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Lognormal MLE Solution with Right Censored Data
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This example compares the Lognormal MLE solution with Fisher matrix bound for right censored data.
Reference Case
The data on page 199 of the book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998 is used.
Data
| Number in State
|
State F or S
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Time to Failure
|
| 1 |
F |
252
|
| 1 |
F |
315
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| 1 |
F |
369
|
| 1 |
F |
403
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| 1 |
F |
474
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| 15 |
S |
500
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Result
- The MLE solution is [math]\displaystyle{ \hat{\mu} = 6.56, \ \hat{\sigma} = 0.543\,\! }[/math].
- The variance and covariance matrix is
- [math]\displaystyle{
\sum =\begin{bmatrix}
0.0581 & 0.0374 \\
0.0374 & 0.0405
\end{bmatrix}\,\! }[/math]
Results in Weibull++
- The MLE solution and the variance/covariance matrix:
- The Fisher matrix bound for parameters:
- For Ln-mu (using normal approximation of Eqn. 8.7 on page 187):
- [math]\displaystyle{ [\hat{\mu}'_{L}, \hat{\mu}'_{U}] = \hat{\mu}' \pm z_{(1-\alpha \setminus 2)}se_{\hat{\mu}'}\,\! }[/math]
- For a confidence level of 0.95, it is:
- [math]\displaystyle{ \begin{alignat}{2}
[\hat{\mu}'_{L}, \hat{\mu}'_{U}] =& \hat{\mu}' \pm z_{(1-\alpha \setminus 2)}se_{\hat{\mu}'}\\
=& 6.564256 \pm 1.96 \times (0.0581)^{0.5}\\
=& [6.0918, 7.0366933]\\
\end{alignat}\,\! }[/math]
- For Ln-Std (using log-normal approximation of Eqn. 8.8 on page 188):
- [math]\displaystyle{ [\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] = \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha \setminus 2)}se_{\hat{\sigma'}}}{\hat{\sigma}'})\,\! }[/math]
- For confidence level of 0.95, it is:
- [math]\displaystyle{ \begin{alignat}{2}
[\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] =& \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha \setminus 2)}se_{\hat{\sigma'}}}{\hat{\sigma}'})\\
=& 0.534049 \times exp(\pm \tfrac{1.96\times 0.040562^{0.5}}{0.534049})\\
=& [0.255, 1.118]\\
\end{alignat} \,\! }[/math]
- The results in Weibull++ are:
