1P-Exponential MLE Solution for Interval Data: Difference between revisions

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Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.
This example validates the calculations for the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data in Weibull++ standard folios.
 


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The cumulative distribution function for an exponential distribution is:  
The cumulative distribution function for an exponential distribution is:  


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The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat\theta}\,\!</math> = 41.72. Therefore the variance is 1740.56. The 95% 2-sided confidence interval for <math>{\theta}\,\!</math> are:
The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat\theta}\,\!</math> = 41.72. Therefore the variance is 1740.56.  
 


The 95% 2-sided confidence interval for <math>{\theta}\,\!</math> are:


* Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on
* Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on


 
::<math>-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = X^{2}_{(0.90,1)}\,\!</math>
::<math>-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = x^{2}_{(0.90,1)}\,\!</math>




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::<math>\begin{alignat}{2}
::<math>\begin{alignat}{2}
[\theta_{L},\theta_{U}]&= \hat{\theta}exp(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\
[\theta_{L},\theta_{U}]&= \hat{\theta}exp\left(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}}\right)\\
&=\left [572.3\times exp(-1.96\times\tfrac{41.72}{572.3}),572.3\times exp(1.96\times\tfrac{41.72}{572.3})\right]\\
&= \left[572.3\times exp \left(-1.96\times\frac{41.72}{572.3}\right),572.3\times exp \left(1.96\times\frac{41.72}{572.3}\right)\right]\\
&= [496,660]\\
&= [496,660]\\
\end{alignat}</math>
\end{alignat}</math>
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The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:
The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:


* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is
* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is:
 
 
::<math>F(t)= 1-e^{-(\frac{t}{\theta})}</math>
 
 
The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat{\theta}}\,\!</math> = 41.72. Therefore the variance is 1740.56.
 
 
The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:


* Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on
[[Image:1PE_interval_data_qcp.png|center|550 px]]


::<math>-2ln \frac{L(\theta)}{L(\hat{\theta})} = X^{2}_{(0.90,1)}</math>


* Based on lognormal approximation (select FM for the bound method), the confidence bounds are:


The two solutions of <math>\theta\,\!</math> in the above equation will be the confidence bounds for <math>\theta\,\!</math>.
[[Image:1PE_interval_data_qcp_FM.png|center|550 px]]

Latest revision as of 16:16, 28 September 2015

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1P-Exponential MLE Solution for Interval Data

This example validates the calculations for the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data in Weibull++ standard folios.


Reference Case

Example 7.1 on page 154 in the book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.


Data

Number in State Last Inspected State F/S State End Time
41 0 F 100
44 100 F 300
24 300 F 500
32 500 F 700
29 700 F 1000
21 1000 F 2000
9 2000 F 4000


Result

The cumulative distribution function for an exponential distribution is:

[math]\displaystyle{ F(t)=1-e^{-\left( \frac{t}{\theta }\right )}\,\! }[/math]


The ML estimate [math]\displaystyle{ \hat{\theta}\,\! }[/math] = 572.3, and the standard deviation is [math]\displaystyle{ se_{\hat\theta}\,\! }[/math] = 41.72. Therefore the variance is 1740.56.


The 95% 2-sided confidence interval for [math]\displaystyle{ {\theta}\,\! }[/math] are:

  • Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on
[math]\displaystyle{ -2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = X^{2}_{(0.90,1)}\,\! }[/math]


The two solutions of [math]\displaystyle{ \theta\,\! }[/math] in the above equation will be the confidence bounds for [math]\displaystyle{ \theta\,\! }[/math].


  • Based on lognormal approximation, the confidence interval is [496, 660]. The calculation is:
[math]\displaystyle{ \begin{alignat}{2} [\theta_{L},\theta_{U}]&= \hat{\theta}exp\left(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}}\right)\\ &= \left[572.3\times exp \left(-1.96\times\frac{41.72}{572.3}\right),572.3\times exp \left(1.96\times\frac{41.72}{572.3}\right)\right]\\ &= [496,660]\\ \end{alignat} }[/math]


Results in Weibull++


The ML estimator for [math]\displaystyle{ \theta\,\! }[/math] and its variance are 572.27 and 1740.52, respectively. They are given below.

1PE interval data.png

The ML estimator for [math]\displaystyle{ \theta\,\! }[/math] and the variance are the same as the values given in the book.


The 95% 2-sided confidence interval for [math]\displaystyle{ \theta\,\! }[/math] are:

  • Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is:
1PE interval data qcp.png


  • Based on lognormal approximation (select FM for the bound method), the confidence bounds are:
1PE interval data qcp FM.png
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