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(Created page with '===Linear Regression (Least Squares)=== The method of least squares requires that a straight line be fitted to a set of data points. If the regression is on <math>Y</math> , the…')
 
 
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===Linear Regression (Least Squares)===
#REDIRECT [[Gompertz_Models#Linear_Regression_.28Least_Squares.29]]
The method of least squares requires that a straight line be fitted to a set of data points. If the regression is on  <math>Y</math> , then the sum of the squares of the vertical deviations from the points to the line is minimized. If the regression is on  <math>X</math> , the line is fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the line is minimized. To illustrate the method, this section presents a regression on  <math>Y</math> . Consider the linear model [2]:
 
::<math>{{Y}_{i}}={{\widehat{\beta }}_{0}}+{{\widehat{\beta }}_{1}}{{X}_{i1}}+{{\widehat{\beta }}_{2}}{{X}_{i2}}+...+{{\widehat{\beta }}_{p}}{{X}_{ip}}</math>
 
<br>
or in matrix form where bold letters indicate matrices:
<br>
::<math>Y=X\beta </math>
<br>
:where:
<br>
::<math>Y=\left[ \begin{matrix}
  {{Y}_{1}}  \\
  {{Y}_{2}}  \\
  \vdots  \\
  {{Y}_{N}}  \\
\end{matrix} \right]</math>
 
::<math>X=\left[ \begin{matrix}
  1 & {{X}_{1,1}} & \cdots  & {{X}_{1,p}}  \\
  1 & {{X}_{2,1}} & \cdots  & {{X}_{2,p}}  \\
  \vdots  & \vdots  & \ddots  & \vdots  \\
  1 & {{X}_{N,1}} & \cdots  & {{X}_{N,p}}  \\
\end{matrix} \right]</math>
<br>
:and:
<br>
::<math>\beta =\left[ \begin{matrix}
  {{\beta }_{0}}  \\
  {{\beta }_{1}}  \\
  \vdots  \\
  {{\beta }_{p}}  \\
\end{matrix} \right]</math>
 
The vector  <math>\beta </math>  holds the values of the parameters. Now let  <math>\widehat{\beta }</math>  be the estimates of these parameters, or the regression coefficients. The vector of estimated regression coefficients is denoted by:
 
::<math>\widehat{\beta }=\left[ \begin{matrix}
  {{\widehat{\beta }}_{0}}  \\
  {{\widehat{\beta }}_{1}}  \\
  \vdots  \\
  {{\widehat{\beta }}_{p}}  \\
\end{matrix} \right]</math>
 
Solving for  <math>\beta </math>  in Eqn. (linear) requires the analyst to left multiply both sides by the transpose of  <math>X</math> ,  <math>{{X}^{T}}</math> :
 
::<math>({{X}^{T}}X)\widehat{\beta }={{X}^{T}}Y</math>
 
Now the term  <math>({{X}^{T}}X)</math>  becomes a square and invertible matrix. Then taking it to the other side of the equation gives:
 
::<math>\widehat{\beta }={{(}^{T}}^{-1}{{X}^{T}}Y</math>

Latest revision as of 02:11, 27 August 2012

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