Exam (elaborations) TEST BANK FOR Econometric Analysis 5th Edition By William H. Greene (Solution manual)

Exam (elaborations) TEST BANK FOR Econometric Analysis 5th Edition By William H. Greene (Solution manual) Solutions Manual Econometric Analysis Fifth 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 moments are computed using 1/(n-1) as the divisor. We use the notation ‘Var[.]’ and ‘Cov[.]’ to indicate the sample variances and covariances. Our information is Var[N] = 1, Var[D] = 1, Var[Y] = 1. Since C = N + D, Var[C] = Var[N] + Var[D] + 2Cov[N,D] = 2(1 + Cov[N,D]). From the regressions, we have Cov[C,Y]/Var[Y] = Cov[C,Y] = .8. But, Cov[C,Y] = Cov[N,Y] + Cov[D,Y]. Also, Cov[C,N]/Var[N] = Cov[C,N] = .5, but, Cov[C,N] = Var[N] + Cov[N,D] = 1 + Cov[N,D], so Cov[N,D] = -.5, so that Var[C] = 2(1 + -.5) = 1. And, Cov[D,Y]/Var[Y] = Cov[D,Y] = .4. Since Cov[C,Y] = .8 = Cov[N,Y] + Cov[D,Y], Cov[N,Y] = .4. Finally, Cov[C,D] = Cov[N,D] + Var[D] = -.5 + 1 = .5. Now, in the regression of C on D, the sum of squared residuals is (n-1){Var[C] - (Cov[C,D]/Var[D])2Var[D]} 5 based on the general regression result Σe2 = Σ(yi - y )2 - b2Σ(xi - x )2. All of the necessary figures were obtained above. Inserting these and n-1 = 20 produces a sum of squared residuals of 15. 12. Using the matrices of sums of squares and cross products immediately preceding Section 3.2.3, compute the coefficients in the multiple regression of real investment on a constant, real GNP and the interest rate. Compute R2. The relevant submatrices to be used in the calculations are Investment Constant GNP Interest Investment * 3.0500 3.9926 23.521 Constant 15 19.310 111.79 GNP 25.218 148.98 Interest 943.86 The inverse of the lower right 3×3 block is (X′X)-1, 7.5874 (X′X)-1 = -7.41859 7.84078 .27313 -. . The coefficient vector is b = (X′X)-1X′y = (-., ., -.)′. The total sum of squares is y′y = .63652, so we can obtain e′e = y′y - b′X′y. X′y is given in the top row of the matrix. Making the substitution, we obtain e′e = .63652 - .63291 = .00361. To compute R2, we require Σi (xi - y )2 = .63652 - 15(3.05/15)2 = ., so R2 = 1 - .00361/. = .77925. 13. In the December, 1969, American Economic Review (pp. 886-896), Nathanial Leff reports the following least squares regression results for a cross section study of the effect of age composition on savings in 74 countries in 1964: log S/Y = 7.3439 + 0.1596 log Y/N + 0.0254 log G - 1.3520 log D1 - 0.3990 log D2 (R2 = 0.57) log S/N = 8.7851 + 1.1486 log Y/N + 0.0265 log G - 1.3438 log D1 - 0.3966 log D2 (R2 = 0.96) where S/Y = domestic savings ratio, S/N = per capita savings, Y/N = per capita income, D1 = percentage of the population under 15, D2 = percentage of the population over 64, and G = growth rate of per capita income. Are these results correct? Explain. The results cannot be correct. Since log S/N = log S/Y + log Y/N by simple, exact algebra, the same result must apply to the least squares regression results. That means that the second equation estimated must equal the first one plus log Y/N. Looking at the equations, that means that all of the coefficients would have to be identical save for the second, which would have to equal its counterpart in the first equation, plus 1. Therefore, the results cannot be correct. In an exchange between Leff and Arthur Goldberger that appeared later in the same journal, Leff argued that the difference was simple rounding error. You can see that the results in the second equation resemble those in the first, but not enough so that the explanation is credible. 6 Chapter 4 Finite-Sample Properties of the Least Squares Estimator 1. Suppose you have two independent unbiased estimators of the same parameter, θ, say θ and θ , with different variances, v ∧ 1 ∧ 2 1 and v2. What linear combination, = c θ ∧ 1θ ∧ 1 + c2θ ∧ 2 is the minimum variance unbiased estimator of θ? Consider the optimization problem of minimizing the variance of the weighted estimator. If the estimate is to be unbiased, it must be of the form c1θ ∧ 1 + c2θ ∧ 2 where c1 and c2 sum to 1. Thus, c2 = 1 - c1. The function to minimize is Minc1L* = c1 2v1 + (1 - c1)2v2. The necessary condition is ∂L*/∂c1 = 2c1v1 - 2(1 - c1)v2 = 0 which implies c1 = v2 / (v1 + v2). A more intuitively appealing form is obtained by dividing numerator and denominator by v1v2 to obtain c1 = (1/v1) / [1/v1 + 1/v2]. Thus, the weight is proportional to the inverse of the variance. The estimator with the smaller variance gets the larger weight. 2. Consider the simple regression yi = βxi + εi. (a) What is the minimum mean squared error linear estimator of β? [Hint: Let the estimator be β = c′y]. Choose c to minimize Var[ β ] + [E( β - β)] ∧ ∧ ∧ 2. (The answer is a function of the unknown parameters.) (b) For the estimator in (a), show that ratio of the mean squared error of to that of the ordinary least β ∧ squares estimator, b, is MSE[ ] / MSE[b] = τ β ∧ 2 / (1 + τ2) where τ2 = β2 / [σ2/x′x]. Note that τ is the square of the population analog to the `t ratio' for testing the hypothesis that β = 0, which is given after (4-14). How do you interpret the behavior of this ratio as τ→∞? First, β = c′y = c′x + c′ε. So E[ β ] = βc′x and Var[ β ] = σ ∧ ∧ ∧ 2c′c. Therefore, MSE[β ] = β ∧ 2[c′x - 1]2 + σ2c′c. To minimize this, we set ∂MSE[β ]/∂c = 2β ∧ 2[c′x - 1]x + 2σ2c = 0. Collecting terms, β2(c′x - 1)x = -σ2c Premultiply by x′ to obtain β2(c′x - 1)x′x = -σ2x′c or c′x = β2x′x / (σ2 + β2x′x). Then, c = [(-β2/σ2)(c′x - 1)]x, so c = [1/(σ2/β2 + x′x)]x. Then, = c′y = x′y / (σ β ∧ 2/β2 + x′x). The expected value of this estimator is E[ β ] = βx′x / (σ ∧ 2/β2 + x′x) so E[ β ] - β = β(-σ ∧ 2/β2) / (σ2/β2 + x′x) = -(σ2/β) / (σ2/β2 + x′x) while its variance is Var[x′(xβ + ε) / (σ2/β2 + x′x)] = σ2x′x / (σ2/β2 + x′x)2 The mean squared error is the variance plus the squared bias, MSE[β ] = [σ ∧ 4/β2 + σ2x′x]/[σ2/β2 + x′x]2. The ordinary least squares estimator is, as always, unbiased, and has variance and mean squared error MSE(b) = σ2/x′x. 7 The ratio is taken by dividing each term in the numerator MSE β ΜΣΕ(β) ∧      = ( ) ( / )/( / ' ) ' ( / ' / ' σ σ σ σ σ 4 2 2 2 2 2 2 2 β β x x x x / x x x x + + ) = [σ2x′x/β2 + (x′x)2]/(σ2/β2 + x′x)2 = x′x[σ2/β2 + x′x]/(σ2/β2 + x′x)2 = x′x/(σ2/β2 + x′x) Now, multiply numerator and denominator by β2/σ2 to obtain MSE[β ]/MSE[b] = β ∧ 2x′x/σ2/[1 + β2x′x/σ2] = τ2/[1 + τ2] As τ→∞, the ratio goes to one. This would follow from the result that the biased estimator and the unbiased estimator are converging to the same thing, either as σ2 goes to zero, in which case the MMSE estimator is the same as OLS, or as x′x grows, in which case both estimators are consistent. 3. Suppose that the classical regression model applies, but the true value of the constant is zero. Compare the variance of the least squares slope estimator computed without a constant term to that of the estimator computed with an unnecessary constant term. The OLS estimator fit without a constant term is b = x′y / x′x. Assuming that the constant term is, in fact, zero, the variance of this estimator is Var[b] = σ2/x′x. If a constant term is included in the regression, then, b′ = i (xi x)(yi y) n − − =1 Σ / i (xi x) n − = 2 1 Σ The appropriate variance is σ2/ i (xi x n − = Σ 2 1 ) as always. The ratio of these two is Var[b]/Var[b′] = [σ2/x′x] / [σ2/ i (xi x) n − = Σ 2 1 ] But, i (xi x n − = Σ 2 1 ) = x′x + n x 2 so the ratio is Var[b]/Var[b′] = [x′x + n x 2]/x′x = 1 - n x 2/x′x = 1 - { n x 2/[Sxx + n x 2]} < 1 It follows that fitting the constant term when it is unnecessary inflates the variance of the least squares estimator if the mean of the regressor is not zero. 4. Suppose the regression model is yi = α + βxi + εi f(εi) = (1/λ)exp(-εi/λ) > 0. This is rather a peculiar model in that all of the disturbances are assumed to be positive. Note that the disturbances have E[εi] = λ. Show that the least squares constant term is unbiased but the intercept is biased. We could write the regression as yi = (α + λ) + βxi + (εi - λ) = α* + βxi + εi *. Then, we know that E[εi *] = 0, and that it is independent of xi. Therefore, the second form of the model satisfies all of our assumptions for the classical regression. Ordinary least squares will give unbiased estimators of α* and β. As long as λ is not zero, the constant term will differ from α. 5. Prove that the least squares intercept estimator in the classical regression model is the minimum variance linear unbiased estimator. Let the constant term be written as a = Σidiyi = Σidi(α + βxi + εi) = αΣidi + βΣidixi + Σidiεi. In order for a to be unbiased for all samples of xi, we must have Σidi = 1 and Σidixi = 0. Consider, then, minimizing the variance of a subject to these two constraints. The Lagrangean is L* = Var[a] + λ1(Σidi - 1) + λ2Σidixi where Var[a] = Σi σ2di 2. Now, we minimize this with respect to di, λ1, and λ2. The (n+2) necessary conditions are ∂L*/∂di = 2σ2di + λ1 + λ2xi, ∂L*/∂λ1 = Σi di - 1, ∂L*/∂λ2 = Σi dixi The first equation implies that di = [-1/(2σ2)](λ1 + λ2xi). Therefore,