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==Confidence Bounds==
#REDIRECT [[The Exponential Distribution]]
In this section, we present the methods used in the application to estimate the different types of confidence bounds for exponentially distributed data. The complete derivations were presented in detail (for a general function) in Chapter 5.
At this time we should point out that exact confidence bounds for the exponential distribution have been derived, and exist in a closed form, utilizing the <math>{{\chi }^{2}}</math> distribution. These are described in detail in Kececioglu [20], and are covered in the section on test design in Chapter 11. For most exponential data analyses, Weibull++ will use the approximate confidence bounds, provided from the Fisher information matrix or the likelihood ratio, in order to stay consistent with all of the other available distributions in the application. The <math>{{\chi }^{2}}</math> confidence bounds for the exponential distribution are discussed in more detail in Chapter 11.
 
===Fisher Matrix Bounds===
 
====Bounds on the Parameters====
 
For the failure rate <math>\hat{\lambda }</math> the upper (<math>{{\lambda }_{U}}</math>) and lower (<math>{{\lambda }_{L}}</math>) bounds are estimated by [30]:
 
 
::<math>\begin{align}
  & {{\lambda }_{U}}= & \hat{\lambda }\cdot {{e}^{\left[ \tfrac{{{K}_{\alpha }}\sqrt{Var(\hat{\lambda })}}{\hat{\lambda }} \right]}} \\
&  &  \\
& {{\lambda }_{L}}= & \frac{\hat{\lambda }}{{{e}^{\left[ \tfrac{{{K}_{\alpha }}\sqrt{Var(\hat{\lambda })}}{\hat{\lambda }} \right]}}} 
\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 variance of <math>\hat{\lambda },</math> <math>Var(\hat{\lambda }),</math> is estimated from the Fisher matrix, as follows:
 
 
::<math>Var(\hat{\lambda })={{\left( -\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} \right)}^{-1}}</math>
 
 
where <math>\Lambda </math> is the log-likelihood function of the exponential distribution, described in Appendix C.
 
Note that no true MLE solution exists for the case of the two-parameter exponential distribution. The mathematics simply break down while trying to simultaneously solve the partial derivative equations for both the <math>\gamma </math> and <math>\lambda </math> parameters, resulting in unrealistic conditions. The way around this conundrum involves setting <math>\gamma ={{T}_{1}},</math> or the first time-to-failure, and calculating <math>\lambda </math> in the regular fashion for this methodology. Weibull++ treats <math>\gamma </math> as a constant when computing bounds, i.e. <math>Var(\hat{\gamma })=0.</math> (See the discussion in Appendix C for more information.)
 
====Bounds on Reliability====
 
The reliability of the two-parameter exponential distribution is:
 
 
::<math>\hat{R}(T;\hat{\lambda })={{e}^{-\hat{\lambda }(T-\hat{\gamma })}}</math>
 
 
The corresponding confidence bounds are estimated from:
 
 
::<math>\begin{align}
  & {{R}_{L}}= & {{e}^{-{{\lambda }_{U}}(T-\hat{\gamma })}} \\
& {{R}_{U}}= & {{e}^{-{{\lambda }_{L}}(T-\hat{\gamma })}} 
\end{align}</math>
 
These equations hold true for the one-parameter exponential distribution, with <math>\gamma =0</math>.
 
 
====Bounds on Time====
 
The bounds around time for a given exponential percentile, or reliability value, are estimated by first solving the reliability equation with respect to time, or reliable life:
 
 
::<math>\hat{T}=-\frac{1}{{\hat{\lambda }}}\cdot \ln (R)+\hat{\gamma }</math>
 
 
The corresponding confidence bounds are estimated from:
 
 
::<math>\begin{align}
  & {{T}_{U}}= & -\frac{1}{{{\lambda }_{L}}}\cdot \ln (R)+\hat{\gamma } \\
& {{T}_{L}}= & -\frac{1}{{{\lambda }_{U}}}\cdot \ln (R)+\hat{\gamma } 
\end{align}</math>
 
 
The same equations apply for the one-parameter exponential with <math>\gamma =0.</math>
 
===Likelihood Ratio Confidence Bounds===
 
====Bounds on Parameters====
 
For one-parameter distributions such as the exponential, the likelihood confidence bounds are calculated by finding values for <math>\theta </math> that satisfy:
 
::<math>-2\cdot \text{ln}\left( \frac{L(\theta )}{L(\hat{\theta })} \right)=\chi _{\alpha ;1}^{2}</math>
 
This equation can be rewritten as:
 
::<math>L(\theta )=L(\hat{\theta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}</math>
 
For complete data, the likelihood function for the exponential distribution is given by:
 
::<math>L(\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\lambda \cdot {{e}^{-\lambda \cdot {{x}_{i}}}}</math>
 
where the <math>{{x}_{i}}</math> values represent the original time-to-failure data.  For a given value of <math>\alpha </math>, values for <math>\lambda </math> can be found which represent the maximum and minimum values that satisfy Eqn. (lratio3). These represent the confidence bounds for the parameters at a confidence level <math>\delta ,</math> where <math>\alpha =\delta </math> for two-sided bounds and <math>\alpha =2\delta -1</math> for one-sided.
 
 
====Example 5====
 
Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming an exponential distribution, the MLE parameter estimate is calculated to be  <math>\hat{\lambda }=0.013514.</math>  Calculate the 85% two-sided confidence bounds on these parameters using the likelihood ratio method.
 
=====Solution to Example 5=====
The first step is to calculate the likelihood function for the parameter estimates:
 
::<math>\begin{align}
  L(\hat{\lambda })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\hat{\lambda })=\underset{i=1}{\overset{N}{\mathop \prod }}\,\hat{\lambda }\cdot {{e}^{-\hat{\lambda }\cdot {{x}_{i}}}} \\
  L(\hat{\lambda })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,0.013514\cdot {{e}^{-0.013514\cdot {{x}_{i}}}} \\
  L(\hat{\lambda })= & 3.03647\times {{10}^{-12}} 
\end{align}</math>
 
where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:
 
::<math>L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
 
Since our specified confidence level, <math>\delta </math>, is 85%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.85;1}^{2}=2.072251.</math> We can now substitute this information into the equation:
 
::<math>\begin{align}
  L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\
  L(\lambda )-3.03647\times {{10}^{-12}}\cdot {{e}^{\tfrac{-2.072251}{2}}}= & 0, \\
  L(\lambda )-1.07742\times {{10}^{-12}}= & 0. 
\end{align}</math>
 
It now remains to find the values of <math>\lambda </math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>\lambda </math> that will satisfy the equation.  These values represent the <math>\delta =85%</math> two-sided confidence limits of the parameter estimate <math>\hat{\lambda }</math>. For our problem, the confidence limits are:
 
::<math>{{\lambda }_{0.85}}=(0.006572,0.024172)</math>
 
 
====Bounds on Time and Reliability====
In order to calculate the bounds on a time estimate for a given reliability, or on a reliability estimate for a given time, the likelihood function needs to be rewritten in terms of one parameter and time/reliability, so that the maximum and minimum values of the time can be observed as the parameter is varied. This can be accomplished by substituting a form of the exponential reliability equation into the likelihood function. The exponential reliability equation can be written as:
 
::<math>R={{e}^{-\lambda \cdot t}}</math>
 
This can be rearranged to the form:
 
::<math>\lambda =\frac{-\text{ln}(R)}{t}</math>
 
This equation can now be substituted into Eqn. (explikelihood) to produce a likelihood equation in terms of <math>t</math> and <math>R\ \ :</math>
 
::<math>L(t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\left( \frac{-\text{ln}(R)}{t} \right)\cdot {{e}^{\left( \tfrac{\text{ln}(R)}{t} \right)\cdot {{x}_{i}}}}</math>
 
The unknown parameter <math>t/R</math> depends on what type of bounds are being determined. If one is trying to determine the bounds on time for a given reliability, then <math>R</math> is a known constant and <math>t</math> is the unknown parameter. Conversely, if one is trying to determine the bounds on reliability for a given time, then <math>t</math> is a known constant and <math>R</math> is the unknown parameter. Either way, Eqn. (expliketr) can be used to solve Eqn. (lratio3) for the values of interest.
 
====Example 6====
For the data given in Example 5, determine the 85% two-sided confidence bounds on the time estimate for a reliability of 90%. The ML estimate for the time at <math>R(t)=90%</math> is <math>\hat{t}=7.797</math>.
 
=====Solution to Example 6=====
In this example, we are trying to determine the 85% two-sided confidence bounds on the time estimate of 7.797. This is accomplished by substituting <math>R=0.90</math> and <math>\alpha =0.85</math> into Eqn. (expliketr). It now remains to find the values of <math>t</math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>t</math> that will satisfy the equation. These values represent the <math>\delta =85%</math> two-sided confidence limits of the time estimate <math>\hat{t}</math>. For our problem, the confidence limits are:
 
::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>
 
 
====Example 7====
For the data given in Example 5, determine the 85% two-sided confidence bounds on the reliability estimate for a <math>t=50</math>. The ML estimate for the time at <math>t=50</math> is <math>\hat{R}=50.881%</math>.
 
=====Solution to Example 7=====
In this example, we are trying to determine the 85% two-sided confidence bounds on the reliability estimate of 50.881%. This is accomplished by substituting <math>t=50</math> and <math>\alpha =0.85</math> into Eqn. (expliketr). It now remains to find the values of <math>R</math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>t</math> that will satisfy the equation. These values represent the <math>\delta =85%</math> two-sided confidence limits of the reliability estimate <math>\hat{R}</math>. For our problem, the confidence limits are:
 
::<math>{{\hat{R}}_{t=50}}=(29.861%,71.794%)</math>

Latest revision as of 06:58, 10 August 2012

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