9
Sep 18

Applications of the diagonal representation II

Applications of the diagonal representation II

4. Square root of a matrix

Definition 1. For a symmetric matrix with non-negative eigenvalues the square root is defined by

(1) A^{1/2}=Udiag[\sqrt{\lambda_1},...,\sqrt{\lambda_n}]U^{-1}.

Exercise 1. (1) is symmetric and satisfies (A^{1/2})^2=A.

Proof. By properties of orthogonal matrices

(A^{1/2})^T=(U^{-1})^Tdiag[\sqrt{\lambda_1},...,\sqrt{\lambda_n}]U^T=A^{1/2}, (A^{1/2})^2=Udiag[\sqrt{\lambda_1},...,\sqrt{\lambda_n}]U^{-1}Udiag[\sqrt{\lambda_1},...,\sqrt{\lambda_n}]U^{-1} =Udiag[\lambda_1,...,\lambda_n]U^{-1}=A.

5. Generalized least squares estimator

The error term e in the multiple regression y=X\beta +e under homoscedasticity and in absence of autocorrelation satisfies

(2) V(e)=\sigma^2I, where \sigma^2 is some positive number.

The OLS estimator in this situation is given by

(3) \hat{\beta}=(X^TX)^{-1}X^Ty.

Now consider a more general case V(e)=\Omega.

Exercise 2. The variance matrix V(e)=\Omega is always symmetric and non-negative.

Proof. V(e)^T=[E(e-Ee)(e-Ee)^T]^T=V(e),

x^TV(e)x=Ex^T(e-Ee)(e-Ee)^Tx=E\|(e-Ee)^Tx\|^2\geq 0.

Exercise 3. Let's assume that \Omega is positive. Show that \Omega^{-1/2} is symmetric and satisfies (\Omega^{-1/2})^2=\Omega^{-1}.

Proof. By Exercise 1 the eigenvalues of \Omega are positive. Hence its inverse \Omega^{-1} exists and is given by \Omega^{-1}=U\Omega_U^{-1}U^T where \Omega_U^{-1}=diag[\lambda_1^{-1},...,\lambda_n^{-1}]. It is symmetric as an inverse of a symmetric matrix. It remains to apply Exercise 1 to A=\Omega^{-1/2}.

Exercise 4. Find the variance of u=\Omega^{-1/2}e.

Solution. Using the definition of variance of a vector

V(u)=E(u-Eu)(u-Eu)^T=\Omega^{-1/2}V(e)(\Omega^{-1/2})^T=\Omega^{-1/2}\Omega\Omega^{-1/2}=I.

Exercise 4 suggests how to transform y=X\beta +e to satisfy (2). In the equation \Omega^{-1/2}y=\Omega^{-1/2}X\beta +\Omega^{-1/2}e the error u=\Omega^{-1/2}e satisfies the assumption under which (2) is applicable. Let \tilde{y}=\Omega^{-1/2}y, \tilde{X}=\Omega^{-1/2}X. Then we have \tilde{y}=\tilde{X}\beta +u and from (2) \hat{\beta}=(\tilde{X}^T\tilde{X})^{-1}\tilde{X}^T\tilde{y}. Since \tilde{X}^T=X^T\Omega^{-1/2}, this can be written as

\hat{\beta}=(X^T\Omega^{-1/2}\Omega^{-1/2}X)^{-1}X^T\Omega^{-1/2}\Omega^{-1/2}y=(X^T\Omega^{-1}X)^{-1}X^T\Omega^{-1}y.

This is Aitken's Generalized least squares estimator.

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