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Econometrics I. Professor William Greene Stern School of Business Department of Economics. Gauss-Markov Theorem. A theorem of Gauss and Markov: Least Squares is the minimum variance linear unbiased estimator (MVLUE) 1. Linear estimator 2. Unbiased: E[ b | X ] = β
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Econometrics I Professor William Greene Stern School of Business Department of Economics
Gauss-Markov Theorem A theorem of Gauss and Markov: Least Squares is the minimum variance linear unbiased estimator (MVLUE) 1. Linear estimator 2. Unbiased: E[b|X] = β Theorem: Var[b*|X] – Var[b|X] is nonnegative definite for any other linear and unbiased estimator b* that is not equal to b. Definition: b is efficientin this class of estimators.
Implications of Gauss-Markov • Theorem: Var[b*|X] – Var[b|X] is nonnegative definite for any other linear and unbiased estimator b* that is not equal to b. Implies: • bk = the kth particular element of b.Var[bk|X] = the kth diagonal element of Var[b|X]Var[bk|X] < Var[bk*|X] for each coefficient. • cb = any linear combination of the elements of b.Var[cb|X] <Var[cb*|X] for any nonzero c and b* that is not equal to b.
Summary: Finite Sample Properties of b • Unbiased: E[b]= • Variance: Var[b|X] = 2(XX)-1 • Efficiency: Gauss-Markov Theorem with all implications • Distribution: Under normality, b|X ~ N[, 2(XX)-1 (Without normality, the distribution is generally unknown.)
Comparação de modelos • Podemos comparar modelos sem usar testes estatsticos, mas sim medidas conhecidas como criterios de informação que traduzem a qualidade de ajustamento de um modelo. • Para estes indicadores a variável principal acaba por ser uma medida do valor absoluto dos erros.
Medida de Ajuste R2 = bXM0Xb/yM0y (Very Important Result.) R2 is bounded by zero and one only if: (a) There is a constant term in X and (b) The line is computed by linear least squares.
Comparing fits of regressions Make sure the denominator in R2 is the same - i.e., same left hand side variable. Example, linear vs. loglinear. Loglinear will almost always appear to fit better because taking logs reduces variation.
Adjusted R Squared • Adjusted R2 (for degrees of freedom) = 1 - [(n-1)/(n-K)](1 - R2) • includes a penalty for variables that don’t add much fit. Can fall when a variable is added to the equation.
Adjusted R2 What is being adjusted? The penalty for using up degrees of freedom. = 1 - [ee/(n – K)]/[yM0y/(n-1)] uses the ratio of two ‘unbiased’ estimators. Is the ratio unbiased? = 1 – [(n-1)/(n-K)(1 – R2)] Will rise when a variable is added to the regression? is higher with z than without z if and only if the t ratio on z is in the regression when it is added is larger than one in absolute value.
Full Regression (Without PD) ---------------------------------------------------------------------- Ordinary least squares regression ............ LHS=G Mean = 226.09444 Standard deviation = 50.59182 Number of observs. = 36 Model size Parameters = 9 Degrees of freedom = 27 Residuals Sum of squares = 596.68995 Standard error of e = 4.70102 Fit R-squared = .99334 <********** Adjusted R-squared = .99137 <********** Info criter. LogAmemiya Prd. Crt. = 3.31870 <********** Akaike Info. Criter. = 3.30788 <********** Model test F[ 8, 27] (prob) = 503.3(.0000) --------+------------------------------------------------------------- Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- Constant| -8220.38** 3629.309 -2.265 .0317 PG| -26.8313*** 5.76403 -4.655 .0001 2.31661 Y| .02214*** .00711 3.116 .0043 9232.86 PNC| 36.2027 21.54563 1.680 .1044 1.67078 PUC| -6.23235 5.01098 -1.244 .2243 2.34364 PPT| 9.35681 8.94549 1.046 .3048 2.74486 PN| 53.5879* 30.61384 1.750 .0914 2.08511 PS| -65.4897*** 23.58819 -2.776 .0099 2.36898 YEAR| 4.18510** 1.87283 2.235 .0339 1977.50 --------+-------------------------------------------------------------
PD added to the model. R2 rises, Adj. R2 falls ---------------------------------------------------------------------- Ordinary least squares regression ............ LHS=G Mean = 226.09444 Standard deviation = 50.59182 Number of observs. = 36 Model size Parameters = 10 Degrees of freedom = 26 Residuals Sum of squares = 594.54206 Standard error of e = 4.78195 Fit R-squared = .99336 Was 0.99334 Adjusted R-squared = .99107 Was 0.99137 --------+------------------------------------------------------------- Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- Constant| -7916.51** 3822.602 -2.071 .0484 PG| -26.8077*** 5.86376 -4.572 .0001 2.31661 Y| .02231*** .00725 3.077 .0049 9232.86 PNC| 30.0618 29.69543 1.012 .3207 1.67078 PUC| -7.44699 6.45668 -1.153 .2592 2.34364 PPT| 9.05542 9.15246 .989 .3316 2.74486 PD| 11.8023 38.50913 .306 .7617 1.65056 (NOTE LOW t ratio) PN| 47.3306 37.23680 1.271 .2150 2.08511 PS| -60.6202** 28.77798 -2.106 .0450 2.36898 YEAR| 4.02861* 1.97231 2.043 .0514 1977.50 --------+-------------------------------------------------------------
Outrasmedidas de Ajuste Para alternativasnãoaninhadas Incluipenalidade de grau de liberdade. Critério de Informação • Schwarz (BIC): n log(ee) + k(log(n)) • Akaike (AIC): n log(ee) + 2k Quando se quer decidir entre dois modelos não aninhados, o melhor é o que produz o menor valor do critério. A penalidade no BIC de incluir algo não relevante é maior que no AIC.
Outrasmedidas de Ajuste • Tanto o AIC quanto o BIC aumentam conforme SQR aumenta. • Além disso, ambos critérios penalizam modelos com muitas variáveis sendo que valores menores de AIC e BIC são preferíveis. • Como modelos com mais variáveis tendem a produzir menor SQR mas usam mais parâmetros, a melhor escolha é balancear o ajuste com a quantidade de variáveis. • A penalidade no BIC de incluir algo não relevante é maior que no AIC.
Log Income Equation ---------------------------------------------------------------------- Ordinary least squares regression ............ LHS=LOGY Mean = -1.15746 Estimated Cov[b1,b2] Standard deviation = .49149 Number of observs. = 27322 Model size Parameters = 7 Degrees of freedom = 27315 Residuals Sum of squares = 5462.03686 Standard error of e = .44717 Fit R-squared = .17237 --------+------------------------------------------------------------- Variable| Coefficient Standard Error b/St.Er. P[|Z|>z] Mean of X --------+------------------------------------------------------------- AGE| .06225*** .00213 29.189 .0000 43.5272 AGESQ| -.00074*** .242482D-04 -30.576 .0000 2022.99 Constant| -3.19130*** .04567 -69.884 .0000 MARRIED| .32153*** .00703 45.767 .0000 .75869 HHKIDS| -.11134*** .00655 -17.002 .0000 .40272 FEMALE| -.00491 .00552 -.889 .3739 .47881 EDUC| .05542*** .00120 46.050 .0000 11.3202 --------+------------------------------------------------------------- Average Age = 43.5272. Estimated Partial effect = .066225 – 2(.00074)43.5272 = .00018. Estimated Variance 4.54799e-6 + 4(43.5272)2(5.87973e-10) + 4(43.5272)(-5.1285e-8) = 7.4755086e-08. Estimated standard error = .00027341.
Interaction Effect ---------------------------------------------------------------------- Ordinary least squares regression ............ LHS=LOGY Mean = -1.15746 Standard deviation = .49149 Number of observs. = 27322 Model size Parameters = 4 Degrees of freedom = 27318 Residuals Sum of squares = 6540.45988 Standard error of e = .48931 Fit R-squared = .00896 Adjusted R-squared = .00885 Model test F[ 3, 27318] (prob) = 82.4(.0000) --------+------------------------------------------------------------- Variable| Coefficient Standard Error b/St.Er. P[|Z|>z] Mean of X --------+------------------------------------------------------------- Constant| -1.22592*** .01605 -76.376 .0000 AGE| .00227*** .00036 6.240 .0000 43.5272 FEMALE| .21239*** .02363 8.987 .0000 .47881 AGE_FEM| -.00620*** .00052 -11.819 .0000 21.2960 --------+------------------------------------------------------------- Do women earn more than men (in this sample?) The +.21239 coefficient on FEMALE would suggest so. But, the female “difference” is +.21239 - .00620*Age. At average Age, the effect is .21239 - .00620(43.5272) = -.05748.
Linear Restrictions Context: How do linear restrictions affect the properties of the least squares estimator? Model: y = X + Theory (information) R - q = 0 Restricted least squares estimator: b* = b - (XX)-1R[R(XX)-1R]-1(Rb - q) Expected value: E[b*] = - (XX)-1R[R(XX)-1R]-1(Rb - q) Variance:2(XX)-1 - 2 (XX)-1R[R(XX)-1R]-1R(XX)-1 = Var[b] – a nonnegative definite matrix < Var[b] Implication: (As before) nonsample information reduces the variance of the estimator.
Interpretation Case 1: Theory is correct: R - q = 0 (the restrictions do hold). b* is unbiased Var[b*] is smaller than Var[b] How do we know this? Case 2: Theory is incorrect: R - q 0 (the restrictions do not hold). b* is biased – what does this mean? Var[b*] is still smaller than Var[b]
Linear Least Squares Subject to Restrictions Restrictions: Theory imposes certain restrictions on parameters. Some common applications Dropping variables from the equation = certain coefficients in b forced to equal 0. (Probably the most common testing situation. “Is a certain variable significant?”) Adding up conditions: Sums of certain coefficients must equal fixed values. Adding up conditions in demand systems. Constant returns to scale in production functions. Equality restrictions: Certain coefficients must equal other coefficients. Using real vs. nominal variables in equations. General formulation for linear restrictions: Minimize the sum of squares, ee, subject to the linear constraintRb = q.
Restricted Least Squares Solution • General Approach: Programming ProblemMinimize for L = (y - X)(y - X) subject to R = qEach row of R is the K coefficients in a restriction.There are J restrictions: J rows • 3 = 0: R = [0,0,1,0,…] q = (0). • 2 =3: R = [0,1,-1,0,…] q = (0) • 2= 0,3 = 0: R = 0,1,0,0,… q = 0 0,0,1,0,… 0
Solution Strategy • Quadratic program: Minimize quadratic criterion subject to linear restrictions • All restrictions are binding • Solve using Lagrangean formulation • Minimize over (,) L* = (y - X)(y - X) + 2(R-q)(The 2 is for convenience – see below.)
Aspects of Restricted LS 1. b* = b - Cm where m = the “discrepancy vector” Rb - q. Note what happens if m = 0. What does m = 0 mean? 2. =[R(XX)-1R]-1(Rb - q) = [R(XX)-1R]-1m. When does = 0. What does this mean? 3. Combining results: b* = b - (XX)-1R. How could b* = b?