|
|
| (3 intermediate revisions by one other user not shown) |
| Line 1: |
Line 1: |
| ===Fisher Matrix Bounds===
| | #REDIRECT [[The Exponential Distribution]] |
| | |
| ====Bounds on the Parameters====
| |
| | |
| For the failure rate <math>\hat{\lambda }</math> the upper (<math>{{\lambda }_{U}}</math>) and lower (<math>{{\lambda }_{L}}</math>) bounds are estimated by [30]:
| |
| | |
| | |
| ::<math>\begin{align}
| |
| & {{\lambda }_{U}}= & \hat{\lambda }\cdot {{e}^{\left[ \tfrac{{{K}_{\alpha }}\sqrt{Var(\hat{\lambda })}}{\hat{\lambda }} \right]}} \\
| |
| & & \\
| |
| & {{\lambda }_{L}}= & \frac{\hat{\lambda }}{{{e}^{\left[ \tfrac{{{K}_{\alpha }}\sqrt{Var(\hat{\lambda })}}{\hat{\lambda }} \right]}}}
| |
| \end{align}</math>
| |
| | |
| where <math>{{K}_{\alpha }}</math> is defined by:
| |
| | |
| | |
| ::<math>\alpha =\frac{1}{\sqrt{2\pi }}\int_{{{K}_{\alpha }}}^{\infty }{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt=1-\Phi ({{K}_{\alpha }})</math>
| |
| | |
| | |
| If <math>\delta </math> is the confidence level, then <math>\alpha =\tfrac{1-\delta }{2}</math> for the two-sided bounds, and <math>\alpha =1-\delta </math> for the one-sided bounds.
| |
| The variance of <math>\hat{\lambda },</math> <math>Var(\hat{\lambda }),</math> is estimated from the Fisher matrix, as follows:
| |
| | |
| | |
| ::<math>Var(\hat{\lambda })={{\left( -\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} \right)}^{-1}}</math>
| |
| | |
| | |
| where <math>\Lambda </math> is the log-likelihood function of the exponential distribution, described in Appendix C.
| |
| | |
| Note that no true MLE solution exists for the case of the two-parameter exponential distribution. The mathematics simply break down while trying to simultaneously solve the partial derivative equations for both the <math>\gamma </math> and <math>\lambda </math> parameters, resulting in unrealistic conditions. The way around this conundrum involves setting <math>\gamma ={{T}_{1}},</math> or the first time-to-failure, and calculating <math>\lambda </math> in the regular fashion for this methodology. Weibull++ treats <math>\gamma </math> as a constant when computing bounds, i.e. <math>Var(\hat{\gamma })=0.</math> (See the discussion in Appendix C for more information.)
| |
| | |
| ====Bounds on Reliability====
| |
| | |
| The reliability of the two-parameter exponential distribution is:
| |
| | |
| | |
| ::<math>\hat{R}(T;\hat{\lambda })={{e}^{-\hat{\lambda }(T-\hat{\gamma })}}</math>
| |
| | |
| | |
| The corresponding confidence bounds are estimated from:
| |
| | |
| | |
| ::<math>\begin{align}
| |
| & {{R}_{L}}= & {{e}^{-{{\lambda }_{U}}(T-\hat{\gamma })}} \\
| |
| & {{R}_{U}}= & {{e}^{-{{\lambda }_{L}}(T-\hat{\gamma })}}
| |
| \end{align}</math>
| |
| | |
| These equations hold true for the one-parameter exponential distribution, with <math>\gamma =0</math>.
| |
| | |
| | |
| ====Bounds on Time====
| |
| | |
| The bounds around time for a given exponential percentile, or reliability value, are estimated by first solving the reliability equation with respect to time, or reliable life:
| |
| | |
| | |
| ::<math>\hat{T}=-\frac{1}{{\hat{\lambda }}}\cdot \ln (R)+\hat{\gamma }</math>
| |
| | |
| | |
| The corresponding confidence bounds are estimated from:
| |
| | |
| | |
| ::<math>\begin{align}
| |
| & {{T}_{U}}= & -\frac{1}{{{\lambda }_{L}}}\cdot \ln (R)+\hat{\gamma } \\
| |
| & {{T}_{L}}= & -\frac{1}{{{\lambda }_{U}}}\cdot \ln (R)+\hat{\gamma }
| |
| \end{align}</math>
| |
| | |
| | |
| The same equations apply for the one-parameter exponential with <math>\gamma =0.</math>
| |