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Econometric Analysis 6th Edition by William H. Greene – Solutions and Applications Manual | Complete Worked Solutions

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This document provides detailed solutions and applied examples for Econometric Analysis (6th Edition) by William H. Greene. It covers key econometric concepts, model estimation, hypothesis testing, regression analysis, and problem-solving techniques used in advanced undergraduate and graduate econometrics courses. The material is ideal for homework support, exam preparation, and strengthening practical understanding of econometric methods.

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Solutions and
Applications Manual



Econometric Analysis
Sixth Edition
or discussing its implications in practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a student‘s ability to identify relevant legal issues, apply appropriate




William H. Greene
New York University




scenarios. Furthermore, students often practice writing answers to past exams under timed conditions to enhance their ability to structure responses
effectively.________________________________________3. Mathematics Exams3.1. Overview of Mathematics EducationMathematics education spans a wide range of topics, from basic
arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science, and
technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their problem-solving abilities, and their aptitude for logical reasoning.3.2.
Structure of Mathematics ExamsMathematics exams are typically focused on problem-solving and application of formulas, theorems, and mathematical concepts. These exams may
include:Problem Sets: Students are given a series of problems that test various mathematical skills, from basic calculations to more complex word problems. The problems may require students
to apply formulas, solve equations, or prove mathematical theorems.Theoretical Questions: In higher-level mathematics exams, students may be asked to demonstrate their understanding of
theoretical concepts, such as the proof of a mathematical theorem or the explanation of a mathematical concept.Multiple Choice Questions (MCQs): While less common, some mathematics
exams use MCQs to test students' quick recall of formulas, definitions, or theorems.3.3. Skills Tested in Mathematics ExamsProblem-Solving Ability: The core skill tested in mathematics exams
is problem-solving. Students must approach complex

,Contents and Notation
This book presents solutions to the end of chapter exercises and applications in Econometric Analysis. There
are no exercises in the text for Appendices A – E. For the instructor or student who is interested in exercises for
this material, I have included a number of them, with solutions, in this book. The various computations in the
solutions and exercises are done with the NLOGIT Version 4.0 computer package (Econometric Software, Inc.,
Plainview New York, www.nlogit.com).
or discussing its implications in practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a student‘s ability to identify relevant legal issues, apply
appropriate or discussing its implications in practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a student‘s ability to identify relevant legal
issues, apply appropriate is problem-solving. Students must approach complex



Chapter 1 Introduction 1
Chapter 2 The Classical Multiple Linear Regression Model 2
Chapter 3 Least Squares 3
Chapter 4 Statistical Properties of the Least Squares Estimator 10
Chapter 5 Inference and Prediction 19
Chapter 6 Functional Form and Structural Change 30
Chapter 7 Specification Analysis and Model Selection 40
Chapter 8 The Generalized Regression Model and Heteroscedasticity 44
Chapter 9 Models for Panel Data 54
Chapter 10 Systems of Regression Equations 67
Chapter 11 Nonlinear Regressions and Nonlinear Least Squares 80
Chapter 12 Instrumental Variables Estimation 85
Chapter 13 Simultaneous-Equations Models 90
Chapter 14 Estimation Frameworks in Econometrics 97
Chapter 15 Minimum Distance Estimation and The Generalized Method of Moments 102
Chapter 16 Maximum Likelihood Estimation 105
Chapter 17 Simulation Based Estimation and Inference 117
Chapter 18 Bayesian Estimation and Inference 120
Chapter 19 Serial Correlation 122
Chapter 20 Models with Lagged Variables 128
Chapter 21 Time-Series Models 131
Chapter 22 Nonstationary Data 132
Chapter 23 Models for Discrete Choice 136
Chapter 24 Truncation, Censoring and Sample Selection 142
Chapter 25 Models for Event Counts and Duration 147
Appendix A Matrix Algebra 155
Appendix B Probability and Distribution Theory 162
Appendix C Estimation and Inference 172
Appendix D Large Sample Distribution Theory 183
Appendix E Computation and Optimization 184
In the solutions, we denote:
 scalar values with italic, lower case letters, as in a,
 column vectors with boldface lower case letters, as in b,
 row vectors as transposed column vectors, as in b,
 matrices with boldface upper case letters, as in M or ,
 single population parameters with Greek letters, as in ,
 sample estimates of parameters with Roman letters, as in b as an estimate of ,
 sample estimates of population parameters with a caret, as in ˆ or ˆ ,
 cross section observations with subscript i, as in yi,
time series observations with subscript t, as in zt and
panel data observations with xit or xi,t-1 when the comma is needed to remove ambiguity.
Observations that are vectors are denoted likewise, for example, xit to denote a column vector of
observations.

These are consistent with the notation used in the text.

,Chapter 1
Introduction
There are no exercises or applications in Chapter 1.



Chapter 2
The Classical Multiple Linear
Regression Model
There are no exercises or applications in Chapter 2.


scenarios. Furthermore, students often pracor discussing its implications in practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a student‘s ability to
identify relevant legal issues, apply appropriate tice writing answers to past exams under timed conditions to enhance their ability to structure responses
effectively.________________________________________3. Mathematics Exams3.1. Overview of Mathematics EducationMathematics education spans a wide range of topics, from
basic arithmetic and algebra to more advanced areas like calculus, statistics, and abstract algebra. Mathematics is crucial in many industries, including engineering, finance, data science,
and technology. The primary goal of mathematics exams is to test students' understanding of mathematical concepts, their problem-solving abilities, and their aptitude for logical
reasoning.3.2. Structure of Mathematics ExamsMathematics exams are typically focused on problem-solving and application of formulas, theorems, and mathematical concepts. These
exams may include:Problem Sets: Students are given a series of problems that test various mathematical skills, from basic calculations to more complex word problems. The problems
may require students to apply formulas, solve equations, or prove mathematical theorems.Theoretical Questions: In higher-level mathematics exams, students may be asked to demonstrate
their understanding of theoretical concepts, such as the proof of a mathematical theorem or the explanation of a mathematical concept.Multiple Choice Questions (MCQs): While less
common, some mathematics exams use MCQs to test students' quick recall of formulas, definitions, or theorems.3.3. Skills Tested in Mathematics ExamsProblem-Solving Ability: The
core skill tested in mathematics exams is problem-solving. Students must approach complex



Chapter 3
Least Squares
Exercises
 1 x1 
1. Let X  ... ...  .
 1 x 
 n 

(a) The normal equations are given by (3-12), X'e  0 (we drop the minus sign), hence for each of the
columns of X, xk, we know that xke = 0. This implies that  e  0 and  x e  0 .
n n
i 1 i i 1 i i

(b) Use  e to conclude from the first normal equation that a  y  bx .
n
i 1 i

(c) We know that  e  0 and  x e  0 . It follows then that  (x  x )e  0 because
n n n
i 1 i i 1 i i i 1 i i

 xe  x e  0 . Substitute ei to obtain
n n
i 1 i i 1 i

n (x  x )( y  a  bx )  0 or n (x  x )( y  y  b(x  x ))  0
i 1 i i i i 1 i i i

 (x  x )( y  y)
n

Then, n (x  x )( y  y)  bn (x  x )(x  x )) so b  i1 i i
.
i 1 i 1
 (x  x )2
i i i i n
i 1

(d) The first derivative vector of ee is -2Xe. (The normal equations.) The second derivative matrix is
2(ee)/bb = 2XX. We need to show that this matrix is positive definite. The diagonal elements are 2n
and 2n x2 which are clearly both positive. The determinant is (2n)( 2n x2 )-( 2n x )2
i 1 i i 1 i i 1 i




3

, = 4nn x2 -4( nx )2 = 4n[(n x2 )  nx 2 ]  4n[(n (x  x )2 ] . Note that a much simpler proof appears after
i 1 i
2. Write c as b + (c - b). Then, the sum of squared residuals based on c is
(y - Xc)(y - Xc) = [y - X(b + (c - b))] [y - X(b + (c - b))] = [(y - Xb) + X(c - b)] [(y - Xb) + X(c - b)]
= (y - Xb) (y - Xb) + (c - b) XX(c - b) + 2(c - b) X(y - Xb).
But, the third term is zero, as 2(c - b) X(y - Xb) = 2(c - b)Xe = 0. Therefore,
(y - Xc) (y - Xc) = ee + (c - b) XX(c - b)
or (y - Xc) (y - Xc) - ee = (c - b) XX(c - b).
The right hand side can be written as dd where d = X(c - b), so it is necessarily positive. This confirms what
we knew at the outset, least squares is least squares.

3. The residual vector in the regression of y on X is MXy = [I - X(XX)-1X]y. The residual vector in the
regression of y on Z is
MZy = [I - Z(ZZ)-1Z]y
= [I - XP((XP)(XP))-1(XP))y
= [I - XPP-1(XX)-1(P)-1PX)y
= MXy
Since the residual vectors are identical, the fits must be as well. Changing the units of measurement of the
regressors is equivalent to postmultiplying by a diagonal P matrix whose kth diagonal element is the scale
factor to be applied to the kth variable (1 if it is to be unchanged). It follows from the result above that this
will not change the fit of the regression.
scenarios. Furthermore, students often practice writing answers to past exams under timed conditions to enhance their ability to structure responses or discussing its implications in
practice.2.3. Skills Tested in Law ExamsLegal Analysis and Application: Law exams test a student‘s ability to identify relevant legal issues, apply appropriate of formulas, definitions, or
theorems.3.3. Skills Tested in Mathematics ExamsProblem-Solving Ability: The core skill tested in mathematics exams is problem-solving. Students must approach complex


4. In the regression of y on i and X, the coefficients on X are b = (XM0X)-1XM0y. M0 = I - i(ii)-1i is
the matrix which transforms observations into deviations from their column means. Since M0 is idempotent
and symmetric we may also write the preceding as [(XM0)(M0X)]-1(XM0)(M0y) which implies that
the
regression of M0y on M0X produces the least squares slopes. If only X is transformed to deviations, we
would compute [(XM0)(M0X)]-1(XM0)y but, of course, this is identical. However, if only y is transformed,
the result is (XX)-1XM0y which is likely to be quite different.

5. What is the result of the matrix product M1M where M1 is defined in (3-19) and M is defined in (3-14)?
M1M = (I - X1(X1X1)-1X1)(I - X(XX)-1X) = M - X1(X1X1)-1X1M
There is no need to multiply out the second term. Each column of MX1 is the vector of residuals in the
regression of the corresponding column of X1 on all of the columns in X. Since that x is one of the columns in
X, this regression provides a perfect fit, so the residuals are zero. Thus, MX1 is a matrix of zeroes which
implies that M1M = M.

6. The original X matrix has n rows. We add an additional row, xs. The new y vector likewise has an

 Xn  y n 
additional element. Thus, X n,s  and y n,s  . The new coefficient vector is
   
x y
 s  s 
bn,s = (Xn,s Xn,s)-1(Xn,syn,s). The matrix is Xn,sXn,s = XnXn + xsxs. To invert this, use (A -66);
1
(X X )1  (X X )1  (X X )1 x x (X X )1 . The vector is
n,s n,s n n 1 n n s s n n
1 x (X X ) x
(Xn,syn,s) = (Xnyn) + xsys. Multiply out the four terms to get

(Xn,s Xn,s)-1(Xn,syn,s) =
1 1
b – (X X )1 x xb + (X X )1 x y  (X X )1 x x (X X )1 x y
1 x (X X )1 x 1 x (X X )1 x
n n n s s n n n s s n n s s n n s s


=
x (X X )1 x  1
4

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