SOLUTIONS TO PROBLEMS

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1 Z Z T Z conometric Theory, 1, 005, Printed in the United States of America+ DI: S SLUTINS T PRBLMS Solutions to Problems Posed in Volume 0(1) and 0() A Hausman Test Based on the Difference between Fixed ffects Two-Stage Least Squares and rror Components Two-Stage Least Squares 1 Solution B+H+ Baltagi ~the poser of the problem! Texas A&M University ~a! Let yi 1 Qy 1 and Z 1 Q, where Q is the fixed effects ~F! transformation given by Q I NT P with P I N i T i ' T 0T+ Baltagi ~1981! showed that dd 1, FSLS ~ Z! 1 Z yi 1 d 1 ~ Z! 1 Z u 1, where P X X~X ' X! 1 X ' with plim~ dd 1, FSLS! d 1 and the asymptotic variance is given by var~ dd 1, FSLS! s n11 ~ Z! 1 + Also, Baltagi ~1981! showed that dz 1, CSLS Z Z 1 Z yi 1 Z yt 1 d 1 X Z 1 Z u 1 ' P F ' P XP u 1, where y 1 Py 1, Z 1 P, and Ts m11 s n11 + Under H 0, plim~ dz 1, CSLS! d 1 and the asymptotic variance of d 1, CSLS is given by var~ dz 1, Z 0s n11 Z ' 1 P XP 0s 111 # 1 + Under H 0, both dd 1, FSLS and d 1, CSLS are consistent, with plim q[ 1 0+ Using the expressions for d 1, CSLS d 1 and dd 1, FSLS d 1, one gets the asymptotic covariance 005 Cambridge University Press $

2 F F P F 484 SLUTINS T PRBLMS cov~ dd 1, FSLS, dz 1, CSLS! plim~ Z! 1 Z ' 1 P F Z Z plim~ Z! 1 plim~ Z! 1 plim~ Z Z X u 1 u 1 ' P XF 1 Z u 1 u u 1 Z u 1 u ' 1 P XP 1 ' P XP Z Z! s 1 n Z # Z Z Z var~ dz 1, CSLS!+ 1 1 This makes use of the following expression: plim~ X F ' ' u 1 u 1 X0n! X ' plim~u 1 u ' 1 0n! XF X ' QV 11 QXF s n11 ~ X F ' X F!, where n NT and V 11 s n11 Q s 111 P+ Also, plim~ X F ' ' u 1 u 1 X0n! X F ' plim~u 1 u ' 1 0n! XP X F ' QV 11 PXP 0, because PQ 0+ Hence cov~ q[ 1, dz 1, CSLS! cov~ dd 1, FSLS, dz 1, CSLS! var~ dz 1, CSLS! 0+ ~b! Using the fact that dd 1, FSLS dz 1, CSLS q[ 1, one gets var~ dd 1, FSLS! var~ dz 1, CSLS! var~ q[ 1! because cov~ q[ 1, dz 1, CSLS! 0+ Therefore var~ q[ 1! var~ dd 1, FSLS! var~ dz 1, CSLS!+ NT 1+ An excellent solution has been proposed independently by Shiferaw Gurmu, Georgia State University+ RFRNC Baltagi, B+H+ ~1981! Simultaneous equations with error components+ Journal of conometrics 17,

3 A Range quality for Block Matrices with rthogonal Projectors 1 Solution Hans Joachim Werner University of Bonn, Germany This solution offers additional insights into the theory of block-tridiagonal Toeplitz matrices+ Block Toeplitz matrices have constant blocks on each block diagonal parallel to the block main diagonal+ A block partitioned matrix is said to be block-tridiagonal if the nonzero blocks occur only on the block subdiagonal, the block main diagonal, and the block superdiagonal+ Block-tridiagonal Toeplitz matrices are particularly nice in that they are inexpensive to investigate+ ur first observation on such particular block Toeplitz matrices is easy to check, and its proof is therefore left to the reader+ THRM 1+ For any given pair of real m k matrices A and B, consider the following block partitioned matrices: AAt BBt ABt BA t AA t BB t L M : L L AB t and BA t SLUTINS T PRBLMS 485 AA t BB tnn N :B A B L L A B, An~n1! where M consists of n n blocks, whereas N has n ~n 1! blocks. Then M NN t, thus particularly showing that the symmetric block-tridiagonal Toeplitz matrix M is a nonnegative definite matrix satisfying R~M! R~N! and N~M! N~N t!, with R~{!, N ~{!, and ~{! t denoting the range (column space), the null space, and the transpose, respectively, of the matrix ~{!. By assuming a disjointness property between the matrices A and B, Theorem 1 can even be strengthened as follows+ THRM + Let A and B be real matrices of the same order satisfying ~A t N~B t!! R~B t! $0% or ~B t N~A t!! R~A t! $0%. LetMandNbe defined in terms of A and B as in Theorem 1. Then

4 486 SLUTINS T PRBLMS N~M! N~N t! N~W! or, equivalently, R~M! R~N! R~W!, where W : AAt BBt L + AA t BB tnn Proof. Theorem 1 tells us that N~M! N~N t!+ Now, let x ~x i! i1,+++,n N~N t! be partitioned in accordance with N t + Then B t x 1 0, A t x n 0, and, for i 1,+++,n 1, A t x i B t x i1 0+ Clearly, if ~A t N~B t!! R~B t! $0% or if ~B t N~A t!! R~A t! $0%, then this set of n 1 equations is satisfied if and only x i N~A t! N~B t! for i 1,,+++,n+ Hence, because of N~AA t BB t! N~A t! N~B t!, N~M! N~W! or, equivalently, as claimed R~M! R~W!; recall that M and W are symmetric matrices and observe that ~N~G!! R~G t! holds for any matrix G, where ~{! indicates the orthogonal complement of ~{! with respect to the usual inner product+ According to Werner ~1986!, two real m k matrices, say, A and B, are said to be weakly complementary to each other if R~A t! R~B t! $0%+ Such matrices are of great importance, for instance, in statistics ~cf+ Werner, 1985b, 1987!+ If A and B are such a pair of weakly complementary matrices, then trivially ~A t N~B t!! R~B t! $0%, and so, by virtue of Theorem, R~M! R~W!+ Next, let A and B be symmetric matrices+ Moreover, let at least one of these matrices, say, A, be nonnegative definite+ Then it is well known that ~AN~B t!! R~B! $0% ~cf+ Werner, 1985a, Lemma 3+; or the theorem in Werner, 003!+ Hence, because of Theorem, again R~M! R~W!+ Finally, as in Tian s posed problem, let A and B be orthogonal projectors+ Note that a real matrix P is an orthogonal projector if and only if P P t and P P, that is, if and only if P is symmetric and idempotent+ Then P P PP t, thus showing that P is also nonnegative definite+ Consequently, by virtue of the previous case, again R~M! R~W!+ In other words, Tian s claim follows as a very special case from our more general result stated in Theorem + Throughout this text we considered only real matrices+ Needless to say, all results can be generalized to complex matrices just by replacing symmetric matrices and transposition by Hermitian matrices and conjugate transposition, respectively+ We conclude with mentioning that Theorem also generalizes IMAG Problem 31-3 by Tian ~003!+ NT 1+ xcellent solutions have been proposed independently by Geert Dhaene ~Katholieke Universiteit, Leuven, Belgium!, Dietmar Bauer ~Technische Universität, Wien, Austria!, and Yongge Tian ~Queens University, Canada!, the poser of the problem+

5 SLUTINS T PRBLMS 487 RFRNCS Tian, Y+ ~003! A range equality for block matrices+ IMAG, The Bulletin of the International Linear Algebra Society 31, 44+ Werner, H+J+ ~1985a! More on BLU estimation in regression models with possibly singular covariances+ Linear Algebra and Its Applications, 67, Werner, H+J+ ~1985b! Weak complementarity and missing observations+ Methods of perations Research 50, Werner, H+J+ ~1986! Generalized inversion and weak bi-complementarity+ Linear and Multilinear Algebra 19, Werner, H+J+ ~1987! C+ R+ Rao s IPM method: A geometric approach+ In M+L+ Puri, J+P+ Vilaplana, &W+Wertz ~eds+!, New Perspectives in Theoretical and Applied Statistics, pp Wiley+ Werner, H+J+ ~003! Product of two Hermitian nonnegative definite matrices+ Solution IMAG, The Bulletin of the International Linear Algebra Society 30, Characterizations of Hermitian Projectors Luc Lauwers ~one of the posers of the problem! Katholieke Universiteit, Leuven, Belgium Let P in C nn be a projector with P H its conjugate transpose+ ~a! P H P is idempotent n P is Hermitian+ By contradiction+ Let P be an oblique projector of rank r+ Choose an orthonormal set of r ~resp+ s n r! eigenvectors with eigenvalue 1 ~resp+ 0!+ These n eigenvectors form a basis+ The matrix C that collects these eigenvectors as its columns satisfies P CJ r C 1 and C H C I r T T H I s, with J r the diagonal matrix with on the diagonal r ones and n r zeros ~the ones are at the top left!; and with I r ~resp+ I s! the identity matrix of dimension r ~resp+ s!+ The matrix T C rs and is nonzero ~as the projector P is oblique!+ In this notation we have P H P ~C 1! H J r C H CJ r C 1 ~C 1! H J r C 1 + Now, the matrix P H P is idempotent if and only if the matrix D C 1 P H PC is idempotent+ Note that D C 1 ~C 1! H J r C 1 C C 1 ~C 1! H J r ~C H C! 1 J r + Hence, D is of rank r and is of the form 0 F0 with in C rr and F in C sr satisfying I r TF I r, T H I s F 0, and hence ~I r TT H!I r +

6 488 SLUTINS T PRBLMS Thus, is nonsingular+ Now, suppose that D is idempotent+ Then, is idempotent+ Hence, I r and TT H 0+ This conflicts with T being nonzero+ Conclude that D, and therefore also P H P, is not idempotent+ ~a! P is Hermitian n P H P is idempotent+ If P P H, then the composition P H P coincides with P and is idempotent+ ~b! P is Hermitian m PP H is idempotent+ The projector P is Hermitian if and only if the projector P H is Hermitian+ Apply condition ~a! upon P H and obtain that the projector P is Hermitian if and only if the composition ~P H! H ~P H! PP H is idempotent+