GLS Efficiency - 經濟

By Quintina
at 2010-10-27T00:39
at 2010-10-27T00:39
Table of Contents
※ 引述《washburn (Back to school.)》之銘言:
: Let Omega^{-1} = P'P, in which P is invertible.
: Let Z = P'^{-1} X.
: var(beta_GLS)^{-1} - var(beta_OLS)^{-1}
: = X' Omega^{-1} X - X'X (X' Omega X)^{-1} X'X
: = X' P' [I - P'^{-1} X (X' P^{-1} P'^{-1} X)^{-1} X' P^{-1}] P X
: = X' P' [I - Z (Z'Z)^{-1} Z'] P X, which is positive semidefinite.
: Q.E.D.
感謝解惑,不過對於 I - Z (Z'Z)^{-1} Z' 為何是 p.s.d. 仍不太清楚
令 A = I - Z (Z'Z)^{-1} Z',則我可以證明 A 是 idempotent
對於一個 idempotent matrix A 與任意一個非零向量 x,
x'Ax = (Ax)'(Ax) 只是 Ax 所有元素的平方加總而已,
所以 x'Ax 恆非負,也就是 A 至少是 p.s.d.(通常會是 p.d.)
這樣應該正確吧?不知有沒有問題?
覺得一個看似應該很直觀的證明卻不是那樣容易......
--
: Let Omega^{-1} = P'P, in which P is invertible.
: Let Z = P'^{-1} X.
: var(beta_GLS)^{-1} - var(beta_OLS)^{-1}
: = X' Omega^{-1} X - X'X (X' Omega X)^{-1} X'X
: = X' P' [I - P'^{-1} X (X' P^{-1} P'^{-1} X)^{-1} X' P^{-1}] P X
: = X' P' [I - Z (Z'Z)^{-1} Z'] P X, which is positive semidefinite.
: Q.E.D.
感謝解惑,不過對於 I - Z (Z'Z)^{-1} Z' 為何是 p.s.d. 仍不太清楚
令 A = I - Z (Z'Z)^{-1} Z',則我可以證明 A 是 idempotent
對於一個 idempotent matrix A 與任意一個非零向量 x,
x'Ax = (Ax)'(Ax) 只是 Ax 所有元素的平方加總而已,
所以 x'Ax 恆非負,也就是 A 至少是 p.s.d.(通常會是 p.d.)
這樣應該正確吧?不知有沒有問題?
覺得一個看似應該很直觀的證明卻不是那樣容易......
--
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經濟
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at 2010-10-28T08:43
at 2010-10-28T08:43

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at 2010-11-01T10:34
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