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ESTIMATION OF STRESS STRENGTH RELIABILITY USING PRANAV DISTRIBUTION
Ankitha Lukose, Chacko V M •
Department of Statistics St.Thomas College(Autonomous) Thrissur, Kerala, 680 001, India [email protected] [email protected]
Abstract
This paper deals with the estimation of stress strength reliability parameter R, which is the probability of Y less than X when X and Y are two independent distribution with different scale parameter and same shape parameter. The maximum likelihood method is used to find an estimator for R. We also obtain the asymptotic distribution of the maximum likelihood estimator ofR. Based on this asymptotic distribution, the asymptotic confidence interval can be obtained. We also propose bootstrap confidence interval for the parameter R. Analysis of a simulated data and a real life data have been presented for illustrative purposes.
Keywords: stress strength model, maximum-likelihood estimator, bootstrap confidence intervals, asymptotic distributions and confidence interval
1. Introduction
One of the significant, challenging, but manageable problems in reliability analysis is the calculation of stress-strength reliability using a variety of distributions. Estimating the stress-strength parameter, R, is very helpful in the statistical literature. For instance, if X represents the strength of a system under a stress, Y, then R is a measure of system performance that arises naturally from the mechanical dependability of a system. Only when the applied stress exceeds the system's strength at any point does the system fail. Many lifespan distributions are utilised in reliability analysis. Terms like exponential, Weibull, log-Normal, and their generalisations are commonly used in dependability analysis. The exponential, Lindley, and Weibull distributions are more often used than the gamma and lognormal distributions because their survival functions can both be expressed in closed forms and do not require numerical integration. While sharing a common parameter, the exponential and Lindley distributions differ in that the hazard rate of the exponential is constant whereas the hazard rate of the Lindley is monotonically dropping. Although the Lindley distribution has been used by many academics to model lifetime data and is crucial for understanding stress-strength reliability modelling, there are numerous instances in which modelling actual lifetime data may not be appropriate from a theoretical or practical standpoint. Recently, a number of academics presented many distributions, with the new ones showing a better fit than the currently popular distributions. In order to fix a problem with new models while utilising better fitted models in stress-strength analysis, one may need to carefully examine the estimation process. If the estimation process fails using the available methodologies, one may not be able to do so. Therefore, it is crucial to estimate multiple reliability factors, and researchers must focus more on estimates while utilising better fitted models.
Ankitha Lukose, Chacko V M RT&A, No 4 (76) ESTIMATION OF STRESS STRENGTH RELIABILITY_Volume 18, December 2023
In the literature, it has been debated how to estimate a stress strength model's reliability or survival probability when X and Y have known distributions. A number of authors have examined the survival probabilities of a single component stress-strength (SSS) model for various distributions, including Raqab and Kundu[12], Kundu and Gupta [9], [6], Constantine and Karson[4], and Downtown[5]. The issue of estimating R has been investigated by a number of authors. Church and Harris [3] developed the MLE of R when X and Y are independently and normally distributed. Awad et al study's on the MLE of R under the condition that X and Y have bivariate exponential distributions was published in 1981. In a simulation research, Awad and Gharraf [2] compared three estimations of R when X and Y are independent Burr random variables with different distributions. Estimates for R where X and Y are Burr Type X distributions were reported by Ahmad et al.[1] and Surles and Padgett[16],[17]. Other papers which describes the same idea on different distributions are Akhila and Chacko [18],Kundu and Raqab [11],saraccouglu,Kinaci and Kundu [14] and Shahsanaei, Fatemeh and Daneshkhah, Alireza [15].
In this paper, we consider the problem of estimating the stress strength parameter R = P[Y < X], when X and Y be independent strength and stress random variables having Pranav distribution with parameters 61 and 62 respectively. Krishna Kumar Shukla [8]introduced Pranav distribution which is a mixture of two distributions, exponential distribution having scale parameter 6 and gamma distribution having shape parameter 4 and scale parameter 6, and their mixing proportion of 6, and gj+g respectively. The probability density function (pdf) of Pranav lifetime distribution can be defined as
g4
f (x, 6) = -&—7 (6 + x3)e-6x, x > 0,6 > 0 (1)
64 + 6
The corresponding cumulative distribution function(c.d.f) is
6x(62x2 + 36x + 6
F(x, 6) = 1 -
1 +
+ 6
e-6x, x > 0,6 > 0 (2)
The estimation of the stress strength parameter R = P[Y < X], when X and Y are both one-parameter Pranav distributions with parameter 61 and 62 respectively, is an unsolved problem. Statistical inference on stress-strength parameters are important in reliability analysis. It is observed that the maximum likelihood estimators can be obtained implicitly by solving two nonlinear equations, but they cannot be obtained in closed form. So, MLE's of parameters are derived numerically. It is not possible to compute the exact distributions of the maximum likelihood estimators, and we used the asymptotic distribution and we constructed approximate confidence intervals of the unknown parameters.
The rest of the paper is organized as follows. In Section 2, the MLE of R is computed. The asymptotic distribution of the MLE's are provided in Section 3. Bootstrap confidence interval is presented in Section 4. In Section 5, simulation study is given. Theoretical results are verified by analyzing data set in Section 6 and conclusions are given in Section 7.
2. Maximum Likelihood Estimator of R
In this section, the procedure of estimating the reliability of R = P[Y < X] models using Pranav distributions, is considered. It is clear that
R = P[Y < X]
p TO
= Jo f (x, 6x)F(x, 62) dx (3)
where f (x,y), is the joint probability density function (pdf) of random variables X and Y, having Pranav distributions. If the r.v's X and Y are independent, then f (x, y) = f (x)g(y), where f (x)
and g(y) are the marginal pdf's of X and Y, so that
r ~ / 04 (
lo V0T+6(
R
1 + x3)e
-e x
1 -
1 +
e x(e2 x2 + 3e x + 6 e4 + 6
ex
dx
(4)
On simplification we get
01
R = 1 -
36O0102 + 108003 + 14402 (01 + 02 )2+'
6(0103 + 04 + 6)(01 + 02 )3+ 6(0102)2 (01 + 02 )4 + 60102 (01 + 02)5 + 01 (04 + 6)(01 + 02 )6
(04 + 6) (04 + 6)(01 + 02)7
(5)
If we have two ordered random samples representing strength (X1,X2...Xn) and stress (Y1,Y2,....,Ym) having sizes n and m respectively, following Pranav distribution with parameters 01 and 02, respectively. A technique for figuring out a model's parameter values is called maximum likelihood estimation. The parameter values were selected to maximise the likelihood that the model's described process resulted in the observed data. The optimal distribution for a collection of data is chosen using the maximum likelihood estimate (MLE). A reliable method for estimating parameters is maximum likelihood. So, a variety of estimation scenarios can apply maximum likelihood estimations. With those parameter values, the likelihood of those observed outcomes is the same as the probability of those observed events. Thus, the likelihood function for the combined random sample can be calculated:
n 04 n ai ?
L = n ^ 0 + x?) e-01 x? n ^ (02 + y!
the log likelihood is
nn
l = logL = 4nlog01 - nlog0 + 6) + £(log(01 + x?)) - 01 £ Xi
(6)
i=0
i=0
+ 4mlog02 - mlog(04 + 6) + £(log(02 + y?)) - 02 £ y (7)
j=1
j=1
The solution of the following non-linear equations yield the MLE of the parameters parameters 01 and 02. Differentiating (7) with respect to parameters 01 and 02, we obtain
dd¡
dl W2
4n 01
4m ~d2
+ £
i
4n0f
(0|y+6 ' i=o 01 + xf i=o 1
4m0f m
(02H6 + ;=o 02 + xf j=0
— £xi i=0 m
— £yj
The second partial derivatives of (7) with respect to parameters 01 and 02, are
n 1
- i=0 W+X)2
dl2 —4n 4n02(18 — 04)
d02 = 02 (04 + 6)2
dl2 -4m 4m 0 2(18 — 04)
d02 022 (024 + 6)2
1
MLE of R is obtained as
01
R = 1 -
-
36001 (02 + 108002 + 14402(01 + 02 )2+ 6(01023 + 024 + 6)(01 + 02 )3 + 6(0102) 2 (01 + 02 )4 + . 60102 (01 + 02 )5 + 01 (024 + 6) (01 + 02)6 . (Or + 6)(024 + 6)(01 + 02)7
(8) (9)
(10) (11)
(12)
e
Where 61 and 62 are the maximum likelihood estimators of 61 and 62 respectively. This is used in estimation of stress strength for given data.
3. Asymptotic Distribution and Confidence Interval
The asymptotic distribution and confidence interval of the MLE of R are obtained in this section. To find an asymptotic variance of the MLE RML, let us denote the Fisher information matrix of 0 = (01,02) as I(0) = [Iij(d); i, j = 1,2], i.e.,
I(0) = E
To establish normality assumption we define
d(0) =
-d21 d02 -d21 d02 d01
-d2l ■ B01B02
d! d02 j
R is in the form
Hence
dR dR ,
( Wi d0>) (d1, d2 )
dR d0i
R = 1 - UP
( U )'P+4
, i = 1,2
U = 01
V = (04 + 6)0 + 6)(01 + 02)7
(13)
(14)
(15)
(16)
(17)
(18) (19)
P = 36001(02 + 108002 + 14402 (01 + 02 )2 + 6(01023 + 024 + 6)(01 + 02 )3 + 6(0102)2 (01 + 02)4+ 60102(01 + 02 )5 + 01 (024 + 6)(01 + 02 )6 The partial derivatives of U and P with respect to 01 is dP
— = 36002 + 28802 (01 + 02 ) + 180i 02 + (01 + 02)2 +
30i
603(01 + 02 )3 + 1802 (01 + 02 )2 + 18(01 + 02)2 +
24(0102 )2(01 + 02 )3 + 120102 (01 + 02 )4 + 300102(01 + 02 )4 +
602 (01 + 02)5 + 601 (04 + 6)(01 + 02 )5 + (04 + 6)(01 + 02)6
( U )'
( V)
■ (04 + 6)(04 + 6)(01 + 02)7- ■ 701 (04 + 6)(04 + 6)(01 + 02)6-404 (024 + 6)(01 + 02 )7
[(04 + 6)0 + 6)(01 + 02 )7 ]2
(20)
(21)
(22)
Then partial derivative of R with respect to 01 becomes
dR
-7010 + 6)(04 + 6)(0i + 02)6 - 404(04 + 6)(0i + 02)7+ (04 + 6)0 + 6)(0i + 02)7
+
((04 + 6)0 + 6)(0i + 02)7)2 3600102 + 108003 + 14402 (01 + 02 )2 + 6(0103 + 04 + 6)(01 + 02)3 + 6(0102)2(01 + 02)4 + 6(0102)2(01 + 02)4 + 60102(01 + 02)5 + 01(04 + 6)(01 + 02)6 36002 + 28802 (01 + 02 ) + 180103 (01 + 02 )2603 (01 + 02 )3 + 18202 (01 + 02)2 + 18(01 + 02)2 + 24(0102 )2(01 + 02 )3 + 120102 (01 + 02 )4 + 300102(01 + 02 )4 +
.602(01 + 02 )5 + 601(04 + 6)(01 + 02)5 + (04 + 6)(01 + 02 )6_ 01
(04 + 6)0 + 6)(01 + 02)7
The partial derivatives of V and P with respect to 02 is dP
— = 7200102 + 324002 + 28802 (01 + 02) + 1440 + 02 )2
d02
+ 180103(01 + 02 )2 + 180102(01 + 02 )3 + 1804(01 + 02 )2 + 2403 (01 + 02 )3 + 18(01 + 02 )2 + 24(0102)2 (01 + 02 )3 + 1202 02 (01 + 02 )4 + 300102 (01 + 02 )4 + 601 (01 + 02)5 + 601 (04 + 6)(01 + 02 )5 + 40103 (01 + 02)6
(U )' = -
(V)
[-701 (04 + 6)(04 + 6)(01 + 02)6 - 40103(04 + 6)(01 + 02)7]
((04 + 6)(04 + 6)(01 + 02)7 )2
Then the partial derivative of R with respect to 02 becomes
dR d02
+
[-701(04 + 6)(04 + 6)(01 + 02 )6 - 40103 (04 + 6)(01 + 02 )7 ] ((04 + 6) (04 + 6)(01 + 02 )7)2 3600102 + 108003 + 14402(01 + 02)2+ 6(0103 + 04 + 6)(01 + 02)3 + 6(0102 )2 (01 + 02 )4 + .6(0102 )2 (01 + 02 )4 + 60102 (01 + 02 )5 + 01 (04 + 6)(01 + 02 )6.
7200102 + 324002 + 28802(01 + 02) + 144(01 + 02)2+ 180103 (01 + 02 )2 + 180102(01 + 02 )3 + 1804 (01 + 02 )2 + 2403(01 + 02 )3+ 18(01 + 02)2 + 24(0102 )2 (01 + 02 )3 + 120202 (01 + 04) + 300102 (01 + 02 )4 + 601 (01 + 02 )5 + 601 (04 + 6)(01 + 02 )5 + 40103 (01 + 02 )6
01
as
(04 + 6) (04 + 6)(01 + 02 )7
We obtain the asymptotic distribution of RML
VnTm(Rml - R) ^ N(0,d(0)I-1 (0)d(0)) 1
AV (R ML)
n + m
d (0)I-1(0)d(0)
i.e, AV (RML) = V (01)d2 + V (02)d2 + 2d1 d2Cov(d\, 02) Asymptotic 100(1 - a) percentage confidence interval for R
RML ± Za/2 y/ AV(RML)
(23)
(24)
(25)
(26)
(27)
(28)
(29)
(30)
x
X
X
Ankitha Lukose, Chacko V M RT&A, No 4 (76) ESTIMATION OF STRESS STRENGTH RELIABILITY_Volume 18, December 2023
4. Bootstrap Confidence Interval
In this section, we use confidence intervals based on the parametric percentile bootstrap methods (we call it from now on as Boot-p) based, [10]. Bootstrapping is a technique for estimating the variability in a statistic by sampling with replacement from observed data. Permutation tests are a type of re-sampling that is linked to re-sampling. The bootstrap is frequently used to evaluate the accuracy of an estimate based on a sample of data from a larger population. The bootstrap's main advantage is that it allows statisticians to construct confidence intervals on parameters without making irrational assumptions. This was one of the first of many discoveries in computational statistics, which has now become the standard method for practically all work. It creates multiple re-samples (with replacement) from a single set of observations, and computes the effect size of interest on each of these re-samples. To estimate confidence intervals of R in this methods, the following steps are used.
1. Estimate 0, say 0, from the sample using MLE method.
2. Generate a bootstrap sample using 0. Using these bootstrap sample obtain the bootstrap estimate of 0, say 0* and compute the bootstrap estimate of R.
!. Repeat Step [2] N-BOOT times to get the parametric bootstrap estimates of R
4. Let CDF = P(0* < x), be the cumulative distribution function of R. Define 0Boot-p(x) = CDF (x) for a given x. The approximate (100 — a)% confidence interval for 0 is given by
(0BOOT-P (a/2), dBOGT-p (1 — (a/2))) (31)
5. Simulation Study
To measure the efficacy of R's estimators, we provide some results based on the inversion method in this section. According to its definition, data simulation is the process of using a significant amount of data to replicate or simulate real-world settings in order to determine the optimal course of action, predict future events, or test a model. The many forms of data simulations are numerous. In simulation statistics, artificially generated data are used to test a hypothesis or statistical method. Every time a new statistical method is created or used, certain assumptions need to be verified. Simulated data is used by statisticians to test their theories. For this purpose, we have generated 1000 samples from independent Pranav(0i) and Pranav(02) distributions. We considered sets of parameter values (1) and (1.99) parameter values. The bias and the mean square error (MSE) of the parameter estimates are calculated. In Table 5.1, the average biases, mean squared errors (MSE) and confidence intervals of the estimates of R is given.
Table 1: Simulation Results
(n,m) R(ML) Bias MSE
(7,7) 0.8982136 0.1017854 0.01036027
(15,15) 0.8982252 0.1017738 0.0103579
(20,20) 0.8982344 0.1017646 0.01035603
(22,22) 0.8985076 0.1014914 0.0103005
(25,25) 0.8985383 0.1014607 0.01029428
AS(CI)
BT(CI)
0.8783155,0.9181117 0.8782696,0.9181808 0.8783453,0.9181235 0.8787993,0.9182159 0.8789089,0.9181677
0.8947541,0.8999972 0.8966189,0.8994737 0.8965753,0.8999412 0.8969416,0.8993731 0.8973348,0.898984
From the simulation results, it is observed that as the sample size (n, m) increases the biases and the MSE decrease. Thus the consistency properties of all the methods are verified.
6. Data Analysis
In this section we present a data analysis of the strength data reported by Badar and Priest(1982). The data represent the strength data measured in GPA, for single carbon fibers and impregnated 1000-carbon fiber tows. Single fibers were tested under tension at gauge lengths of 1, 10, 20 and 50 mm. Impregnated tows of 1000 fibers were tested at gauge lengths of 20, 50, 150 and 300 mm. It is already observed that the Weibull model does not work well in this case. Surles and Padgett[16], [17] and Raqab and Kundu[12] observed that generalized Rayleigh works quite well for these strength data. For illustrative purposes we are also considering the same transformed data sets as it was considered by Raqab and Kundu [9], the single fibers of 20 mm (Data Set I) and 10 mm (Data Set II) in gauge lengths with sample sizes 69 and 63 respectively. They are presented below:
Data Set I: 0.312 0.314 0.479 0.552 0.700 0.803 0.861 0.865 0.944 0.958 0.966 0.997 1.006 1.021 1.027 1.055 1.063 1.098 1.140 1.179 1.224 1.240 1.253 1.270 1.272 1.274 1.301 1.301 1.359 1.382 1.382 1.426 1.434 1.435 1.478 1.490 1.511 1.514 1.535 1.554 1.566 1.570 1.586 1.629 1.633 1.642 1.648 1.684 1.697 1.726 1.770 1.773 1.800 1.809 1.818 1.821 1.848 1.880 1.954 2.012 2.067 2.084 2.090 2.096 2.128 2.233 2.433 2.585 2.585
Data Set II: 0.101 0.332 0.403 0.428 0.457 0.550 0.561 0.596 0.597 0.645 0.654 0.674 0.718 0.722 0.725 0.732 0.775 0.814 0.816 0.818 0.824 0.859 0.875 0.938 0.940 1.056 1.117 1.128 1.137 1.137 1.177 1.196 1.230 1.325 1.339 1.345 1.420 1.423 1.435 1.443 1.464 1.472 1.494 1.532 1.546 1.577 1.608 1.635 1.693 1.701 1.737 1.754 1.762 1.828 2.052 2.071 2.086 2.171 2.224 2.227 2.425 2.595 3.220 These data were first used by Badar and Priest(1982) and later used by Raqab and Kundu[13] and Hassan and Kumaraswamy [7].
Table 2: Data Analysis Results
Length(in mm) MLE K-S statistic P-value
10 1.596362 0.62319 0.7571
20 1.715981 0.79365 0.4375
Table 6.1 gives the result of the goodness of fit test. Maximum likelihood estimates are 1.596362 and 1.715981.The estimated value of the stress strength reliability using these estimates is 0.8850866.The 95% asymptotic confidence interval for R is (0.8530591,0.9171141) and 95% bootstrap confidence interval for R is (0.8748553 ,0.8895613).
7. Conclusion
In this paper, we considered the problem of estimating stress strength reliability using Pranav distribution. The maximum likelihood estimate of stress strength reliability, R is obtained. Also, asymptotic 100(1 - a)% confidence interval for the reliability parameter is computed. Bootstrap confidence interval for the reliability parameter is also computed. When the sample size is increased, mean square error caused by the estimates comes nearer to zero by extensive simulation. Finally, real data sets are analyzed
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