Estimation of the Stress-Strength Reliability for the Dagum Distribution

156 Journal of Advanced Statistics, Vol 1, No 3, September 2016 Estimation of the Stress-Strength Reliability for the Dagum Distribution Ber Al-Zahrani Samerah Basloom Department of Statistics, King Abdulaziz University, POBox 80203, Jeddah 21589, Saudi Arabia Email: bmalzahrani@kauedusa Abstract The paper deals with the problem of estimating the reliability of a single multicomponent stress-strength based on a three-parameter Dagum distribution The maximum likelihood estimators the Bayes estimators of R = P Y < X when X Y are two independent rom variables follow Dagum distribution their asymptotic distributions are obtained Also, the reliability of a multicomponent stress-strength model is estimated using the maximum likelihood Consequently, the asymptotic confidence intervals of the reliability of a single multicomponent stress-strength model are constructed Gibbs Metropolis sampling were used to provide sample-based estimates of the reliability its associated credible intervals Finally, Monte Carlo simulations are carried out for illustrative purposes Keywords: Dagum distribution, maximum likelihood estimator, Bayes estimator, stress-strength models 1 Introduction Dagum distribution was introduced by Dagum 1 for modeling personal income data as an alternative to the Pareto log-normal models This distribution has been extensively used in various fields such as, income wealth data, meterological data, reliability survival analysis The Dagum distribution is also known as the inverse Burr XII distribution, especially in the actuarial literature An important characteristic of Dagum distribution is that its hazard function can be monotonically decreasing, an upside-down bathtub, or bathtub then upside-down bathtub shaped, for details see Domma 2 This behavior of the distribution has led several authors to study the model in different fields In fact, recently, the Dagum distribution has been studied from a reliability point of view used to analyze survival data see Domma et al, 3, Domma Condino 4 Kleiber Kotz 5 Kleiber 6 provided an exhaustive review on the origin of the Dagum model its applications Domma et al 3 estimated the parameters of Dagum distribution with censored samples Shahzad Asghar 7 used TL-moments to estimate the parameter of this distribution Oluyede Ye 8 presented the class of weighted Dagum related distributions Domma Condino 4 proposed the five parameter beta-dagum distribution A continuous rom variable T is said to have a three-parameter Dagum distribution, abbreviated as T Dagumα, β, δ, if its density probability function pdf is given as f t; α, β, δ = βαδt δ 1 1 + αt δ β 1, t > 0, 11 where α > 0 is the scale parameter its two shape parameters β δ are both positive The corresponding distribution function is given by F t; α, β, δ = 1 + αt δ β, t > 0, β, α, δ > 0 12 In this paper we focus our attention on estimating the reliability of a single component stress-strength, R = P Y < X where X Y are two independent rom variables following Dagum distribution with different shape parameter β have the same scale parameter α shape parameter δ Here R is the probability that the strength X of a component during a given period of time exceeds the stress Y We also focus on estimating the reliability of multicomponent stress-strength R s,k, where s k are two intgers s k The problem of estimating R R s,k has attracted the attention of many authors, a

Journal of Advanced Statistics, Vol 1, No 3, September 2016 157 good overview on estimating R can be found in the monograph of Kotz et al 9 See also, Pey Uddin 10 Srinivasa Kantam 11 for details on estimation of R s,k It is worth mentioning that the problem here seems to be clearly an application of well known mathematical methods However, there were some challenges in finding the estimatiors The rest of the paper is organized as follows In Section 2, we obtain the maximum likelihood estimatior MLE of R the asymptotic distribution of ˆR The MLE Bayes estimators of R when α is known is considered Also, simulation studies have been presented for illustrative purposes In section 3, the MLEs are obtained employed to get the asymptotic distribution confidence intervals for R s,k Finally, the conclusion comments are provided in Section 4 2 Estimation of R with Different β Common δ α In this section, we investigate the properties of R when the shape parameters β are different, the scale parameter α the shape parameter δ are constants The other cases, ie the parameters α δ are not constants the general case where all parameters are different, can be studied in a similar way to the case presented in this paper 21 Maximum Likelihood Estimator of R Let X Dagumβ 1, α, δ Y Dagumβ 2, α, δ, where X Y are two independent rom variables All three parameters, β, α δ are unknown to us Then it can be shown that R = P Y < X = 0<y<x fx, ydxdy = β 1 β 1 + β 2 21 To compute the MLE of R, suppose that X 1, X 2,, X n is a rom sample from Dagumβ 1, α, δ Y 1, Y 2,, Y m is another rom sample from Dagumβ 2, α, δ Then the log-likelihood function of the observed sample is where ln Lβ 1, β 2, α, δ = n lnβ 1 + n + m lnα + n + m lnδ n m δ + 1 lnx i + lny j L δ = n+m δ β 1 + 1s 1 x, α, δ + m lnβ 2 β 2 + 1s 2 y, α, δ, s 1 x, α, δ = s 2 y, α, δ = n m ln1 + αx δ i, 22 ln1 + αy δ j 23 The MLEs of β, α δ say ˆβ, ˆα ˆδ, respectively, can be obtained as the solutions of the following equations L β 1 = n β 1 s 1 x, α, δ = 0 L β 2 = m β 2 s 2 y, α, δ = 0 L α = n+m α β 1 + 1 n x δ i β 1+αx δ 2 + 1 m y δ j = 0 1+αy δ i j n lnx i m lny j β 1 + 1 24 n β 2 + 1 m y δ lny j j = 0 1+αy δ j αx δ i lnx i 1+αx δ i

158 Journal of Advanced Statistics, Vol 1, No 3, September 2016 We obtain n ˆβ 1 = ŝ 1 x, α, δ 25 m ˆβ 2 = ŝ 2 y, α, δ 26 The estimators ˆα ˆδ can be obtained as the solutions of the nonlinear equations given in 24 The plug-in estimation of R, say ˆR, is readily computed as ˆR = 22 Asymptotic Distribution of ˆR ˆβ 1ML ˆβ 1MR + ˆβ 2ML 27 In this section, the asymptotic distribution of ˆθ = ˆβ 1, ˆβ 2, ˆα, ˆδ the asymptotic distribution of ˆR are obtained The Fisher information matrix of θ, denoted by Jθ = EI, θ, is giving below, where I = I i,j i,,2,3,4 is the observed information matrix ie, Iθ = α 2 β 1 α β 2 α δ α the elements of Iθ are as follows I 11 = n + m α 2 + β 1 + 1 n α β 1 β1 2 β 2 β 1 δ β 1 α β 2 β 1 β 2 β2 2 δ β 2 α δ β 1 δ β 1 δ δ 2 x δ i 2 1 + αx δ i + β 2 + 1 I 11 I 12 I 13 I 14 = I 21 I 22 I 23 I 24 I 31 I 32 I 33 I 34, I 41 I 42 I 43 I 44 m y δ j 2 1 + αy δ j n X δ m i I 12 = I 21 = 1 + αx δ i, I y δ j 13 = I 31 = 1 + αy δ j n x δ i ln x i n x δ 2 I 14 = I 41 = β 1 + 1 1 + αx δ i i α ln xi 1 + αx δ i 2 m y δ + β 2 + 1 j ln y j m y δ 2 1 + αy δ j j α ln yj 1 + αy δ j 2 I 22 = n β1 2, I 23 = I 32 = 0, I 24 = I 42 = I 33 = m β2 2 I 34 = I 43 = n + m δ 2 + m n αx δ i ln x i 1 + αx δ i αy δ j ln y j 1 + αy δ j β 1 + 1nαx δ lnx2 1 + αx δ β 2 + 1mαy δ lny 2 1 + αy δ + β 2 + 1mα 2 y δ 2 lny 2 1 + αy δ 2 I 44 = n + m n αx δ i lnx i 2 n α 2 β 1 + 1 1 + αx δ i α 2 x δ i 2 lnx i 2 1 + αx δ i 2 m αy δ + β 2 + 1 j lny j 2 m 1 + αy δ j α 2 y δ j 2 lny j 2 1 + αy δ j 2 + β 1 + 1nα 2 x δ 2 lnx 2 1 + αx δ 2

Journal of Advanced Statistics, Vol 1, No 3, September 2016 159 The following integrals can be helpful when finding the elements of the Fisher information matrix Note that ψ is the digamma function, defined as ψx = d/dx lnγ x 0 0 0 0 lnt1 + t β 1 dt = ψβ + γ, β 0 t ln 2 t1 + t β 3 ψ 2 β + γ 2 + π dt = ψβ 6 β + 1β + 2 t 2 lnt1 + t β 3 2ψβ 2γ + 3 dt = ββ 2 + 3β + 2, t lnt1 + t β 2 dt = Ψβ + γ 1, αβ + 1 2 + 2 βγ 1 + 1 ψβ γβ 1 + 1, ββ + 1β + 2 t 2 ln 2 t1 + t β 3 dt = 6ψ2 β + 12ψβγ + 6γ 2 + π 2 18ψβ 18γ + 6ψ1, β + 6 β3β 2 + 9β + 6 The elements of the Fisher information matrix are obtained by taking the expectations of the observed matrix Doing so will result in the following: J 11 = n + m α 2 nβ 1β 1 + 1 α 2 B3, β 1 mβ 2β 2 + 1 α 2 B3, β 2 J 12 = nβ 1 α B2, β 1, J 13 = J 31 = mβ 2 α B2, β 2, where B, is the beta function J 14 = J 41 = nβ 1β 1 + 1 B2, β 1 lnα + ψβ 1 + γ 1 δα αβ 1 + 1 B3, β 1 lnα + 2ψβ 1 2γ + 3 β 1 β1 2 + 3β 1 + 2 mβ 2β 2 + 1 B2, β 2 lnα + ψβ 2 + γ 1 δα B3, β 2 lnα + 2ψβ 2 2γ + 3 β 2 β2 2 + 3β 2 + 2 αβ 2 + 1 Again, the functions B, ψ are the beta function the digamma function J 22 = n β1 2, J 23 = J 32 = 0, J 24 = J 42 = nβ 1 B2, β 1 lnα + ψβ 1 + γ 1 δ β 1 β 1 + 1 J 33 = m β2 2, J 34 = J 43 = mβ 2 δ B2, β 2 lnα + ψβ 2 + γ 1 β 1 β 2 + 1 the element J 44 is defined as follows J 44 = n + m α 2 + nβ 1β 1 + 1 δ 2 B3, β 1 ln 2 α + 2 lnα ψβ 1 + γ 1 β 1 β 1 + 1 + 6ψβ 1 2 + 12ψβ 1 γ + 6γ 2 + π 2 12ψβ 1 12γ + 6ψ1, β 1 6β 1 β 1 + 1 1 ln 2 α 2B3, β 1 + 2 2ψβ 1 2γ + 3 α β 1 β1 2 + 3β 1 + 2 6ψ2 β 1 + 12ψβ 1 γ + 6γ 2 + π 2 18ψβ 1 18γ + 6ψ1, β 1 + 6 β 1 3β1 2 + 9β 1 + 6 mβ 2β 2 + 1 δ 2 B3, β 2 ln 2 α + 2 lnα ψβ 2 + γ 1 β 2 β 2 + 1 + 6ψβ 2 2 + 12ψβ 2 γ + 6γ 2 + π 2 12ψβ 2 12γ + 6ψ1, β 2 6 β 2 β 2 + 1

160 Journal of Advanced Statistics, Vol 1, No 3, September 2016 1 ln 2 α 2B3, β 2 + 2 2ψβ 2 2γ + 3 α β 2 β2 2 + 3β 2 + 2 6ψ2 β 2 + 12ψβ 2 γ + 6γ 2 + π 2 18ψβ 2 18γ + 6ψ1, β 2 + 6 β 2 3β 2 2 + 9β 2 + 6 It is worth mentioning that the Dagum family of distributions satisfies all the regularity conditions, for example see Nadarajah Kotz 12 Now, it turns out that we can formulate the limiting joint distribution of estimators Theorem 1 As n m n m p, where p is positive real constant, then m ˆβ1 β 1, n ˆβ 2 β 2, mˆα α, mˆδ δ N 4 0, A 1 β 1, β 2, α, δ, where the covariance matrix A is given as a 11 a 12 a 13 a 14 Aβ 1, β 2, α, δ = a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 Proof We first give the elements of the covariance matrix A a 11 = J 11 lim n,m m = p + 1 α 2 pβ 1β 1 + 1 α 2 B3, β 1 β 2β 2 + 1 α 2 B3, β 2 a 12 = a 21 = a 13 = a 31 = lim n,m lim n,m J 12 p pβ1 = B2, β 1 n α J 13 m = β 2 α B2, β 2 J 14 a 14 = a 41 = lim n,m m p pβ1 β 1 + 1 = B2, β 1 lnα + ψβ 1 + γ 1 δα αβ 1 + 1 B3, β 1 lnα + 2ψβ 1 2γ + 3 β 1 β1 2 + 3β 1 + 2 β 2β 2 + 1 B2, β 2 lnα + ψβ 2 + γ 1 pαδ αβ 2 + 1 B3, β 2 lnα + 2ψβ 2 2γ + 3 β 2 β2 2 + 3β 2 + 2 a 22 = J 22 lim n,m n = 1 β1 2 a 34 = a 43 = a 24 = a 42 = lim n,m, a 23 = a 32 = 0 J 34 m p = β 2 δ p J 24 lim n,m n = B 1 δ B2, β 2 lnα + ψβ 2 + γ 1 β 2 β 2 + 1 B2, β 1 lnα + ψβ 1 + γ 1 β 1 β 1 + 1 a 44 = J 44 lim n,m n = 1 α 2 p + β 1β 1 + 1 δ 2 B3, β 1 ln 2 α + 2 lnα ψβ 1 + γ 1 β 1 β 1 + 1

Journal of Advanced Statistics, Vol 1, No 3, September 2016 161 6ψβ1 2 + 12ψβ 1 γ + 6γ 2 + π 2 12ψβ 1 12γ + 6ψ1, β 1 + 6β 1 β 1 + 1 1 ln 2 α 2B3, β 1 + 2 2ψβ 1 2γ + 3 α β 1 β1 2 + 3β 1 + 2 6ψ2 β 1 + 12ψβ 1 γ + 6γ 2 + π 2 18ψβ 1 18γ + 6ψ1, β 1 + 6 β 1 3β1 2 + 9β 1 + 6 β2 β 2 + 1 pδ 2 B3, β 2 ln 2 α + 2 lnα ψβ 2 + γ 1 β 2 β 2 + 1 + 6ψ2 β 2 + 12ψβ 2 γ + 6γ 2 + π 2 12ψβ 2 12γ + 6ψ1, β 2 6β 2 β 2 + 1 1 ln 2 2ψβ2 2γ + 3 α 2B3, β 2 + 2 α β 2 β2 2 + 3β 2 + 2 6ψ2 β 2 + 12ψβ 2 γ + 6γ 2 + π 2 18ψβ 2 18γ + 6ψ1, β 2 + 6 β 2 3β2 2 + 9β 2 + 6 The proof follows immediately by invoking the asymptotic properties of MLEs the multivariate central limit theorem One of the main results here is Theorem 2, concerning the asymptotic properties of the parameter R Theorem 2 As n, m, n m p, where p is a positive number, then n ˆR R N0, BA, where B A = 1 β U A β 1 + β 2 4 2 β 2 a22 a 33 a 44 a 22 a 2 43 a 33 a 2 42 +β 1 a 21 a 33 a 44 a 21 a 43 + a 31 a 42 a 43 a 33 a 41 a 42 β 1 β 2 a21 a 33 a 44 a 21 a 2 43 + a 31 a 2 42a 43 a 33 a 41 a 42 β 1 a11 a 33 a 44 a 11 a 2 43 a 2 31a 44 + 2a 31 a 41 a 43 a 33 a 2 41 U A = a 11 a 22 a 33 a 44 a 11 a 22 a 2 43 a 11 a 33 a 2 24 a 2 12a 33 a 44 + a 2 12a 2 43 2a 12 a 13 a 24 a 34 2a 12 a 33 a 14 a 24 + a 22 a 2 13a 44 + 2a 22 a 13 a 14 a 34 a 22 a 33 a 2 14 + a 2 13a 2 24 Proof Applying the delta method see Casella Roger, 13 using Theorem 1, we can write the asymptotic distribution of ˆR where ˆR = g ˆβ 1, ˆβ 2, ˆα, ˆδ gβ 1, β 2, α, δ = β 1 /β 1 + β 2 as the following: n ˆR R N0, BA, where b A = R β 1 R β 2 R α R δ B A = b t AA 1 b A, = 1 β 1 + β 2 2 β 2 β 1 0 0,

162 Journal of Advanced Statistics, Vol 1, No 3, September 2016 A 1 = 1 U A v 11 v 12 v 13 v 14 v 21 v 22 v 23 v 24 v 31 v 32 v 33 v 34 v 41 v 42 v 43 v 44 The elements of the A 1 are obtained as follows v 11 = a 22 a 33 a 44 a 22 a 2 43 a 33 a 2 42 v 12 = v 21 = a 21 a 33 a 44 a 21 a 2 43 + a 31 a 42 a 43 a 33 a 41 a 12 v 13 = v 31 = a 21 a 42 a 43 + a 22 a 31 a 44 a 22 a 41 a 43 a 31 a 2 42 v 14 = v 41 = a 21 a 33 a 42 + a 22 a 31 a 43 a 22 a 33 a 41 v 22 = a 11 a 33 a 44 a 11 a 2 43 a 2 31a 44 + 2a 31 a 41 a 43 a 33 a 2 41 v 23 = v 32 = a 11 a 42 a 43 + a 21 a 31 a 44 a 21 a 41 a 43 a 31 a 41 a 42 v 24 = v 42 = a 11 a 33 a 42 + a 21 a 31 a 43 a 21 a 33 a 41 a 2 31a 42 v 33 = a 11 a 22 a 44 a 11 a 2 42 a 2 21a 44 + 2a 21 a 41 a 42 a 22 a 2 41 v 34 = v 43 = a 11 a 22 a 43 a 2 21a 43 + a 21 a 31 a 42 a 22 a 31 a 41 v 44 = a 11 a 22 a 33 a 2 21a 33 a 22 a 2 31 Therefore, B A = b t AA 1 b A = 1 β U A β 1 + β 2 4 2 β 2 a 22 a 33 a 44 a 22 a 2 43 a 33 a 2 42 +β 1 a 21 a 33 a 44 a 21 a 2 43 + a 31 a 42 a 43 a 33 a 41 a 42 β 1 β 2 a 21 a 33 a 44 a 21 a 2 43 + a 31 a 42 a 43 a 33 a 41 a 42 β 1 a 11 a 33 a 44 a 11 a 2 43 a 2 31a 44 + 2a 31 a 41 a 43 a 33 a 2 41 The proof is now completed It should be noted that Theorem 2 can be used to construct the asymptotic confidence intervals for R However, the variance B A is not known it has to be estimated The estimator of B A, say ˆB A, is obtained by replacing β 1, β 2, α δ involved in B A by their corresponding MLEs The 1001 γ% confidence intervals for R are given by ˆBA ˆR Z 1 γ/2, ˆR ˆBA + Z 1 γ/2, 28 n n where Z γ is the γ% percentile of N0, 1 As mentioned in Asgharzadeh et al 14, instead of approximating ˆR R / var ˆR as stard normal variable, it is possible to consider some other normalizing transformation, say gr, of R assume that g ˆR gr / varg ˆR N0, 1, Now we use delta method to approximate the variance of gr, see for example Held Bove 15 It is written as follows var g ˆR = g R 2 var ˆR = g R 2 B A /n

Journal of Advanced Statistics, Vol 1, No 3, September 2016 163 As before, the 1001 γ% confidence intervals for gr are given as g ˆR ± Z 1 γ/2 var g ˆR If the function gr is strictly increasing then approximate 1001 γ% confidence intervals for R are derived to be g g 1 ˆR Z 1 γ/2 var ˆR, g 1 g ˆR + Z 1 γ/2 var ˆR Specifically, we follow Asgharzadeh et al 14 consider the following two transformations R 1 Logit transformation: gr = ln 1 R with g 1 R = R1 R 2 Arcsine transformation: gr = sin 1 R with g 1 R = 2 R1 R More details about these transformations one can refer to Lawless 16 or Mukherjee Maiti 17 23 Bootstrap Confidence Intervals For small sample sizes, confidence intervals carried out based on the asymptotic results are usually expected not to perform well Therefore, we propose to use confidence intervals based on two parametric bootstrap methods These methods are: i the percentile bootstrap method, shortened as Boot-p, based on the idea of Hall 18 ii the bootstrap-t method, shortened as Boot-t, based on the idea of Hall 18 The algorithms for estimating the confidence intervals of R using both methods are summarized below i Boot-p method 1 Use the samples {x 1,, x n } {y 1,, y m } equations 24, 25 26 to compute the estimates ˆβ 1ML, ˆβ 2ML, ˆα ML ˆδ ML 2 Use the estimates ˆβ 1ML, ˆα ML ˆδ ML to generate a bootstrap sample {x 1,, x n } similarly use ˆβ 2ML, ˆα ML ˆδ ML to generate a bootstrap sample {y 1,, y n } Based on {x 1,, x n } {y 1,, y n } compute the bootstrap estimate of R, say R, using ˆR = ˆβ 1ML ˆβ 1ML + ˆβ 2ML 3 Repeat Step 2, NBOOT times 4 Let g 1 x = P ˆR x be the cdf of ˆR define ˆR Bp x = g1 1 x for a given x Then the approximate 1001 γ% confidence intervals for R are given by γ ˆRBp, 2 ˆR Bp 1 γ 2 ii Boot-t method 1 Use the samples {x 1,, x n } {y 1,, y m } equations 24, 25 26 to compute ˆβ 1ML, ˆβ 2ML, ˆα ML ˆδ ML 2 Use ˆβ 1ML, ˆα ML ˆδ ML to generate a bootstrap sample {x 1,, x n } similarly use ˆβ 2ML, ˆα ML ˆδ ML to generate a bootstrap sample {y 1,, y n } Based on {x 1,, x n } {y 1,, y n } compute the bootstrap estimate of R, say R, using ˆR = statistic n T = ˆR ˆR var ˆR ˆβ 1ML ˆβ 1ML + ˆβ 2ML the 3 Repeat Step 2, NBOOT times 4 Let g 2 x = P T x be the cdf of T define ˆR Bt x = ˆR + g2 1 x var ˆR n for a given x Then the approximate 1001 γ% confidence intervals for R are given by γ ˆRBt, 2 ˆR Bt 1 γ 2

164 Journal of Advanced Statistics, Vol 1, No 3, September 2016 24 Bayes Estimation of R In this subsection, we discuss the estimation of the parameter R using Bayes method assuming that the shape parameters β 1, β 2, δ the scale parameter α are rom variables unknown to us It is assumed that β 1, β 2, δ α have density functions Gammaa 1, b 1, Gammaa 2, b 2, Gammaa 3, b 3 Gammaa 4, b 4 respectively Moreover, it is assumed that β 1, β 2, δ α are independent Based on the above assumptions, we obtain the likelihood function of the observed data as follows Ldata β 1, β 2, α, δ = β n 1 α n+m δ n+m n x δ+1 i s 1 x, α, δ m y δ+1 j s 2 y, α, δ, 29 where s 1 x, α, δ s 2 y, α, δ are defined as before in equations 22 23 The joint density of data β 1, β 2, α δ can be obtained as Ldata, β 1, β 2, α, δ = Ldata; β 1, β 2, α, δ π 1 β 1 π 2 β 2 π 3 α π 4 δ, 210 where π 1 β 1, π 2 β 2, π 3 α π 4 δ are gamma prior densities for β 1, β 2, α δ respectively Therefore, the joint posterior density of β 1, β 2, α δ given the data is Ldata β 1, β 2, α, δ = 0 0 Ldata, β 1, β 2, α, δ 0 0 Ldata, β 211 1, β 2, α, δdδdαdβ 2 dβ 1 Equation 211 cannot be written in closed form, thus we apply the Gibbs sampling technique to compute the Bayes estimate of R along with the corresponding credible intervals The posterior pdfs of β 1, β 2, α δ can be obtained readily as follows β 1 β 2, δ, α, data Gamman + a 1, b 1 + s 1 x, α, δ, β 2 β 1, δ, α, data Gammam + a 2, b 2 + s 2 y, α, δ, fα β 1, β 2, δ, data α n+m+a3 1 exp α b 3 + n x i + n y j β 1 + 1s 1 x, α, δ β 2 + 1s 2 y, α, δ, fδ β 1, β 2, α, data δ n+m+a4 1 exp δb 4 β 1 + 1s 1 x, α, δ β 2 + 1s 2 y, α, δ Clearly, the above forms of the posterior density do not lead to explicit Bayes estimates of the model parameters For this reason, we prefer to use the Metropolis-Hasting method with normal proposal distribution The algorithm of Gibbs sampling is summarized below: 1 Start with an initial guess β 0 1, β0 2, α0, δ 0 2 Set t = 1 3 Generate β t 1 from Gamma n + a 1, b 1 + s 1 x, α t 1, δ t 1 4 Generate β t 2 from Gamma n + a 2, b 2 + s 2 y, α t 1, δ t 1 5 Use the Metropolis-Hasting method to generate α t from fα t 1 β 1, β 2, δ, data with the Nα t 1, 05 proposal distribution 6 Use the Metropolis-Hasting method to generate δ t from fδ t 1 β 1, β 2, α, data with the Nδ t 1, 05 proposal distribution 7 Compute R t from equation 21 8 Set t = t + 1

Journal of Advanced Statistics, Vol 1, No 3, September 2016 165 9 Repeat Steps 3-8, T times Now the approximate posterior mean variance of R, respectively, are calculated as ÊR data = 1 T T R t, 212 t=1 var ˆ R data = 1 T T t=1 R t ÊR data 2 213 Using the result in Chen Shao 19, we construct the 1001 γ% highest posterior density HPD credible intervals as R γ T, R 2 1 γ 2 T, where R γ T R 2 1 γ 2 T are the γ 2 T -th smallest integer the 1 γ 2 T -th smallest integer of {R t, t = 1, 2,, T }, respectively It is worthy to point out that the Metropolis algorithm adopts only symmetric proposal distributions Therefore the normal distribution is appropriate It is checked here that the normal proposal distribution with variance σ 2 = 05 is best for the rapid convergence of the Metropolis algorithm 25 Numerical Simulations The comparisons between the the MLEs Bayes estimators of R cannot be done theoretically Thus, we present some simulations to compare the performance of the obtained results We compare the MLEs Bayes estimators in terms of their biases mean squared errors MSE We also compare different confidence intervals, namely; the confidence intervals obtained by using asymptotic distribution of the MLEs, bootstrap confidence intervals the HPD credible intervals in terms of the average confidence lengths The Bayes estimates are computed under the squared error loss function To assess the performance of the methods used in estimation, we use different parameter values, different hyper-parameters different sample sizes Further, we assume two priors to obtain the Bayes estimators HPD credible intervals These priors are typical given as: Prior 1: a j = 00001, b j = 00001, where j = 1, 2, 3, 4, Prior 2: a j = 1, b j = 3, where j = 1, 2, 3, 4 In Table 1, we present the average biases MSEs of the MLEs Bayes estimators based on 1000 replications The results are given under different sample sizes The average confidence/credible lengths the corresponding coverage percentages are given in Table 2 It should be noted that for the two bootstrap methods, we compute the confidence intervals based on 1000 bootstrap iterations The Bayes estimates the corresponding credible intervals are based on T = 1000 samples Also, the confidence level, γ used in finding the confidence intervals or the credible intervals is 095 It is observed that the MLE compares very well with the Bayes estimators in terms of biases MSEs, see Table 1 Also, it is observed that the Bayes estimators based on Prior 2 perform better than those obtained based on Prior 1 Table 2 shows that, the confidence intervals based on the asymptotic distributions of the MLEs work well when n m are getting larger, the Boot-p confidence intervals perform better than the Boot-t confidence intervals, the bootstrap method provide the smallest average lengths, the HPD credible intervals are wider than the other confidence intervals

166 Journal of Advanced Statistics, Vol 1, No 3, September 2016 Table 1 Biases MSEs of MLE Bayes estimators of R for different values of parameters BS BS n, m Method MLE prior1 prior2 MLE prior1 prior2 β 1 = 15, β 2 = 2, α = δ = 1 β 1 = 2, β 2 = 15, α = δ = 1 15,15 MSE 00106 00161 00161 00108 00176 00173 Bias -00054 05012 05011-00030 05489 05410 15,25 MSE 00089 00160 00156 00087 00175 00170 Bias -00023 04980 04877-00009 05462 05305 15,50 MSE 00083 00158 00152 00081 00173 00166 Bias -00017 04934 04768-00002 05411 05213 25,15 MSE 00091 00158 00162 00091 00173 00174 Bias -00156 04932 05048-00133 05417 05452 25,25 MSE 00071 00157 00157 00070 00172 00170 Bias -00143 04908 04914-00095 05383 05337 25,50 MSE 00065 00157 00154 00063 00172 00168 Bias -00076 04926 04846-00051 05394 05285 50,15 MSE -00081 00161 00166 00082 00176 00179 Bias -00046 05029 05214-00020 05516 05619 50,25 MSE 00067 00160 00162 00066 00174 00175 Bias -00015 05013 05096 00001 05484 05516 50,50 MSE 00056 00159 00159 00055 00174 00173 Bias 00012 05002 05000 00026 05478 05448 Table 2 Average confidence credible length the coverage percentage MLEs Boot BS n, m Untransformed Logit Arcsin boot-p boot-t prior1 prior2 β 1 = 15, β 2 = 2, α = δ = 1 15,15 Len 06355 05421 04353 02291 02487 03482 03340 CP 09930 09830 09610 08570 08700 09780 09920 15,25 Len 06052 05279 03779 01951 01966 03140 03045 CP 09990 09960 09660 09470 09480 09840 09920 15,50 Len 06329 05516 03594 01839 01843 02847 02881 CP 09990 09990 09780 09250 09260 09820 09920 25,15 Len 05097 04543 03194 02121 02159 03160 03071 CP 09940 09790 09250 05550 05630 09940 09940 25,25 Len 04923 04454 02705 01774 01806 02752 02658 CP 09960 09940 09280 07560 07610 09900 09920 25,50 Len 04627 04258 02225 01444 01465 02401 02380 CP 09990 09990 09190 05730 05790 09780 09900 50,15 Len 03624 03354 02074 01923 01990 02893 02924 CP 09590 09500 07960 07810 07890 09860 09820 50,25 Len 03626 03400 01744 01601 01649 02426 02387 CP 09880 09840 08070 06330 06450 09860 09820 50,50 Len 03481 03300 01376 01255 01257 02007 01940 CP 09990 09990 08110 09180 09190 09860 09860 β 1 = 2, β 2 = 15, α = δ = 1 15,15 Len 06673 05629 04549 02405 02494 03476 03338 CP 09960 09890 09710 07630 07720 09780 09860 15,25 Len 05528 04864 03451 02301 02305 03163 03074 CP 09970 09890 09360 09650 09660 09900 09920 15,50 Len 04006 03689 02292 02125 02130 02888 02929 CP 09850 09680 08430 09520 09530 09800 09780 25,15 Len 05920 05186 03702 01909 01920 03142 03045 CP 09980 09970 09610 05100 05130 09900 09980 25,25 Len 05169 04653 02839 01863 01868 02756 02658 CP 09980 09950 09300 08640 08650 09880 09900 25,50 Len 03960 03694 01905 01748 01752 02435 02389 CP 09970 09940 08500 07810 07820 09880 09860 50,15 Len 04595 04213 02637 01335 01346 02860 02887 CP 09960 09940 09040 08340 08360 09880 09960 50,25 Len 04335 04023 02087 01353 01356 02407 02387 CP 09990 09970 08970 06420 06430 09860 09920 50,50 Len 03655 03460 01446 01317 01320 02014 01941 CP 09990 09990 08210 09180 09190 09760 09800

Journal of Advanced Statistics, Vol 1, No 3, September 2016 167 3 Multicomponent Stress-Strength Model In this section, we study the multicomponent stress-strength reliability for Dagum distribution when both stress strength variates follow the same population We also obtain the asymptotic confidence intervals for the multicomponent stress-strength reliability using MLE We run a simulation study to compare the estimators Let the rom samples Y, X 1, X 2,, X k be independent, Gy be the continuous distribution function of Y F x be the common continuous distribution function of X 1, X 2,, X k The reliability in a multicomponent stress-strength model developed by Bhattacharyya Johnson 20 is R s,k = P at least s of the x 1, x 2,, x k exceed Y k k = 1 Gy i Gy k i df y 31 i i=s 31 Maximum Likelihood Estimator of R s,k Let X Dagumβ 1, α, δ Y Dagumβ 2, α, δ, where X Y are two independent rom variables with unknown shape parameters β 1 β 2 common scale parameters δ shape parameter α The reliability in multicomponent stress-strength for Dagum distribution using 11 results in: k k R s,k = 1 αy δ β1 i 1 + αt δ β1 k i β 2 αδt δ 1 1 + αt δ β2 1 dt i i=s k k = v 1 t i t k i+v 1 dt i i=s k k = v Bk + v i, i + 1 32 i i=s where t = 1 + αy δ β1, v = β 2 /β 1 B, is the incomplete beta function After some simplification equation 32 is reduced to The MLE of R s,k becomes ˆR s,k = ˆv R s,k = v k i=s k i=s k! k i! k! k i! k k + v j j=i k k + ˆv j j=i 1 1 33, where ˆv = ˆβ 1 ˆβ 2 34 To obtain the asymptotic confidence intervals for R s,k, we proceed as follows: The asymptotic variances AV of the ˆβ 1 ˆβ 2 are given by AV ˆβ 1 = E β 1 1 = β2 1 n, AV ˆβ 2 = E β 2 1 = β2 2 m The asymptotic variance of an estimate of R s,k is given by For details see Rao 21: AV ˆR s,k = AV ˆβ 1 Rs,k β 1 2 + AV ˆβ 2 Rs,k β 2 2 35 Thus from equation 35, the asymptotic variance of ˆR s,k can be obtained To avoid the difficulty of derivation of R s,k, we obtain ˆR s,k their derivatives for s, k = 1, 3 2, 4 separately They are

168 Journal of Advanced Statistics, Vol 1, No 3, September 2016 given as 3 ˆR 1,3 = ˆv + 3, ˆR2,4 = ˆR 1,3 = 1 3ˆv β 1 β 1 ˆv + 3 2, ˆR 2,4 β 1 = 1 β 1 12ˆv2ˆv + 7 ˆv + 4ˆv + 3 2 12 ˆv + 4ˆv + 3,, ˆR 1,3 = 1 3 β 2 β 1 ˆv + 3 2, ˆR 2,4 = 1 β 2 β 1 122ˆv + 7 ˆv + 4ˆv + 3 2 Therefore, AV ˆR 1,3 = 9ˆv2 1 ˆv + 3 4 n + 1, m AV ˆR 2,4 = 144ˆv2 2ˆv + 7 2 1 ˆv + 4ˆv + 3 4 n + 1 m As n m, we have R s,k ˆR s,k /AV ˆR s,k N0, 1 for s, k = 1, 2, The asymptotic 95% confidence interval CI of system reliability R s,k is ˆR s,k 196 AV ˆR s,k Thus, the asymptotic 95% confidence intervals for R 1,3 R 2,4 are, respectively, given by 32 Simulation Study 3ˆv 1 ˆR 1,3 196 ˆv + 3 2 n + 1 m, 12ˆv2ˆv + 7 1 ˆR 2,4 196 ˆv + 4ˆv + 3 2 n + 1 m We generate 3000 rom samples of size 10, 15, 20, 25, 30 each from stress strength populations for β 1, β 2 = 30, 15, 25, 15, 20, 15, 15, 15, 15, 20, 15, 25 15,30 as proposed by Bhattacharyya Johnson 20 The MLEs of β, α δ, say ˆβ, ˆα ˆδ are estimated by solutions of the nonlinear equation These ML estimators of β 1 β 2 are then substituted in v to get the multicomponent reliability for s, k = 1, 3, 2, 4 The average bias average MSE of the reliability estimates over the 3000 replications are given in Table 3 Average confidence length of the simulated 95% confidence intervals of, R s,k are given in Table 3 The true values of reliability in multicomponent stress-strength with the given combinations for s, k = 1, 3 are 0857, 0833, 0800, 0750, 0692, 0643, 0600 for s, k = 2, 4 are 0762, 0725, 0674, 0600, 0519, 0454, 0400 Thus the true value of reliability in multicomponent stress-strength decreases as β 2 increases for a fixed β 1 whereas reliability in multicomponent stress-strength increases as β 1 increases for a fixed β 2 in both cases of s, k Therefore, the true value of reliability decreases as v increases vice versa The average bias average MSE decrease as sample size increases for both situations of s, k Also, it is noted that the bias is negative relatively small in all the combinations of the parameters in both situations of s, k which leads MLE to underestimate the parameters, thus the R s,k This generally proves the consistency property of the MLE of R s,k Whereas, among the parameters the absolute bias MSE decrease as β 1 increases for a fixed β 2 in both cases of s, k The absolute bias MSE increase as β 2 increases for a fixed β 1 in both cases of s, k It is also clear that as the sample size increases the length of CI decreases in all cases

Journal of Advanced Statistics, Vol 1, No 3, September 2016 169 Table 3 In each cell the first row represents the average bias, the second row represents the average MSE the third row represents the average confidence length of the simulated 95% confidence intervals of R s,k using MLE β 1, β 2 s, k n, m 30,15 25,15 20,15 15,15 15,20 15,25 15,30 10,10-00100 -00106-00119 -00143-00171 -00177-00176 00043 00053 00069 00094 00124 00147 00167 02197 02466 02809 03247 03633 03856 03976 15,15-00039 -00045-00056 -00080-00108 -00121-00122 00027 00033 00044 00060 00080 00094 00106 01754 01983 02276 02655 02994 03200 03318 1,3 20,20-00011 -00015-00025 -00045-00071 -00091-00097 00019 00024 00031 00043 00057 00069 00079 01504 01705 01964 02299 02604 02792 02901 25,25-00003 -00009-00018 -00037-00061 -00081-00092 00015 00018 00024 00033 00044 00053 00061 01344 01526 01760 02063 02339 02512 02614 30,30-00007 -00011-00021 -00042-00066 -00086-00097 00013 00016 00020 00028 00037 00044 00050 01232 01398 01612 01891 02145 02304 02398 10,10-00131 -00129-00133 -00139-00141 -00114-00081 00101 00117 00143 00173 00196 00205 00208 03376 03713 04098 04506 04742 04765 04675 15,15-00044 -00048-00055 -00073-00088 -00083-00064 00065 00076 00093 00115 00132 00139 00140 02728 03022 03363 03735 03962 04004 03945 2,4 20,20-00006 -00009-00018 -00036-00055 -00064-00056 00047 00055 00068 00085 00099 00106 00107 02353 02614 02921 03258 03470 03516 03468 25,25-00004 -00002-00012 -00031-00051 -00064-00065 00036 00048 00053 00066 00077 00082 00084 02108 02345 02626 02934 03130 03175 03132 30,30-00003 -00008-00019 -00042-00063 -00077-00079 00031 00037 00045 00055 00064 00069 00070 01933 02150 02407 02692 02872 02914 02876 4 Concluding Remarks In this paper, we have considered the problem of estimation of a single multicomponent stress-strength reliability for a three-parameter Dagum distribution when both stress strength variates follow the same distribution The maximum likelihood Bayes estimators of the stress-strength parameter, R have been derived We have compared the MLEs Bayes estimators in terms of their biases mean squared errors It may be noted, from Table 1 that the maximum likelihood estimates have the smallest mean squared errors as compared with their corresponding Bayes estimates We observed the Bayesian method is sensitive to the choice of the priors Generally the Bayes estimators based on Prior 2 perform better than the Bayes estimators based on Prior 1 For all sample sizes n, m, the bootstrap methods provide the smallest average lengths It is observed that boot-p confidence intervals perform better than those obtained by the boot-t methods Further, the coverage probability is quite close to the given value in all sets of parameters For the multicomponent stress-strength reliability, we have estimated the asymptotic confidence intervals The simulation result in Table 3 indicates that the averages of bias MSE decrease as sample size increases Also, the absolute bias MSE decrease increase as β 1 increases β 2 increases in both situations of s, k Acknowledgments The authors are grateful to the anonymous referees for their valuable comments suggestions which improved the presentation of the paper This study is a part of the Master Thesis of the second named author whose work was supervised by the first named author References 1 C Dagum, A New Model for Personal Income Distribution: Specification Estimation, Economic Applique, 33 1977 413 437

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