计量经济分析(第六版)答案 Midterm05-answers.docx
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1、NYUtSTERNDepartment of EconomicsNEW YORK UNIVERSITY LEONARD N. STERN SCHOOL OF BUSINESSECONOMETRICS IFall 2005 - Tuesday, Thursday, 1-00 - 2-20Professor William GreeneOffice: KMC 7-78Office Hours: OpenPhone: 212.998.0876Home page:www.stern.nyu.edu/-wgreeneEmail: wgreenestern.nyu.eduURL for course we
2、b page:www.stern.nyu.edu/-wgreene/Econometrics/Econometrics.htmMidtermIn the classical regression model,yi = Xi邛i + Xi2、2 + /, i - L E8i|X = 0, Var8i|X=o2;Xi is Ki variables and X2 is K2 variables. There are two possible estimators of pi, the first Ki coefficients in the long regression, of y on X)a
3、nd X2 and the Ki coefficients in the short regression of y on X). Let X = Xi,X2j. We will assume that plim(l/n)XfXJ = Q, a positive definite matrix.a. 5 points Assume that plim(l/n)X/X2丰 0. Is either estimator unbiased? Is either estimator consistent?The long regression estimator is unbiased and con
4、sistent in all cases. We showed unbiasedness early on - it doesnt depend on X/X2. We need plim(l/n)Xr = 0 for consistency, and we have plim(l/n)XrX = Q in the problem.The short regression is biased and inconsistent. We showed in class, when you leave variables out of a regression, the estimator isbi
5、 = pl + (XiXi 尸 XJX2 + (XiXi)XigAs long as X/X? is not zero, the short estimator is biased. As for consistency, even though plim(l/n)Xze would imply plim(l/n)X/e = 0, the middle term is (divide then multiply by n) is not going to go away. The second term doesnt go to zero, so the short regression es
6、timator is inconsistent.b. 5 points Assume that plim(l/n)X/X2 = 0. Is either estimator unbiased? Is either estimator consistent?Using the results above, the long regression estimator is still unbiased and consistent.The short regression is still biased. We didnt assume that X/X2 = 0. But the bias go
7、es away as X/X? goes to zero, so it is consistent.c. 5 points Explain the difference between consistency and unbiasedness. Does either imply the other? Explain.Done in detail in class and on the practice exam.d. 5 points Suppose the assumption in a. is correct. The estimator we will use is the follo
8、wing: We will compute the long regression. F is the conventional F statistic for testing the null hypothesis that 02 is zero. If F 2, we will use the long estimator. If F 20 based on a t statistic this low.4. Suppose the conditional distribution of y|x is Poissonf(y|x) = exp(-Xy) yk/y!where X = a +
9、px. We are interested in estimating a and & The Poisson distribution has the property that the mean equals the variance, and both equal X. I propose the following two estimators of a and p:l. ) Linear regression of y on (l,x)m. ) Linear regression of (y y)2 on (l,x).n. 10 points is the first estimat
10、or unbiased? Consistent? Justify your answer.Yes. Its a linear regression model.The model is a linear regression model. It may look a little weird, but the crucial assumption is Efy|x = a + px. Unbiasedness and consistency dont relate the the variance. All our usual results can be used here. This is
11、 just one way a linear model can arise.o. 5 points Is the second estimator unbiased? Consistent? Justify your answer.Maybe. To be discussed in class. This one is hard. Since the conditional variance of y is a + px also, one might think you can use the expected squared deviation to form the regressio
12、n. The trouble with the proposal is that your left hand variable is the unconditional variance, basedon y-bar, not Ey|x. So, perhaps not. Well discuss it in class. In terms of grading, however, any clear thought about what might be the right answer is worth 5 points.5. 15 points In a recent (real) e
13、lection case in Pennsylvania, it was alleged that theabsentee ballots in a certain state senators race had been tampered with. Orley Ashenfelter (the same Orley Ashenfelter who studied twins in Twinsburg with Alan Krueger) was asked to analyze the data to help the judge decide what to do with the el
14、ection results. On the basis of a regression of 21 previous elections absentee ballots totals on the corresponding machine ballot totals, Ashenfelter formed a prediction interval for this absentee ballot total and determined that it looked like an outlier (statistically outside the expected range).
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