Lecture #10

10.1

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Lecture #10Studenmund (2006): Chapter 10

Objectives

• What is heteroscedasticity?

• What are the consequences?

• How is heteroscedasticity be identified?

• How is heteroscedasticity be corrected?

10.2


The probability density function for Yi at three different levels of family income, X1 i , are identical.

Homoscedasticity Case

.

.

x1ix11=80 x13=10

0

Y if(Yi)

expe

nditu

re

income

Var(i) = E(i2)= 22

.

x12=90

10.3


...

....

...

...

..

..

. .. ... . .

. .. .. .

....

...

..

.

xi

yi

0

Homoscedastic pattern of errors

The scattered points spread out quite equally

10.4


The variance of Yi increases as family income, X1i, increases.

Heteroscedasticity Case

.

x1ix11 x12

Y if(Yi)

expe

nditu

re

x13

..

income

Var(i) = E(i2)= ii

22

10.5


Heteroscedastic pattern of errors.

... .

. .. .

..

.

.

.

.

..

..

.

..

.

..

...

.

..

..

..... ..... .

..

xi

yi

0The scattered points spread out quite unequally

Small i associated with small value of Xi

large i associated with large value of Xi

10.6


Definition of Heteroscedasticity:

Two-variable regression: Yi = 0 + 1 X1i + i

1 = = wi Yi = wi (0 + 1Xi + i) xy

x2

^

=> 1 = 1 + wi i E(1) = 1 unbiased^ ^

if 12 = 2

2 = 32 = …

i.e., homoscedasticity 22

xi2

Var (1) = ^ xi

2

if 12 2

2 32 …

i.e., heteroscedasticity= ii

22

= ii2 x2

( xi2)2

Var (1) = E (1 - 1)2 = E (wi i)2

= E (w12 1

2 + w22 2

2 + …. + 2w1w2 1 2 + …)

= w12 1

2 + w22 2

2 + … …..+ 0 + ...

= ii2 wi

2

^ ^

Var(ii) = E(ii2) = ii

22 22

Refer to lecture notesSupplement #03A

10.7


Consequences of heteroscedasticity

1. OLS estimators are still linear and unbiased

2. Var ( i )s are not minimum. ^

=> not the best => not efficiency => notnot BLUEBLUE

4. 2 = is biased, E(E(22) ) 22i

2^ ^

n-k-1 ^

5. t and F statistics are unreliableunreliable. Y = 0+ 1 X +

Cannot be min.

SEE = RSS = i2^ ^

3. Var ( 1) = instead of Var( 1)= i

2

x2

^ ^ 2

x2

Two-variable case

10.8


Detection of heteroscedasticity

1. Graphical method :

plot the estimated residual ( i ) or squared (i 2 ) against the predicted dependent Variable (Yi) or any independent variable(Xi).

^ ^^

Observe the graph whether there is a systematic pattern as:

Y

2^

Yes, heteroscedasticity exists

10.9


Detection of heteroscedasticity: Graphical method

Y

2^ yes

Y

2^ yes

Y

2 yes

Y

2 yes

Y

2^ yes

Y

2^no heteroscedasticity

10.10


Yes, heteroscedasticityYes, heteroscedasticity

no heteroscedasticityno heteroscedasticity



10.11


Park test procedures:

1. Run OLS on regression: Yi = 1 + 2 Xi + i , obtain i

2. Take square and take log : ln ( i2)^

4. Use t-test to test H0 : 2* = 0 (Homoscedasticity)

If t* > tc ==> reject H0 ==> heteroscedasticity exists

If t* < tc ==> not reject H0 ==> homoscedasticity

3. Run OLS on regression: ln ( i2) = 1

* + 2* ln Xi + vi

^

Statistical test: (i) Park test

H0 : No heteroscedasticity exists i.e., Var( i ) = 2

(homoscedasticity)

H1 : Yes, heteroscedasticity exists i.e., Var( i ) = i2

Suspected variablethat causes

heteroscedasticity

10.12


Example: Studenmund (2006), Equation 10.21 (Table 10.1), pp.370-1

Park TestPractical

In EVIEWS

Procedure 1:

PCON: petroleum consumption in the ith stateREG: motor vehicle registrationTAX: the gasoline tax rate

May misleading

10.13


Graphical detectionGraphical detection

10.14


Procedure 2: Obtain the residuals, take square and take log

10.15


ˆˆˆˆ

YYYY

Scatter plot

Horizontal variable

10.16


EGvs.

REGvs.ˆ2

10.17


Yvs

Yvs

Yvs

i

i

i

ˆ.)ˆlog(

ˆ.ˆ

ˆ.ˆ

2

10.18


Procedure 3 & 4

Refers to Studenmund (2006), Eq.(10.23), pp.373

If | t | > tc => reject H0

=> heteroscedasticity

**1

*0

2 )ˆlog( REG

10.19


(ii)Breusch-Pagan test, or LM test

(4) Compare the W and 2df (where the df is #(q) of regressors in (2))

if W > 2df ==> reject the Ho

H0 : homoscedasticity Var ( i ) = 2

H1 : heteroscedasticity Var ( i ) = ii22

(3) Compute LM=W= nR2Or F=

R2u / q

(1 - R2u) / n-k

if F*> Fcdf ==> reject the Ho

Test procedures:(1) Run OLS on regression: Yi = 0 + 1X1i + 2X2i +...+ qXqi + i ,

obtain the residuals, i^

(2) Run the auxiliary regression: i

2 = 0 + 1 X1i + 2X2i +… +qXqi + vi^

10.20


Yi = 0 + 1X1i + 2X2i + 3X3i + i

10.21


WW=

BPG test for a linear modellinear modelPCON=0+1REG+2Tax+The WW-statistic indicates thatthe heteroscedasticity is existed.

FC(0.05, 5, 44) = 2.45

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

Decision rule: WW > 2df ==> reject the Ho

10.22


FC(0.05, 5, 44) = 2.45

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

Decision rule: WW > 2df ==> reject the Ho

WW=

The BPG test for a transformed log-log model:log(PCON)=0+1log(REG)+2log(Tax)+

The WW-statistic indicates that the heteroscedasticity is still existed.

Therefore, a double-log transformation may notnot necessarilynecessarily remedy theHeterocsedasticity.

10.23


(iiia) White’s general heteroscedasticity test (no cross-termno cross-term) (The White Test)(The White Test)

(4) Compare the W and 2df (where the df is # of regressors in (2))




(3) Compute W (or LM) = nR2

Test procedures:(1) Run OLS on regression: Yi = 0 + 1 X1i + 2 X2i + i ,



2 = 0 + 1 X1i + 2 X2i + 3 X21i + 4 X2

2i + vi^

10.24


(4) Compare the W and 2df (where the df is # of regressors in (2))




(3) Compute W (or LM) = nR2

Test procedures:(1) Run OLS on regression: Yi = 0 + 1 X1i + 2 X2i + i ,



2 = 0 + 1 X1i + 2 X2i + 3 X21i + 4 X2

2i + 5 XX1i 1i XX2i2i + vi^

(iiib) White’s generalgeneral heteroscedasticity test (with cross-termswith cross-terms) (The White Test)(The White Test)

Cross-termCross-term

10.25


No cross-termWith cross-term

10.26


W=

White test for a linear modellinear modelPCON=0+1REG+2Tax+

The WW-statistic indicates thatthe heteroscedasticity is existed.

FC(0.05, 5, 44) = 2.45

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

Decision rule: W > 2df ==> reject the Ho

10.27


FC(0.05, 5, 44) = 2.45

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

Decision rule: W > 2df ==> reject the Ho

W=

The White test for a transformed log-log model:log(PCON)=0+1log(REG)+2log(Tax)+

The WW-statistic indicates that the heteroscedasticity is still existed.

Therefore, a double-log transformation may notnot necessarilynecessarily remedy theHeterocsedasticity.

10.28


Another example 8.4 (Wooldridge(2003), pp.258)

The White test for a linear modelPCON=0+1REG+2Tax+The test statistic indicates heteroscedasticity is existed.

2(0.05, 9) = 16.92

2(0.10, 9) = 14.68

W =

Decision rule: W > 2df => reject the H0

10.29


Testing the log-log model

The White test for a log-log modelThe test statistic indicates heteroscedasticity is not existed

Using the log-log transformationin some cases may remedy the heteroscedasticity, (But not necessaryBut not necessary).

2(0.05, 9) = 16.92

2(0.10, 9) = 14.68

Decision rule: WW < 2df => not reject H0

W=

10.30


RemedyRemedy :Weighted Least SquaresWeighted Least Squares (WLSWLS)

Suppose : Yi = 0 + 1 X1i + 2 X2i + i

E(i) = 0, E(i j )= 0 i j

Vqr (i2) = i

2 = 2 [f (ZZX2i)] = 2ZZi2

If all ZZii = 1 (or any constant), homoscedasticity returns. But ZZii can be any value, and it is the proportionality factor.

In the case of 2 was known :To correct the heteroscedasticity

Transform the regression:

Yi 1 X1i X2i i =0 + 1 + 2 +ZZii ZZii Z Zii Z Zii Z Zii

=> YY** = = 00 X X00** + + 11 X X11

** + + 22 X X22* + * + ii**

If Var(i2)=2ZZii

Then each term divided by ZZii

10.31


In the transformed equation where

(i) E ( ) = E (i) = 0i

ZZii ZZii

1

(ii) E ( )2 = E (i2) = ZZii

222 = 2i

ZZii ZZii22

1 1ZZii

22

(iii) E ( ) = E ( i j ) = 0i j

ZZii ZZjj ZZiiZZjj

1

These three results satisfy the assumptions of classical OLSThese three results satisfy the assumptions of classical OLS..

Why the WLS transformation can remove the heteroscedasticity?Why the WLS transformation can remove the heteroscedasticity?

10.32


Therefore, we might expect i2 = Zi

22 Zi2 = X2i

2 =>Zi =X2i

These plots suggest variance is increasing proportional to X2i2.

The scattered plots spreading out as nonlinear patternnonlinear pattern.

=> Yi* = 1 X0

* + 2 X1* + 3 + *

Now this becomes

the interceptcoefficient

If the residuals plot against X2i are as following :

X2

i

+0-

^

X2

2

Yi 1 X1i X2i i

X2i X2i X2i X2i X2i

= 0 + 1 + 2 +

Hence, the transformed equation becomes

Where **ii satisfies the assumptions of classical OLS

10.33


Example: Studenmund (2006), Eq. 10.24, pp.374

C.V.=0.3392

10.34


The correction is built in EVIEWS

Use the weight (1/REG)(1/REG)to remedy the

heteroscedasticity

10.35


Refers to Studenmund (2006), Eq.(10.28), pp.376

OLS result

WLS result

10.36


FC(0.05, 5, 44) = 2.45

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

W=W=

Decision rule: WW < 2

df ==> not reject Ho

Now, after the reformulationNow, after the reformulationThe test statistic value indicates that the heteroscedasticity is not existed.

10.37


Alternative remedy of heteroscedasticity: reformulate with per capitaper capita

iii

i

i

i TAXPOP

REG

POP

PCON 210

10.38


W=

Now, after the reformulationNow, after the reformulationThe test statistic value indicates thatthe heteroscedasticity is not existed.

FC(0.05, 5, 44) = 2.45

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

Decision rule: W < 2df ==> not reject Ho

10.39


WW < 2df => not reject Ho

10.40


This plot suggests a variance is increasing proportional to X2i. The scattered plots spreading out as a linear patternlinear pattern

If the residuals plot against X2i are as following :

X2

i

+0-

^

X2

2

Therefore, we might expect i2 = Zi2 hi

2 = X2i => hi = X2i

=> Yi* = 1 X0* + 2 X1

* + 3 X2* + *

= 0 + 1 + 2 +Yi 1 X1i X2i i

X2i X2i X2i X2i X2i

The transformed equation is

10.41


Transformation: divided by the squared root term --“@sqrt(@sqrt(XX))”

= 0 + 1 + 2 +Yi 1 X1i X2i i

X2i X2i X2i X2i X2i

10.42


calculate: the C.V. = 0.36928

2(0.05, 5) = 11.07

2(0.10, 5) = 9.24

W < 2df => not reject Ho

Compare to the transformationdivided by the REG, the CV ofThat one is smaller.

10.43


Simple OLS result :R&D = 192.99 + 0.0319 Sales SEE = 2759 (0.194) (3.830) C.V. = 0.9026

Example: Gujarati (1995), Table 11.5, pp.388

10.44


White Test for heteroscedasticity

2(0.05, 2)= 5.9914

2(0.10, 2)= 4.60517

W=

10.45


Observe the shape pattern of residuals: linear or nonlinear?linear or nonlinear?

10.46


( ) = -246.67 + 0.036 Xi

Yi

Xi

1

Xi(-0.64) (5.17)

^

=>(1) R&Di = -246.67 + 0.036 Sales SEE = 7.25 ^(-0.64) (5.17) C.V. = 0.8195C.V. = 0.8195

1.

Transformation equations:

( ) = 0 + 1

Yi

Xi

1 Xi

Xi Xi

(2) R&D = -243.49 + 0.0369 Sales SEE = 0.021^

(-1.79) (5.52) C.V. = 0.7467C.V. = 0.7467

2.Compare the C.V.

To determine which weight is appropriated

1

ˆˆ,

ˆ..

22

knwhere

YVC

10.47


Transformation: divided by the squared root term --“@sqrt(@sqrt(XX))”

1

Xi

0 Xi+1

Yi

Xi

^=

10.48


1

Xi

0 Xi+1

Yi

Xi

^=

calculate: the C.V. = 0.8195

10.49


After transformation by @sqrt(x)@sqrt(x),, the W-statistic indicates there is no heteroscedasticity

2(0.05, 2)= 5.9914

2(0.10, 2)= 4.60517


10.50


After transformation by @sqrt(Xi), residuals still spread out

10.51


Transformation: divided by the suspected variable (Xi)

1

Xi

0 + 1Yi

Xi

^=

10.52


1

Xi

0 + 1Yi

Xi

^=

Calculate the C.V. = 0.7467

10.53


After transformation divided by the suspected X, the W-statistic indicates there is no heterose\cedasticity

2(0.05, 2)= 5.9914

2(0.10, 2)= 4.60517


10.54


After transformation divided by Xi, residuals spread out more stable

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