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| ===Bounds on Projected Failure Intensity===
| | #REDIRECT [[Crow Extended Confidence Bounds]] |
| ====Fisher Matrix Bounds====
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| The projected failure intensity <math>{{\lambda }_{P}}(T)</math> must be positive, thus <math>\ln {{\lambda }_{P}}(T)</math> is approximately treated as being normally distributed as well:
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| ::<math>\frac{\ln {{{\hat{\lambda }}}_{P}}(T)-\ln {{\lambda }_{P}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{P}}(T)})}\sim N(0,1)</math>
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| ::<math>CB={{\hat{\lambda }}_{P}}(T){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{P}}(T))}/{{{\hat{\lambda }}}_{P}}(T)}}</math>
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| <br>
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| where:
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| <br>
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| :• <math>{{\hat{\lambda }}_{P}}(T)=\tfrac{{{N}_{A}}}{T}+\underset{i=1}{\overset{M}{\mathop{\sum }}}\,(1-{{d}_{i}})\tfrac{{{N}_{i}}}{T}+\overline{d}\tfrac{M}{T}\bar{\beta }</math> when there are no BC modes.
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| :• <math>{{\hat{\lambda }}_{P}}(T)={{\widehat{\lambda }}_{EM}}={{\widehat{\lambda }}_{CA}}-{{\widehat{\lambda }}_{BD}}+\underset{i=1}{\overset{M}{\mathop{\sum }}}\,(1-{{d}_{i}})\tfrac{{{N}_{i}}}{T}+\overline{d}\widehat{h}(T|BD)</math> when there are BC modes.
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| :• <math>{{N}_{i}}</math> is the total failure number of the <math>{{i}^{th}}</math> distinct BD mode.
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| <br>
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| You can then get:
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| ::<math>Var({{\lambda }_{P}}(T))\approx Var({{\hat{\gamma }}_{GP}})+\mu _{d}^{2}Var(h(T))\approx \frac{{{{\hat{r}}}_{GP}}}{T}+\mu _{d}^{2}Var(h(T))</math>
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| :where:
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| ::<math>\begin{align}
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| & \hat{h}(T)= & \frac{M}{T}\bar{\beta } \\
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| & Var(\hat{h}(T))= & {{(\frac{M}{T})}^{2}}Var(\bar{\beta })={{(\frac{M}{T})}^{2}}{{(\frac{M}{M-1})}^{2}}Var(\hat{\beta })=\frac{{{M}^{4}}}{{{T}^{2}}{{(M-1)}^{2}}}Var(\hat{\beta })
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| \end{align}</math>
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| The <math>Var(\hat{\beta })</math> can be obtained from Fisher Matrix based on <math>M</math> distinct BD modes.
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| <br>
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| ====Crow Bounds====
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| ::<math>\begin{align}
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| & {{[{{\lambda }_{P}}(T)]}_{L}}= & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot {{C}^{2}}+\frac{{{C}^{4}}}{4}} \\
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| & {{[{{\lambda }_{P}}(T)]}_{U}}= & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot \ \,{{C}^{2}}+\frac{{{C}^{4}}}{4}}
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| \end{align}</math>
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| where <math>C=\tfrac{{{z}_{1-\alpha /2}}}{\sqrt{T}}</math> .
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| <br>
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