Bai Giang CPPDB [Compatibility Mode]

8/2/2019 Bai Giang CPPDB [Compatibility Mode]

1/238

Chng 1. M U


2/238

1.1 Khi nim v phn loi d bo1.2 Cc nguyn tc d bo

1.3 Tng quan v tiu chun la chnphng php d bo

1.4 Cc phng php nh gi d bo

Chng 1

M U


3/238

1.1.1 Khi nim d bo

Chng 1

1.1 Khi nim v phn loi d bo

Ting Hy Lp progrosic ngha l bit trc

T c xa d bo c p dng trong cucsng hng ngy nhng mang nng mu scthn b tn gio.


4/238

Chng 1


Cc hin tng t nhin

Cc hin tng x hi

Cc hin tng v i sng x hi

Thi Hy lp c chia chia thnh cc lnh vc:


5/238

Chng 1


n th k XVI, XVII khi cc khoa hc t nhin pht trin c bit l s xut hin cc hcthuyt ca Marx th d bo t thn b kinhnghim pht trin thnh b mn khoa hc clp.


6/238

Chng 1


Ngy nay vai tr ca d bo ngy cng ckhng nh v tng ln ng k trong mi lnhvc ca i sng x hi.


7/238

Chng 1


D bo l s tin on c cn c khoa hc,mang tnh cht xc sut v mc , ni dung,cc mi quan h, trng thi, xu hng phttrin ca i tng nghin cu hoc v cch

thc v thi hn t c cc mc tiu nhtnh ra trong tng lai


8/238

Chng 1


D bo bao gi cng c thi gian xc nh hay

tm xa ca d bo. Tm xa d bo l khongcch ti a t hin ti n thi im pht biud bo


9/238

1.1.2 Phn loi d bo

Chng 1


- D bo kinh t- D bo tin b khoa hc cng ngh

- D bo dn s v ngun nhn lc- D bo x hi- Do mi trng sinh thi

a. Theo i tng


10/238

Chng 1


- D bo ngn hn: c tm xa d bo t 1 3nm, lm cn c cho vic xy dng v iuchnh cc k hoch ngn hn. Sai s cho phpnn < 5%

b. Theo tm xa

- D bo tc nghip: c tm xa ngn, c th lgi, ngy, tun, thng n di nm. Sai sthng < 3%


11/238

Chng 1


- D bo trung hn: c tm xa t 5 - 7 nm, lmcn c cho vic lp cc k hoch trung hn, khoch 5 nm. L d bo c vai tr rt quantrng trong nn kinh t quc dn


12/238

Chng 1


- D bo di hn: c tm xa d bo t 10 20

nm, nhm cung cp thng tin v d liu chovic hoch nh cc chin lc pht trin kinht x hi di hn hay hnh thnh mt tm nhnc tnh nh hng trong s pht trin lu di

ca t nc- Ngoi ra cn c cc d bo siu di hn


13/238

Chng 1


- D bo vi m: l cc d bo cp n v nhhay cc doanh nghip

c. Theo quy m, cp ca i tng

- D bo vm: l cc d bo v cc ch tiuln mang tnh tng hp bao hm ton b nnkinh t, cc vng kinh t, cc ngnh


14/238

1.2.1 Nguyn tc lin h bin chng

Chng 1

1.2 Cc nguyn tc d bo

Yu cu xem xt mi vn trong mt iukin v hon cnh c th, tnh n s phthuc ln nhau gia cc mt ca vn nghin

cu. ng thi phi tnh n c mi quan hvi cc lnh vc khc...


15/238

1.2.2 Nguyn tc k tha lch s

Chng 1


S nghin cu su sc trng thi ca itng d bo trong qu kh v hin ti s lc s tt, vng chc d bo cc xu hng

vn ng v pht trin ca i tng d botrong tng lai


16/238

1.2.3 Nguyn tc tnh c th v bn cht

ca i tng d bo

Chng 1


Phi tnh n c im, c th v bn chtca i tng d bo, tnh c th v nhng

quy lut pht trin ca n, gi tr tuyt i v gitr tnh ton ca gii hn pht trin trong tnglai


17/238

1.2.4 Nguyn tc m t ti u i tngd bo

Chng 1


i hi m t i tng d bo nh th no nhm m bo s xc thc v chnh xc cho

trc ca d bo vi chi ph d bo thp nht


18/238

1.2.5 Nguyn tc v tnh tng t cai tng d bo

Chng 1


i hi khi phn tch phi thng xuyn sosnh nhng tnh cht ca i tng d bo vinhng i tng tng t bit v vi cc

m hnh ca cc i tng nhm tm ra itng tng t hoc mt s yu t ca mhnh... ng dng vo d bo cho cc itng mi nhm tit kim chi ph v thi gian...


19/238

1.3.1 Tng quan v phng php d bo

Chng 1

1.3 Tng quan v tiu chunla chn phng php d bo

L cc phng php da trn mi lin h xy dng m hnh d bo

Vic xy dng m hnh ch yu da trn 2phng php c bn l phn tch hi quy vchui thi gian

a. Nhm cc phng php m hnh ha


20/238

Chng 1


c s dng trong trng hp cc s liuthc nghim khng p ng c yu cu

nghin cu hoc i tng d bo khng thnh lng hoc c th nh lng nhng rttn km

b. Nhm cc phng php chuyn gia


21/238

Chng 1


c thc hin trn c s thu thp x l cc kin nh gi d bo ca cc chuyn gia ri

a ra kt qu d bo. Trong c s dngcc cng c ton hc v logic m bo sthng nht khch quan


22/238

Chng 1


c tin hnh theo 1 quy trnh lp nhm thchin vic xch li gn nhau gia cc phng n

nhn c t vic m phng theo m hnh trnmy tnh vi cc kin ca cc chuyn gia choti khi t c s thng nht chp nhn c

c. Nhm cc phng php kt hp


23/238

1.3.2 Tiu chun la chn pp d bo

Chng 1


chnh xc ca d bo

Chi ph ca d bo

Tnh tng hp v kh nng ng dng ca pp

Thi gian d bo

C s d liu d bo


24/238

1.4.1 nh gi trc d bo

Chng 1

1.4 Cc phng php nh gid bo

Kim tra thng tin v tnh y , chnh xc,ph hp vi mc tiu d bo, di chuiquan st, cu trc chui quan st


25/238

1.4.1 nh gi trc d bo

Chng 1


Kim tra cc bin tham s i din cho cc miquan h ca chng ti i tng d bo

Kim tra dng hm hoc m hnh d bo sdng


26/238

1.4.2 nh gi sau d bo

Chng 1


: l gi tr thc t thi im t.K hiu

: l gi tr d bo thi im t.

: l sai s d bo thi im t.

tY

tY

te

ttt YYe


27/238

a. Sai s trung bnh

Chng 1


n

YY

n

e

ME

n

t

tt

n

t

t

11


28/238

b. Sai s tuyt i trung bnh

Chng 1


n

YY

n

e

MAE

n

t

tt

n

t

t

11


29/238

c. Sai s bnh phng trung bnh

Chng 1


n

YY

n

e

MSE

n

t

tt

n

t

t

1

2

1

2


30/238

d. Sai s phn trm tuyt i trung bnh

Chng 1


n

Y

YY

n

Y

e

MAPE

n

t t

tn

t t

t

11


31/238

e. Cn bc hai ca sai s bnh phng trung bnh

Chng 1


n

YY

n

e

MSERMSE

n

t

tt

n

t

t

1

2

1

2


32/238

Chng 1


f. H s khng ngang bng Theils U

2

1

2

tt

tt

NAIVE

NEW

YY

YY

RMSE

RMSEU

: RMSE ca m hnh d bo: RMSE ca m hnh d bo th

NAIVE

NEW

RMSE

RMSE


33/238

Chng 1


2

1

2

tt

tt

NAIVE

NEW

YY

YY

RMSE

RMSEU

: Tt hn MH d bo th: Ngang vi MH d bo th: Km MH d bo th1

1

1

U

U

U


34/238

n Yt Yt^ et |et| et^2 |et/Yt|

1 30

2 39 30 9 9 81 0.230769

3 36 39 -3 3 9 0.083333

4 42 36 6 6 36 0.142857

5 38 42 -4 4 16 0.105263

6 57 38 19 19 361 0.333333

7 63 57 6 6 36 0.095238

8 59 63 -4 4 16 0.067797

9 70 59 11 11 121 0.157143

Tng 434 364 40 62 676 1.215734


35/238

Chng 1

1.5 Cc bc tin hnh d bo

Bc 1: Xc nh mc tiu d bo (i tngd bo, khu vc d bo v khong thi giancn d bo)

Bc 2: Phn tch cc yu t nh hng ni lng cn d bo


36/238

Chng 1


Bc 3: Thu thp v phn loi d liuBc 4: Phn tch xu hng tin trin ca ilng cn d bo

Bc 5: Xc nh k thut (phng php) dbo s s dng v tnh ton cc gi tr d bo


37/238

Chng 1


Bc 6: Kim tra tnh ph hp ca m hnh

Bc 7: Xc inh cc gi tr d bo


38/238

Chng 2

D BO BNGPHN TCH HI QUY


39/238

2.1 Cc khi nim c bn2.2 M hnh hi quy nhiu bin

2.3 c lng v kim nh gi thit

2.4 Phn tch hi quy v d bo2.5 D bo bng MHHQ vi bin gi

Chng 2

D BO BNG PHN TCH HI QUY


40/238

2.1.1 Phn tch hi quy

Chng 2

2.1 Cc khi nim c bn

Nghin cu mi lin h ph thuc gia gi trca mt bin Y- gi l bin ph thuc hay binc gii thch vi gi tr ca mt hoc nhiu

bin khc Xj (j=1,..,m) cc bin ny gi l ccbin c lp hay bin gii thch


41/238

Ta thng gi thit

Chng 2

2.1 Cc khi nim c bn

Bin ph thuc Yl bin ngu nhin, c quylut phn phi xc sut xc nh

Cc bin c lp Xjkhng phi l bin ngunhin, gi tr ca chng l xc nh


42/238

Phn tch hi quy gip ta:

Chng 2

2.1 Cc khi nim c bn

-c lng gi tr ca bin ph thuc Y khi bit gi tr ca (cc) bin c lp Xj

- Kim nh gi thit v s ph thuc

- D bo gi tr trung bnh hoc c bit ca binph thuc khi bit gi tr ca (cc) bin clp


43/238

2.1.2 M hnh hi quy tng th v m hnhhi quy mu

Chng 2

2.1 Cc khi nim c bn

( 2.1 )

M hnh hi quy tng th (hm tng th - PRF)l hm c dng tng qut

)()/( j ij i XfXYE


44/238

Nu (2.1) biu din mi quan h gia bin phthuc Yv mt bin gii thch Xth (2.1) c

gi l m hnh hi quy n hay m hnh hi quy2 bin

Chng 2

2.1 Cc khi nim c bn

Nu s bin gii thch nhiu hn 1 th (2.1)

c gi l m hnh hi quy bi (hi quy nhiubin)


45/238

Chng 2

2.1 Cc khi nim c bn

( 2.2 )

l c lng ca E(Y / Xji)

l c lng ca f

M hnh hi quy mu (hm hi quy mu - SRF)c th c biu din nh sau

)( jii XfY

iY

f


46/238

2.1.3 Sai s ngu nhin

Chng 22.1 Cc khi nim c bn ca kinh t

lng

Ui = Yi E ( Y / X ji), j=1,..,m; i=1,..,n

Ui c gi l sai s ngu nhin (nhiu ngu

nhin), biu th nh hng ca cc yu t khcngoi cc bin gii thch Xj ti gi tr ca bin Y


47/238

2.1.3 Sai s ngu nhin

Chng 22.1 Cc khi nim c bn ca kinh t

lng

Khi hm hi quy tng th (2.1) c th biudin di dng

ij ii UXfY )(


48/238

2.2.1 M hnh hi quy nhiu bin

Chng 22.2 M hnh hi quy nhiu bin v

phng php bnh phng nh nht

ikikiii UXXXY ...33221

Trong :Yi: gi tr ca bin ph thuc Y ( )

h s chn (h s t do)h s gc (h s hi quy ring) ca bingii thch Xj ( j= )

Ui: sai s ngu nhin

ni ,1

1j

2,k


49/238

M hnh hi quy mu xy dng da trn mungu nhin kch thc n



Trong :

c lng ca Yi ( )c lng ca h s hi quy tng th

ni ,1

kikiii XXXY ... 33221

niXXXY kiiii ,1),,...,,,( 32

iY

j

),1( kjj


50/238

Ta k hiu



Th m hnh hi quy tng th c th biu dindi dng ma trn:

nY

Y

Y

Y...

2

1

k

...

2

1

nU

U

U

U...

2

1

knnn

k

k

XXX

XXX

XXX

X

...1

...............

...1

...1

32

23222

13121

U XY


51/238

Tng t, nu ta k hiu



Th m hnh hi mu cng c th biu dindi dng ma trn nh sau:

nY

Y

Y

Y

...

2

1

k

...

2

1

XY


52/238

Cc gi thit c bn ca MHHQ nhiu bin



Gi thit 1. Cc bin gii thch Xj (j = ) khngphi bin ngu nhin, gi tr ca chng l xcnh

2,k


53/238



Gi thit 2. K vng ton ca cc sai s ngunghin U

ibng khng

Gi thit 3.

)(0

)().(

2

ji

jiUUE ji

)(0)/()( iXUEUE ii


54/238



Gi thit 4. Hng ma trn X bng k

rg(X) = kGi thit ny c ngha gia cc bin Xj khng chin tng cng tuyn hay cc ct ca ma trn

X c lp tuyn tnh

Gi thit 5. )(),0(~ 2 iNUi


55/238

2.2.2 Phng php bnh phng nh nht

Xt hm hi quy tng th v hm hi quy mu

hoc dng ma trn





U XYXY


56/238

Ta k hiu cc phn d ei:

Cc phn d ny cng c th biu din didng ma trn nh sau:



iii YYe

XYYY

...

......e 2

1

2

1

2

1

nnn Y

YY

Y

YY

e

ee


57/238

Theo phng php bnh phng nh nht, khi

xy dng hm hi quy mu, cc h s hi quymu phi c xc nh sao cho tng bnhphng cc phn d t gi tr nh nht, tc l:



min2

ie

j


58/238

Ta c

Gii phng trnh trn ta c:



eeeT

i 2

0

)(min2

eeeT

i

)3.2(. 1 YXXX TT


59/238

Cng thc (2.3) l cng thc xc nh h s hiquy mu theo phng php bnh phng nh

nht v cc c lng c xc nh theocng thc (2.3) c gi l cc c lng bnhphng nh nht.



)3.2(.

1

YXXXTT

j

j


60/238

Ma trn XTX c xc nh nh sau:



knn

k

k

knkk

n

XX

XX

XX

XXX

XXX

...1

............

...1

...1

...

............

...

1...11

XX

2

222

121

21

22221T

2

32

232

2

22

32

...

...............

...

...

kiikiikiki

kiiiiii

kiii

XXXXXX

XXXXXX

XXXn


61/238

Ma trn XTY cng c xc nh tng t:



kii

ii

i

nknkk

n

XY

XYY

Y

YY

XXX

XXX

......

...

............

...1...11

YX22

1

21

22221T


62/238

Nghin cu mi quan h ph thuc gia doanhs bn ra vi chi ph dnh cho qung co vgi bn, ngi ta thu thp c cc s liu sauy ti 10 ca hng cng kinh doanh mt loimt hng:

V D 2.1

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


63/238

Trong :

Yi: doanh s bn ra trong mt thng ca cahng th i (triu ng)

Xi: chi ph dnh cho qung co trong mt thngca ca hng th i (triu ng)

Zi: gi bn ca ca hng th i (ngnng/1sn phm)

Bng phng php bnh phng nh nht v davo s liu trn, hy xy dng hm hi quy mudi dng sau:

1 2 3

i i i Y X Z


64/238

p s:

1082

12746

7766

i

T

i i

i i

Y

X Y YX

YZ

2

2

10 114 73

114 1356 816

73 816 541

i i

T

i i i i

i i i i

n X Z

X X X X XZ

Z ZX Z

1 9 4 4TX X

67740 2106 5964

2106 81 162

5964 162 564

T

A A A

X X A A A

A A A

11 21 31

12 22 32

13 23 33


65/238

1

67740 2106 59641 1

2106 81 16219445964 162 564

T T

TX X X XX X

1

67740 2106 5964 1082 69 537041

2106 81 162 12746 6 083331944

5964 162 564 7766 4 20370

T T

, X X .X Y ,

,

6 9 5 3 7 0 4 6 0 8 3 3 3 4 2 0 3 7 0i i iY , , X , Z


66/238

: Khi gi bn khng i, chi phdnh cho qung co tng ln 1 triu ng, thdoanh s bn ra trung bnh ca ca hng tng

ln 6.08333 triu ng.

: Khi chi ph dnh cho qung cokhng i, gi bn tng ln 1ngn ng/ 1 vsn phm, th doanh s bn ra trung bnh caca hng gim xung 4.2037 triu ng.

ngha ca cc h s hi quy

2 6 08333.

3 4 2037.


67/238

Cc tnh cht ca c lng BPNN

1. ng hi quy mu i qua im trung bnhmu , tc l:

trong :



kk XXY ... 221

),...,( 2 kXXY

iYnY1

),2(1

kjXn

X jij


68/238

2. Gi tr trung bnh ca cc gi tr c xc nh

theo hm hi quy mu bng gi tr trung bnh cabin ph thuc, tc l:

3. Tng cc phn d ca hm hi quy mu bng 0:



YYn

Y i 1

0 ie

iY


69/238

4. Cc phn d ei khng tng quan vi :

5. Cc phn d ei khng tng quan vi :



iY

j iX

0 iiYe

),2(0 kjXe jii


70/238

6. (nh l Gauss Markov): Vi cc gi thit

ca m hnh hi quy tuyn tnh c in th ccc lng bnh phng nh nht l cc clng tuyn tnh, khng chch v c phng sainh nht trong lp cc c lng tuyn tnh,khng chch ca .



j

),1( kjj


71/238

Xt hm hi quy tng th v hm hi quymu

Chng 22.3 Khong tin cy v kim nh gi

thit v cc h s hi quy




72/238



C th chng minh c rng:

trong Ajj l phn t th j trn ng chochnh ca ma trn

jjTjA

XXVar

2

)(

*)( XX

T


73/238



Trong thc hnh khi s dng cng thc ny, dophng sai cha bit, nn ngi ta thngthay bng c lng khng chch ca n l:

knei

2

2

2


74/238



c xc nh bng cng thc :

Nu khai trin cng thc trn, ta c:

YXYYeeeTTTT

i 2

kiikiiiii XYXYYYe ... 22122

2ie


75/238

2.3.1 Khong tin cy ca cc h s hi quy



Do ta cha bit m phi thay bng c lngkhng chch ca n l , nn

),1()(~)(

kjknTseT j

jj

22


76/238



),1()(~)(

kjknT

seT

j

jj

1)(

)(

2

kntse

Pj

jj

1)().()().(22

jjjjj sekntsekntP

)().(;)().(

22

jjjj sekntseknt


77/238

2.3.2 Kim nh gi thit v cc h s hi quy



Gi s vi mc ngha cho trc ta cn kimnh gi thit:

),(:

:

***

1

*

0

jjjjjj

jj

H

H


78/238

2.3.2 Kim nh gi thit v cc h s hi quy



Ta xy dng tiu chun kim nh

)(

*

j

jj

seT

Nu H0 ng th T~T(n-k)


79/238



Loi gt H0 H1 WHai pha

Tri

Phi

*

jj

*

jj

*

jj

*

jj

*

jj

*

jj

)(:2/ kntttW tntn

)(: kntttW tntn

)(: kntttW tntn


80/238

2.4.1 H s xc nh bi

Chng 22.4 Phn tch phng sai v kim nh

gi thit ng thi

Xt hm hi quy tng th v hm hi quy mu:




81/238

Tng t trng hp hi quy 2 bin ta c hthc sau:


gi thit ng thi

2)(.1 YYTSS i

22)()(.2 YYYYESS ii

2)(.3 ii YYRSS

TSS ESS RSS


82/238

nh ngha: H s xc nh bi R2

c nh ngha nh sau:

R2 =


gi thit ng thi

1E SS RSST SS T SS


83/238

Trong thc hnh ta c th s dng cng thc:


gi thit ng thi

2T

2TT

2

YnYY

YnYXR

22

2

2212...

YnY

YnXYXYYR

i

kiikiii

Nu khai trin ta c


84/238

1. 0 R2 1

- Nu R2 = 1, hm hi quy c th coi l hon ho- Nu R2 = 0, hm hi quy a ra l khng ph hpV th R2 c dng lm thc o mc ph hpca hm hi quy

2. R2 l hm khng gim, ph thuc vo s bingii thch c trong m hnh


gi thit ng thi

Ch 2


85/238

2.4.2 Kim nh gi thit ng thi

Xt gi thit

hay gi thit tng ng l


gi thit ng thi

),(:

...:

kjshmtnhttH

H

j

k

20

0

1

320

0:

0:2

1

2

0

RH

RH

Ch 2


86/238

Ta xy dng tiu chun kim nh:


gi thit ng thi

1.12

2

kkn

R

RF

Nu H0 ng th F~F(k-1, n-k)

),1(: knkfffW tntn

)),1(( knkfFP


87/238

Xt hm hi quy tng th v hm hi quy mu

hoc dng ma trn

Chng 22.5 Phn tch hi quy v d bo



U XY

XY


88/238

Bi ton t ra: vi cc gi tr cho trc ca bingii thch X2=X20, X3=X30, ..., Xk=Xk0hoc c th

k hiu

cn d bo gi tr trung bnh E(Y/X0) hoc gi trc bit Y=Y0khi X=X0


0

30

20

0

1

kX

X

X

X

...


89/238

2.5.1 D bo gi tr trung bnh


0303202100 .... kkT

XXXXY

Vi tin cy = 1 cn d bo E(Y/X0)

c lng im ca E(Y/X0) l:


90/238

Do cha bit nn thng k


2

)(~)(

)/(

0

00 knTYse

XYEYT

)(2/ knt ta tm gi tr phn v sao cho: 1)(2/ kntTP


91/238


1)(

)()/(

2/

0

00 kntYse

XYEYP

1)().()/()().( 02/0002/0 YsekntYXYEYsekntYP

)().(;)().(02/002/0

YsekntYYsekntY

1)(2/ kntTP


92/238


0

1

0

2

000 ).(.).cov(.)( XXXXXXYVar

TTT

0

1

000 ).()()( XXXXYVarYse

TT

Trong


93/238

2.5.2 D bo gi tr c bit


0303202100.... kk

TXXXXY

Vi tin cy cn d bo gi tr Y=Y0khi X=X0

c lng im ca Y0vn l:


94/238

Hon ton tng t ta xy dng thng k


)(~)(

00

00 knTYYse

YYT

)().(;)().( 002/0002/0 YYsekntYYYsekntY

Bng php bin i tng ng ta cng suy

ra c khong tin cy ca Y0 l


95/238


2

000 )()( YVarYYVar

)

()

( 0000 YYVarYYse

Trong


96/238


V d: Xt tip v d 2. Vi tin cy = 0,98hy d bo doanh s bn ra trung bnh trongmt thng ca cc ca hng c chi ph dnhcho qung co l 10 triu ng/ thng v gi

bn l 8 ngn ng/ .v.


97/238


Xt tip v d 2. Vi tin cy = 0,98 hy dbo doanh s bn ra trong mt thng ca cahng c chi ph dnh cho qung co l 10 triu

ng/ thng v gi bn l 8 ngn ng/ .v.


98/238

2.6.1 Khi nim v bin gi

Chng 22.6 M hnh hi quy vi bin gi

Bin cht lng: Biu th nhng thuc tnhno (v d: gii tnh, ngh nghip)

Bin s lng:Gi tr ca cc bin cbiu th bng s (v d: thu nhp, doanh s)


99/238


biu th mc nh hng ca cc bin

cht lng ti bin ph thuc, ta cn lng hacc tiu thc, thuc tnh ny bng cch sdng bin gi


100/238


VD: biu th gii tnh, ta s dng bin gi Z

v quy c:- Z = 0 Nam- Z = 1 N

Vy bin gi l g? L bin cht lng clng ha, cc gi tr c th c ca bin gi ch

l 2 gi tr 0 v 1. N ch ra c hay khng cmt thuc tnh no .


101/238

2.6.2 MHHQ vi bin cht lng c 2 phm tr


Gi s mt x nghip sn xut c th p dng 2cng ngh sn xut A v B, nng sut ca micng ngh l LNN phn phi theo quy lut

chun c phng sai bng nhau, nhng kvng ton c th khc nhau


102/238


)1.4(21 iii UZY

Trong :Yi: nng sut ca x nghipUi: sai s ngu nhinZ

i: bin gi biu th cng ngh sn xut c

p dng v c th quy c:- Zi = 0 cng ngh sn xut A- Zi = 1 cng ngh sn xut B


103/238


c lng MHHQ (4.1) ta cng s tin hnh

tng t i vi MHHQ 2 bin thng thng

Yi 22 19 18 21 18.5 21 20.5 17 17.5 21.5

Zi 1 0 0 1 0 1 1 0 0 1


104/238


c lng MHHQ (4.1) ta cng s tin hnh

tng t i vi MHHQ 2 bin thng thng2,3

5,2

822

i

ii

z

zy

185,02,36,19

21 xZY

ii ZY 2,318


105/238


Vi gi thit E(Ui) = 0 c tha mn th

)3.4()1/()2.4()0/(

21

1

iiii

ZYEZYE

-(4.3) cho bit khi p dng cng ngh sn xutB, nng sut trung bnh ca x nghip l 1 + 2

-(4.2) cho bit khi p dng cng ngh sn xutA, nng sut trung bnh ca x nghip l 1


106/238



)3.4()1/()2.4()0/(

21

1

iiii

ZYEZYE

- Nh vy 2 l s chnh lch (khc nhau) vnng sut khi chuyn t cng ngh sn xut Asang cng ngh sn xut B


107/238


Tr li v d va ri:

ii ZY 2,318

Nng sut trung bnh khi p dng cng nghsn xut B l 18 + 3,2 = 21,2 (vsp)

Nng sut trung bnh khi p dng cng nghsn xut A l 18 (vsp)


108/238

2.6.3 MHHQ vi bin cht lng c nhiu hn

2 phm tr


Nu k hiu s phm tr l m th s bin gi cna vo m hnh ( lng ha bin cht

lng) s l m 1


109/238


Gi s x nghip n ngoi cng ngh sn xut

A v B cn c th p dng cng ngh sn xutC, khi ta cn s dng 2 bin gi l Z1i v Z2iv m hnh hi quy tng th s c dng sau:

)4.4(23121 iiii UZZY


110/238


Yi: bin ph thuc (nng sut)Ui: sai s ngu nhin

Z1i, Z2i: bin gi biu th cc cngngh sn xut c p dng


Z1i Z2i

A 0 0

B 1 0

C 0 1

Ch 2


111/238



3121

2121

121

)1,0/(

)0,1/(

)0/(

iii

iii

iii

ZZYE

ZZYE

ZZYE

Ch 2


112/238


- Nng sut trung bnh ca x nghip khi pdng cng ngh sn xut B l 1 + 2

-Nng sut trung bnh ca x nghip khi p

dng cng ngh sn xut A l 1

- Nng sut trung bnh ca x nghip khi pdng cng ngh sn xut C l 1 + 3

Ch 2


113/238


- 2: chnh lch (khc nhau) v nng sut khichuyn t CNSX A sang CNSX B

- 3: chnh lch (khc nhau) v nng sut khichuyn t CNSX A sang CNSX C

3121

2121

121

)1,0/(

)0,1/(

)0/(

iii

iii

iii

ZZYE

ZZYE

ZZYE

Ch 2


114/238


H0: 2 = 3 = 0


Nu mc ngha no ta khng bc bc H0 th iu ny c ngha ( mc ngha ) cc cng ngh sn xut khc nhau cho nngsut nh nhau

Ch 2


115/238


Ch

Phm tr c s hiu theo ngha l vic so snh

c tin hnh vi phm tr ny

Phm tr ng vi cc gi tr bng 0 ca cc bingi c gi l phm tr c s.

Ch 2


116/238

2.6.4 MHHQ vi nhiu bin cht lng


Gi s MHHQ c k bin gii thch l bin chtlng, s phm tr ca Xj l mj

Tng s bin gi cn thit biu th cc bin

cht lng cn a vo m hnh s l:

)( 1jm

Ch 2


117/238


Xt m hnh hi quy biu th nhu cu v maymc ca nam, n thanh nin thuc cc nghnghip khc nhau: cng nhn, nng dn, tiuthng, sinh vin.

S bin gi cn a vo m hnh s l:

(2 1) + (4 1) = 4

Ch g 2


118/238


Z1i Z2i Z3i Z4i

Nam SV 0 0 0 0N SV 1 0 0 0

Nam CN 0 1 0 0

N CN 1 1 0 0

Nam ND 0 0 1 0

N ND 1 0 1 0

Nam TT 0 0 0 1

N TT 1 0 0 1

Chng 2


119/238

2.6.5 MHHQ hn hp


L m hnh hi quy m trong c c bin slng ln bin cht lng, bin cht lng s

c biu th bng bin gi theo quy tc nh ni trn.

Chng 2


120/238

2.6.5 MHHQ hn hp


Vic phn tch hi quy mi quan h gia binph thuc vi cc bin gii thch (bao gm cbin s lng v bin gi), c tin hnhtng t nh i vi m hnh hi quy nhiubin


121/238

iu tra v mc chi tiu cho may mc canam, n cng nhn mt nh my, ngi tathu c s liu sau:

V D 3

3.0 4.0 3.5 4.5 4.0 4.5 4.0 5.5

2.5 2.5 3.0 3.0 3.5 3.5 4.0 4.00 1 0 1 0 1 0 1

iY

iX

iZ


122/238

Trong :Yi: mc chi tiu cho may mc trong mt nm

(triu ng)Xi: mc thu nhp trong mt thng (triu ng)Z

i: l bin gi biu th gii tnh, trong :

Zi = 0: nam Zi = 1: nBng phng php bnh phng nh nht v davo s liu trn, hy xy dng hm hi quy mu

di dng sau:

1 2 3

i i i Y X Z



123/238

: Nu cng gii tnh, khi thu nhp

tng thm 1 triu ng/ thng th chi tiu trungbnh cho may mc s tng thm 0,8 triu ng/nm.

: Nu cng thu nhp, th n cngnhn chi tiu cho may mc trung bnh nhiu hnnam l 1,0 triu ng/ nm.


iii ZXY 8,0025,1

8,02

0,13

Chng 2


124/238

2.6.6 ng dng ca MHHQ vi bin gi


So snh 2 hi quy

Phn tch thi v

Hi quy tuyn tnh tng on


125/238

Chng 3CHUI THI GIAN V D BO

TRN CHUI THI GIAN

Chng 3


126/238

3.1 Chui thi gian

3.2 D bo trn chui thi gian

Chng 3CHUI THI GIAN V D BO TRN

CHUI THI GIAN

Ch 3


127/238

3.1.1 Khi nim chui thi gian

Chng 3

3.1 Chui thi gian

Chui thi gian l tp hp cc gi tr ca mtbin ngu nhin (ch tiu thng k) c spxp theo th t thi gian: ngy, tun thng, qunm,

c k hiu bng cc ch ci Yt, Xt, Zt

Ch 3


128/238

Chng 3

3.1 Chui thi gian

Phn tch chui thi gian l s dng ccphng php thng k khc nhau lm rcu trc (cc thnh phn) ca chui thi giantrong s bin ng ca n

Ch 3


129/238

Chng 3

3.1 Chui thi gian

- Thnh phn xu th (Tr T)

Chui thi gian thng cha 4 thnh phn:

- Thnh phn chu k (Cl C)

- Thnh phn ma (Sn S)- Thnh phn ngu nhin (Ir I)

Ch 3


130/238

Chng 3

3.1 Chui thi gian

Chui thi k: l chui s biu hin bin ngca ch tiu nghin cu qua tng thi k

Chui thi im: l chui s liu biu hin

bin ng ca ch tiu nghin cu qua cc thiim nht nh

Ch 3


131/238

Chng 3

3.1 Chui thi gian

Cc mc trong chui thi k c th cng li

vi nhau qua thi gian, phn nh mc cach tiu nghin cu trong 1 thi k di hn (vd:sn lng c ph)

Cc mc trong chui thi im khng th

cng li theo thi gian v con s ny khng ngha (vd: gi vng)

Ch 3


132/238

3.1.2 Cc i lng m t chui thi gian

Chng 3

3.1 Chui thi gian

Vi chui thi k, th:

a. Trung bnh theo thi gian

n

Y

Y

n

i i 1

Ch 3


133/238

3.1.2 Cc i lng m t chui thi gian

Chng 3

3.1 Chui thi gian

Vi chui thi im, th:

a. Trung bnh theo thi gian

12

1...

2

1

21

nYYYY

n

i

n

i

ii

t

tY

Y 1

Ch ng 3


134/238

Chng 3

3.1 Chui thi gian

Lng tng (gim) tuyt i lin hon

b. Lng tng (gim) tuyt i

),2(1

niYY iii

Lng tng (gim) tuyt i nh gc),2(

1niYYii

Chng 3


135/238

Chng 3

3.1 Chui thi gian

),2(1

niYY iii

),2(1

niYYii

n

n

i

i 2 Ta thy

Chng 3


136/238

Chng 3

3.1 Chui thi gian

111

12

n

YY

nn

nn

n

i

i

Lng tng (gim) tuyt i trung bnh

i lng ny ch c ngha khi cc lng tng

(gim) tuyt i lin hon xp x nhau, ngha ltrong sut thi k nghin cu, hin tng tng(gim) vi mt lng tng i u

Chng 3


137/238

Chng 3

3.1 Chui thi gian

Tc pht trin lin hon

c. Tc pht trin

),2(1

niY

Yt

i

ii

Tc pht trin nh gc),2(

1

niY

YT ii

Chng 3


138/238

Chng 3

3.1 Chui thi gian

Ta thy

i

i

i tT

T

1

n

n

i

i Tt 2

Chng 3


139/238

Chng 3

3.1 Chui thi gian

Tc pht trin trung bnh

1

1

1

2

n nnn

i

iY

Ytt

i lng ny ch c ngha khi cc tc phttrin lin hon xp x nhau, ngha l trong sutthi k nghin cu, ch tiu pht trin vi mttc tng i u

Chng 3


140/238

Chng 3

3.1 Chui thi gian

Tc tng (gim) lin hon

d. Tc tng (gim)

),2(111

1 nitYY

YYa i

i

i

i

iii

Tc tng (gim) nh gc),2(1

11

1 niTYY

YYA i

iii

Chng 3


141/238

Chng 3

3.1 Chui thi gian

Tc tng (gim) trung bnh

1 ta

Chng 3


142/238

Chng 3

3.1 Chui thi gian

e. Tr tuyt i ca 1% tng (gim) lin hon

100100

(%)

1

1

1

1

i

i

ii

ii

i

ii

Y

Y

YY

YY

ag

Chng 3


143/238

3.2.1 Cc phng php d bo n gin

Chng 3


Vd: Mt quy bo d kin s ly bao nhiu tbo Tui Tr vo ngy mai c th cn nhctrong ngy hm nay tiu th ht bao nhiu

a. M hnh d bo th n gin

)1.3( 1 tt YY

Chng 3


144/238

Chng 3


M hnh ny c p dng trong trng hpthiu d liu qu kh, v n ch da trn ccthng tin sn c gn nht

Chng 3


145/238

Chng 3


: gi tr d on thi im n + L

: gi tr thc t thi im n: lng tng (gim) tuyt i trung bnh: tm xa d bo

b. D bo bng lng tng (gim) tuyt i trung bnh

)2.3( LYY nLn

L

Y

Y

n

Ln

Chng 3


146/238

Chng 3


Phng php ny c s dng khi bin ng

ca hin tng c lng tng (gim) tuyt ilin hon xp x nhau.

b. D bo bng lng tng (gim) tuyt i trung bnh

)2.3( LYY nLn

Chng 3


147/238

Chng 33.2 D bo trn chui thi gian

: gi tr d on thi im n + L

: gi tr thc t thi im n: tc pht trin trung bnh: tm xa d bo

c. D bo bng tc pht trin trung bnh

)3.3( LnLn tYY

L

t

Y

Y

n

Ln

Chng 3


148/238


Phng php ny c s dng khi bin ng

ca hin tng c tc pht trin lin hon xpx nhau.

c. D bo bng tc pht trin trung bnh

)3.3(L

nLn tYY

Chng 3


149/238


: gi tr d on thi im n + 1: gi tr thc t thi im n: khong trt, thng chn l 3, 4, 5,

d. D bo bng phng php trung bnh trt

)4.3(... 11

1k

YYYY knnnn

k

Y

Y

n

n 1

Chng 3


150/238


Xu th l s vn ng tng hay gim ca d liutrong mt thi gian di. S vn ng ny c thc m t bng mt ng thng hay ngcong ton hc

C th m hnh ha xu th bng cch thc hinmt hm hi quy thch hp gia bin cn d boY v thi gian T

3.2.2 D bo bng cc MH xu th

Chng 3


151/238


tt UTY 21

tt UTTY 2

321

tt UTTTY 3

4

2

321

Mt s dng hm xu th in hnh

Chng 3


152/238


tt UTY )ln(21

tt UTY )1

(21

tUT

t eY 21

tt UTY 21)ln(

Mt s dng hm xu th in hnh

Chng 3


153/238


D bo im vi hm xu th

TYt 21

2

321 TTYt

3

4

2

321 TTTYt

Chng 3


154/238


)ln(21 TYt

)

1

( 21 TYt

2

22

221 )

ln( tYTt eeY

TYt 21 )ln(

D bo im vi hm xu th

Chng 3


155/238

g3.2 D bo trn chui thi gian

L ng dng m rng ca phng php trungbnh trt:

3.2.2 D bo bng pp san m

Trung bnh trt da vo k quan st gn nht

Da vo gi tr trung bnh trt vi trng s gimdn cho tt c cc quan st trong qu kh

Chng 3


156/238


: gi tr d on thi im t + 1

: gi tr d on thi im t: gi tr quan st thi im t: h s san m

a. San m n gin

)5.3(1 1 ttt YYY

t

t

t

Y

Y

Y

1

Chng 3


157/238


a. San m n gin

Phng php san m n gin cho rng gi tr

d bo mi l mt gi tr trung bnh c trng sgia gi tr thc t v gi tr d bo thi im t

)5.3(1 1 ttt YYY

Chng 3


158/238


Nu dy s c nhiu bin i bt thng ta nnchn gn 0 v ngc li nn chn gn 1 nu

mun kt qu d bo kt hp vi nhng thay ign nht trong s liu.

Yu t quan trng nht trong phng php san

m l vic xc nh h s san m .

Chng 3


159/238


b. San m Holt

Hu ht cc chui d liu kinh t him khi theomt xu th c nh

Khi chui thi gian c yu t xu th (cc b) th ta

cn phi d bo c gi tr trung bnh (gi tr sanm) v dc (xu th) hin ti lm c s chod bo tng lai.

Chng 3


160/238


Khi chui thi gian c yu t xu th (cc b) th tacn phi d bo c gi tr trung bnh (gi tr sanm) v dc (xu th) hin ti lm c s chod bo tng lai.

Chng 3


161/238


tng c bn ca phng php Holt l sdng cc h s san m , khc nhau clng gi tr trung bnh v dc ca chui thigian (theo m hnh san m n gin)

Chng 3


162/238


: gi tr san m (mi) thi im t: gi tr quan st thi im t: h s san m ca gi tr trung bnh (0


163/238


: gi tr c lng xu th

: gi tr san m mi: h s san m ca gi tr xu th (0<


164/238


: gi tr d on thi im t+p

: gi tr c lng xu th: gi tr san m mit

t

pt

L

T

Y

D bo p giai on trong tng lai

)3.6.3( ttpt pTLY

Chng 3


165/238


c. San m Holt-Winter

L m rng ca san m Holt i vi cc d liuc cha yu t ma.

Yu t ma trong chui thi gian c th thucdng cng tnh hoc nhn tnh.

Chng 3


166/238


Dng cng tnh: c ngha l yu t ma cc

nm khc nhau c lp i lp li mt cch un.

Dng nhn tnh: c ngha l yu t ma nm

sau c lp i lp li nhng vi mt cng cao hn hoc thp hn so vi tng ma trongnm trc.

Chng 3


167/238


M hnh san m Holt Winter tng qut nht lm hnh dng nhn tnh

Chng 3


168/238


: gi tr san m (mi) thi im t: gi tr quan st thi im t

: h s san m ca gi tr trung bnh (0 < < 1): gi tr c lng xu th: gi tr c lng ca ch s ma vi di s

1.7.3))(1( 11

tt

st

tt TLS

Y

L

st

t

t

t

S

T

Y

L

c lng gi tr trung bnh hin ti

Chng 3


169/238


: gi tr c lng xu th

: gi tr san m mi: h s san m ca gi tr xu th (0<


170/238


: gi tr c lng ca ch s ma: gi tr san m mi: gi tr quan st ti thi im t: h s san m ca ch s ma

t

t

t

Y

L

S

c lng gi tr ch s ma

3.7.3)1( stt

tt S

LYS

Chng 3


171/238


: gi tr d on thi im t+p: gi tr c lng xu th: gi tr san m mi: gi tr c lng ca ch s ma

t

t

t

pt

S

L

T

Y

D bo p giai on trong tng lai

)4.7.3()( pstttpt SpTLY

Chng 3


172/238


Khng phi khi no d bo bng m hnh xu thcng c p dng n gin nh trnh bytrong phn trc

3.2.4 D bo bng pp phn tch

Thc t d liu nhiu khi khng ch tng gimtheo thi gian m cn bin i theo ma

Chng 3


173/238


- Thnh phn xu th (Tr T)

Chui thi gian thng cha 4 thnh phn:

- Thnh phn chu k (Cl C)

- Thnh phn ma (Sn S)- Thnh phn ngu nhin (Ir I)

Chng 33 2


174/238


Khi nghin cu cc thnh phn ca mt chui

thi gian, cn xem xt cc thnh phn ny linquan nh th no vi chui d liu gc Y

C 2 m hnh th hin mi quan h ny l mhnh cng tnh v m hnh nhn tnh

Chng 33 2 D b t h i thi i


175/238


M hnh nhn tnh (Multiplicative compomentsmodels):

Yt = Trt . Clt . Snt . Irt

M hnh cng tnh (Additive compoment models)Yt = Trt + Clt + Snt + Irt



176/238


Yt = Trt . Clt . Snt . Irt

M hnh nhn tnh s ph hp khi s bin thinca chui thi gian tng dn theo th t thigian. C ngha l cc gi tr ca chui tri rng ra

khi xu th tng dn v tp hp cc quan st cdng hnh ci loa hay hnh phu.



177/238


Yt = Trt + Clt + Snt + Irt

M hnh cng tnh c hiu qu khi chui d liuang c phn tch c s bin thin xp x unhau sut di ca chui thi gian. C ngha lcc gi tr ca chui thi gian v c bn nm

trong mt di gi tr c rng l mt hng s vtrung tm ca di ny l ng xu th.

240 9000


178/238

40

80

120

160

200

03Q1 03Q3 04Q1 04Q3 05Q1 05Q3 06Q1 06Q3

Y

4000

5000

6000

7000

8000

1996 1997 1998 1999

Y

m hnh nhn tnh m hnh cng tnh



179/238


Vi m hnh nhn tnh, ta s dng t l trung bnhtrt

Vi m hnh cng tnh, ta s dng chnh lch so

vi trung bnh trt



180/238


a. D bo bng m hnh nhn tnh

Bc 1. Tnh trung bnh trung tm (CMA)

quYYYYY

thngYYY

CMAttttt

ttt

t

45.05.0

12

5.0......5.0

1112

66



181/238


CMA bao gm xu th v chu k kt hp li

quYYYYY

thngYYY

CMAttttt

ttt

t

4

5.05.012

5.0......5.0

1112

66



182/238


Bc 2. Tnh t l

ttt CMAY /

Do Yt = Trt . Clt . Snt . Irt nn

ttt IrSn

Chng 33 2 D bo trn chui thi gian


183/238


Bc 3. Tnh ton cc ch s ma v

Nu chui d liu theo thng, ch s ma im chothng m bng trung bnh ca vi cc quan stch cho nhng thng m (mi nm c 1 thng m)

Nu chui d liu theo qu, ch s ma iq cho quq bng trung bnh ca vi cc quan st ch chonhng qu q (mi nm c 1 qu q)

t

t



184/238


qu

thng

Bc 4. iu chnh cc ch s ma tch cachng bng 1 bng cch tnh cc nhn t ma Sn

nh sau

121 2 3 12

...

mm

iSn

i i i i

41 2 3 4

q

q

iSni i i i



185/238


Bc 5. Chui d liu iu chnh yu t ma:

Yt/Snt = Trt.Clt.Irt

Ta gi nh khng c yu t chu k v yu tngu nhin b trit tiu khi tnh trung bnh nhmtm ra ch s ma bc 3: Clt = Irt = 1

Ta s s dng chui Yt/Snt d on thnhphn xu th trong tng lai

t



186/238


b. D bo bng m hnh cng tnh

Bc 1. Tnh trung bnh trung tm (CMA)

quYYYYY

thngYYY

CMAttttt

ttt

t

45.05.0

12

5.0......5.0

1112

66



187/238


Bc 2. Tnh s khc bit

ttt CMAYd



188/238


Bc 3. Tnh ton cc ch s ma v

Nu chui d liu theo thng, ch s ma im chothng m bng trung bnh ca vi cc quan stcho nhng thng m (mi nm c 1 thng m)

Nu chui d liu theo qu, ch s ma iq

cho quq bng trung bnh ca vi cc quan st ch chonhng qu q (mi nm c 1 qu q)

td

td



189/238


Bc 4. iu chnh cc ch s ma tng cachng bng 0 bng cch tnh cc nhn t ma Sn

nh sau

l trung bnh ca tt c cc ch s ma.Sn

tcho bit thi im t, Y cao hn (hay thp

hn) 1 lng Snt so vi chui d liu iuchnh yu t ma.

iiSn tt

i



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Bc 5. Chui d liu iu chnh yu t ma:

Yt Snt = Trt + Clt + Irt

Ta gi nh khng c yu t chu k v yu tngu nhin b trit tiu khi tnh trung bnh nhmtm ra ch s ma bc 3: Clt = Irt =0

Ta s s dng chui Yt Snt d on thnhphn xu th trong tng lai


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Chng 4D BO BNG PHNG PHP

BOX-JENKINS

Chng 4D BO BNG PHNG PHP


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4.1 Tnh tng quan trong d liu

chui thi gian4.2 Tnh dng ca chui thi gian

4.3. M hnh Box-Jenkins (ARIMA)

4.3. M hnh Box-Jenkins cho chui thi gianc tnh ma v (SARIMA)

BOX-JENKINS

Chng 44.1 Tnh tng quan trong d liu


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4.1.1 H s t tng quan

chui thi gian

)(

),(

t

kttk

YVar

YYCov

Phng trnh trn c gi l hm t tngquan, k hiu l ACF



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chui thi gian

)(

),(

t

ktt

k YVar

YYCov

Tnh cht:1. -1 k 1 k

2. 0 = 1



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chui thi gian

Do thc t ch c d liu mu, nn ta ch c thc lng c h s t tng quan mu thayv h s t tng quan tng th



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chui thi gian

n

t

t

kt

n

kt

t

k

YY

YYYY

r

1

2

1

)(

))((

: gi tr trung bnh mu ca Yt: tr: s quan stn

kY

H s t tng quan mu:



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chui thi gian

n

t

t

kt

n

kt

t

k

YY

YYYY

r

1

2

1

)(

))((

Tng t, rk cng c tnh cht:

1. -1 rk 1 k2. r0 = 1



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chui thi gian

H s t tng quan ring PACFk dng o

lng mc quan h gia Yt v Yt-k khi ccnh hng ca cc tr t 1 n k-1 cloi tr



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4.1.2 Kim tra tnh tng quan

chui thi gian

C 2 cch kim tra t tng quan:

1. Xem xt tng rk = 0 hay khng

2. Xem xt ton b tp cc gi tr rk = 0 hay khng



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chui thi gian

Ta tha nhn kt lun: Nu chui thi gian mbo tnh c lp th rk ~ N (0, 1/n )

T ta c khong tin cy ca rk:

n

ur

n

uk

2/2/

Cch 1. H0: rk = 0



201/238

chui thi gian

Chui thi gian c th c kt lun khng chin tng t tng quan nu tt c cc h s

tng quan mu tnh ton c u nm tronggii hn ny. Chui khi c tnh nhiu trng

n

ur

n

uk

2/2/



202/238

chui thi gian

Cch 2: H0: r1 = r2 = = rk = 0

h

k

krkn

nnQ1

2*.

1)2(

: s quan st

: tr: h s tng quan bc k: tr ln nht c xem xth

r

k

n

k



203/238

chui thi gian

h

k

krkn

nnQ1

2*.

1)2(

Nu H0 ng th Q* ~2(h-m)

)(:2**

mhQQW tntn

Chng 44.2 Tnh dng ca chui thi gian


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4.2.1 Khi nim chui thi gian dng

4.2 Tnh dng ca chui thi gian

Chui thi gian Yt, phi tha mn:constYE t )(

constYVar t 2

)(

constYYEYYCov kkttktt ))((),(



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4.2 Tnh dng ca chui thi gian

Gi s ta dch Y t Yt ti Yt+m. Nu Yt l chui

dng th gi tr trung bnh, phng sai v hipphng sai ca Yt v Yt+m phi bng nhau

)( tYE

2

)( tYVar kkttktt YYEYYCov ))((),(



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d g ca c u g a

Tm li, nu l mt chui dng, th gi tr trung

bnh, phng sai v hip phng sai (c trkhc nhau) bng nhau, khng cn bit ta olng chng ti thi im no

iu ny c ngha cc i lng ny khng itheo thi gian



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g g

Mt chui d liu nh vy s c xu hng tr

v gi tr trung bnh v nhng dao ng xungquanh gi tr trung bnh s nh nhau

Nu mt chui thi gian khng dng theo cchnh ngha trn, ta gi l chui khng dng



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g g

Vi mt chui thi gian khng dng, ta ch

c th nghin cu hnh vi ca n ch trongkhong thi gian c xem xt.

Do vy ta khng th khi qut ha cho cc giaion thi gian khc



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4.2.2 Kho st tnh dng

g g

Ngoi vic chng minh mt chui dng theonh ngha, ta cn c th da vo biu ttng quan xc nh mt chui thi giandng hay khng



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g g

Nu h s t tng quan u tin khc 0nhng cc h s t tng quan tip theo bng

khng c ngha thng k th l mt chuidng.

Nu mt s h s tng t tng quan khc

khng mt cch c ngha thng k th lmt chui khng dng.



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4.2.3 Loi b tnh khng dng

g g

iu kin c bn nht cho vic d bo mtchui thi gian l n phi c tnh dng.

Vy khi gp mt chui khng dng ta cn loi

b tnh khng dng trc khi lm cc phn tchk tip.



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g g

Ta c th lm dng mt chui bng phng

php sai phn.



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g g

Sai phn bc 1

1 ttt YYYSai phn bc 2.

)( 21112

ttttttt YYYYYYY

21

22 tttt YYYY



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g g

Chui sai phn bc 1 s dng nu xu hng

ca chui gc l tuyn tnh v n ch cn (n-1)quan st.

Nu sau khi ly sai phn bc 1 m cc bin

kim tra vn cho thy d liu cha dng thphi ly tip sai phn bc 2, , bc n.



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Vi mt s xu th phi a thc, ta p dng

phng php sai phn bin dng.

Nh s bin thin ca , phng php tnh sai

phn ny c th p dng cho rt nhiu xu thkhc nhau

)10(1

ttt YYY

Chng 44.3 M hnh Box-Jenkins


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M hnh Box-Jenkins l tn gi chung ca mt

h rt nhiu m hnh khc nhau do s tn tiring l hoc kt hp ng thi ca qu trnh thi quy (AR) v trung bnh trt (MA) v c thkt hp c qu trnh ly sai phn nu chui l

khng dng (I).



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4.3.1 Qu trnh t hi quy (AR)

M hnh t hi quy bc 1, AR(1), c dng

Trong :

1. ut: sai s tha mn tnh nhiu trng2. - 1


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Cc MH t hi quy ch ph hp vi chui dng,v h s 0 th hin mc trung bnh ca chui

ttt uYY 110

Nu d liu giao ng xung quanh gi tr 0hoc dng sai phn th khng cn h s 0trong m hnh



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M hnh t hi quy bc p, AR(p), c dng

iu kin mt chui trong m hnh AR(p) ldng nu tha mn iu kin

tptpttt uYYYY ...22110

11

p

i

i



220/238

xc nh s tr p, ta s dng gin ttng quan nh sau:

ACF s c xu hng bng 0 ngay lp tc

H s t tng quan ring PACF c xu hng

khc 0 mt cch c ngha thng k cho n tr p v bng 0 ngay sau tr p



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4.3.2 Qu trnh trung bnh trt (MA)

M hnh trung bnh trt bc 1, MA(1), c dng

Trong : : l hng sut : l sai s tha mn nhiu trng

11 ttt uuY



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M hnh trung bnh trt bc q, MA(q), c dng

Qu trnh ny dng nu -1 < i < 1.

qtqtttt uuuuY ...2211

Thng thng ta hay gp q < 2.



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xc nh s tr q, ta s dng gin ttng quan nh sau:

PACF s c xu hng bng 0 ngay lp tc

ACF c xu hng khc 0 mt cch c nghathng k cho n tr q v bng 0 ngay sau tr q



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4.3.3 Qu trnh phi hp t hi quy trungbnh trt (ARMA)

Kt hp m hnh AR(1) v m hnh MA(1) ta cm hnh ARMA(1,1)

11110 tttt uuYY



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Tng t, nu kt hp m hnh AR(p) v mhnh MA(q) ta c m hnh ARMA(p,q)

qtqtttptpttt uuuuYYYY ...... 221122110



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M hnh ARMA(p,q) ch thch hp vi cc chuidng, ta cn xc nh tr p v q thch hptheo cch trnh by trn: ACF bng 0 ngaylp tc sau tr th q cn PACF bng 0 ngaysau tr p



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4.3.4 M hnh ARIMA

M hnh ARMA ch c th c thc hin khi Ytl chui dng.

Chui thi gian trong kinh t v ti chnhthng l chui khng dng

suy ra chui dng, ta phi kh yu t xu thtrong d liu gc bng phng php sai phn



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Chui thi gian nu dng sai phn bc 1,

c k hiu I(1)M rng ra, nu mt chui dng sai phn bcd, ta k hiu I(d)

M hnh ARIMA c k hiu chung lARIMA(p,d,q)



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Bc 1. Kho st d liu gc

Quy trnh la chn m hnh ARIMA

Tnh ACF v PACF ca d liu gc, kim trachui c dng hay khng, nu dng, chuyn

sang bc 3



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Bc 2. Ly sai phn bc 1 ca Yt

Trong trng hp d liu gc bin ng nhiuta cn ly log ri mi ly sai phn.



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Bc 3. Phn tch gin t tng quan tm p v q, t suy ra cc m hnh c th c

Bc 4. c lng m hnh d kin (cxc nh trong bc 3)



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Bc 5. Phn tch, chun on v la chn MH

Kim tra h s ca tr cao nht nu khngc ngha thng k, gim bt tr p, q

Nu m hnh ng th ACF v PACF khng c

ngha thng kSo snh cc sai s d bo, ph hp cc MH



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Bc 6. Nu c thay i trong m hnh gc,quay tr li bc 4

Ch . Nn so snh gia cc m hnh vi nhauch khng nn phn tch mt cch ring l



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4.3.5 M hnh SARIMA

M hnh SARIMA p dng cho chui thi gianc tnh ma v

Ta s xem xt ring bit thnh phn c tnhma v v khng c tnh ma v bng m hnhARIMA



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Bc 1. Kim tra nu l chui khng dng thloi b tnh dng

Ta cng s dng sai phn bin mt chuithi gian khng dng c tnh ma thnh chuidng



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Bc 2. Xem xt thnh phn khng c tnhma v

Kho st trn ACF v PACF ti cc tr khng lbi s ca di ma L kt lun AR hay MA



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Bc 3. Xem xt thnh phn c tnh ma v

Kho st trn ACF v PACF ti cc tr l bi sca di ma L kt lun AR hay MA



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M hnh tng qut

ARIMA(p,d,q)(P,D,Q)L

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