P Y < X Estimating of ( )in the Exponential case Based on Censored Samples 8 0 0 Abd Elfattah, A. M. 2 n Department of mathematical Statistics, Institute of Statistical Studies and Research a J Cairo University, Cairo, Egypt. 7 ] E M Marwa O. Mohamed . t a t Department of mathematics, Zagazig University, Cairo, Egypt. s [ 1 v Abstract 2 2 In this article, the estimation of reliability of a system is discussed p(y < x)when 9 0 strength,X, and stress, Y , are two independent exponential distribution with differ- . 1 0 ent scale parameters when the available data are type II Censored sample. Different 8 methods for estimating the reliability are applied . The point estimators obtained 0 : v are maximum likelihood estimator, uniformly minimum variance unbiased estimator, i X andBayesianestimatorsbasedonconjugateandnoninformativepriordistributions. A r comparisonoftheestimatesobtainedisperformed. Intervalestimatorsofthereliability a are also discussed. Key Words: Maximum likelihood estimator; Unbiasedness; Consistency ; Uniform mini- mumvarianceunbiasedestimator; Bayesian estimator; Pivotal quantity; Fisherinformation . 1 1 Introduction In life testing, it is often the case that items drawn from a population are put on test and their times of failure are recorded. For a number of reasons, such as budget or limited time, it is often necessary to terminate the test before all the failure times have been observed; In this case the data become available in some ordered manner. The test is usually terminated after a fixed time or a fixed number of failures are observed giving a censored sample see Epstein and Sobel (1953). In stress-strength model, the stress Y and the strength X are treated as random variables and the reliability of a component during a given period (0,T)is taken to be the probability that the strength exceeds the stress during the entire interval. the reliability of a component is P(Y < X). Several authors have considered different studies for stress and strength with exponential distribution with complete sample. Tong(1974) discussed the estimation ofP(Y < X) in the exponential case. Tong(1977) had a look at the estimation of P(Y < X) for exponential families. A good review of the literature can befound in Johnson(1988). Beg(1980a,b and c)estimated the exponential family ,two parameter exponential distribution and truncation parameter distributions. Basu(1981)consideredmaximumlikelihoodestimators(MLE)forP(Y ≤ X) in case of gamma and exponential distributions. Sathe and Shah (1981) studied estimation ofP(Y > X)fortheexponentialdistribution. Chao(1982)providedsimpleapproximations for bias and mean square error of the maximum likelihood estimators of reliability when stress and strength are independentexponentially distributed random variables. Awad and Charraf (1986) studied three different estimators for reliability has a bivariate exponential distribution. Dinh, et al (1991) obtained the MVUE of R when X and Y have the bivariate normal distribution. Moreover, they considered the case when X and Y have the bivariate exponentialdistribution. BaiandHong(1992) estimatedP(Y ≤ X)intheexponentialcase withcommonlocationparameter. KunchurandMounoli(1993)obtainedUMVUEofstress- strength model for multi component survival model based on exponential distribution for parallel system. Siu-Keung Tse and Geoffrey Tso (1996) studied the shrinkage estimation of reliability for exponential distributed lifetime. A note on the UMVUE on P(Y ≤ X) in the exponential case discussed in Cramer and Kamps (1997). Selvavel, et al (2000) studied 2 reliability R when(X,Y)jointly follows atruncated bivariate exponential distributionwith a common parameters. Khayar (2001) discussed the reliability of time dependent stress- strengthmodelsforexponentialandRayleigh distributions. Shrinkageestimation ofP(Y ≤ X) in the exponential case discussed by Ayman and Walid (2003).Tachen (2005)deals with the empirical Bayes testing the reliability of an exponential distribution . In the present article, the reliability, R , is studied when X and Y two independent exponential distribution with different scale parameters. Which can be represented as f(x;α) = 1e−αx, x >0,α > 0. α Different estimators of R are derived, namely, maximum likelihood estimator (MLE), uni- formminimumvarianceunbiasedestimator,(UMVUE),andBayesianestimatorswithmean square error loss functions corresponding to conjugate and non informative priors. A com- prehensive comparison of the various point estimators (MLE,UMVUE, and Bayes) is per- formed on the basis of the mean squared error. Interval estimators of R are also discussed. A numerical comparison of the intervals obtained. 2 Reliability Let X be the strength of a component and Y be the stress acting on it. Let X and Y be exponential independent random variables with parameters α and β, respectively. That is , the probability density functions (pdfs) of X and Y are, respectively, 1 −x f(x;α) = e α , x > 0,α > 0 (2.1), α and 1 −y f(y;β) = e β , y > 0,β > 0 (2.2), β where α and β are unknown parameters . The reliability of the component will be R = P(Y < X) = 1 0∞(1−e−βx)e−αxdx α R = fracαα+β. (2.3) if α and β are known then R is simply calculated using Eq.(2.3). 3 R 3. Point Estimation of 3.1 Maximum Likelihood Estimator of R If α and β are unknown the MLE of, Rˆ , of R is given by 1 Rˆ = fracαˆαˆ+βˆ, (3.1) 1 where αˆ and βˆ are the MLEs of α and β , respectively. For obtaining αˆ and βˆ we argue as follows: Suppose that r components where r ≤ n with strengths X ;i = 1,...,r , each of which 1 1 i 1 havingexponentialdistributionwithparameterαasinEq. (2.1)aresubjected,respectively, the likelihood function will be L = (n−nr!1)! i=1r1α1e−αxi[e−xαr1]n−r1 Q L = n! (1)r1e− αri=11xi−−xr1(αn−r1),eqno(3.2) (n−r1)! α P Taking the logarithm of Eq. (3.2) and find the derivative with respect to α dlnL = −r1 + ri=11xi + (n−r1)xr1 = 0 α α α2 α2 P −r α = r1 x +(n−r )x , 1 i=1 i 1 r1 P r1 x +(n−r )x αˆ = i=1 i 1 r1, (3.3) r P 1 To stress Y ; j = 1,...,r , having exponential distribution with parameter β as in Eq. (2.2), j 2 where r ≤ m . Assuming that X and Y ; i = 1,...,r and j = 1,...,r , are independent , 2 1 j 1 2 the likelihood function will be L = (mm−r!2)! j=1r2β1e−βyj[e−yβr2]m−r2 Q L = m! (1)r2e− rjβ=21yj−−yr2(βm−r2), (3.4) (m−r )! β P 2 Taking the logarithm of Eq. (3.4) and find the derivative with respect toβ dlnL = −r2 + rj=21yj + (m−r2)yr2 = 0 β β β2 β2 P −r β = r2 y +(m−r )y , The MLE of β will be 2 j=1 j 2 r2 P r2 y +(m−r )y βˆ= j=1 j 2 r2, (3.5) r P 2 4 Now we shall study some properties of Rˆ . 1 First, if r = r = r : 1 2 1] Unbiasedness E(Rˆ ) = α [1− (2r−1)(1− α )2] 1 α+ rβ r(r−2) α+ rβ r+1 r+1 limr→∞E(Rˆ1) = R−Rlimr→∞ (1(r−−R1))2 then limr→∞E(Rˆ1) = R then, Rˆ asymptotically unbiased estimator of R . 1 2] Consistency rβ Var(Rˆ ) = (2r−1)[ (r−1)α ]2[ 1 ]2 1 r(r−2) (r+1)αβ 1+ rβ (r−1)β (r−1)α limr→∞Var(Rˆ1) = R2(αβ limr→∞ 1r then limr→∞Var(Rˆ1) = 0 then, Rˆ is a consistent estimator for R. 1 Second, if r 6= r : 1 2 1] Unbiasedness E(Rˆ ) = α [1− (r1+r2−1)(1− α )2] 1 α+ r1β r2(r1−2) α+ r1β r1+1 r1+1 For fixed r , 2 limr1→∞E(Rˆ1)= limr1→∞ α+αr1β [1− (rr12+(rr12−−21))(1− α+αr1β )2] r1+1 r1+1 limr1→∞E(Rˆ1)= R[1− r12(1−R)2] then limr1,r2→∞E(Rˆ1)= R then, Rˆ asymptotically unbiased estimator of R . 1 2] Consistency Var(Rˆ ) = (r1+r2−1)[ (r2r−1β1)α ]2[ 1 ]2 1 r2(r1−2) (r1+1)αβ 1+ r1β (r2−1)β (r1−1)α For fixed r , 2 5 limr1→∞Var(Rˆ1) = limr1→∞ (rr12+(rr12−−21)) limr1 → ∞[((rr12r+−11β1))ααβ]2limr1→∞[1+ 1r1β ]2 (r2−1)β (r1−1)α and, limr1,r2→∞Var(Rˆ1) = R2[αβ]4limr2→∞ r12 then limr1,r2→∞Var(Rˆ1) = 0 then, Rˆ is a consistent estimator for R. 1 3.2 Uniform Minimum Variance Unbiased Estimator of R Let X ,...,X andY ,...,Y betwo independentrandom samples , of size r andr , respec- 1 r1 1 r2 1 2 tively, drawn from exponential distributions with parameters α and β , respectively, Define zi = lnexi,vj = lneyj,i = 1,...,r1andj = 1,...,r2, Z = i= 1r z ,andV = j = 1r v 1 i 2 j ClearPly from Eq. (3.2) anPd (3.4) we see that Z,V is a complete sufficient statistic for α , β. Now, we have E(W) = 1.P(v < z )+0.P(v ≥ Z ) 1 1 1 1 E(W) = P(lneyj < lnexi) E(W) = P(y 0 the likelihood function with type II censored sample is , respectively, n! 1 f(x1,...,xr1|α) = (n−r )!(α)r1e−α1( i=1r1xi+xr1(n−r1)), (3.7) 1 P and f(y1,...,yr2|β) = (mm−!r )!(β1)r2e−β1( j=1r2yj+yr2(m−r2)), (3.8) 2 P Assuming that αandβ are independent having prior gamma distributions, the posterior distributions of αandβ will be gamma distribution also, π2 = f(α|x1,...,xr1) = (v1+ ri=11xi+xr1(n−Γr(1r))u+1+ur1(+α1)1r)1+u1e−(Pri=11xi+αxr1(n−r1)), P 1 1 (3.9) 7 and π = f(β|y ,...,y ) = (v2+ rj2=1yj +yr2(m−r2))u2+r2(β1)r2+u2e−(Prj=21yj+βyr2(m−r2)), 3 1 r2 Γ(r +u +1) P 2 2 (3.10) the joint posterior function, put, r1 r2 ζ = (v + x +x (n−r )),τ = (v + y +y (m−r )) 1 i r1 1 2 j r2 2 i=1 j=1 X X and k = 1 Γ(r1+u1+1)Γ(r2+u2+1) π(α,β|x,y) = kζ(u1+r1)τ(u2+r2)(1)r1+u1(1)r2+u2e−αζe−βτ, (3.11) α β Hence Bayes estimator Rˆ of R will be 3 kζ(u1+r1)τ(u2+r2) 1 R(r1+u1)(1−R)(r2+u2+1) Rˆ = E(R|x,y) = dR, (3.12) 3 Γ(r1+u1+r2+u2+1) Z0 ((1−R)ζ +Rτ)u1+r1+u2+r2 From equation (3.12) there is no explicit form of Rˆ so, The computation of the Bayes 3 estimator Rˆ ,is very complicated ,we will use the MATHCAD program to evaluate the 3 value of Rˆ . 3 3.3.2. Non Informative Prior Distributions Let X ,...,X be a random sample from exponential distribution with parameter α . The 1 r1 prior distribution of α is proportional to I(α) , where I(α) is Fisher’s information of the sample about α , and is given by p 1 I(α) = , (3.13) α2 from that the prior distribution 1 π ∝ , (3.14) 1 α Similarly, if Y ,...,Y is a random sample from exponential distribution with parameter β, 1 r2 the prior distribution of β will be given by: 1 π ∝ , (3.15) 2 β 8 if we have α and β are independent then the posterior joint distribution of α and β ,will be π(α,β|x ,...,x ,y ,...,y )∝ L(x ,...,x |α)L(y ,...,y |β)π (α)π (β), (3.16) 1 r1 1 r2 1 r1 1 r2 1 2 then n! 1 m! π(α,β|x1,...,xr1,y1,...,yr2) = (n−r )!(α)r1e−α1( i=1r1xi+xr1(n−r1))(m−r )! 1 2 P (1)r2e−β1( j=1r2yj+yr2(m−r2)),α,β > 0 β P put δ = ( r x +x (n−r )),ε = ( r y +y (m−r )) 1 i r1 1 2 j r2 2 i=1 j=1 X X n!m!α−(1+r1)β−(1+r2)e−αδe−βε π(α,β|x ,...,x ,y ,...,y ) = ,α,β > 0, (3.17) 1 r1 1 r2 (n−r )!(m−r )! 1 2 under the mean square error, Bayes estimator Rˆ of R will be 4 n!m! 1 R(r2+2)(1−R)(r1+1) Rˆ = E(R|x,y) = dR, (3.18) 4 (n−r1)!(m−r2)!(r1 +r2+3)! Z0 ((1−R)δ+Rε)r1+r2+2 From equation (3.18) there is no explicit form of Rˆ so, The computation of the Bayes 4 estimator Rˆ ,is very complicated ,we will use the MATHCAD program to evaluate the 4 value of Rˆ . 4 R 4. Interval Estimation of 4.1 Approximate confidence interval Let X ,...,X and Y ,...,Y be a random samples with size r ,r from exponential distri- 1 r1 1 r2 1 2 butionwithparameterα,β respectively,wecanshowthatthemaximumlikelihoodfunction ofR,Rˆ isasymptoticallynormaldistributionwithmeanRandvariance-covariancematrices I−1(α,β) where I(α,β) is the Fisher information matrix and given by: r1 0 α2 I(α,β) = 0 r2 ! β2 9 The variance-covariance matrix is obtained by inverting the information matrix with elements that are negatives of the expected values of the second order derivatives of log- arithms of the likelihood functions, and the asymptotic variance-covariance matrix is obtained by replacing values by their maximum likelihood estimators. Hence, the asymptotic variance-covariance matrix will be β2 0 I−1(α,β) = r2 0 α2 ! r1 From (Surles and Padjett (2001)) that the maximum likelihood function of R ,Rˆ is asymp- 1 totically normal distribution with mean R and variance r r (α+β)2 σ2 = 1 2 Rˆ1 α2β2 Hence,(1−α)100 an approximate confidenceinterval for R would be(L ,U ),and the value 1 1 of L ,U is given as: 1 1 L1 = Rˆ1−z1−α2σRˆ1, (4.1) and U1 = Rˆ1+z1−α2σRˆ1, (4.2) where z1−α quantile of the standard normal distribution and Rˆ1 is given by Eq.(3.1). 2 4.2 Exact confidence interval Let X ,...,X and Y ,...,Y be random samples with size r ,r from exponential distri- 1 r1 1 r2 1 2 bution with parameter α,β respectively. Since zi = lnexiand vj = lneyj are independent with gamma distribution with parameters (r1,α) and (r2,β) respectively,2αlne ri=11xi and 2βlne rj=21yj are independentwith chi square distributions with degree of freedPom 2r and 2r hePnce rRˆ = (1+ β)−1 1 2 1 α we know that F = fracVαZβ has F-distribution with (2r ,2r )degrees of freedom ,rRˆ = 1 1 2 1 (1+ βF )−1 ,this equation can be written as follow F = (1−Rˆ1) R α 1 1 Rˆ1 (1−R) using F as a pivotal quantity,we obtain a (1−α) 100% confidence interval for R as 1 V L2 = F1−α(2r2,2r1)(F1−α(2r2,2r1)+ )−1 2 2 Z 10

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