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Least Squares Derivation question

  1. Sep 20, 2011 #1
    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..

    [tex]y_i-mx_i-c[/tex]
    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.

    [tex]\Sigma (y_i-mx_i-c)^2[/tex]
    [tex]\frac{\partial S}{\partial m}=-2\Sigma x_i(y_i-mx_i-c)=0[/tex]
    [tex]\frac{\partial S}{\partial c}=-2\Sigma (y_i-mx_i-c)=0[/tex]

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

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

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

    [tex]m=\frac{\Sigma (x_i-\overline{x})y_i}{\Sigma (x_i-\overline{x})^2}[/tex]
    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 [itex]x_i[/itex]'s turn into [itex](x_i-\overline{x})[/itex]'s?? Did they made c=0 for some reason? Any help would be appreciated!
     
  2. jcsd
  3. Sep 20, 2011 #2

    Stephen Tashi

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    In the first equation, if you replace [itex] c [/itex] by [itex] \overline{y} - m \overline{x} [/itex] then you get m's on both sides and you have to solve for m again. Maybe that's how to do it.
     
  4. Sep 20, 2011 #3
    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

    [tex] y_i = m \, (x_i-\bar{x})[/tex]

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

    [tex] \sum y \, (x_i-\bar{x}) = m \cdot \sum (x_i-\bar{x})^2[/tex]

    Solve for m to yield text answer
     
  5. Sep 20, 2011 #4

    Stephen Tashi

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    [tex]m=\frac{\Sigma x_i y_i -c\Sigma x_i}{\Sigma x_i ^2}[/tex]
    [tex] m = \frac{\Sigma x_i y_i - (\overline y - m \overline x) \Sigma x_i}{\Sigma x_i ^2}[/tex]
    [tex] m = \frac{ \Sigma x_i y_i -\overline y \Sigma x_i + m \overline x \Sigma x_i}{\Sigma x_i^2} [/tex]
    [tex] m \Sigma x_i^2 - m \overline x \Sigma x_i = \Sigma x_i y_i - \overline y \Sigma x_i [/tex]
    [tex] m= \frac{ \Sigma x_i y_i - \overline y \Sigma x_i}{\Sigma x_i^2 - \overline x \Sigma x_i} [/tex]

    [tex] = \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 } [/tex]

    The numerator is equal to [itex] \Sigma x_i y_i - \Sigma y_i (\frac{\Sigma x_i}{N}) [/itex]
    [tex] = \Sigma x_i y_i - \Sigma y_i \overline x [/tex]
    [tex] = \Sigma (x_i - \overline x) y_i [/tex]

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

    [tex] \sigma^2 = \frac { \Sigma (x_i - \overline x)^2}{N} [/tex]
    [tex] = \frac{ \Sigma ( x_i^2 - 2 x_i \overline x + \overline x \overline x)}{N} [/tex]
    [tex] = \frac{ \Sigma x_i^2 - 2 \overline x \Sigma x_i + \Sigma \overline x \overline x }{N} [/tex]
    [tex] = \frac {\Sigma x_i^2}{N} - 2 \overline x \frac{\Sigma x_i}{N} + \frac{ N \overline x \overline x}{N} [/tex]
    [tex] = \frac {\Sigma x_i^2}{N} - 2 \overline x \overline x + \overline x \overline x [/tex]
    [tex] = \frac{ \Sigma x_i^2}{N} - \overline x \overline x [/tex]

    This establishes that
    [tex] \frac {\Sigma (x_i - \overline x)^2}{N} = \frac{\Sigma x_i^2}{N} - \overline x \overline x [/tex]

    So
    [tex] \Sigma (x_i - \overline x)^2 = \Sigma x_i^2 - N \overline x \overline x [/tex]
    [tex] = \Sigma x_i^2 - N ( \frac{\Sigma x_i}{N} \frac{\Sigma x_i}{N}) [/tex]
    [tex] = \Sigma x_i^2 - \frac{\Sigma x_i \Sigma x_i}{N} [/tex]
     
  6. Sep 22, 2011 #5
    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. :]
     
  7. Sep 22, 2011 #6

    Stephen Tashi

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

    Divide both sides by [itex] \Sigma x_i^2 - \overline x \Sigma x_i [/itex]

     
    Last edited: Sep 23, 2011
  8. Sep 25, 2011 #7
    Thank you very much!!
     
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