1.2.7.md

Solution to Review Question

by Qiang Gao, updated at Mar 21, 2017

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Chapter 1 Finite-Sample Properties of OLS

Section 2 The Algebra of Least Squares

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Review Question 1.2.7 (Relation between $$ R_{uc}^2 $$ and $$ R^2 $$)

Show that

$$ 1 - R^2 = \left( 1 + \frac{ n \cdot \bar{y}^2 }{ \sum_{i=1}^n (y_i - \bar{y})^2 } \right) (1 - R_{uc}^2). $$

Hint: Use the identity $$ \sum_i (y_i - \bar{y})^2 = \sum_i y_i^2 - n \cdot \bar{y}^2 $$.

Solution

By definition of $$ R^2 $$, the left side equals

$$ \frac{ \sum_{i=1}^n e_i^2 }{ \sum_{i=1}^n (y_i - \bar{y})^2 }. $$

By definition of $$ R_{uc}^2 $$, the right side equals

$$ \begin{align} & \left( 1 + \frac{ n \cdot \bar{y}^2 } { \sum_{i=1}^n (y_i - \bar{y})^2 } \right) \frac{ \sum_{i=1}^n e_i^2 }{ \sum_{i=1}^n y_i^2 } \ = & \left( \frac{ \sum_i y_i^2 - n \cdot \bar{y}^2

• n \cdot \bar{y}^2 } { \sum_{i=1}^n (y_i - \bar{y})^2 } \right) \frac{ \sum_{i=1}^n e_i^2 }{ \sum_{i=1}^n y_i^2 } \qquad \text{(hint)} \ = & \left( \frac{ \sum_i y_i^2 } { \sum_{i=1}^n (y_i - \bar{y})^2 } \right) \frac{ \sum_{i=1}^n e_i^2 }{ \sum_{i=1}^n y_i^2 } \ = & \frac{ \sum_{i=1}^n e_i^2 }{ \sum_{i=1}^n (y_i - \bar{y})^2 }. \end{align} $$

Left side and right side are equal.

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Copyright ©2017 by Qiang Gao