Optimal Expected Rank in a Two-Sided Secretary Problem

OPERATIOS RESEARCH Vol. 55, o. 5, Septembe Octobe 007, pp. 9 93 issn 0030-364X eissn 56-5463 07 5505 09 infoms doi 0.87/ope.070.0403 007 IFORMS Optimal Expected Rank in a Two-Sided Secetay Poblem Kimmo Eiksson, Jonas Sjöstand, Pontus Stimling Depatment of Mathematics and Physics, Mäladalen Univesity, SE-7 3 Västeås, Sweden {kimmo.eiksson@mdh.se, jonas.sjostand@mdh.se, pontus.stimling@mdh.se} In a two-sided vesion of the famous secetay poblem, employes seach fo a secetay at the same time as secetaies seach fo an employe. obody accepts being put on hold, and nobody is willing to take pat in moe than inteviews. Pefeences ae independent, and agents seek to optimize the expected ank of the patne they obtain among the potential patnes. We find that in any subgame pefect equilibium, the expected ank gows as the squae oot of wheeas it tends to a constant in the oiginal secetay poblem). We also compute how much agents can gain by coopeation. Subject classifications: games/goup decisions: stategic secetay poblem; dynamic pogamming/optimal contol: optimal stopping. Aea of eview: Optimization. Histoy: Received June 005; evisions eceived Mach 006, June 006; accepted June 006. Published online in Aticles in Advance. July 0, 007.. Intoduction Two-sided matching has been studied intensely in the last few decades. This aea of eseach has been motivated by impotant eal-life poblems of matching employes with job seekes, o colleges with potential students, o men with women. In game theoy, the standad appoach is to study stable matchings Roth and Sotomayo 990). Matchings ae stable if no pai of agents would pefe to leave thei cuent patnes fo each othe. Accoding to theoy, a stable two-sided matching can always be found assuming that all pefeences ae known. Howeve, thee ae many situations whee agents do not know thei own pefeences fom the beginning. Fo example, an employe may base his pefeences on inteviews with applicants. If thee ae too many candidates, the employe will not be able to inteview moe than a few of them. Job seekes may have altenative employes to inteview with, and the employe may not be able to put a easonably good candidate on hold fo long, while seaching fo a bette candidate. In these types of situations whee only a small potion of pefeences will eve be evealed, it does not make sense to speak about the best oveall matching. Instead, we focus on how agents optimize thei seach stategies. A simple vesion of this poblem involves a single employe willing to inteview at most secetaial candidates, who all want the job but will not be put on hold. This is the famous secetay poblem in optimal stopping theoy. Fo the histoy of this poblem, see Feguson 989.) The employe knows that if he wee to inteview all candidates, he could ank ode them. Howeve, he must stop at some point and hie this peson without having seen the emaining candidates. When to stop depends on what measue the employe is tying to optimize; we will assume this measue to be the expected ank of the hied secetay among all candidates. Lindley 96) showed that as gows, the optimal expected ank tends to a constant slightly less than fou. Thus, even with a line of thousands of candidates, the employe has a stopping stategy so that he can expect to obtain the fouth anked candidate. The exact fomula fo this constant was poved by Chow et al. 964). In this pape, we extend thei methods to solve Lindley s poblem fo the moe ealistic two-sided case whee both employes and candidates ae choosy Rittaud 005). We find that in this case, the optimal stopping stategy leads to agents obtaining patnes with an expected ank of among the potential patnes any agent is willing to evaluate. Fo example, if evey employe and job seeke is willing to take pat in at most 00 inteviews, then they can expect to obtain a patne at the tenth best pecentile. Of couse, ou esults depend on details of the maket. We make the necessay assumptions simple to have a welldefined and solvable poblem. All agents ae ational and intelligent, i.e., assume othe agents to be ational as well. Pefeences ae independent; the favoite of one agent is no moe likely than anyone else to be favoed by anothe agent. Agents ae eithe coopeative o noncoopeative: if eveyone is coopeative it gives a highe on aveage payoff fo all. We compae the espective outcomes and obtain a measue of the cost to society fo noncoopeation in this kind of game. 9

9 Opeations Reseach 555), pp. 9 93, 007 IFORMS In the following, we will adhee to the popula tadition in two-sided matching of descibing the model in tems of mutual mate selection of males and females cf. Roth and Sotomayo 990). Indeed, mate seach models elated to ous ae aleady studied in theoetical biology, although not in connection with Lindley s poblem. Fo a suvey of such models, see Alpen et al. 005.) The outline of this pape is as follows. In, we descibe the fomal model and summaize ou esults. Sections 3 and 4 descibe the solutions to the coopeative and the noncoopeative vesions of the poblem. In 5, we discuss aspects of the model, the method, and the esults. Thee appendices contain technical pats of the poofs.. Model, Appoach, and Summay of Results We assume a lage univese of men and women who will seach fo a mate fo a maximum of peiods. In each peiod, evey agent will date an agent of the opposite sex. If they both agee on maying, they leave the game. Othewise, they poceed to the next peiod and will neve date each othe again. In the th peiod, agents must may. The pefeences of agents ae geneated by the following pocess. Fom the viewpoint of any agent in peiod, they ank thei cuent date elative to the patnes aleady obseved. The ank of the th patne is a andom vaiable dawn fom a unifom distibution on the set of possible anks to ). Let us call this the unifomity assumption. All these andom vaiables ae assumed to be independent, meaning that how agent A anks agent B caies no infomation about how A anks futue dates, no about the anks of any othe agent. Let us call this the independence assumption. This implies that no conclusions can be dawn fom an agent s histoy of being ejected o poposed to. As in Lindley s 96) vaiant of the secetay poblem, we assume that each agent wants to minimize the expected ank of the mate among the patnes the agent would meet if she completed all peiods. Although the actual set of mates that an agent would meet in the emaining peiods is not known, we can easily compute the expected final ank fo a mate who is anked R among the patnes obseved up to peiod. Accoding to the unifomity and independence assumptions, the expected ank of this mate afte one moe date is R + R R + + R = + + + + R whee the fist tem coesponds to the case whee the next date is wose anked than R and the second tem coesponds to when it is bette anked. It follows immediately by epeating the same computation fo peiods + + ) that the expected final ank of the cuent mate is + + + 3 + + R = + + R A stategy in this two-sided secetay game is a ule that says whethe to popose maiage in peiod to a date of obseved ank R. Remembe that the agent cannot be cetain that the date will agee on maying. If agents on the othe side always agee, then we would be back in the one-sided secetay poblem. Given all agents stategies, we can define R end as the expected final ank of a cetain agent enteing peiod. In the last peiod, eveyone can expect to obtain an aveage mate: R end = + /. Agents want to choose stategies that minimize the expected final ank at the stat of the game: R end... Appoach and Summay of Results Denote pobability and expected value by P and E, espectively. Fom the definition of the model, it is evident that the following fundamental ecuence govens the expected final ank when an agent entes peiod : R end = Pmay + ER may + + Pnot may R end + ) whee R denotes the ank of the date in peiod among the patnes obseved up to that point. This ecuence will be the basis fo ou analysis. In any given peiod, agents face the following basic game. Two agents, Playe, and Playe meet. Thee exist some utilities, u and u, fo Playe of stopping with Playe and vice vesa. These ae based on thei anking of each othe. These values ae pivate infomation. If the playes do not both agee to stop in this peiod, they must go on seaching. By the popeties of the model and assuming that both agents continue to play optimally, they can then expect the same utility c = R end +. See Table. The basic game can be tackled eithe coopeatively o noncoopeatively by the playes. We will discuss below what this will entail and what esults we obtain in the diffeent cases. oncoopeation. The optimal noncoopeative stategy is fo Playe i to accept only if u i c fo i =. In the th and last peiod, evey agent who is still in the game will accept maiage because it is always pefeable to not maying at all). In the next to last peiod, an agent will want to may if his o he cuent date is at least as good as what he o she would expect to obtain in the last peiod. Solutions to games obtained in this way by backwad induction ae called subgame pefect equilibia Selten 975). If at some point an agent is indiffeent between maying a Table. Payoff matix fo the basic game in a given peiod. Accept Don t accept Accept u u cc Don t accept cc cc

Opeations Reseach 555), pp. 9 93, 007 IFORMS 93 mate o poceeding to the next peiod, we get multiple subgame pefect equilibia. Howeve, the expected final ank is the same in both cases, so all subgame pefect equilibia yield the same value of R end. Fo simplicity, we theefoe efe to the paticula subgame pefect stategy whee agents always may if indiffeent as the optimal stategy, which is the same fo all agents in this subgame pefect equilibium. In 4, we will pove the following esult. Theoem. In a subgame pefect equilibium of the noncoopeative two-sided secetay poblem, lim R end = Coopeation. If utilities ae tansfeable and the agents tuthfully communicate the values of u and u to each othe, then they can do bette by ageeing to stop wheneve u + u > c. Because tansfeable utilities and tustwothiness ae vey stong assumptions, we call this stategy stongly coopeative. Howeve, even if utilities ae not tansfeable and communication is absent, it is still possible fo the agents to benefit fom coopeation. The ageement should be fo Playe i to accept wheneve u i ĉ fo some theshold ĉ<c chosen to optimize the expected payoff, i.e., maximizing Pu ĉ Pu ĉ Eu u ĉ + c Pu ĉ Pu ĉ fo Playe and analogously fo Playe. We call this stategy weakly coopeative. In 3, we analyze the two coopeative vesions of the model. The weakly coopeative scenaio is completely solved. Theoem. Unde the optimal stategy of the weakly coopeative two-sided secetay game, R lim end = 7/3 09 We pesent evidence in the stongly coopeative scenaio fo the following conjectue. Conjectue. Unde the optimal stategy of the stongly coopeative two-sided secetay game, lim R end = 3/4 087 Social cost of noncoopeation. This game illustates the game-theoetic concept of a social dilemma. Befoe the game, all agents would like to agee on the coopeative stategy and pomise to be somewhat moe geneous in accepting maiage poposals. Howeve, when an individual agent finds heself in a position whee the stategy calls on he to may a date that she finds below he expectations, she is tempted to eject this patne to optimize he own good instead. If all agents do that, then the expected outcome is wose fo all; accoding to Theoems and, the expected change is about 8% fo the wose. Compaed to the conjectued outcome of the stongly coopeative game, the cost of noncoopeation is about 3%. 3. The Optimal Coopeative Stategies Let us assume that the agents agee on a stategy, i.e., a set of intege thesholds s s s with the ule that maiage must be poposed in peiod if the obseved ank does not exceed the theshold s. Consequently, the pobability that an agent will popose in peiod is s /, due to the unifomity assumption. Hence, the pobability of an ageement to may is Pmay = s ) and the expected obseved ank of the patne, given that an agent poposes, is ER may = s + Plugging these expessions into the fundamental ecuence ) yields the ecuence fo the coopeative poblem: s ) R end = s + + + ) ) s + R end + ) with bounday condition R end = + /. We want to minimize R end. All factos and tems ae nonnegative, so fo each fom to, we simply want to agee on the theshold s that will minimize R end. To find this theshold, we diffeentiate the expession with espect to s and find the geate of the two oots to be t = 3 ) + + R end + 3) The smalle oot is zeo. ow set s to eithe t o t, depending on which value minimizes R end, with the exception that s =t if t < 0. Because R end + + /, we have /3 <t /3 4) which guaantees that 0 s as equied. Fo example, we have t = /3 5) which means that in the next-to-the-last peiod an agent will popose if the cuent date is among the best two thids of all patnes the agent has seen. Fo futue use, let us also investigate when the theshold s becomes zeo, i.e., when an agent will not may anyone. Lemma. The theshold s = 0 if and only if t /3.

94 Opeations Reseach 555), pp. 9 93, 007 IFORMS Poof. It is staightfowad to veify fom ) that zeo is the nonnegative intege value of s that minimizes R end if and only if R end + + / +, which is equivalent to t /3. Appoximating s by t, we can ewite Equation ) using 3) into a ecuence puely in R end, which fo educes to a simple appoximative ecuence: R end R end + R end = + / 6 7 + R end + 3 If we now appoximate the diffeence R end + R end by the deivative, we obtain a simple diffeential equation with the exact solution R end = + 7/3 + 4 /. If this cude appoximation would wok all the way down to the fist peiod, then an agent s expected final ank when enteing the game would be R end 7/3. This tuns out to be tue, but we do not know how to pove it diectly fom the quality of the appoximation. Instead, we analyze the ecuence in detail, using the same appoach as Chow et al. 964), although we will encounte diffeent technical difficulties. Recall that s is eithe t o t. Ou fist measue is to define = s t, so that. ow ewite the ecuence ) as a ecuence in t by eplacing s with t + and eplacing R end and R end + accoding to 3). Afte extensive cancellations, we obtain t = t3 /3 + + t + 3 /3 /3 6) + Because and 3, we have an uppe bound of t T t def = t3 /3 + + t + /3 /3 7) + Similaly, the inequalities 0 and 3 givealowe bound of t t def = t3 /3 + t /3 /3 8) + To pove explicit lowe and uppe bounds fo t we need to examine the functions T t and t. Lemma. Fo sufficiently lage, the following holds:if /3 x /3 and x<y< 3/ +, then T x < T y and x < y. Poof. Diffeentiating the cubic polynomial T t with espect to t eveals that T t has a local minimum at t = + and a local maximum at t = +. This means that T t is an inceasing function in the inteval + +. Thus, if y +, it follows diectly that T x<t y. The case when y> + emains. In this case, because T t is deceasing in the inteval + 3/ +,wehave T y > T 3/ + ) = 6/4 + o while T x T /3 = 46/8 + o Fotunately, 6/4 > 46/8, which shows that T x < T y fo lage. Fo t, the easoning is completely analogous. Lemma 3. Fo sufficiently lage, the uppe bound t 3/ + 3 holds fo all in the inteval. Poof. By backwads induction. The lemma holds fo = because by 5), t = /3 3/ + 3 fo all. ow assume that the lemma holds fo a given >, and let 3/ + y = 3 By the induction hypothesis, t y. Because > which is lage, it follow fom Lemma with x = t, togethe with 4), and the fact that y< 3/ +, that T t T y. Combined with 7), this yields 3/ + t T t T ) 3 To conclude the induction step, we only need to pove that 3/ + T ) 3/ + 3 + 3 wheneve fo sufficiently lage. This can be veified though Stum sequences and seveal steps of computations in Maple see Appendix A). The lowe bound follows a simila patten. Lemma 4. Fo sufficiently lage, the lowe bound t 3/ + 3 holds fo all in the inteval.

Opeations Reseach 555), pp. 9 93, 007 IFORMS 95 Poof. Following the poof of the pevious lemma, using 8) instead, we only need to pove that 3/ ) 3/ + 3 + + 3 wheneve fo sufficiently lage. Again, this can be veified though Stum sequences and seveal steps of computations in Maple see Appendix A). Fo a peiod such that t < /3, we have s = 0by Lemma, so no patnes ae accepted and the expected final ank is the same as in the next peiod: R end = R end +. Define cit to be the geatest with t < /3 o let cit = 0 if thee is no such ). Then, R end = R end = =R end cit +. Lemma 5. cit / 3/7 and t cit /3 as. Poof. Let = 3/7 /3 and = 3/7. Fom Lemma 3 with =, we obtain t < /3 fo sufficiently lage. Fom Lemma 4 with =, we obtain t /3 fo sufficiently lage. We conclude that cit < and hence cit / 3/7 as. Once again using Lemma 3 and Lemma 4, we obtain t /3 and t /3, and hence t cit /3 as. We can now finish the poof of Theoem. By 3), ) 3 R end + = t + + / + 9) Thus, we have R end = R end cit + = 3/t cit + + / cit + = 3/t cit + + / cit / + / 3/ /3 + 3/7 = 7/3 3.. The Stongly Coopeative Case In the stongly coopeative case, both agents accept maiage if thei aveage ank of each othe is bette than the ank they can expect by poceeding to the next peiod. This will yield the best possible aveage outcome fo agents. We conjectue that this outcome satisfies R end / 3/4. We will motivate this conjectue, as well as discuss why the method fo the weakly coopeative case does not wok completely in the stongly coopeative case. Suppose that two agents on a date agee to may if the sum of thei obseved anks is at most s + fo some theshold value s to be optimized. Fom the bounday condition, it is easy to deduce that the optimal thesholds will satisfy s fo evey peiod. Hence, out of the possible combinations of anks in peiod, the numbe of combinations with a sum of at most s + is+ + +s = s s + /. We obtain the pobability of maiage to be Pmay = s s + Given that two agents agee to may i.e., given that thei ank sum is at most s + ), the expected ank of each date is ER may = s + s + +s s s + / = s + 3 The fundamental ecuence ) now takes the fom R end = s s + s + + 6 + + s ) s + R end + with the invaiant bounday condition R end = +/. Diffeentiating the ight-hand expession with espect to s, we find the smalle of the two oots to be negative while the lage oot is t = + + R end + ) + + + R end + + + R end ++/3 = + + R end + 3 + whee is an eo tem that tends to zeo as + / + R end + gows lage. This eo tem makes it difficult to poceed with the same method as in the pevious section. In addition to a tem to account fo the ounding, we now must handle the two tems and + to account fo the above eo in t, espectively, t +. The algeba becomes moe complicated but is still wokable. Unfotunately, the appoximations needed to deive uppe and lowe bounds of the exact ecuence become so cude that the bounds do not convege. Howeve, we can just ignoe the small eo tems to deive the same kind of appoximative ecuence as befoe. We obtain R end R end + 3 + R end + 3 R end = + / Again, the coesponding diffeential equation is solvable, and the asymptotics of the solution is R end / 3/4. This is the theoetical suppot fo ou conjectue. Futhemoe, it is possible to compute R end / fo lage by compute, and these values indeed seem to appoach 3/4 087.

96 Opeations Reseach 555), pp. 9 93, 007 IFORMS 4. The Optimal oncoopeative Stategy The thesholds that define the subgame pefect equilibium stategy in the noncoopeative case ae detemined by what agents can expect to achieve if they decline a poposal. An agent should agee to may a mate of ank R in peiod if the expected final ank achieved by maying now does not exceed the expected final ank in case of not maying: + + R R end + The theshold s should be the lagest intege value of R satisfying the above inequality, so that s =t with def t = + + R end + Fo example, we have t = 0) which means that in the next-to-the-last peiod we shall popose if and only if ou cuent date is among the best half of all patnes we have seen including the median if we have seen an odd numbe of patnes). To find the subgame pefect equilibium, we assume that all othe agents eason in the same way, so that we have Pmay = s / as in the pevious section. Consequently, we can just plug this new value of s into the ecuence ). The same appoximations now yield the ecuence R end R end + R end = + / + R end + 3 whose appoximating diffeential equation has the solution R end = + / + 4. If we could tust this appoximation, we would obtain R end. The poof of this esult will mimic the poof of the pevious section, with one new difficulty appeaing at one stage. Fo convenience, we euse the symbols T and fom the pevious section fo the coesponding functions in this new context. Fist, we ewite the ecuence ) as t = s s + + s t t + = ) Fo any, define = t s. Then, 0 <, and ) becomes t = t t + + t t ) + We now see that t + t + t 0. If we add this numbe to the numeato of ), we obtain the uppe bound t T t def = t3 + t + t 3) + Similaly, if we subtact t + t + t + 0 fom the numeato of ), we obtain the lowe bound t t def = t3 + t + t 4) + Lemma 6. Let f be any eal function with lim f= Fo sufficiently lage, the uppe bound t + + 3 holds fo all in the inteval f. Poof. By backwads induction. The lemma holds fo = because t = / + 4 fo all. ow assume that the lemma holds fo a given >f. It is easy to see that T t is an inceasing function in the inteval 0 t /3. Thus, by the induction hypothesis and the facts that t > 0 and + / + 3 < /3, we obtain + t T t T ) + 3 whee the fist inequality stems fom 3). Hence, to conclude the induction step, we only need to pove that ) + T + + 3 + 4 holds fo all in the inteval f fo sufficiently lage. This can be veified though seveal steps of computations in Maple see Appendix B). The lowe bound is tickie. We have not been able to find a lowe bound fo which the induction step woks in the whole inteval that we need, but by adding a constant tem of 0.48 to the denominato, we obtain a lowe bound that woks fom and downwads. Lemma 7. Fo sufficiently lage, the lowe bound t + + 3 + 048 holds fo all in the inteval +. Poof. Fo lage, we can compute an appoximation of t by pefoming steps of the ecuence, keeping only the most significant tem in each step: If t k = a k + o, wehaves k = a k + o too, so

Opeations Reseach 555), pp. 9 93, 007 IFORMS 97 ou ecuence ) gives t k = a 3 k + a k/ + o. Computation shows that t 0947 0945 40795 = + + 3 + 048 Again, we poceed by backwads induction. Assume that the lemma holds fo a given > +. It is easy to see that t is an inceasing function in the inteval 0 t /3. Thus, by the induction hypothesis and the facts that t > 0 and +/ + 3+048< /3, we obtain + t t ) + 3 + 048 whee the fist inequality stems fom 4). It emains fo us to pove that ) + + 3 + 048 + 4 + 048 fo lage. Again, seveal steps in Maple veify this inequality see Appendix B). We can now finish the poof of Theoem in much the same way as we poved Theoem. In the case t <, we always have s = 0, so no patnes ae eve accepted. Let cit be the geatest with t < o put cit = 0 if thee is no such ), so that R end = R end = =R end cit +. Lemma 7 gives that, fo lage, t + + + + > + 3 + 048 so cit < +. Lemma 6 with f= /3 gives, fo lage, t / /3 / /3 + / /3 / /3 +3 = 5/6 + o 5/6 / + o / < so cit + /3. Because /3 cit + < +, Lemma 6 with f= / /3 gives that t cit + cit ++ cit + cit + ++ + + as. Then, s cit + =, and by ), we get t cit. By definition of t,wehaver end cit + = t cit + / cit +. Thus, we have R end = R end cit + = + cit + t cit as because t cit and / /3 cit < +. 5. Discussion We will biefly discuss thee points about ou model, method, and esults. A unifying esult? We have computed the asymptotic behavio of the expected final ank of two vesions of the two-sided secetay poblem. In both cases, the answe was suggested immediately fom an appoximating diffeential equation, but to pove the coectness of the answes, we needed a much moe detailed analysis equiing seveal technical steps specific to each case. Pehaps some unifying appoximation esult of the following kind can be found: Given a ecuence t = t + bt 3 / + Q 3 t 3 + Q t + Q t + Q 0 t = c whee b and c ae constants and Q i fo i = 0 3, ae ational functions satisfying cetain conditions. The appoximating diffeential equation t = bt 3 / t= c satisfies that t/t as. Although confident that some geneal esult of this type must hold, we have not been able to see how to sufficiently bound the eo of the appoximation. We have consideed vaious subtleties of ou poofs fo specific cases. These indicate that a geneal esult might be difficult to obtain. Anothe way of looking at the noncoopeative poblem? An unexpected consequence of adopting the optimal noncoopeative stategy is that maiages will be appoximately unifomly distibuted ove the peiods. Fo evey agent who entes peiod, the pobability p of maiage and leaving the game in this peiod is p = s /. In the solution to the noncoopeative case, we obtained s / +, and hence p / +. Thus, the initial pobability of maying in peiod is p p p p + + + = Seeing this simple solution, we cannot help but wonde if thee is some a pioi agument showing that the optimal stategy must be one that makes optimal use of all the peiods. A model with eplacement. Suppose that the total numbe of agents, say U, is finite and not vey lage compaed to ). Then, we would have to allow fo the possibility that the same agents may come to date seveal times. We can think of it as unmaied dates being put back in the pool. Such a model would be much like the game studied in the laboatoy with human subjects by Eiksson and Stimling 005). How fa does this vaiation with eplacement take us fom the model studied in this pape? In the model with eplacement, you gain some infomation about the pool of

98 Opeations Reseach 555), pp. 9 93, 007 IFORMS emaining agents each time you do not may a date. This infomation may not change decisions vey much, but it fundamentally changes the complexity of the poblem. Futhemoe, the initial assumption of independence of pefeences will be compomised as the game poceeds, with an inceasing bias towad emaining agents not liking each othe vey much. Hence, the unifomity assumption will not hold, so we cannot even define the expected final ank of a date. Moe conceptual wok is needed to make the model with eplacement well defined. We leave it as an open poblem. Appendix A. Lemmas fo Theoem Hee we pove two lemmas that ae needed in the poof of Theoem. Lemma 8. Let T be the function defined in 7). Fo sufficiently lage, the inequality 3/ + T ) 3/ + 3 + 3 holds fo all. Poof. Afte the substitution z =, ou desied inequality tansfoms to 3/ + T ) z 3 3/ + A) z + 3 and the inteval tansfoms to z, which implies z We will pove that, fo sufficiently lage, inequality A) holds fo z see Figue A.). A lage implies a lage, so the lemma will follow. Evaluating T, inequality A) becomes gz def 3/ + = z + 3 3/ + 3 z 3 3/ + ++ z + 0 ) 3 3 Fist, investigating the case of z =, we obtain g = 3 4 + o )+ 3 3 ) Figue A.. Lage numbe We show that inequality A) holds in a egion like this. z > lage numbe which is positive fo lage. Because gz is continuous, it suffices to show that fo lage, gz 0 fo z. The zeos of gz ae the same as the zeos of 3/ + z + ) 3 [ 3/ + + 3 z 3/ + + + z 3 ) 3 3 ) + ] 3 We multiply this by 3 3 8 + z + z 6 and obtain a polynomial of degee 8 in z: q z def = 5z 8 576 69 + 5 + / z 7 + 57464 + 87488 / + 9 4 + 0 7/ + 9 3 34668 + 3/ + 5 ) z 6 65488 / + 3 7/ + 64849 + 8 / + 4 / + 3/ + 880 + / z 5 + 87488 / + 45 5 + 36 9/ 4 4 7/ 6 3 4 5/ 4346 34668 3/ +z 4 + 6644 5 9/ + 4 4 + 8 7/ +3433 3 65 5/ 46 8 3/ z 3 + 877 6 + 90 / + 8 5 36 9/ 03 4 70 7/ 33 3 6 5/ 4 z + 6644 / + 5 5/ z 9683 / + 6 3 We must show that fo lage, q z 0 fo z. Using a compute pogam like Maple, this can be done z

Opeations Reseach 555), pp. 9 93, 007 IFORMS 99 accoding to the following pocedue: Compute the fouth deivative q 4 z = a 4 z 4 + a 3 z 3 + a z + a z + a 0. Using Stum sequences, veify that fo lage, the polynomial q 4 z has no eal oots in the inteval z. Check that q 4 >0 fo lage, so that q 4 z > 0 eveywhee in the inteval z. Check that q, q, q, and q ae positive fo lage. This shows that fo sufficiently lage, q z > 0 in the inteval z. Lemma 9. Let be the function defined in 8). Fo sufficiently lage, the inequality 3/ ) 3/ + 3 + + 3 holds fo all. Poof. Afte the substitution z =, ou inequality tansfoms to 3/ z + ) 3/ 3 z + + A) 3 and the inteval tansfoms to z, which implies z< Evaluating in A) and multiplying by 364 + z + 3 z + + >0 yields the equivalent inequality p z def = 8 6 z 3 4z z + z + 4z + z + ) 3 + 34z + z + + ) +8 6 9z 3 +3z +9z+5 4z+ z + ) 84z + 3 z + + ) 0 whee p z is clealy a cubic polynomial in. Let us fist show that the leading coefficient of p z is positive, i.e., z + 4z + z + > z 3 + 4z + z + Afte squaing, expanding, and collecting, we obtain 4z 3 + 9z + 4z>0, which is tue fo all z. Using Stum sequences, we show that thee is a ẑ such that fo all z>ẑ, the polynomial p z has no eal oots geate than z, see Appendix C). Because the leading coefficient of p z is positive, we conclude that p z > 0 wheneve ẑ<z<. It emains to pove positivity of p z fo z ẑ. Fo each z, let + z be the lagest absolute value of the thee complex oots of the polynomial p z. Because + z is a continuous function of z, it attains a supemum on evey compact inteval, so we can define ˆ = sup + z z ẑ see Figue A.). Figue A.. ˆ The gaph of + z. We know that p z > 0 above the cuve.) z > ˆ zˆ Consequently, fo any >ˆ, it is tue that p z > 0 wheneve z<. A lage implies a lage because n, so the lemma follows. Appendix B. Lemmas fo Theoem Hee we pove the two lemmas that ae needed in the poof of Theoem. Lemma 0. Let T be the function defined in 3), and let f be any eal function with lim f=. Then, fo sufficiently lage, the inequality ) + T + 3 + + 4 holds fo all f. Poof. Afte the substitution z = + 3, ou inequality tansfoms to ) + T + B3) z z + and the condition tansfoms to z. We will pove that inequality B3) holds fo all z and sufficiently lage. A lage implies a lage because f, so the lemma will follow. Evaluating T in B3) yields gz def = + z + + 3 + + + + z 3 z z >0 + We see that [ ) g = 3 + 5/ + + + ) + 3 3/ + ) ] z

930 Opeations Reseach 555), pp. 9 93, 007 IFORMS which is positive fo lage. Because gz is continuous, it suffices to show that if is lage, gz 0 holds fo all z. The zeos of gz ae the same as the zeos of ) + z + [ + 3 + + + ] + + z 3 z z We multiply this by + z + z 6 and obtain a polynomial of degee 6 in z: q z def = 4 5 + 4 3 / 4 3) z 6 8 5 + 3 9/ + 3 4 + 7/ z 5 +4 / +5 5 +4 9/ 4 7/ 6 3 4 5/ z 4 + 4 5 9/ + 4 4 + 8 7/ + 5 3 + 5/ z 3 + 3 6 + 0 / + 9 5 4 9/ 5 4 7/ 5 3 6 5/ 4 z + 4 5 + 5 9/ + 0 4 + 0 7/ + 5 3 + 5/ z 6 +6 / +5 5 +0 9/ +5 4 +6 7/ + 3 We must show that if is lage, q z 0 holds fo all z. Using a compute pogam like Maple, this is just a matte of veification: Compute the fouth deivative q 4 z = a z + a z + a 0. Veify that a /a a 0 /a < 0 fo lage. This implies that q 4 z has no eal oots. Check that q 4 0>0 fo lage so that q 4 z > 0 eveywhee. This is immediately evident fom the expession fo q z above. Check that q, q, q, and q ae positive fo lage. This shows that fo sufficiently lage, the inequality q z > 0 holds fo all z. Lemma. Let be the function defined in 4). Fo sufficiently lage, the inequality ) + + 3 + 048 + 4 + 048 holds fo all in the inteval +. Poof. Let = 048. Afte the substitution + 3 +, ou inequality tansfoms to z = ) + z z + + B4) The inteval + tansfoms to 5 z + 3 B5) Evaluating in B4) and multiplying by z 3 z + + > 0 yields p z def = z z + z 3 + z ) + z z + + ) z z + z + + ) 0 a quadatic polynomial in. Let us fist show that the leading coefficient of p z is positive, i.e., z z + > z 3 + z Afte squaing, expanding, and collecting, we get 4z 3 3z z + > 0. The zeos of this cubic polynomial in z ae z 059, z 0559, and z 3 500, all of which ae less than 5 +. Thee ae always eal oots to p z because its constant tem is negative fo z 5 +. Fo each z, let + z be the geate of the two oots of p z. We can, of couse, wite down an explicit fomula fo + z, and with a compute pogam like Maple, it is easy to check that lim z + z z =. Thus, fo any >0, thee is a ẑ such that p z > 0 when >z+ +and z>ẑ. We choose = /4. In the inteesting inteval B5), we have z + + /4 + 3 + 5/4 <, whee the last inequality follows fom + 3 < 5/4, which is tue if 5. In the inteval ẑ /4 <z+ + 3, we know that p z > 0, so we only have to woy about the inteval 5+ z ẑ /4. Because + z is a continuous function, we can define ˆ = sup + z 5+ z ẑ /4. Then, p z > 0 in the inteval B5) fo all >ˆ. A lage implies a lage because +, so the lemma follows. Appendix C. Stum Sequences Given an n-degee polynomial Px with eal coefficients, the Stum sequence of Px is a finite sequence of polynomials P 0 x P xp n x defined ecusively by P 0 x = Px P x = P x P k x = emp k x P k x fo k n Hee empx Qx is the emainde when Px ae divided by Qx. Let Va 0 a a n be the numbe of sign vaiations in the list a 0 a a n, i.e., Va 0 a a n = i a i a i+ < 0 The following famous theoem counts the eal oots of Px. Fo a poof, we efe to Pestel and Delzell 00). Theoem 3 Stum s Theoem). If a<b ae eal numbes which ae not oots of Px, then the numbe of eal

Opeations Reseach 555), pp. 9 93, 007 IFORMS 93 oots of Px in the inteval a b is VP 0 ap n a VP 0 bp n b A geneal cubic polynomial Px= bx 3 + cx + dx + e has the Stum sequence P 0 x = bx 3 + cx + dx + e P x = 3bx + cx + d P x = 9b c 6bdx + cd 9be P 3 x = 9b 4bd3 + c d + 8bcde 4c 3 e 7b e 43bd c povided that the denominatos do not vanish. ow let Px be the polynomial p z in the poof of Lemma 9, identifying x with. Staightfowad computation in Maple yields lim P iz =+ and z lim lim P ib =+ z b fo 0 i 3. Theefoe, lim lim VP 0zP n z VP 0 bp n b z b = 0 0 = 0 and Stum s theoem poves that thee is a ẑ such that if z>ẑ, then p z has no eal oots geate than z. ow conside a geneal fouth-degee polynomial Px = ax 4 + bx 3 + cx + dx + e with Stum sequence P 0 xp 4 x. With Maple, we can compute these polynomials explicitly, but thee is no point in displaying them hee. Let Px be the polynomial q z in the poof of Lemma 8, identifying x with z. Fo 0 i 4, define a i = lim P i b i = lim P i Staightfowad computation in Maple yields a 0 a a a 3 a 4 = + + + b 0 b b b 3 b 4 = + + + + Theefoe, lim VP 0P n VP 0 P n = =0 and Stum s theoem poves that fo sufficiently lage, q z has no eal oots in the inteval. Acknowledgments This eseach was patially suppoted by the Swedish Reseach Council and the Euopean Commission s IHRP Pogamme, Algebaic Combinatoics in Euope gant HPR- CT-00-007). Refeences Alpen, S., I. Katantzi, D. Reynies. 005. Mathematical models of mutual mate choice. Recent Reseach Developments in Expeimental and Theoetical Biology. Tanswold Reseach etwok, Tivandum, Keala, India, 77 00. Chow, Y. S., S. Moiguti, H. Robbins, S. M. Samuels. 964. Optimal selection based on elative ank the secetay poblem ). Isael J. Math. 8 90. Eiksson, K., P. Stimling. 005. How unstable ae matchings fom decentalized mate seach? Pepint. Feguson, T. 989. Who solved the secetay poblem? Statist. Sci. 4 8 96. Lindley, D. 96. Dynamic pogamming and decision theoy. Appl. Statist. 0 39 5. Pestel, A., C.. Delzell. 00. Positive Polynomials Fom Hilbet s 7th Poblem to Real Algeba. Spinge-Velag, Belin, Gemany. Rittaud, B. 005. Quand les assistantes choisissent leu paton. La Recheche 38 6-7. Roth, A. E., M. A. O. Sotomayo. 990. Two-Sided Matching. Cambidge Univesity Pess, ew Yok. Selten, R. 975. Reexamination of the pefectness concept fo equilibium points in extensive games. Intenat. J. Game Theoy 4 5 55.