# Least Squares Derivation question

1. Sep 20, 2011

### Xyius

So I am learning how to use the method of least squares to find the best fit straight line and there is this one part in the derivation I do not fully understand and I was wondering if anyone could help me out.

So basically we start out with the residuals of the equation of a straight line..

$$y_i-mx_i-c$$
And now we take the root mean square of these residuals and try to find the minimum points of m and c by taking the partial derivatives.

$$\Sigma (y_i-mx_i-c)^2$$
$$\frac{\partial S}{\partial m}=-2\Sigma x_i(y_i-mx_i-c)=0$$
$$\frac{\partial S}{\partial c}=-2\Sigma (y_i-mx_i-c)=0$$

Now from the second equation it is easy to see that
$$c=\overline{y}-m\overline{x}$$
since
$$\overline{y}=\frac{1}{n}\Sigma y_i$$
and
$$\overline{x}=\frac{1}{n}\Sigma x_i$$

Solving the first equation for m we get..
$$m=\frac{\Sigma x_i y_i -c\Sigma x_i}{\Sigma x_i ^2}$$

The part I do not understand is the book says that m is equal to..

$$m=\frac{\Sigma (x_i-\overline{x})y_i}{\Sigma (x_i-\overline{x})^2}$$
I feel like I must be missing something simple due to the books lack of explaination. Can anyone help me get this formula from the one I got for m? Why did the $x_i$'s turn into $(x_i-\overline{x})$'s?? Did they made c=0 for some reason? Any help would be appreciated!

2. Sep 20, 2011

### Stephen Tashi

In the first equation, if you replace $c$ by $\overline{y} - m \overline{x}$ then you get m's on both sides and you have to solve for m again. Maybe that's how to do it.

3. Sep 20, 2011

### TheoMcCloskey

Not quite right - looks like they assumed a model with zero y-mean. Another way of looking at this is a model, in x about x-mean, through the origin. With this later terminology, you only have one parameter, the slope (m), to estimate

$$y_i = m \, (x_i-\bar{x})$$

Hence, setting the derivative of S(m) to zero yields

$$\sum y \, (x_i-\bar{x}) = m \cdot \sum (x_i-\bar{x})^2$$

Solve for m to yield text answer

4. Sep 20, 2011

### Stephen Tashi

$$m=\frac{\Sigma x_i y_i -c\Sigma x_i}{\Sigma x_i ^2}$$
$$m = \frac{\Sigma x_i y_i - (\overline y - m \overline x) \Sigma x_i}{\Sigma x_i ^2}$$
$$m = \frac{ \Sigma x_i y_i -\overline y \Sigma x_i + m \overline x \Sigma x_i}{\Sigma x_i^2}$$
$$m \Sigma x_i^2 - m \overline x \Sigma x_i = \Sigma x_i y_i - \overline y \Sigma x_i$$
$$m= \frac{ \Sigma x_i y_i - \overline y \Sigma x_i}{\Sigma x_i^2 - \overline x \Sigma x_i}$$

$$= \frac{ \Sigma x_i y_i - \frac{\Sigma y_i}{N} \Sigma x_i}{\Sigma x_i^2 - (\frac{\Sigma x_i}{N}) \Sigma x_i }$$

The numerator is equal to $\Sigma x_i y_i - \Sigma y_i (\frac{\Sigma x_i}{N})$
$$= \Sigma x_i y_i - \Sigma y_i \overline x$$
$$= \Sigma (x_i - \overline x) y_i$$

To deal with the denominator, consider an estimator for the sample variance:

$$\sigma^2 = \frac { \Sigma (x_i - \overline x)^2}{N}$$
$$= \frac{ \Sigma ( x_i^2 - 2 x_i \overline x + \overline x \overline x)}{N}$$
$$= \frac{ \Sigma x_i^2 - 2 \overline x \Sigma x_i + \Sigma \overline x \overline x }{N}$$
$$= \frac {\Sigma x_i^2}{N} - 2 \overline x \frac{\Sigma x_i}{N} + \frac{ N \overline x \overline x}{N}$$
$$= \frac {\Sigma x_i^2}{N} - 2 \overline x \overline x + \overline x \overline x$$
$$= \frac{ \Sigma x_i^2}{N} - \overline x \overline x$$

This establishes that
$$\frac {\Sigma (x_i - \overline x)^2}{N} = \frac{\Sigma x_i^2}{N} - \overline x \overline x$$

So
$$\Sigma (x_i - \overline x)^2 = \Sigma x_i^2 - N \overline x \overline x$$
$$= \Sigma x_i^2 - N ( \frac{\Sigma x_i}{N} \frac{\Sigma x_i}{N})$$
$$= \Sigma x_i^2 - \frac{\Sigma x_i \Sigma x_i}{N}$$

5. Sep 22, 2011

### Xyius

The only thing that confuses me Stephen Tashi, is line 4 and 5. Where do you get those relations and how does "m" turn into the expression in line 5? Otherwise, everything after it is fine and I understand completely. :]

6. Sep 22, 2011

### Stephen Tashi

Multiply both sides of the equation by $\Sigma x_i^2$ and then subtract $m\overline x \Sigma x_i$ from both sides.
$$m( \Sigma x_i^2 - \overline x \Sigma x_i) = \Sigma x_i y_i - \overline y \Sigma x_i$$

Divide both sides by $\Sigma x_i^2 - \overline x \Sigma x_i$

Last edited: Sep 23, 2011
7. Sep 25, 2011

### Xyius

Thank you very much!!