Non-Parametric Bayesian - Expert Opinion - Revision history

Lisa Hacker at 18:55, 18 September 2023

2023-09-18T18:55:42Z

← Older revision		Revision as of 18:55, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the [~~[Reliability_Test_Design\|~~Life ~~Data Analysis Reference book]~~]''.		''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''.

	</noinclude>		</noinclude>

Kate Racaza at 17:14, 9 December 2015

2015-12-09T17:14:38Z

← Older revision		Revision as of 17:14, 9 December 2015
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	Again, the above beta distribution equation for the system reliability can be utilized. The figure below shows the result from Weibull++. The results show that the required sample size is 103. Weibull++ always displays the sample size as an integer.		Again, the above beta distribution equation for the system reliability can be utilized. The figure below shows the result from Weibull++. The results show that the required sample size is 103. Weibull++ always displays the sample size as an integer.

	[[Image:Bayesian Prior Expert Solve for n.png\|center\|~~650px~~\| ]]		[[Image:Bayesian Prior Expert Solve for n.png\|center\|800px\| ]]

Richard House at 19:51, 26 September 2012

2012-09-26T19:51:03Z

Richard House at 03:14, 30 August 2012

2012-08-30T03:14:31Z

← Older revision		Revision as of 03:14, 30 August 2012
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	::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}=0.000803 </math>		::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}=0.000803 </math>

	~~<br>~~ These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha_{0}</math> and <math>\beta_{0}</math> used in the beta distribution for the system reliability, as given next:		These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> used in the beta distribution for the system reliability, as given next:

	::<math> \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=127.0794</math>		::<math> \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=127.0794</math>
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	::<math> \beta\,\!_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=20.40153</math>		::<math> \beta\,\!_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=20.40153</math>

	With <math>\alpha_{0}</math> and <math>\beta_{0}</math> known, any single value of the four quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other three using the beta distribution function:		With <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> known, any single value of the four quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other three using the beta distribution function:

	::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)</math>		::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)</math>

Chris Kahn at 05:40, 22 August 2012

2012-08-22T05:40:54Z

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	</noinclude>		</noinclude>

	You can use the non-parametric Bayesian method to design a test using prior knowledge about a system's reliability. For example, suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:		You can use the non-parametric Bayesian method to design a test using prior knowledge about a system's reliability. For example, suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:

Chris Kahn at 05:15, 22 August 2012

2012-08-22T05:15:56Z

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	</noinclude>		</noinclude>
	~~'''Example: Non-Parametric Bayesian with Prior Information from Expert Opinion'''~~

	~~Suppose~~ you wanted to know the reliability of a system and you had the following prior knowledge of the system:		You can use the non-parametric Bayesian method to design a test using prior knowledge about a system's reliability. For example, suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:

	:*Lowest possible reliability: ''a'' = 0.8		:*Lowest possible reliability: ''a'' = 0.8

Lisa Hacker at 01:28, 15 August 2012

2012-08-15T01:28:15Z

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			<noinclude>{{Banner Weibull Examples}}
			''This example appears in the [[Reliability_Test_Design\|Life Data Analysis Reference book]]''.

			</noinclude>
	'''Example: Non-Parametric Bayesian with Prior Information from Expert Opinion'''		'''Example: Non-Parametric Bayesian with Prior Information from Expert Opinion'''

Lisa Hacker: Created page with ''''Example: Non-Parametric Bayesian with Prior Information from Expert Opinion''' Suppose you wanted to know the reliability of a system and you had the following prior knowledg…'

2012-08-15T01:27:26Z

Created page with ''''Example: Non-Parametric Bayesian with Prior Information from Expert Opinion''' Suppose you wanted to know the reliability of a system and you had the following prior knowledg…'

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'''Example: Non-Parametric Bayesian with Prior Information from Expert Opinion'''

Suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:

:*Lowest possible reliability: ''a'' = 0.8

:*Most likely reliability: ''b'' = 0.85

:*Highest possible reliability: ''c'' = 0.97

This information can be used to approximate the expected value and the variance of the prior system reliability.

::<math> E\left(R_{0}\right)=\frac{a+4b+c}{6}=0.861667 </math>

::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}=0.000803 </math>

<br> These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha_{0}</math> and <math>\beta_{0}</math> used in the beta distribution for the system reliability, as given next:

::<math> \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=127.0794</math>

::<math> \beta\,\!_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=20.40153</math>

With <math>\alpha_{0}</math> and <math>\beta_{0}</math> known, any single value of the four quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other three using the beta distribution function:

::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)</math>

'''Solve for System Reliability ''R'''''

Given ''CL'' = 0.9, ''n'' = 20, and ''r'' = 1, using the above prior information to solve ''R''.

First, we get the number of successes: s = n – r = 19. Then the parameters in the posterior beta distribution for ''R'' are calculated as:

::<math> \alpha\,\!=\alpha\,\!_{0}+s=146.0794</math>

::<math> \beta\,\!=\beta\,\!_{0}+r=21.40153</math>

Finally, from this posterior distribution, the system reliability ''R'' at a confidence level of ''CL''=0.9 is solved as:

::<math> R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.838374 </math>

'''Solve for Confidence Level ''CL'''''

Given ''R'' = 0.85, ''n'' = 20, and ''r'' = 1, using the above prior information on system reliability to solve for ''CL''.

First, we get the number of successes: s = n – r = 19. Then the parameters in the posterior beta distribution for ''R'' are calculated as:

::<math> \alpha\,\!=\alpha\,\!_{0}+s=146.07943</math>

::<math> \beta\,\!=\beta\,\!_{0}+r=21.40153</math>

Finally, from this posterior distribution, the corresponding confidence level for reliability ''R''=0.85 is:

::<math> CL=\text{Beta}\left(R,\alpha,\beta\right)=0.81011 </math>

'''Solve for Sample Size ''n'''''

Given ''R'' = 0.9, ''CL'' = 0.8, and ''r'' = 1, using the above prior information on system reliability to solve the required sample size in the demonstration test.

Again, the above beta distribution equation for the system reliability can be utilized. The figure below shows the result from Weibull++. The results show that the required sample size is 103. Weibull++ always displays the sample size as an integer.

[[Image:Bayesian Prior Expert Solve for n.png|center|650px| ]]

← Older revision		Revision as of 19:51, 26 September 2012
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	This information can be used to approximate the expected value and the variance of the prior system reliability.		This information can be used to approximate the expected value and the variance of the prior system reliability.

	::<math> E\left(R_{0}\right)=\frac{a+4b+c}{6}=0.861667 </math>		::<math> E\left(R_{0}\right)=\frac{a+4b+c}{6}=0.861667 \,\!</math>

	::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}=0.000803 </math>		::<math>Var({{R}_{0}})={{\left( \frac{c-a}{6} \right)}^{2}}=0.000803 \,\!</math>

	These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> used in the beta distribution for the system reliability, as given next:		These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> used in the beta distribution for the system reliability, as given next:

	::<math> \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=127.0794</math>		::<math> \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=127.0794\,\!</math>

	::<math> \beta\,\!_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=20.40153</math>		::<math> \beta\,\!_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=20.40153\,\!</math>

	With <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> known, any single value of the four quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other three using the beta distribution function:		With <math>\alpha_{0}\,\!</math> and <math>\beta_{0}\,\!</math> known, any single value of the four quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other three using the beta distribution function:

	::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)</math>		::<math>1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right)\,\!</math>

	'''Solve for System Reliability ''R'''''		'''Solve for System Reliability ''R'''''
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	First, we get the number of successes: s = n – r = 19. Then the parameters in the posterior beta distribution for ''R'' are calculated as:		First, we get the number of successes: s = n – r = 19. Then the parameters in the posterior beta distribution for ''R'' are calculated as:

	::<math> \alpha\,\!=\alpha\,\!_{0}+s=146.0794</math>		::<math> \alpha\,\!=\alpha\,\!_{0}+s=146.0794\,\!</math>


	::<math> \beta\,\!=\beta\,\!_{0}+r=21.40153</math>		::<math> \beta\,\!=\beta\,\!_{0}+r=21.40153\,\!</math>

	Finally, from this posterior distribution, the system reliability ''R'' at a confidence level of ''CL''=0.9 is solved as:		Finally, from this posterior distribution, the system reliability ''R'' at a confidence level of ''CL''=0.9 is solved as:

	::<math> R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.838374 </math>		::<math> R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.838374 \,\!</math>

	'''Solve for Confidence Level ''CL'''''		'''Solve for Confidence Level ''CL'''''
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	First, we get the number of successes: s = n – r = 19. Then the parameters in the posterior beta distribution for ''R'' are calculated as:		First, we get the number of successes: s = n – r = 19. Then the parameters in the posterior beta distribution for ''R'' are calculated as:

	::<math> \alpha\,\!=\alpha\,\!_{0}+s=146.07943</math>		::<math> \alpha\,\!=\alpha\,\!_{0}+s=146.07943\,\!</math>

	::<math> \beta\,\!=\beta\,\!_{0}+r=21.40153</math>		::<math> \beta\,\!=\beta\,\!_{0}+r=21.40153\,\!</math>

	Finally, from this posterior distribution, the corresponding confidence level for reliability ''R''=0.85 is:		Finally, from this posterior distribution, the corresponding confidence level for reliability ''R''=0.85 is:

	::<math> CL=\text{Beta}\left(R,\alpha,\beta\right)=0.81011 </math>		::<math> CL=\text{Beta}\left(R,\alpha,\beta\right)=0.81011 \,\!</math>

	'''Solve for Sample Size ''n'''''		'''Solve for Sample Size ''n'''''