2014 Preliminary Examination
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1 014 reliminary Examination 1) Standard error consistency and test statistic asymptotic normality in linear models Consider the model for the observable data y t ; x T t n Y = X + U; (1) where is a k 1 parameter vector. Let X be the n k matrix of covariates and let U be the n 1 vector of latent errors, where E UU T jx = I n. Let ^ n denote the OLS estimator of. For this model, consider the following conditions. C1 n 1 X T X! M; where M is a nonsingular matrix. C n 1 X T U! 0; where 0 is the zero vector of dimension k. C3 n 1 X T U N (0; S) ; where C4 where > 0. denotes convergence in distribution and S is a k k matrix. n 1 U T U! ; a) Establish consistency for ^ n under C1-C3. b) Derive S. c) Establish n 1 N (0; V ) under C1-C3. Derive V. d) Let ^U = Y X ^ n and let ^ = n 1 ^U T ^U. Establish consistency of the variance estimator bv n = n 1 X T X 1 n 1 X T ^ I n X n 1 X T X 1 ; under C1-C4.
2 roposed Answer to Question 1 a) First ^ n = X T X 1 X T Y = + X T X 1 X T U For consistency, ^ n = n 1 X T X 1 n 1 X T U Therefore, by C1, C and the continuous mapping theorem! 0; which implies ^ n!. b) We have If E UU T jx = I n, then S = lim n!1 n 1 V ar X T U = lim n!1 n 1 E X T UU T X lim n 1 E X T UU T X = lim n 1 E X T E UU T jx X = lim n 1 E X T X = M n!1 n!1 n!1 c) For asymptotic normality, n 1 = n 1 X T X h 1 n 1 X T U i Therefore, by C1, C3 and Slutzky s Theorem n 1 N 0; M 1 SM 1 recisely, the limit distribution has variance M 1 S M T 1, where M is a symmetric matrix. Because S = M, V = M 1. d) First, observe that the estimator is simpli ed as bv n = ^ n 1 X T X 1 n 1 X T X n 1 X T X 1 = ^ n 1 X T X 1
3 We then have ^U T ^U = h(y X) X i T h i (Y X) X = U T U U T X T X T U + T X T X Taking each term separately, n 1 U T U! ; by C4; by C, result 1 that n 1 U T X! ;! 0, and the continuous mapping theorem; T n 1 X T U; by C, result 1 that! 0, and the continuous mapping theorem; and T n 1 X T X ; by C1, result 1 that! 0, and the continuous mapping theorem. Hence ^! and bv n = ^ n 1 X T X 1! M 1
4 ) Identi cation, Consistency, Variance Estimation, Consistency in Intercept Model You are an analyst who needs to estimate how many ihones are being produced. The structure In each month, J ihones are produced and each phone carries a unique serial number indexed by j, with j f1; ; ; Jg. The data For each month t, where t = 1; ; n, you observe one phone, selected at random from the J phones produced in that month. Thus your data set consists of T independent observations fy t g n. The model Y t = + U t where is a scalar parameter. Let ^ n denote the OLS estimator of. Let U be the n 1 vector of latent errors. a) Interpret. What assumption is needed to identify? b) What is the distribution of U t? Verify that the distribution you derive satis es the identi cation assumption from part a. c) Derive ^ n. Explain why the OLS estimator is also a method-of-moments estimator. Establish that ^ n is a consistent estimator of. d) Establish that ^ n is asymptotically normal and derive the variance of the asymptotic distribution. Construct an estimator of this variance and establish that your variance estimator is consistent. Extra Credit) How could ^ n be used to form an estimator of J? What is the variance of this estimator?
5 roposed Answer to Question a) The coe cient captures the location of the distribution of serial numbers. As we typically model the mean of the dependent variable, in the typical setting is the mean of the distribution of serial numbers. For the case at hand, Y t is a discrete random variable that takes the values 1; ; J, so JX E (Y ) = J 1 j = J j=1 1 J (J + 1) = (J + 1) For identi cation, E (Y ) = + E (U) ; so the identi cation condition is E (U) = 0. b) Because the potential values of U t are U t = Y t ; 1 E (Y ) ; ; J E (Y ) ; each of which is equally likely under the fact that the number is chosen randomly, and so occurs with probability J 1. By direct calculation, the mean of the distribution of U t is E (U t ) = J 1 JX JX (j E (Y )) = J 1 j E (Y ) = 0; j=1 which satis es the identi cation assumption. c) The OLS estimator is de ned as ^ n = arg min b for which the rst-order condition is j=1 (y t b) ; y t = 0 ) ^ n = n 1 y t
6 The OLSE is the sample mean, which is in turn the sample analog of the population mean, and so the OLSE is a method-of-moments estimator. To establish consistency = n 1 y t E (Y t ) Because the data are i.i.d., a law of large numbers implies n 1 y t! E (Y t ) which implies ^ n!. d) To establish asymptotic normality p n = p! n n 1 (y t ) = n 1 u t Because the data are i.i.d., a central limit theorem implies n 1 u t N 0; ; where = V ar (U t ). Hence p n asymptotic distribution is. The estimator is ^ n with ^ n = n 1 N (0; ) and the variance of the y t. For consistency of the estimator of the asymptotic variance, y t = u t ; so n 1 y t = n 1 u t n 1 u t +
7 Because U t is an i.i.d. random variable, n 1 u t! and n 1 u t! 0; and by the consistency of ^ n for,! 0 Hence the continuous mapping theorem implies n 1 u t! 0 and! 0; so ^ n!. extra credit) Because an estimator of J is E (Y t ) = J + 1 ; ^J = ^ n 1 The estimator is consistent, by consistency of ^ n, asymptotically normal, by the asymptotic normality of ^ n, and the asymptotic variance of the estimator is 4 =n (which can be consistently estimated with use of ^ n).
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