applying regression using e views (with commands)

7/30/2019 APPLYING REGRESSION USING E VIEWS (WITH COMMANDS)

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1 | E C O N O M E T R I C S 2 T E R M P A P E R

APPLIED ECONOMICS RESEARCH CENTRE

2013-14

ECONOMETRICS 2

TERM PAPER

SAFIA ASLAM


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CONTENTS

Chapter8

Multiple Regression Analysis: The Problem of Inference

Rest r i ct ed Least Squares: Testing Linear Equality RestrictionsWa ld Test

Testing For Structural or Parameter Stability of Regression Models

O The Chow Test

Computer Application using EViews

Chapter10

Multicollinearity: What Happens If The Regressors Are Correlated?

The Nature of Multicollinearity

Sources of Multicollinearity

Practical Consequences of Multicollinearity

Detection of Multicollinearity

O High R2 but

Few Significant T-Ratios

O High Pair-wise Correlations among Regressors

O Examination of Partial Correlations

O Auxiliary Regressions

O Tolerance and Variance Inflation Factor

O Remedial Measures

O Do Nothing Or

O Follow Some Rules Of Thumb.

O A Priori Information

O Combining Cross-Sectional and Time Series Data

O Dropping a Variable(S) and Specification Bias

O Transformation of Variables

O First Difference Form

O Ratio Transformation


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Ad di ti on al or New Data

O Computer Application using EViews

Chapter12

Autocorrelation: What Happens If the Error Terms Are Correlated?

The Nature of Autocorrelation

Sources of Autocorrelation

O Inertia.

O Specification Bias: Excluded Variables Case

O Specification Bias: Incorrect Functional Form

O Cobweb Phenomenon

O Lags.

O Manipulation Of Data

O Data TransformationPractical Consequences of Autocorrelation


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2

i

Detection of Autocorrelation Gr ap hi ca l Method

The Runs Test DurbinWatson D Test

A General Test of Autocorrelation: The BreuschGodfrey (Bg) Test Remedial Measures of Autocorrelation

NeweyWest Method. Genera l i zed Least-Square (G|LS) Method.

Correcting For (Pure) Autocorrelation: The Method of Generalized LeastSquares (GLS)

O When Is Known:

O When Is Not Known:The First-Difference Method:

O BerenbluttWebb Test,

Based On DurbinWatson D Statistic:

Estimated From The Residuals:

Theil-Nagar Estimate Based On D Statistic:

Estimating : The CochraneOrcutt (CO) Iterative

Procedure: Estimating : Durbins Two-Step Method:

The Durbin h Statistic


Chapter11

Heteroscedasticity: What Happens If the Error Variance Is Non constant?

The Nature of Heteroscedasticity

Sources of Heteroscedasticity

Error-Learning Models

Data Collecting Techniques Improve Is Likely To Decrease.

Outliers.

Other Sources of Heteroscedasticity:

O Incorrect Data Transformation (E.G., Ratio or First Difference

Transformations)

O Incorrect Functional Form (E.G., Linear Versus LogLinear Models).

Practical Consequences of Heteroscedasticity

Detection of Heteroscedasticity

Informal Methods


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Nature Of The Problem

Graphical Method

Formal Methods

O Park Test

O Glejser Test

O Goldfeld-Quandt Test

Whites General Heteroscedasticity Test Koenker Bassett (Kb) Test

Remedial Measures

1. When i2

is knownThe method of weighted least square

2. When i2is not known

Whites heteroscedasticity-Consistent Variances and Standard Errors

Plausible Assumptions about Heteroscedasticity Pattern.

Assumption 1: The Error Variance Is Proportional To Xi2

Assumption 2: The Error Variance Is Proportional ToXi. The Square Root

Transformation

Assumption 3: The Error Variance Is Proportional To the Square Of The

Mean Value OfY.

Assumption 4: A Log Transformation

3. Computer Application using EViews

Chapters18 To 20

Simultaneous Regression Models

1. The Nature Of Simultaneous-Equation Models2. The Identification Problem3. Rules For Identification

The Order Condition Of Identifiability

The Rank Condition Of Identifiability Hausman Specification Test

The Method Of Indirect Least Squares (ILS): A Just Identified Equation

The Method Of Two-Stage Least Squares (2SLS): An Over identified


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Equation

The Method Of Three-Stage Least Squares (3SLS) Using EViews

The Granger Test


Chapter9

Dummy Variable Regression Models

The Nature of Dummy Variable

Caution In the Use of Dummy VariablesAnalysis of Variance (ANOVA) Model

Analysis of Covariance (ANCOVA) Models.

Computer Application using EView


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7 | P a g e

CHAPTER 8

MULTIPLE REGRESSION ANALYSIS: THE PROBLEM OF

INFERENCE

RESTRICTED LEAST SQUARES: TESTING LINEAR EQUALITY RESTRICTIONS

WALD Test: Wald test is used to test the validity of the linear restriction imposed on the

parameters. There are occasions where economic theory may suggest that the coefficients in a

regression model satisfy some linear equality restrictions.

For instance, consider the CobbDouglas production function

Y=KLe

Where Yi = output, Li = labor input, and Ki = capital input.

Written in log form, the equation becomes

LnYi = 1 + 2 ln Ki + 3 ln Li + i (8.1)

Now if there are constant returns to scale (equi proportional change in output for an

equiproportional change in the inputs), economic theory would suggest that

2 + 3 = 1

This is an example of a linear equality restriction.

TESTING LINEAR RESTRICTIONS IN EVIEWS:

Step1: First Regress the Model (8.1) equation.

Quickestimate equationwrite: log(Y) C log(K) log(L)

Dependent Variable: LOG(Y)

Method: Least Squares

Date: 08/21/13 Time: 13:41


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Sample: 1960 2012

Included observations: 53

VariableCoefficien

t Std. Error t-Statistic Prob.

C 10.94926 0.420685 26.02720 0.0000

LOG(K) 3.734688 0.138653 26.93548 0.0000

LOG(L) -1.271933 0.214719 -5.923718 0.0000

R-squared 0.996592 Mean dependent var 18.36495

Adjusted R-squared 0.996456 S.D. dependent var 0.432456

S.E. of regression 0.025746 Akaike info criterion -4.426138

Sum squared resid 0.033143 Schwarz criterion -4.314612

Log likelihood 120.2927 F-statistic 7310.653

Durbin-Watson stat 0.054050 Prob(F-statistic) 0.000000

Step2: First Regress the Model (8.1) equationRegression window

ViewCoefficient TestsWald-Coefficients Restrictions: Write: C(2) + C(3) = 1

Wald Test:

Equation: Untitled

Test Statistic Value df Probability

F-statistic 319.2578 (1, 50) 0.0000

Chi-square 319.2578 1 0.0000

Null Hypothesis Summary:


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Normalized Restriction (= 0) Value Std. Err.

-1 + C(2) + C(3) 1.462755 0.081865

Restrictions are linear in coefficients.

WALD TEST MANUALLY (USING EVIEWS):

Unrestricted model: lnYi = 0 + 2 ln Ki + 3 ln Li + i

STEP 1:

First Regress the (8.1) equation.

Quickestimate equationwrite: log(Y) C log(K) log(L)

And Obtain RSSUR



Date: 08/21/13 Time: 13:41

Sample: 1960 2012


VariableCoefficien

t Std. Error t-Statistic Prob.

C 10.94926 0.420685 26.02720 0.0000

LOG(K) 3.734688 0.138653 26.93548 0.0000

LOG(L) -1.271933 0.214719 -5.923718 0.0000




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Restricted model: ln (Yi /Li) = 0 + 2 ln (Ki / Li) + i

STEP 2:

First Regress the (8.2) equation.

Quickestimate equationwrite: log(Y/L) C log(K/L)

And Obtain RSSR

Dependent Variable: LOG(Y/L)


Date: 08/21/13 Time: 13:48

Sample: 1960 2012


Variable Coefficient Std. Error t-Statistic Prob.

C 18.42422 0.119150 154.6300 0.0000

LOG(K/L) 5.936631 0.170981 34.72092 0.0000







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STEP 3:

Apply the F Test of RSS Version.

F = { (RSSRRSSUR) /M} / {RSSUR/ (N-K)}

And compare it with F critical Values at m and (n-k) degrees of freedom;

if | FCal | > | FCritical | then Restriction is invalid and vice versa

F = {0.244764 0.033143 / 1} / {0.033143 / (53-3)}

F = 319.69

The above example shows that F test is significant 1% level of significance.

TESTING FOR STRUCTURAL OR PARAMETER STABILITY OF REGRESSIONMODELS: THE CHOW TEST

When we use a regression model involving time series data, it may happen that there is a

structural change in the relationship between the regressand Y and the regressors. By structural

change, we mean that the values of the parameters of the model do not remain the same

through the entire time period.

USING E VIEWS:

Step 1: First Regress the Model (8.3) equation (with n = 34);

Quickestimate equationwrite: Sav C Yd

Dependent Variable: SAV



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Date: 08/21/13 Time: 14:08

Sample: 1960 2012



C -15.73656 1.331322 -11.82025 0.0000

YD 0.771388 0.022667 34.03118 0.0000



S.E. of regression 0.880589 Akaike info criterion 2.620554

Sum squared resid 39.54728 Schwarz criterion 2.694904

Log likelihood -67.44468 F-statistic 1158.121


Step 2: First Regress the Model (8.3) equation. Regression windowViewStability TestsChow

Breakpoint Test:

Write Breakpoint Year in the Box: 1980

Chow Breakpoint Test: 1980

F-statistic 303.2648 Probability 0.000000

Log likelihood ratio 137.4620 Probability 0.000000

The above example shows that F test is significant 1% level of significance


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MODEL

DETERMINANTS OF LIFE EXPECTANCY IN PAKISTAN:

Y=B1+B2X1+B3X2++B4X3+B5X4+B6X5

X1: POPULATION

X2: GDP

X3: UNEMPLOYEMENT

X4: URBAN POPULATION

X5: HEALTH EXPENDITURE

REGRESSION RESULTS:

Dependent Variable: Y


Date: 08/19/13 Time: 13:57

Sample: 1960 2012



C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150




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CHAPTER 10

MULTICOLLINEARITY: WHAT HAPPENS IF THE

REGRESSORS ARE CORRELATED?

THE NATURE OF MULTICOLLINEARITY

The term multicollinearity is due to Ragnar Frisch. Originally it meant the existence of a

perfect, or exact, linear relationship among some or all explanatory variables of aregression model.

X1 + X2 . + Xk = 0 (10.1.1)

X1 + X2 . + Xk + i = 0 (10.1.2)

Why does the classical linear regression model assume that there is no multicollinearity among

the Xs? The reasoning is this: If multicollinearity is perfect in the sense of (10.1.1), the

regression coefficients of the X variables are indeterminate and their standard errors are

infinite. If multicollinearity is less than perfect, as in (10.1.2), the regression coefficients,

although determinate, possess large standard errors (in relation to the coefficients themselves),

which means the coefficients cannot be estimated with great precision or accuracy.


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SOURCES OF MULTICOLLINEARITY

There are several sources of multicollinearity. As Montgomery and Peck note, multicollinearity

may be due to the following factors;

1. The data collection method employed, for example, sampling over a limited range of the valuestaken by the regressors in the population

2. Constraints on the model or in the population being sampled. For example, in the regressionof electricity consumption on income (X2) and house size (X3) there is a physical constraint in

the population in that families with higher incomes generally have larger homes than families

with lower incomes.

3. Model specification, for example, is adding polynomial terms to a regression model, especiallywhen the range of the X variable is small.

4. An over-determined model. This happens when the model has more explanatory variables thanthe number of observations. This could happen in medical research where there may be a smallnumber of patients about whom information is collected on a large number of variables.

An additional reason for multicollinearity, especially in time series data, may be that theregressors included in the model share a common trend, that is, they all increase or decrease

over time.

PRACTICAL CONSEQUENCES OF MULTICOLLINEARITY

In cases of near or high multicollinearity, one is likely to encounter the following consequences:

a) Although BLUE, the OLS estimators have large variances and co variances, making precise estimationdifficult.

b) Because of consequence 1, the confidence intervals tend to be much wider, leading to the acceptance of thezero null hypothesis (i.e., the true population coefficient is zero) more readily.

c) Also because of consequence 1, the t-ratio of one or more coefficients tends to be statistically insignificant.d) Although the t-ratio of one or more coefficients is statistically insignificant, R2, the overall measure of

goodness of fit, can be very high.

e) The OLS estimators and their standard errors can be sensitive to small changes in the data.

DOING REGREESION USING EVIEWS FOR MULTICOLLINEARITY

DETECTION METHODS:


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1. HIGH R2 BUT FEW SIGNIFICANT t- RATIO:

MODEL: Yi = 1 + 2 X1i + 3 X2i + 4 X3i +5X4i + 6X5i +i

STEP:

First regress the above equation.

Open the File containing dataQuickestimate equationwrite: Y C X1 X2 X3 X4 X5

And check the R2 and t-ratios.



Date: 08/19/13 Time: 13:57

Sample: 1960 2012



C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150







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The above regression results shows that R2

= 0.997288(high).X1, X2, X3, X4 are significant while X5 is insignificant.

There may be multicollinearity here.

1. HIGH PAIR WISE CORRELATION AMONG REGRESSORS:

STEP:

First open the File containing data QuickGroup StatisticsCorrelation

Write: X1 X2 X3 X4 X5

X1 X2 X3 X4 X5

X1 1.000000 0.898710 0.900580 0.988885 -0.071599

X2 0.898710 1.000000 0.713399 0.863979 -0.222654

X3 0.900580 0.713399 1.000000 0.902906 -0.159905

X4 0.988885 0.863979 0.902906 1.000000 -0.053507

X5 -0.071599 -0.222654 -0.159905 -0.053507 1.000000

The above correlation matrix shows that X1 and X4 are highly correlated.

2. EXAMINATION OF PARTIAL CORRELATION:

STEP:

First regress the equations. Open the File containing data Quickestimate

equationwrite: Y C X1, Y C X2, Y C X3, Y C X4, Y C X5 individually and compare their r2 with R2 of overall

regression.

REGRESS Y ON X1:


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Date: 08/19/13 Time: 14:02

Sample: 1960 2012



C 46.13325 0.681312 67.71239 0.0000

X1 1.20E-07 6.11E-09 19.57344 0.0000


R2

0.997288

REGRESS Y ON X2:



Date: 08/19/13 Time: 14:03

Sample: 1960 2012



C 54.61037 0.659574 82.79642 0.0000

X2 7.42E-11 8.68E-12 8.554942 0.0000


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R2 0.997288

REGRESS Y ON X3:



Date: 08/19/13 Time: 14:03

Sample: 1960 2012



C 47.82694 0.889198 53.78662 0.0000

X3 3.221181 0.245909 13.09909 0.0000


R2

0.997288

REGRESS Y ON X4:



Date: 08/19/13 Time: 14:03

Sample: 1960 2012




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C 22.00695 1.082998 20.32039 0.0000

X4 1.241685 0.036487 34.03118 0.0000


R2

0.997288

REGRESS Y ON X5:



Date: 08/19/13 Time: 14:04

Sample: 1960 2012



C 58.94157 5.968745 9.875036 0.0000

X5 -0.146728 1.927387 -0.076128 0.9396


R2

0.997288

The above regression shows data had no multicollinearity

3. AUXILLARY REGRESSIONS:

STEP:

First regress the equations. Open the File containing data Quickestimate equationwrite: X2 C X3 X4,

write: X3 C X2 X4, write: X4 C X2 X3 individually and compare the r2 with R2.


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REGRESS X1 ON X2 X3 X4 X5:

Dependent Variable: X1


Date: 08/19/13 Time: 14:06

Sample: 1960 2012



C -1.22E+08 11339153 -10.78366 0.0000

X2 0.000186 2.70E-05 6.900419 0.0000

X3 4481648. 1162156. 3.856322 0.0003

X4 6366302. 583130.7 10.91745 0.0000

X5 4542256. 1906829. 2.382100 0.0212

R-squared 0.989170 Mean dependent var 1.03E+08

R2

0.997288




Date: 08/19/13 Time: 14:07

Sample: 1960 2012




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C 1.84E+11 7.50E+10 2.451252 0.0179

X1 2677.158 387.9703 6.900419 0.0000

X3 -2.16E+10 3.97E+09 -5.426820 0.0000

X4 -8.45E+09 3.94E+09 -2.141991 0.0373

X5 -2.88E+10 6.42E+09 -4.489499 0.0000


R2

0.997288




Date: 08/19/13 Time: 14:07

Sample: 1960 2012



C 1.445271 2.266943 0.637542 0.5268

X1 5.28E-08 1.37E-08 3.856322 0.0003

X2 -1.76E-11 3.25E-12 -5.426820 0.0000

X4 -0.011486 0.118091 -0.097265 0.9229

X5 -0.757352 0.189557 -3.995372 0.0002



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R2

0.997288




Date: 08/19/13 Time: 14:08

Sample: 1960 2012



C 18.57003 0.745923 24.89536 0.0000

X1 1.12E-07 1.03E-08 10.91745 0.0000

X2 -1.03E-11 4.82E-12 -2.141991 0.0373

X3 -0.017156 0.176382 -0.097265 0.9229

X5 -0.051514 0.267322 -0.192705 0.8480


R2

0.997288




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Date: 08/19/13 Time: 14:08

Sample: 1960 2012



C 2.736547 1.448764 1.888884 0.0650

X1 2.33E-08 9.77E-09 2.382100 0.0212

X2 -1.03E-11 2.29E-12 -4.489499 0.0000

X3 -0.329525 0.082477 -3.995372 0.0002

X4 -0.015007 0.077873 -0.192705 0.8480


R2

0.997288

The above regression shows data had no multicollinearity.

REMEDIAL MEASURES:

A. DROPPING A VARIABLE(S) AND SPECIFICATION BIAS :

STEP:

First Regress the (10.1) equation without X1 (by assuming it causes Multicollinearity).

Open the File containing dataQuickestimate equationwrite: Y C X2 X4

And check the R2 and t-ratios



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Date: 08/19/13 Time: 14:12

Sample: 1960 2012



C 9.097915 1.536073 5.922841 0.0000

X2 -3.84E-11 3.65E-12 -10.50413 0.0000

X3 -0.751619 0.157433 -4.774216 0.0000

X4 1.907816 0.078995 24.15124 0.0000

X5 -0.706191 0.258311 -2.733880 0.0087







After dropping X1, regression shows that multicollinearity has been removed.

B. TRANSFORMATION OF VARIABLE:

I m transforming model into lag form.

Dependent Variable: D(Y)


Date: 08/19/13 Time: 14:18


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Sample (adjusted): 1961 2012

Included observations: 52 after adjustments


C 1.080539 0.091881 11.76017 0.0000

D(X1) -2.20E-07 1.15E-08 -19.15005 0.0000

D(X2) -4.13E-12 1.19E-12 -3.461294 0.0012

D(X3) 0.002617 0.020385 0.128386 0.8984

D(X4) -0.482934 0.278996 -1.730969 0.0902

D(X5) 0.009456 0.021666 0.436466 0.6645







C. ADDITION OR NEW DATA.


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CHAPTER

12

AUTOCORRELATION: WHAT HAPPENS IFTHE ERROR

TERMS ARE CORRELATED?

THE NATURE OF AUTOCORRELATION

The term autocorrelation may be defined as correlation between members of series ofobservations ordered in time [as in time series data] or space [as in cross-sectional data].In theregression context, the classical linear regression model assumes that such autocorrelation

does not exist in the disturbances Ui.

Tintner defines autocorrelation as lag correlation of a given series with itself, lagged by anumber of time units, whereas he reserves the term serial correlation to lag correlationbetween two different series.

SOURCES OF AUTOCORRELATION

1. Inertia. A salient feature of most economic time series is inertia, or sluggishness. Asis well known, time series such as GNP, price indexes, production, employment, and

unemployment exhibit (business) cycles.

2. Specification Bias: Excluded Variables Case3. Specification Bias: Incorrect Functional Form4. Cobweb Phenomenon

5. Lags.6. Manipulation of Data: Another source of manipulation is interpolation or

extrapolation of data.

7. Data Transformation


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PRACTICAL CONSEQUENCES OF AUTOCORRELATION

In the presence of autocorrelation the usual OLS estimators, although linear, unbiased, and

asymptotically (i.e., in large samples) normally distributed, are no longer minimum varianceamong all linear unbiased estimators. In short, they are not efficient relative to other linear and

unbiased estimators. Put differently, they may not be BLUE. As a result, the usual, t, F, and may not be valid.

DOING REGREESION USING EVIEWS FOR AUTOCORRELATION

DETECTION METHOD:

1. GRAPHICAL METHOD:

Step 1: Regress the Model

Open the File containing dataQuickestimate equation

Write:

X C Y

Step 2: obtain the Residuals

From the estimated equation result window Procmake residuals series(name) ok.

Step 3: Plot these residuals and see the pattern for Autocorrelation.

From step2QuickGraphScatter plot

Write: r1(-1) r1


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It is clear from the Graph that the residuals Rt and Rt-1 are serially correlated and positively correlated

2. RUNS TEST:




Write:

Y C X1 X2 X3 X4 X5


From the estimated equation result window Procmake residuals series (name) ok.

Step 3: check the Runs by looking at changing sign of residuals, then find the interval by calculating mean variance

the check whether Runs obtained lies in interval or not for clarity regarding Autocorrelation

(- - - - - - -, + + + + + + + + + + , - - - - - - - - - - - - - -,+, - - -, + + + + + + + + + +, - - - - -, + + )

R=8

N1=23

N2=29

MEAN(R) = 2N1N2 /N +1

= 1334/52+1

= 25.653

VARIANCE () = 2N1N2 (2N1N2N) / N2

(N1)

= 1334*1282/2704*51


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= 1710188/137904

= 12.4012

If the null hypothesis of randomness is sustainable, following the properties of the normal

distribution, we should expect that

PROB [E(R)-1.96 R E(R) +1.96]

PROB [25.653-1.96(12.4012) 8 25.653+1.96(12.4012)]

PROB [1.3466 8 49.959]

Obviously, this interval includes 8. So we can accept the null hypothesis that residuals do not containautocorrelation.

3. DURBIN WATSON d TEST:




Write:

Y C X2X3 X4 X5

In the regression results there is Durbin-Watson dstatistics



Date: 08/19/13 Time: 13:57

Sample: 1960 2012




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C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150







4. A GENERAL TEST OF AUTOCORRELATION:THE BREUSCHGODFREY (BG) TEST



Open the File containing dataQuickestimate equationequation

Specification window

Write:

Y C X2X3 X4 X5


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Step 2: from step 1, in estimated equation windowViewResidual Tests

Serial Correlation LM Test

Breusch-Godfrey Serial Correlation LM Test:


Obs*R-squared 42.88204 Probability 0.000000

Test Equation:

Dependent Variable: RESID


Date: 08/19/13 Time: 14:44

Pre sample missing value lagged residuals set to zero.


C -0.511900 0.658397 -0.777494 0.4413

X1 -5.69E-09 4.60E-09 -1.238405 0.2226

X2 1.71E-12 1.19E-12 1.433777 0.1592

X3 0.035050 0.040369 0.868241 0.3903

X4 0.027397 0.034699 0.789556 0.4343

X5 0.030173 0.062743 0.480893 0.6331

RESID(-1) 0.772369 0.154615 4.995436 0.0000

RESID(-2) 0.109866 0.201820 0.544377 0.5891

RESID(-3) 0.067084 0.201378 0.333126 0.7407

RESID(-4) -0.015117 0.211764 -0.071384 0.9434

RESID(-5) 0.029055 0.218522 0.132961 0.8949

RESID(-6) -0.316267 0.171752 -1.841416 0.0728

R-squared 0.809095 Mean dependent var -2.50E-15


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REMEDIAL MEASURES:

1. NEWEYWEST CONSISTENT STANDARD ERRORS METHOD:


Step: Regress the Model

Open the File containing dataQuickestimate equationequation

Specification window

Write:

Y C X2X3

In equation specification windowoptionsestimation option windowtick

Heteroscedasticity-consistent covariance NeweyWestclick ok



Date: 08/21/13 Time: 11:22

Sample: 1960 2012


Newey-West HAC Standard Errors & Covariance (lag truncation=3)



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C -5.519635 2.390986 -2.308518 0.0254

X1 -1.20E-07 1.43E-08 -8.372859 0.0000

X2 -1.61E-11 2.39E-12 -6.749603 0.0000

X3 -0.215865 0.081656 -2.643598 0.0111

X4 2.668870 0.133068 20.05638 0.0000

X5 -0.163192 0.136790 -1.193012 0.2389







Only Std. Error of coefficients are corrected for autocorrelation

2. GENERALIZED LEAST SQUARES (GLS) METHOD:

When is known: We use that rho to estimate the generalized difference equation through OLSthen these results will be reliable and dont have autocorrelation problem.

When is Unknown:

based on DurbinWatson d Statistic:

1 d/2




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Write:

Y C X1 X2X3 X4 X5

And take Durbin Watson statistics and calculate rho (= p).



Date: 08/19/13 Time: 13:57

Sample: 1960 2012



C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150







1 0.196325/2

0.9018


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Step 2: Regress the equation


Write: Y- 0.9018*d(Y) 1-0.9018 X1-0.9018*d(X1) X5-0.9018*(X5)

Dependent Variable: Y-0.9018*D(Y)


Date: 08/21/13 Time: 11:30




1-0.9018 -54.11289 13.94341 -3.880893 0.0003

X1-0.9018*D(X1) -1.17E-07 9.85E-09 -11.87465 0.0000

X2-0.9018*D(X2) -1.71E-11 2.88E-12 -5.928562 0.0000

X3-0.9018*D(X3) -0.235757 0.089920 -2.621862 0.0118

X4-0.9018*D(X4) 2.659116 0.071015 37.44442 0.0000

X5-0.9018*D(X5) -0.185270 0.137451 -1.347904 0.1843





Log likelihood -6.909378 Durbin-Watson stat 0.187082


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estimated from the Residuals:

t= t-1 + t



Write:

Y C X1 X2X3 X4 X5

Obtain the Residuals, from the estimated equation result window Procmake

Residuals series (name) ok

Step 2: Regress the residual on it lag to get rho to transform model in GDE

Write:

R1R1 (-1)

Dependent Variable: R1


Date: 08/21/13 Time: 11:32




D(R1) 0.356889 0.311125 1.147091 0.2568








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Write: Y- 0.3568*d(Y) 1-0.3568 X1-0.3568*d(X1) X5-0.3568*(X5)

Dependent Variable: Y-.3568*D(Y)


Date: 08/21/13 Time: 11:35




1-.3568 -7.471330 2.116689 -3.529725 0.0010

X1-.3568*D(X1) -1.14E-07 9.48E-09 -12.06953 0.0000

X2-.3568*D(X2) -1.71E-11 2.64E-12 -6.461472 0.0000

X3-.3568*D(X3) -0.247995 0.090371 -2.744203 0.0086

X4-.3568*D(X4) 2.635871 0.070181 37.55831 0.0000

X5-.3568*D(X5) -0.198586 0.146473 -1.355786 0.1818






from Durbin two steps method:

Step 1: Regress the equation given suggested by Durbin for his two step methods



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write: Y C X1 d(X1) X2 d(X2) X3 d(X3) X4 d(X4) X5 d(X5) d(Y)

The coefficient of Yt-1 is estimated rho (=P)



Date: 08/21/13 Time: 11:39




C -17.98627 3.065724 -5.866892 0.0000

X1 -1.47E-07 1.22E-08 -12.00310 0.0000

D(X1) -3.78E-07 2.41E-07 -1.570401 0.1242

X2 -2.07E-11 2.75E-12 -7.517914 0.0000

D(X2) 4.08E-12 8.47E-12 0.481797 0.6326

X3 -0.213495 0.091318 -2.337915 0.0245

D(X3) 0.176647 0.106567 1.657618 0.1052

X4 3.177549 0.142231 22.34072 0.0000

D(X4) -0.389698 1.727353 -0.225604 0.8227

X5 -0.037026 0.130887 -0.282886 0.7787

D(X5) 0.074601 0.113507 0.657232 0.5148

D(Y) 3.333366 1.153424 2.889976 0.0062








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= 3.3333

Step 2: Regress the equation (12.4)


Write: Y- 3.3333*d(Y) 1-3.333 X1-3.3333*d(X1) X5-3.3333*(X5)

Dependent Variable: Y-3.333*D(Y)


Date: 08/21/13 Time: 11:43




1-3.333 3.158922 0.687286 4.596224 0.0000

X1-3.333*D(X1) -1.40E-07 1.09E-08 -12.87266 0.0000

X2-3.333*D(X2) -9.24E-12 3.84E-12 -2.405610 0.0202

X3-3.333*D(X3) -0.060601 0.055325 -1.095376 0.2791

X4-3.333*D(X4) 2.759547 0.086447 31.92176 0.0000

X5-3.333*D(X5) -0.032140 0.061097 -0.526048 0.6014







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BerenbluttWebb test g-statistics for estimated =1:

To test the hypothesis that 1. The test statistic they use is called the g-statistic, which

is defined as follows:

g = et2 / t

2



Write: Y C X1 X2 X3 X4 X5

Obtain residuals, t2

Step 2: Regress the equation First difference


Write: D(Y) C D(X10 D(X2) D (X3) D(X4) D(X5)

Obtain residuals, et2



Date: 08/21/13 Time: 16:27

Sample: 1960 2012



C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152


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Dependent Variable: D(Y)


Date: 08/21/13 Time: 16:29




C 1.080539 0.091881 11.76017 0.0000

D(X1) -2.20E-07 1.15E-08 -19.15005 0.0000

D(X2) -4.13E-12 1.19E-12 -3.461294 0.0012

D(X3) 0.002617 0.020385 0.128386 0.8984

D(X4) -0.482934 0.278996 -1.730969 0.0902

D(X5) 0.009456 0.021666 0.436466 0.6645





g = 0.0311

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150






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We find that dL=1.364 and dU =1.590 (5 percent level)

It is clear that g statistics lies in the 0dL range we can take first difference by assuming =1.

CHAPTER 11

HETEROSCEDASTICITY: WHAT HAPPENS IF THE ERROR

VARIANCE IS NONCONSTANT?

THE NATURE OF HETEROSCEDASTICITY

A critical assumption of the classical linear regression model is that the disturbances i have all

the same variance, 2 if this assumption is not satisfied, there is heteroscedasticity. Hence,there is heteroscedasticity.

E i = i2

Notice the subscript ofi2,

which reminds us that the conditional variances

of i(= conditional variances ofYi) are no longer constant

SOURCES OF HETEROSCEDASTICITY


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1. Following the error-learning models, as people learn, their errors of behavior

become smaller over time. In this case, i2is expected to decrease.

2. As data collecting techniques improve, i2

is likely to decrease.

3. Heteroscedasticity can also arise as a result of the presence ofoutliers. An outlying

observation, or outlier, is an observation that is much different (either very small or

very large) in relation to the observations in the sample.

4. Another source of heteroscedasticity arises from violating Assumption 9 of CLRM,

namely, that the regression model is correctly specified.

5. Another source of heteroscedasticity is skewness in the distribution of one or more

regressors included in the model.

6. Other sources of heteroscedasticity: As David Hendry notes, heteroscedasticity can

also arise because of (1) incorrect data transformation (e.g., ratio or first difference

transformations) and (2) incorrect functional form (e.g., linear versus loglinear models).

PRACTICAL CONSEQUENCES OF HETEROSCEDASTICITY

In the presence of heteroscedasticity the usual OLS estimators, although linear, unbiased, and

asymptotically (i.e., in large samples) normally distributed, are no longer minimum variance

among all linear unbiased estimators. In short, they are not efficient relative to other linear and

unbiased estimators. Put differently, they may not be BLUE. As a result, the usual, t, F, and may not be valid.

DOING REGREESION USING EVIEWS FOR HETEROSCEDASTICITY


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DETECTION OF HETEROSCEDASTICITY:

1. GRAPHICAL METHOD:




Write: Y C X1 X2 X3 X4 X5


From the estimated equation result window Procmake residuals series(name) ok.

Step 3: Plot these residuals squares and X1, X2, X3, X4, X5 & individually and see the pattern for

Hetroscedasticity. Y

From step2QuickGraphScatter plot

Write:

X1 R1^2, X2 R1^2, X3 R1^2, X4 R1^2, X5 R1^2


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It is clear from the graph that there is such pattern seen that causes hetroscedasticity.

2. FORMAL METHODS:

a. PARK TEST:


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Write:

Y C X1X2

And obtain the Residuals from the estimated equation result window Procmake

Residuals series (name) ok.



Date: 08/19/13 Time: 13:57

Sample: 1960 2012



C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150






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Lni =1++ 2lnXi+i

Step 2: Regress the Equation (11.1) suggested by Park.


Write:

log(r1^2) c log(x1)

log(r1^2) c log(x2)

log(r1^2) c log(x3)

log(r1^2) c log(x4)

log(r1^2) c log(x5)

And check the significance of coefficient of explanatory variable

Dependent Variable: LOG(R1^2)


Date: 08/21/13 Time: 12:07




C 39.65786 10.25252 3.868109 0.0003

LOG(X1) -2.361010 0.557693 -4.233531 0.0001


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Date: 08/21/13 Time: 12:07




C 17.17790 5.018112 3.423181 0.0012

LOG(X2) -0.866555 0.207702 -4.172101 0.0001



Date: 08/21/13 Time: 12:07




C -1.586677 0.639137 -2.482532 0.0164

LOG(X3) -1.935249 0.532505 -3.634238 0.0007



Date: 08/21/13 Time: 12:08




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C 19.37869 5.598606 3.461342 0.0011

LOG(X4) -6.848214 1.657294 -4.132166 0.0001



Date: 08/21/13 Time: 12:08




C -5.142800 2.376507 -2.164016 0.0353

LOG(X5) 1.264550 2.120873 0.596240 0.5537

If2s turns out to be statistically significant at 1%, it would suggest that heteroscedasticity is present

in the data in the each case.

b. GLESJER TEST:

MODEL: Yi = 1 + 2X1i + 3 X2i + 4 X3i +5X4i + 6X5i +i




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Write:

Y C X1X2



= 1+2+Xi+i

Step 2: Regress the Equation (11.1a) suggested by Glesjer.


Write:

abs(r1) c x1

abs(r1) c x2

abs(r1) c x3

abs(r1) c x4

abs(r1) c x5


Dependent Variable: ABS(R1)


Date: 08/21/13 Time: 12:16




C 0.462368 0.051704 8.942531 0.0000

X1 -2.31E-09 4.60E-10 -5.020498 0.0000



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Date: 08/21/13 Time: 12:16




C 0.290397 0.029296 9.912631 0.0000

X2 -1.30E-12 3.82E-13 -3.408431 0.0013



Date: 08/21/13 Time: 12:16




C 0.423221 0.050627 8.359624 0.0000

X3 -0.060295 0.013883 -4.343235 0.0001



Date: 08/21/13 Time: 12:17




C 0.914931 0.137492 6.654407 0.0000


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X4 -0.023500 0.004612 -5.094806 0.0000



Date: 08/21/13 Time: 12:17




C 0.123370 0.185112 0.666459 0.5082

X5 0.031866 0.059835 0.532554 0.5967

If2s turns out to be statistically significant at 1%, it would suggest that heteroscedasticity is presentin the data in each case.

= 1+2+Xi+ i

Step 2: Regress the Equation (11.1b) suggested by Glesjer.


Write:

abs(r1) c x1^0.5

abs(r1) c x2^0.5

abs(r1) c x3^0.5

abs(r1) c x4^0.5


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abs(r1) c x5^0.5




Date: 08/21/13 Time: 12:18




C 0.697759 0.093902 7.430750 0.0000

X1^.5 -4.76E-05 9.19E-06 -5.181553 0.0000



Date: 08/21/13 Time: 12:18




C 0.383647 0.041604 9.221475 0.0000

X2^.5 -8.01E-07 1.80E-07 -4.443441 0.0000



Date: 08/21/13 Time: 12:18



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C 0.633219 0.090155 7.023649 0.0000

X3^.5 -0.230573 0.049249 -4.681793 0.0000



Date: 08/21/13 Time: 12:18




C 1.599442 0.269902 5.926016 0.0000

X4^.5 -0.254295 0.049674 -5.119270 0.0000



Date: 08/21/13 Time: 12:19




C 0.004676 0.366416 0.012761 0.9899

X5^.5 0.123829 0.209160 0.592029 0.5565


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in the data in each case.

= 1+2+1/Xi+ i

Step 2: Regress the Equation (11.1c) suggested by Glesjer.


Write:

abs(r1) c 1/x1

abs(r1) c 1/x2

abs(r1) c 1/x3

abs(r1) c 1/x4

abs(r1) c 1/x5




Date: 08/21/13 Time: 12:20




C -0.010629 0.048537 -0.218995 0.8275

1/X1 20352326 3925383. 5.184800 0.0000




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Date: 08/21/13 Time: 12:20




C 0.135861 0.026436 5.139347 0.0000

1/X2 1.42E+09 2.98E+08 4.761296 0.0000



Date: 08/21/13 Time: 12:20




C 0.009848 0.044574 0.220930 0.8260

1/X3 0.577436 0.110386 5.231046 0.0000



Date: 08/21/13 Time: 12:20




C -0.451516 0.132770 -3.400733 0.0013


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1/X4 19.46190 3.801832 5.119086 0.0000



Date: 08/21/13 Time: 12:21




C 0.357066 0.180871 1.974150 0.0539

1/X5 -0.410269 0.541472 -0.757693 0.4522

If2s turns out to be statistically significant at 1%, it would suggest that heteroscedasticity is presentin the data in each case.

= 1+2+1/Xi+ i

Step 2: Regress the Equation (11.1d) suggested by Glesjer.


Write:

abs(r1) c 1/x1^0.5

abs(r1) c 1/x2^0.5

abs(r1) c 1/x3^0.5

abs(r1) c 1/x4^0.5

abs(r1) c 1/x5^0.5



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Date: 08/21/13 Time: 12:22




C -0.245656 0.090575 -2.712173 0.0091

1/X1^0.5 4471.569 848.7235 5.268582 0.0000



Date: 08/21/13 Time: 12:22




C 0.047878 0.037971 1.260883 0.2132

1/X2^0.5 25715.18 4894.805 5.253565 0.0000



Date: 08/21/13 Time: 12:22


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C -0.199286 0.083843 -2.376886 0.0213

1/X3^0.5 0.713478 0.138597 5.147859 0.0000



Date: 08/21/13 Time: 12:22




C -1.135887 0.265139 -4.284128 0.0001

1/X4^0.5 7.318108 1.426138 5.131418 0.0000



Date: 08/22/13 Time: 13:12




C 0.475984 0.362154 1.314313 0.1947

1/X5^0.5 -0.443667 0.629237 -0.705088 0.4840


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in the data in each case

c. WHITES GENERAL HETEROSCEDASTICITY TEST:




Write:

Y C X1X2X3 X4 X5

Step 2: applying Whites General Hetroscedasticity Test

In estimated equation in step1viewResiduals Tests

White Heteroskedasticity Test (cross terms)

or

White Heteroskedasticity Test (no cross term)

White Heteroskedasticity Test: no cross term



As n*R2 (=31.13871) > the critical chi-square (=3.94) value at the 5% level of significance, the

conclusion is that there is heteroscedasticity


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White Heteroskedasticity Test: cross term



As n*R2 (=37.24425) > the critical chi-square (=10.85) value at the 5% level of significance, theconclusion is that there is heteroscedasticity

d. KOENKERBASSETT (KB) TEST:




Write:

Y C X1X2X3 X4 X5

I2= 1+2I

2+i

Step 2: Regress the (11.6a) equation


Write: in

R1^2 C (Y_CAP)^2

Where I are the estimated values from the model (11.6). The null hypothesis is that 2= 0. If this is not rejected,then one could conclude that there is no heteroscedasticity.

The null hypothesis can be tested by the usual ttest or theFtest,


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Dependent Variable: R1^2


Date: 08/21/13 Time: 12:44




C 0.414590 0.072768 5.697453 0.0000

(Y_CAP)^2 -9.74E-05 2.07E-05 -4.714133 0.0000







REMEDIAL MEASURES:

WHEN i2

IS KNOWN: THE METHOD OF WEIGHTED LEAST SQUARES:

Ifi2

is known, the most straightforward method of correcting heteroscedasticity is by

Means of weighted least squares, for the estimators thus obtained are BLUEYi = 1 + 2 X1i + 3 X2i + 4 X3i +5X4i + 6X5i +i

Yi/ = 1 (1/)+ 2 X1i /+ 3 X2i/+ 4 X3i/ +/+6X5i/+i WEIGHTED LEAST SQUARE

Where iare the standard deviations


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WHEN i2

IS NOT KNOWN: Whites Heteroscedasticity-Consistent Variances and Standard Errors.




Write:

Y C X1X2X3 X4 X5



Date: 08/19/13 Time: 13:57

Sample: 1960 2012



C -5.519635 1.350874 -4.085972 0.0002

X1 -1.20E-07 9.29E-09 -12.86161 0.0000

X2 -1.61E-11 2.45E-12 -6.580235 0.0000

X3 -0.215865 0.085649 -2.520339 0.0152

X4 2.668870 0.070082 38.08228 0.0000

X5 -0.163192 0.129846 -1.256813 0.2150


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Write:

Y C X1X2X3 X4 X5

In equation specification windowoptionsestimation option windowtick

Heteroscedasticity-consistent covariance Whitesclick ok



Date: 08/22/13 Time: 14:14

Sample: 1960 2012


White Heteroskedasticity-Consistent Standard Errors & Covariance


C -5.519635 1.416856 -3.895694 0.0003

X1 -1.20E-07 8.29E-09 -14.42540 0.0000

X2 -1.61E-11 1.64E-12 -9.811624 0.0000

X3 -0.215865 0.056197 -3.841198 0.0004

X4 2.668870 0.078266 34.09986 0.0000

X5 -0.163192 0.106943 -1.525967 0.1337

As we can see that in case of heteroscedasticity the OLS standard errors of slope coefficient

of X1 X2 X3 are over estimated and X4 X5 are under estimated. And intercept was under

estimated. After applying Whites Hetroscedasticity-Consistent Variances and Standard Errorsremedial procedure they were now corrected for hetroscedasticity.

Plausible Assumptions about Heteroscedasticity Pattern :


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Assumption 1: The error variance is proportional to XI2:

E(i2)=2Xi

2

We may transform by dividing the original model through byXi

Transforming using X1:

Yi = 1(1/X1) + 2 + 3 X2i/X1 + 4 X3i/X1 + +6X5i/X1 +i TRANSFORMED MODEL

Step 1: Regress the equation (11.7a)

Open the File containing dataQuickestimate equationequation specification window

Write: Y/X1 C 1/X1 X2/X1 X3/X1 X4/X1 X5/X1

Dependent Variable: Y/X1


Date: 08/22/13 Time: 14:31

Sample: 1960 2012




C -1.35E-07 1.30E-08 -10.36097 0.0000

1/X1 -8.939349 1.641135 -5.447055 0.0000

X2/X1 -1.65E-11 3.01E-12 -5.480641 0.0000

X3/X1 -0.360089 0.097640 -3.687939 0.0006

X4/X1 2.859217 0.091651 31.19696 0.0000

X5/X1 -0.206358 0.123891 -1.665644 0.1024

R-squared 0.999641 Mean dependent var 6.51E-07


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Adjusted R-squared 0.999603 S.D. dependent var 2.18E-07

S.E. of regression 4.34E-09 Akaike info criterion -35.56657

Sum squared resid 8.85E-16 Schwarz criterion -35.34352



Similarly for X2 X3 X4 and X5:



Date: 08/22/13 Time: 14:33

Sample: 1960 2012




C -1.60E-11 1.20E-11 -1.337315 0.1876

X1/X2 -1.80E-07 2.77E-08 -6.486804 0.0000

1/X2 -15.75433 2.283040 -6.900594 0.0000

X3/X2 -0.400503 0.187234 -2.139050 0.0377

X4/X2 3.241575 0.140842 23.01573 0.0000

X5/X2 -0.141550 0.103679 -1.365271 0.1787

R-squared 0.999966 Mean dependent var 3.39E-09

Adjusted R-squared 0.999962 S.D. dependent var 3.44E-09





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Date: 08/22/13 Time: 14:34

Sample: 1960 2012




C -0.308845 0.117205 -2.635089 0.0114

X1/X3 -1.49E-07 1.46E-08 -10.18503 0.0000

X2/X3 -1.40E-11 3.08E-12 -4.564873 0.0000

1/X3 -10.39333 1.716490 -6.054987 0.0000

X4/X3 2.948408 0.095104 31.00182 0.0000

X5/X3 -0.215436 0.120129 -1.793374 0.0793








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Date: 08/22/13 Time: 14:35

Sample: 1960 2012




C 2.734611 0.080073 34.15149 0.0000

X1/X4 -1.25E-07 9.49E-09 -13.13093 0.0000

X2/X4 -1.64E-11 1.99E-12 -8.252539 0.0000

X3/X4 -0.255739 0.068636 -3.726028 0.0005

1/X4 -6.702676 1.430325 -4.686120 0.0000

X5/X4 -0.189060 0.115224 -1.640812 0.1075









Date: 08/22/13 Time: 14:35

Sample: 1960 2012


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C -0.107515 0.114543 -0.938644 0.3527

X1/X5 -1.19E-07 7.82E-09 -15.21492 0.0000

X2/X5 -1.56E-11 1.60E-12 -9.744259 0.0000

X3/X5 -0.213828 0.057516 -3.717711 0.0005

X4/X5 2.657934 0.074921 35.47637 0.0000

1/X5 -5.458239 1.338748 -4.077122 0.0002







Notice that in the transformed regression the intercept term 2 is the slope coefficient in the

original equation and the slope coefficient 1 is the intercept term in the original model.

Therefore, to get back to the original model we shall have to multiply the estimated

transformed model byXi.

Assumption 2: The error variance is proportional toXi. The square root transformation:

E (i2) =2Xi

Transforming using X1:


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Yi /X1= 1 (1/X1) + 2X1 + 3 X2i/X1 + 4 X3i/X1 + +6X5i/X1 +i TRANSFORMED

Step 1: Regress the equation (11.8a)

Open the File containing dataQuickestimate equationequation specification window

Write:

Y/x1^0.5 c 1/x1^0.5 x2/x1^0.5 x3/x1^0.5 x4/x1^0.5 x5/x1^0.5

Dependent Variable: Y/X1^0.5


Date: 08/22/13 Time: 14:42

Sample: 1960 2012




C -0.003423 0.000293 -11.69658 0.0000

1/X1^0.5 -3.483622 1.212010 -2.874252 0.0061

X2/X1^0.5 -2.95E-11 1.99E-12 -14.88087 0.0000

X3/X1^0.5 -0.356717 0.062820 -5.678385 0.0000

X4/X1^0.5 3.378221 0.131543 25.68143 0.0000

X5/X1^0.5 -0.166805 0.123406 -1.351681 0.1829








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Similarly do for X2 X3 X4 and X5.

Note an important feature of the transformed model: It has no intercept term.

Therefore, one will have to use the regression-through-the-origin model to

Estimate 1 and 2. Having run (11.8), one can get back to the original model

Simply by multiplying transformed model by Xi.

Assumption 3: The error variance is proportional to the square of the mean value ofY.

E (i2)=2E(Yi)

2

Equation postulates that the variance of i2 is proportional to the square of the expected

Value ofY.

Therefore, if we transform the original equation as follows:

Yi /E (Yi) = 1 (1/ E (Yi)) + 2X1/E (Yi) +3 X2i/ E (Yi) + 4 X3i/ E (Yi) + +6X5i/ E (Yi)+i

The transformation (11.9) is, however, in operational because E(Yi) depends on 1 and 2,

which are unknown. Of course, we know, which is an estimator ofE(Yi)

Transforming using: i

Yi /i = 1 (1/ i) + 2X1i/ i+ 3 X2i/ i + 4 X3i/i + +6X5i/ i + i

Step: Regress the equation (11.8a)

Open the File containing dataQuickestimate equationequation specifications window

Write: Y/Y_CAP 1/Y_CAP X1/Y_CAP X2/Y_CAP X3/Y_CAP X4/Y_CAP X5/Y_CAP

Dependent Variable: Y/Y_CAP


Date: 08/22/13 Time: 14:55

Sample: 1960 2012



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1/Y_CAP -6.465594 1.412492 -4.577437 0.0000

X1/Y_CAP -1.24E-07 9.12E-09 -13.58257 0.0000

X2/Y_CAP -1.62E-11 1.85E-12 -8.803458 0.0000

X3/Y_CAP -0.237086 0.063967 -3.706393 0.0006

X4/Y_CAP 2.721034 0.078917 34.47970 0.0000

X5/Y_CAP -0.182424 0.112227 -1.625497 0.1107


Adjusted R-squared -0.091201 S.D. dependent var 0.005202



Log likelihood 204.3804 Durbin-Watson stat 0.186775

Assumption 4: A log transformation such as:

lnYi = 1 + 2 lnXi+i

very often reduces heteroscedasticity.

Log Transforming:

lnYi = 1 + 2 lnX1i + 3 lnX2i + 4 lnX3i + 5lnX4i+6lnX5i+i

Step: Regress the equation (11.8c)

Open the File containing dataQuickestimate equationequation specifications window

Write: log(y) c log(x1) log(x2) log(x3) log(x4) log(x5)


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Date: 08/22/13 Time: 15:00

Sample: 1960 2012




C 5.341832 0.443418 12.04694 0.0000

LOG(X1) -0.397072 0.050292 -7.895290 0.0000

LOG(X2) -0.040926 0.010557 -3.876748 0.0003

LOG(X3) 0.020821 0.008782 2.370920 0.0219

LOG(X4) 2.058282 0.171616 11.99351 0.0000

LOG(X5) 0.037678 0.012541 3.004484 0.0043







This result arises because log transformation compresses the scales in which the variables

are measured, thereby reducing a tenfold difference between two values to a twofold

difference.

To conclude our discussion of the remedial measures, we reemphasize that all the

transformations discussed previously are ad hoc; we are essentially speculating


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about the nature ofi2.

Which of the transformations discussed previously will

work will depend on the nature of the problem and the severity of

heteroscedasticity.

CHAPTERS 18 TO 20

SIMULTANEOUS REGRESSION MODELS

THE NATURE OF SIMULTANEOUS-EQUATION MODELS

In contrast to single-equation models, in simultaneous-equation models more than onedependent, or endogenous, variable is involved, necessitating as many equations as the

number of endogenous variables. A unique feature of simultaneous-equation models is that the

endogenous variable (i.e., regressand) in one equation may appear as an explanatory variable

(i.e., regressor) in another equation of the system.

Y1i = 10 + 12Y2i + 11 X1i + u1i

Y2i = 20 + 21Y1i + 21 X1i + u2i

Where Y1 and Y2 are mutually dependent, or endogenous, variables and X1 is an exogenous

variable and where u1 andu2 are the stochastic disturbance terms, the variables Y1 and Y2 areboth stochastic.

Therefore, unless it can be shown that the stochastic explanatory variable Y2 in (18.1) is

distributed independently of u1 and the stochastic explanatory variable Y1in (18.2) is

distributed independently of u2, application of the classical OLS to these equations individually

will lead to inconsistent estimates.


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As a consequence, such an endogenous explanatory variable becomes stochastic and is usually

correlated with the disturbance term of the equation in which it appears as an explanatory

variable.

In this situation the classical OLS method may not be applied because the estimators thus

obtained are not consistent, that is, they do not converge to their true population values no

matter how large the sample size. Since simultaneous-equation models are used frequently,

especially in econometric models, alternative estimating techniques have been developed by

various authors.

DOING REGREESION USING EVIEWS SIMULTENENOUS MODELS

MODEL: Yi = 1 + 2 X1i + 3 X2i +4X4i + 5X5i +1i

X1i= 1+2Yi+3X2i+4X3i+2i

Where Yi and X1iare mutually dependent, or endogenous, variables X2i, X3i, X4i and X5i

are an exogenous variable and where 1i and2i are the stochastic disturbance terms, the

variables YiandX1iare both stochastic. Therefore, unless it can be shown that the

stochastic explanatory variable Yi is distributed independently of1i

and the stochastic explanatory variableX1i is distributed independently

of2i, application of the classical OLS to these equations individually will lead to

inconsistent estimates.

THE IDENTIFICATION PROBLEM

RULES FOR IDENTIFICATION

THE ORDER CONDITION OF IDENTIFIABILITY

A necessary (but not sufficient) condition of identification, known as the order condition,

May be stated in two different but equivalent ways as follows (the necessary as well as


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Sufficient condition of identification will be presented shortly):

MK G - 1

M= numbers of variables in the simultaneous model

iK= numbers of variables in the specific equation

G = numbers of endogenous variables or equations

FOR EQUATION 1:

65 2 1

1 1

FOR EQUATION 2:

64 2-1

2 1

Hence equation 1 is just identified and equation 2 is over identified. As we know equation 1 is

just identified, is estimated using ILS and 2SLS and equation 2 is over identified, is estimated

using 2SLS method.

THE RANK CONDITION FOR IDENTIFIABILITY:

Here we can not apply the rank condition there are only two simultaneous equations only

HAUSMAN SPECIFICATION TEST

It is also called test of endogeneity or simultaneity problem

EVI EWS:

Step 1: Regress the equation 1. Run Yi onX1 X2 X3 X4 X5



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Write:

Y C X1 X2 X4 X5





Date: 09/04/13 Time: 12:58

Sample: 1960 2012



C 9.097915 1.536073 5.922841 0.0000

X2 -3.84E-11 3.65E-12 -10.50413 0.0000

X3 -0.751619 0.157433 -4.774216 0.0000

X4 1.907816 0.078995 24.15124 0.0000

X5 -0.706191 0.258311 -2.733880 0.0087







Step 2: Regress the equation 2 using residual of step 1 as explanatory variable


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Write:

X1 C Y X2 X3 R1

Now check the significance of this residual coefficient if it exceeds critical then we can say both the equation are

simultaneous



Date: 09/04/13 Time: 13:01

Sample: 1960 2012



C -1.53E+08 10008321 -15.31155 0.0000

Y 3794853. 203298.8 18.66638 0.0000

X2 0.000291 1.31E-05 22.24633 0.0000

X3 5855713. 671827.1 8.716101 0.0000

RESID01 -10309118 797193.2 -12.93177 0.0000


Adjusted R-squared 0.993859 S.D. dependent var 42305993

S.E. of regression 3315248. Akaike info criterion 32.95555

Sum squared resid 5.28E+14 Schwarz criterion 33.14143



As residual is significant 1% level of significance; w can conclude that both equation are

simultaneous


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Similarly for equation 2 Repeat step 1 now.

Step 2: Regress the equation 1 using residual of step 1 as explanatory variable


Write:

y c x1 x2 x4 x5 r1

Now check the significance of this residual coefficient if it exceeds critical then we can say both the equation are

simultaneous



Date: 09/04/13 Time: 13:05

Sample: 1960 2012



C -0.811293 1.899331 -0.427146 0.6712

X1 -9.78E-08 1.21E-08 -8.112440 0.0000

X2 -1.70E-11 2.24E-12 -7.562896 0.0000

X4 2.397581 0.101523 23.61622 0.0000

X5 -0.049479 0.107277 -0.461224 0.6468

RESID01 0.362655 0.101890 3.559284 0.0009




Sum squared resid 3.659450

applying regression using e views (with commands)

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