Ứng dụng kinh tế lượng trong phân tích , nghiên cứu

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ng dng kinh t lng trong phn tch, nghin [email protected] thiu s lc v KLT nng caoHng dn thc hnh v phn tch m hnhng dng Stata trong phn tchGii thiu v Kinh t lng c bnGii thiu lp hcTng kt, kim tra nh gi [email protected]. Gii thiuKinh t lng v ng [email protected] t lng (KTL) nghin cu nhng mi quan h Kinh t X hi; thng qua vic xy dng, phn tch, nh gi cc m hnh cho ra li gii bng s.

[email protected] te lng la s ket hp:1. Cc l thuyt kinh t2. M hnh ton kinh t3. Xc sut thng [email protected] nh m hnh l thuyt cn phn tchThit lp m hnh ton hcThu thp s liuc lng cc tham s ca m hnhKim nh m hnhPhn tch m hnh v d bo

Cc bc tin [email protected]: Phn tch mi quan h gia tiu dng v thu nhpL thuyt: Thu nhp tng ko theo tiu dng tng.M hnh: C = f(Y)S liu: S thng k hoc VHLSSc lng: S dng Stata, Spss hoc Eviews c lng m hnhKim nh m hnh

Mt v d n gin minh [email protected] s v dPhn tch nh hng ca mt s yu t n cu lao ngL thuyt:M hnh ton: L = f(Y,K/L, wage,..)S liu: iu tra Doanh nghipc lng m hnhKim nhPhn tch m hnh [email protected] cht ca phn tch hi quyL phn tch mi lin h ph thuc gia:Mt bin gi l bin ph thuc, bin c gii thch, bin ni sinhMt hoc mt s bin gii thch (bin c lp, bin ngoi sinh)- c lng trung bnh bin ph thuc trong nhng iu kin xc nh ca bin gii thchM HNH HI QUY, MT VI TNG C BNPhn [email protected] d: ng cong m t quan h tin lng bnh qun v nhu cu lao ng trong DNCu lao ng1086420Tin lng bnh [email protected] hnh hi quy tng th dng tuyn tnh:

Cc h s j cha bit, cn phi c lng

[email protected] s chnh s hi quy ringNu X2 tng 1 n v m X3,..,Xk gi nguyn th gi tr trung bnh ca bin Yi tng 2 n vsai s ngu nhinBin ph thuc(c gii thch)Cc bin c lp(Gii thch)Trung bnh ca Y khi cc bin gii thch bng 0M HNH HI QUY.D liu s dng dng no?Cc h s c c lng nh th no?Mc ngha thng k cho cc h s c lng ngha cc h s c lng l g? Phn tch nh th no

[email protected] liu theo thi gianS liu choS liu hn hpCC LOI S LiU [email protected] s liuS liu theo thi gian: c thu thp mt n v trong cc thi k (thi gian)S liu cho: c thu thp 1 thi im nhng thc hin ti nhiu n v (khng gian) khc nhauS liu hn hp: Bao gm c hai loi trn.Thn trng khi s dng cc loi s liu!

[email protected] d 1Y- Tiu dng ca h gia nhX- Thu nhp ca h gia nhM HNH HI QUY HAI BiN1- HM HI QUY [email protected] nhp v chi tiu ca hH sChi tiuThu [email protected] tuyn tnh ca hi quy tng th E(Y/Xi) = 1 + 2 Xi (2.2) 1 l h s t do.1, 2 l cc h s hi quy. [email protected] cho bit gi tr trung bnh ca bin ph thuc (Y) l bao nhiu khi bin c lp (X) nhn gi tr 0. [email protected] l h s gc2 cho bit gi tr TB ca bin ph thuc (Y) s thay i bao nhiu khi gi tr ca bin c lp (X) tng 1 n v, trong iu kin cc yu t khc khng [email protected] sai s ngu nhin ca tng th ng vi quan st i MH hi quy tng th ngu nhin: Yi = 1 + 2Xi+Ui [email protected] hi quy c xy dng trn s liu mu gi l hm hi quy mu (SRF - the sample regression function)Hm hi quy [email protected] PRF co dang t.tnh th SRF co dang: l c lng khng chch c phng sai nh nht ca E(Y/Xi), 1, 2

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Dng ngu nhin ca SRFei = Yi la c lng iem cua Ui (phan d)

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(Ordinary Least Square) Theo pp OLS, ta phi tm sao cho n cng gn vi gi tr (Yi) cng tt, hay phn d:

PHNG PHP OLSGi s c mt mu gm n quan st (Yi, Xi), (i = 1, 2, . . . , n)[email protected] = Yi = Yi Xi

Cng nh cng tt

[email protected]^[email protected] ei c th dng c th m, nn ta cn tm SRF sao cho tng bnh phng ca cc phn d cc tiu.Tc , , phi tha mn iu kin: Min

[email protected] Yi, Xi (i = 1, 2, . . . , n) bit, nn V vy ta cn tm , sao cho:f( , ) =(Yi - - Xi )2 min xc nh vi s h tr ca phn mm

l hm ca , [email protected]

Bin gii thch l phi n.n K vng ton ca Ui bng 0, hay: E(Ui/Xi) = 0 Cc Ui c p.sai bng nhauCC Gi THIT CA PHNG PHP [email protected] Khng c tng quan gia ccUi, hay cov(Ui, Uj) = 0 (i j) Ui v Xi khng tng quan vi nhau, hay cov(Ui, Xi) = [email protected] tin cy ca 1; 2Vi tin cy 1- , KTC ca 2 l: [email protected]

Khong tin cy ca 1 l:Ch : Cc thng k ny u c phn mm bo co khi c [email protected]

S dng phng php mc ngha (c cung cp khi chy m hnh)Gi s cn kim nh cp gtH0: 2 = 0; H1: 2 0

KIM NH GI THIT VCC H S HI [email protected] dng P_value kim nh

Nu p < th bc b gi thit H0Nu p> th c th chp nhn gi thit H0Vi l mc ngha

[email protected] ngha ca R2 Kim nh cp gi thit sau: H0: R2 = 0; H1: R2 0S dng kim nh F thc hinF = R2(n-2)/(1-R2) (c output)

KiM NH S PH HP CA HM HI [email protected] hnh hi qui biM hnh :M hnh hi qui tuyn tnh k bin (PRF) :E(Y/X2i,,Xki) = 1+ 2X2i ++ kXkiYi = 1+ 2X2i + + kXki + UiTrong :Y - bin ph thucX2,,Xk - cc bin c [email protected] l h s t doj l cc h s hi qui ring, cho bit khi Xj tng 1 v th trung bnh ca Y s thay i j v trong trng hp cc yu t khc khng i (j=2,,k).

Khi k = 3 th ta c m hnh hi qui tuyn tnh ba bin : E(Y/X2, X3) = 1+ 2X2 + 3X3 (PRF)Yi = 1+ 2X2i + 3X3i + [email protected] lng cc tham sa. M hnh hi qui ba bin :Yi = 1+ 2X2i + 3X3i + Ui(PRF)Hm hi qui mu :

Gi s c mt mu gm n quan st cc gi tr (Yi, X2i, X3i). Theo phng php OLS, (j= 1,2,3) phi tho mn :[email protected] gi thit ca m hnh

Cc gi nh ca m hnh (c thm)Khong tin cy, s ph hp ca hm hi quy, hay kim nh s khc 0 ca cc h s c lng ,.. u c th hin trong bng kt qu c [email protected]

Hi qui vi bin giI. Bn cht ca bin gi- M hnh trong cc bin c lp u l bin giBin nh tnh thng biu th cc mc khc nhau ca mt tiu thc thuc tnh no .V d : lng ho c bin nh tnh, trong phn tch hi qui ngi ta s dng k thut bin [email protected] d: Khi quan st tin lng ca ngi lao ng theo gii tnh, theo khu vc hay theo trnh CMKT, nh gi s khc bit v tin lng gia cc nhm ny, chng ta s dng n k thut bin giM hnh : Yi = 1+ 2Zi + UiTrong :Y : Tin lng, D : bin [email protected] :Mt bin nh tnh c m mc (m phm tr) th cn s dng (m-1) bin gi i din cho n.Phm tr c gn gi tr 0 c xem l phm tr c s (vic so snh c tin hnh vi phm tr ny)[email protected]. Hi qui vi bin nh lng v bin nh tnhV d : Hy lp m hnh m t quan h gia thu nhp ca gio vin vi thm nin ging dy v vng ging dy (thnh ph, tnh ng bng, min ni).Gi Y : thu nhp (triu ng/nm) X : thm nin ging dy (nm) Z1, Z2 : bin [email protected] = 1 : thnh ph Z2i = 1 : tnh 0 : ni khc 0 : ni khcTa c m hnh :Yi = 1+ 2Xi + 3Z1i + 4Z2i + Ui ngha ca 2, 3, 4 : V d : Hy lp m hnh m t quan h gia thu nhp ca gio vin vi thm nin ging dy, vng ging dy (thnh ph, tnh ng bng, min ni) v gii tnh ca gio [email protected] hnh :Yi = 1+ 2Xi + 3Z1i + 4Z2i + 5Di + UiTrong : Y, X, Z1i, Z2i ging v d 3. Di ( bin gi) = 1 : nam gii 0 : n gii ngha ca 5 :

[email protected] d : Lp m hnh quan h gia chi tiu c nhn vi thu nhp v gii tnh ca c nhn .Yi = 1+ Xi + 3Zi + Ui(1)Y chi tiu (triu/thng)X thu nhp (triu/thng) Zi = 1 : nam gii 0 : n gii.* M rng m hnh : Vi m hnh trn, khi thu nhp c nhn tng 1 triu ng th chi tiu tng triu ng bt k l nam hay n. [email protected] vi gi thit cho rng nu thu nhp tng 1 triu ng th mc chi tiu tng thm ca nam v n khc nhau th phi l = 2+ 4ZiLc ny m hnh (1) c vit : Yi = 1+ (2+ 4Zi)Xi + 3Zi + UiHay : Yi = 1+ 2 Xi + 3Zi + 4XiZi + Ui (2)Trong : XiZi c gi l bin tng tc gia X v [email protected] Khi Zi =1 : Yi = (1 +3) + (2+ 4)Xi +Uiy l hi qui chi tiu-thu nhp ca nam.- Khi Zi =0 : Yi = 1+ 2 Xi +Uiy l hi qui chi tiu-thu nhp ca n. ngha ca cc h s :1: Khi khng c thu nhp th chi tiu trung bnh ca mt ngi n l 1 triu.2: Khi thu nhp ca mt ngi n tng 1 triu ng th chi tiu ca h tng 2 triu [email protected]: Khi khng c thu nhp th chi tiu trung bnh ca mt ngi nam chnh lch so vi ca mt ngi n l 3 triu (hay chnh lch v h s tung gc gia hm hi qui cho nam v hm hi qui cho n).4: Khi thu nhp ca mt ngi nam tng 1 triu ng th chi tiu ca h tng nhiu hn ca n 4 triu ng (nu 4 > 0) hay tng t hn ca n 4 triu ng (nu 4< 0) (Hay chnh lch v h s dc gia hm hi qui cho nam v hm hi qui cho n).

[email protected] : H0 : 3 = 0 h s tung gc gia hi qui cho nam v cho n l ging nhau.H0 : 4 = 0 h s dc gia hi qui cho nam v cho n l ging nhau.H0 : 3 = 4 = 0 hi qui cho nam v cho n l ging ht nhau ( chi tiu ca nam v ca n l ging nhau)[email protected] CONG TUYENI. Bn cht ca a cng tuyna cng tuyn l tn ti mi quan h tuyn tnh gia mt s hoc tt c cc bin c lp trong m hnh.Xt hm hi qui k bin :Yi = 1+ 2X2i + + kXki + Ui- Nu tn ti cc s 2, 3,,k khng ng thi bng 0 sao cho :

2X2i + 3X3i ++ kXki = 0Th gia cc bin c lp xy ra hin tng a cng tuyn hon ho.- Nu tn ti cc s 2, 3,,k khng ng thi bng 0 sao cho :2X2i + 3X3i ++ kXki + Vi = 0(Vi : sai s ngu nhin)Th gia cc bin c lp xy ra hin tng a cng tuyn khng hon ho.Ta c : X3i = 5X2i c hin tng cng tuyn hon ho gia X2 v X3 v r23 =1 X4i = 5X2i + Vi c hin tng cng tuyn khng hon ho gia X2 v X3 X21015182430X3507590120150X4527597129152V d : Yi = 1+2X2i+3X3i+ 4X4i + UiVi s liu ca cc bin c lp :Khi c a cng tuyn hon ho th khng th c lng c cc h s trong m hnh m ch c th c lng c mt t hp tuyn tnh ca cc h s . Trng hp c a cng tuyn khng hon hoThc hin tng t nh trong trng hp c a cng tuyn hon ho nhng vi X3i = X2i +Vi Vn c th c lng c cc h s trong m hnh. Hu qu ca a cng tuyn1. Phng sai v hip phng sai ca cc c lng OLS ln.2. Khong tin cy ca cc tham s rng 3. T s t nh nn tng kh nng cc h s c lng khng c ngha4. R2 cao nhng t nh.5. Du ca cc c lng c th sai.6. Cc c lng OLS v sai s chun ca chng tr nn rt nhy vi nhng thay i nh trong d liu.7. Thm vo hay bt i cc bin cng tuyn vi cc bin khc, m hnh s thay i v du hoc ln ca cc c lng. Cch pht hin a cng tuyn1. H s R2 ln nhng t s t nh.2. H s tng quan cp gia cc bin gii thch (c lp) cao.V d : Yi = 1+2X2i+3X3i+ 4X4i + UiNu r23 hoc r24 hoc r34 cao c CT. iu ngc li khng ng, nu cc r nh th cha bit c CT hay khng.3. S dng m hnh hi qui ph.Xt : Yi = 1+2X2i+3X3i+ 4X4i + UiCch s dng m hnh hi qui ph nh sau :Hi qui mi bin c lp theo cc bin c lp cn li. Tnh R2 cho mi hi qui ph :

Hi qui X2i = 1+2X3i+3X4i+u2i Hi qui X3i = 1+ 2X2i+ 3X4i+u3i Hi qui X4i = 1+ 2X2i+ 3X3i+u4i Kim nh cc gi thit H0 :- Nu chp nhn cc gi thit trn th khng c a cng tuyn gia cc bin c lp.BIN PHP KHC PHC 1. S dng thng tin tin nghim2. Lai tr mt bin gii thch ra khi MH:B1: xem cp bin GT no c quan h cht ch, chng hn x2, x3.B2: Tnh R2 i vi cc HHQ khng mt mt trong 2 bin .B3:Lai bin no m R2 tnh c khi khng c mt bin l ln hn.Bin php3.Thu thp thm s liu hoc ly mu mi4. S dng sai phn cp mt5. Gim tng quan trong cc hm hi qui a thc

Phng sai thay iI. Bn cht v nguyn nhn phng sai thay iBn cht : Phng sai c iu kin ca Ui khng ging nhau mi quan st.Var (Ui) = (i=1,2,,n)Nguyn nhn :- Do bn cht ca cc mi quan h trong kinh t cha ng hin tng ny.

Do k thut thu thp s liu c ci tin, sai lm phm phi cng t hn. Do con ngi hc c hnh vi trong qu kh.Do trong mu c cc gi tr bt thng (hoc rt ln hoc rt nh so vi cc gi tr khc).Hin tng phng sai khng ng u thng gp i vi s liu cho.II. Hu qu ca phng sai thay iCc c lng OLS vn l cc c lng tuyn tnh, khng chch nhng khng cn hiu qu na.2. c lng phng sai ca cc c lng OLS b chch nn cc kim nh t v F khng cn ng tin cy na.3. Kt qu d bo khng hiu qu khi s dng cc c lng OLS.III. Cch pht hin phng sai thay iPhng php thXt m hnh : Yi = 1+ 2Xi +Ui(1)- Hi qui (1) thu c cc phn d ei.- V th phn tn ca e theo X.- Nu rng ca biu ri tng hoc gim khi X tng th m hnh (1) c th c hin tng phng sai thay i.* Ch : Vi m hnh hi qui bi, cn v th phn d theo tng bin c lp hoc theo .

2. Kim nh Park tng : Park cho rng l mt hm ca X c dng :

Do :V cha bit nn c lng hm trn Park ngh s dng thay cho

Cc bc kim nh Park :c lng m hnh hI qui gc (1), thu ly phn d ei tnh

- c lng m hnhNu m hnh gc c nhiu bin c lp th hi quitheo tng bin c lp hoc theo Kim nh gi thit H0 : 2 = 0 Nu chp nhn H0 mo hnh gc (1) co phng sai khng i.3. Kim nh WhiteXt m hnh : Yi = 1+ 2X2i + 3X3i +UiBc 1 : c lng m hnh gc, thuBc 2 : Hi qui m hnh ph sau, thu h s xc nh ca hi qui ph :

Bc 3 : Kim nh H0 : Phng sai khng i.Nu bc b H0.Vi p l s h s trong m hnh hi qui ph khng k h s t do (tung gc).

n tp v thc hnh- Stata: Nhm lnh m t binNhm lnh to bin v m haCch to bng tnh mt v 2 chiuHi quy vi regKim nh: H s, S ph hp hm hi quy, a cng tuyn, phng sai sai s thay i,..Xy dng m hnh tin lng, cu lao ngCch phn tch mt bo co k thut.

[email protected] dn lm bi tp nhm v kim tra cui khaPhn nhm:1. Xy dng cu lao ng2. Xy dng m hnh tin lngYu cu:-L thuyt-M hnh-S liu, bin s s dng-c lng-Phn tch kt [email protected] ngha ca cc tham s trong RegCc dng:1. [email protected] thiu chui thi gianTrong thc t m hnh ho mt hin tng kinh t c th s dng hai loi m hnh:M hnh cu trc: Biu din s thay i ca mt bin kinh t trong mi lin h ph thuc vi cc bin khc.M hnh hnh vi: Biu din s thay i ca mt bin ch da vo hnh vi ca qu kh ca chnh bn .nh nghaChui thi gian l tp hp cc gi tr ca mt bin ngu nhin c sp xp theo th t thi gian. Chui thi gian cn c gi l dy s thi gian. n v thi gian c th l ngy, tun, thng, qu, nm. . .Phn tch chui thi gian c mc ch l lm r cu trc ca chui thi gian (tc l cc thnh phn ca n) trong s bin ng ca bn thn n.

[email protected] thi gianCc thnh phn ca chui thi gian Bt k chui thi gian no cng cha ng t nht mt trong bn thnh phn (yu t) sau:- Xu th bin ng;-Bin ng theo ma (hoc thi v);-Bin ng theo chu k;-Bin ng ngu nhin (bt quy tc);

[email protected] thi gianTc l c th ni rng cu trc ca chui thi gian s bao gm 4 thnh phn ni trn. K hiu:Tt (Trend)- thnh phn xu th cho bit xu hng bin ng ca chui thi gian trong mt khong thi gian tng i di.a s chui thi gian th hin mt khuynh hng tng hoc gim kh r theo thi gian. VD: GDP, GNP, St (season) - thnh phn ma v cho bit s bin ng ca chui trong hai hay nhiu khong thi gian ( di c th khc nhau) lin nhau c lp i lp li trong sut thi k xem xt.Cc bin ng ma v c th din ra theo qu (GDP), theo thng, thm ch trong tng ngy.

[email protected] thi gianCt (cycle) - thnh phn chu k cho bit mc bin ng ca chui trong mt khon thi gian no (gi l chu k) s c lp i lp li trong sut thi k nghin cu. Thnh phn chu k ny khng lin quan n yu t ma v m bt ngun t chu k kinh doanh cng nh chu k kinh t.It (Irregular) - thnh phn bt quy tc l kt hp ca v s cc nhn t nh hng n hnh vi ca chui, tng t nh cc nhn t hnh thnh nn cc sai s ngu nhin ui trong m hnh hi [email protected] tch xu thCc m hnh ngoi suy gin nM hnh xu th tuyn tnh: Yt = 1 + 2t + ut M hnh trn c dng nu Yt tng ln mt lng khng i qua mi n v thi gian.M hnh dng mM hnh trn c dng nu sau mi n v thi gian Yt tng ln vi mt t l % khng i.

c lng m hnh trn ta bin i v dng tuyn tnh: lnYt = ln + rt + ut

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Chui thi gianM hnh xu th t hi quyYt = 1 + 2Yt-1 + ut Hm bc haiYt = 1 + 2t + 3t2 + ut Nu 2 > 0 v 3 > 0 th Yt lun tng. Nu 2 < 0 v 3 > 0 th ban u Y gim, sau s tng.

[email protected] phng php san chuiPhng php trung bnh trt (Moving average - MA)Trong nhiu chui thi gian, yu t ngu nhin c th rt ln lm lu m xu th ca hin tng. lm r xu th c th dng phng php trung bnh trt. T tng ca phng php ny l thnh phn bt quy tc bt k thi im no cng s c nh hng t hn nu quan st thi im c trung bnh ho vi cc quan st trc v sau n.

[email protected] php trung bnh trtGi s c chui thi gian Yt (t = 1, . . .,n), lc trung bnh trt bc 2m+1, k hiu l MA(2m+1)t tnh bng cng thc:

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San m gin nPhng php san m gin n cng l mt phng php lm trn s liu khng ch gip ta loi b yu t ngu nhin m cn c th d bo ngn hn gi tr tng lai ca chui.Phng php san m gin n thch hp vi cc chui khng c yu t ma v v khng c yu t xu th tng hay gim. C ngha vi chui khng thay i hoc thay i rt chm theo thi gian th c th dng san m gin n.

[email protected] m gin nGi s c chui Yt (t = 1 . . .n). Ta khng th dng Yn+1, Yn+2, . . v n cha ng cc yu t ngu nhin. Ta cng khng ly trung bnh s hc ca Yn, Yn-1, . . . v nh vy coi cc gi tr hin ti v qu kh u c vai tr nh nhau trong tng lai. Phng php san m gin n da vo trung bnh c trng s. Cng gn hin ti th trng s cng ln. = Yt+ (1-)Yt-1 + (1-)2Yt-2 + . . . . cng gn 1 gi tr hin ti c ngha hn v ngc li.

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Ứng dụng kinh tế lượng trong phân tích , nghiên cứu