# Estimation of Failure Probability and Its Applications in Lifetime Data Analysis

Authors: Ming Han
Journal: International Journal of Quality, Statistics, and Reliability
 (2011)

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## Abstract

Since Lindley and Smith introduced the idea of hierarchical prior distribution, some
results have been obtained on hierarchical Bayesian method to deal with lifetime data.
But all those results obtained by means of hierarchical Bayesian methods involve
complicated integration compute. Though some computing methods such as Markov
Chain Monte Carlo (MCMC) are available, doing integration is still very inconvenient
for practical problems. This paper introduces a new method, namedE-Bayesian
estimation method, to estimate failure probability. In the case of one hyperparameter,
the definition of E-Bayesian estimation of the failure probability is provided; moreover,
the formulas of E-Bayesian estimation and hierarchical Bayesian estimation and the
property of E-Bayesian estimation of the failure probability are also provided. Finally,
calculation on practical problems shows that the provided method is feasible and easy
to perform.


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## Body

## 1. Introduction

In the area related to the reliability of industrial products, engineers often deal with the truncated data in life testing of products, where the data sometimes have small sample size or have been censored, and the products of interest have high reliability. In the literature, Lindley and Smith [1] first introduced the idea of hierarchical prior distribution. Han [2] developed some methods to construct hierarchical prior distribution. Recently, hierarchical Bayesian methods have been applied to data analysis [3]. However, complicated integration compute is a hard work by using hierarchical Bayesian methods in practical problems, though some computing methods such as Markov Chain Monte Carlo (MCMC) methods are available [4, 5].Han [6] introduced a new method—E-Bayesian estimation method—to estimate failure probability in the case of two hyperparameters, proposed the definition of E-Bayesian of failure probability, and provided formulas for E-Bayesian estimation of the failure probability under the cases of three different prior distributions of hyperparameters. But we did not provide formulas for hierarchical Bayesian estimation of the failure probability nor discuss the relations between the two estimations. In this paper, we will introduce the definition for E-Bayesian estimation of the failure probability, provide formulas both for E-Bayesian estimation and hierarchical Bayesian estimation, and also discuss the relations between the two estimations in the case of only one hyperparameter. We will see that the E-Bayesian estimation method is really simple.Conduct type I censored life testingm time, denote the censored times as ti(i=1,2,…,m), the corresponding sample numbers as ni, and the corresponding failure sample numbers observed in the testing process as ri(ri=0,1,2,…,ni).In the situation where no information about the life distribution of tested products is available, Mao and Luo [7] introduce a so-called curves fitting distribution method to give the estimation of the failure probability pi at time ti, pi=P(T<ti) for i=1,2,…,m, where T is the life of a product.This paper introduces a new method, calledE-Bayesian estimation method, to estimate failure probability. The definition and formulas of E-Bayesian estimation of the failure probability are described in Sections 2 and 3, respectively. In Section 4, formulas of hierarchical Bayesian estimation of the failure probability are proposed. In Section 5, the properties of E-Bayesian estimation are discussed. In Section 6, an application example introduces given. Section 7 is the conclusions.


## 2. Definition of E-Bayesian Estimation ofpi

Suppose that there areX failures among n samples, then X can be viewed as a random variable with binomial distribution Bin(n,pi). We take the conjugate prior of pi,Beta(a,b), with density function(1)π(pi∣a,b)=pia-1(1-pi)b-1B(a,b),
where 0<pi<1, B(a,b)=∫01ta-1(1-t)b-1dt is the beta function, and hyperparameters a>0 and b>0.The derivative ofπ(pi∣a,b) with respect to pi is (2)dπ(pi∣a,b)dpi=pia-2(1-pi)b-2[(a-1)(1-pi)-(b-1)pi]B(a,b).According to Han [2], a and b should be chosen to guarantee that π(pi∣a,b) is a decreasing function of pi. Thus 0<a<1, b>1.Given0<a<1, the larger the value of b, the thinner the tail of the density function. Berger had shown in [8] that the thinner tailed prior distribution often reduces the robustness of the Bayesian estimate. Consequently, the hyperparameter b should be chosen under the restriction 1<b<c, where c is a given upper bound. How to determine the constant c would be described later in an example.Sincea∈(0,1) and 1/2 is an expectation of uniform on (0, 1), that we take a=1/2. When 1<b<c and a=1/2, π(pi∣a,b) also is a decreasing function of pi.In this paper we only consider the case whena=1/2. Then the density function π(pi∣a,b) becomes (3)π(pi∣b)=pi(1/2)-1(1-pi)b-1B((1/2),b).The definition of E-Bayesian estimation was originally addressed by Han [6] in the case of two hyperparameters. In the case of one hyperparameter, the E-Bayesian estimation of failure probability is defined as follows.Definition 1.
Withp̂iB(b) being continuous,
(4)pîEB=∫Dp̂iB(b)π(b)db
is called the expected Bayesian estimation of pi (briefly E-Bayesian estimation), which is assumed to be finite, where D is the domain of b, p̂iB(b) is the Bayesian estimation of pi with hyperparameter b, and π(b) is the density function of b over D.Definition1 indicates that the E-Bayesian estimation of pi,(5)pîEB=∫Dp̂iB(b)π(b)db=E[p̂iB(b)],
is the expectation of the Bayesian estimation of pi for the hyperparameter.


## 3. E-Bayesian Estimation ofpi

Theorem 2.
For the testing data set{(ni,ri,ti),i=1,…,m} with type I censor, where ri=0,1,2,…,ni, let si=∑j=imnj and ei=∑j=1irj(i=1,2,…,m). If the prior density function π(pi∣b) of pi is given by (3), then, we have the following.  (i)
With the quadratic loss function, the Bayesian estimation ofpi is
(6)p̂iB(b)=ei+(1/2)si+b+(1/2).(ii)
If prior distribution ofb is uniform on (1, c), then E-Bayesian estimation of pi is
(7)p̂iEB=ei+(1/2)c-1ln⁡(si+c+(1/2)si+(3/2)).Proof.
(i) For the testing data set{(ni,ri,ti),i=1,…,m} with type I censor, where ri=0,1,2,…,ni, according to Han [6], the likelihood function of samples is
(8)L(ri∣pi)=(siei)piei(1-pi)si-ei,0<pi<1,
where si=∑j=imnj and ei=∑j=1irj(i=1,2,…,m).
Combined with the prior density functionπ(pi∣b) of pi given by (3), the Bayesian theorem leads to the posterior density function of pi,
(9)h1(pi∣si,b)=π(pi∣b)L(ri∣pi)∫01π(pi∣b)L(ri∣pi)dpi=piei+(1/2)-1(1-pi)si+b-ei-1∫01piei+(1/2)-1(1-pi)si+b-ei-1dpi=piei+(1/2)-1(1-pi)si+b-ei-1B(ei+(1/2),si+b-ei),0<pi<1.
Thus, with the quadratic loss function, the Bayesian estimation ofpi is
(10)p̂iB(b)=∫01pih1(pi∣si,b)dpi=∫01piei+(3/2)-1(1-pi)si+b-ei-1dpiB(ei+(1/2),si+b-ei)=B(ei+(3/2),si+b-ei)B(ei+(1/2),si+b-ei)=ei+(1/2)si+b+(1/2).
(ii) If prior distribution ofb is uniform on (1,c), then, by Definition 1, the E-Bayesian estimation of pi is
(11)p̂iEB=∫Dp̂iB(b)π(b)db=1c-1∫1cei+(1/2)si+b+(1/2)db=ei+(1/2)c-1ln⁡(si+c+(1/2)si+(3/2)).
This concludes the proof of Theorem2.


## 4. Hierarchical Bayesian Estimation

If the prior density functionπ(pi∣b) of pi is given by (3), how can the value of hyperparameter a be determined? Lindley and Smith [1] addressed an idea of hierarchical prior distribution, which suggested that one prior distribution may be adapted to the hyperparameters while the prior distribution includes hyperparameters.If the prior ofpi, π(pi∣b), is given by (3), prior distribution of b is uniform on (1,c), and the density function is π(b)=(1/(c-1)),1<b<c, then, hierarchical prior density function of pi is (12)π(pi)=∫1cπ(pi∣b)π(b)db=1c-1∫1cpi(1/2)-1(1-pi)b-1B((1/2),b)db,0<pi<1.Theorem 3.
For the testing data set{(ni,ri,ti),i=1,…,m} with type I censor, where ri=0,1,2,…,ni, let si=∑j=imnj and ei=∑j=1irj(i=1,2,…,m). If the hierarchical prior density function π(pi) of pi is given by (12), then, using the quadratic loss function, the hierarchical Bayesian estimation of pi is
(13)p̂iHB=∫1c(B(ei+(3/2),si+b-ei)/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db.Proof.
According to the course of the proof of Theorem2, the likelihood function of samples is
(14)L(ri∣pi)=(siei)piei(1-pi)si-ei,0<pi<1,
where si=∑j=imnj and ei=∑j=1irj(i=1,2,…,m).
From the hierarchical prior density function ofpi given by (12), the Bayesian theorem leads to the hierarchical posterior density function of pi,
(15)h2(pi∣si)=π(pi)L(ri∣pi)∫01π(pi)L(ri∣pi)dpi=∫1c(piei+(1/2)-1(1-pi)si+b-ei-1/B((1/2),b))db∫1c(1/B((1/2),b)){∫01piei+(1/2)-1(1-pi)si+b-ei-1dpi}db=∫1c(piei+(1/2)-1(1-pi)si+b-ei-1/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db,0<pi<1.
With the quadratic loss function, the hierarchical Bayesian estimation ofpi is
(16)p̂iHB=∫01pih2(pi∣si)dpi=∫1c(1/B((1/2),b)){∫01piei+(3/2)-1(1-pi)si+b-ei-1dpi}db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db=∫1c(B(ei+(3/2),si+b-ei)/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db.
Thus, the proof is completed.


## 5. Property of E-Bayesian Estimation ofpi

Now we discuss the relations betweenp̂iEB and p̂iHB in Theorems 2 and 3.Theorem 4.
In Theorems2 and 3, p̂iEB and p̂iHB satisfy limsi→∞p̂iEB=limsi→∞p̂iHB.Proof.
According to the course of the proof of Theorem2, we have that
(17)p̂iEB=1c-1∫1cei+(1/2)si+b+(1/2)db.
Whenb∈(1,c), (ei+(1/2)/si+b+(1/2)) is continuous; by the mean value theorem for definite integrals, there is at least one number b1∈(1,c) such that
(18)∫1cei+(1/2)si+b+(1/2)db=ei+(1/2)si+b1+(1/2)∫1cdb=(ei+(1/2))(c-1)si+b1+(1/2).
According to (17) and (18), we have that
(19)lim⁡si→∞p̂iEB=(ei+12)lim⁡si→∞1si+b1+(1/2)=0.
According to the relations of Beta function and Gamma function, we have that(20)B(ei+(32),si+b-ei)=Γ(ei+(3/2))Γ(si+b-ei)Γ(si+b+(3/2))=(ei+(1/2))Γ(ei+(1/2))Γ(si+b-ei)(si+b+(1/2))Γ(si+b+(1/2))=ei+(1/2)si+b+(1/2)B(ei+(1/2),si+b-ei),
where the Gamma function is Γ(a)=∫0∞xa-1e-xdx.
Whenb∈(1,c), we have that B(ei+(1/2),si+b-ei)>0, and (ei+(1/2)/si+b+(1/2)) is continuous; by the generalized mean value theorem for definite integrals, there is at least one number b2∈(1,c) such that
(21)∫1c(B(ei+(3/2),si+b-ei)/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db=∫1c(ei+(1/2)/si+b+(1/2))⋅(B(ei+(1/2),si+b-ei)/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db=ei+(1/2)si+b2+(1/2)⋅∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db=ei+(1/2)si+b2+(1/2).
According to Theorem3, we have that
(22)lim⁡si→∞p̂iHB=lim⁡si→∞∫1c(B(ei+(3/2),si+b-ei)/B((1/2),b))db∫1c(B(ei+(1/2),si+b-ei)/B((1/2),b))db=(ei+12)lim⁡si→∞1si+b2+(1/2)=0.
According to (19) and (22), we have that limsi→∞p̂iEB=limsi→∞p̂iHB.
Thus, the proof is completed.Theorem4 shows that p̂iEB and p̂iHB are asymptotically equivalent to each other as si tends to infinity. In application, p̂iEB and p̂iHB are close to each other, when si is sufficiently large.


## 6. Application Example

Han [6] provided a testing data of type I censored life testing for a type of engine, which is listed in Table 1 (time unit: hour).Table 1
Test data of the engine.
i123456789ti25045065085010501250145016501850ni333344444ri000001011ei000001123si3229262320161284By Theorems2, 3, and Table 1, we can obtain p̂iEB, p̂iHB, and p̂i-B=|p̂iEB-p̂iHB|. Some numerical results are listed in Table 2.Table 2
Results ofp̂iEB and p̂iHB(i=1,2,…,9).
c23456Rangep̂1EB0.0147070.0144970.0142940.0140990.0139110.000796p̂1HB0.0146930.0144580.0142280.0140040.0137880.000905p̂1-B0.0000140.0000390.0000660.0000950.0001230.000109p̂2EB0.0161300.0158780.0156360.0154040.0151810.000949p̂2HB0.0161140.0158320.0155570.0152910.0150350.001079p̂2-B0.0000160.0000460.0000790.0001130.0001460.000130p̂3EB0.0178590.0175510.0172560.0169750.0167050.001154p̂3HB0.0178390.0174950.0171600.0168380.0165300.001309p̂3-B0.0000200.0000560.0000960.0001370.0001750.000155p̂4EB0.0200020.0196180.0192520.0189040.0185720.001430p̂4HB0.0199780.0195480.0191330.0187360.0183580.001620p̂4-B0.0000240.0000700.0001190.0001680.0002140.000190p̂5EB0.0227310.0222370.0217700.0213280.0209090.001822p̂5HB0.0226990.0221480.0216190.0211180.0206420.001620p̂5-B0.0000320.0000890.0001510.0002100.0002670.000235p̂6EB0.0833550.0811600.0791120.0771940.0753940.007961p̂6HB0.0832370.0808560.0786340.0765730.0746610.008576p̂6-B0.0001180.0003040.0004780.0006210.0007330.000615p̂7EB0.1071880.1036130.1003350.0973170.0945240.012664p̂7HB0.1070110.1031770.0996830.0965090.0936200.013391p̂7-B0.0001770.0004360.0006520.0008080.0009040.000727p̂8EB0.2502090.2388190.2286970.2196230.2114280.030781p̂8HB0.2500160.2387210.2291360.2209510.2138980.036118p̂8-B0.0001930.0000990.0004390.0006720.0024700.002277p̂9EB0.5846890.5427710.5078710.4782260.4526400.132050p̂9HB0.5896430.5591050.5281910.5232500.5121870.077456p̂9-B0.0049540.0163340.0203200.0450240.0595470.054594From Table2, we find that for some c (c=2,3,4,5,6), p̂iEB, and p̂iHB(i=1,2,…,9) are very close to each other and satisfy Theorem 4; for different c (c=2,3,4,5,6), p̂iEB and p̂iHB(i=1,2,…,9) are all robust (when i=9, there exists some difference). In application, the author suggests selecting a value of c in the middle point of interval [2, 6], that is, c=4.Whenc=4, some numerical results are listed in Table 3 and Figure 1.Table 3
Results ofp̂iEB and p̂iHB.
i12345p̂iEB0.0142940.0156360.0172560.0192520.021770p̂iHB0.0142280.0155570.0171600.0191330.021619p̂i-B0.0000660.0000790.0000960.0001190.000151i6789p̂iEB0.0791120.1003350.2286970.507871p̂iHB0.0786340.0996830.2291360.528191p̂i-B0.0004780.0006520.0004390.020320Figure 1
Results ofp̂iEB and p̂iHB.Range: in Figure1, * is the results of p̂iEB and o is the results of p̂iHB(i=1,2,…,9).From Table3 and Figure 1, we find that p̂iEB and p̂iHB are very close to each other and consistent with Theorem 4.From Table3, we find that the results of p̂iEB and p̂iHB are very close to the corresponding results of Han [6].According to Han [6], we may assume that the lifetime of these products obeys Weibull distribution with distribution function (23)F(t)=1-exp⁡{-(tη)m},η>0,m>0,t>0.According to Mao and Luo [7], the least square estimates of η and m are, respectively, (24)η̂=exp⁡(μ̂),m̂=1σ̂,
where μ̂=(BC-AD)/(mB-A2),σ̂=(mD-AC)/(mB-A2), A=∑i=1mxi, B=∑i=1mxi2, C=∑i=1myi, D=∑i=1mxiyi, xi=ln{(1-p̂i)-1)}, p̂i is the estimate of pi and yi=lnti, i=1,2,…,m.According to (24), we can obtain the estimate of the reliability at moment t,(25)R̂(t)=exp⁡{-(tη̂)m̂},
where η̂ and m̂ are given by (24).From (24) and Table 3, we can obtain η̂ and m̂. Some numerical results of η̂ and m̂ are listed in Table 4.Table 4
Results ofη̂ and m̂.
Parameter estimatem̂η̂E-Bayesian estimate2.6408468642738.461813Hierarchical Bayesian estimate2.6782763742639.899868Difference between the above two estimates0.03742950698.56194500From Table4, we find that the results of η̂ and m̂ are very close to the corresponding results of Han [6].From (25) and Table 4, we can obtain the estimate of the reliability with some numerical results of R̂EB(t) and R̂HB(t) listed in Table 5 and Figure 2.Table 5
Results ofR̂EB(t) and R̂HB(t).
t2006001000140018002000R̂EB(t)0.9990030.9820190.9324670.8436430.7187960.646553R̂HB(t)0.9990560.9822470.9320590.8409210.7120320.637398R̂-B(t)0.0000530.0002280.0004080.0027220.0067640.009155Figure 2
Results ofR̂EB(t) and R̂HB(t).Note thatR̂EB(t) is the estimate of reliability at moment t with regard to p̂iEB, R̂HB(t) is the estimate of reliability at moment t with regard to p̂iHB, and R̂-B(t)=|R̂EB(t)-R̂HB(t)|.Range: in Figure2, * is the results of R̂EB(t) and o is the results of R̂HB(t).From Table5, we find that R̂-B(t)<0.0092 when t=200,600,1000,1400, 1800,2000, and the results of R̂EB(t) and R̂HB(t) are very close to those of Han [6].


## 7. Conclusions

This paper introduces a new method, calledE-Bayesian estimation, to estimate failure probability. The author would like to put forward the following two questions for any new parameter estimation method: (1) how much dependence is there between the new method and the other already-made ones? (2) In which aspects is the new method superior to the old ones?For the E-Bayesian estimation method, Theorem4 has given a good answer to the above question (1) and, in addition, the application example shows that p̂iEB and p̂iHB satisfy Theorem 4.To question (2), from Theorems2 and 3, we find that the expression of the E-Bayesian estimation is much simple, whereas the expression of the hierarchical Bayesian estimation relies on beta function and complicated integrals expression, which is often not easy.Reviewing the application example, we find that the E-Bayesian estimation method is both efficient and easy to operate.



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