Template:Bounds on Parameters.LRCB.FMB.ED - Revision history

Lisa Hacker: Redirected page to The Exponential Distribution

2012-08-10T06:54:46Z

Redirected page to The Exponential Distribution

← Older revision		Revision as of 06:54, 10 August 2012
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	~~====Bounds on Parameters====~~		#REDIRECT [[The Exponential Distribution]]

	~~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({{t}_{i}};\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\lambda \cdot {{e}^{-\lambda \cdot {{t}_{i}}}}</math>~~

	where the <math>{{t}_{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 the above likelihood ratio equation. 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:'''~~
	{{Exponential Distribution ~~Example: Likelihood Ratio Bound for lambda}}~~

Harry Guo at 18:27, 9 February 2012

2012-02-09T18:27:13Z

← Older revision		Revision as of 18:27, 9 February 2012
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	where the <math>{{t}_{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 the above likelihood ratio equation. 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.		where the <math>{{t}_{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 the above likelihood ratio equation. 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:'''		'''Example 5:'''
	{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}		{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}

Harry Guo: /* Bounds on Parameters */

2012-02-09T18:26:56Z

Bounds on Parameters

← Older revision		Revision as of 18:26, 9 February 2012
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	where the <math>{{t}_{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 the above likelihood ratio equation. 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.		where the <math>{{t}_{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 the above likelihood ratio equation. 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:'''
	{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}		{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}

Harry Guo at 00:29, 8 February 2012

2012-02-08T00:29:52Z

← Older revision		Revision as of 00:29, 8 February 2012
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	{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}		{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}

	~~====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====~~

	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>~~

	~~====Example 6: Likelihood Ratio Bound for Time====~~
	For the data given in Example 5: Likelihood Ratio Bound for <math>\lambda </math>, 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 the likelihood ratio bound equation. 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: Likelihood Ratio Bound for Reliability====~~
	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 the likelihood ratio bound equation. 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>~~

Harry Guo: /* Bounds on Parameters */

2012-02-08T00:28:29Z

Bounds on Parameters

← Older revision		Revision as of 00:28, 8 February 2012
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	where the <math>{{t}_{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 the above likelihood ratio equation. 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.		where the <math>{{t}_{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 the above likelihood ratio equation. 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.

	{{Exponential Distribution Example: Likelihood Ratio Bound for ~~<math>\~~lambda ~~</math>~~}}		{{Exponential Distribution Example: Likelihood Ratio Bound for lambda}}

	====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====		====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====

Harry Guo: /* Example 5: Likelihood Ratio Bound for \lambda */

2012-02-08T00:25:02Z

Example 5: Likelihood Ratio Bound for \lambda

← Older revision		Revision as of 00:25, 8 February 2012
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	where the <math>{{t}_{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 the above likelihood ratio equation. 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.		where the <math>{{t}_{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 the above likelihood ratio equation. 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.

			{{Exponential Distribution Example: Likelihood Ratio Bound for <math>\lambda </math>}}

	====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====		====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====

Harry Guo: /* Solution to Example 6 */

2012-02-08T00:01:23Z

Solution to Example 6

← Older revision		Revision as of 00:01, 8 February 2012
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	=====Solution to Example 6=====		=====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:		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 the likelihood ratio bound equation. 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>		::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>

Harry Guo: /* Example 7 */

2012-02-08T00:01:00Z

Example 7

← Older revision		Revision as of 00:01, 8 February 2012
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	::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>		::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>

	====Example 7====		====Example 7: Likelihood Ratio Bound for Reliability====
	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>.		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=====		=====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:		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 the likelihood ratio bound equation. 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>		::<math>{{\hat{R}}_{t=50}}=(29.861%,71.794%)</math>

Harry Guo: /* Example 6 */

2012-02-07T23:58:48Z

Example 6

← Older revision		Revision as of 23:58, 7 February 2012
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	::<math>{{\lambda }_{0.85}}=(0.006572,0.024172)</math>		::<math>{{\lambda }_{0.85}}=(0.006572,0.024172)</math>

	====Example 6====		====Example 6: Likelihood Ratio Bound for Time====
	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>.		For the data given in Example 5: Likelihood Ratio Bound for <math>\lambda </math>, 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=====		=====Solution to Example 6=====
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	::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>		::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>


	====Example 7====		====Example 7====

Harry Guo: /* Example 5 */

2012-02-07T23:56:30Z

Example 5

← Older revision		Revision as of 23:56, 7 February 2012
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	where the <math>{{t}_{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 the above likelihood ratio equation. 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.		where the <math>{{t}_{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 the above likelihood ratio equation. 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====		====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====

	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.		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.