by Qiang Gao, updated at Mar 26, 2017
...
Verify that $$ \mathbf{b}
(1) According to (1.2.5'),
$$ \mathbf{b} = \mathbf{S}{\mathbf{x} \mathbf{x}}^{-1} \mathbf{s}{\mathbf{x} \mathbf{y}}. $$
(2)
$$
\begin{align}
\mathrm{SSR} & = \mathbf{e}' \mathbf{e}
&&
\text{(definition (1.2.12))}
\ & =
(\mathbf{y} - \mathbf{X} \mathbf{b})'
(\mathbf{y} - \mathbf{X} \mathbf{b})
&&
\text{(definition (1.2.4))}
\ & =
\mathbf{y}'\mathbf{y} - \mathbf{y}' \mathbf{X}
\mathbf{b} - \mathbf{b}' \mathbf{X}' \mathbf{y} +
\mathbf{b}' \mathbf{X}' \mathbf{X} \mathbf{b}
\ & =
\mathbf{y}'\mathbf{y} - \mathbf{y}' \mathbf{X}
\mathbf{b} - \mathbf{b}' \mathbf{X}' \mathbf{y} +
\mathbf{b}' \mathbf{X}' \mathbf{y}
&&
\text{(definition (1.2.5))}
\ & =
\mathbf{y}'\mathbf{y} - (\mathbf{X}' \mathbf{y})' \mathbf{b}
\ & =
n \cdot \mathbf{y}' \mathbf{y} / n - n \cdot
\mathbf{s}{\mathbf{x} \mathbf{y}}'
\mathbf{S}{\mathbf{x} \mathbf{x}}^{-1}
\mathbf{s}{\mathbf{x} \mathbf{y}}
\ & =
n \cdot ( \mathbf{y}' \mathbf{y} / n -
\mathbf{s}{\mathbf{x} \mathbf{y}}'
\mathbf{S}{\mathbf{x} \mathbf{x}}^{-1}
\mathbf{s}{\mathbf{x} \mathbf{y}} ).
\end{align}
$$
(3)
$$ s^2 = \frac{\mathrm{SSR}}{n - K} = \frac{n}{n - K} \cdot ( \mathbf{y}' \mathbf{y} / n - \mathbf{s}{\mathbf{x} \mathbf{y}}' \mathbf{S}{\mathbf{x} \mathbf{x}}^{-1} \mathbf{s}_{\mathbf{x} \mathbf{y}} ). $$
(4)
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