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1、ECOM023 Econometrics Exercise Sheet 1: Solutions R.G. Pierse Q1. (a) q0q = ? 241 ? 2 4 2 4 1 3 5= 2 ? 2 + 4 ? 4 + 1 ? 1 = 21 (b) qq0= 2 4 2 4 1 3 5 ? 241 ? = 2 4 2 ? 22 ? 42 ? 1 4 ? 24 ? 44 ? 1 1 ? 21 ? 41 ? 1 3 5 = 2 4 482 8164 241 3 5 (c) tr(qq0) = 4 + 16 + 1 = 21 (d) rank(q) = 1 since q is of dim
2、ension 3?1 and so has only one linearly independent column. Therefore rank(qq0) = 1 since rank(AB) 6 min(rank(A);rank(B): Multiplying one matrix by another cannot increase the rank of a ma- trix. Q2. (a) (i) M = In? X(X0X)?1X0 M0=I0n? X00(X0X)?1 0X0 =In? X(X0X)?1X0 since Inand (X0X)?1are both symmet
3、ric and A00= A. (ii) MM=(In? X(X0X)?1X0)(In? X(X0X)?1X0) =InIn? InX(X0X)?1X0? X(X0X)?1X0In +X(X0X)?1X0X(X0X)?1X0 =In? 2X(X0X)?1X0+ X(X0X)?1X0= M: (b) tr(M)=tr(In? X(X0X)?1X0) =tr(In) ? tr(X(X0X)?1X0) =tr(In) ? tr(X0X(X0X)?1) =tr(In) ? tr(Ik) =n ? k: 2 Q3. (a)To show: AA = A. AA= ? 0:50:5 0:50:5 ? 0:
4、50:5 0:50:5 ? = ? 0:5 ? 0:5 + 0:5 ? 0:50:5 ? 0:5 + 0:5 ? 0:5 0:5 ? 0:5 + 0:5 ? 0:50:5 ? 0:5 + 0:5 ? 0:5 ? = ? 0:50:5 0:50:5 ? = A (b) tr(A) = a11+ a22= 0:5 + 0:5 = 1 ? ? ? ? ? a11a12 a21a22 ? ? ? ? = a11a22? a12a21= 0:5 ? 0:5 ? 0:5 ? 0:5 = 0 (c) The eigenvalues of A are found by solving the equation
5、 jA ? ?Ij = 0 for ?. jA ? ?Ij= ? ? ? ? ? a11? ?a12 a21a22? ? ? ? ? ? =(a11? ?)(a22? ?) ? a12a21= 0: In this case (0:5 ? ?)(0:5 ? ?) ? 0:5 ? 0:5 =0:25 ? ? + ?2? 0:25 =?2? ? = 0: Clearly there are two solutions to this equation: ? = 1 and ? = 0 so these are the two eigenvalues of A: ?1and ?2. Note tha
6、t we have that ?1+ ?2= 1 = trA 3 and ?1?2= 0 = jAj as expected. These are properties of any idempotent matrix which always has eigenvalues that are either 0 or 1 in magnitude. Q4. (a) f(x) x = x x0Ax + b0x + x0b + c: Taking the elements term by term and using results from lecture notes we have x x0A
7、x=Ax + A0x =2Ax since A is symmetric, and x b0x = (b0)0= b: Considering the third term, note that x0b is a scalar so that x0b = (x0b)0= b0x and so x x0b = x b0x = b: Therefore, putting all the terms together, we have f(x) x = 2Ax + 2b:(1) (b) Solving f(x) x = 2Ax + 2b = 0 4 and using the result that
8、 A is nonsingular, we have x?= ?A?1b: (c) Dierentiating (1) again with respect to x gives 2f(x) xx0 = x 2Ax + 2b =2A0= 2A: If the matrix A is positive-denite, then the solution x?is a minimum (the function is bowl shaped). If the matrix A is negative-denite, then the solution x?is a maximum (the fun
9、ction is in the shape of an inverted bowl). Otherwise, the solution x?is a saddle point. 5 ECOM023 Econometrics Exercise Sheet 2: Solutions R.G. Pierse Q1.(a) = Dependent Variable: LC Method: Least Squares Date: 10/24/02Time: 14:32 Sample(adjusted): 1949 1985 Included observations: 37 after adjustin
10、g endpoints = Variable Coefficient Std. Error t-Statistic Prob. = C1.2678440.07861816.12674 0.0000 LY0.8772520.007317119.8891 0.0000 INF-0.1264840.045418-2.784851 0.0087 = R-squared0.998401Mean dependent var10.88271 Adjusted R-squared 0.998307S.D. dependent var0.258500 S.E. of regression 0.010636Aka
11、ike info criterion-6.171579 Sum squared resid0.003846Schwarz criterion-6.040964 Log likelihood117.1742F-statistic10615.94 Durbin-Watson stat 0.952491Prob(F-statistic)0.000000 = Q1 (b) (i)The estimate of the income elasticity is 0:877252. A 95% condence interval for the income elasticity is given by
12、0:877252 ? 2:03 ? 0:007317 = (0:8624;0:8921) where 2:03 is the 5% critical value for a two-tail test (2.5% in each tail) of the T-distribution with 34 degrees of freedom: t34. (ii)A t-statistic for the hypothesis that ?2= 1 is ? ? ? ? ?2? 1 b ?2 ? ? ? ? = ? ? ? ? 0:877252 ? 1 0:007317 ? ? ? ? = 16:7
13、757: The 5% critical value for a one-tail test of the T-distribution t34is 1:69 so the hypothesis is strongly rejected. The hypothesis that ?2= 1 is the hypothesis of a unit income elasticity of consumption. (iii)An F test for the joint hypothesis that ?2= 0 and ?3= 0 can be constructed using the Wa
14、ld coe cient restrictions option in the coe cient tests menu in EViews. The restrictions are dened as C(2) = 0 , C(3) = 0 The resulting EViews output is = Wald Test: Equation: Untitled = Test Statistic ValuedfProbability = F-statistic10615.94(2, 34)0.0000 Chi-square21231.8920.0000 = The test statistic is reported in two forms: as an F-statistic with degrees of freedom 2 and 34 and value 10615:94 and as a Chi-squared statistic with degrees of freedom 2 and value 21231:89. (The s
