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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

    321 ...