Solutions to Review Question

by Qiang Gao, updated at Mar 22, 2017

Review Question 1.2.8

Show that

$$R_{uc}^2 = \frac{ \mathbf{y}' \mathbf{P} \mathbf{y} } { \mathbf{y}' \mathbf{y} }.$$

Solution

$$\begin{align} R_{uc}^2 & = 1 - \frac{ \mathbf{e}' \mathbf{e} } { \mathbf{y}' \mathbf{y} } \qquad \text{(definition (1.2.16))} \\ & = \frac{ \mathbf{y}' \mathbf{y} - \mathbf{e}' \mathbf{e} } { \mathbf{y}' \mathbf{y} } \\ & = \frac{ \mathbf{y}' \mathbf{I} \mathbf{y} - \mathbf{y}' \mathbf{M} \mathbf{y} } { \mathbf{y}' \mathbf{y} } \qquad \text{($ \mathbf{e} = \mathbf{M} \mathbf{y} $, $\mathbf{M}$ idempotent)} \\ & = \frac{ \mathbf{y}' (\mathbf{I} - \mathbf{M}) \mathbf{y} } { \mathbf{y}' \mathbf{y} } \\ & = \frac{ \mathbf{y}' \mathbf{P} \mathbf{y} } { \mathbf{y}' \mathbf{y} }. \qquad (\mathbf{M} \equiv \mathbf{I} - \mathbf{P}) \end{align}$$

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