Template:Bayesian test design

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The nonparametric analysis performed in the last section made no assumptions about the underlying distribution of the data and was performed with only the immediate data at hand. However, if prior information regarding system performance is available, it can be incorporated into a Bayesian nonparametric analysis. This subsection will demonstrate how to incorporate prior information about system reliability and also how to incorporate prior information from subsystem tests.

Determining system performance

The values of system reliability R, confidence level CL, number of units tested n, and number of failures r are related by the equation

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

where Beta is the incomplete Beta function. If [math]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \beta\,\!_{0} }[/math] are known, then any quantity of interest can be calculated using the remaining three. The next two examples demonstrate how to calculate [math]\displaystyle{ \alpha\,\!_{0} }[/math] and [math]\displaystyle{ \beta\,\!_{0} }[/math] depending on the type of prior information available.

Template:Btd w info on reliability

Example – Demonstrate with prior information from subsystem tests

Given prior information from subsystem tests, the system reliability can be estimated. Assume a system of interest is composed of 3 subsystems A, B and C. Prior information from tests of these subsystems is given in the table below.

Subsystem Number of Units (n) Number of Failures (r)
A 20 0
B 30 1
C 100 4

This data can be used to calculate expected value and variance of the subsystems’ reliability.

[math]\displaystyle{ E\left(R_{i}\right)=\frac{n-r}{n+1} }[/math]
[math]\displaystyle{ Var\left(R_{i}\right)=\frac{\left(n-r\right)\left(r+1\right)}{\left(n+1\right)^{2}\left(n+2\right)} }[/math]

The results of these calculations are given in the table below.

Subsystem Mean Variance
A 0.952380952 0.002061
B 0.935483871 0.001886
C 0.95049505 0.000461

These values can then be substituted back into equations (2) and (3) to find the prior system reliability.

[math]\displaystyle{ E\left(R_{0}\right)=\left(0.95238095\right)\left(0.935483871\right)\left(0.95049505\right)=0.846831227 }[/math]
[math]\displaystyle{ \text{Var}\left(R_{0}\right)=\left(0.952380952^{2}+0.002061\right)\left(0.935483871^{2}+0.001886\right)\left(0.95049505^{2}+0.000461\right)-\left(0.952380952^{2}\right)\left(0.935483871^{2}\right)\left(0.95049505^{2}\right)=0.003546663 }[/math]

Calculating and as before yields

[math]\displaystyle{ \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{\text{Var}\left(R_{0}\right)}-1\right]=30.12337003 }[/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]=5.448499634 }[/math]

which can then be used to calculate system reliability R, confidence level CL, number of units n, and number of failures r as discussed in the previous subsection, Prior Information on System Reliability.