Exam (elaborations) TEST BANK FOR Econometric Analysis 5th Edition By William H. Greene (Solution manual) Solutions Manual Econometric Analysis Fifth
... [Show More] Edition William H. Greene New York University Prentice Hall, Upper Saddle River, New Jersey 07458 Contents and Notation Chapter 1 Introduction 1 Chapter 2 The Classical Multiple Linear Regression Model 2 Chapter 3 Least Squares 3 Chapter 4 Finite-Sample Properties of the Least Squares Estimator 7 Chapter 5 Large-Sample Properties of the Least Squares and Instrumental Variables Estimators 14 Chapter 6 Inference and Prediction 19 Chapter 7 Functional Form and Structural Change 23 Chapter 8 Specification Analysis and Model Selection 30 Chapter 9 Nonlinear Regression Models 32 Chapter 10 Nonspherical Disturbances - The Generalized Regression Model 37 Chapter 11 Heteroscedasticity 41 Chapter 12 Serial Correlation 49 Chapter 13 Models for Panel Data 53 Chapter 14 Systems of Regression Equations 63 Chapter 15 Simultaneous Equations Models 72 Chapter 16 Estimation Frameworks in Econometrics 78 Chapter 17 Maximum Likelihood Estimation 84 Chapter 18 The Generalized Method of Moments 93 Chapter 19 Models with Lagged Variables 97 Chapter 20 Time Series Models 101 Chapter 21 Models for Discrete Choice 1106 Chapter 22 Limited Dependent Variable and Duration Models 112 Appendix A Matrix Algebra 115 Appendix B Probability and Distribution Theory 123 Appendix C Estimation and Inference 134 Appendix D Large Sample Distribution Theory 145 Appendix E Computation and Optimization 146 In the solutions, we denote: • scalar values with italic, lower case letters, as in a or α • column vectors with boldface lower case letters, as in b, • row vectors as transposed column vectors, as in b′, • single population parameters with greek letters, as in β, • sample estimates of parameters with English letters, as in b as an estimate of β, • sample estimates of population parameters with a caret, as in αˆ • matrices with boldface upper case letters, as in M or Σ, • cross section observations with subscript i, time series observations with subscript t. These are consistent with the notation used in the text. Chapter 1 Introduction There are no exercises in Chapter 1. 1 Chapter 2 The Classical Multiple Linear Regression Model There are no exercises in Chapter 2. 2 Chapter 3 Least Squares 1. (a) Let . The normal equations are given by (3-12), , hence for each of the columns of X, x = n x x X 1 . . 1 1 X′e = 0 = i i i k, we know that xk’e=0. This implies that Σ = 0and x e . i i e Σ 0 (b) Use Σ = 0 to conclude from the first normal equation that i i e a = y − bx . (c) Know that Σ = 0 and . It follows then that i i e Σ = 0 i i i x e Σ ( − ) = 0 i i i x x e . Further, the latter implies ( )( ) = 0 i Σ − bx i − − a i i x x y or (x − x)(y − y − b(x − x))= 0 i i i i Σ from which the result follows. 2. Suppose b is the least squares coefficient vector in the regression of y on X and c is any other Kx1 vector. Prove that the difference in the two sums of squared residuals is (y-Xc)′(y-Xc) - (y-Xb)′(y-Xb) = (c - b)′X′X(c - b). Prove that this difference is positive. 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. Consider the least squares regression of y on K variables (with a constant), X. Consider an alternative set of regressors, Z = XP, where P is a nonsingular matrix. Thus, each column of Z is a mixture of some of the columns of X. Prove that the residual vectors in the regressions of y on X and y on Z are identical. What relevance does this have to the question of changing the fit of a regression by changing the units of measurement of the independent variables? 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. 4. In the least squares regression of y on a constant and X, in order to compute the regression coefficients on X, we can first transform y to deviations from the mean, y , and, likewise, transform each column of X to deviations from the respective column means; second, regress the transformed y on the transformed X without a constant. Do we get the same result if we only transform y? What if we only transform X? 3 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. We can extend the result in (6-24) to derive what is produced by this computation. In the formulation, we let X1 be X and X2 is the column of ones, so that b2 is the least squares intercept. Thus, the coefficient vector b defined above would be b = (X′X)-1X′(y - ai). But, a = y - b′ x so b = (X′X)-1X′(y - i( y - b′ x )). We can partition this result to produce (X′X)-1X′(y - i y )= b - (X′X)-1X′i(b′ x )= (I - n(X′X)-1 x x ′)b. (The last result follows from X′i = n x .) This does not provide much guidance, of course, beyond the observation that if the means of the regressors are not zero, the resulting slope vector will differ from the correct least squares coefficient vector. 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. Adding an observation. A data set consists of n observations on Xn and yn. The least squares estimator based on these n observations is b (X ) 1 n n n n = ′X − X′ .ny Another observation, xs and ys, becomes available. Prove that the least squares estimator computed using this additional observation is 1 , 1 1 ( ) ( 1 ( ) n s n n n s s s n s n n s − y − = + ′ − ′ + ′ ′ b b XX x xb x X X x ). Note that the last term is es, the residual from the prediction of ys using the coefficients based on Xn and bn. Conclude that the new data change the results of least squares only if the new observation on y cannot be perfectly predicted using the information already in hand. 7. A common strategy for handling a case in which an observation is missing data for one or more variables is to fill those missing variables with 0s or add a variable to the model that takes the value 1 for that one observation and 0 for all other observations. Show that this ‘strategy’ is equivalent to discarding the observation as regards the computation of b but it does have an effect on R2. Consider the special case in which X contains only a constant and one variable. Show that replacing the missing values of X with the mean of the complete observations has the same effect as adding the new variable. 8. Let Y denote total expenditure on consumer durables, nondurables, and services, and Ed, En, and Es are the expenditures on the three categories. As defined, Y = Ed + En + Es. Now, consider the expenditure system Ed = αd + βdY + γddPd + γdnPn + γdsPs + εγd En = αn + βnY + γndPd + γnnPn + γnsPs + εn Es = αs + βsY + γsdPd + γsnPn + γssPs + εs. Prove that if all equations are estimated by ordinary least squares, then the sum of the income coefficients will be 1 and the four other column sums in the preceding model will be zero. 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 4 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. Prove that the adjusted R2 in (3-30) rises (falls) when variable xk is deleted from the regression if the square of the t ratio on xk in the multiple regression is less (greater) than one. The proof draws on the results of the previous problem. Let RK 2 denote the adjusted R2 in the full regression on K variables including xk, and let R1 2 denote the adjusted R2 in the short regression on K-1 variables when xk is omitted. Let and denote their unadjusted counterparts. Then, RK 2 R1 2 = 1 - e′e/y′M RK 2 0y R1 = 1 - e 2 1′e1/y′M0y where e′e is the sum of squared residuals in the full regression, e1′e1 is the (larger) sum of squared residuals in the regression which omits xk, and y′M0y = Σi (yi - y )2 Then, RK 2 = 1 - [(n-1)/(n-K)](1 - ) RK 2 and R1 2 = 1 - [(n-1)/(n-(K-1))](1 - R1 ). 2 The difference is the change in the adjusted R2 when xk is added to the regression, RK 2 - R1 2 = [(n-1)/(n-K+1)][e1′e1/y′M0y] - [(n-1)/(n-K)][e′e/y′M0y]. The difference is positive if and only if the ratio is greater than 1. After cancelling terms, we require for the adjusted R2 to increase that e1′e1/(n-K+1)]/[(n-K)/e′e] > 1. From the previous problem, we have that e1′e1 = e′e + bK 2(xk′M1xk), where M1 is defined above and bk is the least squares coefficient in the full regression of y on X1 and xk. Making the substitution, we require [(e′e + bK 2(xk′M1xk))(n-K)]/[(n-K)e′e + e′e] > 1. Since e′e = (n-K)s2, this simplifies to [e′e + bK 2(xk′M1xk)]/[e′e + s2] > 1. Since all terms are positive, the fraction is greater than one if and only bK 2(xk′M1xk) > s2 or bK 2(xk′M1xk/s2) > 1. The denominator is the estimated variance of bk, so the result is proved. 10. Suppose you estimate a multiple regression first with then without a constant. Whether the R2 is higher in the second case than the first will depend in part on how it is computed. Using the (relatively) standard method, R2 = 1 - e′e / y′M0y, which regression will have a higher R2? This R2 must be lower. The sum of squares associated with the coefficient vector which omits the constant term must be higher than the one which includes it. We can write the coefficient vector in the regression without a constant as c = (0,b*) where b* = (W′W)-1W′y, with W being the other K-1 columns of X. Then, the result of the previous exercise applies directly. 11. Three variables, N, D, and Y all have zero means and unit variances. A fourth variable is C = N + D. In the regression of C on Y, the slope is .8. In the regression of C on N, the slope is .5. In the regression of D on Y, the slope is .4. What is the sum of squared residuals in the regression of C on D? There are 21 observations and all mom [Show Less]