tng
7.1. Bn cht, nguyn nhn, hu qu
ca hin tng
7.1. Bn cht, nguyn nhn, hu qu
ca hin tng
7.2. Pht hin hin tng
7.2.1. R2 cao, t s t thp
7.2.2. Nhn t phng i phng sai (VIF)
7.2.3. Hi quy ph
7.2.4. Tng quan gia cc bin c lp
7.2.1. R2 cao, t s t thp
R2 > 0.8
Tn ti |tj|< t(n-k)
/2 hoc P-value >
Kt lun: c xy ra a cng tuyn
Ngc li, nu khng tha mn 1 trong 2 iu kin
trn th khng xy ra HT
7.2.2. Nhn t phng i phng sai (VIF)
7.2.3. Hi quy ph
Xt MHHQ ca mt bin c lp theo cc bin cn
li, Nu MH ph hp (KGT ng thi) th
Kt lun: c xy ra a cng tuyn
7.2.4. Tng quan cp gia cc bin c lp
X LOG(X)/SQR(Z) Z(-1)
X 1.000000 0.960269 -0.731193
LOG(X)/SQR(Z) 0.960269 1.000000 -0.737812
Z(-1) -0.731193 -0.737812 1.000000
pht hin hin tng a cng tuyn
Dependent Variable: Y^2
Method: Least Squares Date: 09/26/13 Time: 14:16
Sample: 1 10
Included observations: 10 Variable Coefficient Std. Error t-Statistic Prob.
C 5277.625 6313.316 0.835951 0.4352
SQR(X) 6566.296 15295.18 0.429305 0.6827
ABS(X-Z) 1181.473 279.0028 4.234627 0.0055
LOG(X^2) -4227.578 11376.48 -0.371607 0.7230 R-squared 0.994098 Mean dependent var 12017.20
Adjusted R-squared 0.991146 S.D. dependent var 4096.183
S.E. of regression 385.4258 Akaike info criterion 15.03575 Sum squared resid 891318.3 Schwarz criterion 15.15678
Log likelihood -71.17874 Hannan-Quinn criter. 14.90297
F-statistic 336.8429 Durbin-Watson stat 3.438842 Prob(F-statistic) 0.000000
pht hin hin tng a cng tuyn ( bin
ph thuc l Y^2)
Dependent Variable: SQR(X)
Method: Least Squares Date: 10/02/13 Time: 05:42
Sample: 1 11
Included observations: 11 Variable Coefficient Std. Error t-Statistic Prob.
C -0.272524 0.104676 -2.603501 0.0314
ABS(X-Z) 0.014708 0.003632 4.049691 0.0037
LOG(X^2) 0.739602 0.024752 29.88082 0.0000
R-squared 0.999504 Mean dependent var 3.340110
Adjusted R-squared 0.999380 S.D. dependent var 0.357814
S.E. of regression 0.008913 Akaike info criterion -6.375713
Sum squared resid 0.000635 Schwarz criterion -6.267196 Log likelihood 38.06642 Hannan-Quinn criter. -6.444118
F-statistic 8055.007 Durbin-Watson stat 1.913985
Prob(F-statistic) 0.000000
pht hin hin tng a cng tuyn ( bin
ph thuc l Y^2)
7.3. Khc phc hin tng
Bi ton: Gi s MH gc
Yi = 1+ 2X2i++ k Xki +Ui
xy ra hin tng a cng tuyn. Khc phc hin
tng trn
7.3. Khc phc hin tng
? Lit k nhng phng php dng khc phc
hin tng a cng tuyn? Hiu c tng ca
mi phng php ny
8.1 Cc thuc tnh ca 1 m hnh tt
8.2 Cc loi sai lm thng mc
8.3 Pht hin v kim nh cc sai lm ch nh
8.4 Mt s m hnh kinh t thng dng
Chng 8CHN M HNH V KIM NH VIC CHN M HNH
Chng 8
8.1 Cc thuc tnh ca m hnh tt
Tnh Kim
ng nht
Ph hp
Bn vng v mt l thuyt
C kh nng d bo tt
Chng 8
8.2 Cc loi sai lm khi chn m hnh
B st bin thch hp
a vo m hnh bin khng thch hp
Chn dng hm khng ng
8.2.1 B st bin gii thch
Chng 8
8.2 Cc loi sai lm khi chn m hnh
Gi s m hnh ng:
Yt = 1 + 2 X2t + 3X3t + Ut
Nhng ta chn m hnh:
Yt = 1 + 2X2t + Vt
Chng 8
8.2 Cc loi sai lm khi chn m hnh
2Nu X2 tng quan X3 th , khng phi l UL vng v l c lng chch v ca 1, 2
1
tt XY 221
2Nu X2 khng tng quan X3 th l UL vng v l c lng khng chch v ca 2, nhng
vn l UL chch ca 11
Phng sai ca sai s c lng t m hnhng v phng sai ca sai s c lng cam hnh ch nh sai s khng nh nhau.
Chng 8
8.2 Cc loi sai lm khi chn m hnh
Khong tin cy thng thng v cc th tckim nh gi thit khng cn ng tin cna.
8.2.2 a bin khng thch hp vo m hnh
Chng 8
8.2 Cc loi sai lm khi chn m hnh
Gi s m hnh ng:
Yt = 1 + 2 X2t + Ut
Nhng ta chn m hnh:
Yt = 1 + 2X2t + 3X3t +Vt
Hm hi quy mu ca m hnh sai:
Chng 8
8.2 Cc loi sai lm khi chn m hnh
Cc c lng BPNN l c lng khng chch v vng nhng khng hiu qu dn n khong tin cy s rng hn
ttt XXY 33221
j
c lng ca 2 l c lng vng
8.2.3 Chn dng hm khng ng
Chng 8
8.2 Cc loi sai lm khi chn m hnh
Cc kt qu thu c t vic phn tch hi quytrong m hnh sai s khng ng vi thc tv dn n cc kt lun sai lm.
8.3.1 Pht hin bin khng cn thit trong MH
Chng 8
8.3 Pht hin v K cc sai lm ch nh
Yi = 1 + 2X2i + 3X3i +4X4i + 5X5i +Ui
H0 : 5 = 0
H0 : 4 = 5 = 0
8.3.2 Kim nh cc bin b b st
Chng 8
8.3 Pht hin v K cc sai lm ch nh
Yt = 1 + 2 X2t + Ut
Nu c s liu ca Z ta ch cn UL m hnh
Yt = 1 + 2Xt + 3Zt +Vt
H0: 3 = 0
Chng 8
8.3 Pht hin v K cc sai lm ch nh
Nu khng c s liu ca Z ta c th s dng mt trong cc kim nh sau
Chng 8
8.3 Pht hin v K cc sai lm ch nh
tY
3tY
2tY
2tY
3tY
Bc 1. Hi quy Yt theo Xt ta c v R2
old
a. Kim nh RESET ca RAMSEY
Bc 2. Hi quy Yt theo Xt, , c R2
new
v kim nh cc h s ca , bng 0
Bc 3. Kim nh c iu kin rng buc:
Chng 8
8.4 Pht hin v K cc sai lm ch nh
m : s bin mi c a vo MHk : s h s ca m hnh mi
Khi n ln ta c F ~ F(m,n-k)
)/()1(
/)(2
22
knR
mRRF
new
oldnew
b. Kim nh d (Durbin-Watson)
Chng 8
8.4 Pht hin v K cc sai lm ch nh
Bc 1. c lng m hnh :
Yi = 1 + 2X2i + Ui
Bc 2. Sp xp ei theo th t tng dn ca
bin b st Z, nu Z cha c s liu th sp xp
ei theo X
Bc 3.
Chng 8
8.4 Pht hin v K cc sai lm ch nh
Bc 4. H0 : Dng hm ng (khng c TTQ)
n
t
te
n
t
tt ee
d
1
2
2
2
1 )(
Bc 2. c lng MH sau thu c R2:
Chng 8
8.4 Pht hin v K cc sai lm ch nh
tY
t
p
tpttt VYYXe .....
2
221
c. Phng php nhn t Lagrange(LM)
Bc 1. Hi quy m hnh gc thu c v et
Chng 8
8.4 Pht hin v K cc sai lm ch nh
Vi n kh ln 2 = nR2 c phn phi 2(p) t
ta kt lun bi ton.
8.3.3 Kim nh tnh PP chun ca sai s NN
Chng 8
8.3 Pht hin v K cc sai lm ch nh
8.4.1 Hm sn xut Cobb-Douglas
Chng 88.4 Mt s m hnh kinh t lng thng dng
Yi : sn lngXi : lng lao ng (lng vn)1,2 : cc tham s ca m hnh (2 1)
Hm sn xut vi 1 yu t u vo:
iu
ii eXY2.1
Chng 88.1 Mt s m hnh kinh t lng thng dng
Logarit ha (8.1) ta c
lnYi = ln1 + 2lnXi +Ui (8.2)
t Yi = lnYi I = ln1 Xi = lnXi
Yi=I + 2 Xi+Ui (8.3)
Chng 88.1 Mt s m hnh kinh t lng thng dng
th ca hm sn xut
Hm sn xut vi nhiu yu t u vo
Chng 88.1 Mt s m hnh kinh t lng thng dng
)4.8(. 321iu
iii eLKY
Yi : sn lngKi : lng vnLi : lng lao ng s dngUi : sai s ngu nhin
Chng 88.1 Mt s m hnh kinh t lng thng dng
2 : co gin ring ca sn lng i vi vn
3: co gin ring ca sn lng /vi lng
Logarit ha (8.4) ta c
lnYi = ln1 + 2lnKi + 3lnLi + Ui (8.5)
Tng (2 + 3) nh gi hiu qu vic tng quy m sn xut
Chng 88.1 Mt s m hnh kinh t lng thng dng
- (2 + 3)< 1 tng quy m km hiu qu
- 2 =0 hoc 3 =0 pht trin khng hiu qu
- (2 + 3)> 1 tng quy m c hiu qu
Chng 88.1 Mt s m hnh kinh t lng thng dng
Hm tng trng kinh t c dng:
Yt = Y0(1+r)t
t : thi gian
8.1.2 Hm tng trng kinh t:
Chng 88.1 Mt s m hnh kinh t lng thng dng
lnYt = lnYt + t*ln(1+r)
t Yt = lnYt, 1 = lnY0, 2 =ln(1+r)
Yt = 1 + 2t
Ta c:
Chng 88.1 Mt s m hnh kinh t lng thng dng
2 : t s thay i tng i ca Y vi thay i tuyt i t
dt
Yd tln2dt
ydy
dt
dyy /)/(
1
8.1.3 M hnh Hyperbol (M hnh nghch o)
Chng 88.1 Mt s m hnh kinh t lng thng dng
M hnh phi tuyn vi X, tuyn tnh vi 1 2
i
i
i UX
Y 1
21
Chng 88.1 Mt s m hnh kinh t lng thng dng
a. 1,2 >0
Thng dng khi phn tch chi ph X sn xut ra 1 sn phm
Chng 88.1 Mt s m hnh kinh t lng thng dng
b. 1 >0, 2
Chng 88.1 Mt s m hnh kinh t lng thng dng
c. 1 0
Mi quan h gia t l thay i tin lng v t l tht nghip
8.1.4 M hnh hi quy a thc
Chng 88.1 Mt s m hnh kinh t lng thng dng
Yi = 1 + 2 Xi + 3 Xi2 + Ui
Nghin cu mi quan h gia tng chi ph Yi vi tng sn phm sn xut Xi
Chng 88.1 Mt s m hnh kinh t lng thng dng
Dng th
Gi tr ti u X0 tng chi ph Y nh nht
M hnh hi quy a thc tng qut
Chng 88.1 Mt s m hnh kinh t lng thng dng
i
k
ikiii UXXXY 12
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