Open Journal of Statistics, 2012, 2, 443-446
http://dx.doi.org/10.4236/ojs.2012.24055 Published Online October 2012 (http://www.SciRP.org/journal/ojs)
A Note on Detecting “More IFR-ness” Property
of Life Distributions
Parameshwar V. Pandit1, Sujatha Inginashetty2
1Department of Statistics, Bangalore University, Bangalore, India
2Department of Statistics, Gulbarga University, Gulbarga, India
Email: panditpv12@gmail.com, sujathasi_gug@rediffmail.com
Received August 25, 2012; revised September 28, 2012; accepted October 11, 2012
ABSTRACT
In this paper, a problem of testing whether one life distribution possesses “more IFR” property than the other is consid-
ered. A new test procedure is proposed and the distribution of the test statistic is studied. The performance of the pro-
cedure is evaluated in terms of Pitman asymptotic relative efficiency. The consistency property of the test procedure is
established. It is observed that the new procedure is better than the existing proced ure in the literature.
Keywords: “More IFR” Property; U-Statistic; Pitman ARE
1. Introduction
A life is represented by a non-negative random variable
X with distribution function F and survival function
F
1F . Classes of life distributions based on notion
of ageing have been introduced in the literature. One of
the earliest and most important classes is the class of
“Increasing Failure Rate” (IFR). We define IFR class
below.
Definiti on 1.1. A distribution F is said to be increasing
failure rate (IFR), if


F
xt
x
,,,
F
is decreasing in x, for t
0.
Proschan and Pyke [1] proposed a test for testing ex-
ponentiality against IFR alternatives followed by Barlow
and Proschan [2], Bickel and Doksum [3], Bickel [4] and
many among others.
In practice, one might be interested in comparing two
life distributions with respect to their possessing positive
ageing property, particularly, IFR. Hollandar, Park and
Proschan [5] developed a test procedure for testing the
null hypothesis that two life distributions F and G are
equal versus the alternative hypothesis that F is more
NBU than G. Tiwari and Za lkikar [6] propo sed a test for
testing the null hypothesis that two life distributions F
and G are identical versus the alternative hypothesis that
F is “More increasing failure rate average” than G. Re-
cently, Lim, Kim and Park [7] developed a class of test
procedures for testing the null hypothesis that two life
distributions F and G are equal against the alternative
that F is “more NBU at specified age” than G. However,
the only test available for testing the null hypothesis that
two life distributions F and G are identical against the
alternative that F is more IFR than G is due to Pandit and
Gudaganavar [8].
In this paper, we develop a simple test procedure for
testing the null hypothesis that two life distributions F
and G are equal against the alternative that F is more IFR
than G. The paper is organized as below: a test statistic is
proposed for the problem of testing whether F is more
IFR than G and its asymptotic distribution is established
in Section 2. Section 3 contains the asymptotic relative
efficiencies of the test proposed with the test due to Pan-
dit and Gudaganavar [8] and some remarks and conclu-
sions are presented in Section 4.
2. The Proposed Two Sample “More IFR”
Test
Let 12 m
X
XX,,,YY Y and 12 n denote two ran-
dom samples from continuous life distributions F and G
respectively. We want to develop test statistic for testing
the null hypothesis.
H0:F = G (the common distribution is not specified);
Versus H1:F is “more IFR” than G based on the two
independent random samples.
Consider the parameter

,
F
GFG



,
where
 
2dd
2
xt
F
FFxFt






C
opyright © 2012 SciRes. OJS
P. V. PANDIT, S. INGINASHETTY
444
and
  
ddGxGt
2
2
xt
GG






Here
F
and can be considered as the
measure of degree of the IFR-ness. Mugadi and Ahmed
[9] test used this measure as basis in the construction of
test statistic for testing exponentiality against the IFR
alternatives in one sample setting. If F and G belong to
IFR class then

G

,
G
0

,
can be taken as a measure for
deciding whether F is “more IFR” than G or not. Under
H0, and it is strictly greater than zero un-
der H1.
,FG
An unbiased estimator for
G
,12
which is given
by
mn
UUU

,
where
 

1
112 i
ijk2
jk
x
x
x





Um
m mI




and
 

1
212 i
ijk
Un
n nIy



 2
jk
yy




1if
0otherwise
ab
,
where

Ia b

The asymptotic normality of the test Um,n is presented
in the following theorem.
Theorem: As n,


,NU FG

,mn

is as-
ymptotically normal with mean zero and variance
22
12
1
2


where
 
dxFx
1
2
2
11
0
1
0
9Var 2
2d
2
X
FX
Xx
F
Fx F





and

 
dyGy
yG
1
2
2
21
0
1
0
9 Var2
2d
2
Y
GY
Yy
GG





Under H0: 222
12



and is given by
2
21
1
Proof: Proof follows from Hoeffding [10] U-statistics
theory that the limiting distribution of
.
,NU FG
,mn
is asymptotic normal with mean 0
and variance 2
, where N = m + n is the combined
sample size, 22
212
1

and lim
N
m
N

and it is
assumed 0 <
< 1.
Here it is to be noted that the asymptotic mean of
,mn is zero, independent of unspecified common
distribution F0. However, the null asymptotic variance
NU
291F
10

does depend on F0 through
10
F
2
and must be estimated from the data. To esti-
mate
, one possible way is to obtain consistent esti-
mator for 2
. For that,
Since

111
Var
F
x



,
where
111 2 311
,,
x
EXXXXx


and


123
123 213
312
,,
122
3
2
XXX
IXXXIXX X
IXXX



,
we get

2
111
1
1ˆ
1
m
i
i
FxU
m





,
where

11
1
ˆ,,
1
m
iijk
i
jki
x
xx x
m


 
In similar fashion, we define

112
1
1ˆ
1
n
j
j
GyU
n





,
where

11
1
ˆ,,
1
m
iijk
i
jki
yyyy
n


Then it is easily verified that

11
1
1ˆ
m
i
i
Ux
m
and

21
1
1ˆ
n
j
n
j
Uy.
1
1G
are consistent estimators of
F
and
1
G
1, and this consistent estimator 2
ˆ
N
F
and is
obtained by replacing
1
F

G
and 1 by
1
F
and
G
2
1, respectively in the expression of
.
Hence the estimator for 2
N
is obtained as
Copyright © 2012 SciRes. OJS
P. V. PANDIT, S. INGINASHETTY 445




11
99NGn

2
ˆ
2
ˆNNFm

.
By the consistency of
N
and Slutsky’s theorem
1
ˆ
N
NV
is asymptotically N(0, 1) under H0.
The approximate
-level test of H0 versus H1 rejects
H0 in favour of H1 if





12
1
n z
,1
99
mn
NUNF mNG


z
where
is the upper
-percentile point of the normal
distribution. This ensures the consistency of the two
sample IFR test against the class of (F, G) pairs satisfy-
ing .
 
0FG



,
3. Asymptotic Relative Efficiency
We study the asymptotic efficacy of Um,n, for two pairs of
distribution ,i
F
G

F
. Here, we assume that G is an
exponential distribution with mean one. The different
distributions considered here for ,i
are given below:
1) Linear Failure Rate Distribution

2
2. exp 2
x
Fx x









, 0, 0
x


2) Makeham Distribution


3. expe 1
x
Fxx x



, 0, 0x


The Pitman asymptotic efficacy is given by



0
2
.
,
i
F G
2
,. 0d
,, d
mn i
Eff UFGv




.
The mean of the proposed test is

 
 
d
dd
.00
00
,d
2
2
i
xy
F
GF
xy










FxFy
GGxGy

since G is exponential, 4
9
G
.
The asymptotic efficacies of the proposed test for
Linear Failure Rate distribution and Makeham
distributio n are

0.8681
1
and

1.4062
1
respectively.
Next, we compute ARE of the two sample test based
on Um,n proposed here, by considering a sequence of
alternatives
,
F
G

where
F
is LFR or Makeham
distribution with parameter
, wh ere 1a
N
 , “a
being arbitrary positive constant, and G
is LFR or
Makeham with parameter
. It is to be noted that as
, the sequence of alternatives converges to the
null hypothesis. That is G
N
F

1 when
. The
efficacy of the V test is given by



0
1
,
,
,
mn
Hmn
FG
eff UU
,
,0
H
where mn
U
is null asymptotic standard deviation
of Um,n, and

1
,d,
1d
FG FG

.
m,n
U
The AREs of Um,n relative to V, the test due to Pandit
and Gudaganavar [8] for different values of
are pre-
sented in Table 1.
4. Some Remarks
1) A simple test procedure for testing the null hy-
pothesis that two life time distributions are identical
against the alternative th at one possesses more IFR prop-
erty than another. The test proposed is based on the
measure considered by Mugadi and Ahmad [9] for one
sample problem.
2) The only test available for the above stated problem
in the literature is due to Pandit and Gudaganavar [8].
Hence, only Pitman AREs relative to the test due to Pan-
dit and Gudaganavar [8] are computed for the few alter-
natives by specifying the common null distribution to be
exponential. However, similar testing problems with
more NBU/IFRA/NBU-t0 property are considered by few
authors, namely, Hollander, Park and Proschan [5], Ti-
wari and Zalkikar [6] and Lim, Kim and Park [7].
3) We have also computed Pitman AREs relative to
Pandit and Gudaganavar [8] for those alternatives with
common null distributions being Makeham and Linear
failure rate distributions.
4) It is observed that the new test proposed here is
better than Pandit and Gudaganavar [8] test for the dis-
tributions considered in this paper. However, the per-
formance of Pandit and Gudaganavar [8] test is better for
Weibull distribution.
5. Acknowledgements
Authors thank the Editorial Board and reviewers for their
Table 1. AREs of relative to V.
θ LFR Makeham
2 1.7814 1.5386
3 1.9358 1.4612
4 1.9483 1.4475
5 1.9822 1.4493
Copyright © 2012 SciRes. OJS
P. V. PANDIT, S. INGINASHETTY
Copyright © 2012 SciRes. OJS
446
valuable comments which lead to this revised version of
the paper.
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