Template:Bounds on instantaneous failure intensity camsaa-cb: Difference between revisions

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(Created page with '===Bounds on Instantaneous Failure Intensity=== ====Fisher Matrix Bounds==== The instantaneous failure intensity, <math>{{\lambda }_{i}}(t)</math> , must be positive, thus <mat…')
 
 
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===Bounds on Instantaneous Failure Intensity===
#REDIRECT [[Crow-AMSAA_-_NHPP#Bounds_on_Instantaneous_Failure_Intensity]]
====Fisher Matrix Bounds====
The instantaneous failure intensity,  <math>{{\lambda }_{i}}(t)</math> , must be positive, thus  <math>\ln {{\lambda }_{i}}(t)</math>  is treated as being normally distributed.
 
::<math>\frac{\ln {{{\hat{\lambda }}}_{i}}(t)-\ln {{\lambda }_{i}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{i}}(t)})}\text{ }\tilde{\ }\text{ }N(0,1)</math>
 
The approximate confidence bounds on the instantaneous failure intensity are then estimated from:
 
::<math>CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}}</math>
 
:where
 
::<math>{{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}</math> 
 
::<math>\begin{align}
  & Var({{{\hat{\lambda }}}_{i}}(t))= & {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\
&  & +2\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) 
\end{align}</math>
 
The variance calculation is the same as Eqn. (variance1) and:
 
<br>
::<math>\begin{align}
  & \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}\ln t \\
& \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }= & \hat{\beta }{{t}^{\hat{\beta }-1}} 
\end{align}</math>
 
====Crow Bounds====
The Crow instantaneous failure intensity confidence bounds are given as:
 
::<math>\begin{align}
  & {{\lambda }_{i}}{{(t)}_{L}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\
& {{\lambda }_{i}}{{(t)}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} 
\end{align}</math>

Latest revision as of 04:09, 24 August 2012