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===Fisher Matrix Confidence Bounds===
#REDIRECT [[The_Normal_Distribution]]
====Bounds on the Parameters====
The lower and upper bounds on the mean,  <math>\widehat{\mu }</math> , are estimated from:
 
::<math>\begin{align}
  & {{\mu }_{U}}= & \widehat{\mu }+{{K}_{\alpha }}\sqrt{Var(\widehat{\mu })}\text{ (upper bound),} \\
& {{\mu }_{L}}= & \widehat{\mu }-{{K}_{\alpha }}\sqrt{Var(\widehat{\mu })}\text{ (lower bound)}\text{.} 
\end{align}</math>
 
 
Since the standard deviation,  <math>{{\widehat{\sigma }}_{T}}</math> , must be positive,  <math>\ln ({{\widehat{\sigma }}_{T}})</math>  is treated as normally distributed, and the bounds are estimated from:
 
::<math>\begin{align}
  & {{\sigma }_{U}}= & {{\widehat{\sigma }}_{T}}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{T}})}}{{{\widehat{\sigma }}_{T}}}}}\text{ (upper bound),} \\
& {{\sigma }_{L}}= & \frac{{{\widehat{\sigma }}_{T}}}{{{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{T}})}}{{{\widehat{\sigma }}_{T}}}}}}\text{ (lower bound),} 
\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 variances and covariances of  <math>\widehat{\mu }</math>  and  <math>{{\widehat{\sigma }}_{T}}</math>  are estimated from the Fisher matrix, as follows:
 
 
::<math>\left( \begin{matrix}
  \widehat{Var}\left( \widehat{\mu } \right) & \widehat{Cov}\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right)  \\
  \widehat{Cov}\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) & \widehat{Var}\left( {{\widehat{\sigma }}_{T}} \right)  \\
\end{matrix} \right)=\left( \begin{matrix}
  -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\mu }^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \mu \partial {{\sigma }_{T}}}  \\
  {} & {}  \\
  -\tfrac{{{\partial }^{2}}\Lambda }{\partial \mu \partial {{\sigma }_{T}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \sigma _{T}^{2}}  \\
\end{matrix} \right)_{\mu =\widehat{\mu },\sigma =\widehat{\sigma }}^{-1}</math>
 
 
<math>\Lambda </math>  is the log-likelihood function of the normal distribution, described in
Chapter 3 and Appendix C.
 
====Bounds on Reliability====
The reliability of the normal distribution is:
 
 
::<math>\widehat{R}(T;\hat{\mu },{{\hat{\sigma }}_{T}})=\int_{T}^{\infty }\frac{1}{{{\widehat{\sigma }}_{T}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\widehat{\mu }}{{{\widehat{\sigma }}_{T}}} \right)}^{2}}}}dt</math>
 
 
Let  <math>\widehat{z}(t;\hat{\mu },{{\hat{\sigma }}_{T}})=\tfrac{t-\widehat{\mu }}{{{\widehat{\sigma }}_{T}}},</math>  then  <math>\tfrac{dz}{dt}=\tfrac{1}{{{\widehat{\sigma }}_{T}}}.</math>  For  <math>t=T</math> ,  <math>\widehat{z}=\tfrac{T-\widehat{\mu }}{{{\widehat{\sigma }}_{T}}}</math> , and for  <math>t=\infty ,</math>  <math>\widehat{z}=\infty .</math>  The above equation then becomes:
 
::<math>\hat{R}(\widehat{z})=\int_{\widehat{z}(T)}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz</math>
 
 
The bounds on  <math>z</math>  are estimated from:
 
::<math>\begin{align}
  & {{z}_{U}}= & \widehat{z}+{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \\
& {{z}_{L}}= & \widehat{z}-{{K}_{\alpha }}\sqrt{Var(\widehat{z})} 
\end{align}</math>
 
:where:
 
::<math>Var(\widehat{z})={{\left( \frac{\partial z}{\partial \mu } \right)}^{2}}Var(\widehat{\mu })+{{\left( \frac{\partial z}{\partial {{\sigma }_{T}}} \right)}^{2}}Var({{\widehat{\sigma }}_{T}})+2\left( \frac{\partial z}{\partial \mu } \right)\left( \frac{\partial z}{\partial {{\sigma }_{T}}} \right)Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right)</math>
 
:or:
 
::<math>Var(\widehat{z})=\frac{1}{\widehat{\sigma }_{T}^{2}}\left[ Var(\widehat{\mu })+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{T}})+2\cdot \widehat{z}\cdot Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \right]</math>
 
 
The upper and lower bounds on reliability are:
 
::<math>\begin{align}
  & {{R}_{U}}= & \int_{{{z}_{L}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (upper bound)} \\
& {{R}_{L}}= & \int_{{{z}_{U}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (lower bound)} 
\end{align}</math>
 
====Bounds on Time====
The bounds around time for a given normal percentile (unreliability) are estimated by first solving the reliability equation with respect to time, as follows:
 
 
::<math>\hat{T}(\widehat{\mu },{{\widehat{\sigma }}_{T}})=\widehat{\mu }+z\cdot {{\widehat{\sigma }}_{T}}</math>
 
:where:
 
::<math>z={{\Phi }^{-1}}\left[ F(T) \right]</math>
 
:and:
 
::<math>\Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z(T)}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz</math>
 
 
The next step is to calculate the variance of  <math>\hat{T}(\widehat{\mu },{{\widehat{\sigma }}_{T}})</math>  or:
 
 
::<math>\begin{align}
  Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \mu } \right)}^{2}}Var(\widehat{\mu })+{{\left( \frac{\partial T}{\partial {{\sigma }_{T}}} \right)}^{2}}Var({{\widehat{\sigma }}_{T}}) \\
  & +2\left( \frac{\partial T}{\partial \mu } \right)\left( \frac{\partial T}{\partial {{\sigma }_{T}}} \right)Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) \\
  Var(\hat{T})= & Var(\widehat{\mu })+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{T}})+2\cdot z\cdot Cov\left( \widehat{\mu },{{\widehat{\sigma }}_{T}} \right) 
\end{align}</math>
 
 
The upper and lower bounds are then found by:
 
 
::<math>\begin{align}
  & {{T}_{U}}= & \hat{T}+{{K}_{\alpha }}\sqrt{Var(\hat{T})}\text{ (upper bound)} \\
& {{T}_{L}}= & \hat{T}-{{K}_{\alpha }}\sqrt{Var(\hat{T})}\text{ (lower bound)} 
\end{align}</math>
 
====Example 4====
Using the data of Example 2 and assuming a normal distribution, estimate the parameters using the MLE method.
 
=====Solution to Example 4=====
In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function,  <math>\Lambda ,</math>  are given by:
 
 
::<math>\begin{align}
  \frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{T}_{i}}-\mu )=0 \\
  \frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{T}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0 
\end{align}</math>
 
 
(The derivations of these equations are presented in Appendix C.) Substituting the values of  <math>{{T}_{i}}</math>  and solving the above system simultaneously, we get  <math>\widehat{\sigma }=29.58</math>  hours <math>,</math>  <math>\widehat{\mu }=45</math>  hours <math>.</math>
 
The Fisher matrix is:
 
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000  \\
  {} & {} & {}  \\
  \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500  \\
\end{matrix} \right]</math>
 
 
Using Weibull++ , the MLE method can be selected from the Set Analysis page.
 
[[Image:weibullfolio1.png|thumb|center|400px| ]]
 
The plot of the solution for this example is shown next.
 
[[Image:weibullfolioplot1.png|thumb|center|400px| ]]

Latest revision as of 09:05, 3 August 2012

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