Template:Normal distribution rank regression on X: Difference between revisions

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===Rank Regression on X===
#REDIRECT [[The Normal Distribution]]
 
As was mentioned previously, 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 fitted line is minimized.
 
Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: 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 for the data, for regression on X, is the straight line:
 
::<math>x=\widehat{a}+\widehat{b}y</math>
 
The corresponding equations for  <math>\widehat{a}</math>  and  <math>\widehat{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}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right]</math>
 
and:
 
::<math>{{x}_{i}}={{t}_{i}}</math>
 
and the  <math>F({{t}_{i}})</math>  values are estimated from the median ranks. Once  <math>\widehat{a}</math>  and  <math>\widehat{b}</math>  are obtained, solve the above linear equation for the unknown value of  <math>y</math>  which corresponds to:
 
::<math>y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x</math>
 
Solving for the parameters, we get:
 
::<math>a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a}</math>
 
and:
 
::<math>b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b}</math>
 
The correlation coefficient is evaluated as before.
 
<br>
'''Example 3:'''
{{Example: Normal Distribution RRX}}

Latest revision as of 04:15, 13 August 2012

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