Empirical Bayes testing for equivalence

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Journal of Statistical Planning and Inference

Journal of Statistical Planning and Inference 141 (2011) 2670–2681

0378-37

doi:10.1

� Cor

E-m

journal homepage: www.elsevier.com/locate/jspi

Empirical Bayes testing for equivalence

Lee-Shen Chen a,�, Ming-Chung Yang b,c

a Applied Statistics and Information Science, Ming Chuan University, Taoyuan, Taiwanb Graduate Institute of Statistics, National Central University, Chung-Li, Taiwanc Department of Banking and Finance, Kainan University, Taoyuan, Taiwan

a r t i c l e i n f o

Article history:

Received 30 August 2010

Received in revised form

22 January 2011

Accepted 11 February 2011Available online 18 February 2011

Keywords:

Asymptotically optimal

Empirical Bayes test

Equivalence

Regret

58/$ - see front matter & 2011 Elsevier B.V. A

016/j.jspi.2011.02.019

responding author.

ail address: [email protected] (L.-S. Ch

a b s t r a c t

For a fixed point y0 and a positive value c0, this paper studies the problem of testing the

hypotheses H0 : jy�y0jrc0 against H1 : jy�y0j4c0 for the normal mean parameter yusing the empirical Bayes approach. With the accumulated past data, a monotone

empirical Bayes test is constructed by mimicking the behavior of a monotone Bayes

test. Such an empirical Bayes test is shown to be asymptotically optimal and its regret

converges to zero at a rate ðlnnÞ2:5=n where n is the number of past data available, when

the current testing problem is considered. A simulation study is also given, and the

results show that the proposed empirical Bayes procedure has good performance for

small to moderately large sample sizes. Our proposed method can be applied for testing

close to a control problem or testing the therapeutic equivalence of one standard

treatment compared to another in clinical trials.

& 2011 Elsevier B.V. All rights reserved.

1. Introduction

Since Robbins (1956, 1964), the empirical Bayes approach to statistical decision problems has generated considerableinterests among the researchers, and such a procedure has been extensively studied in the literature. For the recentdevelopment of empirical Bayes procedures, see Pensky (2002, 2003), Gupta and Li (2006), Li and Gupta (2002), and Liang(2002, 2006a) and the references cited in these papers. Although there are many works on empirical Bayes procedures,two-tailed tests, which have great applications to practical problems, have unfortunately drawn a little attention inempirical Bayes literatures. We mention a few research work in this direction. Wei (1991) and Liang (1999) have studiedempirical Bayes two-tailed tests for one-parameter discrete exponential family, respectively. Wei (1989), Singh and Wei(2000), and Liang (2006b) have, respectively, studied empirical Bayes two-tailed tests for one-parameter continuousexponential families. Liang (2006c) has studied a simultaneous inspection of variable equivalence for finitepopulations. Chen (2009) has studied on empirical Bayes two-tail tests for double exponential distributions. It is surprisingthat normal distribution is an important distribution model but there is no empirical Bayes work done regarding the two-tailed tests for the normal mean.

Consider a normal distribution with mean y and variance s2, where both y and s2 are unknown. Let y0 be a fixed pointand c0 be a positive value. One is interested in testing the hypotheses H0 : jy�y0jrc0 versus H1 : jy�y0j4c0. Let i, i=0,1,denote an action deciding in favor of Hi. For the parameter y and action i, the loss is defined to be

lðy,iÞ ¼ ð1�iÞ½ðy�y0Þ2�c2

0 �Iðjy�y0j�c0Þþ i½c20�ðy�y0Þ

2�Iðc0�jy�y0jÞ, ð1:1Þ

ll rights reserved.

en).

www.elsevier.com/locate/jspi

dx.doi.org/10.1016/j.jspi.2011.02.019

mailto:[email protected]

dx.doi.org/10.1016/j.jspi.2011.02.019

L.-S. Chen, M.-C. Yang / Journal of Statistical Planning and Inference 141 (2011) 2670–2681 2671

where I(x)=1 if x40, and 0 otherwise. It is assumed that the parameter y is a realization of a random variable Y having anunknown nondegenerate prior distribution G over the parameter space O¼ ð�1,1Þ and s2 is a fixed, unknown parameter.

For making a decision, a sample of size mðmZ2Þ random variables X1, . . . ,Xm is taken from a Nðy,s2Þ distribution. Let

Y ¼ ð1=mÞPm

j ¼ 1 Xj and W ¼Pm

i ¼ 1ðXi�YÞ2. Thus, Y follows a Nðy,t2Þ distribution with probability density f ðyjy,t2Þ ¼

1=ðffiffiffiffiffiffi2pp

tÞexpð�ðy�yÞ2=ð2t2ÞÞ, where t2 ¼ s2=m; W=s2 has a w2ðm�1Þ distribution. Obviously, Y and W are independent andY is a sufficient statistic for the parameter y. Since we are interested in Bayes and empirical Bayes tests regarding theparameter y, it suffices to consider tests based on Y.

Let Y be the sample space of Y. A test d is defined to be a mapping on Y into the interval [0, 1], such that for each y in Y.dðyÞ ¼ Pfaccepting H1jY ¼ yg is the probability of accepting H1 when Y=y is observed. Since y0 is known and Y�y0 has aNðy�y0,t2Þ distribution, without loss of generality, it is assumed that y0 ¼ 0.

Let RðG,dÞ denote the Bayes risk of a test d when G is the true prior distribution. Assume thatRy2 dGðyÞ �m2ðGÞo1. Let

fGðyÞ ¼

Zf ðyjy,t2Þ dGðyÞ : the marginal density of Y ,

HGðyÞ ¼

Zðc2

0�y2Þf ðyjy,t2Þ dGðyÞ ¼ c2

0 fGðyÞ�aGðyÞ,

aGðyÞ ¼

Zy2f ðyjy,t2Þ dGðyÞ,jGðyÞ ¼HGðyÞ=fGðyÞ,

cG ¼

Zðy2�c2

0ÞIðy2�c2

0Þ dGðyÞ: ð1:2Þ

By Fubini’s theorem, a straightforward computation yields that

RðG,dÞ ¼ZYdðyÞHGðyÞ dyþcG ¼

ZYdðyÞfGðyÞjGðyÞ dyþcG: ð1:3Þ

A Bayes test dG, which minimizes the Bayes risks among a class of tests, can be obtained from (1.3) as follows:

dGðyÞ ¼ 1 if HGðyÞo0, and 0 otherwise

¼ 1 if jGðyÞo0, and 0 otherwise: ð1:4Þ

The minimum Bayes risk of this testing problem is

RðG,dGÞ ¼

ZYdGðyÞHGðyÞ dyþcG: ð1:5Þ

Since the prior distribution G is unknown, it is not possible to implement the Bayes test dG for the underlying testingproblem. When a sequence of past data is available, we shall study this testing problem using the empirical Bayesapproach of Robbins (1956, 1964).

The rest of the paper is arranged as follows. In Section 2, we construct an empirical Bayes test d�n. Then, the asymptoticoptimality of d�n is studied in Section 3. It is shown that the regret of d�n converges to zero at a rate ðlnnÞ2:5=n where n is thenumber of past data available when the current testing problem is considered. The detailed proof to derive theconvergence rate of the proposed test can be found in Appendix. In Section 4, a Monte Carlo simulation study is givento illustrate the performance of regret for small to moderately large sample sizes of the proposed test. We also use thepercentage of relative regret reduction as a measure of the performance of the proposed empirical Bayes test for anysample size. The results of the simulation study demonstrate that our proposed empirical Bayes test has goodperformance. Our proposed method can be applied for testing close to a control problem or testing the therapeuticequivalence of one standard treatment compared to another in clinical trials.

2. Construction of an empirical Bayes test

2.1. Empirical Bayes framework

In the empirical Bayes framework, let Xi1, . . . ,Xim be a sample of size m arising from a Nðyi,s2Þ distribution at stage i,where yi is a realization of a random variable Yi. Also, let Yi and Wi be the corresponding (Y,W) based on ðXi1, . . . ,XimÞ. It isassumed that ðYi,Wi,YiÞ, i¼ 1,2, . . . are iid copies of ðY ,W ,YÞ, where ðYi,WiÞ, i¼ 1,2, . . . are observable, but Yi,i¼ 1,2, . . . arenot observable. Let YðnÞ ¼ ðY1, . . . ,YnÞ and WðnÞ ¼ ðW1, . . . ,WnÞ. At stage n+1, ðYi,WiÞ,i¼ 1, . . . ,n, denote the n past data, andðYnþ1,Wnþ1Þ � ðY ,WÞ is the present random observation. Let ynþ1 be a realized value of the present random variable Ynþ1.We are interested in testing H0 : jynþ1jrc0 versus H1 : jynþ1j4c0 using the error loss lðynþ1,iÞ of (1.1) with y0 ¼ 0.

An empirical Bayes test dn is a probability function defined on Y=y and (Y(n),W(n)) such that dnðy,YðnÞ,WðnÞÞ � dnðyÞ isthe probability of accepting H1 : jynþ1j4c0. Let RðG,dnjYðnÞ,WðnÞÞ denote the Bayes risk of an empirical Bayes test dn

conditioning on ðYðnÞ,WðnÞÞ, and RðG,dnÞ � EnRðG,dnjYðnÞ,WðnÞÞ the overall Bayes risk of dn, where the expectation En istaken with respect to (Y(n),W(n)). Since RðG,dGÞ is the minimum Bayes risk, which could be achieved if the prior

L.-S. Chen, M.-C. Yang / Journal of Statistical Planning and Inference 141 (2011) 2670–26812672

distribution G were known and the Bayes test dG had been applied. Thus, RðG,dnjYðnÞ,WðnÞÞ�RðG,dGÞZ0 for all ðYðnÞ,WðnÞÞand for all n. Hence, RðG,dnÞ�RðG,dGÞZ0 for all n. The nonnegative difference RðG,dnÞ�RðG,dGÞ, called as the regret of theempirical Bayes test dn, is often used as a measure of performance of the empirical Bayes test dn; for example, see Johnsand Van Ryzin (1972) and van Houwelingen (1976). An empirical Bayes test dn is said to be asymptotically optimal,relative to the prior distribution G, at a rate en if RðG,dnÞ�RðG,dGÞ ¼OðenÞ, where feng is a sequence of positive, decreasingnumbers such that limn-1en ¼ 0.

2.2. An alternative form of the Bayes test

Throughout the paper, it is assumed that the prior distribution G satisfies the following assumption:

Assumption G. [G1] G is nondegenerate and symmetric about the point y0 ¼ 0.

½G2� jGð0Þ404 limy-1

jGðyÞ:

½G3� m2ðGÞ �

Zy2 dGðyÞo1:

Under Assumption [G1], fG(y), HG(y) and jGðyÞ ¼HGðyÞ=fGðyÞ are even functions. Also, jGðyÞ is strictly decreasing in jyj. SincejGðyÞ is a continuous function, by Assumption [G2], there exists a point aG,0oaGo1, such that

jGðyÞ40 for jyjoaG, jGðaGÞ ¼ 0, jGðyÞo0 for jyj4aG: ð2:1Þ

Combining (1.4) and (2.1), the Bayes test dG can be expressed as

dGðyÞ ¼ 1 if jyj4aG, and 0 if jyjraG: ð2:2Þ

aG is thus called a critical point of the Bayes test dG. From (2.2), dG can be viewed as a monotone test of jyj.

2.3. Empirical Bayes estimation

Since an empirical Bayes test often mimics the behavior of a Bayes test, in order to construct an empirical Bayes test,from (1.4) and (2.2), we need to estimate the function HG(y) and the critical point aG. For this purpose, we give arepresentation of HG(y) on which an empirical Bayes estimator Hn(y) for HG(y) is based.

First note that aGðyÞ ¼ t4f ð2ÞG ðyÞþ2t2yf ð1ÞG ðyÞþðt2þy2ÞfGðyÞ, where f ðiÞG ðyÞ denotes the i-th derivative of fG(y) with respectto y and f ð0ÞG ðyÞ ¼ fGðyÞ. Thus,

HGðyÞ ¼ ðc20�t

2�y2ÞfGðyÞ�2t2yf ð1ÞG ðyÞ�t4f ð2ÞG ðyÞ: ð2:3Þ

Hence, we need to estimate f ðlÞG ðyÞ,l¼ 0,1 and 2.We let cfG

ðtÞ, cNðtÞ and cGðtÞ denote the characteristic functions associated with the probability density fG, a Nð0,t2Þ

distribution and the prior distribution G, respectively. Thus, cNðtÞ ¼ expð�t2t2=2Þ and cfGðtÞ ¼cNðtÞcGðtÞ. By the inversion

formula,

f ðlÞG ðyÞ ¼1

2p

Zð�itÞlexpð�ityÞcfG

ðtÞ dt, l¼ 0,1,2: ð2:4Þ

Consider the kernel function KðxÞ ¼ sinx=ðpxÞ. The Fourier transform of K is cK ðtÞ ¼ I½�1,1�ðtÞ. Also,K ð1ÞðxÞ ¼ ðxcosx�sinxÞ=ðpx2Þ, K ð2ÞðxÞ ¼ ð2sinx�2xcosx�x2sinxÞ=ðpx3Þ. Thus, jK ðlÞðxÞjr1 for l=0,1,2, and

RjK ðlÞðxÞj2

dxrk0o1 for some finite k040. We note that the types of estimators of (2.4) have been used in Liang (2006a)and Chen (2009) for constructing empirical Bayes tests.

Estimation of s2 and t2: Note that W1, . . . ,Wn are iid, Wi follows a s2w2ðm�1Þ distribution. Define s2n ¼WðnÞ=ðnðm�1ÞÞ,

WðnÞ ¼W1þ � � � þWn and t2n ¼ s2

n=m. Thus, nðm�1Þt2n=t2 ¼ nðm�1Þs2

n=s2 � w2ðnðm�1ÞÞ, E½t2n� ¼ t2, Varðt2

nÞ ¼ 2t4=ðnðm�1ÞÞand E½t4

n� ¼ t4½1þ2=ðnðmþ1ÞÞ�.Estimation of f ðlÞG ðyÞ,l¼ 0,1,2: Let b¼ 1=

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4mlnnp

. Define cnðtÞ ¼ ð1=nÞPn

j ¼ 1 expðitYjÞ. Note that cnðtÞ is the empiricalcharacteristic function based on Y(n) and is an unbiased and consistent estimator of cfG

ðtÞ. By mimicking the form of fG(l)

(y)given in (2.4), define

f ðlÞn ðyÞ ¼1

2p

Zð�itÞlexpð�ityÞcnðtÞcK ðbsntÞ dt: ð2:5Þ

Note that fn(l)

(y) can also be expressed as

f ðlÞn ðyÞ ¼1

nðbsnÞlþ1

Xn

j ¼ 1

K ðlÞy�Yj

bsn

� �: ð2:6Þ

Estimation of HGðyÞ: Mimicking the form (2.3), we estimate HG(y) by Hn(y), where

HnðyÞ ¼ ðc20�t

2n�y2ÞfnðyÞ�2t2

nyf ð1Þn ðyÞ�t4nf ð2Þn ðyÞ: ð2:7Þ


Estimation of aG: Define cn ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffilnn=ð4m2Þ4

p. Since cn-1 as n-1, thus, for sufficiently large n, 0raGocn. Since jGðyÞ

and HG(y) have the same sign and jGðyÞZ0 for 0ryraG, and jGðyÞo0 for y4aG, the critical point aG can be expressed as:aG ¼

R cn

0 I½jGðyÞ� dy¼R cn

0 I½HGðyÞ� dy. Thus, we estimate aG by ann, where

a�n ¼

Z cn

0I½HnðyÞ� dy: ð2:8Þ

We have

a�n�aG ¼�

Z aG

0I½�HnðyÞ� dyþ

Z cn

aG

I½HnðyÞ� dy: ð2:9Þ

2.4. Monotone empirical Bayes test d�n

By mimicking the behavior of the Bayes test dG given in (2.2), we propose an empirical Bayes test d�n as follows:

d�nðyÞ ¼ 1 if jyj4a�n, and 0 otherwise: ð2:10Þ

The Bayes risk of d�n is

RðG,d�nÞ ¼ZY

En½d�

nðyÞ�HGðyÞ dyþcG: ð2:11Þ

From (2.10), we see that d�n is a monotone test of jyj.

3. Asymptotic optimality

The empirical Bayes test d�n in (2.10) possesses the following asymptotic optimality.

Theorem 3.1. Suppose the prior distribution G satisfies Assumption G. Then, the empirical Bayes test d�n is asymptotically

optimal, and

RðG,d�nÞ�RðG,dGÞ ¼OððlnnÞ2:5=nÞ:

Proof. Since jGðyÞ is continuous and strictly decreasing in jyj and jGðaGÞ ¼ 0, we let aG1ðnÞoaGoaG2

ðnÞ be points such that

jGðaG1ðnÞÞ ¼ 1=

ffiffiffiffiffiffiffiffinb5

pand jGðaG2

ðnÞÞ ¼ �1=ffiffiffiffiffiffiffiffinb5

p. Note that aG1

ðnÞ-aG and aG2ðnÞ-aG as n-1. Let a1=aG/2 and a2=aG+1.

We consider the case where n is sufficiently large so that a1oaG1ðnÞoaGoaG2

ðnÞoa2ocn. From (1.4)–(1.5)

and (2.10)–(2.11), the regret of the empirical Bayes test d�n can be written as

0rRðG,d�nÞ�RðG,dGÞ ¼ 2En

Z aG

a�n

HGðyÞ dy

" #¼ 2En Ið1�ja�n�aGjÞ

Z aG

a�n

HGðyÞ dy

( )þ2En Iðja�n�aGj�1Þ

Z aG

a�n

HGðyÞ dy

( ):

ð3:1Þ

Since HG(aG)=0, by Taylor series expansion, for some yn between ann

and aG,

En Ið1�ja�n�aGjÞ

Z aG

a�n

HGðyÞ dy

( )¼ En Ið1�ja�n�aGjÞH

ð1ÞG ðy

�Þða�n�aGÞ2=2

n orHð1Þ

�

G ðaGÞEnða�n�aGÞ

2, ð3:2Þ

where Hð1Þ�

G ¼maxfj12Hð1ÞðyÞj; jy�aGjr1g.

Also, by Markov inequality

En Iðja�n�aGj�1Þ

Z aG

a�n

HGðyÞ dy

( )rEnða

�n�aGÞ

2Z 1

0jHGðyÞj dy¼mHG

Enða�n�aGÞ

2, ð3:3Þ

where mHG¼R1

0 jHGðyÞj dyo1 by Lemma A4 in the Appendix. Therefore,

0rRðG,d�nÞ�RðG,dGÞrc0Enða�n�aGÞ

2, ð3:4Þ

where c0 ¼Hð1Þ�

G ðaGÞþmHG. From (2.9)

a�n�aG ¼�

Z a1

0I½�HnðyÞ� dy�

Z aG1ðnÞ

a1

I½�HnðyÞ� dy�

Z aG

aG1ðnÞ

I½�HnðyÞ� dy

þ

Z aG2ðnÞ

aG

I½HnðyÞ� dyþ

Z a2

aG2ðnÞ

I½HnðyÞ� dyþ

Z cn

a2

I½HnðyÞ� dy�X6

i ¼ 1

AiðnÞ:

Hence,

Enða�n�aGÞ

2r6X6

i ¼ 1

En½A2i ðnÞ�: ð3:5Þ


Therefore, it suffices to investigate the asymptotic behaviors of En[Ai2(n)] for each i=1,y,6. The asymptotic properties of

Ai2(n) have been evaluated in Appendix. From the Appendix, we have

En½A21ðnÞ�ra2

G½4g1ðnÞþ3g2ðnÞþg3ðnÞþg4ðnÞþg5ðnÞ�, ð3:6Þ

En½A22ðnÞ�r aG½4g1ðnÞþ3g2ðnÞ�j4

Gða1Þþr2

1ðnÞ

ðnb5Þ2þ

64

c41½nðm�1Þ�2

( )nb5

½jð1Þ� �2, ð3:7Þ

En½A23ðnÞ�r

1

nb5½jð1Þ� �2

, ð3:8Þ

En½A24ðnÞ�r

1

nb5½jð1Þ� �2

, ð3:9Þ

En½A25ðnÞ�r a2½4g1ðnÞþ3g2ðnÞ�j4

Gða2Þþr2

1ðnÞ

ðnb5Þ2þ

64

c41½nðm�1Þ�2

( )nb5

½jð1Þ� �2, ð3:10Þ

En½A26ðnÞ�rc2

n ½4g1ðnÞþ3g2ðnÞþg3ðnÞþg4ðnÞþg5ðnÞ�: ð3:11Þ

Here c1,jð1Þ� are positive constants, r1(n), giðnÞ,i¼ 1,2,3,4,5 are functions of n defined in the Appendix. From the Appendix,

we obtain giðnÞ ¼ oðn�2Þ,i¼ 1,2,3,4,5 and r1(n) is a bounded function of n. From (3.1)–(3.11), we conclude that

RðG,d�nÞ�RðG,dGÞ ¼O1

nb5

� �¼O

ðlnnÞ2:5

n

!: &

Table 1

The regrets of empirical Bayes test procedure d�n for a2 ¼ 2 and 4 with various values of s2 and c20 ¼ 0:1þs2=3.

n a2 ¼ 2 4

s2 =0.1 0.2 0.1 0.2

R RR R RR R RR R RR

11 0.002607 (0.000190) 44.6694 0.002385 (0.000183) 19.1452 0.001853 (0.000162) 43.7302 0.001961 (0.000166) 21.4542

21 0.001657 (0.000141) 28.3941 0.001598 (0.000139) 12.8256 0.001475 (0.000137) 34.8216 0.001291 (0.000130) 14.1246

31 0.001360 (0.000126) 23.3008 0.001297 (0.000124) 10.4133 0.000879 (0.000101) 20.7361 0.001066 (0.000110) 11.6635

41 0.000930 (0.000099) 15.9375 0.000962 (0.000100) 7.7243 0.000664 (0.000082) 15.6756 0.000608 (0.000081) 6.6522

51 0.000722 (0.000087) 12.3671 0.000624 (0.000078) 5.0061 0.000661 (0.000086) 15.6065 0.000664 (0.000087) 7.2680

61 0.000592 (0.000073) 10.1417 0.000681 (0.000080) 5.4695 0.000513 (0.000070) 12.1059 0.000631 (0.000081) 6.9059

71 0.000520 (0.000068) 8.9157 0.000625 (0.000078) 5.0173 0.000438 (0.000065) 10.3392 0.000459 (0.000069) 5.0215

81 0.000522 (0.000069) 8.9456 0.000458 (0.000065) 3.6725 0.000552 (0.000074) 13.0237 0.000397 (0.000060) 4.3401

91 0.000335 (0.000056) 5.7359 0.000242 (0.000046) 1.9438 0.000243 (0.000045) 5.7366 0.000304 (0.000052) 3.3253

101 0.000440 (0.000060) 7.5406 0.000381 (0.000057) 3.0601 0.000368 (0.000056) 8.6854 0.000225 (0.000040) 2.4642

111 0.000373 (0.000052) 6.3965 0.000288 (0.000046) 2.3086 0.000308 (0.000048) 7.2650 0.000248 (0.000042) 2.7109

121 0.000282 (0.000045) 4.8390 0.000197 (0.000040) 1.5774 0.000264 (0.000046) 6.2368 0.000266 (0.000046) 2.9117

131 0.000227 (0.000038) 3.8970 0.000139 (0.000028) 1.1167 0.000286 (0.000047) 6.7609 0.000212 (0.000041) 2.3159

141 0.000235 (0.000038) 4.0302 0.000205 (0.000038) 1.6493 0.000271 (0.000045) 6.4067 0.000146 (0.000031) 1.5958

151 0.000205 (0.000036) 3.5061 0.000115 (0.000027) 0.9194 0.000195 (0.000038) 4.5933 0.000141 (0.000031) 1.5472

161 0.000143 (0.000028) 2.4548 0.000170 (0.000031) 1.3644 0.000176 (0.000034) 4.1536 0.000168 (0.000034) 1.8432

171 0.000156 (0.000030) 2.6746 0.000172 (0.000032) 1.3818 0.000159 (0.000030) 3.7642 0.000148 (0.000033) 1.6220

181 0.000148 (0.000026) 2.5362 0.000111 (0.000022) 0.8900 0.000126 (0.000026) 2.9665 0.000178 (0.000035) 1.9513

191 0.000109 (0.000023) 1.8596 0.000149 (0.000030) 1.1947 0.000102 (0.000027) 2.4055 0.000083 (0.000021) 0.9075

201 0.000085 (0.000020) 1.4620 0.000077 (0.000017) 0.6205 0.000076 (0.000019) 1.7984 0.000119 (0.000030) 1.3004

211 0.000092 (0.000020) 1.5837 0.000143 (0.000028) 1.1500 0.000183 (0.000033) 4.3211 0.000112 (0.000028) 1.2225

221 0.000086 (0.000020) 1.4777 0.000114 (0.000026) 0.9180 0.000097 (0.000024) 2.2936 0.000110 (0.000024) 1.2007

231 0.000094 (0.000022) 1.6140 0.000060 (0.000016) 0.4850 0.000060 (0.000018) 1.4169 0.000067 (0.000017) 0.7377

241 0.000060 (0.000015) 1.0342 0.000130 (0.000027) 1.0442 0.000113 (0.000025) 2.6704 0.000081 (0.000023) 0.8858

251 0.000073 (0.000017) 1.2553 0.000114 (0.000026) 0.9147 0.000060 (0.000015) 1.4070 0.000086 (0.000022) 0.9431

261 0.000074 (0.000016) 1.2762 0.000082 (0.000019) 0.6566 0.000069 (0.000020) 1.6245 0.000083 (0.000021) 0.9076

271 0.000079 (0.000017) 1.3496 0.000082 (0.000020) 0.6600 0.000022 (0.000010) 0.5166 0.000046 (0.000015) 0.5024

281 0.000086 (0.000020) 1.4815 0.000074 (0.000019) 0.5915 0.000059 (0.000018) 1.4029 0.000077 (0.000020) 0.8410

291 0.000047 (0.000015) 0.8060 0.000118 (0.000026) 0.9509 0.000053 (0.000016) 1.2617 0.000069 (0.000018) 0.7571

301 0.000050 (0.000017) 0.8496 0.000120 (0.000026) 0.9615 0.000059 (0.000018) 1.3852 0.000077 (0.000022) 0.8411

Note: Estimated standard deviations of regrets are in parentheses.


4. Simulation study

In this section, we design a Monte Carlo simulation to investigate the performance of small to moderately large samplesizes for the proposed empirical Bayes test procedure with the conjugate normal prior. For the (n+1)-st component testingproblem, let X(n) denote the historical data, and Xnþ1 ¼ ðXnþ1,1, . . . ,Xnþ1,mÞ denote the present random observations. Fori=1,y,n+1, (Yi,Wi) is defined, based on (Xi1,y,Xim), as Yi ¼

Pmj ¼ 1 Xij=m and Wi ¼

Pmj ¼ 1ðXij�YiÞ

2. From (2.8), (2.10) and(3.1), we define In � I½a�no jYnþ1joaG�þ I½aGo jYnþ1joa�n� and SRðnÞ � jjGðYnþ1Þj � In=2. Then,

E½SRðnÞ� ¼ E

Z aG

a�n

jGðyÞfGðyÞ dy

" #¼ RðG,d�nÞ�RðG,dGÞ,

where dG is the Bayes test based only on the present observation. Obviously, SR(n) is an unbiased estimator of the regretRðG,d�nÞ�RðG,dGÞ.

Let N and m be fixed positive integers, and let N1=N+1. The simulation scheme is described as follows.

(1)

Tabl

The

n

Note

Generate Yi,i¼ 1, . . . ,N1 from the normal distribution Nð0,a2Þ where a240 is the given value.
(2) At the stage i, given Yi ¼ yi, generate a sample of size m, Xi1, . . . ,Xim from the normal distribution Nðyi,s2Þ where s2 is
the given positive value. Let xi1,y,xim denote the observations at the stage i, and yi ¼Pm

j ¼ 1 xij=m and wi ¼Pmj ¼ 1ðxij�yiÞ

2, i=1,y,N1. Here, (y1,y,yn) and (w1,y,wn) denote the past data, and let (yn + 1,wn + 1) denote the presentobservation. Note that from (2.3), one can derive jGðyÞ ¼HGðyÞ=fGðyÞ ¼ c2

0�t2�y2 þðt4y2þ2t2a2y2þt4a2Þ=ðt2þa2Þ2,

and aG is the root of jGðyÞ. Next, do the following computation for each n=10 (10) 300.
(3) Compute Hn(y) in (2.7). (4) Compute an
nin (2.8).

(5)
Compute in ¼ I½a�no jynþ1joaG�þ I½aGo jynþ1joa�n� and SRðnÞ ¼ jjGðynþ1Þj � in=2.
e 2

regrets of empirical Bayes test procedure d�n for a2 ¼ 6 and 8 with various values of s2 and c20 ¼ 0:1þs2=3.

a2 ¼ 6 8

s2 =0.1 0.2 0.1 0.2

R RR R RR R RR R RR

11 0.001388 (0.000143) 39.0594 0.001442 (0.000141) 19.0655 0.001554 (0.000157) 51.9532 0.001332 (0.000139) 20.5441

21 0.001035 (0.000113) 29.1281 0.001165 (0.000127) 15.4013 0.000809 (0.000101) 27.0421 0.000878 (0.000108) 13.5438

31 0.000758 (0.000094) 21.3189 0.000741 (0.000094) 9.7946 0.000730 (0.000095) 24.4152 0.000633 (0.000087) 9.7659

41 0.000636 (0.000085) 17.8794 0.000650 (0.000085) 8.5878 0.000598 (0.000084) 19.9840 0.000719 (0.000093) 11.0869

51 0.000622 (0.000087) 17.4989 0.000687 (0.000091) 9.0872 0.000571 (0.000082) 19.0725 0.000612 (0.000085) 9.4474

61 0.000527 (0.000075) 14.8180 0.000488 (0.000069) 6.4515 0.000368 (0.000059) 12.3021 0.000491 (0.000071) 7.5807

71 0.000517 (0.000074) 14.5381 0.000545 (0.000074) 7.2064 0.000418 (0.000062) 13.9648 0.000381 (0.000061) 5.8750

81 0.000453 (0.000069) 12.7481 0.000557 (0.000077) 7.3631 0.000492 (0.000070) 16.4297 0.000499 (0.000069) 7.7007

91 0.000335 (0.000057) 9.4255 0.000276 (0.000050) 3.6544 0.000301 (0.000052) 10.0459 0.000313 (0.000055) 4.8279

101 0.000380 (0.000058) 10.7027 0.000333 (0.000054) 4.4030 0.000405 (0.000062) 13.5497 0.000302 (0.000053) 4.6583

111 0.000366 (0.000056) 10.2973 0.000199(0.000040) 2.6250 0.000351 (0.000055) 11.7346 0.000254 (0.000047) 3.9172

121 0.000268 (0.000047) 7.5325 0.000251 (0.000045) 3.3158 0.000210 (0.000042) 7.0087 0.000246 (0.000045) 3.7880

131 0.000235 (0.000041) 6.5991 0.000250 (0.000046) 3.3073 0.000235 (0.000042) 7.8501 0.000180 (0.000038) 2.7789

141 0.000249 (0.000047) 7.0135 0.000190 (0.000040) 2.5165 0.000204 (0.000040) 6.8336 0.000206 (0.000042) 3.1828

151 0.000293 (0.000051) 8.2389 0.000130 (0.000031) 1.7185 0.000226 (0.000042) 7.5696 0.000175 (0.000037) 2.7009

161 0.000219 (0.000040) 6.1700 0.000138 (0.000031) 1.8218 0.000160 (0.000034) 5.3560 0.000175 (0.000036) 2.6990

171 0.000176 (0.000034) 4.9501 0.000152 (0.000033) 2.0072 0.000168 (0.000035) 5.6021 0.000147 (0.000036) 2.2719

181 0.000142 (0.000030) 3.9904 0.000136 (0.000031) 1.7917 0.000131 (0.000031) 4.3670 0.000124 (0.000028) 1.9171

191 0.000131 (0.000030) 3.6969 0.000124 (0.000029) 1.6427 0.000148 (0.000034) 4.9620 0.000134 (0.000031) 2.0725

201 0.000108 (0.000029) 3.0475 0.000105 (0.000026) 1.3875 0.000109 (0.000028) 3.6306 0.000101 (0.000027) 1.5590

211 0.000109 (0.000026) 3.0743 0.000115 (0.000027) 1.5168 0.000061 (0.000020) 2.0390 0.000119 (0.000030) 1.8296

221 0.000119 (0.000028) 3.3426 0.000120 (0.000027) 1.5929 0.000087 (0.000025) 2.9235 0.000080 (0.000025) 1.2306

231 0.000066 (0.000020) 1.8625 0.000105 (0.000030) 1.3827 0.000083 (0.000021) 2.7735 0.000056 (0.000020) 0.8708

241 0.000077 (0.000025) 2.1561 0.000065 (0.000018) 0.8632 0.000122 (0.000029) 4.0864 0.000051 (0.000019) 0.7844

251 0.000050 (0.000014) 1.4056 0.000081 (0.000022) 1.0758 0.000076 (0.000021) 2.5369 0.000070 (0.000020) 1.0729

261 0.000051 (0.000017) 1.4468 0.000079 (0.000022) 1.0408 0.000083 (0.000021) 2.7593 0.000056 (0.000017) 0.8661

271 0.000060 (0.000016) 1.6846 0.000055 (0.000018) 0.7270 0.000127 (0.000030) 4.2323 0.000070 (0.000021) 1.0751

281 0.000035 (0.000012) 0.9725 0.000069 (0.000020) 0.9133 0.000085 (0.000024) 2.8466 0.000070 (0.000022) 1.0786

291 0.000099 (0.000023) 2.7938 0.000072 (0.000018) 0.9541 0.000065 (0.000018) 2.1706 0.000052 (0.000019) 0.8014

301 0.000081 (0.000020) 2.2864 0.000082 (0.000023) 1.0811 0.000057 (0.000018) 1.9169 0.000100 (0.000023) 1.5496

: Estimated standard deviations of regrets are in parentheses.


(6)
Repeat step (1) through step (5) 2000 times. Then take its average denoted by Dnþ1, which is used as estimator of thedifference RðG,d�nÞ�RðG,dGÞ.
The simulation results of the proposed empirical Bayes test procedure for the given m=3, s2 ¼ 0:1,0:2 andc2

0 ¼ 0:1þs2=m, are shown in Table 1 with a2 ¼ 2,4 and in Table 2 with a2 ¼ 6,8, respectively. The estimated standarddeviations of regrets are also reported in Tables 1 and 2. We find that the regret(R) of empirical Bayes test decreasesoscillately as n increases for both tables. Next, we consider the percentage relative regret(RR) of the empirical Bayes testprocedure d�n, which is defined by RRðd�nÞ ¼ 100½RðG,d�nÞ�RðG,dGÞ�=RðG,dGÞ, to evaluate the proposed testing procedure forsmall to moderately large sample sizes. Here the minimum Bayes risk RðG,dGÞ can be obtained from (1.5), and we interpretRR as the percentage regret reduction of d�n in terms of the minimum Bayes risk. The smaller the value of RR is, the betterthe performance of d�n is. These RR’s are reported in Tables 1 and 2. We find that these RR’s decrease oscillately as n

increases. For the small sample size n=11 at different values of a in Table 1, we observe that the RR’s are around 44 percentwhen s2 ¼ 0:1 and around 20 percent when s2 ¼ 0:2. Further, we find d�n always has better performance with largervariance under the same sample size. This phenomenon also occurs in Table 2. For s2 ¼ 0:2 in both tables, the RR’s arearound 20 percent when n=11 and around 10 percent when n=31. Thus, the risk of d�n is relatively close to the minimumBayes risk even for the sample size n=31. Overall, the proposed empirical Bayes test procedure has good performance forsmall to moderately large sample sizes in the simulation study.

Acknowledgments

We would like to thank two referees for the careful review and valuable suggestions which lead to the improvement of thepresentation of the paper. This research was supported by the project NSC 98-2118-M-130-001 of the National Science Councilof ROC.

Appendix A

In order to investigate the asymptotic behavior of En[A2i (n)], we have to study certain properties pertaining to the

estimator Hn(y). From (2.6)–(2.7), Hn(y) can be expressed as

HnðyÞ ¼1

n

Xn

j ¼ 1

Vðy,Yj,tn,bÞ, ðA:1Þ

where

Vðy,Yj,tn,bÞ ¼ðc2

0�y2�t2nÞ

bsnK

y�Yj

bsn

� ��

2yt2n

ðbsnÞ2

K ð1Þy�Yj

bsn

� ��

t4n

ðbsnÞ3

K ð2Þy�Yj

bsn

� �:

Conditioning on tn,Vðy,Yj,tn,bÞ are iid. Since jKðlÞðyÞjr1 for all y and l=0,1,2,

jVðy,Yj,tn,bÞjrðbsnÞ

2ðc2

0þy2þt2nÞþ2b

ffiffiffiffiffimpjyjt3

nþt4n

ðbsnÞ3

¼pðy,tn,bÞ

2ðbsnÞ3

,

where pðy,tn,bÞ ¼ 2fðbsnÞ2½c2

0þy2þt2n�þ2b

ffiffiffiffiffimpjyjt3

nþt4ng. Thus,

jVðy,Yj,tn,bÞ�En½Vðy,Yj,tn,bÞjtn�jrpðy,tn,bÞ

ðbtnÞ3: ðA:2Þ

Let

HnðyjtÞ ¼ ðc20�y2�t2ÞfnðyÞ�2yt2f ð1Þn ðyÞ�t

4f ð2Þn ðyÞ,

p1ðy,tnÞ ¼ ½fnðyÞ�fGðyÞ�þ2y½f ð1Þn ðyÞ�f ð1ÞG ðyÞ�þ½t2nþt

2�½f ð2Þn ðyÞ�f ð2ÞG ðyÞ�,

p2ðy,tnÞ ¼ fGðyÞþ2yf ð1ÞG ðyÞþ½t2nþt

2�f ð2ÞG ðyÞ: ðA:3Þ

Thus,

HnðyÞ�HGðyÞ ¼ fHnðyÞ�En½HnðyjtÞjtn�gþEn½HnðyjtÞ�HGðyÞjtn��ðt2n�t

2ÞEn½p1ðy,tnÞjtn��ðt2n�t

2Þp2ðy,tnÞ: ðA:4Þ

The following Lemmas A.1 and A.2 can be obtained through straightforward computation and the details are omitted.

Lemma A1.

(a)
jEn½fnðyÞ�fGðyÞjtn�jrbsn=ðpt2Þexpð�2s2lnn=s2nÞ.
(b)
jEn½fð1Þn ðyÞ�f ð1ÞG ðyÞjtn�jrexpð�2s2lnn=s2
nÞ=ðpt2Þ.


(c)

(a)

(b)

(c)

(d)

(a)

(b)

jEn½fð2Þn ðyÞ�f ð2ÞG ðyÞjtn�jr

1

pt2exp

�2s2lnn

s2n

� �1

bsnþ

bsn

t2

� �:

(d)
jEn½HnðyjtÞ�HGðyÞjtn�jrB1ðy,sn,bÞ, where
B1ðy,sn,bÞ ¼ exp�2s2lnn

s2n

� �1

pt2bsnðc

20þy2þt2

nÞþ2jyjt2nþt

4n

� �:

(e)
jEn½p1ðy,tnÞjtn�jrB2ðy,sn,bÞ, where
B2ðy,sn,bÞ ¼1

pt2exp

�2s2lnn

s2n

� �bsnþ2jyjþðt2

nþt2Þ

1

bsnþ

bsn

t2

� �� :

(f)

jp2ðy,tnÞjr1

t þ2jyj½jyjþm1ðGÞ�

t3þðt2

nþt2Þ

t2þ2y2þ2m2ðGÞ

t5

� �� B3ðy,snÞ,

where miðGÞ ¼Rjyji dGðyÞ, i¼ 1,2.

Lemma A2.

Var Ky�Yj

bsn

� �tn

� �rk0bsn=t:

Var K ð1Þy�Yj

bsn

� �tn

� �rk0bsn=t:

Var K ð2Þy�Yj

bsn

� �tn

� �rk0bsn=t:

Var Vðy,Yj,tn,bjtnÞ �

rQ ðy,sn,bÞ

ðbsnÞ5

:

where Q ðy,sn,bÞ ¼ ½3ðc20þy2þt2

nÞ2k0ðbsnÞ

4þ12y2t4

nk0ðbsnÞ2þ3t8

nk0�=t.

Lemma A3. Let gG ¼R

expð�y2=ð2t2ÞÞdGðyÞ. Suppose that G is symmetric about the point 0. Then,

(a)
fGðyÞZgGf ðyj0,t2Þ for all y. ffiffiffiffiffiffip ffiffiffiffiffiffiffiffip (b) For sufficiently large n, fGðyÞZgG=ð 2ptÞexpð� lnn=ð8s2ÞÞ � qðnÞ for jyjrcn.
Proof. Since G is symmetric about 0, thus, fGðyÞ=f ðyj0,t2Þ ¼R

expðyy=t2Þexpð�y2=ð2t2ÞÞ dGðyÞ is strictly increasing in y for

yZ0. So, fGðyÞ=f ðyj0,t2ÞZ fGð0Þ=f ð0j0,t2Þ ¼R

expð�y2=ð2t2ÞÞ dGðyÞ ¼ gG. Hence, fGðyÞZgGf ðyj0,t2Þ. Part (b) follows directly

from the definition of the point cn and the function f ðyj0,t2Þ. &

Lemma A4. mHG¼R1

0 jHGðyÞj dyo1, if m2ðGÞ ¼Ry2 dGðyÞo1.

Proof.Z 10jHGðyÞj dy¼

Z 10

Zðc2

0�y2Þf ðyjy,t2Þ dGðyÞ

dyrc2

0

ZY

Z 10

f ðyjy,t2Þ dy dGðyÞþZYy2Z 1

0f ðyjy,t2Þ dy dGðyÞ

rc20þ

Zy2 dGðyÞ ¼ c2

0þm2ðGÞo1: &

The following lemma is from Liang (1997).

Lemma A5.

Pt2

n

t2�14c

� �rexp �

nðm�1Þ

2½c�lnð1þcÞ�

� for c40:

Pt2

n

t2�1oc

� � ¼ 0 if cr�1,

rexp �nðm�1Þ

2½c�lnð1þcÞ�

� for �1oco0:

8><>:

Let jð1Þ� ¼ inff½�jð1ÞG ðyÞ�; a1ryra2g. Since jð1ÞG ðyÞo0 and is continuous for all y40, thus, jð1Þ� 40.


Lemma A6.

(a)

(b)

0oaG�aG1ðnÞr1

jð1Þ�ffiffiffiffiffiffiffiffinb5

p :

0oaG2ðnÞ�aGr1

jð1Þ�ffiffiffiffiffiffiffiffinb5

p :

Proof. We provide proof for part (a) only. Note that jGðyÞ is decreasing in y for y40 and jGðaGÞ ¼ 0. By the definition of

aG1(n) and mean-value theorem, �1=ffiffiffiffiffiffiffiffinb5

p¼jGðaGÞ�jGðaG1ðnÞÞ ¼ ðaG�aG1ðnÞÞjð1ÞG ða

�Þ for some immediate value an

between aG1(n) and aG. Thus, 0oaG�aG1ðnÞ ¼�1=ðjð1ÞG ða�Þ


pÞr1=ðjð1Þ�


pÞ. &

Lemma A7. For aG2ðnÞryrcn,

jHGðyÞjZrðnÞ �f�ffiffiffiffiffiffiffiffinb5

p ,gGffiffiffiffiffiffi2pp

texp �

ffiffiffiffiffiffiffiffilnnp

8s2

!jjGða2Þj

!Z

1ffiffiffinp for sufficiently large n:

where f� ¼minffGðyÞj0ryra2g.

Proof. Note that for aG2ðnÞryrcn,jGðyÞo0, and is strictly decreasing in y. Thus as aG2ðnÞryra2, jHGðyÞj ¼ fGðyÞjjGðyÞj

Z f�jjðaG2ðnÞj ¼ f�=ffiffiffiffiffiffiffiffinb5

p. For a2ryrcn, by Lemma A3, jHGðyÞjZ fGðyÞjjGða2ÞjZgG=ð

ffiffiffiffiffiffi2pp

tÞexpð�ffiffiffiffiffiffiffiffilnnp

=ð8s2ÞÞjjGða2Þj.

Hence, jHGðyÞjZrðnÞZ1=ffiffiffinp

for sufficiently large n. &

For aG2ðnÞryrcn by (A.4),

PfHnðyÞ40g ¼ PfHnðyÞ�HGðyÞ4�HGðyÞgrP HnðyÞ�En½HnðyÞjtn�4�14HGðyÞ

� �þP En½HnðyjtÞ�HGðyÞjtn�4�1

4HGðyÞ� �

þP ðt2n�t

2ÞEn½p1ðy,tnÞjtn�o14HGðyÞ

� �þP ðt2

n�t2Þp2ðy,tnÞo1

4HGðyÞ� �

� Iðy,nÞþ IIðy,nÞþ IIIðy,nÞþ IVðy,nÞ, ðA:5Þ

Iðy,nÞrPfs2n42s2gþP s2

no12s

2� �

þP HnðyÞ�En½HnðyÞjtn�4�14 HGðyÞ,

12s

2rs2nr2s2

� �� I1ðy,nÞþ I2ðy,nÞþ I3ðy,nÞ:

By Lemma A5, we have

I1ðy,nÞ ¼ Ps2

n

s2�141

� rexp �

nðm�1Þ

2ð1�ln2Þ

� � g1ðnÞ, ðA:6Þ

I2ðy,nÞ ¼ Ps2

n

s2�1o�

1

2

� rexp �

nðm�1Þ

2ðln2�1=2Þ

� � g2ðnÞ, ðA:7Þ

and

I3ðy,nÞ ¼ En P HnðyÞ�En½HnðyÞjtn�4�14HGðyÞjtn

� �I 1

2s2rs2

nr2s2 ��

:

By (A.2), Lemma A2(d) and Bernstein inequality,

P HnðyÞ�En½HnðyÞjtn�4�14HGðyÞjtn

� �

rexp �n½HGðyÞ=4�2=2

VarðVðy,Yi,tn,bÞjtnÞþpðy,tn,bÞ

3ðbsnÞ3�jHGðyÞj

4

8>>><>>>:

9>>>=>>>;rexp �

3nðbsnÞ5f 2

G ðyÞj2GðyÞ

8½12Q ðy,sn,bÞþb2s2npðy,tn,bÞfGðyÞjjGðyÞj�

( ):

Note that both Q ðy,sn,bÞ and pðy,tn,bÞ are increasing in sn. Therefore,

I3ðy,nÞrexp�3nb5s5f 2

G ðyÞj2GðyÞ

32ffiffiffi2p½12Q ðy,

ffiffiffi2p

s,bÞþ2b2s2pðy,ffiffiffi2p

t,bÞfGðyÞjjGðyÞj�

( )� I31ðy,nÞ: ðA:8Þ

Combining the preceding results yields

Iðy,nÞrg1ðnÞþg2ðnÞþ I31ðy,nÞ: ðA:9Þ

In (A.5), we have

IIðy,nÞ ¼ P En½HnðyjtÞ�HGðyÞjtn�4�14HGðyÞ

� �rPfs2

n42s2gþP En½HnðyjtÞ�HGðyÞjtn�4�14HGðyÞ,s2

n r2s2� �

� II1ðy,nÞþ II2ðy,nÞ,

ðA:10Þ

where II1ðy,nÞrg1ðnÞ.


For aG2ðnÞryrcn, by Lemma A7, j 14 HGðyÞjZ1=ffiffiffinp

for sufficiently large n. On the event that s2n r2s2, by Lemma A1(d),

jEn½HnðyjtÞ�HGðyÞjtn�jr1

npt2½ffiffiffi2p

bsðc20þc2

nþ2t2Þþ4cns2þ4s4�o1ffiffiffinp :

Thus, II2ðy,nÞ ¼ 0 for sufficiently large n.Combining the preceding results leads to

IIðy,nÞrg1ðnÞ for sufficiently large n: ðA:11Þ

From Lemma A1(e),

IIIðy,nÞrP jt2n�t

2jB2ðy,sn,bÞ4�14HGðyÞ

� �rPft2

n42t2gþPft2no

12t

2g

þP ðt2n�t

2ÞB2ðy,sn,bÞ4�14HGðyÞ,t2rt2

nr2t2� �

þP ðt2n�t

2ÞB2ðy,sn,bÞo14 HGðyÞ,

12t

2rt2nrt2

� �� III1ðy,nÞþ III2ðy,nÞþ III3ðy,nÞþ III4ðy,nÞ,

where III1ðy,nÞrg1ðnÞ,III2ðy,nÞrg2ðnÞ, and by a discussion analogous to the term II2(y,n), we have, for sufficiently large n,III3(y,n)=0 and III4(y,n)=0. Therefore,

IIIðy,nÞrg1ðnÞþg2ðnÞ for sufficiently large n: ðA:12Þ

From Lemma A1(f),

IVðy,nÞrP jt2n�t

2jB3ðy,snÞ4�14HGðyÞ

� �rPft2

n42t2gþPft2no

12t

2gþP ðt2n�t

2ÞB3ðy,snÞ4�14HGðyÞ,t2rt2

nr2t2� �

þP ðt2n�t

2ÞB3ðy,snÞo14 HGðyÞ,

12t

2rt2nrt2

� �� IV1ðy,nÞþ IV2ðy,nÞþ IV3ðy,nÞþ IV4ðy,nÞ,

where IV1ðy,nÞrg1ðnÞ and IV2ðy,nÞrg2ðnÞ. Since B3ðy,snÞ is increasing in jyj and sn, thus,

IV3ðy,nÞ ¼ P t2n�t

24�HGðyÞ

4B3ðy,snÞ,t2rt2

nr2t2

� rP

t2n

t2�14

�HGðyÞ

4t2B3ðy,ffiffiffi2p

sÞ,0r

t2n

t2�1r1

( ),

which is equal to 0 if �HGðyÞ=ð4t2B3ðy,ffiffiffi2p

sÞÞ41. Without loss of generality, we may assume that0o�HGðyÞ=ð4t2B3ðy,

ffiffiffi2p

sÞÞr1. Thus, by the preceding inequality and Lemma A5,

IV3ðy,nÞrexp �nðm�1Þ

2

�HGðyÞ


sÞln �1

HGðyÞ

4t2B3ðy,ffiffiffi2p sÞ

!" #( )� IV31ðy,nÞ,

IV4ðy,nÞrPt2

n

t2�1o

HGðyÞ


sÞ,�1

2ot2

n

t2�1o0

( ),

which is equal to 0, if HGðyÞ=ð4t2B3ðy,ffiffiffi2p

sÞÞo� 12. Without loss of generality, we may assume that

� 12 rHGðyÞ=ð4t2B3ðy,

ffiffiffi2p

sÞÞo0. Hence, by Lemma A5 and the preceding inequality,


2

HGðyÞ


sÞ�ln 1þ

HGðyÞ


sÞ

!" #( )� IV41ðy,nÞ:

Combining the preceding results, we obtain

IV4ðy,nÞrg1ðnÞþg2ðnÞþ IV31ðy,nÞþ IV41ðy,nÞ: ðA:13Þ

Now, combining the results of (A.5), (A.9)–(A.14), we obtain for sufficiently large n,

PfHnðyÞ40gr4g1ðnÞþ3g2ðnÞþ I31ðy,nÞþ IV31ðy,nÞþ IV41ðy,nÞ: ðA:14Þ

Proof of (3.11). A6ðnÞ ¼R cn

a2I½HnðyÞ� dyrcn. By using (A.14), we obtain

En½A26ðnÞ�rcn

Z cn

a2

PfHnðyÞ40g dyrc2n ½4g1ðnÞþ3g2ðnÞ�þcn

Z cn

a2

½I31ðy,nÞþ IV31ðy,nÞþ IV41ðy,nÞ� dy: ðA:15Þ

For a2ryrcn, j2GðyÞZj2

Gða2ÞZj2Gða1,a2Þ, where jGða1,a2Þ ¼minfjjGðaiÞj; i¼ 1,2g, f 2

G ðyÞZg2G=ð2pt2Þexpð�


=ð4s2ÞÞ,

Q ðy,ffiffiffi2p

s,bÞrQ ðcn,ffiffiffi2p

s,bÞ, pðy,ffiffiffi2p

s,bÞrpðcn,ffiffiffi2p

s,bÞ. Substituting these inequalities into I31ðy,nÞ, We obtain

I31ðy,nÞrexp �

3nb5s5g2G=ð2pt2Þexp �


4s2

!j2

Gða1,a2Þ

32ffiffiffi2p½12Q ðcn,

ffiffiffi2p

s,bÞþ2b2s2pðcn,ffiffiffi2p

t,bÞjGða1,a2Þ�

8>>>><>>>>:

9>>>>=>>>>;� g3ðnÞ:


Note that pðcn,ffiffiffi2p

t,bÞ-8t4 and Q ðcn,ffiffiffi2p

s,bÞ-48k0t7 as n-1. Thus, g3ðnÞ ¼ oðn�2Þ. Therefore, we have

cn

Z cn

an

I31ðy,nÞ dyrg3ðnÞc2n ¼ oðn�1Þ: ðA:16Þ

From Lemma A7 �HGðyÞZqðnÞjjGða2Þj for a2ryrcn. Also, B3ðy,ffiffiffi2p

sÞ is increasing in jyj. Thus, on a2ryrcn, �HGðyÞ=

ð4t2B3ðy,ffiffiffi2p

sÞÞZ�qðnÞjGða2Þ=ð4t2B3ðcn,ffiffiffi2p

sÞÞ. Note that qðnÞ=B3ðcn,ffiffiffi2p

sÞ � qðnÞ=c2n � ð1=

ffiffiffiffiffiffiffilnnp

Þexpð�ffiffiffiffiffiffiffilnnp

=ð8s2ÞÞZn�1=4 for

sufficiently large n. For t40, expð�nðm�1Þ=2½t�lnð1þtÞ�Þ is decreasing in t. Thus,


2

�qðnÞjGða2Þ

4t2B3ðcn,ffiffiffi2p

sÞ�ln 1�

qðnÞjGða2Þ


sÞ

!" #( )� g4ðnÞ,

where g4ðnÞ ¼ oðn�2Þ. Hence,

cn

Z cn

an

IV31ðy,nÞ dyrc2ng4ðnÞ ¼ oðn�1Þ: ðA:17Þ

Also note that expð�nðm�1Þ=2½t�lnð1þtÞ�Þ is increasing in t for �1oto0. Thus,


2

qðnÞjGða2Þ


sÞ�ln 1þ

qðnÞjGða2Þ


sÞ

!" #( )� g5ðnÞ,

where g5ðnÞ ¼ oðn�2Þ. Thus,

cn

Z cn

an

IV41ðy,nÞ dyrc2ng5ðnÞ ¼ oðn�1Þ: ðA:18Þ

Combining (A.15)–(A.18) yields En½A26ðnÞ�rc2

n ½4g1ðnÞþ3g2ðnÞþg3ðnÞþg4ðnÞþg5ðnÞ�. &

Proof of (3.10). By Holder inequality,

A5ðnÞrZ a2

aG2ðnÞ

I½HnðyÞ�

j3GðyÞj

ð1ÞG ðyÞ

dy

" #1=2

�

Z a2

aG2ðnÞI½HnðyÞ�j3

GðyÞjð1ÞG ðyÞ dy

� �1=2

,

where

Z a2

aG2ðnÞ

I½HnðyÞ�

j3GðyÞj

ð1ÞG ðyÞ

dyrZ a2

aG2ðnÞ

jð1ÞG ðyÞ

j3GðyÞ½j

ð1ÞG ðyÞ�

2dyr

1

½jð1Þ�G �2

Z a2

aG2ðnÞ

jð1ÞG ðyÞ

j3GðyÞ

dyr1

½jð1Þ�G �2j2

GðaG2ðnÞÞ¼

nb5

½jð1Þ�G �2:

Hence,

En½A25ðnÞ�r

nb5

½jð1Þ�G �2

Z a2

aG2ðnÞPfHnðyÞ40gj3

GðyÞjð1ÞG ðyÞ dy, ðA:19Þ

where by (A.14),Z a2

aG2ðnÞPfHnðyÞ40gj3

GðyÞ djGðyÞra2½4g1ðnÞþ3g2ðnÞ�j4Gða2Þþ

Z a2

aG2ðnÞ½I31ðy,nÞþ IV31ðy,nÞþ IV41ðy,nÞ�j3

GðyÞ djGðyÞ: ðA:20Þ

Let j�Gða2Þ ¼maxðjjGða2Þj,jGð0ÞÞ,

r1ðnÞ ¼32

ffiffiffi2p½12Q ða2,

ffiffiffi2p

s,bÞþ2b2s2pða2,ffiffiffi2p

t,bÞj�Gða2Þ=t�3s5f 2

�

:

Note that r1ðnÞ-384ffiffiffi2p

k0t7=ðs5f 2� Þ as n-1 and pða2,

ffiffiffi2p

t,bÞ-8t8 as n-1. Hence, on aG2ðnÞryra2,

I31ðyÞrexp �3nb5s5f 2

� j2GðyÞ

32ffiffiffi2p½12Q ða2,

ffiffiffi2p

s,bÞþ2b2s2pða2,ffiffiffi2p

t,bÞj�Gða2Þ�

( )¼ expf�nb5j2

GðyÞ=r1ðnÞg:

Thus,

Z a2

aG2ðnÞI31ðy,nÞj3

GðyÞ djGðyÞrZ a2

aG2ðnÞexpf�nb5j2

GðyÞ=r1ðnÞgj3GðyÞ djGðyÞr

Z j2Gða2Þ

j2GðaG2ðnÞÞ

expð�nb5t=r1ðnÞÞt dtrr1ðnÞ

nb5

� �2

:

ðA:21Þ


For aG2ðnÞryra2, �HGðyÞ=ð4t2B3ðy,ffiffiffi2p

sÞÞZ�f�jGðyÞ=ð4t2B3ða2,ffiffiffi2p

sÞÞ ��c1jGðyÞ where c1 ¼�f�=ð4t2B3ða2,ffiffiffi2p

sÞÞ. Since

expð�nðm�1Þ=2½t�lnð1þtÞ�Þ is decreasing in t for t40,


2½�c1jGðyÞ�ln½1�c1jGðyÞ��

� :

Also, for 0oto1, t�lnð1þtÞZt2=4. Thus,Z a2

aG2ðnÞIV31ðy,nÞj3


aG2ðnÞexp �

nðm�1Þ

2½�c1jGðyÞ�ln½1�c1jGðyÞ��

� j3

GðyÞ djGðyÞ

¼1

c41

Z �c1jGða2Þ

�c1jGðaG2ðnÞÞexp �

nðm�1Þ

2½t�lnð1þtÞ�

� t3 dtr

1

2c41

Z �c1jGða2Þ

�c1jGðaG2ðnÞÞexp �

nðm�1Þt2

8

� t2 dtr

32

c41½nðm�1Þ�2

:

ðA:22Þ

Also, expf�nðm�1Þ=2½t�lnð1þtÞ�g is increasing in t for �1oto0, and t�lnð1þtÞZt2=4 for �1oto0. Thus, on

aG2ðnÞryra2.


2½c1jGðyÞ�ln½1þc1jGðyÞ��

� :

Hence,Z a2

aG2ðnÞIV41ðy,nÞj3


aG2ðnÞexp �

nðm�1Þ

2½c1jGðyÞ�ln½1þc1jGðyÞ��

� j3

GðyÞ djGðyÞ

¼1

c41

Z c1jGða2Þ

c1jGðaG2ðnÞÞexp �

nðm�1Þ

2½t�lnð1þtÞ�

� t3 dtr

1

c41

Z c1jGða2Þ

c1jGðaG2ðnÞÞexp �

nðm�1Þt2

2

� t3 dtr

32

c41 ½nðm�1Þ�2

: ðA:23Þ

Combining (A.19)–(A.20) yields

En½A25ðnÞ�r a2½4g1ðnÞþ3g2ðnÞ�j4

Gða2Þþr1ðnÞ

ðnb5Þ2þ

64

c41 ½nðm�1Þ�2

( )nb5

½jð1Þ�G �2¼O

1

nb5

� �: &

Proof of (3.9).

A4ðnÞ ¼

Z aG2ðnÞ

aG

I½HnðyÞ� dyraG2ðnÞ�aG ¼1

jð1Þ�G


p :

Thus,

En½A24ðnÞ�r

1

½jð1Þ�G �2nb5

: &

Proof of (3.6), (3.7), and (3.8). The proofs of (3.6), (3.7), and (3.8) are similar to that of (3.11), (3.10), and (3.9),respectively. The details are thus omitted. &

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Empirical Bayes testing for equivalence