TITLE: help understanding step in derivation of correlation coefficient
QUESTION [5 upvotes]: I'm looking to understand the starred step in the derivation below (also, if someone could help with the LaTex alignment, I'd appreciate it).
The regression line is $y= b_0 + b_1 x$, where $b_0$ and $b_1$ can be found by:
1) taking the difference between each observed value $y_i$ and the expected point regression line, $b_0 + b_1 x_i$
$$\text{ difference } = y_i - b_0 -b_1 x_i$$
2) summing the square of the differences from 1) to get the sum of squares
$$SS = \sum \limits_{i=1}^n (y_i - b_0 -b_1 x_i)^2$$
3) taking the partial derivative with respect to $b_0$ and $b_1$, then solving for each
$$
\begin{align}
\text{ solving for } b_0 \\
SS &= \sum(y_i - b_0 -b_1 x_i)^2\\
SS &= \sum (Y_i ^2 - 2Y_i b_0 - 2 Y_i b_1+ 2b_0 b_1X_i + b_1^2X_i^2+b_0^2) &\text{expand the square}\\
\frac{ \partial }{\partial_{b_0} }SS &= \sum (-2Y_i + 2b_1 X_i + 2b_0) &\text{partial derivative wrt} b_0\\
0 &= \sum 2(-Y_i + b_1 X_i + b_0) &\text{factor out 2 from the sum}\\
0 &= \sum (-Y_i + b_1 X_i + b_0) &\text{divide both sides by 2}\\
0 &= \sum -Y_i + \sum b_1 X_i + \sum b_0 &\text{split summation into parts}\\
\sum Y_i &= \sum b_1 X_i + \sum b_0 \\
\sum Y_i &= b_1 \sum X_i + n b_0 \\
\frac{1}{n}(\sum Y_i - b_1 \sum X_i ) &=  b_0 \\
\bar Y - b_1 \bar X &=  b_0 \text {  rewrite sums as averages since } \frac{1}{n} \sum Y_i = \bar Y\\
\end{align}
$$
$$
\begin{align}
\\
\text{solving for } b_1\\
\frac{ \partial }{\partial_{b_1} }SS &= \sum (-2Y_iX_i + 2b_0 X_i + 2 b_1 X_i^2) &\text{ partial derivative wrt } b_1\\
0 &= 2\sum (-Y_iX_i + b_0 X_i +  b_1 X_i^2) \\
0 &= \sum (-Y_iX_i + b_0 X_i +  b_1 X_i^2) \\
0 &=  -\sum Y_iX_i + b_0 \sum X_i +  b_1 \sum X_i^2 &\text{ split summation}\\
0 &=  -\sum Y_iX_i + (\bar Y - b_1 \bar X) \sum X_i +  b_1 \sum X_i^2 &\text{ substitue } b_0\\
0 &=  -\sum Y_iX_i + (\bar Y \sum X_i - b_1 \bar X \sum X_i)  +  b_1 \sum X_i^2 &\text{ distribute sum}\\
b_1 \bar X \sum X_i - b_1 \sum X_i^2 &=  -\sum Y_iX_i + \bar Y \sum X_i     &\text{ collect } b_1 \text{ terms}\\
b_1 (\bar X \sum X_i - \sum X_i^2) &=  -\sum Y_iX_i + \bar Y \sum X_i   \\
b_1  &= { \bar Y \sum X_i -\sum Y_iX_i \over (\bar X \sum X_i - \sum X_i^2)   }\\
b_1  &= { \frac{1}{n} \sum Y_i \sum X_i -\sum Y_iX_i \over (\frac{1}{n} \sum X_i \sum X_i - \sum X_i^2)   } \biggr ( \frac{-n}{-n} \biggr )\\
b_1  &= { n \sum Y_iX_i - \sum Y_i \sum X_i   \over n \sum X_i^2 -(\sum X_i)^2     } \\
\end{align}
$$
$$
\begin{align}
b_0 &= \frac{1}{n} \sum y_i - b_1 \frac{1}{n} \sum x_i\\\\\\
b_1 &= {n \sum x_i y_i - \sum x_i \sum y_i \over n \sum x_i^2-(\sum x_i)^2}
\end{align}
$$
(derivation shown in http://polisci.msu.edu/jacoby/icpsr/regress3/lectures/week2/5.LeastSquares.pdf)
From this point you can use $b_1$ to get the correlation coefficient as follows:
$$
\begin{align}
b_1 &= {\frac{1}{n} \sum x_i y_i - (\frac{1}{n}\sum x_i) (\frac{1}{n} \sum y_i ) \over (\frac{1}{n} \sum x_i^2) -(\frac{1}{n}\sum x_i)^2}  & \text{ divide top and bottom by } n^2 \\\\
b_1 &= {\frac{1}{n} \sum x_i y_i - (\bar x) (\bar y ) \over (\frac{1}{n} \sum x_i^2) -(\bar x)^2}  & \text{ rewrite product of sums as averages }  \\\\
b_1 &= {\frac{1}{n} \sum (x_i - \bar x)(y_i - \bar y ) \over \frac{1}{n} \sum (x_i - \bar x)^2}  & \color{red} *\text{application of inscrutably arcane magic} \\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum (x_i - \bar x)^2} } & \text{cancel } \frac{1}{n}\text{, factor denominator }\\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2} \sqrt{\sum ( x_i - \bar x)^2} } \biggr({\sqrt{\sum(y_i - \bar y)^2} \over \sqrt{\sum(y_i - \bar y)^2}}\biggr) & \text{multiply by 1 } \\\\
b_1 &= { \sum (x_i - \bar x)(y_i - \bar y ) \over \sqrt{\sum (x_i - \bar x)^2}  \sqrt{\sum(y_i - \bar y)^2}} \biggr({\sqrt{\sum(y_i - \bar y)^2} \over \sqrt{\sum ( x_i - \bar x)^2} }\biggr) & \text{re-arrange } \\\\
b_1 &= R \frac{S_x}{S_y}
\end {align}
$$

REPLY [2 votes]: For an alternative method for the same derivation:
$$
\begin{align}
\text{solving for } b_1\\
\frac{ \partial }{\partial_{b_1} }SS &= \sum (-2Y_iX_i + 2b_0 X_i + 2 b_1 X_i^2) &\text{ partial derivative wrt } b_1\\
0 &= 2\sum (-Y_iX_i + b_0 X_i +  b_1 X_i^2) \\
0 &= \sum(X_iY_i - X_i\bar{Y} + b_1X_i\bar{X}-b_1X_i^2) &\text{replace }b_0 \\ 
 0 &= \sum(X_iY_i - X_i\bar{Y}) - b_1\sum(X_i^2-X_i\bar{X}) &\text{separate into two sums} \\
b_1 &=  \frac{\sum(X_iY_i - X_i\bar{Y})}{\sum(X_i^2-X_i\bar{X})} \\
\end{align}
$$
Noting that:
$$
\begin{align}
\sum X_i\bar{Y} = \sum X_i\left( \frac{1}{n}\sum Y_i \right) = n\left(\frac{1}{n}\sum X_i\right) \left( \frac{1}{n}\sum Y_i \right) = n\bar{X}\bar{Y} \\
\sum X_i\bar{X} = \sum X_i\left( \frac{1}{n}\sum X_i \right) = n\left(\frac{1}{n}\sum X_i\right) \left( \frac{1}{n}\sum X_i \right) = n\bar{X}^2 \: 
\end{align}
$$
$b_1$ can be solved as:
$$
\begin{align}
b_1 &= \frac{\sum(X_iY_i) - n\bar{X}\bar{Y}}{\sum(X_i^2)-n\bar{X}^2}
\end{align}
$$
Finally, to rewrite in a more intuitive form, noting that:
$$
\sum(\bar{X}^2-X_i\bar{X})=0 \: \text{   and   } \: \sum(\bar{X}\bar{Y}-Y_i\bar{X})=0
$$
$b_1$ can be written as the ratio of covariance to predictor variance:
$$
b_1 = \frac{\sum(X_iY_i-X_i\bar{Y}) + \sum(\bar{X}\bar{Y}-Y_i\bar{X})}{\sum(X_i^2-X_i\bar{X}) + \sum(\bar{X}^2-X_i\bar{X})} = \frac{\frac{1}{n}\sum(X_i - \bar{X})(Y_i-\bar{Y})}{\frac{1}{n}\sum(X_i-\bar{X})^2} = \frac{Cov(X,Y)}{Var(X)}
$$
Ref (equations 6-9):
http://seismo.berkeley.edu/~kirchner/eps_120/Toolkits/Toolkit_10.pdf