Template:Bounds on projected failure intensity rga

Bounds on Projected Failure Intensity

Fisher Matrix Bounds

The projected failure intensity [math]\displaystyle{ {{\lambda }_{P}}(T) }[/math] must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{P}}(T) }[/math] is approximately treated as being normally distributed as well:

[math]\displaystyle{ \frac{\ln {{{\hat{\lambda }}}_{P}}(T)-\ln {{\lambda }_{P}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{P}}(T)})}\sim N(0,1) }[/math]

[math]\displaystyle{ CB={{\hat{\lambda }}_{P}}(T){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{P}}(T))}/{{{\hat{\lambda }}}_{P}}(T)}} }[/math]

where:

• [math]\displaystyle{ {{\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.

• [math]\displaystyle{ {{\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.

• [math]\displaystyle{ {{N}_{i}} }[/math] is the total failure number of the [math]\displaystyle{ {{i}^{th}} }[/math] distinct BD mode.

You can then get:

[math]\displaystyle{ 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]

where:

[math]\displaystyle{ \begin{align} & \hat{h}(T)= & \frac{M}{T}\bar{\beta } \\ & 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 }) \end{align} }[/math]

The [math]\displaystyle{ Var(\hat{\beta }) }[/math] can be obtained from Fisher Matrix based on [math]\displaystyle{ M }[/math] distinct BD modes.

Crow Bounds

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{P}}(T)]}_{L}}= & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}-\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot {{C}^{2}}+\frac{{{C}^{4}}}{4}} \\ & {{[{{\lambda }_{P}}(T)]}_{U}}= & {{{\hat{\lambda }}}_{P}}(T)+\frac{{{C}^{2}}}{2}+\sqrt{{{{\hat{\lambda }}}_{P}}(T)\cdot \ \,{{C}^{2}}+\frac{{{C}^{4}}}{4}} \end{align} }[/math]

where [math]\displaystyle{ C=\tfrac{{{z}_{1-\alpha /2}}}{\sqrt{T}} }[/math] .