Template:Rank Regression on Y for Exponential Distribution

Rank Regression on Y for One Parameter Exponential
Performing a rank regression on Y requires that a straight line be fitted to the set of available data points such that the sum of the squares of the vertical deviations from the points to the line is minimized. The least squares parameter estimation method (regression analysis) was discussed in Chapter 3, and the following equations for rank regression on Y (RRY) were derived:


 * $$\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}$$


 * and:


 * $$\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{N}}$$

In our case, the equations for $${{y}_{i}}$$ and $${{x}_{i}}$$ are:


 * $${{y}_{i}}=\ln [1-F({{T}_{i}})]$$

and:


 * $${{x}_{i}}={{T}_{i}}$$

and the $$F({{T}_{i}})$$ is estimated from the median ranks. Once $$\hat{a}$$ and $$\hat{b}$$ are obtained, then $$\hat{\lambda }$$ and $$\hat{\gamma }$$ can easily be obtained from Eqns. (ae) and (be). For the one-parameter exponential, Eqns. (aae) and (bbe) become:


 * $$\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 }}}\,x_{i}^{2}} \end{align}$$

The Correlation Coefficient
The estimator of $$\rho $$ is the sample correlation coefficient, $$\hat{\rho }$$, given by:


 * $$\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}$$