. : \l . ". .‘ -' ' “ . " _. 1,. _l'.,’_:.- - >’ ._. .2 '.’.“_4., . .. . , j. l -‘\ n .~ ~ 5 7: 2-..- '_. . \ r1 ’- 1 ' . ' ' ' “a I .. - I' .\ ' “-l g. 9,, .) .~ .." ." a .' ’3’. -. {233* ./,§../ (n "-5 (J .n u ‘_1 kg, 4" 3:, I -. '9, .w I. A“ ‘1 v.1.WMV'fi‘TIJRWC‘rM‘5B" ‘.‘u-.'.-.;.‘ u b This is to certify that the dissertation entitled Empirical Bayes with Sequential Components presented by Rohana Jith Karunamuni has been accepted towards fulfillment of the requirements for Ph-D- degreein §tafiqrirq 47 {OM/J Major professor Date NOV. 6. 1985 MS U is an Affirmative Action/Equal Opportunity Institution 0-12771 MSU LIBRARIES .—;—. RETURNING MATERIALS: Place in book drop to remove this checkout from your record. ‘FINES will be charged if book is returned after the date stamped below. EMPIRICAL BAYES WITH SEQUENTIAL COMPONENTS by Rohana Jith Karunamuni A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Statistics and Probability 1985 _—'———’ie- ABSTRACT EMPIRICAL BAYES WITH SEQUENTIAL COMPONENTS by Rohana Jith Karunamuni We consider the empirical Bayes decision problem where the component problem is an m—truncated sequential decision problem. The Bayes risk envelope of this component is smaller than that of the fixed m-sample size component when there is a cost for each observation. Empirical Bayes methods are presented that result in empirical Bayes risk approaching envelope risk as the number of components increases. Consider an empirical Bayes decision problem with fixed sample size components. Let (6,5), (61,11m)...(e ,X )... be n —nm i.i.d. with the common distribution being 6 ~ G, with l ~ PexPex...xPe = Pém) conditional on e . An empirical Bayes dec1$ion procedure TS a sequence {tn} where tn = tn(§lm,...,§ ) nm is a function taking values in the set of component decision rules th and EL[e,tn(X X X )(l)] is its n stage risk for making —1m’—2m”"’—nm a decision about 9 . The convergence of this risk to the com- ponent envelope risk whatever be G is termed asymptotic optimality (a.o.). With the sequential components, tn selects both a stopping rule function and a terminal decision function for use in the component with parameter 6'. This results in the observations available at stage n being 51N1’52N2"" X , where N1, N2,...,Nn 9 —nN n are the numbers of component observations taken in the first n components. The random sample size feature adds a degree of complexity to the development of a.o. of empirical Bayes procedures. Our results include the demonstration of a.o. procedures for some finite state components and an empirical Bayes approach to one- step look ahead procedures for some infinite state multiple decision components. ACKNOWLEDGEMENTS I express my sincere thanks and deep gratitude to Professor Dennis C. Gilliland for his continuous encouragement and guidance in the preparation of this thesis. I would like to place my appreciation for the time that he spent in the discussions. I am very grateful to Professor R.V. Erickson, Professor H. Salehi and Professor J.C. Gardiner for their continuing help and for serving on my guidance committee. I would also like to thank Professor James Hannan for many fruitful discussions during my studies at MSU. I wish to acknowledge the Department of Statistics and Probability for providing the financial support during my stay at MSU. Finally, I wish to thank Cathy Sparks for her patience, efficiency and great care in typing this thesis. TABLE OF CONTENTS Chapter Page 1 INTRODUCTION ................ 1 1.1 Introduction ................ l 1.2 The Component Problem ........... 3 1.3 The Empirical Bayes Problem ........ 15 2 TWO ACTION AND MULTIPLE DECISION PROBLEMS . 19 2.1 Two Action Problem ............. 19 2.2 An Empirical Bayes One-step Sequential Decision Procedure for the Two Action Component Case ............. 26 2.3 Asymptotic Results for the Two Action Component Case ............. 30 2.4 Multiple Decision Problem ......... 44 2.5 Final Remarks ............... 61 BIBLIOGRAPHY ................ 62 CHAPTER 1 INTRODUCTION 1.1. Introduction This thesis is concerned with empirical Bayes decision theory with a sequential statistical decision problem as the component. Robbins (1949, 1956) introduces empirical Bayes decision theory, and Robbins (1964) summarizes results for certain components and develops some general methods in regard to estimation of the prior (mixing) distribution. Our development is reasonably self- contained but does presuppose that the reader is somewhat familiar with both sequential analysis and empirical Bayes decision theory. The components to which empirical Bayes methods have been applied are, with few exceptions, given fixed sample size identical statistical decision problems. Exceptions are the varying (non— stochastic) sample size components considered by O'Bryan (1972, 1976), O'Bryan and Susarla (1975, 1976, 1977) and Susarla and O'Bryan (1975). Another exception is found in the works of Laippala (1979, 1980, 1983, 1985). In his case the varying sample sizes are random. In this thesis we define a construct within which we can treat both the random sample size and the usual fixed sample size cases. We believe that the somewhat informal style with which this is accomplished will expedite understanding of a fairly complicated mathematical construct which mixes sequential analysis and empirical Bayes decision theory. In Section 1.2 we introduce the sequential component problem which is the kernel of the empirical Bayes decision problem. In Section 1.3 we give a brief introduction to Robbins' empirical Bayes decision problem and develop new results for some finite parameter space component problems. Finite action linear loss testing and multiple decision component problems are treated in Chapter 2. 1.2. The Component Problem The component problem we consider is the m—truncated sequential decision problem with constant cost per observation. This problem is described in detail in various books concerning decision theory, e.g., Berger (1980, Chapter 7). However, in order to make this presentation reasonably self-contained, we will use this section to develop notations to describe the component problem that is the kernel of the empirical Bayes decision problem that is the subject of our investigation. The parameter space is the measurable space (e,A) and the class of all (prior) probability distributions on A is denoted by G. The parameter a indexes transitions; specifically, P6 is a distri- bution on (X,B) where X is the real line and B is the Borel o-field. Conditional on e, the observable random variables X1""’Xm are iid PG; for k = 1,2,...,m, xk = (x1,...,xk) a X , P6 = P6 x...x Pe (k times) and Bk denotes the Borel o-field k Suppose that the component decision problem has (terminal) action space A and loss function L 3 0 defined on O x A. Let c 3 0 denote the constant cost per observation. For k = O,l,...,m, let Dk denote a set of mappings 6 from m X into A that are constant with respect to the last m-k coordinates and are such that L(e,6) is A x 8m measurable. 00 consists of constant functions. We will regard the domain of k k 6 e D as X when it is convenient to do so, k = 1,2,...,m. -4- We are to sample the Xk sequentially. The observations are taken one at a time, with a decision being made after each observa- tion, to either stop sampling and choose an action in A or to take another observation. Therefore, a sequential decision procedure consists of two components, a stopping rule and a terminal decision rule. The sto in rule 1 is a sequence 1 = (10,rl,...,1m) where To is a constant function representing the probability of making a decision without sampling and, for k = 1,2,...,m rk: Xm + [0,1] is an xk - measurable function representing the conditional probability of stopping at stage k given that sampling did not stop at stages O,l,...,k—l and given the observation xk. With the m-truncated sequential problem, rm E 1. For a nonrandomized stopping rule I, Tk takes values in {0,1} for k = O,l,...,m. Associated with l is the stopping time variable N. We will be concerned only with nonrandomized l and will m take N = min {k: Tk = 1}. Such l partition X into {[N=O], [N=1],..., [N=m]} where [N=k] is a function of xk only. If To = 1, then [N=0] = xm and N20. If =0, then [N=O]=¢ . T0 The terminal decision rule g is a sequence g = (60, O x 01 x...x Dm. The interpretation is that the 51,...,am) c D = D decision function 5k is used on the set [N=k], k = O,l,...,m. A sequential decision procedure is a pair (199), where 1 is a stopping rule and p is a terminal decision rule. We will restrict consideration to the class of procedures T x D, where -5- T denotes the class of all nonrandomized stopping rules 1 such that TO 5 0. (The reason for working with stopping rules such that N 3 1 will be made clear in the next section.) Thus, we may drop first coordinate functions and take I = (11,...,rm), g: (51,...,5m). For a sequential procedure (1:9) the risk (including cost for observations) given 6 is R(e,(l,§)) = f kgl [N=k]{L(e,6k) + ck} 32(dxm) (1.1) m k k r Lge for each 5k 5 X , k = 1,2,...,m. . We suppose that (1.8) r(G) = inf{fL(e,a)G(de):aeA} is attained at each G and that a Bayes decision function 6k(G) attains the infimum posterior Bayes risk, i.e., (1.9) r(Gk) = fL(e,6k(G)) Gk(de) for all xk and k = 1,2,...,m. The one-step look ahead stopping rule is 1L(G) = (T&(G),...,T:(G)) where 15(G) = 1 and, for k = 1,2,...,m—l, . , * L 1 if E r(Gk+1) + c - r(Gk) 3 O (1.10) Tk(G) = * 0 if E r(Gk+1) + c - r(Gk) < O . * . . . . . k _ k WTth E denoting conditional expectation on Xk+l given 5 — x . The one-step look ahead sequential decision procedure is (1L(G), §(G)); i.e., it uses the Bayes terminal decision rule with the stopping rule (1.10). If m = 2, 1L(G) is a Bayes stopping rule. The Bayes risk of (1"(G), _6_(G)) is denoted by RL(G). This is not an envelope risk, rather, the Bayes risk associated with a particularly tractable sequential decision rule. We will consider the empirical Bayes approach to achieving risk RL(G) in Chapter 2 for particular classes of sequential components. We now give three examples to illustrate some of the concepts introduced to this point. The examples are sequential m-truncated components with testing simple versus simple, testing with linear loss and estimation with squared error loss. Example 1.1 (Testing Simple vs. Simple). Let 9 = {0,1}, A = {0,1} and L(0,0) = L(l,1) = 0, L(0,1) = L(1,0) = L > 0, a constant. We identify a prior G on e by the mass n it puts on the state 1 so that G can be identified with the unit interval. Let P0 be N(-1,1) and P1 be N(1,1). (This example is the sequential version of that used by Robbins (1951) to introduce the idea of compound decision theory.) The posterior probability of e = 1 k k 1 x.)} . k . _ given 5 is "k - neXp (i Xj)/{n exp (1 xj) + (l-n) eXp (-2 J and a Bayes decision rule is 1 if “k 3 1/2 (1.11) 5 0 if Wk < 1/2 The event wk 3 1/2 is equivalent to 2: xj 3 C(n) where C(n) (1/2) In ((1-n)/h). Consider the case of truncation at m = 2. In this case, the one-step look ahead procedure is Bayes with respect to n so that RL(n) = RB(n), O gin 5 1. Let the loss for misclassification be L = l and the cost per observation be c = .05. The one-step look ahead stopping rule 3(n) (cf. (1.10)) is defined by i2(n) = 1 and 1 if rl(n1) + .05 - r0(n1) 3 0 (1.12) 11(n) = 0 if r1(n1) + .05 - r0(n1) < 0 Here - «or 51/2]+ (1- n) [n > 1/21, .5 O A d v I and r1(n) = n¢(C(n) - 1) + (1 - n){1 - ¢(C(n) + 1)} IlIllIllllIIIIIIIIII:—————————————————————————————————————- e~ , . -10- where o denotes the standard normal cdf. Calculations show that (1.12) is equivalent to 1 if |h1 - .5) 5 .361567 (1.13) T1(fl) = o if |n1 - .5) > .361567 and 1 if |x1 - c(n)| 3 c(.138433) (1.14) 11(n) 0 if [x1 - c(n)| > c(.138433). The envelOpe risk RB(r) resulting from the Bayes procedure g(i) = (1(n),§(n)) was calculated for selected values of n and is plotted below along with the fixed sample size envelopes R0(n) = r0(n) and Rk(n), k = 1, 2, where (1.15) Rk(n) = n¢((c(n) - k)//E) + (1 - n){1 - ©((C(N) + k)//E)} + .05k. Of course, the optimal fixed sample size risk envelope is RF(n) = min{R1(n), R2(n)}. Note that RB(n) is considerably less than RF(n) for priors n near .5 . Also note that for n's near 0 and l, R0(n) < RF(N). Risk A .20 b .15 P .10 .05 Figure 1.1 Envelope Risk Functions >ii -12- Theorem 1.1 of the next section shows that the Bayes envelope risk RB(n) for the truncated sequential component is achieved in the limit by empirical Bayes decision procedures 9“ = (1h,§n) where In and g" are Bayes with respect to consistent estimates of n. The theorem as stated and proved subsumes the usual fixed sample size component case and the optimal floating fixed sample size case through restriction of the class of component stopping rules. In the next two examples, P6 = N(e,1), e e 0 = (-m,w) and the prior on 0 is G = N(0,02). In this case the posterior distribution Gk of 6 given lk = xk is N(oZSk/(l + koz), k 02/(1 + k02)) where Sk = lej’ k = 1,2,.... In the following examples, “k denotes the conditional mean. Example 1.2 (Testing with Linear Loss). Let 00 e e. Let A = {a0,a1}, where a0 and al correspond to actions "decide 6 i 00" and "decide 6 > 60”, respectively. Let + _ L(e,a0) = (e - 00) and let L(e,a1) = (e — 60) . Let §(G) = (61,...,6m) be the Bayes terminal decision rule where 6k is a Bayes decision rule with respect to G for the fixed k k) sample size k decision problem. If 5k = Pr(a0|§ = x , then we take 1 If pk : 60 (1.16) 6k = 0 If pk > 90 , We now derive the one-step look ahead stopping rule. Posterior minimum Bayes risk at stage k(k 3 1) is given by -13- r(Gk) = igf re L(e,5) Gk(de) re L(e,6k) Gk(de) I00 (0 - 00)Gk(de) if uk 5 00 00 . fm (00 - e)Gk(de) if uk > 00 . Define 0(t) = f:(x - t)¢(x)dx = ¢(t) - t(l - o(t)), where ¢ denotes the density function and 0 denotes the cdf 0f the standard normal distribution. Then ail/2 n (all/2m0 - 1.k)) if pk < 00 _ 2 2 . akl/ W (ai/ (Mk ‘ 90)) If Uk > 90 where ak==(l + k0 /02 . In other words, (1.17) r(Gk) = ail/2 w (at/2190 - ukl). * The conditional Bayes risk EriG given xk = xk is (see DeGroot (1970) p. 286) k+1) from taking Xk+1 E*r(Gk+l) = r(Gk) - ell/2 w (bi/Zleo - ukl) where 2 ) . bk = ak (o ak + 1 Therefore, in defining the stopping rule (1.10), * 2 E r(Gk+1) + c - r(Gk) = c - bi/ v (bi/Zleo - ukl) . Thus, the stopping rule 1L for the m-truncated problem stops sampling for the first k(k = 1,2,...,m-1) for which c 3 bi/Z w (bi/2|00 - ukl) or k = m, whichever is smaller. -14- Example 1.3 (Estimation with Squared Error Loss). Let A = e and let L(0,a) = (e - a)2. Then the Bayes terminal decision rule §(G) = (61,...,6m) for the m-truncated problem is given by the estimators 6k = uk, k = 1,2,...,m. Posterior minimum Bayes risk is the posterior variance r(Gk) = f (e - uk)2Gk(de) = 02/(1 + koz). Therefore, r(Gk) depends only on the number of observations that have been taken and not on the observed values X1,...,Xk. Thus E*r(Gk+1) = 02/(1 + (k+1)02). Therefore E*r(Gk+1) + c - r(Gk) = 02/(1 + (k+1)02) + c - 02/(1 + koz). Then the stopping rule IL for the m-truncated problem stops sampling for the first k (k = 1,...,m-1) for which c 3 04/(1 + k02)(1 + (k+1)02) or for k = m, whichever is smaller. -15- 1.3 The Empirical Bayes Problem In the last section, we introduced the basic idea behind Robbins' empirical Bayes formulation after defining the sequential m-truncated component problem. This component has envelope risk which is no greater than that of any fixed sample size k version, k = 1,2,...,m. By suitable restriction of the class of stopping rules, the component Specializes to the customary fixed sample size component. In its full generality, the sequential component presents some unique problems when it i; the keanel of the empirical Bayes 1 decision problem. Let 5 ,...,l n,... be the observation vectors from the sequence of independent repetitions of the component. For each n, the stopping rule In used in the npp component can depend on (5,1,...,§ n"1). Thus, the sequence of observed random vectors is not the usual iid sequence. This fact raises interesting problems in regard to the efficient use of such random vectors in estimating stopping rules and decision rules for the empirical Bayes application. This thesis will not address these problems. Rather, it will address the issues of the convergence of empirical Bayes risk to envelope risks such as RB(G) and RF(G) and t0 one-step look ahead risk RL(G) for selected empirical Bayes decision procedures based on BnF< (assumed) consistent estimators of G or of 1 G), L(G) and go). 1 -15- We will now indicate the reason why we restrict T to I such that TO = 0. Consider the empirical Bayes problem with the sequential m-truncated component and envelope RB' For specificity, take 0 = {O,l,...,a} and G to be the a—dimensional simplex of probability vectors on o with norm and Borel o-field that induced a+1 from X For certain component loss structure, transitions P e and cost 0, there will exist an open set 60‘; G such that, for GeGO, 10(G) = 1, i.e., G such that a Bayes st0pping rule with respect to G in unrestricted T calls for taking no observations. (See Example 1.1.) Hence, if OH is an a.s. consistent estimator of 686 and if GeGO, then the empirical Bayes strategy based on the estimate' Gn will a.s. cause sampling to terminate after a finite number of repetitions of the component and, therefore, will not enable consistent estimation of the mixing distribution. Hence, it will not be possible to achieve the envel0pe risk RB(G). For this reason, we have restricted consideration to the class of stopping rules T which take at least one observation, i.e., such that T0 = 0. (In this cage note,Nthat the first coordinates of the observation vectors X 1,...,X n,... are iid so that the decision-maker does have standard data available with which to construct consistent estimates of G.) Suppose that the stopping rules are restricted to a class T* g;T. The envelope risk associated with T* is R*(G) inf {R(G,(l,§)): (1,§)€T* X 9} (1.18) inf {R(G,(l,§(G)): 1€T*} -17- * which we assume is attained at l (G), GeG. The nth stage empirical . N N Bayes risk based on the estimate G = Gn(X 1,...,X n-1) and conditional on the data satisfies 0 : R(G,g(é)) - R.(G) : R(G._q(é)) - R(G,d(e)) (1.19) + R(G,g(G)) - R(G,g(é)) for G, Geo, where we have used abbreviations g(H) = (1*(H),§(H))s HeG. Bounds like (1.19) are basic in empirical Bayes analysis and go back at least to Hannan (1957). Hence, (1.20) 0 5 R(G,g(é)) - R*(G) 5 2 sup {|R(é,g) - R(G,g)|: geT*xD}. Thus, if the convergence of the estimator is in the metric defined in RHS (1.20), the empirical Bayes procedure based on G will be asymptotically optimal (a.o.). Theorem 1.1. Suppose that O = {O,l,...,a} and that the loss function L is bounded. Suppose that T* is a specified subset of T, R* denotes the associated envel0pe risk function and that R*(G) is attained by (1*(G),§(G)), GeG. Suppose further that N n n _ ,...,X n 1), n = 2,3,... is an a.s. consistent estimator of 656. (Here we identify G with the a-dimensional simplex of probability vectors on 0 and we denote the sup norm on Xa+1 by II II). Then the empirical Bayes procedure g“ = (1 is a.o. on G. -18- Proof. Since L is bounded, the risk set associated with the component is bounded; specifically, 0 3 R(e,(1,§)) 5 L + c m for 0 = O,l,...,a. Hence, RHS(1.20) 5 2(L + c m)||Gn - G|| so that (1.20) with 0: Gn implies R(G,g_") + R*(G) a.s., 656. N N Taking expectation with respect to X 1,...,X "'1 establishes the L1 convergence, which is sense in which asymptotic optimality is usually defined. D Of course, choices T* = Ti’ T* = TF’ T, = T result in envelopes R* = R R* = R R* = RB defined in the last section i’. and displayed for a particular component in Example 1.1. For a component where T, includes truly sequential stopping rules, the implementation of the empirical Bayes stopping rule 1(Gn) may require considerable calculation involving backward induction. We will not attempt to generalize Theorem 1.1 to cover a general infinite 9 component. In the next chapter we investigate empirical Bayes one-step look ahead sequential procedures for” certain components where the structure is sufficiently tractable to analysis, namely certain finite action components. -19- CHAPTER 2 TWO ACTION AND MULTIPLE DECISION PROBLEMS 2.1 Two Action Problem In this chapter we will establish asymptotic results for empirical Bayes sequential decision procedures in the cases of linear loss testing and the multiple decision problem components. In this section we describe the linear loss testing problem where two composite hypotheses are tested against each other and derive the one-step look procedure gL = (lL(G),§(G)) relative to G for the linear loss two-action problem. An empirical Bayes sequential decision procedure will be constructed in the next section and asymptotic results will be given in Section 2.3. In Section 2.4 we formulate the one-step look ahead procedure for the general multiple decision problem and exhibit an empirical Bayes sequential decision procedure with asymptotic results. Throughout this chapter, we will assume that the para- meter space 0 is a subset of the real line and f6 3 O is a density function of the distribution P with respect to a given 6 o-finite measure u on (X,B). To conserve notation We will also let fe(xk) denote the product fe(xl)...fe(xk) for xkexk, k 3 1. We wish to test the hypothesis H0: 0 3 00 against H1: 0 > 00 where 0050 . Consequently, the action space A consists of two -20- actions only, namely {a0,a1} where a0 and a1 denote the actions of deciding HO and H1, respectively. We assume a linear loss function, specifically, (2.1) L(e,a0) = (e - 00)+, L(e,a1) = (e - 00)', see . We assume that the first moment of e is finite with respect to G, where G is the prior distribution on o. This is sufficient to ensure that the Bayes risk of the m-truncated one-step look ahead procedure defined in Section 1.2 is finite. In the literature many authors have studied the two-action problem from the standard empirical Bayes point of view. Samuel (1963) discussed the two-action problem and exhibited a.o. empirical Bayes tests under various loss structures and, in part, dealt Specifically with certain types of discrete exponential families. Yu (1970), Johns and Van Ryzin (1971, 1972) considered the linear loss two- action problem with exponential families and developed rates of convergence in the regret ER(G,6n) - RB(G) where 5n is an empirical Bayes test for the fixed sample size linear loss two- action problem. Van Houwelingen (1976) proposed monotonizing empir- ical Bayes tests defined in Johns and Van Ryzin (1971, 1972). O'Bryan (1972), O'Bryan and Susarla (1975) treated the testing problem where the sequence of component problems consists of independent but not identical decision problems, all having the same unknown prior dis- tribution. These sequences of decision problems are identical except for the sample size. Laippala (1979, 1980, 1983, 1985) discussed the -21- two-action problem with varying random sample sizes in Binomial and exponential conditional distributions. Now let us determine the m-truncated one-step look ahead procedure gL = (11(G),§(G)) with respect to G for our testing problem (see Section 1.2). The terminal decision rule §(G) is a finite sequence (61,...,6m) where okevk(k=l,...,m) is a Bayes decision function relative to G for the fixed sample size k testing problem (e,A,L) with the loss function (2.1). For k 3 1, 5k can be determined as follows. Let 6k(xk) = Pr {choosing aOIX1 = Xl""’xk = xk} be a randomized decision for the two-action problem with the loss function (2.1) given that the observations are X1 = x1,...,Xk = xk . Then the Bayes risk of decision function 6k relative to the prior distribution G is given by r(e,sk> = ka f9 {L(e.a0>ak(sk) + L(esa1)(1-6k(5k))} P§(dxk) G(de). k Since fe(xk) is a conditional pk-density of x = (x1....,xk), and L(e,a0) - L(e,al) = e - 00 (see (2.1)), one can write (2.2) r(e.sk) = ka ak(sk> 6k(§k) pk (dsk> + Ca where CG = f0 L(e,al)G(de) and (2 3) ak(§k) = f9 (6-60) fe(§k) G(de) . From (2.2) and (2.3) it is clear that a Bayes rule (a minimizer of r(G,6k) for given G) is provided by the nonrandomized rule -22- 1 if ak(xk) < 0 (2.4) (5“) = 5 k . k 0 lf ak(§)>0. Note that we have suppressed the display of dependence of the Bayes rule 6k(xk) on G. Henceforth, the terminal decision rule §(G) = (61,...,6m) for the m-truncated one-step look ahead procedure dL = (1L( G),§(G)) defined in Section 1.2 is given by (2.4) and (2.3), k = 1,2,...,m. Then the minimum posterior Bayes risk r(Gk) with respect to Gk’ for k 3 1, is given by r(Gk) = IO L(e.ak(sk)> ek(de> where Gk is the posterior distribution of 6 given Xk = xk. Thus for the linear loss testing component, k> r(Gk) - (a {L(6.a0)6k(x + L(e.a1)(1-ak(sk))} ek(de> (2.5) x9 {(L(e,a0) - L(0,a1))5k(xk) + L(e,a1)} Gk(de). Since L(e,a0) - L(e,a1) = e - 00, (2.5) can be written as r(Gk) = CG + 0k(xk) f0 (0-00)Gk(d0) (2.6) 1 —————- ; L(e.a )i (x k) G(de) + provided fk(xk) > 0 and r(Gk) = 0 if fk(xk) = 0 where for k 3 1, fk(xk) = f0 fé(xk)G(de), and ak is given by equation (2.3). The expected risk E*r(Gk+1) from taking Xk+1 observation and playing Bayes conditional on Xk = xk is given by -23- ‘k ‘k (2.7) E r(Gk+1) = [X r(Gk+1)f (xk+l)“(dxk+1) where f*(xk+1) is a conditional density of Xk+1 given 5k = 5k . Let fk+1(Xk+l) and fk(Xk) be unconditional marginal k+1 k densities of X, and X respectively. Then a conditional density + k 1Niko") if ik(§k) > 0 and f*(xk+1) = 0 if ik(§k) = 0. Using this fact of X iven Xk = xk is iven b f*(x ) = f (x k+l 9 — — 9 Y k+l k+1 — and changing k to k + 1 in equation (2.6) and then substituting back in (2.7) we get * ' 1 E k(Gk'tl) = m f0 L(6,al)fe(Xk)G(dO) + k ._- (2.8 ) —1— x 5 (W1) (xm) (d k x k+1 - ak+1 — “ xk+1) fk(1 ) provided fk(Xk) > 0 and E*r(G = 0 if fk(Xk) = 0 where k+1) 5k+1 and ak+1 are given by (2.4) and (2.3), reSpectively, with k replaced by k+l. Letting k _ k k (2-9) 01((5 ) ' IX [aki’l : 0] ak+1 U(dxk+l) + ka(_)£ ) ‘ [0k : 0] “k(fi )9 we observe that pk(Xk) = 0 when fk(Xk) = 0 and c if fk(X ) = O . L - ( L L) 11,...,rm of m-truncated one- Hence, the stopping rule I step look ahead procedure (1.10) for our testing problem is defined -24- by T; s 1, and for k = 1,...,m-1 by 1 if pk(5k) > 0 (2.10) it(xk) = 0 if pk(5k) < 0 where we have suppressed the display of dependence on G. Let NL be the stopping time of m-truncated one-step look ahead (imam); then procedure gL NL(Xm) = min {kltt(Xk) = 1} and Since I: “ 1, sampling will be stopped after Xm has been ll observed if it had not been stopped earlier. The risk of this procedure at G is (see (1.1) and (1.2)), m (2.11) R G) = f9 f m [NL=k](L(e,6k(Xk)) + ck) f k=1 X M L( e(5m)G(d6)um(d,>_ 0] for k = 1, ..... m-1 (2.13) Bk = [p1 < 0] .......... [pk_1 < 0][pk = 0] for k = 1, ..... m-l Am = [pl < O] .......... [pm_1 < 0]. Thus, the Bayes risk of m-truncated one-step look ahead procedure L L( g = (l G),§(G)) relative to G for our testing problem can be written in the following form (see (2.12)): m RL(G) = CG + kil fefxm Ak([01k _<_ 0](e - so) + ck) fe(Xm)G(de)pm(dx_m) (2.14) m-l m m m + kEI fefxm Bk([ak 3 O](e - 60) + ck)fe(X )G(d9)u (dX ) . The inequalities defining the indicators correspond to open sets for the Ak and boundary sets for the Bk' In the empirical Bayes application, the functions pk are estimated so that a separate treatment of bound- aries is important in so far as convergences are concerned. -25- 2.2 An Empirical Bayes One-step Sequential Decision Procedure for the Two Action Component Case Suppose that the prior distribution G is unknown but fixed; then the classical Bayes quantities (2.3)-(2.9) of the Section 2.1 are not available to the statistician. However, suppose that we are experiencing independent repetitions of the same component problem. Then applyingthe empirical Bayes approach introduced by Robbins (1956), we may derive empirical Bayes estimates of the classical Bayes quantities (2.3)-(2.9), and, hence, an empirical Bayes one-step sequential decision procedure g” = (In,§n), where In is an empirical Bayes stopping rule and g” is an empirical Bayes terminal decision rule. In order to construct an empirical Bayes sequential decision procedure db = (In,§n), we will make the following assumptions on conditional density fe(x) and the parameter space 0. (A1) For each x, fe(x) is a continuous function of 0. (A2) 9 is compact. At the nth problem of the repetitions, we will have observed N N the random vectors X11,...,anil from the past (n-l) repetitions of the component problem 2.1, where N1,...,Nn_1 are the respective stopping times of the past repetitions. Let {Gn} be a sequence of N N distribution functions on o, where Gn(e) = Gn(6,X1 s~°°$_n_l N N depends only on the random vectors X11,...,XnT11, which converges _27_ weakly to the prior distribution function G with probability one as n + m, that is, Pr {lim Gn(e) = G(e), any continuity point of G} = 1. n+0!) Remark 2.1. Robbins (1964) showed the existence of such a sequence of distribution functions on e = (a,b), - m 3 a < b 3 w, under the following assumptions. (a) For each x, Fe(x) iS a continuous function of e, where Fe(x) is the conditional distribution function of X. (b) If G and G2 are two distributions on e such that I Fell = FGz then G1 = G2 , where FG(x) = f0 Fe(x)G(de) . (c) The limits lim Fe(x) and lim Fe(x) exist for each x G+a G+b and neither lim Fe(x) nor lim Fe(x) is a distribution function. G+a G+b He showed that when 0 is a compact subset of R, condition (c) can be relaxed. Now we define our empirical Bayes sequential decision (EBSD) procedure g” = (ln’én) as follows. Let g" be a finite sequence of functions (69,...,6;), where n 62 is such that 6k(Xk) = Pr {choosing aole = Xk} and, motivated by (2.4) and (2.3), (2.15) afl(_k) = 0 if 01:35“) > 0 and (2.16) aflo") = I, (e - eo)fe(§k>Gn(de) Let in be a stopping rule consisting of a finite sequence of functions (Tn,...,Tn) where, motivated by (2.10) and (2.9), 1 m T; a 1 and, for k = 1,...,m-1, 1 if pE(Xk) > 0 (2.17) T135") = 0 if 02(Xk) < 0 where n k _ n k+1 n k+1 n k n k n k (2-18) “((1 ) ' IX 6k+1 (5 ) ak+1(2(_ )U(ka+l) + ka(5 ) - 61((5 )ak(§) . n k _ k . n z . . with fk(X ) — fepfe(X )Gn(de) . Since Tm — 1, sampling Will be stopped just after Xm has been observed if it had not been stopped earlier. For investigating the risk of the EBSD it is useful to define CE = [pg < 0] .......... [pE_l < 0][pE 3 0] for k = 1,...,m-1 (2.19) n _ n n Cm - [91 < 0] .......... [pm_1 < 0] . Then [Nn = k] = CE for k = 1,...,m-l and [Nn = m] = Cg , where gn = (_Tn’ 6“) Nn denotes the stopping time of the EBSD procedure m m Note that 2 [Nn = k] = 1 implies 2 CE = 1 . k=0 k=1 Let R(G,gn) denote the conditional Bayes risk of g“ = (1n,gn) with respect to G. Then Since the CE partition an, -29- Ill (2.20) k(G,g“) = CG + kEI f9 me cflqafl 5 0](e - 00) + + ck) fe(Xm)G(de)um(de) . In the next section we will treat the difference between the empirical Bayes risk R(G,gn) and the one-step look ahead risk R G) using the decompositions (2.20) and (2.14). L( -30- 2.3 Asymptotic Results for the Two Action Component Case In this section we compare the asymptotic behavior of uncondi- tional Bayes risk of the Section 2.2 EBSD procedure g” with the Bayes risk of m-truncated one-step look ahead procedure gL discussed in Section 2.1. First we prove the following useful lemma. Convergence of sequences of functions on .Xm is understood to be pointwise convergence. Lemma 2.1: Under the assumptions (A1) and (A2) for k 3 1, n pk+pk w.p.1 as n+oo. Proof: From (2.9) and (2.18) with dependence on Xk suppressed, |A IA 0k = (x [“k+1 0J0‘k+1“(d"k+1) + ka ‘ LO‘k 0J0‘k and II n pk ‘ ’x [“k+1 IA A n n n n 0]o¢k+1u(dxk+l) + ka - [01k _ QJOLk . By assumptions (A1) and (A2), fe(Xk) is a bounded and continuous function of e for each Xk. Then recalling the definitions of afl, ak, ffl, fk (see (2.16), (2.3), below (2.18) and below (2.6)) and the assumptions on the sequence {Gn}, we get for k 3 1, (i) aE + ak w.p.1 as n + w and n k + fk w.p.1 as n + m . From (i) it follows that -31- (2.21) [0: 3 OJoE + [ak 3 OJak w.p.1 as n + w , k > 1. n n To Show that IX [ak+1 3 O]ak+1u(dxk+1) + IX [ak+1 3 O]ak+1u(dxk+1) w.p.1 as n + m , we use the Generalized Dominated Convergence Theorem (GDCT). Note that lak+1(X qk+1)[ak+l_ < 0]] : hk+1(_ k 1) where n k+1 _ k+1 k+l _ hk+1(X ) — fle - Golfe(X )Gn(d6) . Then :12 f hk+1(x )u(dxk+l) - . k+1 _ k Tim I fle - eolfe(X )Gn(de)p(dxk+1) - f|e - 00|f6(X )G(d6) w.p.1 n+oo by the assumptions (Al) and (A2). But fle - Golfeqk sk+1) )G(de) is equal to f fl9 - 60|f6( G(d6)u(dX and note that the k+1) following equality is satisfied : k+1 _ . k+1 f f l8 - eOlfeQ )G(de)u(dxk+1) - I Al: I la - 60|f6(§ )Gn(de)u(dxk+1) _ k+1 - I Tim hk+1(x )p(ka+1) w.p.1 . Now use the GDCT and (2.21) to get the required results. D Now we state and prove a theorem which concerns the asymptotic behavior of the conditional Bayes risk of empirical Bayes sequential decision procedure g“ = (1n,§n) defined in the previous section. Theorem 2.1: Under the assumptions (A1) and (A2) (2.22) lim sup R(G,g”) 5 RL(G) w.p.1 n+oo where R(G,gn), and RL(G) are given by the equations (2.20) and (2.14), respectively. -32- Proof: The difference of R(G,gn) and RL(G) can be written as R(G,d") - RL(G) = g g I I CU A.{([a9 3 O] - [o- 3_0])(e - e ) i=1 i=1 ‘ J I 3 O + c(i - i)}fe(xm)e(de)nm(dxm) (LB) +31? waetuw J, Ci Aj 3 [pj < 0][pj > 0]. Then observe that -33- 1 m j-l n . . ldnl _ .E .E f f Lo; 3 OILpi < OIIZIG - eol + cli - 3|}fe(Xm) 3-2 i-1 G(de)nm(dxm) (2.24) m j-I n i Z X f f [0,- < OHIO,- - oil _>_ loi|]{2|e - eol i=2 i=1 + cli - Jllfe(§m)G(d6)um(d§m) . ~— n But fle - 90|G(de) < w, and [pi < OJLlpi - oil 3 [oil] + O w.p.1 as n + w by Lemma 2.1, and hence the DCT gives Idil + 0 w.p.1 as n + w . Similarly, one can Show that ldfil + 0 w.p.1 as n + m . n Since C?, Ai’ a1 and a1 depend only on the first i observa- tions (X1,...,Xi), 0: can be written as n n ' ' f . Ci A1 {[01 3 O] - [oi 3 0]}ai u1(dX1) . . n n BUt for 1:1, |([aI:0]-[a1:0])al )ELIO-i 'aililaiIJIGil so that "MS 3 n . . ldnl s IX, Ila,- - a,l:la,l1la,lu‘(dzs‘). i 1 But I la - 60|G(de) < m implies f 1.loijlpi(dXi) < w, and for i 3 1 , X an + a1 w.p.1 as n + w (see proof of Lemma 2.1) implies for i 3 1 , .i (2.25) [ail [lag - ail :_|ail] +~() w.p.1 as n + w . Now apply the DCT to get |Jg| + 0 w.p.1 as n + w . To derive the asymptotic behaviour of the second double sum in (2.23), observe that it can be written as Ki + Kg + Kg where -34- k1 - m-l 3.1 c” B " 0 n "jiz 121}. f l J {([0‘1 : ] " [015}: 0])(e ‘ 60) + c(i - 1)}fe(xm)e(de)nm(dxm) . 2 m-l n n m Kn = iil I I Ci B1 {([a1 3 0] - [a1 : 01)(6 - 90)}fe(§ ) GIdeIImIdxm) and m-l m (2.26) Ki = z z I I c? Bj {([a? 3 0] - [aj 3 0])(0 — e0) i=1 i=j+1 ' + c(i - 1)}fe(xm)e(de)nm(dxm) . We will show that k; a 0 w.p.1 as n . m , i = 1,2, and lim sup K3 n n+w For i < j, c? Bj < (a? 3 O][pi < 0] so that _— 3 0 w.p.1 . 1 m-l j-I n IR") 3 z z I I [pi 3 0][o. < 0] {2|e - e | + c|i - jl} ._ ._ l 0 3-2 i—l fe(§m)G(de)um(d§m) M ‘H [ l I " I I 3 z 2 I I o. < 0 [ o. - o. 3 lo- ] {2|e - e | j=2 1:1 1 l i i 0 + Cli - J|}fe(§m)G(d6)um(d5m) . This is the bound (2.24) for lJé] which.was shown to converge to zero w.p.1 as n + w . Hence, lKil + O w.p.1 as n + w . Now note that Ki can be written as -35- -1 I o 2 _ m n n i l The above equality follows from the fact that the functions Cg, Bi’ 0? and oi depend only on the first i observations (X1""’Xi) . Thus, 2 m'1 n i i lKnl 31:1] [191' ' “1| 3- '911110‘1111 (d5) - This is the bound for ngl which was Shown to converge to zero w.p.1. We summarize the results concerning the difference (2.21) obtained so far with (2.27) R(G,g”) - R 1 2 -3 1 2 where J”, J”, on, n n have been shown to converge to zero w.p.1. From (2.26) we can write 3 m'1 . . . (2.28) Kn = jEI (L1(n,J) + L2(n,J) + L3(n,J)) where m (2.29) L1(n,i) = 1.=§+1I I c? Bj Ia? s 01(e - 60)fe(§m)G(d6)um(d§m) (2.30) 1201.1) = 3? I I c? 83. Ia, _<_ 01(e - eoiieommideiamws'“) i=j+1 and m (2.31) L3(n,j) = z I I C? B (i - j)fe(§m)G(d9)um(d§m) . i=j+1 . C J -35- A m We define CE = z c - i=k do: for k = j + 1,...,m, j = 1,..., m-l and observe that CE = [of < 0] .......... [p:_1 < 0] for k = J + 1,..., m, and, hence, CE depends only on the first k - l observations (x1....,xk_1) . Now using the definition of GE , k 3 1, we can write Ll(n,j), L2(n,j) and L3(n,j) as follows: in L3(n,j) = c z I I E: B. fe(Xm)G(de)um(de) k=j+1 J m “n k-l k-l k-l (2.32) = c k I 1I ck Bj fk_1(5 )o (d; ) :J+ m = 2 Ak(n,J) k=j+1 where Ak(nsj) = c I 0E BJ. fk_1(Xk-1)uk-1(ka'l) . Also . _ _ n n _ m L2(n,a) — I I (Cj+1 + ... + Cm) BJIaj : 01(9 90)Io(5 ) G(de)um(dxm) (2.33) = -I I 03,1 Bj[oj : O](e - 00)fe(xm)G(de)nm(dxm) = _ “n j i f Cj+1 Bj [OJ : OJOIJ-p (dX) and m n n i i L1(n,j) = z I C B. [o. < OJa.p (dx ) i=j+1 i J i — i — (2.34) 1 m . . . m- A = 2 I GD 3 [a9 < o]a.u‘(dX‘) - z I C? B. i=j+1 l J 1 — i i=j+1 1+1 J [0? OJaip1(dX ) -37- The last equality follows from the fact that C? = C? - C" 1+1 for i=j+1,...,m-1, j=1,...,m-1 and c; = C; . We can write m m-l (2.35) L1(n,i) = Z T,(n,j) - 2 51(n,j) i=j+1 i=j+l where Ti(n,j) = I C? Bi [0? 3 0]0i0i(dXi) and . _ “n n i i . Sj(nsJ) - f C1+1 Bj [oi 3 OJGiu (dX ) . Thus the sum of L1(n,J) , L2(n,j) and L3(n,j) can be written as m m-l L1(n,j) + L2(n,i) + L3(n,j) = 2 Ti(n,i) - 2 51(n,j) i=j+1 i=j+1 A - I c" B [aj 3 0]oij(dXJ) 3+1 3 m + 2 Ak(n,j) k=j+1 = . _ “n j j m + E [T(naj) " 5' (71,3) i=j+2 1 1"]. + A1(n.i)1 M1(n,i) + M2(n,j) where -38- A man MWJ)=BHUm)-IQHBJUJ:W%NNQI+%HWJ) and m (2.38) M2(n.j) = z [Ti(n,J) - Sj_1(n,J) + 4i(n.j)] . i=j+2 Now observe that M1(n,j) is equal to “n n J J (2.39) f Cj+1 Bj {f[aj+1 3 0]aj+1p(dxj+l) - [aj 3 0]aj + ij}p (dX ) . But 0 3 Bj 3 [pj = 0] and by the definition of pj (see (2.9)) from (2.39), |M1(n.j)| I A _ n f ij’ 0](lf [aj+1 : OJGJ+1U(de+1) (2.40) -I Laj+13 01oj+1u(dxj+l)|)p‘j(de) n .+l .+1 |A . n The R.H.S. of (2.40) goes to zero w.p.1 as n + w, Since aj+l + aj+1 w.p.1 (see the proof of Lemma 2.1), I la luj+l(de+l) < m and by 3+1 an application of the DCT. Thus, M1(n,j) + O w.p.1 as n + w . Now it remains to consider M2(n,j) in (2.38). Observe that for I=j+2,...sms Ti(nsj) ' Si_](n,j) + Ai(n9j) IS equal to “n n i i “n n - i-1 i-1 f Cl Bj [01 : 0]a1u (dX ) - f C1 Bj [ai_1 : 0Ja1_1p (dX ) +IC?%Cfi4QPHN4N§4)- The above sum can be written as -39- (2.41) I c? Bj (I [0? 3 O]o1p(dx1) - [0.2.11 3 0]a._ )ui-l(dXi-1) - + Cfi-I — Adding and subtracting the term [ai_1 3 0]ot1._l into the integrand of the above integral (2.41), we get A n n f C1 Bj (f [oi 3 O]oiu(dxi) - [ai_1 3 01ai_1 (2.42) + I 0? Bi ([a._ Now use (2.25), I lai_1|01'l(dX1'1) < w , and the DCT to Show that “n n i-l i-l I C, Bj ([o1._1 3 O]o1._1 - [ai_1 3 0]a1_1)u (d3 ) goes to zero w.p.1 as n + w . First integral in (2.42) can be rewritten as f C? Bj [pj_1 : 01(f [0? : 0191P(dxi) ' [01-1 5 OJO‘i-l + cfi_1)pi'1(dXI'1) (2.43) an n + [C1 Bj [pl-1 > 0](f [OL_i : 0]aiu(dx1) ' [ml-1 : 0]ai_1 i-l i-l + cf1_1)u (dx ) . But 0 3 c? [o,_1 > 0] 3 [o?_1 < O][01_1 > 0] and therefore the -40- absolute value of the second integral in (2.43) is less than or equal to (2°44) f [p?_1 < O][pI-1 > 0]{f l01|U(dX1) + IaI-ll + Cfi_1}U1-l(d§1-l) : and (2.44) is less than or equal to f [lp?_1 ' 01_1| Z lpl-1|][pl-1 > 0]{f|0i|U(dxi) + laI-1|+ (2.45) l-l l-1 + Cf1_1}u (d5 ) ' 1'1) < w and the We use Lemma 2.1, flail gi(dXi) < w ,flai_1|pi-1(dX DCT to conclude that (2.45) goes to zero w.p.1 as n + w , and, hence, the second integral in (2.43) goes to zero w.p.1 as n + w . By the definition of pi-l (see (2.9)), observe that cf1._1 - [ai_1 3 0] o1_1 3 -I [a1 3 OJaip(dXi) on [pi_1 3 O] . Then the first integral in (2.43) is less than or equal to the following expression f C? Bj L91_1 : 0](f [0? E O]aiU(dXi) 'f [a1 : 0]01U(dxi)) (2.46) . . “1-1(dl1-1) 3 and the absolute value of (2.46) s I la,l|[a? 01- [01,: OJIquzi) IA 5 f lailtla? 4,! : |a1|]u‘(dx‘) . which goes to zero w.p.1 by (2.25), I |o1|p1(dx1) < m and the DCT. -41- Now combining (2.39) - (2.46) we get for i=j+2,...,m lim sup (Ti(n,j) - Si_1(n,j) + Ai(n,j)) 3 D w.p.1 . n+w Hence, by the definition of M2(n,j) (see (2.38)) for j=1,...,m-1, lim sup M2(n,j) 3 0 w.p.1 , and by (2.36) lim sup (Ln(n,j) + L2(n,j) n+°° n+oo + L3(n,j)) 3 0 w.p.1, and by (2.28) lim sup Kg 3 0 w.p.1 . This n+oo completes the proof of the Theorem 2.1 . U The next corollary compares the asymptotic behaviour of the unconditional Bayes risk of EBSD procedure g" with RL(G) . Corollary 2.1: Under the assumptions (A1) and (A2) lim sup E R(G,g”) 3 RL(G) , n+oo where E denotes expectation with respect to random vectors XNI XNn-l _1 ,ooo,_n-1 0 Proof. The proof follows from Theorem 2.1 and Fatou's lemma. D Corollary 2.2: If m=2, then lim R(G,d”) = RL(G) w.p.1. and n+m lim E R (e.gn) = R n+oo L(G). Proof: If m=2, then RL(G) = RB(G) and, therefore, R(G,g") 3 RL(G) for all n and G. Hence, lim R(G,gn) = RL(G) w.p.1 follows from Moo (2 22) and then lim E R(G,g") = R n+m L(G) follows from the DCT. D -42- Corollary 2.3: If G is such that I I ije(Xi)uj(de)G(de) = 0 , j=1,...,m-2, then lim R(G,g“) = R n+w L(G) w.p.1 . Proof: From (A2) and (2.3) there exists a constant B such that |a1| 3 Bf, , i = 1,...,m and I|a1|u(dxi) 3 Bf1_1 , i = 2,...,m . Hence, by (2.38) and (2.41) for j = 1,...,m-2 , m . . . |M2(nsj)l < >2 I B]. (28 + c)f1._1(x_1 1111‘ 1(dx‘ 1) . i=j+2 But Bj is 33 measurable so RHS of the above inequality is zero by the hypothesis of the corollary. Combining this result with Tim M1(n,j) = O w.p.1 as n + w , we obtain lim K2 = 0 w.p.1 as n + m so that from (2.27) the proof is complete. D Corollary 2.4: Let Nn be the stopping time associated with the EBSD procedure 9" and let NL denote the stopping time associated with the one-step look ahead procedure 9L. Then w.p.1, Nn is stochastically larger than NL as “.1 w ; Specifically, for i = 1,2,...,m, lim inf I IIN” 3 i]fe(Xm)G(de)nm(de) 3 II [NL 3 i]fe(Xm)G(de)pm(de) w.p.1 n+oo with convergence to the RHS if II ije(XJ)pJ(dXJ)G(de) = 0, j = 1,2,...,m-1 . Proof: By the definitions, [Nn 3 i] = [.2 < 0] .......... [,?_1 < 0] and [NL 3 i] = [01 < O] .......... [01_1 < 0]. Thus, we have [o1 < 0] ........ [o. 1 < 0] 3 lim inf IN“ 3 i] 3 lim sup [Nn 3 i] 1- n+oo n+oo (2.47) : [01$ 0] oooooo [pi-1: O] 3 and the inequality follows from Fatou's Lemma. Observe that .i 0 3 RHS(2.47) — LHS(2.47) 3 1' 1 1 Bj . IIMI Hence, by the hypothesis concerning the Bj, we have lim [Nn 3 i] = [NL 3 i] w.p.1 as n + w , i = 1,2,...,m. -43- The proof is completed by taking expectation and applying the DCT. D -44- 2.4 Multiple Decision Problem In this section we treat a component decision problem that subsumes that treated in the last section. Assume that the action space A = {a0, a1,...,a£} consists of a finite number of distinct actions and let L(e,a) 3 O on o x/\ be the loss function associated with the problem. The m-truncated one-step look ahead sequential decision procedure QL = (3L(G), 3(6)) with reSpect to G, GeG, for this multiple decision problem (e,A,L) can be defined as follows. Let §(G) be the decision rule consisting of a finite sequence of functions (61,...,om) where 5k is a Bayes decision function with respect to G for the fixed sample size k decision problem based on the sample (X1,...,Xk). 5k can be derived as follows (see Van Ryzin and Susarla (1977), Gilliland and Hannan (1977) or Ferguson (1967) Chapter 6). At the stage k of the sequential decision procedure, suppose k) = (tk(0|xk),...,t tk(j|Xk) = Pr{choosing action alek = Xk} and we use a decision rule tk(X Then the Bayes risk of tk(-) w.r. to G is .IIItkIIIBkIPSIdBKIIGIde) r(G,tk) = Z I L(e,aJ I- tk(J|§k){I(L(6.aj) - L(6,a0))fe(Xk)G(d6)}uk(ka) + II L(e,a0)G(de)fe(§k)uk(dxk) . r(G,tk) is minimized by tk(ilxk) = ok(jlxk), j = O,l,...,m where -45- 1 if 3k e Sj (2.48) akII'Ix") = 0 if xk 4 S. — J with k . . . 53' = {X [J = min {t: Ak(at,Xk) = m}n Ak(a.,3k)}} (2.49) k I, (are ) = I(L(e.a,) - L(e.a0))fe(xk)G(de) . When arguments are displayed we will delete the subscripts on 5k and Ak. Therefore, the Bayes terminal decision rule §(G) = (61,...,om) is defined componentwise by (2.48). Then the minimum posterior Bayes risk w.r. to Gk is given by Me I L(6,6k(2<_k))Gk(d6) k) .)G I6(J‘ka)L(e,aJ k(do) (.1. "MN "MN O ‘- A I— A G) \D D.) V I O l K V L: 0': A (_i. X L(e,a0))Gk(de) L4. l o fk(§K) k) = 0 if fk(Xk) = 0. In the above 2 derivations we use the fact that z 6(lek) i=0 to get r(Gk) in the following form: (IKIGIde) I L(e,a0)fe k) provided fk(X > 0 and r(G = 1. Now use (2.49) -45- (2.50) r(G (5k)G(de) . k) I L(e,a0)fe j— o ik(x The stopping rule 3L(G) of gL consists of a finite sequence of functions (T%,...,T;) and stops sampling for the first k (k=1,2,...,m) L L _ l for which tk(Xk) = 1 where rm = and for k=l,...,m-1 , L k 1 if E* r(Gk+1) + c - r(G “((5 ) = 0 if E* r(G + c - r(Gk) < O k+1) and E* denotes the conditional expectation over Xk+1 given Xk = Xk . It is easy to Show that E* r(G ) = 1 I I o(j|xk+l)A(a. xk+1)u(dx ) 1 k + k I L(0,a0)fe(X )G(de) fk(§ ) . k _ ~ k _ prov1ded fk(X ) > O and E* r(Gk+l) - 0 if fk(X ) — O . Then E* r(G ) +~c - r (Gk) 3 0 if and only if k+1 K . k+1 k+1 k z I 6(Jlé )4(a ._ )u(dx ) + cf (1 ) z ._ J k+1 k - J-O J when fk(Xk) > 0 . We define sk+1>aIdxk+1> + ka(§k) "MN aIiIkk)A(a.,3k> O J -47- if fk(3k) > 0 and Bk(3k) = 0 if fk(3k) = 0, and the functions Ak = [81 < O] ..... [Bk-1 < 0][Bk > 0] for k=1,...,m-l Bk = [Bl < 0] ..... [Bk-1 < O][Bk = 0] for k=1,...,m-1 and I> ll m [81 < O] ..... [Sm-1 < 0] . Then [NL = k] = Ak + Bk, k = 1,...,m-1, [NL = m] = A and m m'1 L L kil (Ak + Bk) + Am = l . Therefore, (T1,...,Tm) is defined by 1 If Bk(§k) 3 O L k (2.52) Tk(§ ) = 0 if Bk(§k) < 0 , k=1,...,m-1 and tm : 1 The Bayes risk of gL = (3L(G),§(G)) with respect to G is m L k m m m RL(G) = Z IILN = k1{L(6.6(§ )) + cklfe(§ )G(d6)u (d4 ) k=1 m L K . k m m m = z IILN = k]{ 2 5(Jlx )L(6,a ) + cklf (5 )G(d6)u (d5 ) . k=1 i=0 3 9 Now add and subtract L(e,a0) into the integrand of the above integral 2 m and use the facts 2 o(j|Xk) = 1 and z [NL = j] = 1 to obtain i=0 i=1 m L f k R (G) = 2 II [N = k]{ z 6(Jlx )(L(e,a ) - L(e.a )) L _ ._ J O k-1 3—0 + ckife(xm)e(de)wm(dxm) m L m m m + 2 II [N = k]L(e,a0)fe(x )G(de)u (d5 ), k=1 that is, -43- _ m 2 . k RL(G) - kil I I Ak jED 6(JIX )(L(e.aj) - L(e,a0)) + ck}fe(§m)G(d9)um(d§m) m-l 2 k (2 53) + z I I Bk{ 2 6(ilx )(L(e,aj) - L(6.a0)) k=1 j=0 + cklfe(§m)G(d6)um(dz<_m) + I I L(e.a0)f (Xm)G(de)um(de) 6 Let g“ = (ln’én) denote an empirical Bayes one-step sequential decision procedure for the above multiple decision problem. We shall N construct 9" based on the past data X 1,...,anil and the present data X at the stage n. Assume that there exists a sequence of dis- . N N _ tribution functions {Gn} on o, where Gn(e) = Gn(e,X_1,...,Xn211) N N depends only on the past data X 1,...,an11, which converges weakly to G with probability one as n + w . Let g" be the decision rule consisting of a finite sequence of functions m o?,...,o”) where on(j|Xk) = 1 and, ( 2 o£(Xk) =(on(0|Xk),...,on(£|Xk)) subject to '2 . . n . k .3 k k for J = O,l,...,t, if 5 (Jll ) = Pr{ch005ing ale = X }, then 1 if xk e S. n.k — J (2 54) 6 (J11 ) = k . 0 if X 4 Sj “ _ _ . n k _ . n k SJ - {x [J - min{t:A (at,X ) - min A (ai’5 )} , and l (2.55) I"(a xk) = I (L(e a ) - L(e a )f (xk)G (do) i’— ’ i ’ O B — n -49- The stopping rule in is defined by a finite sequence of functions n) which stops sampling for the procedure g" for the first m k(k = 1,...,m) for which tfl(3k) (.Q,...,. = 1 where t; 1 and for k=1,...,m-1, a“(xk) = k _ n k 0 If Bk(£ ) < 0 9 where (2 56) n(xk) = i I a”( ka+1)I"( xk+1) (dx ) + cfn(xk) ° 8k — i=0 x J 3’ “ k+1 k — Z - z 6"(Jlek)4"(a ask) i=0 3 . n k _ k w1th fk(X ) — I fe(X )Gn(de) . Define CE = [8? < 0] ..... [ofl_1 < 01(82 3 0] for k=1,...,m-1 and n _ n Cm - [81 < O] ..... [Sm-1 < O] . Then [Nn = k] = C: for k=1,...,m where Nn is the stopping time of dn = (3n,§n) for the multiple decision problem. The conditional Bayes risk of g“ = (3n,§n) with respect to G is then equal to m 2 RIG.d“) = z I I GE {.2 a"(i|3k)(L(e,a.) - L(e.a0)) k=1 3-0 3 (2.57) + cklfe(§m)G(d6)um(d§m) + I I L(e,a0)f (Xm)G(de)um(dxm) . O -50- Now we prove a lemma which is used to prove the next theorem on the asymptotic behaviour of conditional Bayes risk of empirical Bayes sequential decision procedure 3" = (in,gn) for the multiple decision problem. Lemma 2.2: Assume L(e,a) is a continuous function of o for each a c A; then under the assumptions (A1) and (A2) for k 3 l Bnk + Bk w.p.1 as n + m where Bk and B: are given by (2.51) and (2.56), respectively. Proof: Observe that for j Z 1, k 1 1,An(a.,Xk) + A(a k) J 1’5 k) k) w.p.1 as n + m by the assumptions in w.p.1 as n + w and f:(X + f k(l . . . n . k . k the lemma. Then by the definitions of 6 (3|X ) and 6(JIX ) we get (2.58) z on(j|X a.,X + z o(j|Xk)A(a.,Xk) w.p.1 as n + w . j=0 J j=0 3 Now to complete the proof of BE(Xk) + Bk(Xk) w.p.1 as n + m , it C remains to prove 2 IX on(j|Xk+l)An(a . o j,xk+l) ( J: dek+l)+.: fxéjl 4(a.,xk+1)u(dxk+1) w.p.1 as n + w . To prove this statement J..— + + . + lexk)14"(a xk 1) + z 6(1l5k 1) we use the GDCT. From (2.58), -,_ 0 J i=0 J "MN k+1) w.p.1 as n + m , and observe that by the definition of k+1 as )|< Aa.,X (J- k+1) (X —k+l) An(aj,X and the boundedness of L(e,a),|A"(a <20 fn k+1 where J is a finite constant and ffl+1(Xk+l) = I fe(Xk+1)Gn(d0) . But I f u dx = fE(Xk) by Fubini's theorem. Then k+1(— k+1) fk(3k) w.p.1 by (A1) and (A2). Therefore, . + n+oo -51- ( k+1) ( I Tim fn Tim I ffl+1 3 (1 dx n+oo (x _k+l)u(dxk+1) w.p.1 follows from k+1): k+1 . + . the equality fk(Xk) = f fk+1(§k l)u(dxk+1) = I Tim fE+1(§k+1)u(dxk+1) n+m w.p.1. It follows from the GDCT that C o + + . k+1 . J ( I ) n+w J 06 J 6.11 0 ”MN k+1 4(aj,X )u(ka+1) w.p.1 . This completes the proof of Lemma 2.2. The next theorem gives us the asymptotic behaviour of the conditional Bayes risk of our empirical Bayes sequential decision procedure g" = (1n,on) for the multiple decision problem. Theorem 2.2: If L(e,a) is continuous function of e for each a c A , then under the assumptions (A1) and (A2), lim sup R(G,gn) 3 RL(G) w.p.1 , n+w where RL(G) and R(G,gn) are given by (2.53) and (2.57), respectively. -52- Proof: Write . m g . R(G,dn) — RL(G) = z z I I c? A { z o”(kIX‘)gk(a) - i=1 j= -1 k= 0 - k§0 6(kl§j>gk(e) + c(i-j)}fe(Xm)G(de)um(d§m) (2.59) m m-l 2 n i + z z I I C? Bj I z o (kIX )gk(e) i=1 j= 1 k= 0 2 - 2 H) «NkaJngI e) + c(i- 1)}f (_ m)G(de) '"(dxmi where gk(o) = L(e,ak) - L(6.60)s k = 1,2,..., 2. The first double sum in (2.59) can be written as J; + dfi + 02 , where 1 m j-1 n I, E. j Jn = z z I I Ci AJ{ 2 n1C(ij)gk( B) -2 o(k|X )gk(o) + i=2 i=1 k=0 k=0 + C(i-J)}fe(5m)G(de)um(dxm) , 2 I'll-'1 m n AZ K ' Jn = z z I I Ci A.{ z n1'(le) gk( 6) -z o(k|X3)gk(e) + J=1 1=j+1 J k=06 k=0 + C(i-J)}fe(§m)G(d6)nm(dfim) . and m 2 - 2 - a3 = z I I c" A. { z a"(ka‘>g (e) -z 6(klx‘)g (enf (5m)G(de)um(dxm) . n ._ i j k _ k e T-l k= 0 k-O Then observe that m j-l 2 Idll : Z X f f C“ A- I 2 l9 (e)l + CIT-Jl}f (1m)G(de)um(dxm) , ” J=2 i=1 1 J k=0 k 9 and -53- m-l m K )JfiI3 z z I I c? Aj{ z ng(e)| + c|i-j|}fe(Xm)G(de)um(de) . i=1 i=j+1 k=0 Recall that for i 3 1 , c? = [3? < 0] ..... [BQ_1 < 0][o? 3 0] and for jil , Aj = [81 < O] ..... [Bj-1< (”[85] > 0], Bj = [81 < 0] ..... [Bj-1< 0] [Bj w.p.1 as n + w and for i > 3, C? A? 3 [89 < O][Bj > O] + O w.p.1 0]. Then by Lemma 2.2 for i < j, c? A? 3 [3? 3 01(81 < 0] . 0 as n + m. Now use the DCT and the assumptions in the theorem to con- clude that J; + O and Jfi + O w.p.1 as n + w . Now to finish the proof that the first double sum in (2.59) goes to zero w.p.1 as n + w , it is enough to Show that J2 goes to zero w.p.1 as n + w , where 2 z k=0 C- II ll [‘43 I I C? Ai{ (6n(k|Xi) - 6(lei))gk(e)}fe(Xm)G(de)um(de) . i 1 From the definition (2.49) of 4(aj,Xk) we can write 0% in the following form: t X OM14!) - oIkIITIIAIak._x_">nl(d3") . k- m n =1 X From (2.58) and An(ak,Xi) + A (ak,Xi) w.p.1 as n + w , k 3 1, i3: 1 , we get for i :_1 , 3 n i i i i (2.60) t 6 (le )A(ak,x ) + z o(k|X )A(ak,X ) w.p.1 as n + w . k=0 Now use I 1.|4(ak,XI))..‘(dXU < e , k = 0,1,...,2 , i 3 1, and the x DCT to conclude that J: + 0 w.p.1 as n + m . The second double sum in (2.59) can be written as Ki + Kg + K: where -54- 1 m-l j-l n 2 n i t j Kn = z z I I Ci Bj{ z 6 (k 5.)9k(9) -z 6(kIX )gk(e) + j=2 i=1 k=0 k=0 C(i-J’)}fe(§m)G(de)um(d§m) . m-I 2 . . Ki = E f f C? Bii Z (6n(k|§1) - 6(kl11))gk(e)Ife(§m)G(de)um(de) , i=1 k=0 and m-l m 2 . K - Kg = 2 z I I C? Bj{ Z on(k|X1)gk(e) -£ o(k|X9)gk(o) + i=1 i=j+1 k=0 k=0 + C(i-J)lfe(§m)G(d6)um(d§m) . K1 2 . . 1 3 . n and Kn have Similar forms as Jn and Jn respectively, and one can Show that K; + 0 and Ki + 0 w.p.1 as n + m using similar arguments as for J; and J3 . We now summarize the results concerning the difference (2.59) obtained so far. n 3 . 3 . (2.61) R(G,g ) - RL(G) = z 0% + 2 kg i=1 i=1 1 2 3 1 where J”, J", J", Kn’ Kn have been shown to converge to zero w.p.1 as n+oo. Now we will Show that lim sup Kg 3 O w.p.1. Write n+oo 3 m-1 (2 62) K = 2 (L (0.1) + L (0.1) + L (n.J)) n J=1 1 2 3 where m n K n T m m m (2.63) L1(n.j) = z I I c, B.( 2 6 i=j+1 J k 0 and m n m m m (2.65) L3(n,j) = z I I C. B. c(i-j)fe(X )G(de)u (dX ) . i=j+l ‘ J “n m n Define Ck = 2 Ci for k = j+1,...,m and j = 1,...,m-1; then i=k n _ “n “n . _ n _ “n . C1 — Ci - Ci+1 for 1 - 1,...,m-l and Cm - cm 0 NOW L1(n’J), L2(n,j) and L3(n,j) can be Simplified into the following form: L1(n.J) = z I 1 c? Bj( z 6n(k|§1)4(ak.§1))ui(dx1) i=j+1 x k=0 m , 2 . . . . = z I 1 c? B.( 2 a”(k|3‘)a(ak,3‘))al(d3‘) i=j+1 x J k=0 (2.66) m-l A 2 . . . . - z I . c? B.( 2 on(k x‘)a(a ,x‘))a‘(dx‘) i=J+l x1 ‘+1 3 k=0 l’ k ‘ ‘ m m-1 = Z T(n:j) -X Si(naJ) i=j+1 i=j+1 where K n l l l l T (0.1) = I C B ( z o (klx )A(a .x ))u (dx ) T l J _ k x k-O and . t Si(n,j) = I . c? B.( 2 a"(ka‘)a(a -55- m 2 L2(n,j) = -2 I I c9 B.( 2 eIkIIkIg (eIIf (5m)G(d6)um(de) i=j+1 ‘ J k=0 k 9 ‘ = -I I B.(cD + 09 . + C")( 2 6(klxi)9 (6)) J 3+1 3+2 m -0 — k (2.67) fe(xm)G(de)um(dxm) , 2 = -I I B]. cg+1(kzo snug)gk(e))fe(3m)e(de)wm(dxm) = -I . B 69 ( z 6(klzj)4(a ,xj))uj(dxj) . x1 3 3+1 k=0 k T 7 and m A m m m L3(n,j) = c 2 I I C BJ fe(x )G(de)u (dX ) i=j+1 m .n i—l i-1 i-I (2.68) = Z c I 1-1 Ci BJ f1_1(X )u (dX ) i=j+l x m = 2 U1(n,J) i=j+l where ° — An I-l- l-l 1-1 Ui(naJ) ' C in-l C1- 8,] fl-1(5 )lJ (d1 ) - The sum Ll(n,j) + L2(n,j) + L3(n,j) now takes the following form: m m-l An 2 . Z Ti(nsJ) " X Si(nnj) 'f ° (3+1 B( Z (“k(éJ) i=j+1 i=j+1 xJ J J k=0 . . . m 4(ak,xJ))uJ(de) + z u,(n.i). i=j+1 that is, -57- . 2 . , _ , ' - . n . J J J J Tj+1(n.J) IXJ CJ+1 BJ(k:0 6(le )A(ak,X ))u (d5 ) (2.69) m The sum of the first three terms in the expression (2.69) is equal to z . . “n n 3+1 3+1 (2.70) IXj (:J.+1 BJ{IX kEO 6 (le )a(ak,x )u(dxj+1) £13“ 331‘ -z 6(kl5 )A(ak.§ ) + cfj(§ )}u (da ) . k=0 By the definition of Bj (see (2.51)), 2 . . . 2 . -2 6(k|XJ)A(ak,XJ) + cf.(XJ) = -I z (lej+l)A )A(ak, J+1) k=0 3 X k=0 u(dxj+1) on [83 = 0] . But 0 3 Bi 3 [83 = 0] so the absolute value of the expression (2.70) is less than or equal to 2 . . (2.71) I j[8j = 0]{|I ( z 6n(k|XJ+1)A(ak,XJ+1) X X k=0 K (k J”) )( J"'11) (d ) j(dj) -k:OA( |X Aak,X u xj+1 |}u X . Expression (2.71) goes to zero w.p.1 as n + w by (2.60) , Z 2 +1) j+1 j+1 |( A .3 )ul (dx ) k=0 xJ+1 < w and the DCT. Finally let us consider the sum 2 {Ti(n9j) - 51_1(nsJ) + U1(n,j)}. i=j+2 Observe that for i=j+2,...,m, Ti(n,j) - S1_1(n,j) + Ui(n,j) is equal to -58- “ I. . I - C” B.( 2 6"(klx1)4(a .x1))u1(dx1) - x1 I J k=0 ’ k ‘ A 2 -_ -_ -_ '- I ._1 c? BJI 2 6”(klx‘ 1)A(ak.§‘ liiwl 1(dx‘ 1) X k=0 (Xi-1)ui-1(dXi-1) . AI"! I C ’ i-l Ci Bj fi-I — — X Rearranging the terms in the above integrals we get I E” B {I ( i a”(k|xl)a(a xl)) (dx ) Xi-I i j k=0 — k’— “ i (2.72) 2 - 2 3"(kI 1‘IIAIa .Il‘l) + cf. (3“1)}al‘l(dr“1> _ k T-l k-O K i-l i-1 Now add and subtract the term 2 o(k|X )A(ak,X ) into the integrand k=0 of (2.72) to get I C” B {I ( f 6n(k|xi)A(a xi)) (dx ) Xi-1 i j x k=0 — k’— “ i 2 . . . . (2.73) - z 3(kal'1)a(a , 1‘1) + cf. ( 1-l)u1-l(dX1-l) _ — k -1 — k-O I + I 1_1 C? B { z o(k|x1 1)A(ak,X1'1) - Z on(k| 1'1) x J k=0 k-o A(ak,§i-1}un-1(d£i-I) . The second integral term in (2.73) goes to zero w.p.1 as n + w from 2 . . ~ . (2.60), z I . 1|n(ak,x“1)|u"1(dx"1) < e , and the DCT. k=0 x“ ‘ ’ -59- The first integral in (2.73) can be rewritten as f En B [8 < O]{f ( é 6n(k|xi)A(a xi)) (dx ) Xi-l i j i-l — x k=0 - k’— “ i i-l) (xi-1)}ui-1(dXi-l) T-l 5(kll ”(3'05 1_ _ K -2 + Cfi- k 0 (2 74) + f E" B [s > 0]{/ ( é 6n(k|xi) - Xi-l i j i-l k=0 — A(ak9£i)U(dxi) z . . . - Z 6(kl51'1)A(ak,§1-1) + cfi 1(51-1)) k=0 Ll1"].(d 1‘1) . But 0 5 E? [ei_1 > 0] 5 [e?_1 < 0][s1._1 > 0] + 0 w.p.1 as n + m by Lemma 2.2 . Therefore, the second term in (2.74) goes to zero w.p.1 as n + m by the DCT and f A(ak,§1)u(dx1) < m , i 3 1 , k 3 1 . The first term in (2.74) is < f En B [ < 0]{f g 6n(k|xi)A(a xi)) (dx ) — Xi-1 i j 8i-1 — k=0 - k,_ u 1 (2.75) i 1 i i-1 i-1 - f (kzo 6(kI5 )A(ak,§,))u(dxj)u (d1 ) . X = The inequality (2.75) follows from the fact that (see 2.51) on [Bi-1 g 0] we have K -2 5(kl51-1)A(ak,§ k=0 0 . z . . i-l) 1'1) z s(k|§‘>a(ak’§‘))u(dx ) - + Cfi-1(5 1 : —fX(k:O Now it is easy to see that the -50- R.H.S. of the inequality (2.75) goes to zero w.p.1 as n + m K . . . from (2.60), z j 1.|A(ak,_)g1)|u1(d§1) < w , and the DCT . Now k=0 X combine (2.74) and (2.75) to get for i=j+2,...,m lim sup {Ti(n,j) - Si_1(n,j) + Ui(n,j)} 3 0 w.p.1 as n + m n+w and then with (2.69) and (2.70), we have for j=1,...,m-1 , lim sup (L1(n,j) + L2(n,j) + L3(n,j)) 5 O w.p.1 . n+oo Therefore, lim sup K: 5 0 w.p.1 follows from (2.62). This completes n+0!) the proof of Theorem 2.2. D Remark 2.2: Corollaries analogous to Corollaries 2.1 - 2.4 of the previous section can be stated and proved in the more general multiple decision problem context as well. -61.. 2.5 Final Remarks Using arguments very similar to those of Sections 2.3 and 2.4, we can show that the result lim sup R(G,gn) 5 RL(G) w.p.1 as n + w also holds for natural empirical Bayes sequential decision procedures for the squared error loss estimation component under (A1) and (A2). A curious feature of the EBSD's in approximating one-step look ahead risk is the inequality in the asymptotic result. (The asymptotic optimality that is typically proved is the convergence to RB(G).) There are examples of two-action linear loss components with Poisson distributions and priors G for which lim sup R(G,gn) < RL(G) on a set of positive probability. BIBLIOGRAPHY -52- BIBLIOGRAPHY Berger, James O. (1980). Statistical Decision Theory. Foundations, Concepts and Methods. Springer-Verlag, New York. DeGroot, Morris H. 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