Template:Exponential Distribution Example: Likelihood Ratio Bound for lambda - Revision history

Lisa Hacker: Redirected page to The Exponential Distribution

2012-08-10T08:41:54Z

Redirected page to The Exponential Distribution

← Older revision		Revision as of 08:41, 10 August 2012
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	~~'''Likelihood Ratio Bound on <math>\lambda </math>'''~~		#REDIRECT [[The Exponential Distribution]]

	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'''~~

	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 the likelihood ratio equation 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>~~

Richard House at 02:02, 8 August 2012

2012-08-08T02:02:30Z

← Older revision		Revision as of 02:02, 8 August 2012
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	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.


	'''Solution'''		'''Solution'''

Harry Guo at 22:44, 2 March 2012

2012-03-02T22:44:14Z

← Older revision		Revision as of 22:44, 2 March 2012
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	\end{align}</math>		\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:		where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:

	::<math>L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>		::<math>L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>

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

2012-02-09T18:26:33Z

Example 5: Likelihood Ratio Bound on \lambda

← Older revision		Revision as of 18:26, 9 February 2012
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	~~=====Example 5:~~ Likelihood Ratio Bound on <math>\lambda </math>~~=====~~		'''Likelihood Ratio Bound on <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.


	'''Solution ~~to Example 5~~'''		'''Solution'''

	The first step is to calculate the likelihood function for the parameter estimates:		The first step is to calculate the likelihood function for the parameter estimates:

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

2012-02-08T22:42:20Z

Example 5: Likelihood Ratio Bound on \lambda

← Older revision		Revision as of 22:42, 8 February 2012
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	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.


	'''Solution to Example 5'''		'''Solution to Example 5'''

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

2012-02-08T22:17:02Z

Example 5: Likelihood Ratio Bound for \lambda

← Older revision		Revision as of 22:17, 8 February 2012
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	=====Example 5: Likelihood Ratio Bound ~~for~~ <math>\lambda </math>=====		=====Example 5: Likelihood Ratio Bound on <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.

Harry Guo: /* Solution to Example 5 */

2012-02-08T22:14:38Z

Solution to Example 5

← Older revision		Revision as of 22:14, 8 February 2012
Line 3:		Line 3:
	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.

	~~======~~Solution to Example 5~~======~~		'''Solution to Example 5'''

	The first step is to calculate the likelihood function for the parameter estimates:		The first step is to calculate the likelihood function for the parameter estimates:

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

2012-02-08T22:09:27Z

Example 5: Likelihood Ratio Bound for \lambda

← Older revision		Revision as of 22:09, 8 February 2012
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	====Example 5: Likelihood Ratio Bound for <math>\lambda </math>====		=====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.

	=====Solution to Example 5=====		======Solution to Example 5======
	The first step is to calculate the likelihood function for the parameter estimates:		The first step is to calculate the likelihood function for the parameter estimates:

Harry Guo: Created page with '====Example 5: Likelihood Ratio Bound for $\lambda$ ==== Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming …'

2012-02-08T00:27:40Z

Created page with '====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 …'

New page

====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>

Template:Exponential Distribution Example: Likelihood Ratio Bound for lambda - Revision history

Lisa Hacker: Redirected page to The Exponential Distribution

Richard House at 02:02, 8 August 2012

Harry Guo at 22:44, 2 March 2012

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

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

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

Harry Guo: /* Solution to Example 5 */

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

Harry Guo: Created page with '====Example 5: Likelihood Ratio Bound for \lambda ==== Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming …'

Harry Guo: Created page with '====Example 5: Likelihood Ratio Bound for $\lambda$ ==== Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming …'