Solutions and Applications Manual - NYU

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., 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.

2 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.

3 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, 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.

4 2 Chapter 3 Least Squares Exercises 1. 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).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[.

5 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). 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.]]

6 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. 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.

7 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 additional element. Thus, ,, and .nnsnsssy == XnyXyx The new coefficient vector is 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); 111,,11( )()() ()1( )nsnsnnnnss nnsnn s = +X XXXXXxx XXxXX x1 . The vector is (Xn,s yn,s) = (Xn yn) + xsys.

8 multiply out the four terms to get (Xn,s Xn,s)-1(Xn,s yn,s) = bn 111()1( )nnssnsnn s +XXxxbxXX x+ 1()nn XX xsys 1111() ()1( )nnss nnsnn s +XXxx XXxXX x xsys = bn + xsys 1()nn XX111()()1( )snn snnsssnn sy +xXX xXXxxXX x 111()1( )nnssnsnn s +XXxxbxXX x bn + 111()1()1( )snn snnsssnn sy + xXX xXXxxXX x 111()1( )nnssnsnn s +XXxxbxXX x bn + 111()1( )nnsssnn sy +XXxxXX x 111()1( )nnssnsnn s +XXxxbxXX x bn + 111()(1( )nns ssnsnn sy +XXxxbxXX x) 7. Define the data matrix as follows: (The subscripts on the parts of y refer to the observed and missing rows of X. We will use Frish-Waugh to obtain the first two columns of the least squares coefficient vector. b=(X M X)-1(X M y). Multiplying it out, we find that []112, and .1011omy ==== yix00 XXXXy112112M2 = an identity matrix save for the last diagonal element that is equal to 0.)

9 X1 M2X1 = . This just drops the last observation. X1 M2y is computed likewise. Thus, the coeffients on the first two columns are the same as if y0 had been linearly regressed on X1. The denomonator of R2 is different for the two cases (drop the observation or keep it with zero fill and the dummy variable). For the first strategy, the mean of the n-1 observations should be different from the mean of the full n unless the last observation happens to equal the mean of the first n-1. 11111 00XX XX0 For the second strategy, replacing the missing value with the mean of the other n-1 observations, we can deduce the new slope vector logically. Using Frisch-Waugh, we can replace the column of x s with deviations from the means, which then turns the last observation to zero. Thus, once again, the coefficient on the x equals what it is using the earlier strategy. The constant term will be the same as well.

10 48. For convenience, reorder the variables so that X = [i, Pd, Pn, Ps, Y]. The three dependent variables are Ed, En, and Es, and Y = Ed + En + Es. The coefficient vectors are bd = (X X)-1X Ed, bn = (X X)-1X En, and bs = (X X)-1X Es. The sum of the three vectors is b = (X X)-1X [Ed + En + Es] = (X X)-1X Y. Now, Y is the last column of X, so the preceding sum is the vector of least squares coefficients in the regression of the last column of X on all of the columns of X, including the last. Of course, we get a perfect fit. In addition, X [Ed + En + Es] is the last column of X X, so the matrix product is equal to the last column of an identity matrix. Thus, the sum of the coefficients on all variables except income is 0, while that on income is 1. 9. Let RK2 denote the adjusted R2 in the full regression on K variables including xk, and letR12denote the adjusted R2 in the short regression on K-1 variables when xk is omitted.