TheThe AnnalsAnnals ofof ProbabilityProbability 1991,1991, Vol.Vol. 19,19, No.No.1, 1, 342-353342-353 MINIMAX-OPTIMAL STOP RULES AND DISTRIBUTIONS ININ SECRETARY PROBLEMS
l BYBy THEODORE P. HILL1HILL AND ULRICH KRENGEL Georgia InstituteInstitute of Technologyand Universityof GdttingenGottingen For the secretary (or(or best-choice)problem with an unknownnumber N of objects,minimax-optimal stop rules and (worst-case) distributions are derived,under the assumptionthat N is a random variable with unknown distribution,but knownupper bound n. Asymptotically,the probabilityof selectingthe best object in this situationis of order of (log n)-Y.n)- 1. For example,even if the only information available is that there are somewhere between1 and 100 objects,there is stilla strategywhich will select the best itemabout one time in five.
1. Introduction. In the classicalsecretary problem, a knownnumber of rankableobjects is presentedone by one in randomorder (all n! possible orderingsbeing equally likely). As each objectis presented,the observermust eitherselect it and stopobserving or rejectit and continueobserving. He may neverreturn to a previouslyrejected object, and his decisionto stop mustbe based solelyon the relativeranks of the objectshe has observedso far.The goal is to maximizethe probabilitythat the best object is selected. This problem,also knownas the marriageproblem or best-choiceproblem, is well known,and the readeris referredto Freeman(1983) and Ferguson (1989) for a history and review of the literature. Suppose now that the total number of objects is not known, but is a random variable N taking values in {I,{1,2, ..... ,n},, n}, where n is a known fixed positive integer. How should the observer play in order to guarantee the highest probability of selecting the best object, what is this probability and what is the worst distribution for N? The main goal of this paper is to determine these minimax-optimal stop rules, values and distributions as a function of n. For example, if n = 5, the strategy "stop with the first object with probability 26/75; otherwise continue and stop with the second object with probability 26/49 provided it is better than the first object; and otherwise stop the first time an object is observed which is better than any previously observed object" is minimax-optimal. This strategy will select the best object with probability at least 26/75 for all distributions of N (~5),(< 5), and that probability is best possible. Conversely, ifif N has the the distribution P(N = 1)1) = 13/75,13/75, P(N = 2) = 2/75, P(N = 5) = 60/75, then no strategy will select the best object with probability greater than 26/75, so this distribution isis alsoalso mini-
Received May 1989; 1989;revised October 1989.1989. 1Research'Researchpartially supportedsupported by a a Fulbright Research GrantGrant andand NSF Grant DMS-89-01267.DMS-89-01267. AMS 19801980 subjectsubjectclassifications. classifications.Primary 60G40; 60G40; secondarysecondary62C20, 62C20, 90D05.90D05. Key words andand phrases. SecretarySecretary problem, best-choice problem, marriage-problem,marriage-problem, minimaxminimax- optimaloptimalstopstop rule,rule, minimax-optimalminimax-optimal distribution,distribution, randomizedrandomized stopstop rule.rule. 342342 MINIMAXSTRATEGIES IN SECRETARY PROBLEMS 343
max.(It is assumed that, given N, all N! orderings are equally likely, and that if an object is rejected and no more objects remain, the game is over and the best object has not been selected.) A numberof results are known for the general situation where the number of objects N is a randomvariable. Presman and Sonin (1972) derive optimal stop rules when N has a known prior distribution and mention the necessarily complex form ("islands") of optimal stop rules for certain prior distributions. Irle (1980) gives a concrete example of such a priorfor which the optimal stop rule has these islands and sufficient conditions for existence of simple "non"non- island" stop rules. Abdel-Hamid, Bather and Trustrum (1982) derive necessary and sufficient conditions for admissibility of randomized stop rules. Extensions to the situation where the interarrival times of the objects are continuous random variables with known distributions have been studied by Presman and Sonin (1972), Gianini and Samuels (1976) and Stewart (1981). More recently, Bruss (1984) and Bruss and Samuels (1987) derive surprising and very general minimax-optimal strategies in this same context and even for more general loss functions. In contrast to the minimax-optimal stop rules derived in this paper, which are based on knowledge of a bound for N, those of Bruss and Samuels are based on knowledge of the distributions of the continucontinu- ous i.i.d. interarrival times; in this sense our results complement theirs. This paper is organized as follows: Section 2 contains notation, results for the classical secretary problem and basic results concerning randomized stop rules;rule~; Section 3 contains the statements of the main results and examples; Sections 4 and 5 contain the proofsof the minimax-optimal stop rules and distributions, respectively; and Section 6 contains remarks and asymptotics.
2. Preliminaries. A well-known equivalent formulation of the classical secretary problem is the following. R1,R I , R2,...,R 2 , ••• , RnR n are independent random c'variablesvariableson a probabilityspace en,(fQ,!T,Y, P), where n is a fixed positive integer and P(RJpeR) = i)i)=j= j-I1forfor all i E{1, {I, 2, ...... , ,j}j} and all j EE{1, {I, 2,2,...,... , n}.n). If ~n-de- ~ notes the stop rules for R1,R I , R2,...,R 2 , ••• , Rn,R n , then the value of a stop rule t E E (given that there are n objects) is
V(t/NV(tIN = n) = P{RP(Rtt = 1 and R)Ri > 1 for all j > t); that is, V{t/NV(tIN = n) is the probabilityof selecting the best object using the stop-rule strategy t, given that there are n objects. The goal is to find a t making V{t/NV(tlN = n) as large as possible,and the solution to this problem is well known[cf. Ferguson (1989) and Freeman (1983)] and is recorded here for ease l of reference.Throughout this paper, Soso = 0, and for j ?~ 1, s)s; = E{=li-E 1i' .
DEFINITION2.1. For each positiveinteger n, kknn is the nOIlnegativenonnegativeinteger satisfying
Sn-1 - Skn-1 2 1 > Sn-1 - Skn 344 T. P. HILL AND U. KRENGEL
PROPOSITION 2.2. The stop rule intnE ~E defined by in4n = min{min{j > kn:k n: R = I}, n} is optimal, that is, Rij = 1}, n} is optimal,that is, V(inlNV(tnIN== n) = sup V(tINV(tIN== n). tE~t E En
In other words, given that there are n objects, the optimal strategy is to observe the first kkn n objects without stopping and then to stop with the first object, if any, that is betterthan any object previously seen. It is well known ~ ~ that n/kn/knn -* e as n - 00,oo,and the next example records a few typical values of n.
EXAMPLE 2.3. kk1=i = kk22 = O,0, kk33 = kk44 = 1, kk5s = kk66 = kk77 = 2 and kgk8 = kgkg== kk10lO = 3.
Next, the above notations will be generalized to the setting where the numberof objects N is a random variable and randomized stop rules are allowed. (In the classical setting of a fixed known number of objects, it is clear that randomization does not help, that is, infnis also optimal among the larger class of randomized stop rules.) For each positive integer n, TIHnn denotes the set of probabilitieson {I,{1, 2,2 ...... , , n}, so Pp EHE TIn n is ofthe form p = (PI'(P1,P2'P2, ...., ' Pn),Pn) wherepiPi ~2 °0 for all i and E'/=IPiE=11pi=1.= 1. N is a random variable with distribution .J?(N)A(N) E TIn'I-IngRR1,... I , ... , RRnn are as above and independent of N and ~5n denotes the set of randomized stop rules E,~ for R1,...,R I , ... , Rn n , that is, t E En means that {t = i} is in the IT-algebraa-algebra genergener- ated by RR1,I , U U1I , ...., , Ri'~'Ri, Ui, where UU1, I , UU2, 2, ... are i.i.d. U[O, 1] random variables which are independent of the {R{Ri}i } process and of N. In other words, the observer is allowed to base his selection rule not only on the observed relative ranks,but also on an independent event, say flipping a coin or using a random number generator. Clearly the only stop rules which are of interest (for the goal of selecting the best object) are those which never stop with an object which is not the best seen so far, so every "reasonable" t E ~En may be described by t = (ql'(q1,q2'q2 ...... , ,qn)qn) E [0, l]n,1]n, where qi is the probabilitythat t = i, given that RiR i = 1 and t > i-I.i - 1. Accordingly, it will be assumed throughout that only such stop rules are used, so ~En is essentially [0, l]n.1]n.The stop rule t = (ql'(q1, q2'q2 ...... , ' qn) describes the selection strategy "stop with the first object with probability qiq1 (i.e., if UU1I ~< ql);q1); otherwise continue observing and if the second object is betterthan the first, stop with probability q2 (i.e., UU22 ~< q2); otherwise continue,continue,..." ... " [see Abdel-Hamid, Bather and Trustrum (1982)]. To relate this to the classical problem, Proposition 2.2 says that if N = n, then an optimal stop rule is (0,(O,... .,,0,1, 0 1,...,... ,1),1), where kkn n zeros precede n + kknn ones. [Formally speaking, the above stop rules (ql'(ql, ... ' qn) are not forced to stop by time n, but since stopping with a relative rank less than 1 is worthnothing, it is easily seen that forcing a stop by time n changes nothing.] MINIMAXMINIMAXSTRATEGIESSTRATEGIES IN IN SECRETARYSECRETARYPROBLEMSPROBLEMS 345345
DEFINITIONDEFINITION 2.4.2.4. ForFor tt == (ql'(q 1,...... ,' qn)qn) EE ~9 andand pp == (PI'(P1, .... **, ' Pn) PO) EE TIn'fn, thethe valuevalueof ofusingusing t t givengiventhatthat thethe distribution distributionofof N N isis p,p, V(t/p),V(tlp),isis givengivenbyby ~ V(t/p)V(tlp) == P{tP(t < NandN and RRtt == 11 andand RRii >> 11
ViEV i E {tft+ 1,t19t+ +292, ...... , N}IJ(N)N) l(N) == p). (Recall(Recallthe theassumption assumptionthatthat if ifthe theobserver observerrejectsrejects thethe jthjth objectobjectandand NN ==jgj, thenthenhehe loses.)loses.)
The next lemmalemma isis foundfoundin Abdel-Hamid, BatherBather andand TrustrumTrustrum(1982)(1982) andand isis recorded here forforcompleteness.completeness. (For(For notational convenience,convenience, the product over an empty set isis taken to be 1.)1.)
LEMMA2.5. For t = (q19q2 ...* qn) E 5n and p = (P1,... ,Pn) E Hn n )i i-Ii-1 1 V(t/p)V(tlp) = EEpj-EPjj-l E qi Hn (1 - m-m-1qm).qm)· )=1j=1 i=1i= m=1m=
PROOF. Using t, the probability that all of the first m objects will be rejected, r(t, m), is ret,r(t, m) = (1 - ql)(lql)(1 - q2/q2/2)2) ... (1 - qm/m) and if N = j, the probabilityof winningwith this rule t is V(tIN = j)j)= = j-lEj=j-1E{=lqir(t, lqir(t, i-I).i - 1). Since V(tlp) = EE>1=J=IPlpjV(tlN j V(tIN =j), this yieldsthe desired equality.Da
3. Main theorems and examples. Recall that Sj = E {= 1i-I and k n is sj Ei=li-l k the "cutoff'"cutoff" forthe optimalrule in the classical secretaryproblem with n ,objects"'objects (Definition 2.1 and Proposition2.2).
DEFINITION3.1. Let a,a 1 = 1, a2a 2 = 1/2 and, forn > 2,
=Sn- - Skn-1 (n - kn)/kn + (sn-1 Skn-l)Skn (See(See Table 1 for anan', n = 3,3,4, 4, 5, 10.) Recallalso that En~ isis the set ofrandomized randomized stop rules for n objects,HfnTI n isis the set of probabilitiesprobabilities on {1,{I, ...... ,,n} n} and V(tlp)V(t/p) isis the probabilityprobability of selecting thethe bestbest objectobject usingusing t,t, givengiven thatthat thethe distributiondistribution ofof thethe numbernumber ofof objectsobjects isis p.p. TheThe followingfollowing threethree theoremstheorems areare thethe mainmain resultsresults of thisthis paper. paper.
THEOREMTHEOREM A.A. supteSUPtEY"n n infpsinfpEllnT, V(tlp)V(t/p) == aan == infp,=-infpElln supretSUPtEY"n V(tlp).
REMARK.REMARK. AlthoughAlthough eacheach ofof theth~ termsterms inin thethe definitiondefinition ofof ana n hashas aa naturalnatural probabilisticprobabilistic interpretationinterpretation (e.g.,(e.g., siSi -- s;Sj isis thethe expectedexpected numbernumber of of relativerelative rankrank 1 1 candidatescandidates occurringoccurring betweenbetween thethe ithith andand jthjth candidates),candidates), thethe authorsauthors 346 T. P. HILL AND U. KRENGEL
know of no intuitive explanation why an should be the minimax constant appearing in Theorem A.
THEOREM B (Minimax-optimal stop rule). If t;tn = (qi,(q1.*,... , q;)q *) E 3n~ is defined by J(ac -1 _s_)-- )-1 fiorj = 1, k* , q~q= ((an - Sj-1 for j = 1, ... , k n , J 1 1 forknfork n THEOREM C (Minimax-optimal distribution for N). If Pn*Pn* = (pi,(pr, ... ,n,P:)p*) E TIfinn is definedby (an(ian(j + 1)-11) forjfor j < kn,k n , - Pj=pI = an(lan(1 - (Sn-l(Sn-1 - Skn_l)-l)skn-1) ) forjforj = kknn , o0 forknfor k n [Verification of the above expression for pn*p; and of the fact that qi*q?' E [0,1][0,1] is left to the reader; this requires only elementary algebra applied to the definitions of an'an, kknn and Sn.sn. For example, to show qi*q ? ~< 1,1 the monotonicity ofthe {Sj} implies thatit is enough to show that an :::;< (1 + Sk -1)-1,1), and using {s)} implies enough an (1 Skn-n using the definition of ana and Sj this is equivalent to (k - 1)(Sn-1 - Sk -1) ~ n - n sj equivalent (knn 1)(snl - Skn-1)n < kn,k n , which clearly holds.] Table 1 lists the minimax values {an},{fa}, and the minimax-optimal stop rules and distributions for several values of n. REMARKS. Irle's (1980) example of an "unpleasant" distribution, that is, a distribution for which no stop rule of the form (0,0, ... ,,0,1,1, 0, 1, 1,...,... ,1)1) is optimal, is p = (0,0.895,0.001,0.001,(0, 0.895,0.001, 0.001, ... ,,0.001,0.1) 0.001,0.1) E TIs,H8, for which he calcucalcu- lates the value of the optimal stop rule (0, 1,0,1,1,1,1,1)1,0, 1, 1, 1, 1, 1) to be approximately TABLE 1 n kn ?h ti;tn* = (q[,(q *,...qn)•.. ,qi;> P.* Pl@@P 1 0 1 (1) (1) 2 0 !2 (1,(1,1)1) (0,(0,o1)1) 3 1 ~3 (~,1,1) (t,o,~) 3 1 (7 7,1,1) - 6 4 1 * (*,1,1,1). (~,O,O,~) 4 26 26( 26, 1, 1 . (2513 2 29)60 5,·)t22 ~ (~,~,1,1,1)U (*,-fs,0,0,~) 5 2 75 75,49'4 , , (73,75'"75 75) 10 3 0.278+ (0.278+,0.386+,0.478+,1,1,(0.278+, 0.386+, 0.478+, 1, 1,...,1)... ,1) (0.139+,0.092+,0.068+,0,0,(0.139+, 0.092+, 0.068+, 0,0,...,0,... ,0,0.698+) 0.698+) MINIMAXSTRATEGIES IN SECRETARYPROBLEMS 347 0.482. Comparisonof this value with those in Table 1 suggeststhat such island distributionsare far frombeing worst-case(i.e.,(Le., minimax-optimal), althougha directproof of this is not knownto the authors. It shouldalso be observedthat the minimax-optimaldistribution for N is not one of the other "naive-guess"distributions such as N == n or N uni-uni formlydistributed on {1, 2, ... .,,n} n} or N = 1 with probabilityp and = n with probability1 - p. As faras the authorsknow, this Pn*Pn* is a newdistribution on n points. ClearlyTheorem A followsfrom Theorems BBand and C. No directproof that supsupinf inf= infinfsupsup is knownto the authors;although V(tlp) is linearin p and TInfln is convexand compact,V(tlp) is neitherconvex nor concave in t, and knowngeneralizations of the classicalminimax theorem of game theory do not seemto apply.(The resultsin thispaper may also be interpretedas a zero-sum two-persongame as follows.Player I picksthe distributionof N, and playerII picksthe stop-ruleor selection-strategyt; if t selectsthe best ofthe N objects, then player I pays player II one dollar; and otherwiseno moneychanges hands.The constantan thenrepresents the value of thisgame.) 4. Proof of Theorem B. The conclusionsof Theorems BandB and C are trivial for n ="= 1 and easy for n = 2, so for the remainder of this paper, n will be a fixedinteger strictly bigger than 2, and to simplify notation, k = kn,k n , 7=`= En~ and TIII = TIn.Hn. (Observe that n > 2 precludesthe degenerate cases where k = 0; see Example 2.3.) LEMMA 4.1. Suppose {aJi=l{a iY11 are real numbers, and j and k are positive integers satisfying n ~? j ~ 2 k. If both (1) (a1 + .a kak)/, 2 (a1 + .+aj)/ and (2) aam2am+l m ~ am + 1 forallmE{+kfor all m E {k + 1, k + 2, ... ,,n),n }, then (a1 + ... +aj)/j ? (a1 + ... +an)/n. PROOF. If j = n, the conclusion is trivial, so assume j < n. Then condicondi- tions (1) and (2), respectively, imply ~ (a(a1l + ...*. +aj)/j 2 (ak+l(ak+1 + *. +aj)/(j --k k) ~? (a(aj+lj + l + * +an)/(n?+an)/(n --j),j), ~ so (a(a, l + ... +aj)/j+ aj)/j 2 (a(a, l + ... +an)/n. +a DE fROPOSITIONPROPOSITION 4.2. sup inf V(tlp) = max min V(tIN =j). ttEE Y5 PeHIIpEn tt=(q1,...,qn)E= (q1, ... , q n) E Y:[7: qq=1=Vi>k£ = 1 't/ i > k jjE{1,2,...,k,n)E {I, 2, ... , k, n} 348348 T.T. P.P. HILLHILL ANDAND U.U. KRENGELKRENGEL PROOF.PROOF. SinceSince V(tlp)V(t/p) isis continuouscontinuous inin bothboth tt andand p,p, andand sincesince !T7 andand Hn areare compact,compact, thethe supsup andand infinfare are attained.attained. MoreoverMoreover (3)(3) infinf V(tlp)V(t/p) == minV(tIN=j),minV(t/N =j), pElIpEnH j5:n Fix t = (ql'(q1, ... ,' qk'qk, 1, ... ,1) E1]= [0, l]n and define real numbers {ai}i=l{ai}in> as folfol- lows: a a,l = qlq1 and ai = qin~2l(1qi~lH-11(1- m-lqm) for i > 1. Since qi = 1 for all i > k and qj E [0,1][0, 1] for all j, (7) amam>am+l> a m+ l forforallm>k.all m > k. By Lemma 2.5, V(tINV(tIN ==j)j) = (a(a,l + ...** +aj)/j for all j E {I,{1,2,...2, ... ,, n}. To establish (6), suppose V(tIN = j) ~< V(tIN = k), that is, ~ (8)(8) (a(a,l + ...*. +ak)/k ? (a(a, l + ... +aj)/j.+aj)lj. By (7) and (8)(8) and Lemma 4.1 (with(withk = k),k)g ~ V(tINV(tIN =j)=j) == (a(a,l ++ ...*D +aj)/j+aj)/j ? (a(a,l ++ ...* +an)/n?+an)/n == V(tINV(tIN== n),n) , whichwhichestablishesestablishes (6).(6). Da LEMMALEMMA4.3.4.3. ForForallall tt== (ql'(q19...... , ,qk' qk9 1,19 1,19 ...... ,,1) 1)E=E !T,A, :~>j'+ V(tINV(tIN== n)n) == knkn-{( [(Sn-1sn-l -- Sk-)(1Sk-l) (1 -- E (J + 1)1) -lV(tINV(tlN==j)) j) ) = -(Sn-l-(Sn-- - Sk-lSk-1 --1)V(tlN1)V(tIN = k)k).j. MINIMAXMINIMAXSTRATEGIESSTRATEGIES IN IN SECRETARYSECRETARYPROBLEMSPROBLEMS 349349 PROOF.PROOF. FirstFirstitit will willbebe shownshownthatthat )i )-1j-1 1 (9) n (1-m-lqm)(1 - m- qm) = -1 - LE (m(m? ++ 1)-lV(tlN=V(t/N = m)m) (9)(9) m=1m=l m=1m=1 --V(tIN=j)V(tIN =j) forforallall j~k.j k. TheThe proofproofof of(9) (9) isis byby inductioninductionon on j. ForFor j == 1,1,(1 (1 -- q1)q1) == 11 -- V(tINV(tIN == 1)1) byby LemmaLemma2.5. 2.5. AssumeAssumethatthat the theequality equalityinin (9)(9) holdsholdsfor forallall jj ~< kk andand calculatecalculate k+lk+1 kk kk (1 - m - 1 m) (1 - m - 1 m) - (k 1) - 1qk+ 1 (1 - m - 1 m) n (1 - mlqm)q == n (1 - mlqm)q - (k ++ 1) qk+l n (1 - mlqm)q m=1H m=11 m=1H k-lk-1 = 1= -1- L,(m(m + 1)-l1) - VV(tlN(t/N = m) - V(t/NV(tJN== k) (10)(10) m=1 - V(t/N= kA + 1) + k(k"" + 1) -1V(t/N= k)" k = 1 - L (m + 1)-lV(tIN = m) - V(tIN = k + 1), m=1 where the second equality in (10) follows by the induction hypothesis and the fact (from Lemma 2.5) that = 1 (fromLemma 2.5) that m=1(m k 1 V(t/NV(tIN= k + 1) = (k(ik+ 1)1)1q~-lqk nIiH (1 1 --m m- q qm)m) +1(s k(k +1)t 1) -lV(tINV(t/N=I), = k), m=1 whichestablishes (9). Since qj = 1 forall j > k, Lemma 2.5 and the definitionof {Sj}{sj} implythat - k ii-I -1 = n-11 - V(tIN=n)V(tlN- n) =n- [ iJ;:lqifll(1-m-lqm)E qi h (1 m-lqm) +k(sn-1+k(Sn-l - Sk-1)Sk-l) mQlk (1 - mlqm)J.m-1qm) l 1 But EL7=lqin~21(1 =lqitlim 1(1 - m-lqm)m- qm) = kV(tlNkV(tiN = k) (Lemma 2.5 again), so (9) (with jj = k) and (11) yieldthe desiredequality. mD Heuristics.Heuristics. Althougha directcalculus-based proof of Theorem B shouldbe possible,possible, the proofproof given below below isis greatlyfacilitated byby Proposition4.2 and LemmaLemma 4.3, whichwhich bothboth also serve as heuristicsheuristics for the structure of the minimax-optimalminimax-optimal stopstop rule. rule. For For example,example, PropositionProposition 4.24.2 says that that any any general general stopstop rule rule cancan bebe replacedreplaced byby a a stop rule rule with with qi q i == 1 for allall ii > k,k, and thatthat with with suchsuch stopstop rules,rules, thethe criticalcritical valuesvalues occuroccur whenwhen NN == jj forfor somesome jj inin {1, 2,...,2, ... ,k,k, n};n}; thatthat is,is, ifif NN = = jj eE {k{k +1....+ 1, ... ,, nn -- 1}, thethe observer'sobserver's probabil-probabil ityity of of selectingselectingthethe bestbest objectobject isis atat leastleast asas highhigh as as thethe minimumminimum ofof the the otherother possiblepossible valuesvalues forfor j.j. (Incidentally(Incidentally thisthis also also suggestssuggests whywhy thethe minimax-optimalminimax-optimal distributiondistribution inin TheoremTheorem CC placesplaces nono massmass onon {k{k ++ 1,1, ...... ,, nn -- 1}.1}. ForFor fixedfixed tt ofof thethe formform knownknown to to bebe optimaloptimal (i.e.,(i.e., qiqi == 1 1 forfor all all ii >> k),k), LemmaLemma 4.34.3 impliesimplies 350 T. P. HILL AND U. KRENGEL that V(tlNV(tIN= n) is a decreasingfunction of V(tINV(tlN = j)j) forj j <~ k. Together with Proposition4.2, this suggestsvia a "Robin Hood principle"(shifting mass to decreasethe maximumand increasethe minimum)that the extremal case occurs when V(tINV(tlN = 1) = V(tINV(tlN = 2) = *... = V(tINV(tlN = k) = V(tINV(tlN = n). Solvingthis set of k equationsfor the k unknownsq1,q l' ...... , qkq k leads to the minimax-optimalstop rule in TheoremB. Once the correctextremal stop rule is guessed,of courseit is then mucheasier to provedirectly that it is inil) fact optimal,without justifying the derivationof the guess. PROOFOF THEOREMB. By (6) it sufficesto show that (12) V(t,iIN=j)=aV(t*lN =j) = a n forforj={1,2,j ={1 ,2 ,...,... ,k,n}.n}. To establish(12), firstcheck by inductionthat qqj*n~-=ll(1 lHi-11(1 - m-lq,*)m-lq~) = a?an for all j ~< k, so Lemma2.5 impliesthat V(t,iIN V(t*lN = j)j) = an forall jj ~< k. To check that V(t,iINV(t*IN = n) = as,an' use Lemma 4.3 and the fact that (Sn-l(sn1 - skl)Sk-l) = k-lan(nk-1an(n - k)(1 - ansk)-lanSk)-1 to calculate a<~>j V(t;:'INV(tn*N= = n) = n -lk[(1k[(Sn-1Sn-l - Skl)(11Sk-l) (1 --an A (j + 1)1)i)-1) -(Sn-l-(Sn-1 - Sk-lsk-1 - 1)a1)an] n] = - -an j~/-l)Ji) = n -lk[(-k [(sn-1Sn-l - Sk-l)Sk-1) (1 - an + anan] ] - - - = = n-1k[ank-1(nn1k[sank 1(n - k)(I-k)(1 anSk)-l(l-anSkl) (1 anSk) + an] = an.a?n D? 5. Proof of Theorem C. As mentioned above, Proposition 4.2 suggests that any minimax-optimal (worst-case for the observer) distribution places no mass on {k + 1, ... , n - I},1), and again a Robin Hood principle leads to a guess which has break-even values for each j in {I,{1,2, 2,...,... , k, n}. For example, clearly piPr ~< an'a?, since otherwise taking t = (1,1,(1 1 ...... ,,1) 1) yields V(tIPV(tlPn*) n*) > an.an As was the case for the optimal stop rule, once a worst-case Gistributiondistribution PPn*n * has been guessed, the check that it is in fact minimax is then much easier. Thus moatmostof the work was hidden in the heuristics which generated the guess for PPn*. n*. FORMAL ARGUMENT. It is enough to show (13) V(tIPV(tlPn*)n*) ~< an for all t = (ql'(q9 *... ,' qk'qk 1,1,1, 1, ...* * ,1)1) E :T,H7 ~ since by (4), V«ql'V((q1,...... ', qn)lp)qn)Ip) < V«q1'V((q, ... , qk'qk, 1,1,...... ,,1)lp) l)Ip) for all {qi}{qJ} E [0,1][0, 1] and all p EEiII. H. [In fact, it will be seen that (13). holds with equality throughout, which says intuftivelyintuitivelythat against PPn*, n*, all"all "reasonable"reasonable" stop rules, Le.,i.e., all stop rules with qi = 1 for all i > k, select the best object with the same probability.] MINIMAX STRATEGIES IN SECRETARY PROBLEMS 351 Fix t = (ql'(ql ...... ' qk'qk, 1, 1,12 ...... ,1)X1) Ee [0, l]n1]n and calculate (k-1 1 i i-1 = anC~: + i~l - V(tIPV(tlPn*)n*) = an E (j(j + 1»1)) -1 E qiqj I:LH (1 - mmlqm)-lqm) j=l i=i m=l k i-1 k l k i-I k ) +k-+k1 i~lE qiqj l!lH (1(- - m-lqm)m-lqm) + m~lE (1 --m-lqm)m-lqm) i=1 m=i m=i k-Ik-Ik-1 k-1 i-Ii-i = ananE i~l EiE (i-Ii1 - (j + 1) -l)qi H (1 - mm-1qm)-lqm) ( (j 1)1)qi JI k i-1 k l k i-I k ) +k-+k 1 i~l qiqj l~[lH (1 -m-lqm)- m -lqm) + l~[lH (1 --m-m -lqm)qm) i=i m=i m=i k i-1 = anL~l (i-I(i- -k-1)qi- k-l)qi l:LH (1 - m-lqm) z=1 ~m=l k i-1 k l k i-I k ) +k-+k i~lE qiqj l!lH (1 -m-lqm)- m-lqm) + J}lH (1 - m-lqm) i=i m=i m=i k i-1 k = an[i~l - = n Ei1qjj(1i-lqi IiI (1 - mm1qm)-lqm) + mQl (1 - mmlqm)j-lqm)] i=1 m=l m=l [k i-1 i-1 = an[i~la[E {((ijlqii-lqi - 1) lilfII(1 - m -lqm)qm) + lilH (1 - mmlqm)}-lqm)} i=1 m=l m=l k lq + mQlH (1 - m-m-lqm)m )] m=i k i -1 = an[i~l {~~(H lq = n {H(i (1-- m-lqm)m q ) - iJlH (1 - m-m qm)}m )} i=1 m=l m=l k lq + mQlH1(1 - m-m-lqm)m )] m=i = aEng where the first equality follows by Lemma 2.5, the second since L E j: I tL i= l1 = LE*-1lE*i7:l L j:l and the third since the first summand disappears for i = k. This completes the proof of (13) and the theorem. DO 6. Asymptotics. Since TIn[In can be viewed as a subset of TI[nIn + 11,, Theorem A shows, indirectly, that the sequence {a{an} n} is nonincreasing. A direct check using the definition of ana n and general observations about kn (e.g., kn + 1 is either k n or kkitn + 1) shows that in fact the {an} are strictly decreasing in n. 352352 T.T. P.P. HILLHILL ANDANDU. U. KRENGELKRENGEL ~ ~ 1 ~ SinceSincesn sn loglognand n and kknn nnee- 1 (where(whereanan bbnn meansmeans limnlimn -+00 an/ba n === 1),1), itit followsfollowseasilyeasily thatthat ~ anatn (logn)-l,(log n) *(f(log n - log j) - 1 forforjj :::;; e-1n, q. ~ (logn-logj)1 In particular, lim n -+ oo an === 0, in contrast to the well-known classical result that for the deterministic case N = n, the probability of selecting the best 1 object (using(using an optimal strategy) decreases monotonically to e-e-1 as n ~-* 00.oc. The optimal "stopping-probabilities" {qj*}{qji} are nondecreasing, which is also intuitively plausible, since if it is optimal to stop with a certain probability at time i (given R'iR-i = 1), then at later times with even more information accrued it should be optimal to stop with at least as high as probability if a rank 1 object is observed. The following alternative possible derivation of the asymptotic result aan n ~ (log n)-ln) - 1 has been given by Samuels (1989). Since the expected number of relativelybest ones ("("records")records") will be about log N, thissuggests that the rule stopwith probability l/log1/log n at each ofthe first log n records will succeed with probabilityabout 1/logl/log n no matterwhat the distributionof N is. (A formal derivationusing this approachseems to requiremore information about the ",actualactual distributionof the numberof recordsthan just its expectation.) Acknowledgments. The questions answeredin this paper arose in a discussionfollowing a lectureby Professor S. Samuelsat theAmerican Mathe-Mathe matical Societymeeting in Atlanta in January1988. The authors are also gratefulto H. J. D6ringDoring for a number of stimulatingconversations and suggestions,to the MathematicsResearch Institute at Oberwolfachfor hospi-hospi talityduring a visit in whichthis researchwas begun and to the Associate Editorand refereefor several suggestions. REFERENCES ABDEL-HAMID, A., BATHER, J. and TRUSTRUM, G. (1982). The secretary problem with an unknown numberof candidates. J. Appl. Probab.19 619-630. . BRUSS,F. (1984).(1984). A unifiedapproach to a class ofbest choice problems with an unknownnumber ofoptions. Ann.Ann. Probab.12 882-889. BRUSS,F. and SAMUELS, S. (1987).(1987). A unifiedapproachapproach toto a class ofoptimal selection problemsproblems with a','if an unknownunknown number of options. Ann.Ann. Probab.15 824-830. FERGUSON,FERGUSON, T. S. (1989).(1989). 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