A nonparametric test for comparison of mean past lives

Journal Pre-proof A nonparametric test for comparison of mean past lives Dhrubasish Bhattacharyya, Ruhul Ali Khan, Murari Mitra

PII: DOI: Reference:

S0167-7152(20)30025-0 https://doi.org/10.1016/j.spl.2020.108722 STAPRO 108722

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Statistics and Probability Letters

Received date : 4 September 2019 Revised date : 15 January 2020 Accepted date : 28 January 2020 Please cite this article as: D. Bhattacharyya, R.A. Khan and M. Mitra, A nonparametric test for comparison of mean past lives. Statistics and Probability Letters (2020), doi: https://doi.org/10.1016/j.spl.2020.108722. This is a PDF file of an article that has undergone enhancements after acceptance, such as the addition of a cover page and metadata, and formatting for readability, but it is not yet the definitive version of record. This version will undergo additional copyediting, typesetting and review before it is published in its final form, but we are providing this version to give early visibility of the article. Please note that, during the production process, errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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A nonparametric test for comparison of mean past lives Dhrubasish Bhattacharyya a , Ruhul Ali Khana , Murari Mitraa Department of Mathematics, Indian Institute of Engineering Science and Technology, Shibpur P.O. - Botanic Garden, Howrah - 711103, West Bengal, India

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a

Abstract

A flexible nonparametric two sample test is developed for comparing Mean Past Lives (MPL) of two life distributions. The test is shown to be consistent having prescribed asymptotic size. We further develop an efficient algorithm for computing the test statistic. Finally, the performance of the test is assessed by means of a simulation study. Keywords: Mean Inactive Time function, Nonparametric test, Asymptotic normality, Monte Carlo simulation AMS 2010 classification code : 62G10; 62G30; 90B25 1. Introduction

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In reliability theory and risk management, one is primarily concerned with the distribution of the residual lifetime of an item which is typically modeled using a non-negative random variable. However, in many real life scenarios, it is reasonable to consider the underlying random variable which is related to the past rather than the future. For instance, suppose the efficiency of a treatment is measured by means of the remission time of a disease that can recur. In this situation, the actual remission time is often unknown since it is difficult to continuously monitor the patients due to the high cost and effort involved. In such scenarios, the actual remission time is estimated using the concept of the Mean Past Life (MPL) function which is also popularly known as the Mean Inactive Time (MIT) function. In fact, this is a well-known reliability measure and has potential applications in various fields including reliability theory, actuarial

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science and survival analysis, see, for instance Chandra and Roy (2001) and Finkelstein (2002). Suppose X is a non-negative random variable which represents the lifetime of an item with cumulative distribution function (cdf) F and survival function F¯ = 1 − F respectively. For any fixed t > 0, let X(t) = [t − X|X ≤ t] be the random variable representing the time elapsed till time t after the failure of the item. Then X(t) is said to be the Inactive Time (IT) and the mean of X(t) , known as the MIT, is given by

µF (t) = E[X(t) ] =

Rt 0

F(x)dx F(t)

, for t > 0.

Email addresses: [email protected] (Dhrubasish Bhattacharyya), [email protected] (Ruhul Ali Khan), [email protected] (Murari Mitra)

Preprint submitted to Statistics & Probability Letters

January 15, 2020

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Because of its potential applications in reliability theory and risk analysis, the MIT function has received widespread attention in the literature, see, for instance, Kayid and Ahmad (2004), Ahmad et al. (2005), Kayid and Izadkhah

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(2014) and the references therein. On the inferential side, the problem of testing exponentiality against alternatives belonging to Increasing Mean Inactive Time (IMIT) class of life distributions is of importance and the works of Kayid and Ahmad (2004) and Zhang and Cheng (2010) are significant in this context.

Based on the comparison of MIT functions of two non-negative random variables, a well-known stochastic order (i.e., MIT order) and its applications to reliability theory has been studied extensively in the literature, see, for example, Nanda et al. (2006), Ahmad and Kayid (2005), Kayid and Izadkhah (2014) and the references therein. Let X and Y be two non-negative random variables with respective cdfs F and G. Then, X is said to be smaller than Y in the MIT order if and only if

Rt 0

F(u)du

0

G(u)du

Rt

or, equivalently,

is non-increasing for t > 0

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µF (t) ≥ µG (t) for all t > 0.

However, the present authors are unaware of the existence of any statistical methodology for comparing MIT functions of two distributions. In this paper, we propose a distribution free test procedure for such comparison based on two independent samples. The main feature of the proposed test is that it has the flexibility of handling crossings of two MIT functions and may be applied to obtain confidence statements of the form “µF (t) > µG (t) for all t ∈ I,” where I is an interval of values determined by the sets of data. We are interested in the following testing problem H0 : µF (t) ≤ µG (t) for some t ∈ [a, b] vs.

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H1 : µF (t) > µG (t) for all t ∈ [a, b],

(1)

In Section 2, we consider a simple testing problem ′

′

H0 : µF (t) ≤ µG (t) vs. H1 : µF (t) > µG (t) for any fixed t > 0.

(2)

Subsequently, we apply the intersection-union principle (cf., Berger and Sinclair (1984), Berger et al. (1988) and Kaur et al. (1994)) for developing a test procedure for the problem given in (1). The organization of the rest of the paper is as follows. In Section 2, we develop the test procedure for the relevant problem. We establish the consistency of the test and obtain a bound on asymptotic size in Section 3. Section 4 contains a nice computational technique for 2

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evaluating the proposed test statistic. Finally, in Section 5, we assess the merit of the test procedure by means of a simulation study. The proofs of all theorems are presented in the appendix.

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2. Development of the test procedure

Suppose X(1) ≤ . . . , ≤ X(m) are the order statistics based on a random sample X1 , . . . , Xm of size m from a continuous distribution function F and Y(1) ≤ . . . ≤ Y(n) are the order statistics based on another random sample Y1 , . . . , Yn of size n from a continuous distribution function G. Now, based on these samples, the empirical cdfs Fm (t) and Gn (t) of F and G respectively are given by

     0, 0 ≤ t < X(1) ,        j X(i) ≤ t < X(i+1) , i = 1, . . . , m − 1 and Gn (t) =    n,         1, t ≥ X(m)

     0,        i Fm (t) =    m,         1,

0 ≤ t < X(1) ,

X( j) ≤ t < X( j+1) , j = 1, . . . , n − 1 .

t ≥ X(n)

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We assume that E[X 2 ] < ∞ and E[Y 2 ] < ∞. For any fixed t > 0, we first consider the testing problem given in (2) and formulate an asymptotic level α test for the problem. Rt Rt Let us choose Λ(t) = G(t) 0 F(u)du − F(t) 0 G(u)du as our measure of departure towards H1 from H0 , since it

is easy to observe that ∆(t) ≤ 0 under H0 and is strictly positive when H1 is true. We now define, for 1 ≤ i ≤ m and 1 ≤ j ≤ n,

T i (t) = (t − Xi )I(t − Xi ) and W j (t) = (t − Y j )I(t − Y j ),

where I(a) = 1 or 0 according as a > 0 or a ≤ 0. Observe that T i ’s are independent and identically distributed and so are W j ’s. Now, let i be the integer such that t ∈ [X(i) , X(i+1) ). Then it follows that t

Z

X(1)

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Z

0

Fm (u)du =

0

Fm (u)du +

Z

X(2)

X(1)

Fm (u)du + · · · +

Z

t X(i)

Fm (u)du

 1 X(2) − X(1) + 2(X(3) − X(2) ) + · · · + i(t − X(i) ) m i m  1X 1 1X = it − (X(1) + · · · + X(i) ) = (t − X(k) ) = T k (t) = T¯ (t). m m k=1 m k=1

=

Similarly, one can obtain

Z

Further, note that E[T i (t)] =

Z

t 0

t 0

¯ Gn (u)du = W(t).

F(u)du and E[W j (t)] =

3

Z

t 0

G(u)du.

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For the testing problem (2), we now propose a natural test statistic Λm,n (t) given by

where 2 S m,T (t) =

m

(3)

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¯ Gn (t)T¯ (t) − Fm (t)W(t) Λm,n (t) = q , 2 (t) + F 2 (t) 1 S 2 (t) G2n (t) m1 S m,T m n n,W n

1X 1X 2 2 ¯ [T i (t) − T¯ (t)]2 and S n,W [W j (t) − W(t)] . (t) = m i=1 n j=1

We define Λm,n (t) to be 0 when both the numerator and denominator in (3) are 0.

Suppose Zα is the (1 − α)-th quantile of the standard normal distribution. For large values of m, n, we propose ′

′

to reject H0 in favor of H1 at a significance level α if and only if Λm,n (t) > Zα . Thus, for the testing problem (1), we propose to reject H0 in favor of H1 for large values of m and n if and only if Λm,n (t) > Zα for all t ∈ [a, b], or, equivalently, if and only if

inf Λm,n (t) > Zα .

t∈[a,b]

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3. Asymptotic size and consistency

We now establish that our proposed test for the testing problem (2) is consistent and has the prescribed asymptotic size α.

Theorem 3.1. Suppose X and Y are two non-negative random variables representing the lifetimes of two items with respective MIT functions µF (t) and µG (t). Let min(m, n) → ∞ in such a way that m/(m + n) → λ (0 < λ < 1). Further assume that both F and G have finite second moments and let D2 (t) = G2 (t)Var(T 1 (t)) +

   0,     lim P[Λm,n (t) > Zα ] =  α   min(m,n)→∞   1

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If D(t) > 0, then

λ F 2 (t)Var(W1 (t)). 1−λ

(4)

if µF (t) < µG (t) if µF (t) = µG (t) if µF (t) > µG (t),

where Zα is the (1 − α)-th quantile of the standard normal distribution. If D2 (t) = 0, then Λm,n (t) = 0 almost surely for all m, n and P[Λm,n (t) > Zα ] = 0. Theorem 3.1 establishes that the proposed test for the testing problem (2) is of asymptotic level α and is consistent. These results about a fixed t ≥ 0 can be extended to the more general problem envisaged in (1). Theorem 3.2. Let m, n → ∞ in such a way that m/(m + n) → λ, 0 < λ < 1. Then (i) for F, G ∈ H0 ,

lim

min(m,n)→∞

"

#

P inf Λm,n (t) > Zα ≤ α t∈[a,b]

4

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(ii) If F, G ∈ H0 with µF (t0 ) = µG (t0 ) for some t0 ∈ (a, b) and µF (t) > µG (t) for all t ∈ [a, b] − {t0 }, then " # lim P inf Λm,n (t) > Zα = α t∈[a,b]

and (iii) for F, G ∈ H1 ,

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min(m,n)→∞

lim

min(m,n)→∞

4. Computation of the test statistic

" # P inf Λm,n (t)] > Zα = 1. t∈[a,b]

As the proposed test statistic does not lend itself to a straightforward computation, we develop an efficient computational technique in order to evaluate the test statistic.

Suppose Z(1) ≤ Z(2) ≤ . . . ≤ Z(m+n) be the ordered observations combining Xi , i = 1, . . . , m and Y j , j = 1, . . . , n. Then one can see that Λm,n (t) = 0 for t < Z(1) . Also for t ≥ Z(m+n) , we have Fm (t) = Gn (t) = 1 and T i (t) = t − Xi , W j (t) = t − Y j for all i, j which imply that ∆m,n (t) =

q

¯ X¯ Y− Pn 2 1 ¯ ¯ 2 i=1 (Xi − X) + 2 j=1 (Y j −Y)

Pm

1 m2

is free of t. Now suppose x p of the

n

Xi ’s and y p of the Y j ’s are less than or equal to Z(p) , p = 1, . . . , m + n − 1. Then the following theorem helps in the

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evaluation of inf t∈[a,b] Λm,n (t).

Theorem 4.1. For Z(p) ≤ t < Z(p+1) , p = 1, . . . , m + n − 1, "

where

and

#3/2 Fm2 (t) 2 G2n (t) 2 d S m,T (t) + S n,W (t) Λm,n (t) = a p t + b p , m n dt

 x  yp p   x p y2 X X  x2p y p p    X(i) − Y( j)   3 2 (m − x p ) + 2 3 (n − y p ) a p =  m n m n i=1 j=1

   xp yp yp xp 2 (n − y ) X  x X    y2 (m − x p ) X X x y p p p p p  . b p =  X + Y y( j) − X(i)   (i) ( j)  2 n3 mn j=1 mn i=1 m3 n2 m i=1 j=1

(5)

(6)

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Theorem 4.1 indicates that in order to evaluate inf t∈[a,b] Λm,n (t), we need to find the minimum of the computed b

values of Λm,n (t) at the points a, b and at the points Z(p) , d p = − a pp (p = 1, . . . , m + n − 1) for which a ≤ Z(p) ≤ b and Z(p) ≤ d p < Z(p+1) hold. 5. A simulation study

In this section we carry out a simulation study in order to assess the merit of the proposed test. The simulation is done using MATLAB on PC platform and the results are reported based on 10,000 replications. We evaluate the empirical power of the proposed test for the following two scenarios : Scenario I : Let X and Y be two independent random variables which follow exponential distributions with means 1 and 8 respectively. Then it follows that µX (t) ≥ µY (t) for all t > 0. Here we generate m realizations from X and n 5

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8 µX (t)

7 µY (t)

5 4 3 2 1 0 0

1

2

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MIT Functions

6

3

4

5

6

7

8

Time (t)

Figure 1: Plots of MIT functions for Scenario II

realizations from Y and present the computed power of the proposed test at significance levels α for various choices

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of [a, b].

Scenario II : Let X and Y be two independent random variables where X follow an exponential distribution with mean 1 and Y follow a gamma distribution with shape parameter 0.4 and scale parameter 8. Then the plots of the MIT functions of X and Y given in Figure 1 indicate that the two MIT functions cross. We generate random observations from X and Y of sizes m and n respectively and present the computed empirical power at significance levels α in Table

Table 1: Empirical Power Estimates for Scenario I

m, n

50,50

α

0.10 0.05 0.01 0.10 0.05 0.01 0.10 0.05 0.01 0.10 0.05 0.01 0.10 0.05 0.01

[3.0,5.0] 0.4615 0.2730 0.0761 0.6192 0.4003 0.1052 0.7131 0.5236 0.1884 0.9079 0.7594 0.3640 0.9337 0.8248 0.4494

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2.

50,80

100,80

100,150

150,150

[4.0,6.0] 0.7806 0.5941 0.2523 0.9050 0.7609 0.3781 0.9615 0.8729 0.5793 0.9966 0.9814 0.8336 0.9982 0.9908 0.9022

6

[a, b] [5.0,7.0] 0.9341 0.8299 0.5066 0.9826 0.9381 0.6948 0.9962 0.9840 0.8781 1.0000 0.9989 0.9810 1.0000 1.0000 0.9935

[4.0,7.0] 0.7799 0.5990 0.2506 0.9039 0.7649 0.3854 0.9584 0.8811 0.5926 0.9962 0.9792 0.8359 0.9983 0.9919 0.8983

[5.0,8.0] 0.9362 0.8347 0.5128 0.9825 0.9413 0.6836 0.9958 0.9838 0.8708 0.9999 0.9996 0.9817 0.9999 0.9997 0.9932

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Table 2: Empirical Power Estimates for Scenario II

40,40

60,60

80,80

100,100

150,150

α 0.10 0.05 0.01 0.10 0.05 0.01 0.10 0.05 0.01 0.10 0.05 0.01 0.10 0.05 0.01

[6.0,8.0] 0.2547 0.1185 0.0181 0.3417 0.1813 0.0346 0.4331 0.2440 0.0544 0.5093 0.3065 0.0740 0.6768 0.4779 0.1563

[7.0,10.0] 0.3768 0.2159 0.0471 0.5259 0.3363 0.0893 0.6493 0.4507 0.1438 0.7412 0.5462 0.2035 0.8871 0.7591 0.3945

[a, b] [8.0,12.0] 0.5218 0.3181 0.0851 0.6826 0.4782 0.1648 0.7961 0.6293 0.2614 0.8758 0.7380 0.3741 0.9632 0.9039 0.6390

[9.0,12.0] 0.6030 0.4011 0.1312 0.7777 0.5988 0.2521 0.8886 0.7534 0.3884 0.9436 0.8502 0.5318 0.9915 0.9659 0.7965

[9.0,15.0] 0.6135 0.4135 0.1298 0.7752 0.6065 0.2516 0.8843 0.7578 0.3938 0.9434 0.8562 0.5339 0.9890 0.9632 0.8076

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m, n

From both the tables it is evident that our test exhibits high power in all the alternative scenarios considered and

Acknowledgements

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the power increases when the difference between the two MIT functions increases.

The authors are grateful to the anonymous reviewers and the Editor for their valuable comments and helpful suggestions which have substantially improved the presentation of the paper. The authors are also grateful to the Council of Scientific and Industrial Research (CSIR), India and University Grant Commission (UGC), India for financial support.

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References

Ahmad, I. and Kayid, M. (2005). Characterizations of the RHR and MIT orderings and the DRHR and IMIT classes of life distributions. Probability in the Engineering and Informational Sciences, 19(4):447–461. Ahmad, I. A., Kayid, M., and Pellerey, F. (2005). Further results involving the MIT order and the IMIT class. Probability in the Engineering and Informational Sciences, 19(3):377–395. Berger, R. L., Boos, D. D., and Guess, F. M. (1988). Tests and confidence sets for comparing two mean residual life functions. Biometrics, 44(1):103–115. Berger, R. L. and Sinclair, D. F. (1984). Testing hypotheses concerning unions of linear subspaces. Journal of the American Statistical Association, 79(385):158–163. 7

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Chandra, N. K. and Roy, D. (2001). Some results on reversed hazard rate. Probability in the Engineering and Informational Sciences, 15(1):95–102.

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Finkelstein, M. S. (2002). On the reversed hazard rate. Reliability Engineering & System Safety, 78(1):71–75. Kaur, A., Rao, B. P., and Singh, H. (1994). Testing for second-order stochastic dominance of two distributions. Econometric Theory, 10(5):849–866.

Kayid, M. and Ahmad, I. (2004). On the mean inactivity time ordering with reliability applications. Probability in the Engineering and Informational Sciences, 18(3):395–409.

Kayid, M. and Izadkhah, S. (2014). Mean inactivity time function, associated orderings, and classes of life distributions. IEEE Transactions on Reliability, 63(2):593–602.

Nanda, A. K., Bhattacharjee, S., and Alam, S. (2006). On upshifted reversed mean residual life order. Communications in Statistics-Theory and Methods, 35(8):1513–1523.

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Zhang, S. and Cheng, H. (2010). Testing for increasing mean inactivity time. Statistics, 44(5):467–476.

Appendix

Proof of Theorem 3.1

Observe that Λm,n (t) can be written as

Λm,n (t) = (Mm,n (t) + Nm,n (t))/Dm,n (t),

Z

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where

Mm,n (t) = Gn (t)

t

0

√

m[Fm (u) − F(u)]du −

Nm,n (t) =

and

√

p

m/nFm (t)

Z

t

√

0

n[Gn (u) − G(u)]du,

" # Z t Z t m Gn (t) F(u)du − Fm (t) G(u)du 0

0

2 D2m,n (t) = G2n (t)S m,T (t) +

m 2 2 F (t)S n,W (t). n m

Now from (7), it follows that Mm,n can be written as i p i √ h √ h ¯ − E[W1 (t)] . Mm,n = Gn (t) m T¯ (t) − E[T 1 (t)] − m/nFm (t) n W(t) 8

(7)

(8)

(9)

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An application of the Central Limit Theorem gives

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i d √ h m T¯ (t) − E[T 1 (t)] → − N (0, Var(T 1 (t))) . a.s

Further, Strong Law of Large Numbers (SLLN) implies Gn (t) −−→ G(t) as n → ∞. Slutsky’s theorem now yields i d   √ h Gn (t) m T¯ (t) − E[T 1 (t)] → − N 0, G2 (t)Var(T 1 (t)) .

  i d √ h d ¯ − E[W1 (t)] − → N 0, F 2 (t)Var(W1 (t)) . Hence, Mm,n (t) → − N(0, D2 (t)), where Similarly, it follows that Fm (t) n W(t) a.s

D2 (t) is given by (4). Another application of SLLN yields D2m,n (t) −−→ D2 (t) as min(m, n) → ∞. Consequently, from

Slutsky’s theorem, we get

Mm,n (t) d → − N(0, 1) as min(m, n) → ∞. Dm,n (t) Also, it follows that

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      −∞, if µF (t) < µG (t)        Nm,n (t) →  0, if µF (t) = µG (t)           ∞, if µF (t) > µG (t).

Hence the results of the theorem now follow from (10) and (11).

(10)

(11)

✷

Proof of Theorem 3.2

(i) Since F, G ∈ H0 , there exists t0 ∈ [a, b] such that G(t0 )

Z

t0

0

F(u)du − F(t0 )

Z

t0 0

G(u)du ≤ 0.

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If D(t0 ) = 0, then it follows that T i (t0 ), W j (t0 ) are degenerate random variables for i = 1, . . . , m and j = 1, . . . , n. Hence, Λm,n (t0 ) = 0 for all m, n and consequently P

"

# inf Λm,n (t) > Zα = 0.

t∈∈[a,b]

We now assume that D(t0 ) > 0. Then from Theorem 3.1, we get "

P inf Λm,n (t) > Zα t∈[a,b]

#

" # Mm,n (t0 )   ≤ P Λm,n (t0 ) > Zα ≤ P > Zα → α, as min(m, n) → ∞. Dm,n (t0 )

Hence the theorem follows. (ii) The conditions of the theorem imply that D(t0 ) > 0, for if D(t0 ) = 0, then both T i (t0 ) and W j (t0 ) are degenerate 9

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random variables which contradicts the assumptions. Next note that inf Λm,n (t) ≤ Λm,n (t0 )

and hence lim

repro of

t∈[a,b]

inf Λm,n (t) ≤

min(m,n)→∞ t∈[a,b]

lim

min(m,n)→∞

Λm,n (t0 ) =

lim

min(m,n)→∞

Mm,n (t0 ) . Dm,n (t0 )

(12)

Since, for any fixed m, n > 0, Λm,n (t) is a step function with a finite number of jumps, inf t∈[a,b] Λm,n (t) is attained for some tm,n ∈ [a, b]. Then, one can show that tm,n → t0 . Thus, inf Λm,n (t) =

t∈[a,b]

and consequently, lim

inf Λm,n (t) ≥

lim

min(m,n)→∞

Mm,n (tm,n ) Mm,n (t0 ) = lim . Dm,n (tm,n ) min(m,n)→∞ Dm,n (t0 )

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min(m,n)→∞ t∈[a,b]

Mm,n (tm,n ) Nm,n (tm,n ) + Dm,n (tm,n ) Dm,n (tm,n )

(13)

Hence, from (12) and (13), we have

lim

inf Λm,n (t) =

min(m,n)→∞ t∈[a,b]

lim

min(m,n)→∞

Mm,n (t0 ) Dm,n (t0 )

and the result now follows from Theorem 3.1. (iii) Note that Mm,n (t) can be written as Z

F(t)

0

Z i p √ h m Fm (F −1 (x)) − x dF −1 (x) − m/nFm (t)

G(t) 0

i √ h n Gn (G−1 (x)) − x dG−1 (x)

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Mm,n (t) = Gn (t)

We define stochastic processes

Ym (x, F) =

i i √ h √ h m Fm (F −1 (x)) − x and Yn (x, G) = n Gn (G−1 (x)) − x .

Clearly, Ym (x, F) and Yn (x, G) are elements of D[0, 1] and by weak convergence of probability measures on D[0, 1], it follows that Ym (x, F) and Yn (x, G) converge weakly to Brownian bridge processes Bi (x), x ∈ [0, 1], i = 1, 2 respectively with E[Bi(x)] = 0 and Cov(Bi (x), Bi (y)) = min(x, y) − xy for all x, y ∈ [0, 1], ∀ i = 1, 2.

10

Journal Pre-proof

Hence Mm,n (t) converges weakly and uniformly to a Gaussian process A(t) with continuous path given by

0

t

Z t p B1 (F(t))dt − λ/(1 − λ)F(t) B2 (G(t))dt.

(14)

0

repro of

A(t) = G(t)

Z

Further, it can also be proved that Dm,n (t) converges almost surely and uniformly to D(t). Also from GlivenkoCantelli theorem, we get

a.s 1 sup √ Nm,n (t) − Λ(t) −−→ 0 as min(m, n) → ∞. m t∈[a,b]

Now F, G ∈ H1 implies D(t) > 0 for all t ∈ [a, b], because D(t0 ) = 0 for some t0 ∈ [a, b] implies G(t0 )

Z

t0

0

¯ F(u)du − F(t0 )

Z

t0

0

¯ G(u)du = 0,

which contradicts the fact that F, G ∈ H1 . Thus, it follows that inf t∈[a,b] D(t) = γ > 0. Also F, G ∈ H1 implies that

lim

Hence we obtain

lim

min(m.n)→∞

which implies

m

# inf Nm,n (t) > 0.

t∈[a,b]

rna lP

min(m.n)→∞

"

−1/2

m−1/2

"

inf

t∈[a,b]

# Nm,n (t) > 0, Dm,n (t)

"

# Nm,n (t) 1/4 P inf ≤m → 0, as min(m, n) → ∞. t∈[a,b] Dm,n (t)

Thus, we have "

P inf Λm,n (t) ≤ Zα

"

# " # Mm,n (t) Nm,n (t) 1/4 1/4 ≤ Zα − m ≤m ≤ P inf + P inf t∈[a,b] Dm,n (t) t∈[a,b] Dm,n (t)

Jou

t∈[a,b]

#

and consequently

lim

min(m,n)→∞

"

P inf Λm,n (t) ≤ Zα t∈[a,b]

#

"

Mm,n (t) ≤ lim P inf ≤ Zα − m1/4 min(m,n)→∞ t∈[a,b] Dm,n (t)

From the fact that D(t) > 0 for all t ∈ [a, b], we have from (14), inf

t∈[a,b]

Mm,n (t) d A(t) → − inf , t∈[a,b] Dm,n (t) D(t)

11

#

(15)

Journal Pre-proof

which is a finite-valued random variable. Hence from (15), we have #

P inf Λm,n (t) ≤ Zα = 0 =⇒ t∈[a,b]

lim

min(m,n)→∞

"

#

P inf Λm,n (t) > Zα = 1 t∈[a,b]

repro of

lim

min(m,n)→∞

"

and this establishes the consistency of the test.

Proof of Theorem 4.1 For Z(p) ≤ t < Z(p+1) , we have

  y     xp  xp yp p   1  X X X  1   X 1 1 ¯ =  (t − Y( j) ) = y p t − X(i)  and W(t) Y( j)  . T¯ (t) =  (t − X(i) ) =  x p t − m i=1 m n j=1 n i=1 j=1

Also we have

2 S m,T (t)

and

y y   yp p p  1 X  X 1 X    . 2 2 2 2 2 ¯ (t) =  Y − 2t ¯ =  (t − Y( j) ) − nW Y + y t − n W (t) ( j) p  ( j)  n n j=1 j=1 j=1

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2 S n,W (t)

   xp  xp xp X  1 X  1 X 2 2  2 2 2 ¯ ¯ =  (t − X(i) ) − mT (t) =  X(i) − 2t X(i) + x p t − mT (t) m i=1 m i=1 i=1

Thus,

xp yp d ¯ d ¯ T (t) = and W(t) = . dt m dt n

Now,

m

′ ′ 2X d 2 S m,T (t) = [T i (t) − T¯ (t)][T i (t) − T¯ (t)] dt m i=1

m

=

′ 2X [T i (t) − T¯ (t)]T i (t) m i=1

Jou

  xp xp X   2 2 2X [T i (t) − T¯ (t)] = (m − x p )T¯ (t) = 2 (m − x p )  x p t − = X(i)  . m i=1 m m i=1

Similarly we get

  yp   X 2 d 2  Y( j)  . S n,W (t) = 2 (n − y p ) y p t − dt n j=1

Since Fm (t) = x p /m and Gn (t) = y p /n for t ∈ [Z(p) , Z(p+1) ) we have Λm,n (t) = q

¯ Gn (t)T¯ (t) − Fm (t)W(t)

2 (t) Fm G 2n (t) 2 2 m S m,T (t) + n S n,W (t)

12

= q

yp ¯ n T (t)

−

xp ¯ m W(t)

y2P 2 x2 2 (t) S (t) + m2pn S n,W mn2 m,T

.

✷

Journal Pre-proof

Hence,

−

y2p 2 S (t) mn2 m,T

+

1/2 h x2p 2 xp ¯ ′ i yp ¯ ′ S (t) n T (t) − m W (t) m2 n n,W

y2p 2 S (t) mn2 m,T

h yp

¯ n T (t) −

i xp ¯ 1 W(t) m 2

which implies



+

x2p 2 S (t) m2 n n,W

repro of

d Λm,n (t) = dt



y2p 2 S (t) mn2 m,T

+

−1/2 x2p 2 d S (t) 2 dt m n n,W

y2p 2 S (t) mn2 m,T

+

x2p 2 S (t) m2 n n,W



y2p 2 S (t) mn2 m,T

+



x2p 2 S (t) m2 n n,W

 2 3/2   2  y p 2   y p 2 x2p 2 x2p 2   y p ′ xp ′  d ¯ ¯ (t)     S (t) + S (t) S + S Λ (t) = T (t) − W    2 m,T m,n m mn2 m,T m2 n n,W  dt mn m2 n n,W  n   d  y2 x2p 2  xp 1  yp ¯  p  2 ¯  S + S  T (t) − W(t) − 2 n m dt  mn2 m,T m2 n n,W 

On simplification, we obtain

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 2 3/2  y p 2  x2p 2 d   S (t) + S (t) Λm,n (t) m,T n,W 2 2 dt mn m n    x  xp yp yp p 2 (n − y )     y2 (m − x p )  X X X X   x p   p p  +  .  x t − y t − X Y X(i) − Y( j)   =  p (i) p ( j)     3 n2 2 n3  m m i=1 j=1 i=1 j=1

(16)

The coefficient of t2 in the R.H.S of (16) is 0 while the coefficient of t and the constant term are given by

   yp xp xp yp 2 (n − y ) X   y2 (m − x p ) X   x X X x y p p p p p  y( j) − X(i)   X + Y b p =  (i) ( j)  2 n3 mn j=1 mn i=1 m3 n2 m i=1 j=1

Jou

and

x   yp p X   x p y2 X  x2p y p p       a p =  X(i) − Y( j)   3 2 (m − x p ) + 2 3 (n − y p ) m n m n i=1 j=1

respectively. Hence the theorem follows.

13

✷

*Author Contributions Section

Journal Pre-proof Author Contribution Statement

Jou

rna lP

repro of

Dhrubasish Bhattacharyya: Conceptualization, Methodology, Software, Investigation, Writing - Original Draft. Ruhul Ali Khan: Methodology, Software, Investigation, Writing - Review & Editing. Murari Mitra: Supervision.