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| ===Bounds on Instantaneous MTBF===
| | #REDIRECT [[Crow-AMSAA_-_NHPP#Bounds_on_Instantaneous_MTBF_2]] |
| ====Fisher Matrix Bounds====
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| The instantaneous MTBF, <math>{{m}_{i}}(t)</math> , must be positive, thus <math>\ln {{m}_{i}}(t)</math> is approximately treated as being normally distributed as well.
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| ::<math>\frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1)</math>
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| The approximate confidence bounds on the instantaneous MTBF are then estimated from:
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| ::<math>CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}}</math>
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| :where:
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| ::<math>{{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}</math>
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| ::<math>\begin{align}
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| & Var({{{\hat{m}}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\
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| & & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda })
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| \end{align}</math>
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| The variance calculation is the same as Eqn. (variances) and:
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| ::<math>\begin{align}
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| & \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\hat{\lambda }{{{\hat{\beta }}}^{2}}}{{t}^{1-\hat{\beta }}}-\frac{1}{\hat{\lambda }\hat{\beta }}{{t}^{1-\hat{\beta }}}\ln t \\
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| & \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}\hat{\beta }}{{t}^{1-\hat{\beta }}}
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| \end{align}</math>
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| ====Crow Bounds====
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| :Step 1: Calculate <math>P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K</math> .
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| :Step 2: Calculate:
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| ::<math>A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{{{\left[ P{{(i)}^{{\hat{\beta }}}}\ln P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{\widehat{\beta }}}\ln P{{(i-1)}^{{\hat{\beta }}}} \right]}^{2}}}{\left[ P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{{\hat{\beta }}}} \right]}</math>
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| :Step 3: Calculate <math>D=\sqrt{\tfrac{1}{A}+1}</math> and <math>W=\tfrac{({{z}_{1-\alpha /2}})\cdot D}{\sqrt{N}}</math> . Thus an approximate 2-sided <math>(1-\alpha )</math> 100-percent confidence interval on <math>{{\hat{m}}_{i}}(t)</math> is:
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| ::<math>MTB{{F}_{i}}={{\widehat{m}}_{i}}(1\pm W)</math>
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