# How to get from $R$ to $R^2$ the hard way

If $r = \frac{S_{X, Y}}{S_X S_Y}$ and $r^2 = \frac{SS_{reg}}{SS_{tot}}$, how do I prove that squaring $r$ really gets me to $r^2$ using those definitions or derivations thereof?

I guess the best way is to start by equating $\left[\frac{\sum{xy} - n\bar{x}\bar{y}}{\sqrt{\sum{x^2} - n\bar{x}^2}\sqrt{\sum{y^2} - n\bar{y}^2}}\right]^2$ and $\frac{\sum{(\hat{y}-\bar{y})^2}}{\sum{(y-\bar{y})^2}}$, but I can't get much past that. How can I further simplify this equation in order to prove that $\left(\frac{S_{X, Y}}{S_X S_Y}\right)^2 = \frac{SS_{reg}}{SS_{tot}}$?

• What's the reason for this question? If it is a homework or self-study question, the tag "self-study" should be added. (I'm asking because... isn't simply squaring R sufficient?) – Patrick Coulombe Jul 29 '13 at 23:16
• @Patrick, It's self study. I've been studying ANOVA and the relationships between the $F$ statistic, $r^2$ and $r$ by extension. I'll add the tag, Thanks for the tip! – Waldir Leoncio Jul 29 '13 at 23:48

You are missing square roots for $S_X$, $S_Y$ and $SS_{reg}$ should be $\sum(\hat{y}-\overline{y})^2$ but I assume that these are just typos.

Let $y=\alpha+\beta x+\varepsilon$ and let us note that

$$r=\beta\frac{S_X}{S_Y}$$ $$r^2=\beta^2\frac{\overline{x^2}-\overline{x}^2}{\overline{y^2}-\overline{y}^2}$$

On the other hand we have

$$\frac{\sum{(\hat{y}-\overline{y})^2}}{\sum{(y-\bar{y})^2}}=\frac{\overline{\hat{y}^2}-2\overline{y}\overline{\hat{y}}+\overline{y}^2}{\overline{y^2}-2\overline{y}^2+\overline{y}}=\frac{\overline{\hat{y}^2}-\overline{\hat{y}}^2}{\overline{y^2}-\overline{y}^2}$$

since $\overline{y}=\overline{\hat{y}}$ (show it). Now we are left to prove that

$$\beta^2(\overline{x^2}-\overline{x}^2)=\overline{\hat{y}^2}-\overline{\hat{y}}^2$$

which can be done by manipulating the rhs, e.g. using $\overline{\hat{y}}=\hat{\alpha}+\hat{\beta}\overline{x}$.

• Thank you for pointing the typos and providing a solution! – Waldir Leoncio Jul 30 '13 at 11:34
• The key there is that insight that $\bar{y} = \bar{\hat{y}}$, which rings true since $(\bar{x}, \bar{y})$ is the fulcrum of the regression line. Nice catch! ;) – Waldir Leoncio Jul 30 '13 at 11:41