Solutions to Review Question

by Qiang Gao, updated at Mar 21, 2017

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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