Template:Likelihood Ratio Confidence Bounds FMB ED - Revision history

Lisa Hacker: Redirected page to The Exponential Distribution

2012-08-10T08:21:06Z

Redirected page to The Exponential Distribution

← Older revision		Revision as of 08:21, 10 August 2012
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	~~===Likelihood Ratio Confidence Bounds===~~		#REDIRECT [[The Exponential Distribution]]

	~~{{Bounds on Parameters.LRCB.FMB.ED}}~~

	~~{{Bounds on Time and Reliability.LRCB.FMB.ED}}~~

Nicolette Young: /* Likelihood Ratio Confidence Bounds */

2012-01-04T16:15:48Z

Likelihood Ratio Confidence Bounds

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	{{Bounds on Parameters.LRCB.FMB.ED}}		{{Bounds on Parameters.LRCB.FMB.ED}}

	~~====~~Bounds on Time and Reliability~~====~~		{{Bounds on Time and Reliability.LRCB.FMB.ED}}
	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.

Nicolette Young at 16:12, 4 January 2012

2012-01-04T16:12:33Z

← Older revision		Revision as of 16:12, 4 January 2012
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	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.		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>~~

Nicolette Young: /* Example 5 */

2012-01-04T16:11:38Z

Example 5

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	{{Bounds on Parameters.LRCB.FMB.ED}}		{{Bounds on Parameters.LRCB.FMB.ED}}

	~~====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====		====Bounds on Time and Reliability====

Nicolette Young: /* Bounds on Parameters */

2012-01-04T16:11:12Z

Bounds on Parameters

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 ===Likelihood Ratio Confidence Bounds===
-====Bounds on Parameters====
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 ====Example 5====

Nicolette Young: Created page with '===Likelihood Ratio Confidence Bounds=== ====Bounds on Parameters==== For one-parameter distributions such as the exponential, the likelihood confidence bounds are calculated b…'

2012-01-04T16:09:52Z

Created page with '===Likelihood Ratio Confidence Bounds=== ====Bounds on Parameters==== For one-parameter distributions such as the exponential, the likelihood confidence bounds are calculated b…'

New page

===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>