Startseite Wirtschaftswissenschaften Multi-component stress-strength model for Weibull distribution in progressively censored samples
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Multi-component stress-strength model for Weibull distribution in progressively censored samples

  • Akram Kohansal ORCID logo EMAIL logo , Shirin Shoaee und Saralees Nadarajah
Veröffentlicht/Copyright: 12. Februar 2022
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Abstract

One of the important issues is risk assessment and calculation in complex and multi-component systems. In this paper, the estimation of multi-component stress-strength reliability for the Weibull distribution under the progressive Type-II censored samples is studied. We assume that both stress and strength are two independent Weibull distributions with different parameters. First, assuming the same shape parameter, the maximum likelihood estimation (MLE), different approximations of Bayes estimators (Lindley’s approximation and Markov chain Monte Carlo method) and different confidence intervals (asymptotic and highest posterior density) are obtained. In the case when the shape parameter is known, the MLE, uniformly minimum variance unbiased estimator (UMVUE), exact Bayes estimator and different confidence intervals (asymptotic and highest posterior density) are considered. Finally, in the general case, the statistical inferences on multi-component stress-strength reliability are derived. To compare the performances of different methods, Monte Carlo simulations are performed. Moreover, one data set for illustrative purposes is analyzed.

MSC 2010: 62F10; 62F15; 62N02

A Appendix

So the log-likelihood function from (2.2) is

( α , θ , λ ) = n ( k + 1 ) log ( α ) - n k log ( θ ) - n log ( λ ) - 1 θ i = 1 n j = 1 k ( R j + 1 ) x i j α - 1 λ i = 1 n ( S i + 1 ) y i α + ( α - 1 ) ( i = 1 n log ( y i ) + i = 1 n j = 1 k log ( x i j ) ) + Constant .

The MLEs of 𝜃, 𝜆 and 𝛼, say θ ^ , λ ^ and α ^ , respectively, can be obtained as follows:

θ = - n k θ + 1 θ 2 i = 1 n j = 1 k ( R j + 1 ) x i j α = 0 ,
λ = - n λ + 1 λ 2 i = 1 n ( S i + 1 ) y i α = 0 ,
α = n ( k + 1 ) α + i = 1 n log ( y i ) + i = 1 n j = 1 k log ( x i j ) - 1 λ i = 1 n ( S i + 1 ) y i α log ( y i ) - 1 θ i = 1 n j = 1 k ( R j + 1 ) x i j α log ( x i j ) = 0 .

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Received: 2020-10-18
Revised: 2021-12-18
Accepted: 2021-12-21
Published Online: 2022-02-12
Published in Print: 2022-05-01

© 2022 Walter de Gruyter GmbH, Berlin/Boston

Heruntergeladen am 21.12.2025 von https://www.degruyterbrill.com/document/doi/10.1515/strm-2020-0030/pdf
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