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===Rank Regression on X for Exponential Distribution===
#REDIRECT [[The Exponential Distribution]]
Similar to rank regression on Y, performing a rank regression on X requires that a straight line be 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.
 
Again the first task is to bring our exponential <math>cdf</math> function into a linear form. This step is exactly the same as in regression on Y analysis. The deviation from the previous analysis begins on the least squares fit step, since in this case we treat <math>x</math> as the dependent variable and <math>y</math> as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter [[Parameter Estimation]]), is the straight line:
 
::<math>x=\hat{a}+\hat{b}y</math>
 
The corresponding equations for <math>\hat{a}</math> and <math>\hat{b}</math> are:
 
::<math>\hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}</math>
 
and:
 
::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}}</math>
 
where:
 
::<math>{{y}_{i}}=\ln [1-F({{t}_{i}})]</math>
 
and:
 
::<math>{{x}_{i}}={{t}_{i}}</math>
 
 
The values of <math>F({{t}_{i}})</math> are estimated from the median ranks. Once <math>\hat{a}</math> and <math>\hat{b}</math> are obtained, solve for the unknown <math>y</math> value, which corresponds to:
 
::<math>y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}}x</math>
 
Solving for the parameters from above equations we get:
 
::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>
 
and:
 
::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>
 
For the one-parameter exponential case, equations for estimating a and b become:
 
::<math>\begin{align}
  \hat{a}= & 0 \\
  \hat{b}= & \frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}} 
\end{align}</math>
 
 
The correlation coefficient is evaluated as before.
 
 
'''Example 3:'''
{{Example: 2 Parameter Exponential Distribution RRX}}

Latest revision as of 08:19, 10 August 2012

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