Counter-examples to “minimax lower bounds of empirical Bayes tests”

Journal of Statistical Planning and Inference 87 (2000) 273–286

www.elsevier.com/locate/jspi

Counter-examples to “minimax lower bounds of empirical Bayes tests” TaChen Liang Department of Mathematics, Wayne State University, Detroit, MI 48202, USA Received 8 March 1999; accepted 15 September 1999

Abstract We point out errors in the paper of Karunamuni (1996, Ann. Statist. 24, 212 –231.). We also provide examples in which the regrets of the constructed empirical Bayes tests converge to zero having rates faster than the convergence rates of the “minimax lower bounds of regrets of c 2000 Elsevier Science B.V. All rights empirical Bayes tests” claimed in Karunamuni (1996). reserved. MSC: 62C12 Keywords: Asymptotically optimal; Empirical Bayes; Minimax lower bound; Monotone test; Rate of convergence; Regret Bayes risk

1. Introduction Let X denote a random variable having a probability density in the exponential family f(x | ) = m(x)h()ex , −∞6a6x6b6 + ∞, where m(x)R is known and positive for x b in (a; b) and  is in the natural parameter space = { | a f(x | ) d x = 1}. Consider the problem of testing the hypotheses H0 : 60 versus H1 :  ¿ 0 with the linear error loss L(; i) = i(0 − )I (0 − ) + (1 − i)( − 0 )I ( − 0 )

(1.1)

where 0 is a known constant, i is the action deciding in favor of Hi ; i = 0; 1, and I (x) = 1(0) if x¿0 (x ¡ 0). It is assumed that the parameter  is a realization of a random parameter  having an unknown prior distribution G over . Let  be the sample space of the random variable X . A test is deÿned to be a mapping from  into [0; 1] so the (x) = P{accepting H1 | X = x}, the probability of accepting H1 when X = x is observed. Let E-mail address: [email protected] (T. Liang). 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 8 8 - 3

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R(G; ) denote R the Bayes risk of the test when G is the “true” prior distribution. Suppose that || dG() ¡ ∞. Thus, the Bayes risk R(G; ) can be expressed as Z (1.2) R(G; ) = (x)[0 − ’G (x)]fG (x) d x + ConstG ; R



where Const G = ( − 0 )I ( − 0 ) dG(), and Z fG (x) = f(x | ) dG(): the marginal pdf of X; Z ’G (x) = EG [ | X = x] =

f(x | ) dG()=fG (x):

(1.3)

Note that ’G (x) is continuous and nondecreasing in x. Then, a Bayes test G , which minimizes the Bayes risks among a class of tests, is clearly given by G (x) = 1

if ’G (x)¿0

and 0 otherwise:

(1.4)

We assume that Assumption A. lim ’G (x) ¡ 0 ¡ lim ’G (x):

x→a

x→b

(1.5)

Under Assumption A, the prior distribution G is nondegenerate and ’G (x) is strictly increasing in x. Since ’G (x) is continuous, there exists a point, say CG , such that ’G (CG ) = 0 ; ’G (x) ¡ 0 , if x ¡ CG and ’G (x) ¿ 0 when x ¿ CG . CG is called the critical point of the Bayes test G . Therefore, G can also be written as G (x) = 1 if x¿CG

and 0 otherwise:

(1.6)

Hence, G is a monotone test. The minimum Bayes risk of this testing problem is Z R(G; G ) = G (x)[0 − ’G (x)]fG (x) d x + ConstG : (1.7) 

When a sequence of past data is available, this testing problem has been treated via the empirical Bayes approach of Robbins (1956, 1964), see Johns and Van Ryzin (1972), Van Houwelingen (1976), Karunamuni and Yang (1995) and Karunamuni (1996), among many others. Let X1 ; : : : ; Xn ; Xn+1 be iid random variables having probability density fG (x), where X (n) = (X1 ; : : : ; Xn ) denotes the n past data and Xn+1 ≡ X stands for the current random observation. An empirical Bayes test n is a test based on the current observation X = x and the past data X (n) so that n (x; X (n)) ≡ n (x) is the probability of accepting H1 . Given X (n), the conditional Bayes risk of n is Z n (x)[0 − ’G (x)]fG (x) d x + ConstG (1.8) R(G; n | X (n)) = 

and the (unconditional) Bayes risk of n is R(G; n ) = En R(G; n | X (n)), where the expectation En is taken with respect to the probability measure Pn generated by X (n). Since R(G; G ) is the minimum Bayes risk, R(G; n | X (n))−R(G; G )¿0 for all X (n)

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and for all n, and therefore, R(G; n ) − R(G; G )¿0 for all n. The nonnegative regret Bayes risk R(G; n ) − R(G; G ) is often used as a measure of performance of the empirical Bayes test n . An empirical Bayes test n is said to be asymptotically optimal if R(G; n ) − R(G; G ) → 0 as n → ∞; an empirical Bayes test n is said to be asymptotically optimal, relative to the prior distribution G, at a rate of convergence of order O(n ) if R(G; n )−R(G; G )=O(n ), where {n } is a sequence of decreasing, positive numbers such that limn→∞ n = 0: Johns and Van Ryzin (1972), Van Houwelingen (1976), Karunamuni and Yang (1995), and Karunamuni (1996) have constructed empirical Bayes tests for the underlying decision problem, and investigated the associated rates of convergence of the regret Bayes risks. Recently, Karunamuni (1996) raised the following questions: In terms of rates of convergence, what are the best empirical Bayes tests? What are the optimal rates of convergence? In order to answer these questions, Karunamuni (1996) attempted to derive an asymptotic minimax lower bound for the regret Bayes risk R(G; n ) − R(G; G ) for any monotone empirical Bayes test n . Based on the claimed asymptotic minimax lower bound, Karunamuni (1996) claimed having obtained the optimal rates of convergence of empirical Bayes tests for normal distribution and exponential distribution models. However, the claimed asymptotic minimax lower bound is incorrect and the claimed optimal rates of convergence is also incorrect. Thus, the questions raised in Karunamuni (1996) are still unsolved. In Section 2, we point out the errors in the statements of Theorem 2:1 and Corollaries 2:1 and 2:2 of Karunamuni (1996) regarding the “asymptotic minimax lower bound” of empirical Bayes tests. In Section 3, we provide examples in which the regrets of the constructed empirical Bayes tests converge to zero at rates faster than the convergence rates of the “minimax lower bounds of regrets of empirical Bayes tests” claimed in Karunamuni (1996).

2. A comment In order to ÿnd out the optimal rate of convergence, Karunamuni (1996) tried to establish a minimax lower bound for the regret Bayes risks of monotone empirical Bayes tests. We describe his approach as follows (also, see p. 216 of Karunamuni (1996)). Let G0 be a distribution function satisfying Assumption A of (1.5) with density function g0 such that g0 () ¿ 0 for all  in . Let {n } be a sequence of positive numbers suchR that n → 0 as n → ∞. Let H be a bounded continuous function on

satisfying H () d = 0: For a ÿxed positive number k, deÿne the function gn on

by gn () = g0 () + kn H (n ):

(K.1)

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Thus, by suitable choice of H and g0 such that gn ()¿0 for all  in for small n , gn will be a density on . Let Gn denote its distribution function. Let C0 denote the critical point of the Bayes test G0 . That is, C0 = CG0 . Karunamuni (1996) claimed the following result. The statement of Theorem K2.1 is essentially the same as that of Theorem 2:1 of Karunamuni (1996). Interested readers are referred to Karunamuni (1996) for the original statement of Theorem 2:1 of Karunamuni (1996). Theorem K2.1 (Karunamuni, 1996). Suppose that functions H; g0 and the sequence {n } are chosen such that gn is nonnegative; and for some real number k ¿ j ¿ 0; Z (K.2) ( − 0 )eC0  h()H (n ) d = O(−j n ) and

Z

[fGn (x) − fG0 (x)]2 =fG0 (x) d x6d1 =n:

(K.3)

for some constant d1 ¿ 0. Then; sup [R(G; n ) − R(G; G )] ¿ d2 n2(k−j)

G∈#∗

(K.4)

for all suciently large n and some positive constant d2 which is independent of n ; where #∗ denotes any set consisting of priors satisfying Assumption A of (1:5) such that {G0 ; Gn } @ #∗ . Note that the statements given in (K.2) – (K.4) correspond to that of (2:17)–(2:19) of Karunamuni (1996), respectively. Since H is a bounded function on , thus, R C0  = ||e h() d and |H ()|6B for all  in

for some positive number B. Let B 1 R B2 = eC0  h() d. In many cases, both B1 and B2 are ÿnite. For example, for each of the two examples given in Corollaries 2:1 and 2:2 of Karunamuni (1996), respectively (also, see Counter-examples 1 and 2 of Section 3), the corresponding B1 and B2 are ÿnite. The notation used in (K.2) is abused, and the meaning of (K.2) is ambiguous. We try to ÿgure out its meaning as follows. Suppose that one reads (K.2) as: Z (K.5) ( − 0 )eC0  h()H (n ) d ∼ C−j n for some number C 6= 0. Then, Z Z  Z C0  C0  ( − 0 )eC0  h()H (n ) d 6 B ||e h() d + | 0 | e h() d = B{B1 + | 0 |B2 }

−j n ;

(K.6)

when n is suciently large since n = o(1). Thus, there is no bounded, continuous function H (·) satisfying (K.5). Hence, Theorem 2:1 of Karunamuni (1996) says nothing

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about the minimax lower bounds of empirical Bayes tests, and the results of Corollaries 2:1 and 2:2 are dubious since the function H (·) in each of Corollaries 2:1 and 2:2 of Karunamuni (1996) does not satisfy (K.5). Suppose that one read (K.2) as: Z ( − 0 )eC0  h()H (n ) d 6C−j (K.7) n for some positive constant C. Based on this interpretation, we present a counter-example in the following. Our construction of the counter-example is conceptually simple: We consider two prior distributions G0 and Gn such that limn→∞ Gn = G0 and CG0 = CGn . Therefore, G0 = Gn . Let M denote the set of all monotone empirical Bayes tests including those ÿxed-point monotone tests n; a for which n; a (x) = 1

if x¿a

and 0 otherwise:

(2.1)

for all n = 1; 2; : : : . Let C ∗ = CG0 = CGn . Thus, R(G; n; C ∗ ) − R(G; G ) = 0 for each G = G0 ; Gn . Hence, sup

G∈{G0 ; Gn }

[R(G; n; C ∗ ) − R(G; G )] = 0

(2.2)

and therefore, inf

sup

n ∈M G∈{G ;G } 0 n

[R(G; n ) − R(G; G )] = 0:

(2.3)

Now, a counter-example is given as follows. √ Let f(x | ) ∼ N(; 1). Then f(x | )=m(x)h()ex , where h()=1= 2 exp(− 2 =2); m(x) = exp(−x2 =2). Let G0 be the prior distribution with the pdf g0 (), where g0 () = c1 (1 +  2 )−‘ ; −∞ ¡  ¡ ∞; 1 ¡ ‘ ¡ 1:5 and c1 is a constant so that g0 () is a probability density. We choose the function H () as: H ()=(5 4 −6 2 +1)=5 if ||61; and 0 otherwise. Deÿne gn ()=g0 ()+kn H (n ); where k is a positive number and n ¿ 0, decreases in n,R and limn→∞ n = 0. For k¿3, when n is suciently large, gn ()¿0 for all , and gn () d = 1: So, gn () is a probability density over (−∞; +∞). We denote the distribution of gn () by Gn . Let 0 = 0: Consider the testing hypotheses H0 : 60 versus H1 :  ¿ 0 , with the linear error loss (1.1). Note that both g0 () and gn () are even functions. It is easy to see that for each G = G0 ; Gn ; ’G (x) ¡ 0 for x ¡ 0, ’G (0) = 0, and ’G (x) ¿ 0 for x ¿ 0. Therefore, CG0 = CGn = 0: According to the preceding discussion, we have inf

sup

n ∈M G∈{G ;G } 0 n

[R(G; n ) − R(G; G )] = 0:

(2.4)

Next, we verify the conditions given in (2:17)–(2:18) of Karunamuni (1996). Note that 0 = 0 and CG0 = 0 and since h()H (n ) is an odd function, Z ( − 0 ) exp(CG0 )h()H (n ) d = 0 = O(−j n )

for any j¿0:

(2.5)

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So, condition (2:17) of Karunamuni (1996) is satisÿed. Note that Z fG0 (x) =

  c1 (x − )2 1 √ exp − d 2 (1 +  2 )‘ 2

and fGn (x) − fG0 (x) =

kn

Z

1=n

−1=n

  (x − )2 1 √ exp − H (n ) d: 2 2

Hence, for k¿3 and n being suciently large, |fGn (x)−fG0 (x)| =fG0 (x)61. By noting that |H (n )|61, we obtain Z

∞

−∞

Z

2

[fGn (x) − fG0 (x)] =fG0 (x) d x 6

−∞

Z 6

∞

−∞

Z =

∞

|fGn (x) − fG0 (x)|d x kn

1=n

−1=n

Z

kn

1=n

−1=n

Z

∞

−∞

f(x | ) d d x f(x | ) d x d

= 2nk−1 = 2n−1

if n = n−1=(k−1) :

(2.6)

Thus, condition (2:18) of Karunamuni (1996) is also satisÿed. If the statement of Theorem 2:1 of Karunamuni (1996) was correct, then, by this theorem, we would have that lim inf inf

sup

n→∞ n ∈M G∈{G ;G } 0 n

n−2(k−j) [R(G; n ) − R(G; G )] ¿ 0

(2.7)

for the ÿxed k¿3 and k ¿ j ¿ 0: However, (2.7) and (2.4) contradict each other. According to our construction, (2.4) is always correct. So, the statement of Theorem 2:1 of Karunamuni (1996) is incorrect. Therefore, the results of Corollaries 2:1 and 2:2 are dubious. We summarize the previous discussion as follows: If (K.2) is read as (K.5), then, there is no bounded function H (·) satisfying (K.5), and therefore, the conclusions of Theorem 2:1 and Corollaries 2:1 and 2:2 of Karunamuni (1996) are dubious; and if (K.2) is read as (K.7), the provided counter-example again points out the errors contained in the statement of Theorem 2:1 and Corollary 2:1 of Karunamuni (1996). It should be noted that the computations involving the condition (K.2) given in Corollaries 2:1 and 2:2 of Karunamuni (1996) are incorrect.

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3. Counter-examples 3.1. Counter-example 1 Consider the exponential distribution having probability density f(x | )=e−x ; x ¿ 0;  ¿ 0: Let G0 be the prior distribution having the probability R density g0 () = C‘ (1R +  2 )−‘ ;  ¿ 0 and 0 otherwise, where ‘ ¿ 1: Then, ’G0 (x) =  2 exp(−x) dG0 ()=  exp(−x) dG0 () is strictly decreasing in x: We let 0 = ’G0 (2) so that CG0 = 2; where CG0 is the critical point of the Bayes test G0 : Deÿne w() to be the probability density of a (p; 1) distribution. That is, w() = 1= (p) p−1 exp(−);  ¿ 0 and 0 otherwise. Let H () = C‘ [w() − w( − 1)]: Let n =n−1=(2k−3) ; where k ¿ 4 is a ÿxed positive integer. Deÿne gn ()=g0 ()+kn H (n ): Note that for n being suciently large, gn () is a probability density. Denote its corresponding distribution by Gn and CGn the critical point of the Bayes test Gn : By the deÿnition of gn (); we have limn→∞ CGn = CG0 = 2: Let #n = {G0 ; Gn }: The set #n has been used by Karunamuni (1996) to establish “a minimax lower bound of regrets of empirical Bayes tests”. Karunamuni (1996) claimed that for any empirical Bayes test n ; sup [R(G; n ) − R(G; G )] ¿ dn−2(k−4)=(2k−3)

G∈#n

(3.1)

for all suciently large n and for some positive constant d which is independent of the prior distributions. Note that the function H () is a bounded function. As pointed out in Section 2, the assumptions of Theorem 2:1 of Karunamuni (1996) are not fulÿlled. Thus, it is not known whether the claimed minimax lower bound of the regrets given in (3.1) holds or not. In the following, we provide a counter-example to this claimed minimax lower bound of regrets R of empirical Bayes tests. Let G (x) = f(x | ) dG(): Note that ’G (x) = G (x)=fG (x); which is decreasing in x for x ¿ 0: Let hG (x) = 0 fG (x) − G (x): Thus, a Bayes test G can be written as G (x) = 1

if hG (x)60

and 0 otherwise:

(3.2)

Let s be a positive integer and let s0 = s and s1 = s + 1: For each i = 0; 1; let Fi (si ) be the family of all Borel-measurable bounded functions vanishing outside the interval [0,1], such that for each Ki in Fi (si ); |Ki (y)|6B for y in [0,1] and  1 if j = i;   Z 1  yj Ki (y) dy = 0 if j 6= i; 06j6si − 1;  0   Bi if j = si :

(3.3)

Let {b=b(n)} be a sequence of positive, decreasing numbers such that limn→∞ b=0; and limn→∞ nb3 = ∞: Let X1 ; : : : ; Xn be iid random variables following the marginal

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R probability density fG (x) = f(x | ) dG(): For x ¿ 0; deÿne   n Xj − x 1 P ; K0 fn (x) = nb j=1 b n 1 P K1 n (x) = − 2 nb j=1



Xj − x b

 ;

hn (x) = 0 fn (x) − n (x):

(3.4)

Note that hn (x) is a consistent estimator of hG (x) = R0 fG (x) − G (x): ∞ We consider only those priors G in #(C1 ; C2 )={G | 0  dG() ¡∞; C1 ¡CG ¡C2 }; where 0 ¡ C1 ¡ 2 ¡ C2 ¡ ∞ are known constants. Note that since CG0 = 2; thus G0 is in #(C1 ; C2 ): Also since limn→∞ CGn = CG0 = 2; for n being suciently large, Gn is in #(C1 ; C2 ): Motivated by the form (3.2) and the assumption that G is in #(C1 ; C2 ); we propose an empirical Bayes test n∗ as follows: For x ¿ 0; ( 1 if (x ¡ C1 ) or (C1 6x6C2 and hn (x)60); ∗ (3.5) n (x) = 0 if (x ¿ C2 ) or (C1 6x6C2 and hn (x) ¿ 0): Liang (1999) has proved the following Theorem. Theorem 3.1 (Liang, 1999). For each prior G in #(C1 ; C2 ); the empirical Bayes test n∗ is asymptotically optimal in the sense that 4bs DG (C1 )(0 + 1) 16B2 QG (C1 ; C2 ) + ; (3.6) fG (C2 ) nb3 fG (C2 ) R where DG (x) = max(DG (x; s); DG (x; s + 1)) and DG (x; s) = B=(s + 1)!  s+1 e−x dG(); and QG (C1 ; C2 ) = maxC1 6x6C2 [fG (x)=(−’(1) G (x))] ¡ ∞: R(G; n∗ ) − R(G; G )6

If b is chosen to be b = n−1=(s+3) ; then, R(G; n∗ ) − R(G; G ) = O(n−s=(s+3) ): ∗

∗

(3.7) ∗

Let D = max06n ¡ ∞ DGn (C1 ); Q = max06n ¡ ∞ QGn (C1 ; C2 ); and f = min06n ¡ ∞ fGn (C2 ): Then, D∗ ¡ ∞; Q∗ ¡ ∞ and f∗ ¿ 0: Since for suciently large n; #n ⊂ #(C1 ; C2 ); from Theorem 3.1, the following holds: sup [R(G; n∗ ) − R(G; G )] 6

G∈#n

4bs D∗ (0 + 1) 16B2 Q∗ + f∗ nb3 f∗

= O(n−s=(s+3) ) if b = n−1=(s+3) :

(3.8)

Note that the choice of s is independent of Gn ; and therefore, independent of the value of k: Hence, we choose s so large such that sk: In such situations, we see that the upper bound convergence rate of the regret R(G; n∗ ) − R(G; G ) (see (3.8)) is faster than the lower bound convergence rate claimed by Karunamuni (1996), (see (3.1)). Thus, this provides a counter-example to the statements claimed by Karunamuni

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(1996) regarding “the minimax lower bound” and “the optimal rate of convergence” of regrets of empirical Bayes tests for exponential distribution case. 3.2. Counter-example 2 Consider a normal distribution N(; 1) with probability density f(x | )√= m(x)h() exp(x); −∞ ¡ ; x ¡ ∞; where m(x) = exp(−x2 =2); h() = exp(− 2 =2)= 2. Let G0 be the prior distribution having the probability density g0 () = C‘ (1 +  2 )−‘ ; −∞ ¡  ¡ ∞; and ‘ ¿ 1: Let 0 = 0: Since g0 () is an even function, a simple algebraic computation yields that CG0 = 0 where CG0 is the critical point of the Bayes test G0 : Deÿne H ()=C‘ [h()−h(+1)]: For a ÿxed integer k ¿ 4; let n =n−1=(2k−4) : Deÿne gn () = g0 () + kn H (n ): Note that when n is large enough, gn () is a probability density. Denote its corresponding distribution by Gn and the critical point of the Bayes test Gn by CGn : By the deÿnition of gn (); we have: limn→∞ CGn = CG0 = 0: Let #n = {Gn ; G0 }: The set #n has been used by Karunamuni (1996) to establish “a minimax lower bound of regrets of empirical Bayes test” for a normal mean case. Karunamuni (1996) claimed that for any empirical Bayes test n ; sup [R(G; n ) − R(G; G )] ¿ dn−2(k−4)=(2k−4)

G∈#n

(3.9)

for all suciently large n; and for some constant d which is independent of the prior distributions. Again, the function H () is a bounded function. Thus, the assumption of Theorem 2:1 of Karunamuni (1996) is not fulÿlled. Therefore, it is not known whether the claimed “minimax lower bound” (see (3.9)) holds or not. In the following, a counter-example is provided to this claimed minimax lower bound of regrets of empirical Bayes tests. R R Note that fG (x)=m(x) G (x) where G (x)= h()ex dG(): Then, G (x) ≡ f(x | ) dG() = m(x) G(1) (x), where G(r) denotes the rth derivative of G : Note that ’G (x) = G(1) (x)= G (x) is increasing in x. Thus, a Bayes test G can be written as: for each x, G (x) = 1

if G(1) (x)¿0

and 0 otherwise:

Let K1 be the kernel function characterized in (3.3) previously. Deÿne   n Xj − x 1 P (1) I (x6Xj 6x + b)=m(Xj ); K1 n (x) = 2 nb j=1 b

(3.10)

(3.11)

where {b = b(n)} is a sequence of positive, decreasing numbers such that limn→∞ b = 0 and limn→∞ nb3 = ∞: From Lemma A.1, we see that n(1) (x) is a consistent estimator of G(1) (x): R We consider only those priors G in #A = {G | ||dG() ¡ ∞ and |CG |6A}; where A is a known, ÿxed positive value. Since CG0 = 0; so G0 is in #A : Also, since limn→∞ CGn = CG0 = 0; for n being suciently large, Gn is in #A : Motivated by the form (3.10) and by the assumption that G is in #A ; we propose an empirical Bayes

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test n as follows: for each x: ( 1 if (x ¿ A) or (|x|6A and n(1) (x)¿0); ◦ n (x) = 0 if (x ¡ − A) or (|x|6A and n(1) (x) ¡ 0): G (x+bt) }; G (A)=max|x|6A { Let m∗ (x)=min06t61 {m(x+bt)}; G (x)=max06t61 { m(x+bt)

( ∗G (A; b)

= min

|x|6A

m∗ (x) 3 × 8 B2 G (x)m∗ (x) + b3 B| G(1) (x)|

)

(3.12) m(x) }, (2) G (x)

;

) ( R1 | 0 ys+1 K1 (y) G(s+1) (x(y; b; s)) dy| ; G (A) = max G (x) |x|6A where x(y; b; s) is some value between x and x + b, depending on (y; b; s). Then,

G (A) ¡ ∞; ∗G (A; b) ¿ 0 and G (A) ¡ ∞. ◦ The empirical Bayes test n has the following asymptotic optimality. The proof of Theorem 3.2 is given in the appendix. ◦

Theorem 3.2. For each prior distribution G in #A ; the empirical Bayes test n is asymptotically optimal in the sense that ◦

R(G; n ) − R(G; G )6

G (A) 4bs G (A) : + ∗G (A; b)nb3 (s + 1)!

(3.13)

If we let b = n−1=(s+1) , then ◦

R(G; n ) − R(G; G ) = O(n−s=(s+3) ):

(3.14)

Let ∗ = max06n ¡ ∞ Gn (A); ∗ = min06n ¡ ∞ ∗Gn (A; b) and ∗ = max06n ¡ ∞ Gn (A). Then, ∗ ¡ ∞; ∗ ¡ ∞ and ∗ ¿ 0. Since for suciently large n; #n @ #A , from Theorem 3.2, the following holds: ◦

sup [R(G; n ) − R(G; G )] 6

G∈#n

∗ 4bs ∗ + ∗ nb3 (s + 1)!

= O(n−s=(s+3) ) if b = n−1=(s+3) :

(3.15)

Since the choice of the value of the integer s is independent of the ÿxed value of k, we choose s so large such that sk: In such situations, we see that the upper bound ◦ convergence rate of the regret R(G; n )−R(G; G ) (see (3.15)) is faster than the lower bound convergence rate claimed by Karunamuni (1996) (see (3.9)). Thus, this provides a counter-example to the statements claimed by Karunamuni (1996) regarding “the minimax lower bound” and “the optimal rate of convergence” of regrets of empirical Bayes tests regarding a normal mean case.

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4. Concluding remark In this paper, we point out errors in the paper of Karunamuni (1996) regarding minimax lower bounds of empirical Bayes tests. From the discussion presented in Section 2, we see that Karunamuni (1996) put too much assumptions in his Theorem 2:1 that in fact cannot be fulÿlled in practical situations. Also, the notation “big O” is abused so that the meaning of (K.2) (or (2:17) of Karunamuni (1996)) is ambiguous. This ambiguity has certain impact when Theorem 2:1 of Karunamuni (1996) is applied to normal and exponential distributions; see Examples 2:1 and 2:2 of Karunamuni (1996). In Karunamuni (1996), the computations presented in (2:30)–(2:32) and (2:39)– (2:42) contain serious errors. The ambiguity of (K.2) may be blamed for the computational errors. Appendix A. Note that n(1) (x) =

n 1P Yj (x; b); n j=1

where

 Yj (x; b) = K1

Xj − x b



I (x6Xj 6x + b)=[m(Xj )b2 ];

(A.1)

j = 1; : : : ; n, are iid, bounded random variable such that |Yj (x; b)|6

B b2 m∗ (x)

(A.2)

where m∗ (x) = min06t61 {m(x + tb)}. We have the following lemma: R1 Lemma A.1. (a) En Yj (x; b)= G(1) (x)+bs =(s+1)! 0 ys+1 K1 (y) G(s+1) (x(y; b; s)) dy where x(y; b; s) denotes some point between x and x + b, depending on (y; b; s). (b) |Yj (x; b) − En Yj (x; b)|6 b2 m2B∗ (x) . (c) Var(Yj (x; b))6 B

2

G (x) b3 ,

G (x+tb) where G (x) = max06t61 { m(x+tb) }. R s 1 s+1 (1) (1) (s+1) b (d) En n (x) = G (x) + (s+1)! 0 y K1 (y) G (x(y; b; s)) dy. 2

G (x) . (e) Var( n(1) (x))6 B nb 3

Proof. Straightforward computations will lead to the results. The details are omitted here. Let

) ( R1 | 0 ys+1 K1 (y) G(s+1) (x(y; b; s)) dy| : G (A) = max G (x) |x|6A

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T. Liang / Journal of Statistical Planning and Inference 87 (2000) 273–286

Note that G (A) ¡ ∞: Since limn→∞ b=0, hence, for n being suciently large, 4bs =(s+ 1)!G (A) ¡ min(|’G (−A)|; ’G (A)). By the strictly increasing property and continuity of ’G (·), there exists points, say C1 (n) and C2 (n) such that −A ¡ C1 (n) ¡ CG ¡ C2 (n) ¡ A and 2bs G (A); −’G (C1 (n)) = ’G (C2 (n)) = (s + 1)! ’G (x)6 − ’G (x)¿

2bs G (A) (s + 1)!

2bs G (A) (s + 1)!

|’G (x)|6

2bs G (A) (s + 1)!

for x6C1 (n);

for x¿C2 (n); for C1 (n)6x6C2 (n):

(A.3)

Lemma A.2. (a) For each −A ¡ x ¡ C1 (n); En n(1) (x)6 12 G(1) (x) ¡ 0. (b) For each C2 (n) ¡ x ¡ A; En n(1) (x)¿ 12 G(1) (x) ¿ 0. Proof. We provide proof for part (a) only. The proof of part (b) is analogous to that of part (a), and hence is omitted here. For each −A ¡ x ¡ C1 (n); ’G (x)6’G (C1 (n))=−2bs G (A)=(s+1)! ¡ 0. From Lemma A.1(d) and by the deÿnition of G (A), it follows that # " R1 bs 0 ys+1 K1 (y) G(s+1) (x(y; b; s)) dy (1) En n (x) = G (x) ’G (x) + (s + 1)! G (x) 

1 bs G (A) 1 6 G (x) ’G (x) + ’G (x) + 2 2 (s + 1)!   1 6 G (x) ’G (x) 2 =

1 (1) (x) ¡ 0: 2 G

Let

( ∗G (A; b)

= min

|x|6A

3m∗ (x) 8[B2 G (x)m∗ (x) + b3 B| G(1) (x)|]

Note that ∗G (A; b) ¿ 0 and lim

n→∞



∗G (A; b)

= min

|x|6A



3 8B2 G (x)

) :

 ¿ 0:

Lemma A.3. (a) For each −A¡x¡C1 (n); Pn { n(1) (x)¿0}6exp{−nb3 ∗G (A; b)[ G(1) (x)]2 }. (b) For each C2 (n) ¡ x ¡ A; Pn { n(1) (x) ¡ 0}6exp{−nb3 ∗G (A; b)[ G(1) (x)]2 }.

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285

Proof. (a) Combining Lemmas A:2(a), A:1(b) and an application of Bernstein inequality, we obtain for each −A ¡ x ¡ C1 (n), Pn { n(1) (x) ¡ 0} 6 Pn { n(1) (x) − En n(1) (x)¿ − 12 G(1) (x)} ) ( n=2[ − G(1) (x)=2]2 6 exp − Var(Yj (x; b)) + 13 × 2B=m∗ (x)| 12 G(1) (x)| ) ( n=8[ G(1) (x)=2]2 6 exp − B2 A (x)b−3 + B| G(1) (x)|=(3m∗ (x)) ) ( 3m∗ (x)[ G(1) (x)]2 3 = exp −nb × 8[B2 G (x)m∗ (x) + b3 B| G(1) (x)| 6 exp{−nb3 ∗G (A; b)[ G(1) (x)]2 }:

(A.4)

(b) The proof of part (b) can be obtained in a similar discussion. Hence, the detail is omitted here. Proof of Theorem 3.2. For a prior distribution G in #A ; according to (3.10) and (3.12), ◦ the regret of the empirical Bayes test n can be written as Z C1 (n) ◦ Pn { n(1) (x)¿0}m(x)[ − G(1) (x)] d x R(G; n ) − R(G; G ) = −A

Z +

CG C1 (n)

Z + +

C2 (n)

CG

Z

Pn { n(1) (x)¿0}m(x)[ − n(1) (x)] d x

A C2 (n)

Pn { n(1) (x) ¡ 0}m(x) G(1) (x) d x

Pn { n(1) (x) ¡ 0}m(x) G(1) (x) d x:

(A.5)

Let G (A) = max|x|6A {m(x)= G(2) (x)}. For each −A ¡ x ¡ C1 (n), by Lemma A.3(a), Z C1 (n) Pn { n(1) (x)¿0}m(x)[ − G(1) (x)] d x −A

Z 6

C1 (n)

exp{−nb3 ∗G (A; b)[ G(1) (x)]2 }m(x)[ − G(1) (x)] d x

−A

Z 6 G (A) 6

C1 (n)

−A

exp{−nb3 ∗G (A; b)[ G(1) (x)]2 }(− G(1) (x)) G(2) (x) d x

G (A) : 2∗G (A; b)nb3

(A.6)

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T. Liang / Journal of Statistical Planning and Inference 87 (2000) 273–286

For each C2 (n) ¡ x ¡ A, from Lemma A.3(b), following a discussion analogous to (A.6), we can obtain Z A

G (A) Pn { n(1) (x) ¡ 0}m(x) G(1) (x) d x6 ∗ : (A.7) 2G (A; b)nb3 C2 (n) Next, by (A.3), we obtain Z Z CG Pn { n(1) (x)¿0}m(x)[ − G(1) (x)] d x 6 C1 (n)

C1 (n)

6 and

Z

C2 (n) CG

CG

Pn { n(1) (x) ¿ 0}m(x) G(1) (x) d x6

m(x) G (x)[ − ’G (x)] d x

2bs G (A) (s + 1)!

2bs G (A) : (s + 1)!

(A.8)

(A.9)

Combining (A.5) – (A.9) leads to ◦

R(G; n ) − R(G; G )6

G (A) 4bs G (A) + : ∗G (A; b)nb3 (s + 1)!

(A.10)

◦

If we let b = n−1=(s+3) , then R(G; n ) − R(G; G ) = O(n−s=(s+3) ). Hence, the proof of Theorem 3.2 is complete. References Johns Jr., M.V., Van Ryzin, J.R., 1972. Convergence rates for empirical Bayes two-action problem. II. Continuous case. Ann. Math. Statist. 43, 934–947. Karunamuni, R.J., 1996. Optimal rates of convergence of empirical Bayes tests for the continuous one-parameter exponential family. Ann. Statist. 24, 212–231. Karunamuni, R.J., Yang, H., 1995. On convergence rates of monotone empirical Bayes tests for the continuous one-parameter exponential family. Statist. Decisions 13, 181–192. Liang, T., 1999. On empirical Bayes tests in a positive exponential family. J. Statist. Plann. Inference, to appear. Robbins, H., 1956. An empirical Bayes approach to statistics. Proceedings of the Third Berkeley Symposium on Mathematics Statistics Probability, Vol. 1, University of California Press, Berkeley, pp. 157–163. Robbins, H., 1964. The empirical Bayes approach to statistical decision problems. Ann. Math. Statist. 35, 1–20. Van Houwelingen, J.C., 1976. Monotone empirical Bayes tests for the continuous one parameter exponential family. Ann. Statist. 4, 981–989.