cÁc cÁch phÁt hiỆn hiỆn hiỆn tƯỢng Đa cỘng tuyẾn(2)

CC CCH PHT HIN HIN HIN TNG A CNG TUYN

A. L THUYT:I.GII THIU V A CNG TUYN:

Thng thng cc bin c lp khng c mi quan h tuyn tnh, nu quy tc ny b vi phm s c hin tng a cng tuyn. Nh vy, a cng tuyn l hin tng cc bin c lp trong m hnh ph thuc ln nhau v th hin c di dng hm s

II. CC CCH PHT HIN HIN TNG A CNG TUYN1. R cao nhng t s t thp

Trong trng hp Rcao (thng R> 0,8) m t s t thp th chnh l du hiu ca hin tng a cng tuyn .2. Tng quan cp gia cc bin gii thch cao

Nu h s tng quan cp gia cc bin gii thch cao (vt 0,8) th c kh nng c tn ti a cng tuyn. Tuy nhin tiu chun ny thng khng chnh xc. C nhng trng hp tng quan cp khng cao nhng vn c a cng tuyn. Th d, ta c 3 bin gii thch X, X, X nh sau

X = (1,1,1,1,1, 0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0)

X= (0,0,0,0,0, 1,1,1,1,1, 0,0,0,0,0, 0,0,0,0,0)

X = (1,1,1,1,1, 1,1,1,1,1, 0,0,0,0,0, 0,0,0,0,0)

R rng X = X + Xngha l ta c a cng tuyn hon ho, tuy nhin tng quan cp l:

r = -1/3 ; r = r =0,59

Nh vy a cng tuyn xy ra m khng c s bo trc cu tng quan cp nhng du sao n cng cung cp cho ta nhng kim tra tin nghim c ch.3. Xem xt tng quan ring

V vn c cp n da vo tng quan bc khng. Farrar v Glauber ngh s dng h s tng quan ring. Trong hi quy ca Y i vi cc bin X, X ,X. Nu ta nhn thy rng r cao trong khi r; r; r tng i thp th iu c th gi rng cc bin X, X v Xc tng quan cao v t nht mt trong cc bin ny l tha. D tng quan ring rt c ch nhng n cng khng m bo rng s cung cp cho ta hng dn chnh xc trong vic pht hin ra hin tng a cng tuyn.4. Hi quy ph

Mt cch c th tin cy c nh gi mc ca a cng tuyn l hi quy ph. Hi quy ph l hi quy mi mt bin gii thch X theo cc bin gii thch cn li. R c tnh t hi quy ny ta k hin R(x1=x2 x3 x4..) Mi lin h gia F v R: F=

F tun theo phn phi F vi k 2 v n-k +1 bc t do. Trong n l , k l s bin gii thch k c h s chn trong m hnh. R l h s xc nh trong hi quy ca bin X theo cc bin X khc. Nu F tnh c vt im ti hn F(k-2,n-k+1) mc ngha cho th c ngha l X c lin h tuyn tnh vi cc bin X khc. Nu F c ngha v mt thng k chng ta vn phi quyn nh liu bin X no s b loi khi m hnh. Mt tr ngi ca k thut hi quy ph l gnh nng tnh ton. Nhng ngy nay nhiu chng trnh my tnh c th m ng c cng vic tnh ton ny.5. Nhn t phng i phng sai

Mt thc o khc ca hin tng a cng tuyn l nhn t phng i phng sai gn vi bin X, k hiu l VIF(X).VIF(X) c thit lp trn c s ca h s xc nh R trong hi quy ca bin X vi cc bin khc nhau nh sau: VIF(X) = (5.15)Nhn vo cng thc (5.15) c th gii thch VIF(X) bng t s chung ca phng sai thc ca trong hi quy gc ca Y i vi cc bin X v phng sai ca c lng trong hi quy m Xtrc giao vi cc bin khc. Ta coi tnh hung l tng l tnh hung m trong cc bin c lp khng tng quan vi nhau, v VIF so snh tnh hung thc v tnh hung l tng. S so snh ny khng c ch nhiu v n khng cung cp cho ta bit phi lm g vi tnh hung . N ch cho bit rng cc tnh hung l khng l tng.

th ca mi lin h ca R v VIF l

Nh hnh v ch ra khi R tng t 0,9 n 1 th VIF tng rt mnh. Khi R =1 th VIF l v hn.

C nhiu chng trnh my tnh c th cho bit VIF i vi cc bin c lp trong hi quy.6. o Theil

Kha cnh ch yu ca VIF ch xem xt n tng quan qua li gia cc bin gii thch. Mt o m xem xt tng quan ca bin gii thch vi bin c gii thch l o Theil. o Theil c nh ngha nh sau:

m = R-( R- R)Trong Rl h s xc nh bi trong hi quy ca Y i vi cc bin X , X X trong m hnh hi quy:

Y = + X + X+ . + X+ U

R l h s xc nh bi trong m hnh hi quy ca bin Y i vi cc bin X , X, ,X, X, ,X

i lng R - Rc gi l ng gp tng thm vo vo h s xc nh bi. Nu X , X X khng tng quan vi nhau th m = 0 v nhng ng gp tng thm cng li bng R. Trong cc trng hp khc m c th nhn gi tr m hoc dng ln. thy c o ny c ngha, chng ta xt trng hp m hnh c 2 bin gii thch X v X. Theo k hiu s dng chng trc ta c: m = R- ( R- r) (R r)

T s t lin h vi tng quan ring r, r

Trong phn hi quy bi ta bit:

R = r + (1- r) r

R = r + (1- r) r

Thay 2 cng thc ny vo biu thc xc nh m ta c:

m = R- (r + (1- r) r - r) - ( r + (1- r) r- r ) = R- ((1- r) r + (1- r) r) t 1- r = w; 1- r = wv gi l cc trng s. Cng thc (5.16) c vit li di dng

m = R- (w r + w r)Nh vy o Theil bng hiu gia h s xc nh bi v tng c trng s ca cc h s tng quan ring.

Nh vy chng ta bit mt s o a cng tuyn nhng tt c u c ngha s dng hn ch. Chng ch cho ta nhng thng bo rng s vic khng phi l l tng.Cn mt s o na nhng lin quan n gi tr ring hoc thng k Bayes chng ta khng trnh by y.

III. Bin php khc phc

1. S dung thng tin tin nghim

Mt trong cac cach tip cn giai quyt vn a cng tuyn la phai tn dung thng tin tin nghim hoc thng tin t ngun khac c lng cac h s ring.

Thi du : ta mun c lng ham san xut cua 1 qua trinh san xut nao o co dang :

Qt =AL

Trong o Qt la lng san phm c san xut thi ky t ; Lt lao ng thi ky t ; Kt vn thi ky t ; Ut la nhiu ;A , (, la cac tham s ma chung ta cn c lng .Ly ln ca 2 v (5.17) ta c :

LnQt = LnA + (lnLt + Kt Ut

t LnQt = Q*t ; LnA = A* ; LnLt = L*t

Ta c Q*t = A* + (L*t + K*t + Ut (5.18)

Gia s L|K va L co tng quan rt cao di nhin iu nay se dn n phng sai cua cac c lng cua cac h s co gian cua ham san xut ln .

Gia s t 1 ngun thng tin co li theo quy m nao o ma ta bit c rng nganh cng nghip nay thuc nganh cso li tc theo quy m khng i nghia la ( + =1 .Vi thng tin nay ,cach x ly cua chung ta se la thay = 1 - ( vao (5.18) va thu c :

Q*t = A* + (L*t + ( 1 - ( )K*t + Ut (5.19)

T o ta c

Q*t K*t = A* + ((L*t K*t ) + Ut

t

Q*t K*t = Y*t va L*t K*t = Z*t ta c

Y*t = A* + ( Z*t + Ut

Thng tin tin nghim a giup chung ta giam s bin c lp trong m hinh xung con 1 bin Z*t

Sau khi thu c c lng cua ( thi tinh c t iu kin = 1

2. Thu thp s liu hoc ly thm mu mi

Vi a cng tuyn la c trng cua mu nn co th co mu khac lin quan n cung cac bin trong mu ban u ma a cng tuyn co th khng nghim trong na. iu nay co th lam c khi chi phi cho vic ly mu khac co th chp nhn c trong thc t . i khi chi cn thu thp them s liu , tng c mu co th lam giam tinh nghim trong cua a cng tuyn .

3. Bo bin Khi co hin tng a cng tuyn nghim trong thi cach n gian nht la bo bin cng tuyn ra khoi phng trinh. Khi phai s dung bin phap nay thi cach thc tin hanh nh sau :

Gia s trong m hinh hi quy cua ta co Y la bin c giai thich con X2 .X3 Xk la cac bin giai thich . Chung ta thy rng X2 tng quan cht che vi X3 .Khi o nhiu thng tin v Y cha X2 thi cung cha X3 .Vy nu ta bo 1 trong 2 bin X2 hoc X3

Khoi m hinh hi quy , ta se giai quyt c vn a cng tuyn nhng se mt i 1 phn thng tin v Y .

Bng phep so sanh R2 va trong cac phep hi quy khac nhau ma co va khng co 1 trong 2 bin chung ta co th quyt inh nn bo bin nao trong bin X2 va X3 khoi m hinh .

Thi du R2 i vi hi quy cua Y i vi tt ca cac bin X1X2X3 Xk la 0.94; R2 khi loai bin X2 la 0.87 va R2 khi loai bin X3 la 0.92 ;nh vy trong trng hp nay ta loai X3

Chung ta lu y 1 han ch cua bin phap nay la trong cac m hinh kinh t co nhng trng hp oi hoi nht inh phai co bin nay hoc bin khac trong m hinh .Trong trng hp nh vy vic loai bo 1 bin phai c cn nhc cn thn gia sai lch khi bo 1 bin cng tuyn vi vic tng phng sai cua cac c lng h s khi bin o trong m hinh .4. S dung sai phn cp 1

Thu tuc c trinh bay trong chng 7 t tng quan .Mc du bin phap nay co th giam tng quan qua lai gia cac bin nhng chung cung co th c s dung nh 1 giai phap cho vn a cng tuyn .

Thi du Chung ta co s liu chui thi gian biu thi lin h gia cac bin Y va cac bin phu thuc X2 va X3 theo m hinh sau :

Yt = 1 + 2 X 2t + 3X 3t+ U t (5.20)

Trong o t la thi gian . Phng trinh trn ung vi t thi cung ung vi t-1 nghia la :

Yt-1 = 2 + 2 X 2t-1 + 3X 3t-1 + U t-1 (5.21)

T (5.20) va (5.21) ta c :

Yt Yt-1 = 2 (X 2t - X 2t-1 ) + 3 (X 3t - X 3t-1) + U t - U t-1 (5.22)

t yt = Yt Yt-1

x2t = X 2t - X 2t-1

x3t = X 3t - X 3t-1

Vt = U t - U t-1

Ta c : yt = 2 x2t + 3 x3t + Vt (5.23)

M hinh hi quy dang (5.23) thng lam giam tinh nghim trong cua a cng tuyn vi du X2 va X3 co th tng quan cao nhng khng co ly do tin nghim nao chc chn rng sai phn cua chng cung tng quan cao.

Tuy nhin bin i sai phn bc nht sinh ra 1 s bn chng han nh s hang sai s Vt trong (5.23) co th khng thoa man gia thit cua m hinh hi quy tuyn tinh c in la cac nhiu khng tng quan .Vy thi bin phap sa cha nay co th lai con ti t hn cn bnh .5.Giam tng quan trong hi quy a thc

Net khac nhau cua hi quy a thc la cac bin giai thich xut hin vi luy tha khac nhau trong m hinh hi quy .Trong thc hanh giam tng quan trong hi quy a thc ngi ta thng s dung dang lch .Nu vic s dung dang lch ma vn khng giam a cng tuyn thu ngi ta co th phai xem xet n ky thut a thc trc giao .

6. Mt s bin phap khac Ngoai cac bin phap a k trn ngi ta con s dung 1 s bin phap khac na cu cha cn bnh nay nh sau :

hi quy thanh phn chinh

S dung cac c lng t bn ngoai

Nhng tt ca cac bin phap a trinh bay trn co th lam giai phap cho vn a cng tuyn nh th nao con phu thuc vao ban cht cua tp s liu va tinh nghim trong cua vn a cng tuyn.B. V D MINH HABi ton: Cho bng s liu sau.

Trong :

Y: sn lng du th (n v: nghn tn)

X: kim ngch xut khu du th (n v: nghn tn)

Z: vn u t khai thc (n v trm triu ng)

Yu cu: Hy pht hin hin tng a cng tuyn v tm bin php khc phc. Cho = 5%.2.997513.039426.444

3.261513.283671.3427

3.953413.6048129.8

5.366913.937230.7305

6.097314.3781341.7524

7.207214.5893481.4634

7.824315.2548601.2952

8.179615.7597696.9732

9.535915.9621863.8135

10.711816.18651003.6598

11.996616.82561144.594

13.993117.61211287.8756

15.954418.27761420.5488

17.197418.83641569.5317

18.450318.88811814.2707

Tin hnh c lng hm hi quy mu ta cDependent Variable: Y

Method: Least Squares

Date: 05/06/10 Time: 19:25

Sample: 1 15

Included observations: 15

VariableCoefficientStd. Errort-StatisticProb.

C12.475490.30109041.434450.0000

X0.2283220.1053222.1678520.0510

Z0.0014310.0009241.5477510.1476

R-squared0.990379 Mean dependent var15.76234

Adjusted R-squared0.988776 S.D. dependent var1.989505

S.E. of regression0.210776 Akaike info criterion-0.099186

Sum squared resid0.533118 Schwarz criterion0.042424

Log likelihood3.743892 F-statistic617.6576

Durbin-Watson stat1.650553 Prob(F-statistic)0.000000

I/ Pht hin hin tng a cng tuyn

Ta c hm hi quy mu:

Cch 1: H s xc nh bi cao nhng t thp.

Nhn xt:

Thng k t ca h s ng vi bin X

T = 2.167852 < 2.179

Thng k t ca h s ng vi bin Z

T = 1.547751 < 2.179

Vy cao nhng t thp. Suy ra c hin tng a cng tuyn.

Cch 2: H s tng quan cp gia cc bin gii thch cao

Ta c.

XZ

X1.0000000.994412

Z0.9944121000000

=> Nh vy ta cng c c s kt lun c hin tng a cng tuyn trong m hnh trn

Cch 3: Hi quy ph

Ta hi quy bin X theo bin Z c kt qu nh sau:

Dependent Variable: X


Date: 05/06/10 Time: 21:05

Sample: 1 15



C2.7174760.24617411.038840.0000

Z0.0087270.00025733.961600.0000



S.E. of regression0.555048 Akaike info criterion1.784043


Log likelihood-11.38032 F-statistic1153.390


Ta c ta i kim nh gi thit

: X khng c hin tng a cng tuyn vi Z

: X c hin tng a cng tuyn vi Z

Nhn xt:

Ta thy gi tr p-value ca thng k F l 0.000000 < =0.05

=> bc b gi thit chp nhn gi thit

Vy cng c c s khng nh m hnh trn c hin tng a cng tuyn

Cch 4: o Theil

Ta c cc h s tng quan gia cc bin Y v X,Z nh YXZ

Y1.0000000.9942130.993283

X0.9942131.0000000.994412

Z0.9932830.9944121.000000

tnh c o Theil ta phi tnh c ,. Theo cng thc bit chng hai ta c

=

Vy m = = 0.99038 2(1-0.98846)0.16636=0.98654

m khc 0 nn chng t c hin tng a cng tuyn sy ra. V mc a cng tuyn l 0.98654

II/ Khc phc hin tng a cng tuynCch 1: B bin

Bc 1: hi quy Y theo X =>

Bc 2: hi quy Y theo Z =>

Bc 3: so snh v trong cc hi quy trn

Bc 4: kt lun.* Bc 1 : Hi quy Y theo X

Dependent Variable: Y


Date: 05/06/10 Time: 22:42

Sample: 1 15



C12.047400.12519996.225800.0000

X0.3904230.01170133.367620.0000







* Bc 2 Hi quy Y theo ZDependent Variable: Y


Date: 05/06/10 Time: 22:44

Sample: 1 15



C13.095950.105953123.60140.0000

Z0.0034230.00011130.951390.0000







* Bc 3 :

T kt qu hi quy trn ta c:

= 0.990379 = 0.988776 = 0.988459 = 0.987571

= 0.986612 = 0.985582

* Bc 4:

Ta tin hnh so snh. V kt lun trong trng hp ny loi bin ZCch 2: S dng sai phn cp 1

Chng ta c s liu chui thi gian biu th lin h gia bin Y v cc bin ph thuc X,Z theo m hnh sau (*)

Vi t l thi gian. Phng trnh trn ng vi t th cng ng vi t-1 ngha l :

(**)

Tr (* ) cho (** ). V t

Ta thu c bng s liu mi

0.24420.26444.8987

0.32120.691958.4573

0.33220.14135100.9305

0.44110.7004111.0219

0.21121.1099139.711

0.66550.6171119.8318

0.50490.355395.678

0.20241.3563166.8403

0.22441.1759139.8463

0.63911.2848140.9342

0.78651.9965143.2816

0.66551.9613132.6732

0.55881.243148.9829

0.05171.2529244.739

Hi quy sai phn cp 1

Dependent Variable: Y


Date: 05/07/10 Time: 00:26

Sample: 1 14



C0.4929190.1568683.1422450.0094

X0.2539560.1182462.1476990.0549

Z-0.0025990.001415-1.8368800.0934




Sum squared resid0.438416 Schwarz criterion-0.060256



Ta c h s tng quan gia cc bin gii thch

1.0000000.582640

0.5826401.000000

Hi quy ph ca bin sai phn theo ta cDependent Variable: X


Date: 05/07/10 Time: 00:52

Sample: 1 14



C0.1206020.3813800.3162260.7573

Z0.0069710.0028072.4833860.0288





Log likelihood-8.724231 F-statistic6.167204


Nhn xt =0.318112< 0.8

= 0.582640

cÁc cÁch phÁt hiỆn hiỆn hiỆn tƯỢng Đa cỘng tuyẾn(2)

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