Solutions and Applications Manual

1 Solutions and Applications Manual Econometric Analysis Sixth Edition William H. Greene New York University Prentice Hall, Upper Saddle River, New Jersey 07458 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 computer package (Econometric Software, Inc.)

2 , Plainview New York, ). In order to control the length of this document, only the Solutions and not the questions from the exercises and Applications are shown here. In some cases, the numerical Solutions for the in text examples shown here differ slightly from the values given in the text. This occurs because in general, the derivative computations in the text are done using the digits shown in the text, which are rounded to a few digits, while the results shown here are based on internal computations by the computer that use all digits.

3 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

4 128 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 or , cross section observations with subscript i, as in yi.

5 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. 2 Chapter 3 Least Squares Exercises 1.

6 Let = X. (a) The normal equations are given by (3-12), =X'e0(we drop the minus sign), hence for each of the columns of X, xk, we know that xk e = 0. This implies that 10niie= =and10niiixe= =. (b) Use to conclude from the first normal equation that 1niie= aybx= . (c) We know that and 10niie= =10niiixe= =. It follows then that 1()niiixxe=0 =because 110nniiiixexe== = =. Substitute ei to obtain 1()()niiiixxyabx= =0 or 1()(())niiiixxyybxx= =0 Then, 11121()(()( ) ()()) so ()nnniiiiiiiiinii).

7 Xxy yxxyy b xxxx bxx==== = = (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 221niix= which are clearly both positive. The determinant is (2n)( 212niix= )-(12niix= )2 = -4(214niinx= nx)2 = 22112()] 4[(()]nniii inxnx n xx== = 4[. Note that a much simpler proof appears after (3-6). 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).)

8 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.]]

9 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. 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.

10 Since M0 is idempotent and symmetric we may also write the preceding as [(X M0 )(M0X)]-1(X M0 )(M0y) which implies that the 3regression 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)?