Template:Rank Regression on Y for Exponential Distribution

Rank Regression on Y

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:

[math]\displaystyle{ \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} }[/math]

and:

[math]\displaystyle{ \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{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In our case, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}=\ln [1-F({{T}_{i}})] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{T}_{i}} }[/math]

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

[math]\displaystyle{ \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} }[/math]

The Correlation Coefficient

The estimator of [math]\displaystyle{ \rho }[/math] is the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math], given by:

[math]\displaystyle{ \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}}}} }[/math]

Example 2

Fourteen units were being reliability tested and the following life test data were obtained (Table 7.1):

Table 7.1 - Life Test Data
Data point index	Time-to-failure
1	5
2	10
3	15
4	20
5	25
6	30
7	35
8	40
9	50
10	60
11	70
12	80
13	90
14	100

Assuming that the data follow a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, [math]\displaystyle{ \rho }[/math], using rank regression on Y.

Solution to Example 2

Construct Table 7.2, as shown next.

[math]\displaystyle{ \overset{{}}{\mathop{\text{Table 7}\text{.2 - Least Squares Analysis}}}\, }[/math] [math]\displaystyle{ \begin{matrix} N & T_{i} & F(T_{i}) & y_{i} & T_{i}^{2} & y_{i}^{2} & T_{i} y_{i} \\ \text{1} & \text{5} & \text{0}\text{.0483} & \text{-0}\text{.0495} & \text{25} & \text{0}\text{.0025} & \text{-0}\text{.2475} \\ \text{2} & \text{10} & \text{0}\text{.1170} & \text{-0}\text{.1244} & \text{100} & \text{0}\text{.0155} & \text{-1}\text{.2443} \\ \text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.2064} & \text{225} & \text{0}\text{.0426} & \text{-3}\text{.0961} \\ \text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.2958} & \text{400} & \text{0}\text{.0875} & \text{-5}\text{.9170} \\ \text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.3942} & \text{625} & \text{0}\text{.1554} & \text{-9}\text{.8557} \\ \text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.5032} & \text{900} & \text{0}\text{.2532} & \text{-15}\text{.0956} \\ \text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.6257} & \text{1225} & \text{0}\text{.3915} & \text{-21}\text{.8986} \\ \text{8} & \text{40} & \text{0}\text{.5349} & \text{-0}\text{.7655} & \text{1600} & \text{0}\text{.5860} & \text{-30}\text{.6201} \\ \text{9} & \text{50} & \text{0}\text{.6046} & \text{-0}\text{.9279} & \text{2500} & \text{0}\text{.8609} & \text{-46}\text{.3929} \\ \text{10} & \text{60} & \text{0}\text{.6742} & \text{-1}\text{.1215} & \text{3600} & \text{1}\text{.2577} & \text{-67}\text{.2883} \\ \text{11} & \text{70} & \text{0}\text{.7439} & \text{-1}\text{.3622} & \text{4900} & \text{1}\text{.8456} & \text{-95}\text{.3531} \\ \text{12} & \text{80} & \text{0}\text{.8135} & \text{-1}\text{.6793} & \text{6400} & \text{2}\text{.8201} & \text{-134}\text{.3459} \\ \text{13} & \text{90} & \text{0}\text{.8830} & \text{-2}\text{.1456} & \text{8100} & \text{4}\text{.6035} & \text{-193}\text{.1023} \\ \text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324} \\ \sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899} \\ \end{matrix} }[/math]

The median rank values ( [math]\displaystyle{ F({{T}_{i}}) }[/math] ) can be found in rank tables or they can be estimated using the Quick Statistical Reference in Weibull++. Given the values in the table above, calculate [math]\displaystyle{ \hat{a} }[/math] and [math]\displaystyle{ \hat{b} }[/math] using Eqns. (aae) and (bbe):

[math]\displaystyle{ \begin{align} \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}{{y}_{i}}-(\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 }}}\,T_{i}^{2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}})}^{2}}/14} \\ \\ \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14} \end{align} }[/math]

or:

[math]\displaystyle{ \hat{b}=-0.02711 }[/math]

and:

[math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{T}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{T}_{i}}}{N} }[/math]

or:

[math]\displaystyle{ \hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748 }[/math]

Therefore, from Eqn. (be):

[math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour} }[/math]

and from Eqn. (ae):

[math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711} }[/math]

or:

[math]\displaystyle{ \hat{\gamma }=10.1365\text{ hours} }[/math]

Then:

[math]\displaystyle{ f(T)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}} }[/math]

The correlation coefficient can be estimated using Eqn. (RHOe):

[math]\displaystyle{ \hat{\rho }=-0.9679 }[/math]

This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on Y (RRY), as shown in the figure on the following page.

The estimated parameters and the correlation coefficient using Weibull++ were found to be:

[math]\displaystyle{ \hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679 }[/math]

The probability plot can be obtained simply by clicking the Plot icon.