# Simple Linear Regression Model

I'm having trouble on this question and was wondering if you guys could hep me out.

I have the linear regression model $Y_i=\alpha+\beta x_i +R_i$, with data $y=(y_1, \ldots,y_n)$ and $x = (x_1, \ldots, x_n)$ from the model. I know that the correlation coefficient between $y$ and $x$ is defined as $r(y,x)=s_{xy}/(\sqrt{s_{yy}}\sqrt{s_{xx}})$. Let $\hat{r}=(\hat{r_1}, \ldots , \hat{r_n})$ where $r_i=y_i-\hat{\alpha} - \hat{\beta} x_i$, $i=1,\ldots,n$ be the residuals from fitting the simple regression model.

I'm can't seem to show that $r(\hat{r}, x) = 0$.

## 1 Answer

We have $\beta = s_{xy}/s_{xx}$ and we have to show $r(\hat{r},x)=0$. I will show that $s_{x\hat{r}}=0$. I use the symbols $\bar{x}, \bar{y}, \bar{r}$ for the mean values. Please note that $\sum{(x_i-\bar{x})}=0$, this will be used a few times.

$s_{x\hat{r}}=\sum{(x_i-\bar{x})(r_i - \bar{r}}) = \sum{(x_i-\bar{x})(y_i-\alpha -\beta x_i- \bar{r})}$

$s_{x\hat{r}}= \sum{(x_i-\bar{x})(y_i-\alpha -\beta x_i)} - \bar{r}\sum{(x_i-\bar{x})}$

$s_{x\hat{r}}=\sum{(x_i-\bar{x})(y_i-\alpha -\beta x_i)}$

$s_{x\hat{r}}=\sum{(x_i-\bar{x})((y_i-\bar{y})+ (\bar{y}-\alpha) - (\beta x_i- \beta \bar{x}}+ \beta\bar{x}))$

$s_{x\hat{r}}=\sum{(x_i-\bar{x})(y_i-\bar{y})}+ (\bar{y}-\alpha) \sum{(x_i-\bar{x})}- \beta\sum{(x_i-\bar{x})^2}- \beta\bar{x} \sum{(x_i-\bar{x})}$

Again, since $\sum{(x_i-\bar{x})}=0$ this simplifies to

$s_{x\hat{r}}=\sum{(x_i-\bar{x})(y_i-\bar{y})} - \beta\sum{(x_i-\bar{x})^2}$

$s_{x\hat{r}}=s_{xy} - \beta s_{xx} = 0$