Solution to Review Question by Qiang Gao, updated at May 8, 2017
Chapter 1 Finite-Sample Properties of OLS Section 3 Finite-Sample Properties of OLS ...
Prove part (d) of Proposition 1.1, under Assumptions 1.1—1.4, $$ \mathrm{Cov} ( \mathbf{b}, \mathbf{e} \mid \mathbf{X} ) = \mathbf{0} $$, where $$ \mathbf{e} \equiv \mathbf{y} - \mathbf{X} \mathbf{b} $$.
By definition of covariance ,
$$
\begin{align}
\mathrm{Cov} ( \mathbf{b}, \mathbf{e} \mid \mathbf{X} )
& =
\mathrm{E} { [ \mathbf{b} - \mathrm{E} ( \mathbf{b} \mid \mathbf{X} ) ][ \mathbf{e} - \mathrm{E} ( \mathbf{e} \mid \mathbf{X} ) ]' \mid \mathbf{X} }
\ & =
\mathrm{E} { [ \mathbf{A} \boldsymbol{ \varepsilon } ][ \mathbf{M} \boldsymbol{ \varepsilon } ]' \mid \mathbf{X} }
&&
( \mathbf{A} \equiv (\mathbf{X}' \mathbf{X})^{-1} \mathbf{X}', \mathbf{M} \equiv \mathbf{I} - \mathbf{X}
( \mathbf{X}' \mathbf{X} )^{-1} \mathbf{X}' )
\ & =
\mathbf{A} \mathrm{E} ( \boldsymbol{ \varepsilon } \boldsymbol{\varepsilon}' \mid \mathbf{X} ) \mathbf{M}'
\ & =
\sigma^2 \mathbf{A} \mathbf{M}'
&&
( \mathbf{M} \mathbf{X} = \mathbf{0} )
\ & = \mathbf{0}.
\end{align}
$$
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