1P-Exponential MLE Solution for Interval Data

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Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.

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(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\ &=\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]\\ &= [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.

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

[math]\displaystyle{ F(t)= 1-e^{-(\frac{t}{\theta})} }[/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 \frac{L(\theta)}{L(\hat{\theta})} = 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].

1P-Exponential MLE Solution for Interval Data

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