Mot So Van de Co Ban Ve Xac Suat Thong Ke

BI GING 3: N TP XC SUT THNG K

ThS Phng Thanh Bnh

1

BI GING 3

MT S VN C BN

V XC SUT THNG K

TRONG KINH T LNG

MC TIU BI GING:

1. K hiu tng

2. Php th, khng gian mu v bin c

3. Bin ngu nhin

4. Xc sut

5. Bin ngu nhin v hm phn phi xc sut

6. Hm mt xc sut a bin

7. c im ca cc phn phi xc sut

8. Mt s phn phi xc sut quan trng

9. Mt s php ton ma trn

10. Suy din thng k

I TNG BI GING:

1. Ti liu bi ging cho sinh vin i hc

2. Ti liu tham kho n tp cho hc vin cao hc

K HIU TNG

K hiu tng

K t (sigma) c thng nht s dng ch tng:

n21

n

1iii X...XXXX

(3.1)

Thao tc vi Eviews

Trn ca s lnh ca Eviews ta nhp: scalar sumX=@sum(x)


ThS Phng Thanh Bnh

2

Tnh cht ca php ton tng

1. Khi k l mt hng s

nkkn

1i

(3.2)

2. Khi k l mt hng s

n

1ii

n

1ii XkkX (3.3)

3. Tng ca tng hai bin Xi v Yi

iiii YX)YX( (3.4)

4. Tng ca mt hm tuyn tnh

ii Xbna)bXa( (3.5)

PHP TH, KHNG GIAN MU, V BIN C

Php th

Mt php th c hai c tnh:

1) Khng bit chc kt qu no xy ra

2) Nhng bit c cc kt qu c th xy ra

Khng gian mu hay tng th

Tp hp tt c cc kt qu c th xy ra ca mt php th

c gi l tng th hay khng gian mu.

Bin c

Mt bin c l mt nhm cc kt qu c th xy ra c mt

php th. Ni cch khc, l mt tp hp con ca khng

gian mu.

Cc php tnh v bin c:

Bin c hi (AB): A xy ra hay B xy ra

Bin c giao (AB): A xy ra v B xy ra

Bin c ph (A ):A xy ra, A khng xy ra

Bin c xung khc: AB =


ThS Phng Thanh Bnh

3

BIN NGU NHIN

V d, tung hai ng xu, quan st v lp thnh bng kt

qu ca cc php th nh sau:

BNG 3.1: nh ngha khi nim bin ngu nhin

ng xu th

nht

ng xu th

hai

S mt nga

T

T

T

H

H

T

H

H

T

H

0

1

1

1

2

Ngun: Gujarati, 2006, trang 25

Ta gi bin s mt nga l mt bin ngu nhin. Ni mt

cch tng qut, mt bin m gi tr (bng s) ca n c

xc nh bi kt qu ca mt php th c gi l mt

bin ngu nhin. Nh vy, bin ngu nhin l bin m gi

tr ca n c xc nh mt cch ngu nhin.

Mt bin ngu nhin c th c gi tr ri rc hoc

lin tc. Mt bin ngu nhin ri rc ch c mt s gi

tr hu hn (hoc v hn c th m c). Mt bin ngu

nhin lin tc l mt bin ngu nhin c bt k gi tr

no trong mt khong gi tr no .

XC SUT

Xc sut ca mt bin c: nh ngha c in

Nu mt php th c th c n kt qu loi tr nhau v c

kh nng xy ra nh nhau, v nu m kt qu t php th

ny hp thnh bin c A, th P(A), xc sut A xy ra,

l t s m/n.

n

m)A(P (3.6)

Xc sut ca mt bin c: Tn sut tng i

gii thiu khi nim ny, ta xem v d sau y. D

liu trong bng 3.1 l phn phi im im thi m kinh t

vi m ca 200 sinh vin. y l mt v d v phn phi


ThS Phng Thanh Bnh

4

tn sut cho bit cc im ngu nhin c phn phi nh

th no. Cc con s trong ct 3 l cc tn sut tuyt

i, ngha l s ln xy ra ca mt bin c nht nh.

Cc con s trong ct 4 c gi l cc tn sut tng

i, ngha l s tn sut tuyt i chia tng s ln xy

ra.

BNG 3.2: Phn phi im KTL ca 200 sinh vin

im im gia ca

khong

Tn sut

tuyt i

Tn sut tng

i

0-9

10-19

20-29

30-39

40-49

50-59

60-69

70-79

80-89

90-99

5

15

25

35

45

55

65

75

85

95

0

0

0

10

20

35

50

45

30

10

Tng 200

0

0

0

0.050

0.100

0.175

0.250

0.225

0.150

0.050

1.000


PHN PHI XC SUT

Phn phi xc sut ca mt bin ngu nhin ri rc

Gi s X l mt bin ngu nhin ri rc vi cc gi tr

x1, x2, ... th hm f c xc nh bi

f(X=xi) = P(X=xi) i = 1, 2, (3.7)

=0 nu x xi

c gi l hm phn phi xc sut ca bin ngu nhin X,

k hiu l PMF hay PF, trong , P(X=xi) l xc sut X c

gi tr xi. Hm PMF c cc tnh cht sau:

0 f(xi) 1 (3.8)

n

1ii 1)x(f (3.9)

V d, bin X l s mt nga khi tung hai ng xu, ta xt

bng sau y:


ThS Phng Thanh Bnh

5

0.25

0.5

0.25

0 1 2

Hnh 3.1: PMF ca bin ngu nhin ri rc

BNG 3.3: PMF ca bin ngu nhin ri rc


Phn phi xc sut ca bin ngu nhin lin tc

V d, gi X l bin chiu cao ca mt ngi, c o

bng mt. Gi s ta mun tnh xc sut chiu cao ca

mt ngi trong khong 1.56m n 1.80m.

Hnh 3.2: PDF ca mt bin ngu nhin lin tc

0.04924276

0.54924276

1.04924276

1.54924276

2.04924276

2.54924276

3.04924276

3.54924276

4.04924276

1.4 1.44 1.48 1.52 1.56 1.6 1.64 1.68 1.72 1.76 1.8 1.84 1.88 1.92 1.96

Xc sut chiu cao ca mt c nhn nm trong khong t

1.56m n 1.80m l din tch di dng phn phi gia

hai gi tr 1.56 v 1.80. i vi mt bin ngu nhin

lin tc X, th hm mt xc sut f(X) nh sau:

P(x1 X x2) = 2

1

x

x

dx)x(f (3.10)

Hm mt xc sut ca mt bin ngu nhin X c cc tnh

cht sau y:

S mt nga

X

PMF

f(X)

0 1 2

Tng 1.00

Xc sut chiu cao trong

khong 1.56 n 1.8


ThS Phng Thanh Bnh

6

Tng din tch di ng f(x) bng 1

P(x1 X x2) l din tch di ng f(x) gia x1 v x2, vi x2 > x1.

V xc sut mt bin ngu nhin nhn mt gi tr nht nh bng khng, nn cc cng thc di y l

tng ng nhau:

P(x1 X x2) = P(x1 X x2) = P(x1 X x2) = P(x1 X x2) (3.11)

Hm phn phi tch ly ca mt bin ngu nhin

Lin quan n PMF hay PDF ca mt bin ngu nhin X l

hm phn phi tch ly ca bin , c xc nh nh

sau:

F(X) = P(X x) (3.12)

P(X x) ngha l xc sut mt bin ngu nhin X c gi tr nh thua hoc bng x, vi x bit. CDF c cc

tnh cht nh sau:

F(-) = 0 v F(+) = 1

F(x) l mt hm khng gim, ngha l nu x2 > x1, th

F(x2) F(x1)

P(X k) = 1 F(k)

P(x1 X x2) = F(x2) F(x1)

BNG 3.4: Hm phn phi xc sut tch ly ca mt bin ngu nhin

S mt nga

(X)

PDF CDF

X PDF X CDF

0 0 X < 1 1/16 X 0 1/16

1 1 X < 2 4/16 X 1 5/16

2 2 X < 3 6/16 X 2 11/16

3 3 X < 4 4/16 X 3 15/16

4 4 X 1/16 X 4 16/16


Nh vy, CDF ch l tch ly hay n gin l tng ca cc

PDF ca cc gi tr X nh thua hoc bng x.

Cc hm mt xc sut a bin

V d, mt i l bn l my tnh bn hai loi thit b

l my tnh c nhn v my in. S my tnh v my in c


ThS Phng Thanh Bnh

7

bn thay i gia cc ngy khc nhau, nhng gim c i

l thu thp doanh s ca 200 ngy qua nh trong bng

sau.

BNG 3.5: Phn phi tn sut ca hai bin ngu nhin X v Y

S my in c bn

(Y)

S my tnh c bn (X) Tng

0 1 2 3 4

0 6 6 4 4 2 22

1 4 10 12 4 2 32

2 2 4 20 10 10 40

3 2 2 10 20 20 54

4 2 2 2 10 30 46

Tng 16 24 48 48 64 200


Bng trn cho thy trong 200 ngy c 30 ngy i l bn

c 4 my tnh v 4 my in, c 2 ngy bn c 4 my

tnh nhng khng bn c my in no. Gii thch tng t

cho cc con s cn li. y l mt v d v phn phi tn

sut kt hp. Nu chia tng con s trong bng trn cho

200, ta s c cc tn sut tng i.

BNG 3.6: Phn phi xc sut ca hai bin ngu nhin X v Y

S my in c bn

(Y)

S my tnh c bn (X) Tng

0 1 2 3 4

0 0.03 0.03 0.02 0.02 0.01 0.11

1 0.02 0.05 0.06 0.02 0.01 0.16

2 0.01 0.02 0.01 0.05 0.05 0.23

3 0.01 0.01 0.05 0.10 0.10 0.27

4 0.01 0.01 0.01 0.05 0.05 0.23

Tng 0.08 0.12 0.24 0.24 0.32 1.00


Do hai bin X v Y l cc bin ngu nhin ri rc, nn

bng 3.6 c gi l hm phn phi xc sut kt hp ca

hai bin ngu nhin.

f(X,Y) = P(X = x v Y = y) (3.13)

= 0 khi X x v Y y

Hm xc sut kt hp c cc tnh cht sau:

f(X,Y) 0

x y

1)Y,X(f


ThS Phng Thanh Bnh

8

Hm xc sut bin

Xc sut X nhn mt gi tr nht nh bt k Y nhn gi

tr g c gi l xc sut bin ca X, v phn phi ca

cc xc sut ny c gi l hm phn phi xc sut bin.

BNG 3.7: Phn phi xc sut bin ca X v Y

X f(X) Y f(Y)

0

1

2

3

4

0.08

0.12

0.24

0.24

0.32

0

1

2

3

4

0.11

0.16

0.23

0.27

0.23

Tng 1.00. 1.00


T bng xc sut kt hp gia X v Y ta c th tnh cc

hm xc sut bin nh sau:

f(X) = y

)Y,X(f

f(Y) =x

)Y,X(f

Nu hai bin X v Y l hai bin ngu nhin lin tc th

ta s thay k hiu tng thnh k hiu tch phn.

Hm xc sut iu kin

Gi s ta mun tm xc sut c 4 my in c bn nu bit

c 4 my tnh c bn trong ny, v chnh l xc sut

c iu kin. Hm phn phi xc sut c iu kin ca mt

bin ngu nhin c th c nh ngha nh sau:

F(YX) = P(Y=yX=x) (3.14)

F(XY) = P(X=xY=y) (3.15)

Mt cng thc n gin tnh hm phn phi xc sut c

iu kin s nh sau:

F(YX) = )X(f

)Y,X(f (3.16)

F(XY) = )Y(f

)Y,X(f (3.17)


ThS Phng Thanh Bnh

9

CC C IM CA PHN PHI XC SUT

Gi tr k vng: Thc o nh tm

Gi tr k vng ca mt bin ngu nhin ri rc, k hiu

l E(X), c nh ngha nh sau:

E(X) = X = x

)X(xf (3.18)

Gi tr k vng ca mt bin ngu nhin l trung bnh c

trng s ca cc gi tr c th c ca bin , vi xc

sut ca cc gi tr ny, f(X), ng vai tr nh cc

trng s. Gi tr k vng ca mt bin ngu nhin cng

c gi l gi tr trung bnh, mc d chnh xc hn l

gi tr trung bnh tng th.

Tnh cht ca gi tr k vng

E(b) = b (3.19)

E(X+Y) = E(X) + E(Y) (3.20)

E(X/Y) )Y(E

)X(E (3.21)

E(XY) E(X)E(Y) (3.22)

Nu X v Y l hai bin ngu nhin c lp, th

E(XY) = E(X)E(Y) (3.23)

E(X2) [E(X)]2 (3.24)

E(aX) = aE(X) (3.25)

E(aX+b) = aE(X) + b (3.26)

Phng sai: Thc o phn tn

Gi tr k vng ca mt bin ngu nhin n gin ch cho

bit trng tm ca bin u ch khng cho bit cc

gi tr ring l ca bin phn tn nh th no xung

quanh gi tr trung bnh. Thc o ph bin nht cho s

phn tn ny l phng sai, v c nh ngha nh sau:

var(X) = 2x = E(X-x)

2 (3.27)

var(X) = )X(f)X(2

x (3.28)


ThS Phng Thanh Bnh

10

Phng sai cho bit cc gi tr X ring l c phn phi

hay phn tn xung quanh gi tr trung bnh nh th no.

Nu cc gi tr X phn tn rng quanh gi tr trung bnh

th phng sai s tng i ln (xem Hnh 3.3). Cn bc

hai ca phng sai l lch chun, k hiu l x.

Hnh 3.3: PDF ca cc bin ngu nhin lin tc cng gi tr k vng

Tnh cht ca phng sai

Phng sai ca mt hng s bng khng.

Nu X v Y l hai bin ngu nhin c lp, th

var(X+Y) = var(X) + var(Y) (3.29)

var(X-Y) = var(X) var(Y)

Nu b l hng s, th

var(aX) = a2var(X) (3.30)

Nu a v b l hng s, th

var(aX+b) = a2var(X) (3.31)

Nu X v Y l hai bin c lp v a v b l hng s, th

var(aX+bY) = a2var(X) + b

2var(Y) (3.32)

Phng sai

qu nh

Phng sai

qu ln

X


ThS Phng Thanh Bnh

11

tin li cho vic tnh ton, cng thc phng sai cng c th c vit li nh sau:

var(X) = E(X2) [E(X)]2 (3.33)

H s bin thin

Lu rng, v lch chun (hay phng sai) ph thuc

vo cc n v o lng khc nhau, cho nn s kh cho

vic so snh gia cc lch chun nu chng c cc

thc o khc nhau. gii quyt vn ny, ta c th

s dng h s bin thin (V) nh sau:

V = 100.x

x

(3.34)

Hip phng sai

Gi s X v Y l hai bin ngu nhin vi E(X) = x v E(Y)

= y, th hip phng sai (cov) gia hai bin s nh sau:

Cov(X,Y) = E[(X-x)(Y-y)]

= E(XY) - xy (3.35)

Hip phng sai gia hai bin c th dng, m, hoc bng

khng. Nu hai bin vn ng theo cng chiu, th hip

phng sai s dng, nu khc chiu, th hip phng sai

s m. Nu hip phng sai gia hai bin bng khng, th

c ngha l khng c mi quan h tuyn tnh no gia hai

bin .

Ta c th tnh hip phng sai theo cng thc sau

y:

cov(X,Y) = x y

yx )Y,X(f)Y)(X(

= x y

yx)Y,X(XYf (3.36)

= E(XY) - xy

Tnh cht ca hip phng sai

Nu X v Y l hai bin ngu nhin c lp, hip phng sai ca chng bng khng v khi E(XY) =

E(X)E(Y) = xy.

cov(a+bX, c+dY) = bdcov(X,Y) (3.37)

cov(X,X) = var(X) (3.38)


ThS Phng Thanh Bnh

12

Nu X v Y l hai bin ngu nhin nhng khng nht thit phi c lp, th cng thc tnh phng sai

(3.29) c vit li nh sau:

var(X+Y) = var(X) + var(Y) + 2cov(X,Y) (3.39)

var(X-Y) = var(X) + var(Y) 2cov(X,Y) (3.40)

H s tng quan

H s tng quan l thc o mi quan h tuyn tnh gia

hai bin ngu nhin, ngha l n cho bit hai c quan

h vi nhau nh th no: mnh hay yu. H s tng quan

tng th (, rho) c xc nh nh sau:

= yx

)Y,Xcov(

(3.36)

Tnh cht ca h s tng quan

Ging hip phng sai, h s tng quan c th m hoc dng.

H s tng quan l mt thc o mi quan h tuyn tnh gia hai bin.

-1 1 (3.37)

H s tng quan l mt con s thun ty khng c n v o lng.

Nu hai bin c lp, h s tng quan bng khng.

H s tng quan khng hm mi quan h nhn qu.

K vng c iu kin

Mt khi nim thng k khc c bit quan trng trong

phn tch hi qui l khi nim k vng c iu kin.

E(XY=y) = X

)yY/X(Xf (3.38)

nghing v nhn

nghing v nhn cho ta bit iu g v hnh dng

ca phn phi xc sut. nghing (S) l mt thc o s

mt cn xng ca th phn phi xc sut, v nhn

(K) l mt thc o cao hay thp ca th phn phi

xc sut.

M men th ba: E(X-x)3 (3.39)

M men th t: E(X-x)4 (3.40)


ThS Phng Thanh Bnh

13

S = 3x

3x )X(E

(3.41)

Hnh 3.4: nghing ca phn phi

C ba kh nng xy ra nh sau:

Nu S = 0, PDF i xng quanh gi tr trung bnh

Nu S > 0, PDF b nghing phi

Nu S < 0, PDF b nghing tri

K = 2 2x

4x

])X(E[

)X(E

(3.42)

C ba kh nng xy ra nh sau:

Nu K = 3, PDF c nhn chun v c gi l mesokurtic

Nu K < 3, PDF c ui ngn v c gi l platykurtic

Nu K > 3, PDF c ui di v c gi l leptokurtic

X

Nghing phi Nghing tri

i xng


ThS Phng Thanh Bnh

14

Hnh 3.5: nhn ca phn phi

T TNG TH N MU

Trung bnh mu

Trung bnh mu ca mt bin ngu nhin X c n quan st

c k hiu l X (c l X ngang) v c nh ngha nh sau:

n

1i

i

n

XX (3.43)

Trung bnh mu c xem l mt c lng ca E(X), t

trung bnh tng th. Mt c lng n gin l mt qui

tc, mt cng thc, hay mt thng k cho ta bit lm sao

c lng mt i lng ca tng th. Gi s X c 7

quan st vi cc gi tr nh sau: 8, 9, 10, 11, 12, 13,

14. Vy X = 11, v con s 11 ny c gi l mt gi tr c lng ca trung bnh tng th.

Thao tc vi Eviews

Trn ca s lnh ca Eviews ta nhp: scalar

meanX=@mean(x)

ui ngn

ui di

nhn chun

X


ThS Phng Thanh Bnh

15

Phng sai mu

Phng sai mu c k hiu bng 2xS , l c lng ca

phng sai tng th 2x . Phng sai mu c nh ngha

nh sau:

n

1i

2i2

x1n

)XX(S (3.44)

n-1 c gi l s bc t do (d.f.). Bc t do l s

ngun thng tin (piece of information) v mt bin ngu

nhin. hiu khi nim ny, ta xt v d sau y.

BNG 3.8: nh ngha khi nim bc t do

Quan st X (X- )X (X- )X2

1 8 -3 9

2 9 -2 4

3 10 -1 1

4 11 0 0

5 12 1 1

6 13 2 4

7 14 3 9

Tng 0 28

Ngun: Tc gi

Ta bit rng tng lch lun lun bng khng1, nn

xem lch ca cc gi tr X so vi gi tr trung bnh

ta phi ly lch bnh phng. Tng ca 7 lch bnh

phng l 28, nhng thc s con s 28 ny ch do 6

ngun ng gp, v quan st th t trng vi gi tr

trung bnh. Nh vy, xem lch trung bnh ta ch ly

28 chia cho s ngun thc s to ra n, tc 7-1 = 6. Vy

phng sai l 4.67 (l mt gi tr c lng ca phng

sai tng th) v cn bc hai ca phng sai mu c gi

l lch chun mu (s.d.). lch chun (2.16) c

xem nh mt thc o sp x cho trung bnh ca 6 lch

tuyt i trn. M rng cho trng hp mt bin ngu

nhin lin tc.

Thao tc vi Eviews

Trn ca s lnh ca Eviews ta nhp: scalar varX=@var(x)

1 Chng minh: 0XXXnXXX)XX(


ThS Phng Thanh Bnh

16

Hip phng sai mu

Hip phng sai mu gia hai bin ngu nhin X v Y l

c lng ca hip phng sai tng th, v c nh

ngha nh sau:

Cov(X,Y) = 1n

)YY)(XX( ii

(3.45)

Thao tc vi Eviews


covXY=@cov(x,y)

H s bin thin mu

H s bin thin mu ca X c xc nh bng cng

thc sau y:

V = 100.X

Sx (3.46)

Thao tc vi Eviews

Trn ca s lnh ca Eview ta nhp: scalar

cvX=@stdev(x)/@mean(x)

H s tng quan mu

H s tng quan mu gia hai bin ngu nhin X v Y l

c lng ca h s tng quan tng th, v c nh

ngha nh sau:

)Y.(d.s)X.(d.s

)1n/()YY)(XX(r ii

(3.47)

Thao tc vi Eviews


corXY=@cor(x,y)

nghing v nhn mu

tnh nghing v nhn mu, ta s dng cc m men

mu th ba v th t nh sau:

M men th ba: )1n(

)XX( 3

(3.48)

M men th t: )1n(

)XX( 4

(3.49)


ThS Phng Thanh Bnh

17

Thao tc vi Eviews

Trn ca s lnh ca Eviews ta nhp:

scalar skewX=@skew (x)

scalar kurtX=@kurt(x)

MT S PHN PHI XC SUT QUAN TRNG

Phn phi chun

Kinh nghim cho thy rng phn phi chun l mt m hnh

hp l cho mt bin ngu nhin lin tc vi gi tr ca

n ph thuc vo nhiu yu t, nhng mi yu t ch c

nh hng tng i nh ln gi tr ca bin s . Phn

phi chun ca mt bin ngu nhin X c th hin thng

qua hai tham s c bn l gi tr trung bnh v phng

sai. C th nh sau:

X ~ N(x,2x ) (3.50)

Hnh 3.6: th phn phi chun

-3 -2 -1 0 1 2 3

-

khong 68%

-2 2 -3 3

khong 99.7%

khong 95%


ThS Phng Thanh Bnh

18

Tnh cht ca phn phi chun

ng phn phi chun i xng quanh gi tr trung

bnh x.

Hm phn phi xc sut PDF ca mt bin ngu nhin theo phn phi chun cao nht ti gi tr trung bnh

nhng nh dn v cc cc tr ca n. Ngha l, xc

sut c mt gi tr ca mt bin ngu nhin theo

phn phi chun cng xa gi tr trung bnh cng nh.

Theo kinh nghim, khong 68% din tch di ng

phn phi chun nm gia gi tr xx, khong 95%

din tch nm gia x2x, v khong 99.7% din tch

nm gia x3x.

Mt phn phi chun c nh ngha hon ton bi hai

tham s x v 2x . Mt khi bit c hai tham s ny

th ta c th tnh c xc sut ca X nm trong mt

khong nht nh theo cng thc sau:

f(X) =

2

x

x

x

X

2

1-exp

2

1 (3.51)

Mt kt hp (hay mt hm) tuyn tnh ca hai hay nhiu bin ngu nhin theo phn phi chun s theo

phn phi chun y l mt tnh cht c bit quan

trng ca phn phi chun trong kinh t lng.

i vi phn phi chun, th nghing S l 0 v nhn K l 3.

Phn phi chun ha

Mc d mt phn phi chun hon ton c xc nh bng

hai tham s, gi tr trung bnh v phng sai tng th,

nhng cc phn phi chun c th khc nhau hoc gi tr

trung bnh, hoc phng sai, hoc c hai.


ThS Phng Thanh Bnh

19

Hnh 3.7: So snh cc phn phi chun c trung bnh v phng sai khc nhau

-3 -2 -1 0 1 2 3 4

Ta khng th so snh cc phn phi chun c cc tnh cht

khc nhau. Cho nn, ngi ta qui v cng mt bin chun

ha Z nh sau:

x

xXZ

(3.52)

Theo tnh cht ca phn phi chun, nu X l mt bin

ngu nhin c trung bnh l x v phng sai l x, X ~

N(X, 2X), th Z l mt k hp tuyn tnh ca X s l mt

bin ngu nhin c phn phi chun vi trung bnh l

khng v phng sai l mt, Z ~ N(0, 1)2.

Nh vy, bt k mt bin ngu nhin theo phn phi

chun vi mt gi tr trung bnh v phng sai nht nh

u c th c chuyn i thnh mt bin chun ha, iu

ny gip n gin ha rt nhiu vic tnh xc sut.

hiu vai tr ca phn phi chun ha, ta xem xt v d

sau y.

2 Chng minh: E(Z) = E 0)xX(E

x

1

x

xX

do E(X-x) = E(X) E(x) = x - x = 0. V Var(Z) =

E[Z-E(Z)]2 = E(Z

2), do E(Z) = 0, vy E(Z2) = E 12x2

x

12)xX(E2x

12

x

xX

1 2


ThS Phng Thanh Bnh

20

Gi s X, s lt khch du lch quc t hng ngy ca mt

cng ty du lch, theo phn phi chun vi gi tr trung

bnh l 70 v phng sai l 9; ngha l, X ~ N(70,9). Hy

tnh xc sut cho mt ngy bt k cng ty c s khch du

lch quc t nhiu hn 75 khch?

Ta thy, do X theo phn phi chun vi gi tr trung

bnh v phng sai bit, n ta c:

67.1 3

7075Z

s theo phn phi chun ha vi trung bnh bng 0 v

phng sai bng 1. Thay v tm P(X > 75), ta c th tm

P(Z > 1.67). Lu , trong cc sch thng k v kinh t

lng thng c km ph lc bng thng k gi tr hm

phn phi xc sut tch ly (CDF) hay gi tr xc sut

tch ly ca phn phi chun ha gia cc gi tr Z = -3

v Z = 3 (ti sao?). Theo bng thng k ny th xc sut

Z nm t -3 n 1.67 l 0.95253. Cho nn,

P(Z > 1.67) = 1 P(Z < 1.67) = 1 0.9525 = 0.0475

Vy xc sut mt ngy bt k cng ty c s lt khch

du lch nhiu hn 75 ngi l 4.75%.

Tm li, mt bin ngu nhin bt k m gi tr ca n

ph thuc vo rt nhiu yu t, nhng khng c yu t no

c nh hng quyt nh gi tr , th bin ngu nhin

s theo phn phi chun4. V bt k mt bin X c phn

phi chun vi gi tr trung bnh v phng sai bit

th u c th chuyn c sang bin chun ha Z c gi

tr trung bnh l 0 v phng sai l 1.

Thao tc vi Eviews


scalar probm167=1-@cnorm(1.67) = 0.0475

scalar probs167=@cnorm(1.67) = 0.9525

scalar probs_167=@cnorm(-1.67) = 0.0475

scalar Zval09525=@qnorm(0.9525) = 1.67

3 Nu qu v ang s dng my vi tnh m li ci tra bng thng k th c y nh bn nhn qua ci khc kht .

Hy m Excel ra l lm th ny: = NORMDIST(X, Mean, Standard_dev, Cumulative). Trong , X l gi tr cn tnh xc sut tch ly (1.67), Mean v Standard_dev y ln lt l trung bnh (0) v lch chun (1) ca bin X, v Cumulative c hai la chn l True (ng tnh xc sut tch ly) v False (khng tnh xc sut tch ly). trng hp ang xt, ta chn True. Ngc li, nu ta bit xc sut tch ly, gi tr trung bnh v phng sai th ta d dng tnh gi tr ca bin nh sau: =NORMINV(0.9525,0,1) = 1.67. 4 y l c s quan trng cho vic gi nh rng hn nhiu ui c phn phi chun (s c ni n bi ging

6).


ThS Phng Thanh Bnh

21

Phn phi xc sut ca trung bnh muX

Gi s ta chn ngu nhin mt mu vi n quan st gm cc

gi tr X1, X2, , Xn t mt tng th c cng hm phn

phi xc sut. Nu ta thc hin m mu nh th th gi tr

trung bnh mu X s l mt bin ngu nhin. Nh vy, vn

t ra l X s c phn phi nh th no?

BNG 3.9: nh ngha bin trung bnh mu v phng sai mu

Mu Gi tr ca mu Gi tr trung bnh mu X Phng sai mu 2xS

1

2

3

.

.

M

X11 X12 . . . X1n

X21 X22 . . . X2n

X31 X32 . . . X3n

.

.

Xm1 Xm2 . . . Xmn

1X

2X

3X

.

.

mX

21xS

22xS

23xS

. 2xnS

V d, mt tng th c phn phi chun vi gi tr trung

bnh l 10 v phng sai l 4, tc N(10,4). T tng th

ny ta thu thp 20 mu ngu nhin vi 20 quan st/mu.

Nh vy ta s c cc gi tr trung bnh, X nh sau.

BNG 3.10: Phn phi xc sut ca trung bnh mu

Cc trung bnh

mu ( X )

Khong ca trung bnh mu

Tn sut tuyt i

Tn sut tng i

9.641

10.040

9.174

10.840

10.480

11.386

9.740

9.937

10.250

10.334

10.134

10.249

10.321

10.399

9.404

8.621

9.739

10.184

9.765

10.410

8.5 8.9

9.0 9.4

9.5 9.9

10.0 10.4

10.5 10.9

11.0 11.4

Tng

1

1

5

8

4

1

20

0.05

0.05

0.25

0.40

0.20

0.05

1.00


Tng ca 20 gi tr trung bnh l 201.05, 052.10n

XX

i

,

v var(X) = 339.019

)XX( 2

.


ThS Phng Thanh Bnh

22

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

8.75 9.25 9.75 10.25 10.75 11.2

Hnh 3.8: Phn phi ca 20 gi tr trung bnh mu t tng th c N(10,4)

L thuyt thng k cho rng, nu X1, X2, , Xn l mt mu

ngu nhin t mt tng th c phn phi chun vi trung

bnh x v phng sai 2x , th trung bnh mu, X ,cng theo

phn phi chun vi trung bnh x nhng phng sai n

2x 5.

Ngha l,

X ~ N(x,n

2x ) (3.53)

Cn bc hai ca phng sai trung bnh mu, n

x , c gi

l sai s chun (se) ca X , tng t nh khi nim

lch chun. Lu , cn bc hai ca phng sai ca mt

5 Chng minh: Do

n

1iiX

n

1X nn ta c:

x)xn(n

1]x...xx[

n

1)]nX(E...)2X(E)1X(E[

n

1)X(E

n

2x)2xn(2n

1

)]nXvar(...)2Xvar()1X[var(2n

1

n

nX...2X1Xvar)Xvar(


ThS Phng Thanh Bnh

23

bin ngu nhin c gi l lch chun (s.d.), v cn

bc hai ca mt c lng c gi l sai s chun (se).

nh l gii hn trung tm

Nh ta va phn tch, trung bnh mu ca mt mu rt ra

t mt tng th phn phi chun cng theo phn phi chun

(bt k c mu bao nhiu). Vn t ra l nu cc mu

rt ra t cc tng th khc khng theo phn phi chun

th sao? nh l gii hn trung tm cho rng nu X1, X2,

, Xn l mt mu ngu nhin t bt k tng th no vi

trung bnh l x v phng l 2x , th trung bnh mu X s

c xu hng theo phn phi chun vi trung bnh l x v

phng sai l n

2x khi c mu tng ln v cng6.

Hnh 3.9: nh l gii hn trung tm: Cc mu c rt ra t mt tng th chun hay

khng chun u c phn phi chun

Phn phi mu ca gi

tr trung bnh

Tng th c phn

phi chun

Tng th khng c phn

phi chun

Phn phi t

Phn phi xc sut c s dng rt nhiu trong phn kinh

t lng cn bn l phn phi t, cng c gi l phn

phi t Student.

6 Trn thc t, cho d phn phi xc sut nn tng l g, trung bnh mu ca mt c mu t nht c 30 quan st s

c th xp x chun (Gujarati, 2006, pp.88).


ThS Phng Thanh Bnh

24

Nu X ~ N(x,n

2x ), th bin chun ha Z c nh ngha

nh sau: Z =

n

)X(

x

x

~ N(0,1) nu c hai tham s x v

n

2x u c bit. Nhng gi s ta ch bit x v gi tr

c lng ca 2x bi c lng mu

1n

)XX(S

2i2

x

. Nh vy,

nu thay x bng xS ta s c mt bin mi nh sau:

t =

n

S

)X(

x

x (3.54)

L thuyt thng k cho rng bin t s theo phn phi t

vi s bc t do l (n-1), y l tham s duy nht ca

phn phi t.

-4 -3 -2 -1 0 1 2 3 4

Phn phi chun

Phn phi t vi df=1

Phn phi t vi df=4

Phn phi t vi df=10

Hnh 3.10: Phn phi t vi mt s bc t do khc nhau

t


ThS Phng Thanh Bnh

25

Tnh cht ca phn phi t

Ging nh phn phi chun, phn phi t i xng quanh gi tr trung bnh.

Trung bnh ca phn phi t, ging nh phn phi chun ha, l khng, nhng phng sai l k/(k-2), vi k l

s bc t do. V vy, phng sai ca phn phi t ch

c xc nh khi s bc t do d.f. > 2.

mimh ha ng dng ca phn phi t trn thc t ta xt

tip v d v s lt khch du lch quc t ti mt cng

ty du lch nh cp. Bit rng, trong giai on 15

ngy qua, s lt khch quc t trung bnh mt ngy l 72

v phng sai mu l 4. Hy tnh xc sut c c s

lt khch trung bnh , bit rng gi tr trung bnh

thc l 70 khch mt ngy?

Nu bit lch thc ca tng th () th ta c th d dng s dng phn phi chun ha tnh xc sut

trn. Nhng y ta c S, l mt c lng ca , nn ta c th s dng phn phi t nh sau:

153

7072t

=1.9365

s theo phn phi chun ha vi trung bnh bng 0 v

phng sai bng 1.17. Thay v tm )72X(P , ta c th tm

P(t > 1.9365). p dng hm phn phi t7 cho trng hp mt

ui ta c:

P(t > 1.9365) = 1 P(t < 1.9365) = 0.0366

Vy xc sut s lt khch trung bnh mt ngy ca

cng ty du lch ny l 3.66%.

Thao tc vi Eviews


scalar probm19365=1-@ctdist(1.9365,14) = 0.0366

scalar probs19365=@ctdist(1.9365,14) = 0.9634

scalar probs_19365=@ctdist(-1.9365,14) = 0.0366

scalar tval09634=@qtdist(0.9634,14) = 1.9365

7 Hm phn phi xc sut t trn Excel l: =TDIST(X, Deg_freedom, Tails). X ngha l gi tr t cn tnh xc

sut (1.9365), ngha l din tch di ng phn phi t t t n + (ta s bit y chnh l vng bc b gi thit H0). Deg_freedom l s bc t do (14). Tails c hai la chn: 1 (mt ui), v 2 (hai ui). Gi tr xc sut ta tnh c t cng thc ny chnh l P-Value (s c gii thiu bi ging 4). Nu ta bit mc ngha (s c trnh by bi ging 4) v s bc t do, ta s tm c gi tr t theo cng thc sau: =TINV(Probability, Deg_freedom). V d, =TINV(3.66%,14) = 1.9365. Lu , Ph lc B cui bi ging 3 s hng dn cch v th phn phi t bng Excel.


ThS Phng Thanh Bnh

26

Phn phi xc sut Chi bnh phng (2)

Ta xc nh phn phi xc sut ca trung bnh mu, X ,

vy cn phng sai mu, 1n

)XX(S

2i2

s c phn phi nh

th no? Phn phi xc sut cn cho mc ch ny chnh l

phn phi xc sut Chi bnh phng (2), cng l mt phn phi c mi quan h rt gn vi phn phi chun. Lu ,

ging nh trung bnh mu, phng sai mu cng thay i t

mu ny qua mu khc. Cho nn, ging nh trung bnh mu,

phng sai mu cng l mt bin ngu nhin.

Ta bit rng nu mt bin ngu nhin X theo phn phi

chun vi trung bnh l x v phng sai l 2x ,

X~N(x,2x ), th bin chun ha Z~N(0,1). L thuyt thng

k cho rng bnh phng ca mt bin chun ha c phn

phi Chi bnh phng (2) vi mt bc t do. K hiu nh sau:

Z2 ~ 2(1) (3.55)

Ging nh phn phi t, bc t do l tham s ca phn phi

Chi bnh phng (2). phng trnh (3.55) ch c mt bc t do v ta ang xt bnh phng ca mt bin chun ha8.

Gi s Z1, Z2, , Zk l cc bin chun ha c lp

(mi bin Z l mt bin ngu nhin c phn phi chun vi

trung bnh bng 0 v phng sai bng 1). Nu ta ly bnh

phng tng bin ny, th tng ca cc bin Z bnh phng

ny cng theo phn phi Chi bnh phng vi k bc t do.

2)k(

21

21

21

2i ~Z...ZZZ (3.56)

8 Bc t do trong phn Chi bnh phng bng s bin ngu nhin c ly bnh phng


ThS Phng Thanh Bnh

27

0 2 4 6 8 10 12 14 16 18

vi df = 4 vi df = 6 vi df = 2

Hnh 3.11: Phn phi chi bnh phng vi cc bc t do khc nhau

Tnh cht ca phn phi 2

Khc phn phi chun, phn phi Chi bnh phng9 ch c gi tr dng t 0 n v cng.

Khc phn phi chun, phn phi Chi bnh phng l mt phn phi nghing, nghing ca phn phi ph

thuc vo s bc t do. Khi bc t do thp, phn phi

Chi bnh phng b nghing phi, nhng khi bc t do

tng ln, phn phi s i xng v dn v phn phi

chun.

Gi tr trung bnh ca mt bin ngu nhin theo phn phi Chi bnh phng l k v phng sai l 2k (k l

s bc t do). y l mt tnh cht ng ch ca

phn phi Chi bnh phng v phng sai gp i gi

tr trung bnh.

9 Hm phn phi xc sut Chi bnh phng trn Excel l: =CHIDIST(X, Deg_freedom). X ngha l gi tr 2

cn tnh xc sut (v d 6), ngha l din tch di ng phn phi Chi bnh phng t 2 n + (ta s bit y chnh l vng bc b gi thit H0). Deg_freedom l s bc t do (v d 2). Gi tr xc sut ta tnh c t cng thc ny (4.98%) chnh l P-Value. Nu ta bit mc ngha (s c trnh by bi ging 4) v s bc t do,

ta s tm c gi tr 2 theo cng thc sau: =CHIINV(Probability, Deg_freedom). V d, =CHIINV(4.98%,2)

=6. Lu , Ph lc C cui bi ging 3 s hng dn cch v th phn phi 2 bng Excel.


ThS Phng Thanh Bnh

28

Nu Z1 v Z2 l hai bin c phn phi Chi bnh phng c lp vi k1 v k2 bc t do, th (Z1+Z2) cng l

mt bin c phn phi chi bnh phng vi bc t do

l (k1+k2).

Tm li, y l mt phn phi rt hay s dng trong phn

tch kinh t lng cho cc bin ngu nhin dng bnh

phng nh kim nh JB, kim nh phng sai ca hn

nhiu, cc kim nh phn d s dng phng trnh hi qui

ph (nR2),

Thao tc vi Eviews


scalar probm6=1-@cchisq(6,2) = 0.0498

scalar probs6=@cchisq(6,2) = 0.9502

scalar chival09502=@qchisq(0.9502,2) = 6

Phn phi F

Mt phn phi xc sut khc cng rt quan trng trong

kinh t lng l phn phi F vi tng nh sau. Gi s

X1, X2, , Xm l mt mu ngu nhin vi c mu m t mt

tng th c phn phi chun vi trung bnh X v 2X; v

Y1, Y2, , Yn l mt mu ngu nhin vi c mu n t mt

tng th phn phi chun vn trung bnh Y v phng sai

2Y. Gi s hai mu c ny lp nhau v c ly t hai tng th c phn phi chun. Gi s ta mun xem phng

sai ca hai tng th trn c ging nhau hay khng (2X =

2Y). Do ta khng th quan st trc tip phng sai tng th nn ta suy ra t cc c lng phng sai nh sau:

S2X =

1m

)X(X 2i

(3.57)

S2Y =

1n

)Y(Y 2i

(3.58)

By gi ta xt t s sau y:

F =

)1n/()YY(

)1m/()XX(

S

S2

i

2i

2Y

2X (3.59)

Cc phng sai cng khc nhau th t s F cng ln. Phn

phi F ph thuc vo hai tham s l bc t do ca t (m-

1) v bc t do ca mu (n-1).


ThS Phng Thanh Bnh

29

c im ca phn phi F

Ging phn phi chi bnh phng, phn phi F10 cng b nghing phi v nm trong khong t 0 n v cng.

Ging phn phi t v phn phi chi bnh phng, phn phi F s dn v phn phi chun khi k1 v k2 tng

ln v cng.

Bnh phng ca mt bin ngu nhin c phn phi t vi k bc t do s c phn phi F vi 1 v k bc t

do.

k,12k F~t (3.60)

y l mt tnh cht c ngha rt quan trng trong phn

phn tch hi qui bi.

Cng nh mi quan h gia phn phi t v F, cng c mi quan h gia phn phi F v phn phi chi bnh

phng nh sau.

F(m,n) = m

2 khi n (3.61)

Tm li, phn phi F rt quan trng trong kinh t lng

khi chng ta thc hin phn tch phng sai (ANOVA) v

kim nh cc bin di dng t s gia cc phng sai v

d kim nh gi thit ng thi, kim nh Wald, kim

nh Chow, kim nh nhn qu Granger,

Ngoi ra, qu v c th tham kho thm cc loi phn

phi khc nh phn phi beta, phn phi nh thc, phn

phi Gama, phi Poisson, phn phi Weibull, trong cc

ti liu chuyn v thng k, v d Hong Trng (2007), v

phn tr gip trong Eviews (statistical distribution

functions). Do cc loi phn phi ny t s dng trong

chng trnh kinh t lng c bn nn cun sch ny s

khng cp.

10 Hm phn phi xc sut F trn Excel l: =FDIST(X, Deg_freedom1, Deg_freedom2). X ngha l gi tr F

cn tnh xc sut (v d 4), ngha l din tch di ng phn phi F t F n + (ta s bit y chnh l vng bc b gi thit H0). Deg_freedom1 l s bc t do ca t s (v d 2). Deg_freedom2 l s bc t do ca mu s (v d 14). Gi tr xc sut ta tnh c t cng thc ny (4.23%) chnh l P-Value (s c gii thiu bi ging 4). Nu ta bit mc ngha v s bc t do ca t v mu s, ta s tm c gi tr F theo cng thc sau: =FINV(Probability, Deg_freedom1, Deg_freedom2). V d, =FINV(4.23%,2,14) = 4.


ThS Phng Thanh Bnh

30

0.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

0 0.5 1 1.5 2 2.5 3 3.5 4

vi df1=3, df2=10 vi df1=10, df2=30 vi df1=50, df2=50

Hnh 3.12: Phn phi F vi cc bc t do khc nhau

Thao tc vi Eviews


scalar probm4=1-@cfdist(4,2,14) = 0.0423

scalar probs4=@cfdist(4,2,14) = 0.9577

scalar fval09577=@qfdist(0.9577,2,14) = 4


ThS Phng Thanh Bnh

31

I S MA TRN

Mt s khi nim

Ma trn

Ma trn A c M dng v N ct thng c k hiu gn l

[aij] c th c th hin nh sau:

MN3M2M1M

N2232221

N1131211

ij

a ... a a a

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

a ... a a a

a ... a a a

]a[A

Trong aij l thnh t dng th i v ct th j ca ma

trn A. Trong m hnh hi qui c in, ma trn X c gi

l ma trn M dng v N ct. d dng nhn bit s bc

ca mt ma trn, ngi ta thng k hiu s dng v ct

ca ma trn nh sau:

3 1 6

5 3 2A

3x2

11 9 8

4 0 1

7 5 1

B3x3

Thao tc vi Eviews

to ma trn A, th trn ca s lnh ca Eviews ta nhp: matrix(2,3) matrixA

Sau nhp gi tr cc thnh t ca ma trn A nh sau:

matrixA.fill 2, 6, 3, 1, 5, 3

Vect ct

Mt ma trn c M dng v ch mt ct c gi l mt

vect ct, v c minh ha nh sau:

9

5

4

3

Y4x1

Trong m hnh hi qui c in, cc ma trn dng vect ct

l ma trn bin ph thuc Y, ma trn h s hi qui B, v

ma trn hng nhiu u.

Thao tc vi Eviews

to vect ct Y, th trn ca s lnh ca Eviews ta nhp: vector(4) vectorY

Sau nhp gi tr cc thnh t ca vect Y nh sau: vectorY.fill 3, 4, 5, 9


ThS Phng Thanh Bnh

32

Vect dng

Mt ma trn c mt dng v N ct c gi l mt vect

dng, v c minh ha nh sau:

17 10 4- 5 2Z5x1

Thao tc vi Eviews

to vect dng Z, th trn ca s lnh ca Eviews ta

nhp: rowvector(5) vectorZ

Sau nhp gi tr cc thnh t ca vect Z nh sau:

vectorZ.fill 2, 5, -4, 10, 17

Ma trn chuyn v

Ma trn chuyn v ca ma trn A vi M dng v N ct, c

k hiu l A l mt ma trn c N dng v M ct c c

bng cch thay i dng v ct ca ma trn A nh sau:

0 5

1 3

5 4

A2x3

0 1 5

5 3 4A

3x2

Thao tc vi Eviews

to ma trn chuyn v (AT) ca ma trn A, th trn ca

s lnh ca Eviews ta nhp:

matrix AT @transpose(A)

Do vect l mt loi ma trn c bit, nn chuyn v mt

vect dng l mt vect ct v ngc li.

9

5

4

3

Y1x4

9 5 4 3A4x1

Ma trn con

Ma trn A c 3 dng, 3 ct, gi s b i dng v ct th

3 ca ma trn A, ta s c ma trn B vi 2 dng, 2 ct. B

c gi l ma trn con ca ma trn A.

11 9 8

4 0 1

7 5 1

A3x3

0 1

5 1 B

2x2


ThS Phng Thanh Bnh

33

Cc loi ma trn

Ma trn vung

Ma trn c s dng v s ct bng nhau c gi l ma

trn vung, nh ma trn A v ma trn B trn.

Ma trn ng cho

Mt ma trn vung vi t nht mt thnh t khc khng

trn ng cho chnh v cc thnh t khc c gi tr

khng c gi l ma trn ng cho.

1 0 0

0 5 0

0 0 2

A3x3

3 0

0 2B

2x2

Ma trn v hng

Mt ma trn ng cho vi cc thnh t trn ng cho

u ging nhau c gi l ma trn v hng. Ma trn v

hng hay gp trong kinh t lng l ma trn phng sai -

hip phng sai ca hn nhiu trong m hnh hi qui tuyn

tnh c in nh sau:

2

2

2

2

2

5x5

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

cov(u)var

Ma trn n v

Mt ma trn vi cc thnh t trn ng cho u bng 1

c gi l ma trn n v, c k hiu l I. Ma trn

n v l mt trng hp c bit ca ma trn v hng.

1 0 0

0 1 0

0 0 1

I3x3

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

I4x4

Ton t ma trn

Php cng ma trn

Ta c A = [aij] v B = [bij]. Nu A v B l cc ma trn c

cng bc, th php cng ma trn c nh ngha nh sau:

A + B = C


ThS Phng Thanh Bnh

34

Trong , C c cng s bc vi A v B v c tnh nh

sau cij = aij + bij cho tt c cc i v j; ngha l, C c

tnh bng cch cng cc thnh t tng ng ca A v B.

V d:

9 8 7 6

5 4 3 2A

4x2

5 1 0 2

3 1- 0 1 B

4x2

V C = A + B th:

14 9 7 4

8 3 3 3C

4x2

Thao tc vi Eviews

thc hin php cng hai ma trn A v B, th trn ca

s lnh ca Eviews ta nhp:

matrix matrixC=A+B

Php tr ma trn

Php tr ma trn cng c thc hin theo mt nguyn tc

nh php cng ma trn (C = A B); ngha l, ta tr cc

thnh t ca A cho cc thnh t tng ng ca C.

Thao tc vi Eviews

thc hin php tr hai ma trn A v B, th trn ca s

lnh ca Eviews ta nhp:

matrix matrixC=A-B

Tch v hng

nhn mt ma trn A vi mt v hng , ta nhn mi

thnh t ca ma trn A vi nh sau:

A = [aij]

V d, nu = 2 v

0 1

5 1 A

2x2

, th A =

0 2

10 2

Thao tc vi Eviews

thc hin php nhn mt s k v ma trn A, th trn

ca s lnh ca Eviews ta nhp: matrix matrixD=k*A

Php nhn ma trn

Gi s ta c A l mt ma trn M dng, N ct, v B l mt

ma trn N dng, P ct. Th ma trn C = AB s l mt ma

trn c M dng v P ct nh sau:


ThS Phng Thanh Bnh

35

N

1kkjikij bac i = 1, 2, M v j = 1, 2, , P

Ngha l, thnh t dng th i v ct th j ca ma trn

C c tnh bng cch nhn cc thnh t ca dng th i

ca ma trn A vi cc thnh t tng ng ca ct j ca ma

trn B v cng tt c cc tch li.

Xem v d sau y11:

1 6 5

7 4 3A

3x2

2 6

5 3

1 2

B2x3

A*B =

(1x2) (6x5) (5x1) (1x6) (6x3) (5x2)

(7x2)(4x5)(3x1) (7x6)(4x3)(3x2)C

2x2

37 43

37 06

Thao tc vi Eviews

nhn ma trn A v B, th trn ca s lnh ca Eviews

ta nhp: matrix AB=A*B

Lu :

AB BA

Tch ca mt vect dng v mt vect ct l mt v hng

uu = [u1 u2 un]

n

2

1

u

.

.

.

u

u

= 2iu

Tch ca mt vect ct v mt vect dng l mt ma trn

uu =

n

2

1

u

.

.

.

u

u

[u1 u2 un] =

s

s

uu u 2121

11

Thc hin php nhn ma trn trn Excel nh sau: Gi s trong bng tnh Excel ma trn A ang v tr A1:C2, v B ang v tr A4:B6. Bc mt, ta chn khi v tr s t ma trn C (gi s D4: E5). Bc hai, ti u tin ca ma trn C (tc D4) ta nhp cng thc =MMULT(A1:C2, A4:B6). Bc ba, nhn phm F2, sau gi ng phm Ctrl+Shift, v cui cng l nhn phm Enter.


ThS Phng Thanh Bnh

36

Nu AMN, BNP, CPK, th (AB)C = A(BC) v A(B + C) = AB + AC v (B+C)A = BA + CA

Php chuyn v ma trn

Php chuyn v ma trn c cc tnh cht nh sau:

(A) = A

C = A + B v C = (A + B) = A + B

(AB) = BA v (ABCD) = DCBA

I = I

=

(A) = A = A = A

Nu A l mt ma trn vung v A = A, th A l mt ma trn i xng

Php nghch o ma trn

Nghch o ca mt ma trn vung A, k hiu l 1A c

nh ngha nh sau:

A1A = 1A A = I

Xem v d sau y12:

8 6

4 2A

2x2

v

4

1-

8

62

1 1-

A 1

2x2

th

1 0

0 1AA 1

Thao tc vi Eviews

tm ma trn nghch o (AI) ca ma trn A, th trn

ca s lnh ca Eviews ta nhp: matrix AI=@inverse(A)

Ma trn nghch o c cc tnh cht sau y:

(AB)-1 = B-1A-1

(A-1) = (A)-1

12 Thc hin php nhn ma trn trn Excel nh sau: Gi s trong bng tnh Excel ma trn A ang v tr A1:C2.

Bc mt, ta chn khi v tr s t ma trn A-1 (gi s A4: B5). Bc hai, ti u tin ca ma trn A-1 (tc A4) ta nhp cng thc =MINVERSE (A1:C2). Bc ba, nhn phm F2, sau gi ng phm Ctrl+Shift, v cui cng l nhn phm Enter.


ThS Phng Thanh Bnh

37

nh thc ma trn

Tng ng vi mi ma trn vung, A, l mt con s nht

nh c gi l nh thc ca ma trn. nh thc ca ma

trn A thng c k hiu l det A hay A . V d,

11 9 8

4 0 1

7 5 1

A v

11 9 8

4 0 1

7 5 1

A

Vic tnh gi tr nh thc i vi ma trn 2 dng 2 ct

tng i n gi bng cch ly tch ca hai thnh t

trn ng cho chnh tr tch ca hai thnh t trn

ng cho ph (a11a22 - a12a21). Tuy nhin, vic tnh gi

tr nh thc ca cc ma trn nhiu dng v nhiu ct (v

d 3 dng 3 ct, hay 5 dng 5 ct) s tr nn kh khn v

lun lm nhiu sinh vin cm thy s hi! Tuy nhin, vic

tnh gi tr nh thc trn Excel thc ra l mt vic ht

sc nh nhng. V d, trn bng tnh Excel gi tr cc

thnh t ca ma trn A nm trong khi A1:C3, ta ch cn

nh cng thc =MDETERM(A1:C3) vo mt no th ta s

bit gi tr ca nh thc A l 116.

Thao tc vi Eviews

tnh nh thc ca ma trn A, th trn ca s lnh

Eviews ta nhp:

scalar detA=@det(A)

Cc tnh cht cn lu ca nh thc ma trn:

Mt ma trn c gi tr nh thc l khng c gi l

ma trn suy bin v mt ma trn c gi tr nh thc

khc khng c gi l ma trn khng suy bin. D

nhin, mt ma trn suy bin th khng th c ma trn

nghch o. Tnh cht ny rt quan trng khi chng ta

phn tch trng hp a cng tuyn hon ho. Mt ma

trn suy bin c ngha l dng (hay ct) ny l mt

hm tuyn tnh ca dng (hay ct) khc trong ma trn

.

Nu tt c cc thnh t ca bt k mt dng no ca

ma trn u bng khng, th gi tr nh thc s bng

khng.

AA

o v tr ca hai dng hay hai ct bt k no s lm

thay i du ca nh thc.


ThS Phng Thanh Bnh

38

Nu mi thnh t ca mt dng hay mt ct ca ma trn

c nhn vi mt v hng , th gi tr nh thc

s tng thn ln.

Nu hai dng hay hai ct ca ma trn l ging nhau,

th gi tr nh thc s bng khng.

Nu mt dng hay ct ca ma trn l bi s ca mt

dng hay mt ct khc th gi tr nh thc s bng

khng (ni cch khc, nu mt dng hay mt ct l mt

kt hp tuyn tnh ca cc dng hay cc ct khc th

gi tr nh thc s bng khng).

BAAB

Tm nghch o ca ma trn vung

iu kin: 0A

)adjA(A

1A 1

(adjA) l chuyn v ca ma trn ph hp ca A

y ta khng trnh by chi tit (adjA) hay ma trn ph

hp, m ch mun lm r tng c bn rng nu ma trn A

c gi tr nh thc bng khng th ta khng th xc nh

c ma trn nghch o ca n c.


ThS Phng Thanh Bnh

39

SUY DIN THNG K L G?

Suy din thng k l vic nghin cu mi quan h gia mt

tng th v mt mu c ly ra t tng th . hiu

r khi nim ny, ta xem xt v d c th sau y v ROE

(sut sinh li trn vn ch s hu, %) trn S giao dch

chng khon TP.HCM.

BNG 3.11: ROE ca 30 cng ty trn S giao dch chng khon TP.HCM

Cng ty ROE Cng ty ROE

ABT 13.7 VFC 13.2

SC5 29.2 VGP 10.5

SFI 41.9 VHC 32.5

BHS 14.1 VHG 16.8

BMP 22.8 VIC 14.6

DHA 17.4 VIS 12.6

DHG 19.7 TSC 44.3

DQC 29.3 VNE 16.6

ITA 11.1 VNM 22.3

PAC 22.9 VPL 8.1

REE 13.0 VSC 26.1

SGH 17.6 VSH 12.7

STB 19.0 BMC 56.2

TAC 34.8 VTB 11.2

UNI 32.5 VTO 30.2

Trung bnh = 22.2 Phng sai = 131.9 lch = 11.5

Ngun: Tc gi tng hp, 2008

Gi s quan tm chnh ca ta khng phi l ROE ca

mt cng ty nht nh no , m l ROE trung bnh ca

ton b chng khon nim yt trn th trng TP.HCM. Trn

nguyn tc, vic thu thp d liu t s ROE ca 152 cng

ty nim yt tnh ROE trung bnh l hon ton c th

thc hin c, nhng thc t, vic lm ny rt tn thi

gian v chi ph (gi s trong tng lai c thm rt nhiu

cng ty nim yt th sao?). Cho nn liu ta c th s

dng ROE trung bnh ca 30 cng ty nh mt gi tr c

lng ca ROE trung bnh tng th hay khng. C th, nu

ta t X = ROE ca mt chng khon v X l ROE trung bnh ca 30 chng khon, th liu ta c th ni g v gi tr

k vng ca ROE, E(X) ca ton b th trng chng khon

TP.HCM hay khng. Qu trnh khi qut ha t gi tr mu,

v d X, cho gi tr tng th, E(X), l ni dung ch yu


ThS Phng Thanh Bnh

40

ca suy din thng k. Ta s tho lun chi tit ch

ny.

Vic thu thp thng tin v tt c cc ROE ca c th

trng chng khon tnh ROE trung bnh l rt tn km,

nn ta c th thu thp mt mu ngu nhin mt s chng

khon tnh ROE trung bnh mu, X. X l c lng ca ROE trung bnh tng th, E(X), v cng c gi l tham

s tng th. Mt gi tr bng s ca c lng c gi

l mt gi tr c lng. V th, c lng l bc u

tin ca suy din thng k. Sau khi c gi tr c

lng ca tham s, ta cn phi tm hiu xem gi tr c

phi l mt gi tr c lng tt cho tham s tng th

hay khng, v mt gi tr c lng c th khng bng gi

tr tham s thc.

Bc th hai ca suy din thng k l kim nh gi

thit. Trong kim nh gi thit ta c th c phn

on hay k vng trc v gi tr ca mt tham s nht

nh. V d, da vo kin thc c hoc kin chuyn

gia ta c th bit ROE trung bnh thc ca th trng l

17.4. Nh vy, gi tr 22.2 t mu 30 cng ty c khc v

mt thng so vi gi tr 17.4 hay khng.

C LNG CC THAM S

Thng thng, mt bin ngu nhin X c cho rng s theo

mt phn phi nht nh, nhng ta khng bit gi tr cc

tham s ca phn phi . Chng hn, nu X theo phn phi

chun, ta mun bit gi tr ca hai tham s trung bnh

E(X) = x v phng sai 2x . c lng cc tham s ny,

qui trnh thng thng l ly mt mu ngu nhin vi n

quan st t mt phn phi xc sut bit v s dng mu

c lng cc tham s cha bit. V th, ta c th s

dng trung bnh mu nh mt gi tr c lng ca trung

bnh tng th v phng sai mu nh mt gi tr c lng

ca phng sai tng th. Qui trnh ny c gi l vn

c lng. Vn c lng c chia thnh hai loi: c

lng im v c lng khong.

c th ha tng ny, gi s bin ngu nhin X

(ROE) l mt bin c phn phi chun vi mt gi tr

trung bnh v mt gi tr phng sai nht nh, nhng ta

khng bit gi tr ca cc tham s ny. Gi s ta c mt


ThS Phng Thanh Bnh

41

mu ngu nhin gm 30 ROE t mt tng th c phn phi

chun nh Bng 3.11.

Ta c th s dng mu ny tnh gi tr trung bnh

tng th x = E(X) v phng sai tng th 2x nh th no.

C th hn, gi s quan tm ca ta by gi l tm gi tr

trung bnh x. Ta thy rng gi tr trung bnh mu, X, l

22.2. Ta gi 22.2 l gi tr c lng im ca x, v

cng thc 30

XX i

c dng tnh gi tr c lng im

c gi l c lng im. Lu rng c lng im l

mt bin ngu nhin v gi tr ca n s khc nhau cc

mu khc nhau. V th, lm sao c th tin mt gi tr c

th nh 22.2 l gi tr c lng ca x. Ni cch khc, lm th no ta c th ch da vo mt gi tr c lng

ca trung bnh tng th. Tuy nhin, c v tt hn nu cho

rng mt khong gi tr no c cha trung bnh tng

th. l tng ca khi nim c lng khong. c

lng khong l mt khong cc gi tr s cha gi tr

thc ca tng th vi mt mc tin cy nht nh.

tng nn tng ca c lng khong l khi nim

phn phi mu ca mt c lng. Gi s, mt bin ngu

nhin X c phn phi chun, X~ N(X, 2), th

X ~ N(X,n

2x ) (3.62)

Z =

n/

X

x

x

~ N(0, 1) (3.63)

iu ny c ngha rng phn phi mu ca trung bnh

muX cng theo phn phi chun. Nu X khng theo phn phi

chun, th theo nh l gii hn trung tmX s theo phn

phi chun nu c mu ln.

Tuy nhin, do khng bit gi tr phng sai2x , nhng

ta c th s dng c lng ca n l 1n

)XX(S

2x2

x

th

t =

n/S

X

x

X ~ td.f.=(n-1) (3.64)

theo phn phi t vi (n-1) bc t do, (d.f.).


ThS Phng Thanh Bnh

42

bit phng trnh (3.64) c s dng nh th no cho

c lng khong ca gi tr trung bnh tng th x ca ROE, ta s dng Bng A2 vi bc t do l 29 nh sau:

P(-2.045 t 2.045) = 0.95 (3.65)

Hnh 3.13: Khong tin cy 95% ca X vi s bc t do l 29

Vi d.f. = 29, xc sut l 0.95 (hay 95%), th khong (-

2.045, 2.045) s cha gi tr t tnh t cng thc (4.3).

Cc gi tr t trn c gi l cc gi tr t ph phn

(critical t values) s cho bit phn trm din tch di

ng phn phi t gia hai gi tr ph phn ny. Trong

, t = -2.045 c gi l gi tr t ph phn chn di

(lower critical t value) v t = 2.045 c gi l gi tr

t ph phn chn trn (upper critical t value).

Th gi tr t t (3.64) vo (3.65) ta c:

P(-2.045

n/S

X

x

X 2.045) = 0.95 (3.66)

P(X - 2.045n

Sx X X + 2.045n

Sx ) = 0.95 (3.67)

-2.045 2.045


ThS Phng Thanh Bnh

43

Hnh 3.14a: Khong tin cy 95% ca X vi s bc t do l 29

Hnh 3.14b: Khong tin cy 99% ca X vi s bc t do l 29

Phng trnh (3.67) l mt c lng khong (interval

estimator) ca trung bnh tng th x.

Trong thng k, phng trnh (3.67) c gi l 95%

khong tin cy (confidence interval) cho gi tr thc

nhng khng bit trung bnh tng th x v 0.95 c gi l h s tin cy (confidence coefficient). Ni cch khc,

phng trnh (3.66) cho bit 95% khong ngu nhin

( )n/S045.2X x cha gi tr trung bnh thc x.

( )n/S045.2X x c gi l gii hn di ca khong v

( )n/S045.2X x l gii hn trn.

n

xS 045.2X

n

S 045.2X x

X

n

S 756.2X x

n

xS 756.2X X


ThS Phng Thanh Bnh

44

Lu rng khong phng trnh (3.67) l mt khong

ngu nhin v n ph thuc vo X v n

Sx , m hai i lng

ny khc nhau cc mu khc nhau. Nhng trung bnh tng

th x, d khng bit, nhng thc ra l mt con s c nh v v th khng ngu nhin. Cho nn, s khng ng nu

ni rng xc sut 95% x nm trong khong ny, m phi ni

rng xc sut 95% khong ngu nhin cha x.

Quay li v d trn vi n = 30, X = 22.2, v Sx = 11.5 th ta xc nh khong ngu nhin nh sau:

22.2 - 30

)5.11)(045.2(2.22

30

)5.11)(045.2(x

17.91 26.49 (3.68)

Phng trnh (3.68) cho rng nu ta xy dng c cc

khong phng trnh (3.68), th trong 100 ln, c 95

khong nh vy s cha gi tr x thc. Hy cn thn, khng th ni rng xc sut l 95% mt khong nht nh

trong phng trnh (3.68) cha x.

Thao tc vi Eviews (trn ca s lnh)

Gi tr trung bnh mu ca ROE: scalar ROEmean=@mean(roe) = 22.2

Gi tr t ph phn: scalar tc95=@qtdist(0.975,29) = 2.045

lch chun ca ROE: scalar stdevROE=@stdev(roe) = 11.5

S quan st ca mu: scalar obs=@obs(roe) = 30

Gi tr chn di ca ROE: scalar ROE_lb=ROEmean-tc95*stdevROE/obs

Gi tr chn trn ca ROE: scalar ROE_ub=ROEmean+tc95*stdev/obs

* Lu : Khi qu v quen cc hm trn Eviews, qu v c th xy dng mt cc trc tip ch khng cn thc hin tng thao tc ring nh th ny.

KIM NH GI THIT

Sau khi xem xt nhnh c lng ca suy din thng k,

by gi chng ta s xem xt chi tit hn nhnh th hai

ca n l kim nh gi thit. Tr li v d v ROE

trn, thay v tm khong tin cy cho x, gi s ta gi

thit rng gi tr thc X bng mt gi tr bng s c

th, v d X = 17.4. Cng vic ca ta by gi l kim


ThS Phng Thanh Bnh

45

nh gi thit ny. Ta s kim nh gi thit ny nh th

no l ng h hay bc b n?

Theo ngn ng kim nh gi thit th mt gi thit

nh x = 17.4 c gi l gi thit khng v c k hiu

l H0. Nh vy, H0: x = 17.4. Gi thit khng lun c kim nh ngc li vi gi thit khc, k hiu l H1. Gi

thit khc c th di mt s hnh thc khc nhau nh

sau:

H1: x > 17.4, y l gi thit khc mt pha hay mt ui

H1: x < 17.4, cng l gi thit khc mt pha hay mt ui

H1: x 17.4, y c gi l gi thit khc kt hp, hai pha, hay hai ui.

kim nh H0, ta s dng d liu mu v l thuyt thng

k xy dng cc qui tc quyt nh nhm xem chng c

ca mu c ng h gi thit khng hay khng. Nu chng c

mu ng h gi thit khng, ta khng bc b H0, ngc li,

ta bc b H0, iu ny cng c ngha ta chp nhn gi

thit H1.

Vn c ra l chng ta xy dng cc qui tc quyt

nh ny nh th no? C hai cch tip cn kim nh c

tnh b sung cho nhau: (i) Khong tin cy, v (ii) Kim

nh ngha. Chng t s dng v d v ROE minh ha

cho cc phng php kim nh ny. Gi s ta c cc gi

thit kim nh nh sau:

H0: x = 17.4

H1: x 17.4

Kim nh da vo khong tin cy

Kim nh da vo khong tin cy c ngha l ta c gng

xy dng khong tin cy cho mt c lng, ri kim tra

xem gi tr thc c gi nh theo gi thit H0 s nm

trong hay ngoi khong tin cy . V trn c s ta s

c quyt nh bc b hay khng bc b gi thit H0.

kim nh gi thit H0, gi s ta c d liu mu

nh trong Bng 3.11. T d liu ny, ta tnh c gi tr

trung bnh mu l 22.2. Ta bit rng trung bnh mu c

phn phi chun vi trung bnh l x v phng sai l

n/2x . Nhng do ta khng bit gi tr phng sai thc ca

tng th, nn ta thay th phng sai ny bng phng sai


ThS Phng Thanh Bnh

46

mu, v nh th th trung bnh mu s theo phn phi t.

Da vo phn phi t ta xy dng c khong tin cy 95%

cho X nh sau:

17.91 X 26.49 (3.69)

Ta bit rng cc khong tin cy a ra mt khong cc gi

tr c th bao gm gi tr x thc vi mt mc tin cy nht nh, chng hn 95%. V th, khong tin cy ny

khng gi tr gi thit khng, x = 17.4, ta c th bc b gi thit H0 ny khng?

Nh vy, khong tin cy v kim nh gi thit l hai

ch c quan h mt thit vi nhau. Theo ngn ng kim

nh gi thit, khong tin cy 95% c gi l vng chp

nhn v vng ngoi vng chp nhn l vng ph phn hay

vng bc b13 gi thit H0. Gi tr chn di v gi tr

chn trn ca vng chp nhn c gi l cc gi tr ph

phn14. Theo ngn ng thng k, nu vng chp nhn c cha

gi tr tham s gi thit H0, ta khng bc b H0 (ngha

l chp nhn H0 l ng). Nhng nu ri ngoi vng chp

nhn (tc nm trong vng bc b), ta bc b H0. v d

ang xt, ta bc b gi thit H0: x = 17.4 v vng chp nhn, nh phng trnh (3.69), khng cha gi tr gi

thit khng ny. .

Cc bc thc hin:

Xc nh gi thit H0 v H1

Chn mc ngha vi phn phi t hai ui

Xy dng khong tin cy

Kim tra xem khong tin cy c cha gi tr ca gi thit H0 hay khng

a ra quyt nh

13

Acceptance region, critical region, v the region of rejection. 14

Critical value. Lu , cc sch thng k hoc kinh t lng Vit Nam dch thut ng critical l ph phn hoc ti hn. Tuy nhin, iu quan trng l ta nn hiu bn cht ca thut ng ny. V d, ta thng ni a critical decision c ngha l mt quyt nh c ngha sng cn, theo kiu thng lm vui, thua lm gic. Trong thng k, cc ranh gii ca vng chp nhn c gi l cc gi tr ph phn, v cc ranh gii ny l ng phn chia gia vic chp nhn v bc b gi thit H0. Ngoi ra, cc gi tr ny l cc gi tr chun c tnh da trn mt phn phi nht nh (v trc y chng c nh km dng cc bng thng k phn ph lc cc sch gio khoa) gip sinh vin tra v so snh, cho nn i khi chng c gi l gi tr tra bng. Ngy nay, chng ta thng tra nhanh cc gi tr ny bng cc cng thc nh TINV, FINV, CHIINV, trn Excel hoc cc hm @qtdist, @fqdist, @qchisq, trn Eviews.


ThS Phng Thanh Bnh

47

Sai lm loi I v loi II

Trong v d v ROE ta bc b gi thit H0: x = 17.4 bi

v chng c t mu ca ta X = 22.2 dng nh khng ph hp vi gi thit ny. C phi iu ny c ngha l mu

30 chng khon nh trong Bng 3.11 khng c ly t mt

tng th c phn phi chun vi gi tr trung bnh l

17.4? Chng ta hon ton khng chc chn, v khong tin

cy y l 95% ch khng phi 100%. Nu iu ny l

ng, ta c th b mc sai lm bc b gi thit H0: x = 17.4. y l loi sai lm loi I: sai lm trong vic bc

b mt gi thit ng. Tng t, gi s H0: x = 20, v theo phng trnh (3.69) ta s khng bc b gi thit

khng ny. Nhng rt c th mu 30 chng khon khng c

ly t mt tng th c phn phi chun vi gi tr trung

bnh l 20. Nh vy, ta c th mc sai lm loi II: sai

lm trong vic chp nhn gi thit sai.

BNG 3.12: Hai loi sai lm trong thng k

Bc b H0 Khng bc b H0

H0 l ng H0 l sai

Sai lm loi I Quyt nh sai

Quyt nh ng Sai lm loi II


Tht l tng nu ta c th ti thiu ha c hai loi sai

lm ny. Tuy nhin, vi mt c mu nht nh ta khng th

no ti thiu ha ng thi hai loi sai lm ny. Theo

Gujarati (2006, 116), cch duy nht gim sai lm loi II

m khng lm tng sai lm loi I l tng c mu. Nhng

y khng phi l iu lun lun d dng. Cch tip cn

c in cho rng sai lm loi I nghim trng hn sai lm

loi II. Cho nn, ngi ta c gng gim thiu sai lm

loi I cng nh cng tt, v d khong 0.01 hay 0.05, v

c gng gim thiu sai lm loi II. Theo l thuyt thng

k, xc sut chp nhn sai lm loi I c qui c l , c gi l mc ngha, v xc sut chp nhn sai lm

loi II c qui c l . Nh vy,

Sai lm loi I = = xc sut bc b H0 (H0 l ng)

Sai lm loi II = = xc sut chp nhn H0 (H0 l sai)

Xc sut khng chp nhn sai lm loi II, ngha l, bc

b H0 khi H0 l sai, l (1-) v c gi l sc mnh ca kim nh.


ThS Phng Thanh Bnh

48

Theo cch tip cn c in v kim nh gi thit th thng c cho cc mc 1%, 5%, ri c gng gim thiu

. Trong thc t, cch tip cn c in thng xc nh

gi tr m khng cn quan tm nhiu n . Nhng lu rng, khi ra quyt nh lun tn ti s nh i gia mc

ngha v sc mnh kim nh. iu ny c ngha rng,

vi mt c mu nht nh, nu ta c gng gim xc sut

ca sai lm loi I, ta c th tng xc sut sai lm loi

II v v th gim sc mnh kim nh. Chng hn, thay v

s dng = 5%, ta s dng = 1%, chng ta c th rt t tin khi bc b H0, nhng chng ta c th s khng t tin

lm khi khng bc b H0.

Quay li v d ROE nhng vi = 1% v ta s tnh c khong tin cy 99% nh sau:

16.41 x 27.99 (3.70)

So vi (3.69), th khong tin cy 99% rng hn khong tin

cy 95%. V khong tin cy 99% ny c cha gi tr gi

thit l 17.4, nn ta khng bc b gi thit khng. iu

ny ni ln iu g? Bng cch gim sai lm loi I t 5%

xung 1%, chng ta tng xc sut chp nhn sai lm

loi II. Ngha l, vic khng bc b gi thit khng theo

cng thc (3.70), chng ta c th ang chp nhn mt cch

sai lm gi thit cho rng gi tr x thc l 17.4. Cho nn, hy nh rng lun lun c s nh i gia sai lm

loi I v sai lm loi II.

Kim nh da vo ngha

Kim nh ngha l mt cch kim nh khc c tnh b

sung v ngn gn hn so vi kim nh da vo khong tin

cy15. Xin nhc li rng t = n/S

)X(

x

X c phn phi t vi

(n-1) bc t do. Cc thng tin X , n, v S bit t mu, nn nu c gi tr x di gi thit khng, ta s tnh c gi tr thng k t (cn c gi l gi tr t

tnh ton). Hn na, vi (n-1) bc t do ta c th xc

nh c cc gi tr t ph phn16.

15

V cc phn mm kinh t lng u cung cp cc thng tin gi tr thng k t v gi tr xc sut tng ng, nn phng php kim nh ny c s dng ph bin.

16 Bit rng t=

n/xS

)XX( theo phn phi t vi bc t do l (n-1), nn ta c th d dng tnh c gi tr xc sut

tng ng ca n bng cng thc =TDIST(x,Deg_Freedom,Tails). V d, =TDIST(2.045,29,2) = 5%. Ngoi ra,


ThS Phng Thanh Bnh

49

Nu khc bit gia X v x l nh (t s s nh v mu s khng i), th gi gi tr tuyt i ca t tnh ton s

nh. Nu X = x, gi tr t tnh ton s bng khng, v chc chn ta s khng bc b gi thit H0. Cho nn, khi

gi tr tuyt i ca t tnh ton cng khc khng, ta s

c xu hng bc gi thit H0. Ni cch khc, khi gi tr

tuyt i ca t tnh ton cng ln, ta cng c xu hng

bc b gi thit H0. Tuy nhin, quyt nh bc b hay chp

nhn H0 ty thuc vo mc ngha c chn. Nu gi tr t

tnh ton nm gia cc gi tr t ph phn chn di v

chn trn th ta chp nhn H0 ( 2/tt ). Ngc li, nu gi

tr thng k t nm ngoi cc gi tr t ph phn chn di

v chn trn th ta bc b H0 ( 2/tt ).

V d, X = 22.2, Sx = 11.5, v n = 30. Gi s H0: x = 17.4 v H1: x 17.4. Ta c:

t = 30/5.11

)5.182.22( = 2.286

Vi s bc t do l 29, cc gi tr t ph phn vi mc

ngha 5% l -2.045 v 2.045. Nh vy, t tnh ton bng

2.286 nm pha ui phi vng bc b ca phn phi t,

nn ta bc b gi thit H0. Ngoi ra, ta d dng tnh xc

sut 286.2t ch l 2.974%.

Cc bc thc hin:

Xc nh gi thit H0 v H1

Tnh gi tr thng k t

Chn mc ngha vi phn phi t hai ui v xc nh cc gi tr t ph phn chn di v chn trn

So snh gi tr thng k t v gi tr t ph phn

a ra quyt nh

Theo ngn ng ca kim nh ngha ta thng gp hai

thut ng sau y:

Kim nh c ngha thng k17

Kim nh khng c ngha thng k

tnh gi tr t ph phn ta dng cng thc =TINV(Probability,Deg_Freedom). V d, =TINV(5%,29) = 2.045. Hoc ta cng c th s dng hm @qtdist(0.975,29) = 2.045 hoc 1-@ctdist(2.045,29) = 5%. 17

Khi ni mt kim nh/kt qu nghin cu c ngha thng k, ngha l ta c th bc b gi thit H0. Ngc li, khi ta khng bc b gi thit H0, ta ni rng kim nh/kt qu nghin cu khng c ngha thng k.


ThS Phng Thanh Bnh

50

Kim nh mt ui hay hai ui?

Cho n y tt c cc v d u xc nh gi thit H1 l

hai ui. V th nu ROE bng 17.4 di gi thit H0, th

di gi thit H1 th ROE c th hoc ln hn hoc nh

thua 17.4. Nh vy, nu thng k kim nh ri vo bt k

ui no ca phn phi (vng bc b) ta bc b H0.

Hnh 3.15a: Kin nh mc ngha t: Hai ui

95%

2.5%2.5%

Tuy nhin, cng c trng hp gi thit khng v gi

thit khc l mt ui, hay mt pha. V d, H0: x 17.4

v H1: x > 17.4, tc gi thit khc l mt ui. Ta s kim nh gi thit ny nh th no?

Th tc kim nh hon ton ging cch kim nh

cc v d trn ngoi tr vic thay v tm hai gi tr ph

phn, by gi ta ch xc nh mt gi tr ph phn duy

nht ca thng k kim nh.

Lu , khi tra bng A2, trng hp kim nh hai pha

ta chn mc ngha dng di, ngc li, kim nh mt

pha ta chn mc ngha dng trn. Khi s dng cng

thc TINV trn Excel, th mc ngha c chn lun l

. Cho nn, nu kim nh hai pha vi = 5%, th TINV(5%,d.f.), ngc li, nu kim nh mt pha cng vi

= 5%, th ta phi chn TINV(10%,d.f.).

-2.045 2.045

Vng chp

nhn H0


ThS Phng Thanh Bnh

51

BNG 3.13: Tm tt kim nh t

Gi thit

khng, H0

Gi thit khc,

H1

Vng ph phn/bc b

H0 nu

x = 0

x = 0

x = 0

x > 0

x < 0

x 0

t = n/S

X

x

0 > t,d.f.

t = n/S

X

x

0 < t, d.f.

t = n/S

X

x

0 > t/2,d.f.


Hnh 3.15b: Kin nh mc ngha t: Mt ui

95%

5%

Lu , v mt t ng, ta nn s dng cch pht biu bc

b hoc khng bc b gi thit H0, hn l bc b hoc

chp nhn mt gi thit. Theo Gujarati (2006), vic ta

khng bc b gi thit khng khng nht thit c ngha

rng gi thit ng, bi v mt gi thit khng khc

c th cng tng thch vi d liu.

Mc ngha v gi tr xc sut p

Gi tr xc sut p (p-value) cng c gi l mc ngha

chnh xc ca thng k kim nh (v d thng k t). Gi

1.699

Vng chp

nhn H0


ThS Phng Thanh Bnh

52

tr xc sut p c th c nh ngha l mc ngha thp

nht ti gi thit H0 c th b bc b. Qui tc quyt

nh vi gi tr xc sut p nh sau: Gi tr xc sut p

cng nh, th bng chng bc b gi thit H0 cng mnh.

Cc phn mm kinh t lng u c bo co gi tr xc

sut p. Chng ta s phn tch tht chi tit vn gi

tr xc sut p Bi ging 7.

Mi quan h gia hai phng php kim nh

T phng trnh (3.64) ta thy rng bin ngu nhin t l

mt hm tuyn tnh ca bin ngu nhin X. Di gi thit

H0 ta c gi tr x, nn s tm c gi tr ca t theo

X. Mi quan h c th hin qua th phn phi xc sut sau y:

Hnh 3.16: Mi quan h gia hai phng php kim nh

X

t -2.045 2.045

17.91 26.49

2.5% 2.5%

2.5% 2.5%

Vng chp

nhn H0

Vng chp

nhn H0


ThS Phng Thanh Bnh

53

Kim nh ngha 2

Nh bit, nu S2 l phng sai mu t mt mu c ly

ngu nhin vi n quan st t mt tng th c phn phi

chun vi phng sai l 2, th

(n-1)2

)1n(2

2

~S

(3.71)

Ngha l, t s ca phng sai mu/phng sai tng th

c nhn vi s bc t do (n-1) c phn phi 2 vi s

bc t do l (n-1). Lu rng, (n-1) v 2 ch l cc con s nht nh, ring bn thn S2 l mt bin ngu nhin v

gi tr ca S2 s thay i t mu ny qua mu khc (tng

t nh X). Do X l mt bin ngu nhin c phn phi

chun, th S2 c xem gn nh mt 2

X , nn theo nh

ngha bi ging 3, S2 s c phn phi 2. Hn na, v tri ca phng trnh (3.71) l mt bin tng ca (n-1)

bin 22S

, nn s bc t do s l (n-1).

BNG 3.14: Tm tt kim nh 2

Gi thit khng, H0 Gi thit khc, H1 Vng ph phn/bc b H0 nu

20

2x

20

2x

20

2x

20

2x

20

2x

20

2x

2)1n(,2

0

2xS)1n(

2)1n(),1(2

0

2xS)1n(

2)1n(),2/1(

2)1n(,2/2

0

2x

hay

S)1n(


V d, t mt mu ngu nhin nh Bng 3.11 ta c phng

sai mu l S2 = 131.9, ta kim nh xem gi tr ny c

khc g mc ngha = 5% so vi gi tr phng sai thc ca tng th l 99.1 hay khng.

H0: 2 = 99.1 v H1:

2 99.1

Lu , qu v cn xem y l loi gi thit g trn ba

trng hp c trnh by Bng 3.14 xc nh gi tr

2 ph phn cho ng. Ta c n = 30 nn thay vo phng


ThS Phng Thanh Bnh

54

trnh (3.70) ta s c gi tr 2 tnh ton l (30-1)1.99

9.131 =

38.598 vi s bc t do l 29. T cng thc

=CHIINV(2.5%,29) = 45.72 > 38.598 > =CHIINV(97.5%,29) =

16.05, ta khng bc b gi thit H0 mc ngha 5%.

Ngoi ra, ta c th tnh xc sut c c gi tr 2 bng hoc ln hn 38.598 (vi 29 bc t do) theo cng

thc =CHIDIST(38.598,29) = 0.1096 hay 10.96%. V xc sut

ny ln hn mc ngha c chn l 5%, nn ta khng bc

b gi thit H0 cho rng phng sai thc l 99.1.


Gi tr phng sai mu ca ROE: scalar ROEvar=@var(roe) = 131.9

hoc scalar ROEvar=@stdev(roe)^2 = 131.9

Gi tr 2 ph phn chn trn: scalar chisqc975=@qchisq(0.975,29) = 45.72

Gi tr 2 ph phn chn di: scalar chisqc0025=@qchisq(0.025,29) = 16.05

S quan st ca mu: scalar obs=@obs(roe) = 30

Gi tr 2 tnh ton: scalar chival=(obs-1)*(ROEvar/99.1) = 38.598

Kim nh ngha F

Ta bit, nu ta c hai mu ngu nhin t hai tng th

c phn phi chun X v Y, vi m v n quan st, th

)1n/()YY(

)1m/()XX(

S

SF

2i

2i

2Y

2X (3.72)

s theo phn phi F vi (m-1) v (n-1) bc t do. Nu ta

c gi thit H0: 2y

2x th ta kim nh da vo bng sau:

BNG 3.15: Tm tt thng k F

Gi thit

khng, H0

Gi thit khc,

H1

Vng ph phn/bc b H0

nu

22

21

22

21

22

21

22

21

ddf,ndf,22

21 F

S

S

ddf,ndf),2/1(

ddf,ndf,2/22

21

F hay

FS

S



ThS Phng Thanh Bnh

55

V d, gi s ta c hai mu ngu nhin 30 cng ty nim

yt trn hai th trng TP.HCM v th trng H Ni (ngy

26 thng 5 nm 2008). T hai mu ny ta thy ROE trung

bnh hai th trng gn bng nhau (X=22.2 v Y=22.39). By gi ta mun xem phng sai ca hai th trng ny c

ging nhau hay khng. T hai mu ta c phng sai ln

lt l 131.9 v 79.6. Vi gi thit H0: 2y

2x v H1:

22

21 , vi mc ngha = 5%. Mt ln na xin qu v hy

lu loi gi thit chng ta a ra l g xc nh

gi tr F ph phn cho ph hp. Ta c gi tr F tnh ton

l 131.9/79.6 = 1.657, c phn phi F vi s bc t do

ln lt l 29 v 29. Do gi tr F ph phn tnh t cng

thc =FINV(5%,29,29) = 1.86 > 1.657, vy ta khng bc b

gi thit H0 mc ngha 5% (ngha l hai phng sai

tng th l ging nhau). Ta cng c th kt lun tng t

bng cch so snh mc ngha 5% vi gi tr xc sut p

=FDIST(1.657,29,29) = 8.99%. Tuy nhin, nu mc ngha

10% th kt lun ca ta s thay i.


Gi tr phng sai mu ca ROE, TP.HCM: scalar

ROE1var=@var(roe1) = 131.9

Gi tr phng sai mu ca ROE, H Ni: scalar

ROE2var=@var(roe2) = 99.6

Gi tr F tnh ton: scalar Fval=ROE1var/ROEvar2

Gi tr F ph phn: fc095=@qfdist(0.95,29,29) = 1.86

By gi ta c th tm tt cc bc trong vic kim nh

mt gi thit thng k bt k nh sau:

Bc 1: Pht biu gi thit H0 v gi thit H1

Bc 2: Chn la thng k kim nh thch hp (v d

trung bnh, phng sai, hay so snh phng sai, )

Bc 3: Xc nh phn phi xc sut thch hp ca

thng k kim nh (v d t, 2, hay F)

Bc 4: Chn mc ngha ( l xc sut chp nhn sai lm loi I)

Bc 5: Chn phng php kim nh thch hp (xy

dng khong tin cy hay kim nh ngha)

Mot So Van de Co Ban Ve Xac Suat Thong Ke

Documents

Transcript of Mot So Van de Co Ban Ve Xac Suat Thong Ke