Template:Weibull parameters rank regression on x

Rank Regression on X
Performing a rank regression on X is similar to the process for rank regression on Y, with the difference being that the horizontal deviations from the points to the line are minimized rather than the vertical. Again, the first task is to bring our function, Eqn. (EQNREF Fw ), into a linear form. This step is exactly the same as in the regression on Y analysis and Eqns. (EQNREF logw ), (EQNREF yw ), (EQNREF aw ) and (EQNREF bw ) apply in this case too. The derivation from the previous analysis begins on the least squares fit part, where in this case we treat as the dependent variable and  as the independent variable. The best-fitting straight line to the data, for regression on X (see Chapter 3), is the straight line:


 * $$ x= \hat{a}+\hat{b}y $$ EQNREF xlinew

The corresponding equations for $$ \hat{a} $$ and $$ \hat{b} $$ are:


 * $$ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\sum\limits_{i=1}^{N}x_{i}}{N} -\hat{b}\frac{\sum\limits_{i=1}^{N}y_{i}}{N} $$


 * and


 * $$ \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}y_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}y_{i}\right) ^{2}}{N}}} $$


 * where:


 * $$ y_{i}=\ln \left\{ -\ln [1-F(T_{i})]\right\} $$ and:

xi = ln(Ti) and the F(Ti) values are again obtained from the median ranks.

Once $$ \hat{a} $$ and $$ \hat{b} $$ are obtained, solve Eqn. (EQNREF xlinew ) for y, which corresponds to:


 * $$ y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}}x $$ Solving for the parameters from Eqns. (EQNREF aw ) and (EQNREF bw ) we get


 * $$ a=-\frac{\hat{a}}{\hat{b}}=-\beta \ln (\eta )$$ EQNREF awx


 * and


 * $$ b=\frac{1}{\hat{b}}=\beta EQNREF bwx $$ The correlation coefficient is evaluated as before using Eqn. (EQNREF RHOw ).