Template:Example: Bayesian Test Design with Prior Information from Expert Opinion

Bayesian Test Design 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]\displaystyle{ E\left(R_{0}\right)=\frac{a+4b+c}{6}=\frac{0.8+4(0.85)+0.97}{6}=0.861667 }[/math]

[math]\displaystyle{ Var\left(R_{0}\right)=\frac{c-a}{6}=\frac{0.97-0.8}{6}=0.028333 }[/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].

[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]=0.861667\left[\frac{0.861667-\left(0.861667\right)^{2}}{0.028333}-1\right]=2.763331 }[/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]=\left(1-0.861667\right)\left[\frac{0.861667-\left(0.861667\right)^{2}}{0.028333}-1\right]=0.44363 }[/math]

With [math]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \beta\,\!_{0} }[/math] known, any single value of the 4 quantities system reliability R, confidence level CL, number of units n, or number of failures r can be calculated from the other 3.

System reliability R can be found if confidence level CL, number of units n, and number of failures r are known. Given the following data

CL = 0.8

n = 20

r = 1

the number of successes s is

[math]\displaystyle{ s = n – r = 19 }[/math]

and the posterior distribution is calculated as

[math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 }[/math]

[math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 }[/math]

[math]\displaystyle{ R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 }[/math]

The confidence level CL can be found if system reliability R, number of units n, and number of failures r are known. Given the following data

R = 0.9

n = 20

r = 1

the number of successes s is

[math]\displaystyle{ s = n – r = 19 }[/math]

and the posterior distribution is calculated as

[math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 }[/math]

[math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 }[/math]

[math]\displaystyle{ CL=\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.812164 }[/math]

The number of units n can be found if system reliability R, confidence level CL, and number of failures r are known. Given the following data

R = 0.9

CL = 0.8

r = 1

the Number of Units utility in the Non-Parametric Binomial tab of the Design a Reliability Demonstration Test window can be used to solve for n.

File:Drt8.jpg

The figure above shows that, in this case, n = 28.925085. The posterior distribution can now be calculated as [math]\displaystyle{ s=n-r=27.925085 }[/math] [math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=2.763331+27.925085=30.688416 }[/math] [math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 }[/math]

which results in a confidence level of [math]\displaystyle{ CL=1-1-\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.91775 }[/math] Since the confidence level (0.91775) is greater than that which is required (0.8), we can reduce the number of units n until the calculated confidence level is close to the required value of 0.8. This results in the number of units n = 19.31.