bài 9- hồi quy đa biến

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BI GING THNG K SINH HCPHN TCH HI QUY A BIN

THNG K SINH HCnguyn bn ting Anh: Nguyn Vn Tun

Ch9: PHN TCH STNG QUAN

II. PHN TCH HI QUY A BIN

DON L KHOA HC?

Mt vi nm trc y, ti c mt cun sch trong c cuc tr chuyn sau y (khngng nguyn vn): "Mt ngi phc vhi Andy Capp1: Nu c chn gia tin bc, quyn

lc, hnh phc, hoc khnng tin on tng lai th ng schn ci no?. Andy trli: tinon tng lai v theo cch ti c thkim ra tin. Tin smang li cho ti sc mnh, vsau ti sc hnh phc".

C ll cng bng khi ni c m c thdon c tng lai th cng xa ta nh bncht con ngi. Nhiu ngi trong chng ta thng coi thng khi nim "bi ton", iu hi lv bn thn khoa hc xoay quanh cc phng php tin on tng lai. Tht ra,chng ta chsdng tvng khc nhau. Tri ngc vi bi ton, chng ta ni vtnh tonthay v tin on, quy lut thay v vn s, v bin ng thng k thay v bin ng tnh c. Tuynhin, mc ch ca phng php khoa hc l nh nhau. Tnhng quan st cc skin qua, chng ta rt ra cc quy lut m khi kim chng xong chng cho php chng ta don

nhng kt qutrong tng lai.

Ly v d, khi nim cho rng tt ccc ng vt cht vo cng mt tui nghe c vkhng c lnu chng ta tnh tui theo nm, thng hay ngy, nhng strnn kh hp l nu chng tadng snhp p ca tim. Chnh nhp p tim m ng vt ny khc vi ng vt khc.Nhng con th nh, nh chut, sng khong 3 nm nhng nhp tim ca chng rt nhanhchng. Nhng con c kch ctrung bnh, chng hn nh th, ch, cu, vv c nhp tim pchm hn v sng trong khong t12 n 20 nm. Voi sng hn 50 nm, nhng c nhp timchm. Khng phi l iu ngc nhin khi mt gio s ni ting tuyn brng "cho n lccht hu ht ng vt c v sng tdo trong tnhin (khng phi trong nh hoc vn th)

cp nht c trung bnh khong mt tnhp tim". C vnh chng ta c thdon tui thca ng vt tnhp tim ca chng. Nhng, chng ta vn cn c mt cch c hthng lmiu ny. Khoa hc hin i cho chng ta k thut phn tch hi quy t c mc tiuny. Chng ta stho lun vmt skha cnh thc tin ca k thut ny trong chny.

I. GII THIU

1Nhn vt trong truyn hi bng tranh ca nh vtranh Anh Reg Smythe (19171998).
http://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQhttp://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQhttp://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQhttp://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQhttp://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQhttp://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQhttp://www.google.com/url?q=http%3A%2F%2Fykhoanet.com%2Fbaigiang%2FTopic09.%2520Multiple%2520regression.pdf&sa=D&sntz=1&usg=AFQjCNEqDTnSms9LpX2YEAbnJVVgoUSgVQ

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Trong chtrc chng ta kho st m hnh hi quy vi mt bin c lp. Kh thng xuyntrong cc phn tch dliu, chng ta mun nghin cu sphthuc ca mt bin ngu nhinY trn nhiu binx1x2, ... , xp.Trong chny, chng ta smrng tng ny bao gmnhiu hn mt bin c lp trong phng trnh hi quy. K thut ny c gi l hi quy tuyntnh a bin .

Vi mc ch minh ho, by gichng ta hy xt mt tp dliu sthu c tmt nghincu kim tra nhit lng pht sinh trong qu trnh ng cng ca xi mng Portland. Nhitlng ny c ginh l mt hm ca cc thnh phn ha hc, cc bin sau y co:

x1:slng tricalcium aluminatx2:slng tricalcium silicatx3:slng ferit nhm tetracalciumx4:slng dicalcium silicatY : nhit lng tora tnh bng calo trn mi gram xi mng.

Bng 1: Dliu quan st

i x1 x2 x3 x4 Y1 7 6 26 60 78,52 1 26 15 52 74,3

3 11 56 8 20 104,34 11 31 8 47 87,6

5 7 52 6 33 95,96 11 55 9 22 109,27 3 71 17 6 102,78 1 31 22 44 72,59 2 54 18 22 93,8

10 21 47 4 26 115,911 1 40 23 34 83,812 11 66 9 12 113,313 10 68 8 12 109,4

Cc binx1, x2, x3vx4c tnh theo phn trm trng lng ca clanhke lm ra xi mng. Vigithuyt cho rng nhit lng sinh ra trong qu trnh ng cng l mt hm tuyn tnh theobn bin, chng ta tin ho m hnh:

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Y= Y(x1, x2, x3, x4)= 0+ 1x1+ 2x2+ 3x3+ 4x4

M hnh ny a ra tin rng ti im (x1i, x2i, x3i, x4i) (hng thi ca ma trn dliu), gi tr

mong i (hay trung bnh) ca nhit lng bng 0

1x1i

2x2i

3x3i

4x4i. Nh vy

gi tro c yicoi nh l mt hin thc ho ca mt bin ngu nhin Yi,bao gm trung bnhni trn cng vi mt lch ngu nhin ei:

0 1x1i 2x2i 3x3i 4x4i eiCc sai sngu nhin eithng c ginh l c lp ln nhau v theo phn phi bnhthng vi trung bnh bng 0 v phng sai chung bng 2.Ni chung, cc hs0, 1, 2, 3v 4 l cha bit. Phng php hi quy c sdng c tnh v kim nghim githuyt vcc hsny. Hschn0l gi trca Yti im

(0, 0, 0, 0).1, 2, 3v 4 c gi l hshi quy tng phn.Chng c thc hiunh sau:Ytng mt lngjkhixjtng thm 1 trong khi tt ccc binxi(i j) khng thayi. Thng thng, 0c gi l hng s, v xut pht tthc tl0c thc xem nhl mt hshi quy tng phn cho mt bin x0m bin ny lun c gi trkhng ixi0= 1.Chng ta thch gi0l hschnhn.

Mt sthut ng: Y c gi l bin phthuchaybin p ng.Ccxjc gi l binhi quyhoc bin c lp(mc d chng khng cn phi c c lp vi nhau theo nghathng k). iu quan trng l lu rng khng c coi xjl bin ngu nhinm l bin c gitr c ngi iu tra quy nh. Nuxjl ngu nhin th cc kt lun l c iu kin trn

cc gi trc hin thc ho.Trong thc tin, m hnh tuyn tnh hu nh khng c hiu lc chnh xc. Tuy nhin, trongnhiu trng hp, trong lnh vc xem xt, n l mt xp xtt cho thgii thc y phc tp.Hn na, him khi c thbit hoc xt ti tt ccc i lng nh hng n trung bnhY.Cc lch eitm hnh tuyn tnh c thc xem nh tng ca nhiu nh hng chabit hoc khng kim sot c v cng c thl sai so lng.

By gichng ta tng kt m hnh nh sau:

(a) Trung bnhYca bin ngu nhin Yphthuc tuyn tnh vi cc bin hi quyx1, x2, x3v

x4:Y= Y(x1, x2, x3, x4)

= 0+ 1x1+ 2x2+ 3x3+ 4x4(b) mi im (x1i, x2i, x3i, x4i), cc lch ei so viYc phn bbnh thng vi trung bnhl 0 v phng sai 2khng i:

ei~ N (0, 2)

(c) Cc lch eil c lp ln nhau.

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II. C LNG BNH PHNG NHNHT V SDI

2.1. KT QUTIU BIU TMT CHNG TRNH MY TNH

Phng thc c lng cc tham strong phn tch hi quy nhiu bin l phc tp. Ngy nay,

nhim vc hc ny thng c xl bng mt chng trnh my tnh. Cc kt qusau yc cho ra bi hthng phn tch thng k SAS:

Trong cc phn sau y, chng ta stho lun vtm quan trng cng nh ngha ca ccgi trsny..

2.2. C LNG BNH PHNG NHNHT

Cc hscha bit0, 1, 2, 3v 4 c c lng nh sau: i vi mi quan st yi,chng ta tm lch so vi trung bnh cha bitYti im (x1i, x2i, x3i, x4i). Tng ca nhiubnh phng sau:

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ni1

(yi01x1i2x2i3x3i4x4i)2c gi l tng cc bnh phng. c xem nh l mt hm ca p + 1 = 4 +1 = 5 thams0, 1, 2, 3v 4 (p l scc bin c lp) v, theo nguyn tc bnh phng nhnht,tng tnh trong trng hp mt bin chtrc, chng ta cn xc nh siu phng 4chiu2c bit (tc l xc nh hs0, 1, 2, 3v 4 trong phng trnh y =0+ 1x1+2x2+ 3x3+ 4x4ca siu phng) lm cho S nhnht. Cc gi trb0, b1, b2, b3v b4lm cc tiu Sl cc c lng bnh phng nhnhtca0, 1, 2, 3v 4. Nh vy chng ta c:

E ni1 (yi-b0-bx1i-b2x2i-b3x3i-b4x4i)2 E c gi mt cch a dng l tng cc bnh phng sai shoc tng cc bnhphng sdihoc tng cc bnh phng nhnht.Vic tnh ton cc hsbji hiphi gii mt hphng trnh tuyn tnh theo 5 bin. (c c chng hn bng cch ly ohm ring ca S theo cci i=0, 1, 2, 3, 4 xem mc 8.1.4. Ch1) Tuy nhin, chng ta skhng tho lun kha cnh k thut ny y.

By gi, sau khi c c bj, chng ta c thtnh ton cc gi trquan st c lng (thngc gi l gi trdon hoc gi trn khp):

y i b0bx1ib2x2ib3x3ib4x4icng thc ny chra gi trc lng y ca im trn siu phng ti im (x1i, x2i, x3i, x4i).

2.3. PHN TCH SDI

Chng ta cng stnh cc sdi, l cc lch ca cc gi tro c yiso vi gi trd

on tng ng y i: ei yiy iTheo E c thvit l:

E ni1

ei2 n

i1(yiy i)2

Gism hnh hp l, nunln th chng ta c thxem cc sdi e i(i= 1, 2, 3,..., n) xp xnh mt mu ngu nhin tmt phn bbnh thng vi trung bnh bng 0. Phng sai cacc sdi c c tnh bng:

s2 Enp1tc l lch chun ca cc sdil:

2Trong khng gian n chiu, mt phng trnh tuyn tnh n bin sxc nh mt siu phng n -1 chiu, c bit siu

phng trong mt phng (2 chiu) chnh l ng thng v siu phng trong khng gian vt l (3 chiu) sl mtphng.

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s Enp1

trong p l sbinx(trong v dca chng ta, p = 4).

s cng c gi l sai schun ca hi quy.

Trong v dca chng ta, chng ta c phng trnh:

Y 00 00-0Cn cvo phng trnh ny, chng ta c thtnh gi trdonca y (k hiu l y ) cho bt kgi trcho sn no cax1, x2, x3vx4. Cc gi trdon cng vi gi trquan st c trnh

by di dng bng nh sau: (bng 2)

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Bng 2: Cc gi trquan st, don ca y v sdi.

i yi y i e i1 78,5 78,495 0,005

2 74,3 72,789 1,511

3 104,3 105,971 -1,671

4 487,6 89,328 -1,728

5 95,9 95,648 0,252

6 109,2 105,274 3,926

7 102,7 104,148 -1,448

8 72,5 75,676 -3,1769 93,1 91,722 1,378

10 115,9 115,619 9,281

11 83,8 81,810 1,990

12 113,3 112,326 0,974

13 109,4 111,693 -2,293

vi tng bnh phng cc sdi l:

E

n

i1ei

2

47

945

v lch chun ca sdi l:s = 2,45.

Phng php bnh phng nhnht cho mt siu phng "ph hp tt nht" d cc ginh nitrn c thohay khng. Tuy nhin, nu nhng ginh l hp l, cc c lng bnh phngnhnht c mt sc tnh quan trng:

- cc i lng b0, b1, b2, b3v b4 l cc c lng khch quan ca cc tham s0, 1, 2, 3v 4.- trong stt ccc c lng khch quan l hm tuyn tnh caxith cc c lng bnh

phng nhnht c phng sai nhnht.- i lng s2l mt c lng khng thin vca 2. Hn na, tginh rng cc ei clp v phn bbnh thng, theo cc c lng bnh phng nhnht b0, b1, b2, b3v b4cng phn bbnh thng.

xc minh cc ginh a ra, vic xem xt cc sdi l rt quan trng. Mt biu cacc sdi l mt cng chtrtt pht hin bng mt cc vi phm c thc vginh

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tnh bnh thng. Mt thphn tn ca cc cp (y ie i), tc l ca cc sdi theo cc gi trdon, thng pht hin m hnh tuyn tnh ang xt l chnh xc hay khng.

Nu cc ginh c tha mn, cc sdi phi nm ri xung quanh 0 v cng phi khngphthuc vo gi trdon ca y. Phng sai ca cc sdi khng l hng s, sphthuc

phi tuyn tnh v cc hnh thc vi phm khc vcc ginh ca m hnh c thc phthin qua cng cthn gin ny.

Bng 2 cng a ra mt lit k cc gi trdon ca y v cc sdi tnh c tphngtrnh thc lin hy vi 4 bin x. Mt biu ca sd v gi trdon ca y (y i ei) cchotrong hnh 1. Hnh nh khng c mi quan hgia hai bin ny, v do ginh cc sdingu nhin c vc tho.

Thnh thong, thphn tn ca cc sdi so vi bin c lp x ring lcng c thcung cpthng tin vcc phthuc khng tuyn tnh. Nu sdi l v cng ln vgi trtuyt i th

chng ta c thnghi ngti mt ngoi l, hoc m hnh tuyn tnh c thkhng hp lchoim cth. Chng ta skhng i su hn vo cc chi tit ca phn tch sdi v n l mtchrng ln ca nghin cu thng k.

Hnh : thphn tn ca cc sdi theo cc gi trdon ca y

III. PHN TCH PHNG SAI

Phn tch phng sai sphn ngun cc bin thin tng thgia cc quan st Ythnh binthin do hi quy vXv cc bin thin di hay cha gii thch c. V vy, chng ta c thni:

Tng bin thin quanh Trung bnh = Bin thin gii thch do m hnh hi quy + Bin thin di

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Theo cch k hiu ANOVA, chng ta c thvit tng ng:

SSTO = SSR + SSE

hoc,

ni1

yiy2 n

i1y iy2

n

i1(yiy i)2

SSTO lin kt vi n -1bc tdo. i vi R, c nm tham s(b0, b1, b2, b3v b4) trong m

hnh, nhng rng buc ni1 y i-y 0lm mt i 1 bc tdo, do cui cng n c 4 bc tdo.i vi SSE, c n sdi (ei), tuy nhin, 5 bc tdo bmt do nm rng buc ln cc eilinkt vi vic c lng cc tham s0, 1, 2, 3v 4. th5 phng trnh (xem Ch8).

Chng ta c thtp hp nhng dliu ny trong mt bng ANOVA nh sau:

________________________________________________________________Ngun Df SS MS

________________________________________________________________

Hi quy 4 R ni1 y i-y2 MSR = SSR / 4Sdi n - 5 E ni1 (yi-y i)2 MSE = SSE / (n-5)Tng n - 1 TO ni1 yi-y2

________________________________________________________________

Trong v dca chng ta, cc gi trstng ng cho phn tch l:

Bng : Phn tch phng sai ca m hnh

Y 00 00-0________________________________________________________________Ngun Df SS MS

________________________________________________________________Hi quy 4 2.667,90 666,97Sdi 8 47,86 5,98Tng 12 2.715,76

________________________________________________________________

Trong v dny, tng cc bnh phng quanh trung bnh l 13i1 yi-y2 271576v tng cc bnh phng "gii thch" bng phng trnh hi quy l 13i1 y i-y2 266790cn li mt tng cc bnh phng di l 13i1 (yi-y i)2 4786.

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Do , tlca tng bin thin ca yquy cho m hnh l 2.667,90 / 2.715,76 = 0,9824 hoc98,24%. Chng ta stho lun v hsR2iu chnh" trong mt phn sau ca chny.

V tp cc dliu c 13 quan st, do c 12 bc tdo lin kt vi tng cc bnh phng.tng th Mt khc, m hnh ny c 5 tham s(bao gm chschn), do sbc tdo

(df) kt hp vi tng cc bnh phng hi quy l 4, do bnh phng trungbnh do m hnh lMSR = 2.667,90 / 4 = 666,97.iu trn cng lm cho sbc tdo lin kt vi tng bnhphng di l 8 v v thbnh phng trung bnh ca n l MSE = 47,86 / 8 = 5,98. Theo nhngha, kim nghim Fcho ngha ca m hnh sl F = 666,97 / 5,98 = 111,479, gi trnyc ngha thng k cao, so vi gi trl thuyt ca F bng 3,84 mc = 0,05 vi 4 bc tdotv 8 bc tdo mu.

Trung bnh bnh phng di (MSE) c thc coi l mt c lng cho phng sai ca m

hnh. Do , cn (bc hai) ca MSE", tc l: 598 245c thc coi l c lng calch chun ca m hnh. Cn phng sai quan st ca y l

13i1 y i-y2

12 27157612 22631au khi xt n m hnh (vi 4 bin s), phng sai chl 5,98, gim 97,3%.y l mt con sng n!

3.1. KIM NGHIM NGHA TNG TH

M hnh tuyn tnh viphshi quy tng phn chc ngha nu t nht c mt hskhc 0(c ngha). Chng ta so snh m hnh y :

Y= 0+ 1x1+ 2x2+ 3x3+ 4x4

vi m hnh thu gn;

Y= 0

c c bng cch ng thi loi btt cpbin, tc l bng cch cho:

1= 2= 3= 4 = 0

c lng bnh phng nhnht ca0 trthnh trung bnh mu ca Y, tc l b0

; SSE

tng ng l:

E0 n

i1yiy2

i lng ny tht ra c gi l SS"tng th" (Total SS) trong bng ANOVA.

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Thth cu hi t ra l "bn bin x c gii thch mt cch c ngha sbin thin quan stca dliu hay khng? C thgii quyt cu hi ny bng cch tnh thng k tng thF, l:F= 666,97 / 5,98 = 111,29. Gi trny so snh (mc = 0,05) vi phn trm 95 ca phn bF c 4 v 8 bc tdo tng ng. M, F0,95; 4; 8= 3,84. V vy, c bng chng mnh mni

rng t nht mt trong cc hshi quy tng phn l khc 0, v chng ta khng thth bttccc bin hi quy.

Tt ccc chng trnh my tnh ln cho phn tch hi quy u tnh ton kim nghim nghatng thv thng k F lin quan trong bng ANOVA. Tri vi thut ngca chng ta, cc SS(tng bnh phng) trong bng ANOVA trong bn in my tnh hu nh c ghi vi nhn :"TOTAL" (cho m hnh thu gnY= 0), "REIDUAL" (cho m hnh y Y= 0+ 1x1+

2x2+ 3x3+ 4x4) v "REGRESSION" (cho tng cc bnh phng lin kt vi m hnh thugn). Thut ngca chng ta c li thl nhc nhngi sdng rng cc m hnh ton hcthc sang c so snh chkhng chl tng cc bnh phng.

3.2. HSXC NH

Ngoi gi trtng thF, ngi ta thng tnh hsxc nh R2, c nh ngha nh sau:

R2 E0EpE0

1 ni1 (yiy i)2ni1 yiy2 (Lu : chsk trong SSEchSSEquy cho k bin trong m hnh, do , SSEpchcc SSEdop bin trong m hnh, trong khi SSE0cp n SSEcam hnh khng c bt k bin c lp no).

i lng ny thng c hiu l tlbin thin ca Ygii thch do hi quy vx1, x2,. . . , xp.Trong v dca chng ta, R2= 0,982 ni rng 98,2% bin thin (khng phi phng sai) ca Yc quy cho cc bin c lp x1n x4. Nu khng c sphthuc tuyn tnh th R

2gn vi0; tuy nhin trong trng hp c mt sphthuc tuyn tnh mnh mth n sgn vi 1. Tuy

nhin, khi lun gii vhsxc nh cn thn trng, bi v i vin p +1, ta c y i yi, v do R2= 1. Lun gii R2chc ngha khi n tht ln sop.

Cn bc hai R R2c gi l hstng quan a bingia Yvx1, x2,. . . , xp. R bngvi hstng quan ca cc cp (y i yi). Trong v dca chng ta, R = 0,991.3.3. R

2

IU CHNH

Mt snh thng k cho rng hsxc nh cn c iu chnh c thnhn ra cscc bin c lp trong m hnh. L do l hsny ni chung c thblm ln hn nu ccbin c lp c thm vo m hnh. thy iu ny, lu rng khi SSTO gicnh th cmi bin c lp thm vo SSEc xu hng trnn nhhn. Mt php o nhn ra c s

bin c lp trong m hnh c gi lhsxc nh iu chnhv k hiu lRa2

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Ra2 1n1

np ETOTrong v dca chng ta, Ra

2iu chnh l 0,974, khng sai bit ln so vi R2= 0,982.

IV. XY DNG M HNH V PHN TCH PHNG SAI

Mt khi tm c phng trnh hi quy, chng ta mun bit liu c cn a tt cccbin c lp vo trong phng trnh hay khng. Chng ta c thnu ra hai trng hp trong cc bin hi quy c thc ly bt ra khi m hnh m khng lm mt thng tin:

(a) Khng c quan hgiaxjv Y;(b)nh hng caxjtrn Ybchi phi qua cc bin khc. Khnng ny c thc minh h

n gin nht bng mt v dm phng. Gism hnh sau l hp l:

y = x1+2 x2

ng thi, gisrng binx3= x1+ x2cng c o. V thchng ta cng c thvit mhnh di dng nh:

y = x2+ x3

hoc nh: y = 2x3- x1

cui cng, chng ta cng c thm tynh l mt hm tuyn tnh ca cba bin hi quy, vdnh:

y = 2x1+ 3x2- x3

Trong m hnh ny, mt bin r rng l tha - ngay ckhi y phthuc vmt hm svox1, x2vx3.Trong thc hnh, thng kh nhn ra nh sphthuc hm sgia cc bin hi quy.Quan hhm thng ln ln vi sai so c, hoc n khng ng l tuyn tnh.

Trong chai trng hp, chng ta u ni n sd tha: binxjtrong trng hp (1) hocmt trong cc bin x1, x2vx3trong trng hp (2) c thc ly ra khi m hnh m khng

lm mt thng tin, tc l n l khng cn thit. Chng ta cgng n gin ha m hnh thngqua vic loi bcc bin hi quy tha. Tuy nhin, nh khnng 2 cho thy, tha khng cngha l cc bin hi quy khng c nh hng.

Vic loi bbinxj khi m hnh c thc hin bng cch cho tham sj= 0. n gintrong cch k hiu, chng ta ginh rng cc bin c xp thttheo cch m m cc thamsc cho bng khng l nhng tham scui cng. M hnh vi tt cpbin hi quy:

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Y=0+ 1x1+ 2x2+ ... + pxp

sc gi lm hnh y (hoc m hnh thay th). Bng cch c gi l rng buctuyn tnh, tc l bng cch t, chng hn:

r+1

=....= p

= 0 (vi r < p)

ta c mt m hnh n gin ho:

Y=0+ 1x1+ ...+ rxr

m hnh ny c gi lm hnh thu gnhay m hnh khng.

Cc m hnh y v thu gn c thc so snh nh sau. i vi chai m hnh ngi tac lng cc tham sv tnh ton cc tng bnh phng di tng ng. Tng ny c khiu bng SSEp(m hnh y ) vSSEr(m hnh thu gn). V siu chnh siu phng trnn ti thn do vic loi b, ta lun c SSEp< SSEr. By gichng ta lp tl:

F ErEpprEpnp1 Theo githuytr+1=....= p = 0, thng k ny c phn btheo phn bF vi (p - r)bc tdo tv (n - p -1)bc tdo mu. Nu tlFtnh c nhhn phn trm (1 - )ca phnbFvi (p - r) v (n - 1 - p) bc tdo, th githuyt c chp nhn. Nu khng, chng ta gili m hnh y .

Tng cc bnh phng di v kim nghim F thng c tm tt trong mt bng phn tchphng nh sau:

_______________________________________________________M hnh SS nhnht DF

_______________________________________________________M hnh thu gn SSEr n - r - 1

0+ 1x1+ ...+ rxr

M hnh y SSEp n - p - 10+ 1x1+ 2x2+ ... + pxp

Gim SSEr- SSEp p - rr+1=....= p = 0_______________________________________________________

Cn lu rng tnh chnh xc ca kim nghim-Fphthuc vo cc ginh ca m hnh.Trong nhiu trng hp, chnn xem gi tr- Fn thun l mt thc o m tcho skhcbit gia hai m hnh m thi. iu ny c bit ng nu mt lot cc so snh m hnh cthc hin chtrn cng mt tp dliu. Chng ta strli vn ny trong mt phn ktip.

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KIM NGHIM GITHUYT TNG PHN

Trong trng hp c bit thhai r = p - 1, cp phn trn, m hnh thu gn l:

Y=0+ 1x1+ 2x2+ ... + p-1xp-1

thu c tm hnh y bng cch p t hn chp= 0. n gin chng ta li gisrng chnh tnh d tha ca chnh bin thp cn c kim nghim. Nh trong githuyttng th, chng ta tnh hai tng bnh phng nhnht Sp min(m hnh y ) v S(p-1) min(mhnh thu gn). Bng cc dng ts:

F Fp 0 Ep1EpEp

n

p

1

gi l thng k-Ftng phn, chng ta kim nghim tnh d tha caxptrong m hnh hi quy.Theo tnh hp lca cc githuyt khng p= 0, thng k kim nghim c phn bFvi 1bc tdo tv (n - p - 1)bc tdo mu.

Bng cch so snh gi trFthu c vi phn trm (1 - )ca F: nu F < F1-, , n-p-1,chng tachp nhn m hnh thu gn n gin hn. Tuy nhin, t khng thkt lun rngxpkhng cnh hng n Y, v nh hng caxpc thc biu hin thng qua mt scc bin hiquy khc. iu duy nht chng ta c thni l ton bcc bin hi quy khng m tmi quanhtuyn tnh tt hn so vi bbin thu gn.

Nu nhiu gi tr- Ftng phn c tnh ton, chng chnn c dng n thun cho mcch m tthi. c bit, skhng ng nu tmt shshi quy tng phn khng c ngha kt lun rng c thloi bng thi nhng bin hi quy tng ng ra khi m hnh.Tht vy, vic loi bmt bin hi quy duy nht c thnh hng mnh mbi cc hskhc.

Nhiu chng trnh my tnh cho thng k-F tng phn vi n - p - 1 bc tdo,chkhng phithng k-Ftng phn kim nghim tnh d tha ca duy nht mt bin. Cc quan hgiahai thng k ny l n gin F = t2, phn nh skin l F vi 1 bc tdo tv mbc tdomu th ging nh bnh phng ca t vi m bc tdo.

SAI SCHUN CA HSHI QUY

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Tgi tr-Ftng phn, sai schun ca bjc thc c tnh nh sau:

Ebj bj

F

j

0

Do , trong v dca chng ta, ngi ta c thbsung phng trnh hi quy c lng bngmt danh sch cc gi trF tng phn v sai schun ca cc hshi quy tng phn nhtrong bng sau.

Bng 4: Hshi quy c lng, sai schun v kim nghim-F tng phn_______________________________________________j bj SE (bj) F (j= 0)_______________________________________________0 62,418 71,59 0,760

1 1,551 0,745 4,3312 0,510 0,724 0,4963 0,102 0,755 0,0184 -0,144 0,710 0,041

_______________________________________________

Cc SEkh ln (so vi gi trtuyt i ca cc hs) v cc gi trF nhlm ta nghi ngrngtp cc bin c thgim bt i theo mt cch no . Mt cch tip cn loi bcc bin dtha sc tho lun trong phn tip theo.

VI. LA CHN MT TP CON CC BIN HI QUY

Vic la chn mt tp con thch hp ca cc bin hi quy thng rt kh khn. Cc thut tonkhc nhau c thmang li cc kt qukhc nhau. Cng c thxy ra l cc tp con khc nhau

c kch thc bng nhau li cho kt qucng cht lng trn thc t. Trong v dca chngta, chng ta c thloi bmt trong bn bin c lp v sc c bn m hnh thu gn vihsxc nh nh cho trong bng sau y:

Bng 5: So snh cc m hnh khc nhau bng cch sdng R2_______________________________M hnh khng cha R2

_______________________________

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x1 0,973x2 0,981x3 0,982x4 0,982

_______________________________

V khng c mt m hnh no vi ba bin hi quy trong cc m hnh trn sinh ra tmt trongnhng m hnh khc thng qua mt hn chtuyn tnh, nn chng khng thso snh vi nhauqua phn tch phng sai. Da theo hsxc nh, ba trong bn m hnh cha ba bin hiquy nu trn u tt gn nh nhau.

Tnh hung nh thny c bit dxy ra khi cc cc bin hi quy lin quan vi nhau mtcch mnh m. Cc quan hgia ccxjthng c m tbng mt ma trn tng quan,mc d gi trm ccxjly c coi l cnh cho mc ch c lng bnh phng nhnht.

Do skhng tng thch gia cc m hnh c cng sbin hi quy ngi ta thng sdng

thut ton phn cp la ra mt tp hp con ca cc bin c lp. Chng ta m typhng php loi bli. Bt u vi m hnh y vipbin, trc ht chng ta loi bcc bin vi gi trFtng phn nhnht. Trong cc bin cn li, chng ta so snh m hnhgmp -1bin vi tt ccc m hnh c c do loi thm mt bin na. Mt ln na, bin vigi trF tng phn nhnht sbloi ra ... v cthtip tc cho n khi cui cng tt ccccc bin hi quy c loi ra khi m hnh. p dng cho v dca chng ta, vic loi bli chokt qunh sau (MPF = minimum partialFvalue, gi trF tng phn nhnht).

Bc 1: M hnh y ______________________________________________Bin bi SE(bj) F (j = 0)

______________________________________________x1 0,745 4,33 1,551x2 0,510 0,724 0,50x3 0,102 0,755 0,02 - MPFx4 0,04 0,710 -0,144H schn 62,418

______________________________________________R2= 0,982, Sai schun ca sdi = 2,448

Bc 2: Loi bx3

______________________________________________Bin bi SE(bj) F (j = 0)

______________________________________________x1 1,452 0,117 154,01x2 0,416 0,186 5,03x4 -0,237 0,173 1,86 - MPFH schn 71,648

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______________________________________________R2 =0,982, Sai schun ca sdi = 2,309

Bc 3: Loi bx4_____________________________________________

Bin bi SE(bj) F (j = 0)______________________________________________x1 1,468 0,121 146,52 - MPFx2 0,662 0,046 208,58H schn 52,577

______________________________________________R2= 0,979, Sai schun ca sdi = 2,406

Bc 4: Loi bx1______________________________________________Bin bi SE(bj) F (j = 0)

______________________________________________x2 0,789 21,96 0,168H schn 57,424

______________________________________________R2= 0,666, Sai schun ca sdi = 9,077

Loi bx2trong bc 5 cui cng cho ta c lng ca b0= y= 95,423 v sai schun ca s

di = lch chun ca Y= 15,044.

By gi, tuchng ta a ra quyt nh quan trng l sly bao nhiu bin hi quy v lnhng bin no m chng ta mun a vo m hnh. Hu ht cc chng trnh c khnng

chra mt gi trti hn ca Fmin. Thut ton dng li ngay khi khng cn mt gi trF tngphn nhhn Fmin.Tuy nhin trong chui thbc ca cc kim nghim ta khng thxem ccgi trF tng phn nh c lp. Do , vic sdng gi trFtng phn chtrong ngha m tcho mc ch xc nh mt thttrong cc bin c lp. Mt tiu chun c thp dng chovic dng li l mt thay i t ngt trong R2.Trong v dca chng ta, hsxc nh gimt ngt sau ba bc, chcong gt ny l mt du hiu cho thy hai bin hi quy loi butin l tha. M hnh cn li chda trnx1vx2m thi; phng trnh hi quy (phn trongngoc l sai schun ca bj) c lng l:

52577 1468x1 0662x2(0,121) (0,046)

By gita c thso snh m hnh ny vi m hnh y . Bng ANOVA nh c a radi y cho gi trca cc thng k kim nghim:

F 9962479458 083

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gi trny nm bn di phn trm 0,95 ca phn bF vi 2 v 8 bc tdo [F0.95, 2,8 = 4,46].Vic n gin ha m hnh n 2 bin hi quy c v c bin minh.

__________________________________________________M hnh SSE Df

__________________________________________________0+ 1x1+ 2x2 57,904 100+ 1x1+ 2x2+ 3x3+ 4x4 47,945 8St gim 9,960 2

__________________________________________________

Cn nhn mnh rng li gii tm c nh vy khng hon ton c ngha l l li gii duynht ng. Mt m hnh vi (x1, x4) thay v (x1, x2) chng hn thc tscho ra cng hsxcnh. Tc l, vic la chn cc bin hi quy m tbin phthuc vn cn tungi sdng quyt nh cn cvo kin thc ca mnh vti.

Hn na, ni mt cch nghim tc, kim nghim a ra trn khng hon ton chnh xc, vn

vi phm nguyn tc rng githuyt khng nn c lp ra v kim nghim trn cng mtdliu.

Bng di y cho mt danh sch cc quan st v gi trdon ca bin phthuc (yiv y i),v cc sdi ei yi-y icho m hnh vi (x1, x2).Tng quan nhiu bin cho m hnh ny c vthnh thca cc cp yiv y i. thnycng ng thi cho php kim tra cc sdi: khong cch theo chiu ngang (hoc thng ng)ca (yiv y i) n ng thng ( y i yi) (dc = 1, gc = 45o) tng ng chnh xc vi sdie i yi-y i. Khng pht hin svi phm no tcc ginh.

Bng 6: Gi trdon v sd chom hnhY=0+ 1x1+ 2x2__________________________________________i yi y i e i

__________________________________________1 78,5 80,074 -1,5742 74,3 73,251 1,0493 104,3 105,815 -1,5154 87,6 89,258 -1,6585 95,9 97,292 -1,3936 109,2 105,152 4,0487 102,7 104,002 -1,3028 72,5 74,575 -2,075

9 93,1 91,275 1,82510 115,9 114,538 1,36211 83,8 80,536 3,26412 113,3 112,437 0,86313 109,4 112,293 -2,893

__________________________________________

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Hnh : Tng quan nhiu bin trong m hnh Y= 0+ 1x1+ 2x2

VII. CC CU HI THNG GP V TRLIHi:Ti phi lm g nu cc ginh ca m hnh (phn bbnh thng ca lch) khngc tho?

Trli:Trong thc t, ngi ta thng gp nhng bi ton trong cc ginh cho kimnghim thng k khng c tho. Tt nhin ngi ta vn c thtnh ton mt cch thun tutrn cc con stm ra mt siu phng "ph hp nht". Nhng cc kim nghim thng kkhng cn ng na. V trong chny chng ta xem xt hi quy nhiu bin chn thunnh mt cng cnn chng ta cn tham kho thm cc ti liu c lin quan.

Hi: Ngoi phng php loi bli, cn c phng cch no khc la chn tp hp con cc

bin hay khng?

Trli: Trong th vin cc chng trnh thng k, cn c cc thut ton sau y thng csdng:

(a)La chn tin: Trong phng php ny, bc 1 ngi ta tm mt bin u tin, khi lyring mt mnh scho ra gi trFcao nht. bc 2, trong stt ccc cp cc bin hi quyc cha bin chn, ngi ta tm ra mt bin lm gi trFtng thln nht. Tng t,

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bc 3 mt bin thba c thm vo hai bin c chn theo cng cch, v cthtiptc.

(b)La chn tng bc: Phng php ny l mt hn hp ca hai phng php nu. Bc1 v 2 lm nh trong phng php la chn tin. Tuy nhin tip theo , trc khi la chnthm mt bin, ta lun lun xem xt gi trFtng phn ca tt ccc bin chn. Nu mtgi trnh thgim xung di mt gii hn nht nh no (do ngi sdng p t) thbin c vn sbloi b, v ri mt bin khc c thm vo, v cthtip tc. Bng diy cho mt danh sch ca tt c15 tp con c thc ca cc bin hi quy. Da vo bngny, chng ta c ththeo di cba thut ton.

_____________________________________________Sbc Li Tin Tng bc

_____________________________________________1 x1x2x3x4 x4 x42 x1x2x4x1x4 x1x4 x1x43 x1x2 x1x2x4 x1x2x4

4 x2 x1x2x3x4 x1x25 x1x2x4

_____________________________________________

Mt so snh gia loi bli v la chn tin cho thy hin tng sau, kh him gp trong thct, l so ngc ca la chn tin trong tt ccc giai on khng cho ra cng mt tp conca cc bin hi quy nh loi bli. Thtc dng li bc 5, v bc ktipx4sbloibthm mt ln na.

Trong phn sau, chng ta slun lun sdng loi bli, v ngay khi bt u ta tin hnh kimnghim tng th. Nu kim nghim ny cho thy khng c ngha, th quyt nh cc hs1,

2,. . . , pu bng 0. Trong trng hp ny, vic la chn cc tp con l khng cn thit,trong khi cc phng thc khc, tmt scc bin hi quy hon ton d tha, ta c thlaly mt vi bin mc d rng trong thc tchng khng c nh hng.

Ngy nay, nhc my tnh nhanh chng cng c thsdng ci gi l thut ton "hi quymi tp con ". Nu c p bin hi quy, mt tm kim vt cn bao hm rng (2p-1) tp hp conphi c phn tch - mt nhim vhu nh rt xa vi, ngay cnup nhbng 10. Tt nhin,c th hn chslng cc tp con nhkinh nghim, tuy nhin slng tnh ton vn l rtln.

Hi:Trong loi bli, ngi ta c thny ra mt tiu chun dng tng v do c thtrnhphi i theo thut ton cho n tn cng hay khng?

Trli: Trong hu ht cc chng trnh my tnh, iu ny tht ra c thlm c. Tuy nhinthtc loi bli thc hin tu n cui c thcung cp thng tin bsung vtm quantrng ca tng bin hi quy ring l.

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Hi:C phi cc ginh trong v dca chng ta khng bvi phm qu nhiu khng? Nh cthnhn thy tcc dliu ma trn, c phi cc bin tx1 n x4chc chn khng phi phnbbnh thng khng?

Trli:y l mt trong nhng vn bhiu lm nht trong scc nh phn tch hi quy

hoc ngi sdng. Cc ginh vbnh thng c p dng cho cc sdieikhng phichoxi, Ccxic ginh l bin cnh (khng ngu nhin).

Hi:Cn schnh xc bng sca cc kt qul nh thno? Nh c bit theo kinhnghim, kt quca cc chng trnh hi quy khc nhau khng lun lun ng nht tt ccc chsthp phn?

Trli: Vic tnh ton cc hschyu bao gm vic tm nghch o ca ca mt ma trn.Trong mi lin kt ny, mt vi vn vmt shc nht nh c thpht sinh, c bit l khicc bin hi quy lin quan cht ch vi nhau. Do , cng mt bi ton hi quy ngi takhuyn nn phn tch vi cc chng trnh khc nhau. Nu kt qukhng thng nht, mtcch c thc gii quyt ca vic ny l xem xt cc thnh phn chyu cc bin hi quynh cc bin c lp thay v xem xt cc bin hi quy tht s. Mt schng trnh phn tchhi quy chuyn su cng c cho thm la chn ny..Hi:Ti sao chng ta khng sdng m hnh y vi tt cp bin? Ti sao chng ta lmsph hp ti thn bi vic loi bcc bin? V R2lun tng nu chng ta bao gm nhiubin hn, n chc chn khng thbtn thng nu chng ta sdng tt ccc don csn?

Trli:Theo quan im khng-thng k, iu nhn xt ny l ng - a vo thm mt bin

slun lun ci thin mc ph hp nh c o bng hsxc nh (R2

). Vn mang tnhthng k l liu sci thin ny c phi chl thun tu do ngu nhin hay khng. Ngoi ra cnc cc l do khc chn mt tp hp con ca cc bin hi quy. V d, chng ta c thlp lunrng mt m hnh n gin l ng chn hn mt m hnh phc tp, min l chai m hnhhot ng tt nh nhau - y l mt quy tc ng n khng chcho cc m hnh thng k. Mtl do rt quan trng l sbt n ca cc c lng tham snu bao gm nhiu bin qu - sbt n theo ngha km chnh xc shc ln tng sai schun ca cc c lng tham s.

iu ny c ththy kh r trong tp dliu trong v dca chng ta: nu chng ta theo diqu trnh loi bli trong bc 1, tt ccc hsrt khng n nh (theo ngha cc sai schun cao), trong khi sau khi loi bx3vx4, cc hscn li l kh n nh. Hin tng nythng xy ra khi cc cc bin hi quy lin quan cht chvi nhau. Nh c tha nhn, vicchn la tp hp con l cch duy nht c thc xl vn ny. Mt k thut th vl "nhhi quy", k thut ny nh i sbin thin cao ca cc c lng tham svi mt mcthin lch no (hi vng khng ng k). K thut ny khng c trnh by trong kha hcca chng ta, nhng c giquan tm c ththam kho thm ti liu ph hp vti ny.

Hi: Chng ta tho lun vvai tr ca cc bin skhc nhau trong mt phng trnh hiquy, v chng ta thy rng mt sbin c thquan trng hn nhng bin khc trong d

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on Y. Chng ta khng thlm mt iu tng tcho cc quan st hay sao? Tc l c thoc mc m mi quan st c nh phng trnh hi quy hay khng?

Trli: Phng php nh gi nh hng ca cc quan st ring lln hi quy thc sc quan tm nhiu trong nhng nm gn y. Trong thc t, chng ta xem xt ngn gn

cc k thut ny trong Ch8 (quan st c nh hng , thng k Cook, vv.) Tuy nhin, nh cp trong phn phn tch sdi, mt cch tip cn n gin chng hn l bqua quanst iv tnh ton li cc phng trnh hi quy da trn n-1 dliu im cn li. au nhhng ca quan st thi c thc nh gi bi nhng thay i xy ra trong cc hshiquy. Mt dn nhp tt p vcc k thut ny, cc bn c ththam kho mt bi bo ca Efronv Gong (1983).

Bit r ngn ngun ca cc dliu l rt hu ch xc nh quan st no cn c xem xtcn khn tm nh hng ca n. iu ny cng ng cho dliu trong v dca chng ta,trong nhng kin thc chi tit ca th nghim c thc sdng xy dng mt m hnhthch hp v chn cc quan st.

Hi:Ta c nn da vo hsc nh (R2) nh gi mc tt xu vtnh ph hp cam hnh hay khng?

Trli: lun gii R2, c thhu ch lu kt qul thuyt sau: khi cc bin-x khng ciu kin (tc l gi trca chng c thngu nhin theo bt k quy lut no, hoc chng c thc cnh), v nu cc bin ngu nhin Ykhng phthuc vo cc bin-x th theo cc ginh kh yu vsphn bca Y (c bit l tnh i xng v khng tng quan ca cc thc

hin ca yi) ta c gi trmong i caR2lE

R2

p

n-1

. Tc l, i vi cmun = 21vp

= 10 bin-x, chng ta c thhi vng c R2= 0,5 thun do may ri! V vy, nh mt quy tc thchnh, chcc tlR2vt qu gi trmong i mi cn c lun gii.

Lin quan vi kt quny, cng ng ch l kim nghim tng thFcho ngha ( lin quancht chn R2vmt hm s) cng ng theo iu kin yu tng t. iu ny c ngha cbit l kim nghim Ftng thkhng yu cu tnh bnh thng chnh xc ca cc sdi mcn mt phn bi xng ni theo cch rng ri. i vi cc kim nghimFtng phn, ngita cha bit tnh ng n ca chng phthuc vo tnh bnh thng nghim ngt n mc no. Tuy nhin, nhn chung, cc kim nghim ny l ng tin cy hn khi cmun ln.

Hi:Nu chng ta chc mt hi quy , nhng hm hi quy l mt hm bc hai ca x. Bi tonny c thc tip cn bng k thut hi quy tuyn tnh hay khng?

Trli:trli cu hi ny, chng ta trc ht phi lm r cc thut ng. Trong m hnh hiquy tuyn tnh ,Y=0+ 1x1+ 2x2+ ... + pxp, T"tuyn tnh" thc ra ni ti skin Ylmt hm tuyn tnh theo cc tham sj. M hnh m cu hi mun ni ti c dng Y0

, c thc xl trong khun khca hi quy tuyn tnh vi hai cc bin hi quy,

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cthl,x1= xvx2= x2. Mt khc, m hnh Y0 khng tuyn tnh theo

tham s2v k thut hi quy tuyn tnh khng p dng c.

VIII. MT S KIN

Phn tch hi quy nhiu bin l mt trong nhng k thut thng k c sdng thng xuynnht (v sdng qu l), c bit l trong cc y vn. Mt sngi dng nh bqua ccnguyn tc khoa hc v gt da cc dliu cho ph hp vi githuyt hto ra. Cc kinsau y c trch tcun sch ca D. Altman ':

"Kh c ththo lun chi tit nhiu vn quan trng c nh hng n phn tch hi quynhiu bin v lun gii n, nhng cc nhn xt sau y chra cc khu vc kh khn cn quantm:

Khi c mt slng ln cc bin c tim nng gii thch chng ta mong i mt strong ccbin c ngha chdo may ri. Khng c cch no hon ton tha ng tm kim mt mhnh ph hp nht m khng phi trgi bng mt cu trli qu lc quan. Vi nhiu bin ngtuyn a vo m hnh, mt snh nghin cu sdng cc kt quphn tch n bin quyt nh nhng bin cn c khai thc trong phn tch a bin. Chin lc ny chng titkim c g vi hi quy tng bc tin, nhng c thct gim ng kthi gian tnh ton (vchi ph) cho hi quy tng bc li hay hi quy mi tp con. Ti khng khuyn vic la chntrc, nhng nu sdng, vic la chn phi da trn mt tiu chun lng lo, chng hn p

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