Estimation of Reliability in the Two-Parameter Geometric Distribution

0.18 MB·English

by  Sudhansu S. Maiti

Estimation of Reliability in the Two-Parameter Geometric
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Sudhansu S. Maiti∗, Sudhir Murmu
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Department of Statistics, Visva-Bharati University

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A Santiniketan 731 235, India
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G. Chattopadhyay
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Department of Statistics, Calcutta University
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2 35, B. C. Road, Kolkata 700 019, India
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Abstract
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In this article, the reliabilities R(t) = P(X ≥ t), when X follows two-parameter geo-
:
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Xi metric distribution and R= P(X ≤ Y), arises under stress-strength setup, when X and Y
r
a assumed to follow two-parametergeometric independently have been found out. Maximum
Likelihood Estimator (MLE) and an Unbiased Estimator (UE) of these have been derived.
MLE andUE ofthe reliabilityofk-out-of-msystemhavealsobeenderived. The estimators
have been compared through simulation study.
AMS Subject Classification: 62N05; 62F10.
Key Words: k-out-of-m system; Maximum Likelihood Estimator; Stress-strength reliability;
Unbiased estimator.
∗
Corresponding author. E-mail Address: dssm1@rediffmail.com
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1 Introduction
Various lifetime models have been proposed to represent lifetime data. Most of these models
assume lifetime to be a continuous random variable. However, it is sometimes impossible or
inconvenient to measure the life length of a device on a continuous scale. In practice, we come
across situations where lifetimes are recorded on a discrete scale. Discrete life distributions
have been mentioned by Barlow and Proschan [1]. Here one may consider lifetime to be the
number of successful cycles or operations of a device before failure. For example, the bulb in
xerox machine lights up each time a copy is taken. A spring may breakdown after completing
a certain number of cycles of ‘to-and-fro’ movements.
The study of discrete distributions in lifetime models is not very old. Yakub and Khan [2]
considered the geometric distribution as a failure law in life testing and obtained various para-
metric and nonparametric estimation procedures for reliability characteristics. Bhattacharya
and Kumar [3] discussed the parametric as well as Bayesian approach to the estimation of the
mean life cycle and reliability for this model for complete as well as censored sample. Krishna
and Jain [4] obtained classical and Bayes estimation of reliability for some basic system con-
figurations. Modeling in terms of two-parameter geometric and estimation of its parameters
and related functions are of special interest to a manufacturer who wishes to offer a minimum
warranty life cycle of the items produced.
The two-parameter geometric distribution [abbreviated as Geo(r,θ)] given by
P(X = x) = (1−θ)θx−r ; x = r,r+1,r+2,... 0< θ < 1, r ∈ {0,1,2,...}, (1.1)
is the discrete analog of two-parameter exponential distribution. If X follows a two-parameter
exponentialdistribution, [X], theinteger partof X, has atwo-parameter geometric distribution
[see Kalbfleish and Prentice [5, Ch. 3]]. The reliability of a component when X follows two-
2
parameter geometric distribution is given by
R(t) =θt−r; t = r,r+1,r+2,.... (1.2)
Laurent[6]andTate[7]obtainedtheuniformlyminimumvarianceunbiasedestimator(UMVUE)
of the reliability function for the two-parameter exponential model. Different estimators of this
reliability function have been discussed in Sinha [8].
If a system consists of m identical components each follows two-parameter geometric distribu-
tion, then the reliability of k-out-of-m system is given by
m
m
R (t) = P(X ≥ t)= R(t)i[1−R(t)]m−i; t = r,r+1,r+2,... (1.3)
s (m−k+1)
i
i=k(cid:18) (cid:19)
X
Special cases of R (t) give series system (for k = m) and parallel system (for k = 1).
s
In the stress-strength setup, R = P(X ≤ Y) originated in the context of the reliability of
a component of strength Y subjected to a stress X. The component fails if at any time the
applied stress is greater than its strength and there is no failure when X ≤ Y. Thus R is a
measureofthereliability ofthecomponent. Manyauthorsconsideredtheproblemofestimation
of R in continuous setup in the past. Particularly, for the two-parameter exponential set up,
Beg [9] derived the MLE and the UMVUE of R. In the discrete setup, the reference list is
very limited. Maiti [10] has considered stress (or demand) X and strength (or supply) Y as
independently distributed geometric random variables, whereas Sathe and Dixit [11] assumed
as negative binomial variables, and derived both MLE and UMVUE of R. Maiti and Kanji
[12] has derived some expressions of R using a characterization and Maiti [13,14] considered
MLE, UMVUE and Bayes Estimation of R for some discrete distributions useful in life testing.
All the above mentioned works have been concentrated on one-parameter family of discrete
distributions.
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If X and Y follow two-parameter geometric distributions with parameters (θ ,r ) and
1 1
(θ ,r ) respectively, then
2 2
R = ρθδ for δ >0
2
= 1−(1−ρ)θ−δ for δ < 0, (1.4)
1
where ρ= 1−θ1 and δ = r −r .
1−θ1θ2 1 2
Here we are interested to see whether similar estimators are obtained in case of the two-
parameter geometric distribution, the discrete analog of the two-parameter exponential dis-
tribution. Then, it might be straightforward to use the two-parameter geometric distribution
in the discrete life testing problem where a minimum warranty life cycle of the item is offered.
In this article, we have found out some estimators of both R(t) and R (t) for complete as well
s
as censored sample. Some estimators of R have also been provided. The estimators have been
compared through simulation study.
The paper is organized as follows. In section 2, we have derived MLE and UE of both R(t)
and R (t). We have also derived MLE of these reliability functions for type-I censored sample.
s
MLE and an unbiased estimator of R have been found out in section 3. In section 4, simulation
results have been reported. Section 5 concludes.
2 Estimation of R(t) and R (t)
s
Let (X ,X ,...,X ) be a random sample from Geo(r,θ) and (X ,X ,...,X ) be ordered
1 2 n (1) (2) (n)
sample. Maximum Likelihood Estimator of r and θ are X and S respectively, where
(1) n+S
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S = n X −X = n X −X . ML Estimators of R(t) and R (t) are given by
i=1 i (1) i=1 (i) (1) s
P (cid:0) (cid:1) P (cid:0) (cid:1)
Rˆ (t) = 1 for t ≤ X
M (1)
t−X
S (1)
= for t > X (2.5)
(1)
n+S
(cid:20) (cid:21)
and
Rˆ (t) = 1 for t ≤ X
sM (1)
m
m i m−i
= Rˆ (t) 1−Rˆ (t) for t > X (2.6)
i M M (1)
Xi=k(cid:18) (cid:19)h i h i
respectively.
Suppose we record the observations (X ,X ,...,X ),p ≤ n that are failed before a pre-
(1) (2) (p)
specified number of cycles c and remainings survive beyond c. Then, MLE of r and θ are X
(1)
and S∗ respectively, where S∗ = p X −X +(n − p){(c +1) −X }. Hence ML
p+S∗ i=1 (i) (1) (1)
P (cid:0) (cid:1)
Estimators of R(t) and R (t) are given by
s
Rˆ∗ (t) = 1 for t ≤ X
M (1)
S∗ t−X(1)
= for t > X (2.7)
p+S∗ (1)
(cid:20) (cid:21)
and
Rˆ∗ (t) = 1 for t ≤ X
sM (1)
m
m i m−i
= Rˆ∗ (t) 1−Rˆ∗ (t) for t > X (2.8)
M M (1)
i
Xi=k(cid:18) (cid:19)h i h i
respectively.
Theorem 2.1 X ,S is sufficient statistic for (r,θ).
(1)
(cid:0) (cid:1)
Proof: Let X = (X ,X ,...,X ), we have to prove that P(X = x|X = u, S = s) does not
1 2 n (1)
depend on r and θ.
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Given X = u and S = s, X is an n-dimensional random variable with domain A =
(1) u,s
{x|x = u, n x = s+(n−1)u}.
(1) i=2 i
P
For x ∈ A ,
u,s
P(X = x)
P(X = x|X = u, S = s)=
(1)
P(X = y)
y∈Au,s
P
and
n
P(X = x)= P(X = x )= (1−θ)nθs+n(u−r).
i i
i=1
Y
Thus P(X = x|X = u, S = s) = 1 , where |A | is the number of elements in A . The
(1) |Au,s| u,s u,s
number of elements in A is the number of possible n-uplets (x ,x ,...,x ) such that x = u
u,s 1 2 n (1)
and n x = s+(n−1)u which clearly does not depend on θ and r.
i=2 i
P
But X ,S is not complete as it is to be seen from the following counter example.
(1)
(cid:0) (cid:1)
Counterexample 2.1 Let us define g(., .) as
g X ,S = 1 if X = r+2, S = 0
(1) (1)
(cid:0) (cid:1)
= −1 if X = r+1, S = n
(1)
= 0 otherwise.
Now X and S can take values r + 2 and 0 with the probability (1 − θ)nθ2n (for X =
(1) (2)
r + 2, ..., X = r + 2), and X and S can take values r + 1 and n with the probability
(n) (1)
(1−θ)nθ2n (one such particular situation is X = r+2, ..., X = r+2, X = r+3).
(2) (n−1) (n)
Therefore, it is found that E g X ,S = 0 but g X ,S 6= 0.
r, θ (1) (1)
(cid:2) (cid:0) (cid:1)(cid:3) (cid:0) (cid:1)
The upcoming theorem will demonstrate the conditional distribution of X for given X ,S .
(1)
(cid:0) (cid:1)
Theorem 2.2 The conditional distribution of X given X ,S is as following:
(1)
(cid:0) (cid:1)
For n =1,
P X = x/X ,S = 1
(1)
(cid:0) (cid:1)
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For n =2,
1
P X = x/X ,S = if x = X
(1) (1)
2
(cid:0) (cid:1) 1
= if x = X +S
(1)
2
For n ≥3, S < n,
S − x−X +n−2 S +n−1
(1)
P X = x/X ,S = / if X ≤ x ≤ X +S
(1) (1) (1)
S − x−X S
(cid:18) (cid:0) (cid:1)(1) (cid:19) (cid:18) (cid:19)
(cid:0) (cid:1)
= 0 (cid:0) other(cid:1)wise.
For n ≥3, S ≥ n,
S +n−2 S +n−1 S −1
P X =x/X ,S = / − if x = X
(1) (1)
S S n−1
(cid:18) (cid:19) (cid:26)(cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
(cid:0) (cid:1)
S − x−X +n−2 S − x−X −1 S +n−1 S −1
= (1) − (1) / −
((cid:18) S(cid:0) − x−X(cid:1)(1) (cid:19) (cid:18) (cid:0) n−2 (cid:1) (cid:19)) (cid:26)(cid:18) S (cid:19) (cid:18)n−1(cid:19)(cid:27)
(cid:0) (cid:1) if X < x ≤ X +S −(n−1)
(1) (1)
S − x−X +n−2 S +n−1 S −1
(1)
= / −
S − x−X S n−1
(cid:18) (cid:0) (cid:1)(1) (cid:19) (cid:26)(cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
(cid:0) (cid:1) if X +S −(n−1) < x ≤ X +S
(1) (1)
= 0 otherwise.
Proof: Joint distribution of X , X , ..., X is given by
1 2 n
P(X1, X2, ..., Xn)= (1−θ)nθPni=1(Xi−r) = (1−θ)nθS+n(X(1)−r), r ≤ X(1).
Now,
P(X = x, X , ..., X /X , S)
P X = x/X ,S = X2, X3, ..., Xn 2 n (1)
(1)
P(X , X , ..., X /X , S)
PX1, X2, X3, ..., Xn 1 2 n (1)
(cid:0) (cid:1)
1
= P (X2, X3, ..., Xn/X(1), S) .
1
P(X1, X2, X3, ..., Xn/X(1), S)
P
Here the denominator is equivalent to finding out the total number of ways in which S indis-
tinguishable balls can be placed in n cells so that at least one cell remains empty. In general, if
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there are r indistinguishable balls to be placed randomly in k cells, then the number of distin-
guishable distributions is k+r−1 whereas the number of distinguishable distributions in which
r
(cid:0) (cid:1)
no cells remains empty is r−1 . Therefore, we get the denominator as S+n−1 − S−1 if S ≥ n
k−1 S n−1
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and if S < n, it will be S+n−1 . Similarly, the numerator is equivalent to finding out the total
S
(cid:0) (cid:1)
number of ways in which S − x−X indistinguishable balls can be placed in n−1 cells so
(1)
that at least one cell remains e(cid:0)mpty and(cid:1)hence, we get it as S−(x−X(1))+n−2 − S−(x−X(1))−1
S−(x−X ) n−2
(1)
(cid:0) S−(x−X (cid:1))+n−(cid:0)2 (cid:1)
if S− x−X ≥ n−1 and if S− x−X < n−1, it will be (1) . Hence the
(1) (1) S−(x−X )
(1)
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
theorem follows.
Since X ,S is sufficientbutnotcomplete statistic for (r,θ), wearehandicappedof searching
(1)
(cid:0) (cid:1)
theUMVUE ofanyestimablefunctionoftheseparametersusingtheLehmann-Scheffe´theorem.
Hence, we will find an improved estimator of the reliability functions using the Rao-Blackwell
theorem.
Define
Y = 1 if X ≥ t
1
= 0 otherwise.
Then R(t) = E(Y) = P (X ≥ t). Using the Rao-Blackwell theorem, an unbiased estimator of
1
R(t) is given as follows:
for n = 1,
R˜ (t) = 1 if t ≤ X
U (1)
= 0 if t >X .
(1)
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for n = 2,
R˜ (t) = 1 if t ≤ X
U (1)
1
= if X < t ≤ X +S
(1) (1)
2
= 0 if t > X +S.
(1)
for n ≥ 3 and S < n,
R˜ (t) = 1 if t ≤ X
U (1)
X +S
(1) S − x−X +n−2 S +n−1
(1)
= / if X < t ≤ X +S
(1) (1)
S − x−X S
x=t (cid:18) (cid:0) (cid:1)(1) (cid:19) (cid:18) (cid:19)
X
= 0 if t >(cid:0)X +S(cid:1).
(1)
It can also be written as
R˜ (t) = 1 if t ≤ X
U (1)
X(1)+S n−1 n−2 X +S +n−x−1−j
(1)
= if X < t ≤ X +S
(1) (1)
X +S +n−1 X +S +n−1−j
x=t (1) j=1 (cid:0) (1) (cid:1)
X Y
= 0 if t > X +S. (cid:0) (cid:1)
(1)
For n ≥ 3, S ≥ n,
R˜ (t) = 1 if t ≤ X
U (1)
X +S−(n−1)
(1) S − x−X +n−2 S − x−X −1 S +n−1 S −1
= (1) − (1) / −
x=t ((cid:18) S(cid:0) − x−X(cid:1)(1) (cid:19) (cid:18) (cid:0) n−2 (cid:1) (cid:19)) (cid:26)(cid:18) S (cid:19) (cid:18)n−1(cid:19)(cid:27)
X
X(1)+S S −(cid:0) x−X (cid:1) +n−2 S +n−1 S −1
(1)
+ / −
S − x−X S n−1
x=X +S−(n−1)+1(cid:18) (cid:0) (cid:1) (1) (cid:19) (cid:26)(cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
(1) X
(cid:0) (cid:1)if X < t ≤ X +S −(n−1)
(1) (1)
X +S
(1) S − x−X +n−2 S +n−1 S −1
(1)
= / −
S − x−X S n−1
x=t (cid:18) (cid:0) (cid:1)(1) (cid:19) (cid:26)(cid:18) (cid:19) (cid:18) (cid:19)(cid:27)
X
(cid:0) (cid:1) if X +S −(n−1) < t ≤ X +S
(1) (1)
= 0 otherwise.
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Inother way theestimator R˜ (t)istobeUMVUE ifitisuncorrelated withallunbiasedestima-
U
torofzero. WetakeaclassofunbiasedestimatorofzeroasU = {u: n c X =u, n c =
0 i=1 i i i=1 i
1}. If R˜ (t) is UMVUE, then Cov(U , R˜ (t)) = 0 i.e. Cov(1000.UP, 1000.R˜ (t)) =P0. Ana-
U 0 U 0 U
lytical derivation seems to be intractable. We go for simulation study taking some particular
choices of (c , c , ..., c ) and different t, and 1000 covariances have been calculated and their
1 2 n
averages have been shown in Tables 7-8. It is noticed that they are not uncorrelated and hence
R˜ (t) is not UMVUE.
U
Thevarianceofthisunbiasedestimatorwillbesmallerthantheunbiasedestimator Pni=1I(Xi≥t),
n
where I(.) is the indicator function.
To study the asymptotic behavior of R˜ (t) we conduct a simulation study taking different
U
values of parameters. 10000 estimates of R˜ (t) and Rˆ (t), their variances, 95% confidence
U M
limits and coverage probability (CP) have been shown in table 9. Histogram of R˜ (t) for
U
n = 20,r = 15,t = 25,θ = 0.96 has been shown in Figure 1. In this set up the true reliability,
R(t) =0.6648326. The figure is near normal. From the table 9, it is also evident from coverage
probability point of view, Rˆ (t) is better if 0.02 < R(t)< 0.5, otherwise R˜ (t) is better. From
M U
R(t)(1−R(t))
the table it is observed that asymptotic variance is approximately .
2n
Define
Z = 1 if at least k of X ’s among X , X , ..., X are greater than or equal to t
i 1 2 m
= 0 otherwise.
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