Non-Parametric Bayesian - Expert Opinion: Difference between revisions
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<noinclude>{{Banner Weibull Examples}} | <noinclude>{{Banner Weibull Examples}} | ||
''This example appears in the [ | ''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference]''. | ||
</noinclude> | </noinclude> | ||
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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: | |||
::<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''''' | ||
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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| | [[Image:Bayesian Prior Expert Solve for n.png|center|800px| ]] | ||
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This example appears in the Life data analysis reference.
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
- 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]\displaystyle{ E\left(R_{0}\right)=\frac{a+4b+c}{6}=0.861667 \,\! }[/math]
- [math]\displaystyle{ 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]\displaystyle{ \alpha_{0}\,\! }[/math] and [math]\displaystyle{ \beta_{0}\,\! }[/math] used in the beta distribution for the system reliability, as given next:
- [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \alpha_{0}\,\! }[/math] and [math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=146.0794\,\! }[/math]
- [math]\displaystyle{ \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]\displaystyle{ 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]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=146.07943\,\! }[/math]
- [math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=21.40153\,\! }[/math]
Finally, from this posterior distribution, the corresponding confidence level for reliability R=0.85 is:
- [math]\displaystyle{ 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.
