Applied EconometricsWilliam Greene Department of Economics Stern School of BusinessApplied Econometrics12. Asymptotics for the Least Squares Estimator in the Classical Regression ModelSettingThe least squares estimator is (X X)-1Xy = (X X)-1ixiyi = + (X X)-1ixii So, it is a constant vector plus a sum of random variables. Our finite sample results established the behavior of the sum according to the rules of statistics. The question for the present is how does this sum of random variables behave in large samples? Well Behaved RegressorsA crucial assumption: Convergence of the moment matrix XX/n to a positive definite matrix of finite elements, QWhat kind of data will satisfy this assumption? What wont? Does stochastic vs. nonstochastic matter? Various conditions for “well behaved X”Probability LimitMean Square ConvergenceEb|X= for any X. Varb|X0 for any specific X b converges in mean square to Probability LimitCrucial Assumption of the ModelConsistency of s2Asymptotic DistributionAsymptoticsAsymptotic DistributionsoFinding the asymptotic distributionob in probability. How to describe the distribution?nHas no limiting distributionoVariance 0; it is O(1/n)oStabilize the variance? Varn b 2Q-1 is O(1)oBut, En b= n which divergesnn (b - ) a random variable with finite mean and variance. (stabilizing transformation)nb apx. +1/ n times that random variableLimiting Distributionn (b - )= n (XX)-1X= n (XX/n)-1(X/n) Limiting behavior is the same as that of n Q-1(X/n) Q is a fixed matrix. Behavior depends on the random vector n (X/n)Limiting NormalityAsymptotic DistributionAsymptotic PropertiesoProbability Limit and ConsistencyoAsymptotic VarianceoAsymptotic DistributionRoot n ConsistencyoHow fast does b ?oAsy.Varb =2/n Q-1 is O(1/n)nConvergence is at the rate of 1/n nn b has variance of O(1)oIs there any other kind of convergence?nx1,xn = a sample from exponential population; min has variance O(1/n2). This is n convergentnCertain nonparametric estimators have variances that are O(1/n2/3). Less than root n convergent.Asymptotic ResultsoDistribution of b does not depend on normality of oEstimator of the asymptotic variance (2/n)Q-1 is (s2/n) (XX/n)-1. (Degrees of freedom corrections are irrelevant but conventional.)oSlutsky theorem and the delta method apply to functions of b.Test StatisticsWe have established the asymptotic distribution of b. We now turn to the construction of test statistics. In particular, we based tests on the Wald statistic FJ,n-K = (1/J)(Rb - q)R s2(XX)-1R-1(Rb - q) This is the usual test statistic for testing linear hypotheses in the linear regression model, distributed exactly as F if the distirbances are normally distributed. We now obtain some general results that will let us construct test statistics in more general situations. Wald StatisticsGeneral approach to the derivation based on a univariate distribution (just to get started). A. Core result: Square of a standard normal variable chi-squared with 1 degree of freedom. Suppose z N0,2, i.e., variance not 1. Then (z/)2 satisfies A. Now, suppose zN,2. Then (z - )/2 is chi-squared with 1 degree of freedom. This is the normalized distance between z and , where distance is measured in standard deviation units. Suppose zn is not exactly normally distributed, but (1) Ezn = , (2) Varzn = 2, (3) the limiting distribution of zn is normal. Then by our earlier results, (zn - )/ N0,1, though again, this is a limiting distribution, not the exact distribution in a finite sample.ExtensionsIf the preceding holds, thenn2 = (zn - )/2 N0,12, or 21. Again, a limiting result, not an exact one. Suppose is not a known quantity, and we substitute for a consistent estimator of , say sn. plim sn = . What about the behavior of the “empirical counterpart,” tn = (zn - )/sn? Because plim sn = , the large sample behavior of this statistic will be the same as that of the original statistic using instead of sn. Therefore, under our assumptions, tn2 = (zn - )/sn2 converges to chi-squared 1, just like n2 . tn and n converge to the same random variable.Full Rank Quadratic formA crucial distributional result (exact): If the random vector x has a K-variate normal distribution with mean vector and covariance matrix , then the random variable W = (x - )-1(x - ) has a chi-squared distribution with K degrees of freedom. Proof of Full Rank Q-F ResultProof: (Short, but very important that you understand and are comfortable with all parts. Details appear in Section 3.10.5 of your text.) Requires definition of a square root matrix: 1/2 is a matrix such that 1/2 1/2 = . Then, V = (1/2)-1 is the inverse square root, such that V V = -1/2 -1/2 = -1. Let z = (x - ). Then z has mean 0, covariance matrix , and