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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.
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| 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 ([[Parameter Estimation|see Chapter 4)]], is the straight line:
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| ::<math>x=\hat{a}+\hat{b}y</math>
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| The corresponding equations for <math>\hat{a}</math> and <math>\hat{b}</math> are:
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| ::<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>
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| and:
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| ::<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>
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| where:
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| ::<math>{{y}_{i}}=\ln [1-F({{t}_{i}})]</math>
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| and:
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| ::<math>{{x}_{i}}={{t}_{i}}</math>
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| 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:
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| ::<math>y=-\frac{\hat{a}}{\hat{b}}+\frac{1}{\hat{b}}x</math>
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| Solving for the parameters from above equations we get:
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| ::<math>a=-\frac{\hat{a}}{\hat{b}}=\lambda \gamma \Rightarrow \gamma =\hat{a}</math>
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| and:
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| ::<math>b=\frac{1}{\hat{b}}=-\lambda \Rightarrow \lambda =-\frac{1}{\hat{b}}</math>
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| For the one-parameter exponential case, equations for estimating a and b become:
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| ::<math>\begin{align}
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| \hat{a}= & 0 \\
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| \hat{b}= & \frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}}
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| \end{align}</math>
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| The correlation coefficient is evaluated as before. | |
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| ====Example 3: 2 Parameter Exponential Distribution RRX====
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| Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.
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| ===== Solution to Example 3=====
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| Table 7.2 constructed in Example 2 applies to this example also. Using the values from this table, we get:
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| ::<math>\begin{align}
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| \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{14}} \\
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| \\
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| \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{22.1148-{{(-13.2315)}^{2}}/14}
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| \end{align}</math>
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| or:
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| ::<math>\hat{b}=-34.5563</math>
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| and:
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| ::<math>\hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}}{14}-\hat{b}\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}</math>
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| or:
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| ::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>
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| Therefore:
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| ::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>
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| and:
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| ::<math>\hat{\gamma }=\hat{a}=12.3406</math>
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| The correlation coefficient is found to be:
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| ::<math>\hat{\rho }=-0.9679</math>
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| Note that the equation for regression on Y is not necessarily the same as that for the regression on X. The only time when the two regression methods yield identical results is when the data lie perfectly on a line. If this were the case, the correlation coefficient would be <math>-1</math>. The negative value of the correlation coefficient is due to the fact that the slope of the exponential probability plot is negative.
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| This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on X (RRX) methods for analysis, as shown below.
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| The estimated parameters and the correlation coefficient using Weibull++ were found to be:
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| ::<math>\begin{array}{*{35}{l}}
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| \hat{\lambda }= &0.0289 \text{failures/hour} \\
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| \hat{\gamma}= & 12.3395 \text{hours} \\
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| \hat{\rho} = &-0.9679 \\
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| \end{array}</math>
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| [[Image:weibullfolio1.png|thumb|center|400px|]]
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| The probability plot can be obtained simply by clicking the '''Plot''' icon.
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| [[Image:weibullfolioplot1.png|thumb|center|400px|]]
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