On testing More IFRA Ordering-II

0.16 MB·English

by  Muhyiddin Izadi

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On testing More IFRA Ordering-II
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S Muhyiddin Izadi1 Baha-Eldin Khaledi
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a Department of Statistics, Razi University, Kermanshah, Iran
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Chin-Diew Lai
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9 Institute of Fundamental Sciences - Statistics and Bioinformatics,
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0 Massey University, Palmerston North, New Zealand
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1Corresponding author
Abstract
Suppose F and G are two life distribution functions. It is said that F is more IFRA than
G (written by F G) if G 1F(x) is starshaped on (0, ). In this paper, the problem of
−
≤∗ ∞
testing H : F = G against H : F G and F = G is considered in both cases when
0 1
∗ ≤∗ 6 ∗
G is known and when G is unknown. We propose a new test based on U-statistics and
obtain the asymptotic distribution of the test statistics. The new test is compared with
some well known tests in the literature. In addition, we apply our test to a real data set
in the context of reliability.
Keywords: Asymptoticnormality, starorder,increasingfailurerateaverage, Pitman’s
asymptotic efficiency, U-statistic.
1 Introduction
Let X be a lifetime of an appliance with density function f, distribution function F and
survival function F¯. Let also denote F 1 as the right continuous inverse function of F. X
−
x
is said to be IFRA (increasing failure rate average) if r˜ (x) = R0 rF(t)dt is nondecreasing in
F x
x 0 which is equivalent to that logF¯(x) is nondecreasing in x 0 where r (x) = f(x). It
≥ − x ≥ F F¯(x)
is of considerable interest to producers and users of the appliances to evaluate the severity
of average failure risk at a particular point of time and to see if r˜ (x) is either increasing
F
or decreasing in time. That is, it is of practical importance to characterize the aging class
of underlying random lifetimes. In particular, since the IFRA class of aging is one of
the most important aging classes, testing that the distribution F has a constant hazard
rate against the hypothesis that F is IFRA has been studied extensively in the literature;
see for example, Deshpande (1983), Kochar (1985), Link (1989), Ahmad (2000) and El-
Bassiouny (2003) among others. In fact, F is IFRA if and only if Eλ−1F(x) is nondecreasing
x
in x 0 or equivalently r˜F(F−1(u)) is nondecreasing in u (0,1) where E is an exponential
≥ r˜E(Eλ−1(u)) ∈ λ
distribution with mean λ. This implies that F ages faster than E, i.e., F is more IFRA
than E .
λ
In order to evaluate the performance of an appliance, we need to compare its aging
behavior with some distributions other than exponential distribution such as the Weibull,
gamma, linear failure rate or even an unknown distribution G. The notion of the star
order that establishes an equivalent class of distributions is one of the useful tools for this
comparison. Let Y be another non-negative random variable with distribution function
G. We say that X is less than Y with respect to the star order (written by X Y or
≤∗
F G ) if G 1F(x) is starshaped on [0, ); that is, G−1F(x) is nondecreasing in x 0.
≤∗ − ∞ x ≥
It is known that
r˜ (F 1(u))
F −
F G is nondecreasing in u (0,1), (1.1)
≤∗ ⇔ r˜G(G−1(u)) ∈
where r˜ and r˜ are failure rate average functions of F and G, respectively. Using (1.1),
F G
the relation X Y is interpreted as X ages faster than Y and it is said that X is more
≤∗
IFRA than Y (cf. Kochar and Xu, 2011 ). It is obvious that if F G and G F then
≤∗ ≤∗
F(x) = G(ax) for all x 0 and some a > 0. In this case, we say F = G.
≥ ∗
1
Izadi and Khaledi (2012) have considered the problem of testing the null hypothesis
H : F = G against H : F G and F = G. They proposed a test based on kernel
0 1
∗ ≤∗ 6 ∗
density estimation. In this paper, we further study this problem of testing in the one-
sample as well as the two-sample problem and propose a new simple test based on a
U-statistic. In both cases, we compare the new proposed test with some well known tests
in the literature. It is found that our test is comparable to the others.
To establish our new test we need the following lemma.
Lemma 1.1 Let X ,X (Y ,Y ) be two independent copies of the random variable X (Y)
1 2 1 2
(2) (2)
with distribution function F (G) and let µ = E[max X ,X ] (µ = E[max Y ,Y ])
F { 1 2} G { 1 2}
where E[.] is the expectation operator. If F is more IFRA than G, then
(2) (2)
µ µ
F G
µ ≤ µ
F G
where µ (µ ) is the expectation of F (G).
F G
Proof: We know that more IFRA order is scale invariant. Thus, X Y implies µY X
≤∗ µX ≤∗
Y. Now, the required result follows from Theorem 7.6 of Barlow and Proschan (1981, page
122).
Remark 1.1 The above lemma has been proved by Xie and Lai (1996) under the condition
that F is more IFR than G (for definition, see Shaked and Shantikumar, 2007, p. 214)
which is stronger than more IFRA order.
µ(2) µ(2)
Now, let δ = F , δ = G and
F µF G µG
δ(F,G) = δ δ . (1.2)
F G
−
It is obvious that if F = G, then δ(F,G) = 0 and if F G and F = G, then it follows
∗ ≤∗ 6 ∗
from Lemma 1.1 that δ(F,G) < 0. That is, δ(F,G) can be considered as a measure of
departure from H : F = G in favor of H : F G and F = G. So, our test statistic is
0 1
∗ ≤∗ 6 ∗
based on the estimation of δ(F,G).
The organization of this paper is as follows. In Section 2, we propose the new test for
the case when G is known. The case when G is unknown is studied in Section 3. In Section
4, the performance of our test is evaluated and compared.
2
2 The One-Sample Problem
Let G be a known distribution function and X ,...,X be a random sample from an
0 1 n
unknown distribution F. Now by using the measure (1.2), the test statistic
ˆ ˆ
δ(F,G ) = δ δ
0 F − G0
is used for testing
H : F = G
0 0
∗
against
H : F G and F = G
1 0 0
≤∗ 6 ∗
where
max X ,X
i j
{ }
δˆ = Xi6=Xj (2.3)
F n(n 1)X¯
−
and X¯ is the mean of the random sample. In the next theorem, we obtain the asymptotic
ˆ
distribution of δ(F,G ) by using the standard theory of U-statistics.
0
Theorem 2.1 Suppose E[max X ,X δF(X +X )]2 < . As n , n1/2[δˆ(F,G )
{ 1 2}− 2 1 2 ∞ → ∞ 0 −
δ(F,G )] is asymptotically normal with mean 0 and variance
0
4 δ
σ2 = Var XF(X)+ ∞tdF(t) FX . (2.4)
F µ2 × − 2
F (cid:18) ZX (cid:19)
Under H , n1/2δˆ(F,G ) is asymptotically normal with mean 0 and variance σ2 = σ2 .
0 0 0 G0
Proof: First note that
δ
F
max X ,X (X +X )
i j i j
{ }− 2
δˆ δ = Xi6=Xj (cid:20) (cid:21)
F − F n(n 1)X¯
−
φ(X ,X )
i j
i=j
= X6 X ,
n(n 1)X¯
−
3
where
δ
F
φ(X ,X ) = max X ,X (X +X ).
i j i j i j
{ }− 2
Let define
φ(X ,X )
i j
i=j
T∗ = X6 X .
n(n 1)
−
By the standard theory of U-statistics, if E[φ2(X ,X )] < , as n
1 2
∞ −→ ∞
√nT
n∗ d N(0,1)
σ →
∗
where
σ2 = 4 Var(φ (X))
1
×
∗
and
φ (x) = E[φ(x,X)].
1
Now by the strong law of large numbers we have X¯ a.s. µ and hence, by Slutsky theorem
F
→
√n[δˆ(F,G ) δ(F,G )] is asymptotically normal with mean 0 and variance σ2 = σ∗2.
0 − 0 F µ2F
Under H , δ(F,G ) = 0 and σ2 = σ2 . (cid:3)
0 0 0 G0
ˆ
A small value of δ(F,G ) indicates that testing H against H is significant. Thus,
0 0 1
we reject H at level α if n1/2δˆ(F,G )/σ < z , where z is αth quantile of the standard
0 0 G0 α α
normal distribution.
In the case G (x) = E (x) = 1 exp( λx), x 0 and λ > 0, the problem is testing
0 λ
− − ≥
the null hypothesis H : F is an exponential distribution against the alternative hypothesis
0
H : F is IFRA and not exponential. It can be shown that δ = 3 and σ2 = 1 . By
1 Eλ 2 Eλ 12
the above theorem, under H , √n(δˆ 3/2) is asymptotically normal with mean 0 and
0 F
−
variance 1 . Thus we reject H in favor of H if √12n(δˆ 3/2) < z .
12 0 1 F − α
ˆ
In the following we find the exact distribution of δ under the hypothesis F is an
F
ˆ
exponential distribution. First, note that we can rewrite δ as
F
n n
2 (i 1)X c D
(i) i:n i
−
δˆ = Xi=1 = Xi=1 (2.5)
F n(n 1)X¯ n
− D
i
i=1
X
4
where X is the ith order statistic of X ’s,
(i) i
n
2 (j 1)
−
j=i
D = (n i+1)(X X ), c = X
i (i) (i 1) i:n
− − − (n 1)(n i+1)
− −
and assuming X = 0. Now, by the same arguments as in Langenberg and Srinivasan
(0)
(1979), we will get the following result.
Theorem 2.2 Let F be an exponential distribution, then
n n
c x
ˆ i:n
P δ x = 1 − I(x < c ) (2.6)
F i:n
{ ≤ } − c c
i:n j:n
i=1 j=1 −
XY
j=i
6
where I(.) is the usual indicator function.
By using Theorem 2.2, we tabulate the critical point of √12n(δˆ 3/2) under expo-
F
−
nentiality for small sample sizes ( 40) in Table 1. So, for small sample sizes, we reject
≤
exponentiality in favor of IFRA-ness if √12n(δˆ 3/2) is smaller than the critical point
F
−
in Table 1 corresponding with the level of significance chosen.
El-Bassiouny(2003)hasconsideredtheproblemoftestingexponentialityagainstIFRA-
ness in the alternative and proposed a class of test. His test is based on the test statistics
Xr+1
2 min Xr+1,Xr+1 i
{ i j }− 2
∆ˆ = Xi<Xj (cid:18) (cid:19)
r+1 n(n 1)X¯r+1
−
and large values of ∆ˆ are significant for the considered problem of testing. If r = 0,
r+1
X
i
2 min X ,X
i j
{ }− 2
∆ˆ = Xi<Xj (cid:18) (cid:19)
1 ¯
n(n 1)X
−
2 min X ,X
i j
{ }
1
i<j
= XX (2.7)
n(n 1)X¯ − 2
−
5
Table 1: Critical values of √12n(δˆ 3/2) for small sample sizes
F
−
n α: Lower Tail α: Upper Tail
0.01 0.05 0.1 0.1 0.05 0.01
2 -2.400500 -2.204541 -1.959592 1.959592 2.204541 2.400500
3 -2.575752 -2.051328 -1.658360 1.658360 2.051328 2.575752
4 -2.560006 -1.918143 -1.516280 1.516280 1.918143 2.560006
5 -2.517587 -1.846175 -1.458997 1.458997 1.846175 2.517587
6 -2.482569 -1.807959 -1.424446 1.424446 1.807959 2.482569
7 -2.458901 -1.781575 -1.400767 1.400767 1.781575 2.458901
8 -2.441786 -1.762473 -1.383974 1.383974 1.762473 2.441786
9 -2.428500 -1.748106 -1.371312 1.371312 1.748106 2.428500
10 -2.417939 -1.736865 -1.361442 1.361442 1.736865 2.417939
11 -2.409356 -1.727862 -1.353531 1.353531 1.727862 2.409356
12 -2.402239 -1.720450 -1.347047 1.347047 1.720450 2.402239
13 -2.396243 -1.714193 -1.341635 1.341635 1.714193 2.396243
14 -2.391124 -1.708937 -1.337050 1.337050 1.708937 2.391124
15 -2.386703 -1.704422 -1.333116 1.333116 1.704422 2.386703
16 -2.382846 -1.700502 -1.329703 1.329703 1.700502 2.382846
17 -2.379451 -1.697066 -1.326714 1.326714 1.697066 2.379451
18 -2.376441 -1.694029 -1.324074 1.324074 1.694029 2.376441
19 -2.373754 -1.691327 -1.321727 1.321727 1.691327 2.373754
20 -2.371340 -1.688906 -1.319625 1.319625 1.688906 2.371340
21 -2.369160 -1.686725 -1.317732 1.317732 1.686725 2.369160
22 -2.367182 -1.684749 -1.316018 1.316018 1.684749 2.367182
23 -2.365378 -1.682952 -1.314460 1.314460 1.682952 2.365378
24 -2.363726 -1.681309 -1.313036 1.313036 1.681309 2.363726
25 -2.362209 -1.679803 -1.311730 1.311730 1.679803 2.362209
26 -2.360810 -1.678415 -1.310529 1.310529 1.678415 2.360810
27 -2.359516 -1.677134 -1.309419 1.309419 1.677134 2.359516
28 -2.358316 -1.675947 -1.308392 1.308392 1.675947 2.358316
29 -2.357199 -1.674844 -1.307437 1.307437 1.674844 2.357199
30 -2.356158 -1.673817 -1.306548 1.306548 1.673817 2.356158
31 -2.355185 -1.672857 -1.305718 1.305718 1.672857 2.355185
32 -2.354273 -1.671960 -1.304942 1.304942 1.671960 2.354273
33 -2.353418 -1.671118 -1.304214 1.304214 1.671118 2.353417
34 -2.352613 -1.670326 -1.303529 1.303529 1.670326 2.352612
35 -2.351855 -1.669581 -1.302885 1.302885 1.669581 2.351854
36 -2.351140 -1.668878 -1.302278 1.302278 1.668878 2.351138
37 -2.350454 -1.668214 -1.301704 1.301704 1.668214 2.350461
38 -2.349821 -1.667586 -1.301161 1.301161 1.667586 2.349821
39 -2.349184 -1.666990 -1.300647 1.300647 1.666991 2.349213
40 -2.348688 -1.666426 -1.300159 1.300159 1.666426 2.348636
6
On the other hand, using the fact that δ = 3,
Eλ 2
max X ,X
i j
{ }
δˆ(F,E ) = Xi6=Xj δ
λ n(n 1)X¯ − Eλ
−
2 [(X +X ) min X ,X ]
i j i j
− { }
3
i<j
= XX
n(n 1)X¯ − 2
−
2 (X +X ) min X ,X
i j i j
− { }
" # 3
i<j i<j
= XX XX
n(n 1)X¯ − 2
−
n
2 (n 1) X min X ,X
i i j
− − { }
" # 3
i=1 i<j
= X XX
n(n 1)X¯ − 2
−
2 min X ,X
i j
{ }
1
i<j
= XX
2 − n(n 1)X¯
−
= ∆ˆ .
1
−
That is, for the case when r = 0 and G is an exponential distribution, the proposed test
0
is equivalent to that of El-bassiouny (2003).
It is worth to mention that our test is consistent; that is, if β (F) is the power of our
n
test, thenunder thealternative hypothesis, lim β (F) = 1 which followsfromTheorem
n n
→∞
2.1 and Problem 2.3.16 in Lehmann (1999),
3 The Two-Sample Problem
Inthissection, weconsiderthetwo-sampleproblemwhenGisunknown. LetX ,X ,...,X
1 2 n
and Y ,Y ,...,Y be two independent random samples from unknown distribution func-
1 2 m
ˆ ˆ
tions F and G, respectively, and N = n + m. Assume that δ is as in (2.3) and δ is
F G
defined similarly in terms of Y ,...,Y and Y¯. The test statistic
1 m
ˆ ˆ ˆ
δ(F,G) = δ δ
F G
−
which is the estimate of the measure in (1.2) is used for testing the null hypothesis
H : F = G (3.8)
0
∗
7
against the alternative hypothesis
H : F G and F = G. (3.9)
1
≤∗ 6 ∗
ˆ
Small values of δ(F,G) are significant for testing H against H . In the following theorem
0 1
ˆ
we obtain the asymptotic distribution of δ(F,G).
Theorem 3.1 If E[max X ,X δF(X +X )]2 and E[max Y ,Y δG(Y +Y )]2 are
{ 1 2}− 2 1 2 { 1 2}− 2 1 2
finite and n and m such that n c, c (0, 1], then √N(δˆ(F,G) δ(F,G)) is
→ ∞ N → ∈ 2 −
asymptotically normal with mean 0 and variance
N N
σ2 = σ2 + σ2
F,G n F m G
where σ2 is given in (2.4) and σ2 is defined similarly in terms of Y.
F G
Proof: It is easy to see that
√N(δˆ(F,G) δ(F,G)) √mσ
− = F [√n(δˆ δ )/σ ]
F F F
σ mσ2 +nσ2 −
F,G
F G
√nσ
p G [√m(δˆ δ )/σ ].
G G G
− mσ2 +nσ2 −
F G
From the result of Theorem 2.1, as bothpn and m , we have that
→ ∞
√n(δˆ δ )/σ d N(0,1) and √m(δˆ δ )/σ d N(0,1).
F F F G G G
− → − →
ˆ ˆ
Since δ and δ are independent, the required result follows from the fact that convergence
F G
indistributionisclosedundertheconvolutionofindependentsequencesofrandomvariables
(cf. Theorem 6.6 of Gut (2009), page 169).
In practice σ2 is unknown, but it can be estimated by the consistent estimator
F,G
N N
σˆ2 = σˆ2 + σˆ2 (3.10)
F,G n F m G
where
n
φˆ2(X )
i
σˆ2 = 4 Xi=1 ,
F × nX¯2
n ˆ
1 δ
ˆ F
φ(X ) = [max X ,X (X +X )]
i i j i j
n 1 { }− 2
− j=1
X
j=i
6
8