Testing mean residual alternatives by dispersion of residual lives

Testing mean residual alternatives by dispersion of residual lives

Journal of Statistical Planning and Inference 86 (2000) 113–127 www.elsevier.com/locate/jspi Testing mean residual alternatives by dispersion of res...

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Journal of Statistical Planning and Inference 86 (2000) 113–127

www.elsevier.com/locate/jspi

Testing mean residual alternatives by dispersion of residual lives ( Felix Belzunce ∗ , Jose Candel, Jose M. Ruiz Departamento de Estadstica  e I.O., Universidad de Murcia, 30100 Espinardo (Murcia), Spain Received 7 April 1999; received in revised form 1 September 1999; accepted 16 September 1999

Abstract Given a nonnegative random variable X , which represents the lifetime of an item, the mean residual live, (t) ≡ E[X − t | X ¿ t], is the expected lifetime for an item which has survived up to time t: The comparison at di erent times of (t) de nes two important classes, known as decreasing (increasing) mean residual life (DMRL (IMRL)) and new better (worse) than used in expectation (NBUE (NWUE)). The problem of testing H0 : X is exponential against H1 : X is DMRL (IMRL) or NBUE (NWUE), has been considered by several authors. In this paper we provide new tests for these ageing classes, based on dispersion of residual lives. We give the exact and asymptotic distributions of these tests. Also we show the performance of these tests c 2000 Elsevier Science B.V. All by comparing them with existing tests for ageing classes. rights reserved. MSC: 62G10, 62E20; secondary 62N05 Keywords: Asymptotic normality; Decreasing mean residual life; Dilation order; Mean residual life; New better than used; Normalized spacings; Order statistics; Pitman’s asymptotic eciency; Total time on test transform

1. Introduction Let X denotes the random lifetime of an item with distribution function FX (t) (t¿0), the mean residual life is de ned as  R∞  t F X (x) d x if F X (t) ¿ 0; (t) ≡ E[X − t | X ¿ t] = F X (t)  0 otherwise; where F X (t) ≡ 1 − FX (t). For items of age t, (t) is the expected remaining lifetime (for properties and applications please see Bryson and Siddiqui, 1969; Guess and Proschan, 1988; Hall and ( ∗

This research was supported by D.G.E.S. (Ministerio de Educacion y Cultura) PB 96-1105. Corresponding author.

c 2000 Elsevier Science B.V. All rights reserved. 0378-3758/00/$ - see front matter PII: S 0 3 7 8 - 3 7 5 8 ( 9 9 ) 0 0 1 6 7 - 6

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Wellner, 1981; Yang, 1978). In reliability and survival analysis two important classes are de ned by comparing the mean residual life at di erent ages. A random lifetime X is said to be decreasing (increasing) mean residual life (X is DMRL (IMRL)) if (t) is decreasing (increasing) and X is said to be new better (worst) than used in expectation (X is NBUE (NWUE)) if (t)6(¿)(0) = E(X ) for all t ¿ 0. However, as we see below, these classes not only compare the expected remaining life but the dispersion of remaining life too. The classi cation of a lifetime distribution is useful, for example, to derive bounds for the survival distribution (see Cai, 1995; Cheng and He, 1989; Korzeniowski and Opawski, 1976). Also it is a tool to reject some parametric models, for the data. For example, if we know that the underlying distribution is Weibull, with shape parameter , and NBUE, then  should be greater than 1. In the literature several orderings of distributions have been de ned to describe when one random variable is more dispersed than another. If we denote by FX−1 (p) ≡ inf {x: FX (x)¿p}; then the random variable Y is said to be more dispersed than the random variable X (X ¡disp Y ) if (see Lewis and Thompson, 1981; and Shaked and Shanthikumar, 1994) FY−1 (q) − FY−1 (p)¿FX−1 (q) − FX−1 (p)

for all 0 ¡ p ¡ q ¡ 1;

this ordering formalizes the notion that Y is more dispersed than X . Hickey (1986) gives a more general concept of dispersive ordering. A random variable Y is more dispersed in dilation than a random variable X (X ¡dil Y ) if E[’(X − E[X ])]6E[’(Y − E[Y ])]; for every convex function ’. Taking ’(x) = x2 , we get X ¡dil Y ⇒ Var(X )6Var(Y ). The dilation order is weaker than the dispersive order, i.e., holds the implication ¡disp ⇒ ¡dil : For a study of the previous orderings, relations with other orderings and applications, see Shaked and Shanthikumar (1994) and the references therein. It is clear from the de nitions of dispersive and dilation orderings that these orders are location-free. Two interesting properties of the dilation order are related to the DMRL (IMRL) and NBUE (NWUE) ageing classes, it can be seen (Belzunce et al., 1996) that X is DMRL (IMRL) if ; and only if ; Xs ¿dil (¡dil ) Xt

for all s ¡ t

(1)

for all t;

(2)

and (Belzunce et al., 1997) X is NBUE (NWUE) if ; and only if ; Xt ¿dil (¡dil )Y

where Xt ≡ {X − t | X ¿ t} is the remaining life at age t and Y is an exponential random variable with mean E(X ). Let us recall that the exponential random variable has the non-ageing property. These properties, as we have mentioned previously, show that DMRL (IMRL) and NBUE (NWUE) ageing classes not only compares the mean remaining life but the dispersion of remaining lives, too. The purpose of this paper is to develop tests for DMRL (IMRL) and NBUE (NWUE) alternatives from the characterizations (1) and (2).

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For this purpose, in Section 2, we address the problem of testing the dilation order. The test is motivated by a characterization of the dilation order that we provide in Section 2. Later in Section 3 we develop the tests for DMRL (IMRL) and NBUE (NWUE) alternatives. We give the exact and asymptotic distributions, and prove the consistency of the test. We also compare these tests with other existing tests for ageing classes in terms of Pitman’s asymptotic eciency. Some conclusions and remarks are given in Section 4.

2. Testing the dilation order The purpose of this section is to provide a test statistic for testing, given independent samples of X and Y , the null hypothesis H0 : X =dil Y against the alternative H1 : X ¡dil Y

and

X 6=dil Y:

The same problem for the dispersive ordering (¡disp ) has been considered by several authors (Aly, 1990; Marzec and Marzec, 1991). Our test is motivated by the following characterization of dilation order. Theorem 2.1. Let X and Y be two random variables with nite left endpoints of the support lX and lY ; such FX (lX ) = 0 = FY (lY ); then X ¡dil Y if; and only if; Z p Z p FX−1 (t) dt¿ FY−1 for allp ∈ [0; 1]: −E(X ) −E(Y ) (t) dt 0

0

Proof. First, we note, that Young’s inequality (Taillie, 1981) holds for all distribution function FX , with nite left-end point lX of the support, i.e., For all x¿lX and p ∈ (0; 1) Z x Z p xp6 FX (x) d x + FX−1 (u) du: lX

0

Now, Theorem 2 of Taillie (1981) can be extended in the following way: Let X and Y be random variables, with nite left-end points of the supports, lX and lY respectively, X ; lY }, the following conditions are equivalents: R x then for a =R min{l x (i) −∞ FX (t) dt6 −∞ FY (t) dt for all x¿a: Rp Rp (ii) 0 FX−1 (u) du¿ 0 FY−1 (u) du for all p ∈ (0; 1): Also, if E(X ) = E(Y ) then, Z +∞ (F X (t) − F Y (t)) dt = 0: −∞

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Let us see the result. It can be seen that (Shaked and Shanthikumar, 1994) X ¡dil Y if, and only if, Z ∞ Z ∞  F X −E(X ) (t) dt6 F Y −E(Y ) (t) dt for all x ∈ R; x

x

where FX −E(X ) (x) = FX (x + E(X )) and FY −E(Y ) (x) = FY (x + E(Y )) for all x ∈ R: Let x¿lX − E(X ); lY − E(Y ), then Z x Z x FY −E(Y ) (t) dt − FX −E(X ) (t) dt −∞

=

−∞

Z

+∞

−∞

(F X −E(X ) (t) − F Y −E(Y ) (t)) dt

 Z +∞   F X −E(X ) (t) dt − F Y −E(Y ) (t) dt − x x Z ∞ Z ∞ = F Y −E(Y ) (t) dt − F X −E(X ) (t) dt¿0: Z

+∞

x

x

Therefore, by the equivalence between (i) and (ii), Z p Z p FX−1 (t) dt¿ FY−1 for all p ∈ [0; 1]: −E(X ) −E(Y ) (t) dt 0

0

Then, if we consider the measure Z 1 Z p Z −1 dil (X; Y ) ≡ 2 FX −E(X ) (t) dt − 0

0

0

p

FY−1 −E(Y ) (t) dt

 dp;

(3)

by Theorem 2.1 under H0 ; dil (X; Y ) = 0, but under H1 , dil (X; Y ) ¿ 0: Thus this measure can be taken as measure of deviation for testing H0 against H1 . A computation shows that Z ∞  Z 1Z p 1  2X (t) dt − E(X ) : F FX−1 (t) dt dp = −E(X ) 2 0 0 lX Therefore

Z

dil (X; Y ) =



lX

2 F X (t) dt − E(X ) −

Z



lY

2 F Y (t) dt + E(Y ):

The sample analog of this measure forms the basis of our test. We observe that this measure is equal to the measure of deviation proposed by Aly (1990) for testing the null hypothesis H0 : X =disp Y against the alternative H1 : X ¡disp Y

and

X 6=disp Y;

thus the test proposed by Aly (1990) is consistent against the weaker dilation order. The asymptotic properties of the test can be found in Aly (1990).

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3. Testing DMRL (IMRL) and NBUE (NWUE) alternatives Motivated by the measure dil and the characterizations (1) and (2), in this section, we propose tests for DMRL (IMRL) and NBUE (NWUE) alternatives. The problem we consider is to test the null hypothesis H0 : X is exponential (FX (t) = 1 − exp(−t);  unspeci ed) against the alternative H1 : X ∈ A and is not exponential; where A is an ageing class (DMRL (IMRL) and NBUE (NWUE)). This problem has been considered by several authors and for other ageing classes (Ahmad, 1975; Bandyopadhyay and Basu, 1990; Chaudhuri, 1997; Deshpande and Kochar, 1983; Ebrahimi, 1997; Hollander and Proschan, 1972, 1975, 1976; Klefsjo, 1983a, 1983b; Langenberg and Srinivasan, 1979; Proschan and Pyke, 1967). 3.1. The test for DMRL (IMRL) alternatives Given a random sample X1 ; : : : ; Xn of X; with distribution function FX , we consider the problem of testing the null hypothesis H0 : X is exponential (FX (t) = 1 − exp(−t);  unspeci ed) against the alternative H1 : X is DMRL (IMRL) and is not exponential: Motivated by (1) and the de nition of dil in previous section, we consider the following measure of departure, from H0 to H1 : Z 2 2 I (s ¡ t)F X (s)F X (t)(dil (Xt ; Xs )) dFX (t) dFX (s); DMRL (X ) ≡ R2

where I (A) denotes the indicator of the event A. We get that under H0 ; DMRL (X ) = 0, but under H1 ; DMRL (X ) ¿ (¡)0. The sample analog of DMRL (X ) forms the basis of our test. Replacing, in DMRL (X ); the distribution function by the empirical distribution function, we obtain the following statistic: P n−1 P 1 n−2 DMRL (n) ≡ 6 (n − i)2 (n − j)2 (n (j) − n (i)); n i=0 j=i+1 where X(0) ≡ 0 and n (i) ≡

1 5n ; (n − i)2 i

being 5ni ≡

n P =i+1

i X( ) ;

(4)

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where i = (n − 2 + i + 1) and X(1) ; : : : ; X(n) are the order statistics of the sample. We include the value X(0) to use the information in the sample concerning lifelength measured from time 0 (Hollander and Proschan, 1975). Now for large (low) values of DMRL (n) we will reject H0 in favor of H1 (X is DMRL (IMRL)). Since DMRL (n) is not scale invariant, we consider the following test statistic which makes our test scale invariant: ∗DMRL (n) ≡ DMRL (n)= X : Next we see the exact distribution of ∗DMRL (n), which allow us to give the critical values of ∗DMRL (n). To obtain the exact distribution we see rst that ∗DMRL (n) can be written, after some algebraic manipulations, in terms of the normalized spacings, D = (n − + 1)(X( ) − X( −1) ); in the following way: Pn j=1 ej; n Dj ∗ DMRL (n) = Pn j=1 Dj where

 5  4    3   1 1 1 j 1 j j 1 1 + − + ej; n = − + + 30 n n 6n 6 n 3n2 3n 3      2  1 5 5 11 3 1 j 1 2 j + + : + + + + + − n 15n3 12n2 12n 6 n 3n4 12n3 4n2 6n

Then by Theorem 2:4 of Box (1954) and taking  = 12 (see also Langenberg and Srinivasan, 1979), we get, under H0 ; that   n Q n P ei; n − x ∗ P(DMRL (n)6x) = 1 − I (x ¡ ei; n ); ei; n − ej; n i=1 j=1 i6=j

for ei; n 6= ej; n , for all i 6= j, for xed n. For large values of n we provide also the asymptotic distribution of ∗DMRL (n). The result is obtained by expressing DMRL (n) as a linear combination of order statistics, in the following way n 1P ((n − i + 1)ei; n − (n − i)ei−1; n )X(i) n i=1 n 1P = in X(i) ; n i=1

DMRL (n) =

where

 5  4    3   1 i 2 i i 1 2 +1 − + 2 + + 5 n n 2n n 3n2 n   2  3 7 9 1 i + + 2+ + 3 n 2n 4n 4n 2   7 1 11 2 i 2 + + : + + − n 15n4 12n3 12n2 3n 3

in = −

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Now, if X is a continuous non negative random variable with E[X 2 ] ¡ + ∞ and let Z (J; FX ) = xJ (FX (x) dFX (x)) R

and 2 (J; FX ) =

Z Z R2

J (FX (x))J (FX (y))[FX (min(x; y) − FX (x)FX (y)] d x dy

where J (u) = − 15 u5 + u4 − 2u3 + 32 u2 − 13 u;

(5)

then we can apply Theorems 2 and 3 of Stigler (1974), and the limiting distribution of   DMRL (n) − (J; FX ) n1=2 (J; FX ) is N (0; 1): By Slutsky’s theorem we get for ∗DMRL (n), that the limiting distribution of   (J; FX ) 1=2 ∗ DMRL (n) − n E(X ) is N (∗ ; (∗ )2 ) where ∗ ≡ (J ∗ ; FX )=E(X ), (∗ )2 ≡ 2 (J ∗ ; FX )=E(X ) and J ∗ (u) ≡ J (u) − (J; FX )=E(X ): We see now the asymptotic distribution under H0 . Since the statistic ∗DMRL (n) is scale invariant, we can suppose that X follows an exponential distribution with  equal to 1, and we get that (J; FX ) = 0 and 2 (J; FX ) = 19=166 320. Thus under H0 , the limiting distribution of (166 320n=19)1=2 ∗DMRL (n) is N (0; 1): For low values of n we can use the exact distribution to give critical values of the test statistic (166 320n=19)1=2 ∗DMRL (n) and to compute the p-value of the test. In Table 1 we give the critical values for several signi cance levels and for several sample sizes. For large values of n, we rejects the null hypothesis, i.e., X is DMRL (IMRL) but it is not exponential, when (166 320n=19)1=2 ∗DMRL (n) ¿ z (¡ z1− ), z is the (1− )-quantile of the standard normal distribution function. 3.2. The test for NBUE (NWUE) alternatives Given a random sample X1 ; : : : ; Xn of X , with distribution function FX , we consider now the problem of testing the null hypothesis H0 : X is exponential (FX (t) = 1 − exp(−t);  unspeci ed) against the alternative H1 : X is NBUE (NWUE) and is not exponential: By the characterization (2), we consider as a measure of departure, from H0 to H1 : Z 2 NBUE (X ) ≡ F X (t) (dil (Xt ; Y )) dFX (s) R

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Table 1 Critical values of (166 320n=19)1=2 ∗DMRL (n) n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 45 50 55 60

: Lower tail

: Upper tail

0.01

0.05

0.1

0.1

0.05

0.01

0.0413 0.2109 −0:2790 −1:0322 −1:4617 −1:7185 −1:8810 −1:9898 −2:0662 −2:1226 −2:1656 −2:1994 −2:2263 −2:2480 −2:2659 −2:2806 −2:2928 −2:3033 −2:3124 −2:3412 −2:3561 −2:3645 −2:3693 −2:3722 −2:3743 −2:3745 −2:3608

0.2067 0.4715 0.0418 −0:04161 −0:6850 −0:8515 −0:9646 −1:0476 −1:1122 −1:1635 −1:2020 −1:2392 −1:2677 −1:2920 −1:3128 −1:3309 −1:3469 −1:3610 −1:3735 −1:4203 −1:4508 −1:4723 −1:4885 −1:5012 −1:5113 −1:5199 −1:5258

0.4135 0.6669 0.2514 −0:0641 −0:2653 −0:4003 −0:4999 −0:5783 −0:6407 −0:6910 −0:7324 −0:7670 −0:7966 −0:8221 −0:8443 −0:8634 −0:8813 −0:8969 −0:9109 −0:9646 −1:0012 −1:0279 −1:0485 −1:0649 −1:0784 −1:0897 −1:0994

3.7214 2.5176 2.2138 2.0051 1.9034 1.8343 1.7803 1.7401 1.7085 1.6825 1.6608 1.6424 1.6265 1.6126 1.6002 1.5892 1.5793 1.5703 1.5621 1.5295 1.5064 1.4887 1.4747 1.4633 1.4537 1.4455 1.4384

3.9281 2.7569 2.4148 2.2086 2.1204 2.0497 1.9986 1.9624 1.9341 1.9116 1.8933 1.8781 1.8653 1.8543 1.8447 1.8363 1.8289 1.8222 1.8163 1.7934 1.7779 1.7664 1.7575 1.7504 1.7444 1.7394 1.7352

4.0953 3.0761 2.7037 2.5637 2.4682 2.3980 2.3591 2.3300 2.3087 2.2932 2.2814 2.2724 2.2655 2.2601 2.2560 2.2526 2.2500 2.2480 2.2464 2.2429 2.2430 2.2445 2.2466 2.2488 2.2511 2.2532 2.2553

where Y is an exponential distribution with mean E(X ). We get that under H0 ; NBUE (X ) = 0, but under H1 ; NBUE (X ) ¿ (¡) 0. As in Section 3.1, the sample analogue of this measure, is Pn   P X(i) 1 n−2 ; (n − i)2 n n (i) + i=1 NBUE (n) ≡ 4 n i=0 2n where X(0) ≡ 0; X(1) ; : : : ; X(n) are the order statistics and n (i) is de ned in (4), and forms the basis of our test. Since the test statistic is not scale invariant, we consider the following test statistic which makes our test scale invariant: ∗NBUE (n) ≡ NBUE (n)= X : To give the exact distribution of ∗NBUE (n), we proceed as in Section 3.1. Thus, rst, we write the statistic in terms of the normalized spacings, D ; in the following way: ∗NBUE (n) =

n P j=1

ej; n Dj =

Pn

j=1 Dj

F. Belzunce et al. / Journal of Statistical Planning and Inference 86 (2000) 113–127

where 1 ej; n = − 2

121

 2     1 1 1 1 i 1 i + + + + − 3; 2 n n 2n 6 4n 12n 2n

and we get, under H0 ; that P(∗NBUE (n)6x)

=1−

n Q n P i=1

j=1 i6=j



ei; n − x ei; n − ej; n

 I (x; ei; n )

for ei; n 6= ej; n , for all i 6= j, for xed n. Now, we derive the asymptotic distribution of ∗NBUE (n). First we obtain NBUE (n) as a linear combination of order statistics, in the following way: NBUE (n) = where 3 in = − 2

n 1P in X(i) ; n i=1

 2     1 1 1 1 i 1 i +1 + + + + − 3: n n 2n 6 4n 12n2 2n

Now if X is a continuous nonnegative random variable with E[X 2 ] ¡ + ∞ and let (J; FX ) and 2 (J; FX ) as in Section 3.1 where J (u) = − 32 u2 + u + 16 ;

(6)

then we can apply Theorems 2 and 3 of Stigler (1974) to obtain that the limiting distribution of   NBUE (n) − (J; FX ) 1=2 n (J; FX ) is N (0; 1). Again for ∗NBUE (n) we get, that the limiting distribution of n1=2 (∗NBUE (n)−(J; FX )= E(X )) is N (∗ ; (∗ )2 ), where ∗ ; (∗ )2 and J ∗ (u) are de ned as in Section 3.1. We see now the asymptotic distribution under H0 . Now supposing that X follows an exponential distribution with  equal to 1, we get that (J; FX )=0 and 2 (J; FX )=1=45. Thus under H0 , the limiting distribution of (n45)1=2 ∗NBUE (n) is N (0; 1). For low values of n, we proceed as in Section 3.1, and in Table 2 we give the critical values for several signi cance levels and for several sample sizes. For large values of n, we rejects the null hypothesis, i.e. X is NBUE (NWUE) but it is not exponential, when (n45)1=2 ∗DMRL (n) ¿ z (¡ z1− ). 3.3. Consistency and asymptotic relative eciency In this section we see the consistency of our proposed tests and we compare our tests with other existing tests on the basis of Pitman’s asymptotic eciency. First we see the consistency of ∗DMRL (X ). Since (J; FX ) = 0 when FX follows the exponential distribution, the consistency follows if and only if (J; FX ) ¿ (¡)0;

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Table 2 Critical values of (45n)1=2 ∗NBUE (n) n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 45 50 55 60

: Lower tail

: Upper tail

0.01

0.05

0.1

0.1

0.05

0.01

0.0237 −0:7596 −1:1819 −1:4157 −1:5599 −1:6588 −1:7328 −1:7906 −1:8369 −1:8748 −1:9064 −1:9332 −1:9562 −1:9762 −1:9937 −2:0092 −2:0231 −2:0355 −2:0468 −2:0901 −2:1195 −2:1410 −2:1575 −2:1706 −2:1813 −2:1903 −2:1979

0.1186 −0:3687 −0:6071 −0:7498 −0:8522 −0:9317 −0:9937 −1:0434 −1:0842 −1:1184 −1:1476 −1:1728 −1:1948 −1:2142 −1:2315 −1:2470 −1:2610 −1:2737 −1:2853 −1:3311 −1:3636 −1:3880 −1:4072 −1:4228 −1:4358 −1:4468 −1:4562

0.2371 −0:0758 −0:2473 −0:3694 −0:4667 −0:5414 −0:6001 −0:6478 −0:6875 −0:7211 −0:7498 −0:7748 −0:7969 −0:8164 −0:8338 −0:8495 −0:8638 −0:8767 −0:8886 −0:9360 −0:9699 −0:9957 −1:0161 −1:0328 −1:0468 −1:0586 −1:0689

2.1345 2.0900 2.0159 1.9471 1.8917 1.8469 1.8102 1.7796 1.7536 1.7313 1.7118 1.6946 1.6793 1.6657 1.6533 1.6421 1.6318 1.6224 1.6138 1.5786 1.5528 1.5328 1.5167 1.5034 1.4922 1.4825 1.4741

2.2531 2.2971 2.2300 2.1816 2.1373 2.1021 2.0729 2.0486 2.0279 2.0101 1.9946 1.9809 1.9688 1.9579 1.9480 1.9391 1.9309 1.9234 1.9165 1.8884 1.8677 1.8517 1.8338 1.8280 1.8190 1.8112 1.8043

2.3480 2.5735 2.5662 2.5449 2.5299 2.5160 2.5047 2.4952 2.4872 2.4804 2.4744 2.4692 2.4646 2.4605 2.4568 2.4535 2.4504 2.4476 2.4450 2.4345 2.4267 2.4205 2.4155 2.4114 2.4078 2.4047 2.4019

when X is DMRL (IMRL) but it is not exponential. This follows by observing that DMRL (X ) = (J; FX ), as we proof in the next theorem. Theorem 3.1. Let X be a continuous nonnegative random variable and let us suppose that limx→+∞ xF X (x) = 0. Then; for J given in (5); it follows that DMRL (X ) = (J; FX ): Proof. From the de nition of DMRL (X ) and integrating by parts we get Z DMRL (X ) = −

0

Z −

+∞

0

+∞

Z +∞  2 F X (u) F X (u) du + du 30 6 0 Z +∞  6 3 F X (u) F X (u) du + du: 6 30 0

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123

For (J; FX ) where J is given in (5) we get Z +∞  2 Z +∞  F X (u) F X (u) du + du (J; FX ) = − 30 6 0 0 Z +∞  6 Z +∞  3 F X (u) F X (u) du + du; − 6 30 0 0 thus DMRL (X ) = (J; FX ). For ∗NBUE (X ) the consistency follows, as previously, by proving that NBUE (X ) = (J; FX ). This result is stated in the next theorem. Theorem 3.2. Let X be a continuous nonnegative random variable; and let us suppose that limx→+∞ xF X (x) = 0; then for J (u) given in (6) we have NBUE (X ) = (J; FX ): Proof. Follows as in previous theorem. Now we compare our tests with other well known tests for ageing classes. Our comparisons are made on the basis of Pitman’s asymptotic relative eciency (ARE). If we consider a sequence of alternative distributions functions F, indexed by the parameter n = 0 + cn−1=2 , where c is an arbitrary positive constant and 0 corresponds to the exponential distribution, then the Pitman’s eciency of a statistic T is given by ] 2 [ @E[T @ |=0 ] : n→∞ Var 0 [n1=2 T ]

eF (T ) = lim

From the asymptotic properties of ∗DMRL (n) and ∗NBUE (n), we can obtain the Pitman’s eciency in the form eF (T ) =

( @(@J;F) |=0 )2 ; 2 (J; F)=0

for T = ∗DMRL (n) and ∗NBUE (n). We have calculated eF (T ) for the following models: (a) Linear failure rate (LFR): F1 (x) = 1 − exp(−(x + x2 =2)), for x¿0; ¿0. (b) Makeham: F2 (x) = 1 − exp(−(x + (x + exp(−x) − 1))) for x¿0; ¿0. (c) Weibull: F3 (x) = 1 − exp(−x ) for x¿0; ¿0. We have compared the eciency of our tests with the eciency, provided by others authors, of some well known tests for ageing classes. In particular we have considered the tests statistics A2 and V for IFR (DFR) ageing classes given by Klefsjo (1983a) and Proschan and Pyke (1967), respectively. The tests statistics Jn , Un and Sn for NBU (NWU) ageing classes given by Hollander and Proschan (1972), Ahmad (1975) and Deshpande and Kochar (1983), respectively. The test statistic B for IFRA (DFRA) ageing classes given by Klefsjo (1983a). The tests statistics V ∗ and Vn (k) for DMRL (IMRL) ageing classes given by Hollander and Proschan (1975) and Bandyopadhyay and Basu (1990), respectively. And the test statistic K ∗ for HNBUE (HNWUE) ageing classes given by Hollander and Proschan (1975) and Klefsjo (1983b).

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Table 3 ARE of ∗DMRL (n)

A2 V Jn Un Sn B V∗ Vn (0:01) Vn (0:05) K∗ ∗NBUE (n)

LFR

Makeham

Weibull

2.0592 1.3284 2.2246 2.4512 19.2510 2.9652 0.9153 1.0093 1.0047 1.0010 0.8649

0.5911 0.5514 0.1021 0.6187 7.0329 0.5911 0.5911 0.4167 0.4286 0.4152 0.4411

0.5367 0.3619 0.2896 0.5344 6.6820 0.3109 0.5579 0.2864 0.3218 0.2712 0.4411

LFR

Makeham

Weibull

2.3808 1.5358 2.5720 2.8340 22.2578 3.4283 1.0582 1.1669 1.1616 1.1574 1.1562

1.3401 1.2500 0.2315 1.4026 15.9439 1.3401 1.3401 0.9447 0.9717 0.9413 2.2670

1.6511 1.1135 0.8908 1.6439 20.5516 0.9561 1.7164 0.8811 0.9900 0.8343 3.0763

Table 4 ARE of ∗NBUE (n)

A2 V Jn Un Sn B V∗ Vn (0:01) Vn (0:05) K∗ ∗DMRL (n)

Tables 3 and 4 show for ∗DMRL (n) and ∗NBUE (n) the asymptotic relative eciencies (ARE) of our tests with respect to the tests mentioned previously. For the class of statistics Vn (k) we have considered k = 0:01 and k = 0:05, which, for practical considerations, are the values chosen in Bandyopadhyay and Basu (1990). As can be seen from Table 3 the statistic ∗DMRL (n) performs well for the LFR model. From Table 4 we see that ∗NBUE (n) has similar or superior eciency than the others tests for the LFR, Makeham and Weibull, except for the statistic Jn for the Makeham model. 4. Conclusions and remarks The ideas developed previously can be used to propose a test for the HNBUE (HNWUE) ageing class. Recall that a random variable X is HNBUE (HNWUE) (harmonic new better (worse) than used in expectation) if, and only if, (see Klefsjo (1982)) Z ∞ F X (t) dt6(¿)E(X )exp(−x=E(X )) for all x¿0: x

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This notion can be de ned in a di erent, but equivalent, manner. A random variable X is HNBUE (HNWUE) if, and only if, X 6dil Y , where Y is an exponential random variable with mean E(X ). As a measure of departure from exponential hypothesis to the HNBUE (HNWUE) hypothesis, we can consider the measure HNBUE (X ) ≡ dil (X; Y ); where Y is an exponential random variable with mean E(X ). After some calculations this measure became Z ∞ HNBUE (X ) = F X (t) dt − E(X )=2: 0

However this measure is equal to the measure proposed by Hollander and Proschan (1975) for testing the NBUE (NWUE) ageing class. So, the test proposed by Hollander and Proschan (1975) is consistent against the weaker HNBUE (HNWUE) ageing class (Klefsjo, 1983b). Therefore, in this paper, we provide a uni ed approach for testing ageing alternatives based on the dilation order. We apply, now, our tests to the data set provided by Bryson and Siddiqui (1969). This data set consists of n = 43 survival times of patients, su ering from chronic granulocytic leukemia. In order to apply a test for an ageing class we need to have some statistical evidence that our sample belongs to the ageing class. This can be done for most of the ageing classes by means of the total time on test (TTT) transform plot. Given an ordered random sample X(1) ; X(2) ; : : : ; X(n) of a random variable X , the total time on test transform is given by (Barlow et al., 1972) TTTn (p) ≡ Hn−1 (p)= X where X is the sample mean and  for 06p61=n;  nX(1) p i  Hn−1 (p) ≡ P (n−j+1) i i  n (X(j) − X(j−1) ) + p − n (X(i+1) − X(i) ) for n 6p6(i+1)=n: j=1

For the DMRL (IMRL) ageing class, it is reasonable to expect that the (1-TTTn (p))= (1 − p) behaves decreasingly (increasingly) for all p ∈ (0; 1). For the NBUE (NWUE) it should be expected to hold (1-TTTn (p))=(1 − p)61 for all p ∈ [0; 1] (Klefsjo, 1983a). Additionally for an exponential distribution (1-TTTn (p))=(1 − p) is close to 1. We have used this technique for the data, as a previous step, before testing the ageing class. For the data set of Bryson and Siddiqui (1969), from the TTT plots (Fig. 1) a NBUE is to be expected. The mean p residual life shows a nonstrictly decreasing trend. In fact for the DMRL test we get 166320 · n=19∗DMRL (n) = 1:3904 and p-value=0:1186, which does not con rm the DMRL hypothesis. For the NBUE test we get √ 45 · n∗NBUE (n) = 1:8577 and p-value=0:0470, which con rms the NBUE hypothesis. Hollander and Proschan (1975) obtain the same conclusions, however, as we

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Fig. 1. TTT plot of data set of Bryson and Siddiqui (1969).

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