Da Cong Tuyen

5/20/2018 Da Cong Tuyen

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BI BO CO

NHM 15 V 21


2/29

A CNG TUYNL G?


3/29

Trong m hnh hiquibi:

Cc bingii thch khng c tngquan nhau(tcl khng c hintngacngtuyn)

Nuc 1bingiithch no tngquan vi1 sbin gii thch khc => c hin tng acngtuyn

...1 2 2

Y X X U i k ki i


4/29

Khng c acngtuyn acngtuynthp

acngtuynva

acngtuyncao


5/29

V d:

Y: nhu cuvcaf Brazil X1: gi caf Brazil

X2: thu nhpngitiu dng Brazil

Do Brazil l ncsnxutnhiucaf nnthu nhpngitiu dng (X2) liphthucvogi caf (X1)

=> C hintngacngtuyn


6/29

* Nu tn ti cc s 1, 2, , k khng ngthibng0, sao cho:

=> Ta c hin tng a cng tuyn honho (1 bingii thch no l hm tuyn tnhcaccbincn li)

*Nutnticc s1, 2, , ksao cho:

vie: sai sngunhin

=> Ta c hin tngacng tuynkhnghon ho (1 bin gii thch no tng quanchtchvi1 sbingiithch khc).

... 01 1 2 2X X X

i i k ki

... 01 1 2 2X X X e

i i k ki


7/29

Xt v d:

X1 X2 X3 e

10 50 52 2

15 75 75 0

18 90 97 7

24 120 129 9

30 150 152 2


8/29

Ta c: X3= X2+ e

* X2= 5X1=> acngtuynhon hogiaX1v X2(X1vX2c tngquan tuyntnh hon hov = 1)

* X3= 5X1+ e=> X1v X3c acngtuynkhng hon ho

( = 0.9959 nn X1v X3c acngtuyncao,

gnhon ho)

12r

13r


9/29

CCH PHT HINA CNG TUYN


10/29

1. R2

cao v thngk t thp:* Khi R2cao => githitH0bbcb(H0:j= 0)(j = 2,3,,k), chpnhngithitc t nht1 h

shiqui ring 0 (H1: j0, j = 2,3,,k)* Khi t thp=> c xu hngchpnhngithithshiqui ring = 0

=> Nghch l => C hin tng a cng tuynnhngchthhinmccao.


11/29

2. Hstngquan ( r ) giacc cpbinc

lpcao:Hstngquan:

X, Z l 2bingiithch no trong m hnh

* Khi r > 0.8 th xyra acngtuyn.Nhngctrnghpr thpnhngvnc acngtuyn.

2 2

i i

XZ

i i

X X Z Zr

X X Z Z


12/29

V d:

Xt m hnh hiqui c 3binclpX1, X2,X3:

X1 = (1,1,1,1,1, 0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0)

X2 = (0,0,0,0,0, 1,1,1,1,1, 0,0,0,0,0,0, 0,0,0,0,0)

X3 = (1,1,1,1,1, 1,1,1,1,1, 0,0,0,0,0,0, 0,0,0,0,0)

* C acngtuynhon hov X3= X1+ X2

*Nhng: r12= -0.333r13= r23= 0.59 (khng cao)


13/29

3. Sdngm hnh hiqui ph: l hiqui c1 bin gii thch X no theo cc bin cn

li.

n: squan st

k: tham strong m hnh hiquiph

* Kimnhgi thit: H: R2= 0 (X tngngkhng tng quan tuyn tnh vi cc bin cnli)

NuchpnhnH => khng c acngtuyn.

2

21 1

R n kF

R k


14/29

V d:

X1: chi ph cho hng

X2: chi ph qungcoX1= 42.012976 + 0.387191X2

R2= 0.22922 F = 2.9738

= 0.05 => F0.025(1,10)= 4.96v F < F0.025(1,10)=> chpnhnH0:R2= 0

=> khng c acngtuyn


15/29

4. S dng nhn t phng i phng sai(VIF):

: hsxc nhcahm hiquiph* tinv1 => acngtuyngiaXjviccbinclpcn licao => VIFjcng ln()

* Khi VIFj> 10 > 0.9 => acngtuynmccao

2

1

1J

j

VIFR

2

jR

2

jR

2

jR


16/29

* Nhng gi tr VIFj cao khng c ngha l

phngsai v cc sai scacc clngcao:

(j = 2,3,,k)

phthucvo , v

=> cao chaxc nhphng sai caclngl cao.

2 2^

2 2 2

1var

1j j

ji ji

VIFx R x

^

var j

2 2jix jVIF

j

VIF


17/29

NGUYN NHN

XY RAA CNG TUYN


18/29

1. Do phngphp thu thpdliu: mukhngctrngcho tngth

V d: ngic thu nhpcao hnsc khuynhhng c nhiu ca ci hn. iu ny c thngvimum khng ngvitngth.

2. Do bn cht ca cc mi quan hgia ccbinngmchahintngacngtuyn

V d: trong miquan hgialnginnng

tiu th(Y) theo thu nhp(X1) v dintch nh (X2) c nchaacng tuynv thng thngnhnggia nhc thu nhpcao th c nh rnghnnhnggia nhc nhpthp.


19/29

3. Do ctrngcam hnhV d:khi b sung nhngbinc ly thabccao vo m hnh, cbit khi phm v d liu

cabinclpl nh.4. Mtm hnh xc nhqu mc: xyra khi sbingii thch nhiuhncmu=> khng xc

nhcduy nhtcc hshiqui.


20/29

CCH KHC PHC


21/29

1. Sdng thng tin tin nghim: t cc ktqutrcc t acngtuynV d:

Y: tiu dng

X1: thu nhp

X2: sgiu c

Ta c thbit tc ng ca X2 ln Y chbng1/10 tc ngcaX1ln Y


22/29

3 2

2

0.10

1 2 2 3 30.101 2 2 3

( 0.10 )1 2 2 3

1 2

i

i

i

i

Y X X U

i i iY X X U i i i

Y X X U i i i

Y X Ui i

Vi

* Khi clngc ta c thclngc

0.102 3

X X Xi i i

^

2^ ^

23 0.10


23/29

2. Loitr1 bingiithch khim hnh

* B1: Xem cpbingiithch no c tngquanchtch

GisX2, X3, . Xklbinclp

Y lbinphthucX2, X3c tngquan chtch

* B2: Tnh R2C mt2bin

Khng c 1 trong 2bin* B3: Loigi trR2tnh ckhi khng c mtbinl lnhn


24/29

V d:

(c mt2binX2, X3)

(khng c mtX2)(khng c mtX3)

=> loiX3ra khim hnh

2

23 0.94R 2

3 0.87R 2

2 0.92R


25/29

3. Thu thpthm sliuhoclymumi:

Vnacngtuynl 1 ctnh camu. C

thtrong 1 mukhc, ccbincngtuync thkhng nghim trng. V vy, tngcmuc thgimbtvncngtuyn.


26/29

V d: Trong m hnh 3bin:

Khi cmutng=> tng

ginhhs khng i=> gim

=> gim=> gip ta clngv kimnhgithitlin quan ti chnh xc hn.

2^

2 2 2

2 23

var1ix r

22ix

23r ^

2var

^

2se

^2


27/29

4. Sdngsai phn cp1:

Sai phn cpc thlm vnacngtuyncthnhi.

V d:

(t: thigian)

1 2 2 3 3t t t t Y X X U


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* ngvit-1:

t:

=>

=> M hnh ny thng lm gim tnh nghim

trng ca a cng tuyn v X2, X3 tng quancao khng c ngha l sai phn ca chng cngtngquan cao.

1 1 2 2 1 3 3 1 1t t t t Y X X U

1 2 2 2 1 3 3 3 1 1( ) ( )t t t t t t t t Y Y X X X X U U

1

2 2 2 1( )

t t t

t t t

y Y Y

x X X

2 2 3 3t t t t y x x V

3 3 3 1

1

( )t t t

t t t

x X X

V U U


29/29

CM N C V CC

BN LNG NGHE!

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