ReferencesFIRST MIDTERM EXAM ECON 7801 SPRING 2001ECONOMICS DEPARTMENT, UNIVERSITY OF UTAHProblem 1. 2 points Let y be a n-vector. (It may be a vector of observations of arandom variable y, but it does not matter how the yiwere obtained.) Prove that thescalar α which minimizes the sum(1) (y1− α)2+ (y2− α)2+ · · · + (yn− α)2=X(yi− α)2is the arithmetic mean α = ¯y.Answer. Use (??). Problem 2.• a. 2 points Verify that the matrix D = I −1nιι>is symmetric and idempotent.Date of exam Tuesday, February 20, 9–10:30 am.12 U OF U ECONOMICS• b. 1 point Compute the trace tr D.Answer. tr D = n − 1. One can see this either by writing down the matrix element by element, oruse the linearity of the trace plus the rule that tr(AB) = tr(BA). tr I = n and tr(ιι>) = tr(ι>ι) =tr n = n. • c. 1 point For an y vector of observations y compute Dy.Answer. Element by element one can write(2) D y =1 −1n−1n. . . −1n−1n1 −1n. . . −1n.........−1n−1n. . . 1 −1ny1y2...yn=y1− ¯yy2− ¯y...yn− ¯yThere is also a more elegant matrix theoretical proof available • d. 1 point Is there a vector a 6= o for which Da = o? If so, give an example ofsuch a vector.Answer. ι is, up to a scalar factor, the only nonzero vector with Dι = o. • e. 1 point Show that the sample variance of a vector of observations y can bewritten in matrix notation as(3) The sample variance of y is1nX(yi− ¯y)2=1ny>DyFIRST MIDTERM EXAM ECON 7801 SPRING 2001 3Answer. Let’s get rid of the factor1nwhich appears on both sides: we have to show that(4)X(yi− ¯y)2= y>D y = y>D>D yThis is the squared length of the vector Dy which we computed in part c. Problem 3. 1 point Compute the matrix product1 2 40 3 34 02 13 8Answer.1 2 40 3 3"4 02 13 8#=1 · 4 + 2 · 2 + 4 · 3 1 · 0 + 2 · 1 + 4 · 80 · 4 + 3 · 2 + 3 · 3 0 · 0 + 3 · 1 + 3 · 8=20 3415 27 Problem 4. 2 points Assume that X has full column rank. Show that ˆε = M ywhere M = I − X(X>X)−1X>. Show that M is symmetric and idempotent.Answer. By definition, ˆε = y − Xˆβ = y − X(X>X)−1Xy =I − X(X>X)−1Xy. Idempotent,i.e. M M = M :MM =I − X(X>X)−1X>I − X(X>X)−1X>= I − X(X>X)−1X>− X(X>X)−1X>+ X(X>X)−1X>X(X>X)−1X>= I − 2X(X>X)−1X>+ X(X>X)−1X>= I − X(X>X)−1X>= M(5)4 U OF U ECONOMICS Problem 5. 2 points We are in the multiple regression model y = Xβ + εεε withintercept, i.e., X is such that there is a vector a with ι = Xa. Define the row vector¯x>=1nι>X, i.e., it has as its jth component the sample mean of the jth independentvariable. Using the normal equations X>y = X>Xˆβ, show that ¯y =¯x>ˆβ (i.e., theregression plane goes through the center of gravity of all data points).Answer. Premultiply the normal equation by a>to get ι>y − ι>Xˆβ = 0. Premultiply by 1/n toget the result. Problem 6. 2 points Let θ be a vector of possibly random parameters, andˆθ anestimator of θ. Show that(6) MSE[ˆθ; θ] =V[ˆθ − θ] + (E[ˆθ − θ])(E[ˆθ − θ])>.Don’t assume the scalar result but make a proof that is good for vectors and scalars.Answer. For any random vector x followsE[xx>] =E(x −E[x] +E[x])(x −E[x] +E[x])> =E(x −E[x])(x −E[x])> −E(x −E[x])E[x]> −EE[x](x −E[x])> +EE[x]E[x]> =V[x] − O − O +E[x]E[x]>.Setting x =ˆθ − θ the statement follows. FIRST MIDTERM EXAM ECON 7801 SPRING 2001 5Problem 7. Consider two very simple-minded estimators of the unknown nonran-dom parameter vector φ =hφ1φ2i. Neither of these estimators depends on any ob-servations, they are constants. The first estimator isˆφ = [1111], and the second is˜φ = [128].• a. 2 points Compute the MSE-matrices of these t wo estimators if the true valueof the parameter vector is φ = [1010]. For which estimator is the trace of the MSEmatrix smaller?Answer.ˆφ has smaller trace of the MSE -matrix.ˆφ − φ =11MSE [ˆφ; φ ] =E[(ˆφ − φ)(ˆφ − φ)>]=E[111 1 ] =E[1 11 1] =1 11 1˜φ − φ =2−2MSE [˜φ; φ ] =4 −4−4 46 U OF U ECONOMICSNote that both MSE-matric es are singu lar, i.e., b ot h estimators allow an error-free look at certainlinear combinations of the parameter vector. • b. 1 point Give two vectors g = [g1g2] and h =h1h2 satisfying MSE[g>ˆφ; g>φ] <MSE[g>˜φ; g>φ] and MSE[h>ˆφ; h>φ] > MSE[h>˜φ; h>φ] (g and h are not unique;there are many possibilities).Answer. With g =1−1 and h =11 for instance we get g>ˆφ − g>φ = 0, g>˜φ − g>φ = 4 ,h>ˆφ; h>φ = 2, h>˜φ; h>φ = 0, therefore MSE[g>ˆφ; g>φ] = 0, MSE[g>˜φ; g>φ] = 16, MSE[h>ˆφ; h>φ] =4, MSE[h>˜φ; h>φ] = 0. An alternative way to compute this is e.g.MSE [h>˜φ; h>φ] =1 −1 4 −4−4 41−1= 16 • c. 1 point Show that neither MSE[ˆφ; φ]−MSE[˜φ; φ] nor MSE[˜φ; φ]−MSE[ˆφ; φ]is a nonnegative definite matrix. Hint: you are allowed to use the mathematical factthat if a matrix is nonnegative definite, then its determinant is nonnegative.Answer.(7) MSE [˜φ; φ] − MS E[ˆφ; φ] =3 −5−5 3Its determinant is negative, and the determinant of its negative is also negative. FIRST MIDTERM EXAM ECON 7801 SPRING 2001 7Problem 8.• a. 2 points Show that the sampling error of the OLS estimator is(8)ˆβ − β = (X>X)−1X>εεε• b. 2 points Derive from this thatˆβ is unbiased and that its MSE-matrix is(9) MSE[ˆβ; β] = σ2(X>X)−1Problem 9.• a. 2 points Show thatSSE = εεε>Mεεε where M = I − X(X>X)−1X>(10)Answer. SSE = ˆε>ˆε, where ˆε = y − Xˆβ = y − X(X>X)−1X>y = M y where M = I −X(X>X)−1X>. Fro m M X = O follows ˆε = M (Xβ + εεε) = Mεεε. Since M is idempotent andsymmetric, it follows ˆε>ˆε = εεε>Mεεε. • b. 1 point Is SSE observed? Is εεε observed? Is M observed?• c. 3 points Under the usual assumption that X has full column rank, show that(11) E[SSE] = σ2(n − k)Answer. E[ˆε>ˆε] = E[tr εεε>Mεεε] = E[tr Mεεεεεε>] = σ2tr M = σ2tr(I − X(X>X)−1X>) = σ2(n −tr(X>X)−1X>X) = σ2(n − k). 8 U OF U ECONOMICSProblem 10. The prediction problem in the Ordinary Least Squares model can beformulated as follows:(12)yy0=XX0β +εεεεεε0E[εεεεεε0] =ooV[εεεεεε0] = σ2I OO I.X and X0are known, y is observed, y0is