Use of Bayesian Techniques for Reliability

Introduction

The world of statistics is traditionally divided into two mutually exclusive camps: Classical and Bayesian. They are divided on basically a principles issue, as will be explained.

The Classical statistician believes that all distribution parameters are fixed values (i.e., Mean Life is 100 hours). On the other hand, Bayesians believe that parameters are random variables and have a distribution of their own (i.e., Mean Life can range from 90 to 110 hours following some pattern). These two principles exclude each other.

In this START sheet, we present the Bayesian approach and some applications to reliability, leaving the subjective evaluation and interpretation of this concept to the reader. Perhaps the best way to introduce this concept is via an example. Let's assume that the life X of a device follows the Exponential distribution with parameter (Mean) θ. The density f(x) of X is:

f(x)=(1 / θ) Exp(-x / θ)

For the Classical statistician, the parameter Mean Life (θ) is a fixed number, say 100 hours. For the Bayesian statistician, in turn, the parameter Mean Life (θ) is a random variable having a (possibly different) statistical distribution with density f(θ).

For example, assume its distribution is Uniform (90, 110):

f(θ)=1/(110-90) = 1/20; for 90 < θ < 110
= 0 for all other values of θ

In practice, this means that the Mean Life θ of the device is not fixed but varies about value 100 as a Uniform random variable does, being as low as 90 hours and as high as 110. See Figure 1.

Figure 1. Representation of the Prior Function f(θ,) known as the Prior Distribution. The probabilities associated with the Prior are referred to as "subjective probabilities" because they quantify some belief as to where the parameter Mean Life θ lays and how it's value varies. (Click to Zoom)

Then, given a random sample of "n" device lives (denoted x1, ..., xn) the conditional distribution of these data, given the parameter Mean Life (θ) is defined and denoted:

f(x1,...,xn | θ) = f(x1 | θ)...f(xn | θ) = 1 / θn Exp(-Σx / θ)

for some value of θ

We then calculate the Joint Distribution f (x1, ..., xn, ) of both the data and the Mean parameter θ, by multiplying their Conditional and Prior distributions:

f(x1,...,xn θ) = f(x1,...,xn | θ) x f(θ)

= (1 / θn) Exp(-Σx / θ) x (1/20) = (1/200n) Exp(-Σx / θ)

for: 90 < θ < 110

We now integrate the Joint Distribution, for all possible values of the parameter Mean Life (θ) to obtain the "marginal" distribution of the sample data:

We can now obtain the "Posterior" distribution of parameter θ, which is the distribution of the Mean Life θ, given the sample x1, ..., xn that we have drawn:

The practical use of all of this theory in reliability studies is as follows. Assume that we believe that a parameter θ, the Mean Life of a device, lies in some range of values, with some given pattern. We can then define such range and pattern (its distribution) via the Prior f(θ) of this parameter. We can draw a sample of actual device lives and obtain via the above approach, an expression for the Posterior distribution of θ. Finally, this Posterior distribution can be continually refined with more incoming data and used for estimation. We can summarize this as a step-wise process:

1. Decide which distribution parameter (e.g., Mean ) you want to study.
2. Define a Prior Distribution f(θ) for this parameter (e.g., Uniform (90, 110)).
3. Establish the Conditional Distribution f(X | θ) for the variable X (e.g., for the device life).
4. Obtain the Joint distribution for the sample data (x1, ..., xn) and θ by multiplying their Prior and Conditional distributions, obtained in steps 2 and 3.
5. Obtain the Marginal distribution by integrating out the θ on the above Joint distribution.
6. Calculate the Posterior Distribution f(θ | x1, ..., xn) of θ, by dividing the Joint by the Marginal distributions and use it for estimation and testing.
In applying this theory, the analyst encounters some practical problems. First, as seen in the preceding example, the closed form of the Posterior distribution f(θ | x1, ..., xn) is not always easy to obtain. Sometimes the Prior distributions are defined using mathematical arguments, rather than statistical ones. This allows the analyst to obtain a closed form solution for the resulting Posterior distribution. However, if the Prior distribution is postulated just for analytical convenience and it is inappropriate, then the inferences made about the parameter, based on such inappropriate Priors, will also be erroneous.

A Numerical Example

The following numerical example helps to illustrate the Bayesian approach. Let's assume that a device has a mission time, T = 100 hours, and that past experience has shown that it fails with probability ρ (and survives with probability = 1 - ρ). Let's assume that such prior experience has also taught us that there are only three different device failure probabilities, ρi, and that these ρi, can occur with the probabilities shown in Figure 2 (denoted f(ρ)). This means that a device failure probability of 0.01 occurs 20% of the time, that a failure probability of 0.02 occurs 50% of the time and that a failure probability of 0.03 occurs 30% of the time.

Figure 2. Prior Distribution of ρ (Click to Zoom)

Assume we want to find the Bayes estimate for the true failure probability, ρ, of this device. To achieve this, we test a random sample of five operating devices and observe one failure.

Let's define the number of failed devices in our sample as X. All (n = 5) five test devices are randomly selected and each fails independently, with the same probability ρ. Hence, each device outcome is a Bernoulli trial and their sum (i.e., total number of failures X = x) is distributed as a Binomial, with n = 5 and probability ρ(denoted B(x; n,ρ)):

P(X = x | ρ) = f(x | ρ) = B(x;n,p)

=Cnx ρx (1-p)n-x; x = 0,1,2,3,4,5

For example, when the true failure probability is ρ = 0.01, the probability that we observe one failure in the random sample of five devices is:

f(x = 1 | ρ=0.01) = B(1,5,0.01)

C51 0.011(1 - 0.01)5 - 1 = 0.04803

By the same token, and applying the above formula with ρ = 0.02 and 0.03, respectively, we get:

f(1|0.02) = B(1;5,0.02) = 0.09224

f(1|0.03) = B(1;5,0.03) = 0.13279

From the previous section we know that the Joint distribution f(x,ρ) = f(x|ρ)*f(ρ), where the occurrence of the device failure probability is denoted f(ρ). Therefore, we can write:

 ρ f(1,ρ) 0.01 0.02 0.03 0.009606 0.04612 0.03984

For example, for the case of one failure (x = 1) in the sample (our case of interest) when the true failure rate is = 0.01 and its probability of occurrence is 0.2, we obtain:

f(x = 1, ρ = 0.01) = f(1|0.01)*f(0.01) = 0.04803*0.2 = 0.009606

The Marginal distribution g(1) of observing one failure (x = 1) in a random sample of five (irrespective of whether the true failure probability = 0.01, 0.02, or 0.03) is the sum of the probability f(,x) for the number of failures observed (x = 1):

g(1) = Σρ f(1,ρ) = 0.0096 + 0.0461 + 0.0398 = 0.095563

We can then obtain the Posterior distribution f (ρ | x = 1) of the proportion of defectives ρ, in the case of testing and observing a single failure (x = 1) in a random sample of five devices (n = 5). We do this via the Posterior distribution formula (i.e., dividing the Joint distribution by the Marginal distribution):

We tabulate the values of the probabilities f(ρ |x = 1):

 ρ f(ρ | x = 1) 0.01 0.02 0.03 0.1 0.483 0.417

By definition, the Bayes Point Estimate (ρ*) of a parameter is nothing else but the weighted average (expected value) of the Posterior Distribution of this parameter:

ρ* = Σ ρf(ρ|x).

Hence, the Bayes point estimate of the true device probability of failure ρ, based on a test that resulted in only one failure, from an experiment with five operating devices, when postulating the prior distribution f(ρ) for the probability of failure ρ, is obtained by the following equation.

ρ* = Σ ρ x f(ρ|x = 1)

= 0.01 x 0.1005 + 0.02 x 0.4826 + 0.03 x 0.4169

= 0.023164

We can then compare the Bayes point estimator with the "classical" estimator (the sample proportion x/n = 1/5 = 0.2).

To provide an illustrative contrast, assume now that no failures were observed in the above-mentioned test (x = 0). In such case, the corresponding probability of obtaining no failures, given the failure probabilities ρ = 0.01 is:

P(X = 0 |ρ) = B(x = 0; n = 5, ρ = 0.01) = 0.9501

Applying the same formula for ρ = 0.02 and 0.03, we obtain:

f(x = 0, ρ = 0.02) = 0.9039 f(x = 0, ρ = 0.03) = 0.8587

The corresponding Joint distribution for zero failures is now:

 ρ f(0,ρ) 0.01 0.02 0.03 0.1902 0.452 0.2576

The Marginal distribution for zero failures is then:

g(0) = Σρ f(0,ρ) = 0.1902 + 0.4520 + 0.2576 = 0.8998

Finally, the Posterior distribution for zero failures f(ρ | x = 0) is:

 ρ f(ρ |x = 0) 0.01 0.02 0.03 0.2114 0.5023 0.2863

One interpretation of these results is that the original Prior probabilities f (ρ) have been "updated" via the experiment, and the "new Prior" is now the Posterior probability f (ρ | x).

We can also try to compare the point estimator ρ* (from this second test, that has no failures) with the "Classical" point estimator (x/n). But since now there are no failures, the sample proportion is zero and will not provide a useful estimate of .

In Table 1, we complete the Bayes analyses results for all the other cases (x = 2, 3, 4, and 5 failures) in this experiment.

Table 1. Table of Joint and Marginal Probabilities for the Experiment
Fail. Prob. Occur-FP f(0, ρ) f(1, ρ) f(2, ρ) f(3, ρ) f(4, ρ) f(5, ρ)
0.01 0.200000 0.190198 0.009606 0.000194 0.000002 0.000000 0.000000
0.02 0.500000 0.451961 0.046119 0.001883 0.000039 0.000001 0.000000
0.03 0.300000 0.257620 0.039838 0.002464 0.000076 0.000000 0.000000
Marginal Probs. 0.899779 0.095563 0.004541 0.000117 0.000001 0.000000

We can see, from Table 1, that X = 4 and 5 cases are nil. Table 2 shows the posterior distributions and the Bayes estimates of ρ.

Table 2. Table of Posterior distributions and Bayes Estimates of ρ
F.P. f(ρ | x = 0) f(ρ | x = 1) f(ρ | x = 2) f(ρ | x = 3)
0.01 0.211383 0.10052 0.042725 0.017138
0.02 0.502302 0.482599 0.414584 0.329906
0.03 0.286315 0.41688 0.542692 0.652956
Bayes (*) 0.020749 0.023164 0.025000 0.026358

Summarizing: in the absence of any particular information about the parameters in question, the Classical estimate is not only the best one but the only one available. When the Prior distribution is an accurate reflection of reality, Bayes estimate is more efficient than the Classical. However, if the Prior distribution is inappropriate or inaccurate, the Bayes estimate may also be significantly in error.

Normal Distribution Case

In practice, the Prior distribution f(θ) used to describe the variation pattern (or belief) of the unknown parameter θ, is often the Normal distribution. When applicable, the Normal greatly simplifies all the preceding derivations, as we will show through another example.

Assume that the unknown parameter of interest θ is now the mean of a Normal (?, σ2) population. Assume that the Prior distribution f(θ) of parameter is, itself, also Normal with mean μ and known standard deviation . Assume that we have obtained the sample average from a random sample of size "n" from the population of interest (i.e., of the Normal (θ, σ2)).

Then, the Posterior Distribution of the population mean is also the Normal, and its parameters μ* and σ*, respectively, will be given by the formulas:

Graphical depictions of the Prior and Posterior distributions θ of are shown in Figure 3.

Figure 3. Prior and Posterior Distributions for the parameter Population Mean (Click to Zoom)

Therefore, a 100(1 - α)% Bayesian confidence interval (CI) for the true mean θ (say, an 95% CI for an α = 0.05 and zα/2 = 1.96) can be constructed from the Posterior distribution of using the following equation.

μ* ± zα/2 x σ* = (μ* - zα/2 x σ* , μ* + zα/2 x σ*)

We will now illustrate the use of all this information with a practical reliability example. Assume that a manufacturer of an electronic amplifier wants to estimate the length of the life X of its device. Assume that, for historical reasons, it is believed that the distribution of the life is Normal, with unknown mean and known standard deviation σ = 150 hours.

Assume that, based on the same historical reasons, this electronic amplifier manufacturer believes that the Prior Distribution for the unknown mean life is also Normal, with known mean μ = 1000 hours and standard deviation δ = 100 hours.

Assume now that the manufacturer measures the lives from a random sample of n = 25 electronic amplifiers (if the sample size is n ≥ 30, then the distribution of life X doesn't even need to be Normal and the unknown variance σ2 can now be approximated by the sample variance s2). Assume that this sample yields an average life x = 900 hours. Using all the information, we derive a 100(1 - α)% Bayesian CI for the amplifier mean life θ, as:

μ* - zα/2 x σ* < θ < μ* + za/2 x σ*

According to the preceding, the resulting Normal Posterior Distribution will have the following parameters.

Therefore, for α = 0.05 and zα/2 = 1.96, a 95% Bayesian CI for the unknown mean life θ is:

μ* ± zα/2 x σ* = 908.26 ± 1.96 √825.69 = 908.26 ± 56.32

= (851.94, 964.58)

Had we chosen to ignore (or had no access to) this information about the Prior Distribution, we could have still obtained a 95% Classic CI for θ as:

x ± zα/2 x σ / √n = 900 ± 1.96 x (150 /√25 = 9000 ± 58.8

= (841.2, 958.8)

As we have advised several times in the previous sections, if the assumed Prior Distribution is appropriate and the Prior's parameters are accurate, then the Bayesian CI is more efficient. Such efficiency becomes evident in the fact that the Bayesian CI is narrower (notice its smaller half-width, 56.32) than the Classical CI (half-width is 58.8). This reduction in the CI width occurs because the Bayesian approach uses more information than the Classical (e.g., the shape of the Prior distribution as well as its parameters).

In addition, the Bayesian point estimator (908.26) of the parameter (θ) differs from the Classical point estimator, given by the sample average (900). This difference is due to Bayes weighing the data via the Prior standard deviation and mean.

However, if any or all of the information regarding Bayes Prior (either the distribution or its parameters) which are used for deriving the Bayesian CI, is inaccurate or incorrect, then the CI obtained with such information will also be incorrect.

In such case, the Bayes point estimator (908.26) will be biased. In addition, the resulting CI, with its narrower half width and centered on this biased point estimator, will most likely not cover the true parameter.

Summary

Through the discussion and the examples presented here, we have seen how the Bayesian approach is applied to data analysis and how Bayesian point estimators and confidence intervals are obtained and updated. In particular we have seen how they are implemented to a life data analysis example.

Bayesian techniques are widely used by some practitioners in industrial statistics and can become very useful analysis tools. However, it is always necessary to bear in mind this large dependency that exists between the applicability of the postulated prior distributions and its parameters, and the results obtained, through them (e.g., our estimations).

We have just touched on a few relevant concepts of Bayesian statistics, a topic that is very extensive and parallels Classical statistics. In For Further Study, we list the usual background readings plus several references that can serve as starting points for those interested in studying in this topic at a more advanced level.

For Further Study

1. Probability and Statistics for Engineers and Scientists, Walpole and Myers, Prentice Hall, NJ, 1988.

2. An Introduction to Probability Theory and Mathematical Statistics, Rohatgi, V.K. Wiley, NY, 1976.

3. Methods for Statistical Analysis of Reliability and Life Data, Mann, N., R. Schafer, and N. Singpurwalla, John Wiley, NY, 1974.

4. Reliability and Life Testing Handbook, (Vol. 2), Kececioglu, D., Editor, Prentice Hall, NJ, 1994.

5. Practical Statistical Tools for Reliability Engineers, Coppola, A., RIAC, 1999.

6. A Practical Guide to Statistical Analysis of Material Property Data, Romeu, J.L. and C. Grethlein, AMPTIAC, 2000.

7. Statistical Confidence, Romeu, J.L., RIAC START, Volume 9, Number 4, http://theriac.org/DeskReference/viewDocument.php?id=223&Scope=reg