OPERATIONS RESEARCH
informs
Vol. 55, No. 5, September–October 2007, pp. 921–931 issn 0030-364X eissn 1526-5463 07 5505 0921
®
doi 10.1287/opre.1070.0403 © 2007 INFORMS
Optimal Expected Rank in a Two-Sided Secretary Problem Kimmo Eriksson, Jonas Sjöstrand, Pontus Strimling
Department of Mathematics and Physics, Mälardalen University, SE-721 23 Västerås, Sweden {
[email protected],
[email protected],
[email protected]}
In a two-sided version of the famous secretary problem, employers search for a secretary at the same time as secretaries search for an employer. Nobody accepts being put on hold, and nobody is willing to take part in more than N interviews. Preferences are independent, and agents seek to optimize the expected rank of the partner they obtain among the N potential partners. We find that in any subgame perfect equilibrium, the expected rank grows as the square root of N (whereas it tends to a constant in the original secretary problem). We also compute how much agents can gain by cooperation. Subject classifications: games/group decisions: strategic secretary problem; dynamic programming/optimal control: optimal stopping. Area of review: Optimization. History: Received June 2005; revisions received March 2006, June 2006; accepted June 2006. Published online in Articles in Advance. July 20, 2007.
1. Introduction
point and hire this person without having seen the remaining candidates. When to stop depends on what measure the employer is trying to optimize; we will assume this measure to be the expected rank of the hired secretary among all N candidates. Lindley (1961) showed that as N grows, the optimal expected rank tends to a constant slightly less than four. Thus, even with a line of thousands of candidates, the employer has a stopping strategy so that he can expect to obtain the fourth ranked candidate. The exact formula for this constant was proved by Chow et al. (1964). In this paper, we extend their methods to solve Lindley’s problem for the more realistic two-sided case where both employers and candidates are choosy (Rittaud 2005). We find that in this case, the optimal stopping strategy leads √ to agents obtaining partners with an expected rank of N among the N potential partners any agent is willing to evaluate. For example, if every employer and job seeker is willing to take part in at most 100 interviews, then they can expect to obtain a partner at the tenth best percentile. Of course, our results depend on details of the market. We make the necessary assumptions simple to have a welldefined and solvable problem. All agents are rational and intelligent, i.e., assume other agents to be rational as well. Preferences are independent; the favorite of one agent is no more likely than anyone else to be favored by another agent. Agents are either cooperative or noncooperative: if everyone is cooperative it gives a higher on average payoff for all. We compare the respective outcomes and obtain a measure of the cost to society for noncooperation in this kind of game.
Two-sided matching has been studied intensely in the last few decades. This area of research has been motivated by important real-life problems of matching employers with job seekers, or colleges with potential students, or men with women. In game theory, the standard approach is to study stable matchings (Roth and Sotomayor 1990). Matchings are stable if no pair of agents would prefer to leave their current partners for each other. According to theory, a stable two-sided matching can always be found assuming that all preferences are known. However, there are many situations where agents do not know their own preferences from the beginning. For example, an employer may base his preferences on interviews with applicants. If there are too many candidates, the employer will not be able to interview more than a few of them. Job seekers may have alternative employers to interview with, and the employer may not be able to put a reasonably good candidate on hold for long, while searching for a better candidate. In these types of situations where only a small portion of preferences will ever be revealed, it does not make sense to speak about the best overall matching. Instead, we focus on how agents optimize their search strategies. A simple version of this problem involves a single employer willing to interview at most N secretarial candidates, who all want the job but will not be put on hold. This is the famous secretary problem in optimal stopping theory. (For the history of this problem, see Ferguson 1989.) The employer knows that if he were to interview all N candidates, he could rank order them. However, he must stop at some 921
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In the following, we will adhere to the popular tradition in two-sided matching of describing the model in terms of mutual mate selection of males and females (cf. Roth and Sotomayor 1990). Indeed, mate search models related to ours are already studied in theoretical biology, although not in connection with Lindley’s problem. (For a survey of such models, see Alpern et al. 2005.) The outline of this paper is as follows. In §2, we describe the formal model and summarize our results. Sections 3 and 4 describe the solutions to the cooperative and the noncooperative versions of the problem. In §5, we discuss aspects of the model, the method, and the results. Three appendices contain technical parts of the proofs.
2. Model, Approach, and Summary of Results We assume a large universe of men and women who will search for a mate for a maximum of N periods. In each period, every agent will date an agent of the opposite sex. If they both agree on marrying, they leave the game. Otherwise, they proceed to the next period and will never date each other again. In the N th period, agents must marry. The preferences of agents are generated by the following process. From the viewpoint of any agent in period r, they rank their current date relative to the r − 1 partners already observed. The rank of the rth partner is a random variable drawn from a uniform distribution on the set of possible ranks (1 to r). Let us call this the uniformity assumption. All these random variables are assumed to be independent, meaning that how agent A ranks agent B carries no information about how A ranks future dates, nor about the ranks of any other agent. Let us call this the independence assumption. This implies that no conclusions can be drawn from an agent’s history of being rejected or proposed to. As in Lindley’s (1961) variant of the secretary problem, we assume that each agent wants to minimize the expected rank of the mate among the N partners the agent would meet if she completed all N periods. Although the actual set of mates that an agent would meet in the remaining periods is not known, we can easily compute the expected final rank for a mate who is ranked R among the r partners observed up to period r. According to the uniformity and independence assumptions, the expected rank of this mate after one more date is R r +1−R r +2 R + 1 + R= R r +1 r +1 r +1 where the first term corresponds to the case where the next date is worse ranked than R and the second term corresponds to when it is better ranked. It follows immediately (by repeating the same computation for periods r + 1 r + 2 N ) that the expected final rank of the current mate is r +2 r +3 N +1 N +1 · ··· R= R r +1 r +2 N r +1
Operations Research 55(5), pp. 921–931, © 2007 INFORMS
A strategy in this two-sided secretary game is a rule that says whether to propose marriage in period r to a date of observed rank R. Remember that the agent cannot be certain that the date will agree on marrying. If agents on the other side always agree, then we would be back in the one-sided secretary problem. Given all agents’ strategies, we can define Rend r as the expected final rank of a certain agent entering period r. In the last period, everyone can expect to obtain an average mate: Rend N = N + 1/2. Agents want to choose strategies that minimize the expected final rank at the start of the game: Rend 1. 2.1. Approach and Summary of Results Denote probability and expected value by P and E, respectively. From the definition of the model, it is evident that the following fundamental recurrence governs the expected final rank when an agent enters period r: N +1 · E R marry r +1 + P not marry · Rend r + 1
Rend r = P marry ·
(1)
where R denotes the rank of the date in period r among the r partners observed up to that point. This recurrence will be the basis for our analysis. In any given period r, agents face the following basic game. Two agents, Player 1, and Player 2 meet. There exist some utilities, u1 and u2 , for Player 1 of stopping with Player 2 and vice versa. These are based on their ranking of each other. These values are private information. If the players do not both agree to stop in this period, they must go on searching. By the properties of the model and assuming that both agents continue to play optimally, they can then expect the same utility c = Rend r + 1. See Table 1. The basic game can be tackled either cooperatively or noncooperatively by the players. We will discuss below what this will entail and what results we obtain in the different cases. Noncooperation. The optimal noncooperative strategy is for Player i to accept only if ui c for i = 1 2. In the N th and last period, every agent who is still in the game will accept marriage (because it is always preferable to not marrying at all). In the next to last period, an agent will want to marry if his or her current date is at least as good as what he or she would expect to obtain in the last period. Solutions to games obtained in this way by backward induction are called subgame perfect equilibria (Selten 1975). If at some point an agent is indifferent between marrying a Table 1.
Accept Don’t accept
Payoff matrix for the basic game in a given period. Accept
Don’t accept
u1 u2 c c
c c c c
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mate or proceeding to the next period, we get multiple subgame perfect equilibria. However, the expected final rank is the same in both cases, so all subgame perfect equilibria yield the same value of Rend 1. For simplicity, we therefore refer to the particular subgame perfect strategy where agents always marry if indifferent as the optimal strategy, which is the same for all agents in this subgame perfect equilibrium. In §4, we will prove the following result. Theorem 1. In a subgame perfect equilibrium of the noncooperative two-sided secretary problem, Rend 1 lim √ = 1 N → N Cooperation. If utilities are transferable and the agents truthfully communicate the values of u1 and u2 to each other, then they can do better by agreeing to stop whenever u1 + u2 > 2c. Because transferable utilities and trustworthiness are very strong assumptions, we call this strategy strongly cooperative. However, even if utilities are not transferable and communication is absent, it is still possible for the agents to benefit from cooperation. The agreement should be for Player i to accept whenever ui cˆ for some threshold cˆ < c chosen to optimize the expected payoff, i.e., maximizing P u1 c ˆ · P u2 c ˆ · E u1 u1 c ˆ + c1 − P u1 c ˆ · P u2 c ˆ for Player 1 and analogously for Player 2. We call this strategy weakly cooperative. In §3, we analyze the two cooperative versions of the model. The weakly cooperative scenario is completely solved. Theorem 2. Under the optimal strategy of the weakly cooperative two-sided secretary game, Rend 1 lim √ = 27/32 ≈ 092 N → N We present evidence in the strongly cooperative scenario for the following conjecture. Conjecture 1. Under the optimal strategy of the strongly cooperative two-sided secretary game, Rend 1 lim √ = 3/4 ≈ 087 N → N Social cost of noncooperation. This game illustrates the game-theoretic concept of a social dilemma. Before the game, all agents would like to agree on the cooperative strategy and promise to be somewhat more generous in accepting marriage proposals. However, when an individual agent finds herself in a position where the strategy calls on her to marry a date that she finds below her expectations, she is tempted to reject this partner to optimize her own good instead. If all agents do that, then the expected outcome is worse for all; according to Theorems 1 and 2, the expected change is about 8% for the worse. Compared to the conjectured outcome of the strongly cooperative game, the cost of noncooperation is about 13%.
3. The Optimal Cooperative Strategies Let us assume that the agents agree on a strategy, i.e., a set of integer thresholds s1 s2 sN −1 with the rule that marriage must be proposed in period r if the observed rank does not exceed the threshold sr . Consequently, the probability that an agent will propose in period r is sr /r, due to the uniformity assumption. Hence, the probability of an agreement to marry is P marry =
2 sr r
and the expected observed rank of the partner, given that an agent proposes, is E R marry =
sr + 1 2
Plugging these expressions into the fundamental recurrence (1) yields the recurrence for the cooperative problem: Rend r =
2 sr s + 1N + 1 · r r 2r + 1 2 s + 1− r · Rend r + 1 r
(2)
with boundary condition Rend N = N + 1/2. We want to minimize Rend 1. All factors and terms are nonnegative, so for each r from 1 to N − 1, we simply want to agree on the threshold sr that will minimize Rend r. To find this threshold, we differentiate the expression with respect to sr and find the greater of the two roots to be 2 2r + 1 tr = (3) · Rend r + 1 − 1 3 N +1 The smaller root is zero. Now set sr to either tr or
tr , depending on which value minimizes Rend r, with the exception that sr = tr if tr < 0. Because 1 Rend r +1 N + 1/2, we have −2/3 < tr 2r/3
(4)
which guarantees that 0 sr r as required. For example, we have tN −1 = 2/3N − 1
(5)
which means that in the next-to-the-last period an agent will propose if the current date is among the best two thirds of all partners the agent has seen. For future use, let us also investigate when the threshold sr becomes zero, i.e., when an agent will not marry anyone. Lemma 1. The threshold sr = 0 if and only if tr 2/3.
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Proof. It is straightforward to verify from (2) that zero is the nonnegative integer value of sr that minimizes Rend r if and only if Rend r + 1 N + 1/r + 1, which is equivalent to tr 2/3. Approximating sr by tr , we can rewrite Equation (2) using (3) into a recurrence purely in Rend , which for r 1 reduces to a simple approximative recurrence: Rend r ≈ Rend r + 1 −
If we now approximate the difference Rend r + 1 − Rend r by the derivative, we obtain a simple differential equation with the exact solution Rend r = N + 1 27/32 · N − r + 4 1/2 . If this crude approximation would work all the way down to the first period, then an agent’s expected final √ rank when entering the game would be Rend 1 ≈ 27N /32. This turns out to be true, but we do not know how to prove it directly from the quality of the approximation. Instead, we analyze the recurrence in detail, using the same approach as Chow et al. (1964), although we will encounter different technical difficulties. Recall that sr is either tr or tr . Our first measure is to define r = sr − tr , so that r ∈ −1 1. Now rewrite the recurrence (2) as a recurrence in tr by replacing sr with tr + r and replacing Rend r and Rend r + 1 according to (3). After extensive cancellations, we obtain −tr3 /3 + 2r + r 2 tr + 23r /3 − 2r/3 rr + 1
(6)
Because 2r 1 and 3r 1, we have an upper bound of def
tr−1 Tr tr =
−tr3 /3 + 1 + r 2 tr + 2/3 − 2r/3 rr + 1
(7)
Similarly, the inequalities 2r 0 and 3r −1 give a lower bound of 3 2 def −tr /3 + r tr
tr−1 r tr =
− 2/3 − 2r/3 rr + 1
Tr y > Tr
3/2r +
√ √ r = 6r/4 + or
while Tr x Tr 2r/3 = 46r/81 + or √ Fortunately, 6/4 > 46/81, which shows that Tr x < Tr y for large r. For r t, the reasoning is completely analogous.
16 R r + 13 27N + 12 end
Rend N = N + 1/2
tr−1 =
remains. √ In this case, √ because √ Tr t is decreasing in the interval 1 + r 2 3/2r + r , we have
(8)
To prove explicit lower and upper bounds for tr we need to examine the functions Tr t and r t. Lemma 2. For sufficiently large holds: If √ r, the following √ −2/3 x 2r/3 and x < y < 3/2r + r, then Tr x < Tr y and r x < r y. Proof. Differentiating the cubic polynomial Tr t with respect√ to t reveals that Tr t has a local√minimum at t = − 1 + r 2 and a local maximum at t = 1 + r 2 . This means √that Tr t in the inter√ is an increasing function √ val − 1 + r 2 1 + r 2 . Thus, if y 1 + r 2 , it √ follows directly that Tr x < Tr y. The case when y > 1 + r 2
Lemma 3. For sufficiently large N , the upper bound √ √ 3/2r + r 2 tr − √ 3 N −r holds for all r in the interval
√ N r N − 1.
Proof. By backwards induction. The lemma holds for r = N − 1 because by (5), √ tN −1 = 2N − 1/3
3/2N − 1 + √ 1
√ N − 1
−
2 3
for all √ N 1. Now assume that the lemma holds for a given r > N , and let √ √ 3/2r + r 2 y= − √ 3 N −r By the induction √ hypothesis, tr y. Because r > N which is large, it follow from Lemma 2 with x =√ tr , together with (4), and the fact that y < √ 3/2r + r, that Tr tr Tr y. Combined with (7), this yields √ tr−1 Tr tr Tr
√ 3/2r + r 2 − √ 3 N −r
To conclude the induction step, we only need to prove that √ √ √ 3/2r + r 2 3/2r − 1 + r − 1 2 Tr − − √ √ 3 3 N −r +1 N −r √ whenever N r N − 1 for sufficiently large N . This can be verified through Sturm sequences and several steps of computations in Maple (see Appendix A). √
The lower bound follows a similar pattern. Lemma 4. For sufficiently large N , the lower bound tr √
√ 3/2r N −r +1
−
2 3
holds for all r in the interval
√ N r N − 1.
Eriksson, Sjöstrand, and Strimling: Optimal Expected Rank in a Two-Sided Secretary Problem Operations Research 55(5), pp. 921–931, © 2007 INFORMS
Proof. Following the proof of the previous lemma, using (8) instead, we only need to prove that √ √ 3/2r 3/2r − 1 2 2 r √ √ − − 3 N −r +1+1 3 N −r +1 √ whenever N r N − 1 for sufficiently large N . Again, this can be verified through Sturm sequences and several steps of computations in Maple (see Appendix A). For a period r such that tr < 2/3, we have sr = 0 by Lemma 1, so no partners are accepted and the expected final rank is the same as in the next period: Rend r = Rend r +1. Define rcrit to be the greatest r with tr < 2/3 (or let rcrit = 0 if there is no such r). Then, Rend 1 = Rend 2 = · · · = Rend rcrit + 1. √ √ Lemma 5. rcrit / N → 32/27 and trcrit → 2/3 as N → . √ √ Proof. Let r1 = 32N /27 − N 1/3 and r2 = 32N /27 . From Lemma 3 with r = r1 , we obtain tr1 < 2/3 for sufficiently large N . From Lemma 4 with r = r2 , we obtain tr2 2/3 for sufficiently√large N . We conclude that r1 √ rcrit < r2 and hence rcrit / N → 32/27 as N → . Once again using Lemma 3 and Lemma 4, we obtain tr1 → 2/3 and tr2 → 2/3, and hence trcrit → 2/3 as N → . We can now finish the proof of Theorem 2. By (3), 3 tr + 1 N + 1/2r + 2 (9) Rend r + 1 = 2 Thus, we have R r + 1 Rend 1 = end √crit √ N N 3/2trcrit + 1N + 1/2rcrit + 2 = √ N 3/2trcrit + 11 + 1/N = √ √ 2rcrit / N + 2/ N 3/2 · 2/3 + 1 √ 2 32/27 = 27/32
→
3.1. The Strongly Cooperative Case In the strongly cooperative case, both agents accept marriage if their average rank of each other is better than the rank they can expect by proceeding to the next period. This will yield the best possible average outcome for√agents.√We conjecture that this outcome satisfies Rend 1/ N → 3/4. We will motivate this conjecture, as well as discuss why the method for the weakly cooperative case does not work completely in the strongly cooperative case. Suppose that two agents on a date agree to marry if the sum of their observed ranks is at most sr + 1 for some
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threshold value sr to be optimized. From the boundary condition, it is easy to deduce that the optimal thresholds will satisfy sr r for every period r. Hence, out of the r 2 possible combinations of ranks in period r, the number of combinations with a sum of at most sr + 1 is 1 + 2 + · · · + sr = sr sr + 1/2. We obtain the probability of marriage to be sr sr + 1 2r 2 Given that two agents agree to marry (i.e., given that their rank sum is at most sr + 1), the expected rank of each date is 1 · sr + 2sr − 1 + · · · + sr · 1 sr + 2 E R marry = = sr sr + 1/2 3
P marry =
The fundamental recurrence (1) now takes the form Rend r =
sr sr + 1sr + 2N + 1 6r 2 r + 1 s s + 1 + 1− r r 2 · Rend r + 1 2r
with the invariant boundary condition Rend N = N + 1/2. Differentiating the right-hand expression with respect to sr , we find the smaller of the two roots to be negative while the larger root is tr =
r +1 · R r + 1 − 1 N + 1 end 2 r +1 r +1 + ·Rend r +1 − ·R r +1+1/3 N +1 N +1 end
=2
r +1 3 · R r + 1 − + r N + 1 end 2
where r is an error term that tends to zero as r + 1/ N + 1 · Rend r + 1 grows large. This error term makes it difficult to proceed with the same method as in the previous section. In addition to a term to account for the rounding, we now must handle the two terms r and r+1 to account for the above error in tr , respectively, tr+1 . The algebra becomes more complicated but is still workable. Unfortunately, the approximations needed to derive upper and lower bounds of the exact recurrence become so crude that the bounds do not converge. However, we can just ignore the small error terms to derive the same kind of approximative recurrence as before. We obtain 2 Rend r ≈ Rend r + 1 − R r + 13 3N + 12 end Rend N = N + 1/2 Again, the corresponding differential equation is solvable, √ and √ the asymptotics of the solution is Rend 1/ N → 3/4. This is the theoretical support for our conjecture. √ Furthermore, it is possible to compute Rend 1/ N for large N by √ computer, and these values indeed seem to approach 3/4 ≈ 087.
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4. The Optimal Noncooperative Strategy The thresholds that define the subgame perfect equilibrium strategy in the noncooperative case are determined by what agents can expect to achieve if they decline a proposal. An agent should agree to marry a mate of rank R in period r if the expected final rank achieved by marrying now does not exceed the expected final rank in case of not marrying: N +1 · R Rend r + 1 r +1
def
N tN −1 = 2
(10)
which means that in the next-to-the-last period we shall propose if and only if our current date is among the best half of all partners we have seen (including the median if we have seen an odd number of partners). To find the subgame perfect equilibrium, we assume that all other agents reason in the same way, so that we have P marry = sr /r2 as in the previous section. Consequently, we can just plug this new value of sr into the recurrence (2). The same approximations now yield the recurrence 1 R r + 13 2N + 12 end
Rend N = N + 1/2 whose approximating differential equation has the solu√ could trust this tion Rend r = N + 1/ N − r + 4. If we √ approximation, we would obtain Rend 1 ≈ N . The proof of this result will mimic the proof of the previous section, with one new difficulty appearing at one stage. For convenience, we reuse the symbols Tr and r from the previous section for the corresponding functions in this new context. First, we rewrite the recurrence (1) as sr2 sr + 1 + 2r 2 − sr2 tr 2rr + 1
tN −1 =
N 2
(11)
For any r, define r = tr − sr . Then, 0 r < 1, and (11) becomes tr−1 =
tr − r 2 tr − r + 1 + 2r 2 − tr − r 2 tr (12) 2rr + 1
We now see that r tr + 1 − r tr2 + r tr − r 0. If we add this number to the numerator of (12), we obtain the upper bound def
tr−1 Tr tr =
−tr3 + 2tr2 + 2r 2 tr 2rr + 1
(14)
Lemma 6. Let f N be any real function with
For sufficiently large N , the upper bound √ r+ r tr √ N −r +3 holds for all r in the interval f N r N − 1.
For example, we have
tr−1 =
−tr3 + tr2 + 2r 2 − 1tr 2rr + 1
lim f N =
r +1 · R r + 1 N + 1 end
Rend r ≈ Rend r + 1 −
def
tr−1 r tr =
N →
The threshold sr should be the largest integer value of R satisfying the above inequality, so that sr = tr with tr =
Similarly, if we subtract r tr − 12 + r tr − r + tr − r + 2r 0 from the numerator of (12), we obtain the lower bound
(13)
Proof. By backwards induction. The lemma holds for r = N − 1 because √ N −1+ N −1 tN −1 = N /2 √ 4 for all N 2. Now assume that the lemma holds for a given r > f N . It is easy to see that T√ r t is an increasing function in the interval 0 t 2/3r. Thus, by the induction √ √ hypothesis√and the facts that tr > 0 and r + r/ N − r + 3 < 2/3r, we obtain √ r+ r tr−1 Tr tr Tr √ N −r +3 where the first inequality stems from (13). Hence, to conclude the induction step, we only need to prove that √ √ r+ r r −1+ r −1 Tr √ √ N −r +3 N −r +4 holds for all r in the interval f N r N − 1 for sufficiently large N . This can be verified through several steps of computations in Maple (see Appendix B). The lower bound is trickier. We have not been able to find a lower bound for which the induction step works in the whole interval that we need, but by adding a constant term of 0.148 to the denominator, we obtain a lower bound that works from N − 22 and downwards. Lemma 7. For sufficiently large N , the lower bound tr √
r +1 N − r + 3 + 0148
holds for all r in the interval
√ N + 1 r N − 22.
Proof. For large N , we can compute an approximation of tN −22 by performing 21 steps of the recurrence, keeping only the most significant term in each step: If tN −k = ak N + oN , we have sN −k = ak N + oN too, so
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our recurrence (11) gives tN −k−1 = −a3k + 2ak /2N + oN . Computation shows that tN −22 ≈ 019427N 019425N − 407925 N − 22 + 1 =√ 22 + 3 + 0148 Again, we proceed by backwards induction. Assume that √ the lemma holds for a given r > N + 1. It is easy to see √ that r t is an increasing function in the interval 0 t 2/3r. Thus, by the√induction hypothesis and √ the facts that tr > 0 and r + 1/ N − r + 3 + 0148 < 2/3r, we obtain r +1 tr−1 r tr r √ N − r + 3 + 0148 where the first inequality stems from (14). It remains for us to prove that r +1 r r √ √ N − r + 3 + 0148 N − r + 4 + 0148 for large N . Again, several steps in Maple verify this inequality (see Appendix B). We can now finish the proof of Theorem 1 in much the same way as we proved Theorem 2. In the case tr < 1, we always have sr = 0, so no partners are ever accepted. Let rcrit be the greatest r with tr < 1 (or put rcrit = 0 if there is no such r), so that Rend 1 = Rend 2 = · · · = Rend rcrit + 1. Lemma 7 gives that, for large N , √ N +1 +1 √ t N +1 > 1 √ N− N + 1 + 3 + 0148 √ √ so rcrit < N + 1. Lemma 6 with f N = N − N 1/3 gives, for large N , N 1/2 − N 1/3 + N 1/2 − N 1/3 t N 1/2 −N 1/3 N − N 1/2 − N 1/3 + 3 N − 2N 5/6 + oN 5/6 = < 1 N − N 1/2 + oN 1/2 √ 1/3 . so rcrit + 1 √ N −N √ 1/3 Because N − N rcrit + 1 < N + 2, Lemma 6 with f N = N 1/2 − N 1/3 gives that √ √ √ N +2+ N +2 rcrit +1+ rcrit +1 →1 1 trcrit +1 √ √ N −rcrit +2 N − N +1 as N → . Then, srcrit +1 = 1, and by (11), we get trcrit → 1. By definition of tr , we have Rend rcrit + 1 = trcrit N + 1/ rcrit + 1. Thus, we have Rend 1 Rend rcrit +1 N +1 trcrit · √ →1 = = √ √ rcrit +1 N N N because trcrit → 1 and N 1/2 − N 1/3 − 1 rcrit
large number
Large number
Appendix A. Lemmas for Theorem 2 Here we prove two lemmas that are needed in the proof of Theorem 2. Lemma 8. Let Tr be the function defined in (7). For sufficiently large N , the inequality √ √ √ 3/2r + r 2 3/2r − 1 + r − 1 2 Tr − − √ √ 3 3 N −r +1 N −r √ holds for all N r N − 1. √ Proof. After the substitution z = N − r, our desired inequality transforms to √
√
√ 3/2r + r 2 − Tr z 3 √ √ 3/2r − 1 + r − 1 2 − (A1) √ 3 z2 + 1 √ and the interval N r N − 1 transforms to 1 z √ r 2 − r, which implies 1 z r We will prove that, for sufficiently large r, inequality (A1) holds for 1 z r (see Figure A.1). A large N implies a large r, so the lemma will follow. Evaluating Tr , inequality (A1) becomes √ √ 3/2r − 1 + r − 1 2 − gr z = √ 3 z2 + 1 √ √ 1 3/2r + r 2 3 − − − 3 z 3 √ √ 2 2 3/2r + r 2 2 +1+r − + − r z 3 3 3
z 1
which is positive for large r. Because gr z is continuous, it suffices to show that for large r, gr z = 0 for 1 z r. The zeros of gr z are the same as the zeros of √ √ 3/2r − 1 + r − 1 2 2 − √ 3 z2 + 1 √
√ 1 1 3/2r + r 2 3 − − − rr + 1 2 3 z 3 √ √ 2 3/2r + r 2 2 2 + 1 + r − + 1 − r z 3 3 We multiply this by 23 38 rr + 12 z2 + 1z6 and obtain a polynomial of degree 8 in z: √ def qr z = −512z8 − 576 69r 2 + 5r + r 1/2 z7 √ + 157464r 2 − 1r + 1r 2 r − 1 − 874818r 11/2 + 19r 4 + 20r 7/2 + 19r 3 − 34668r 2 + 2r 3/2 + r − 512 z6 √ − 652488r 1/2 + 13 r 7/2 + 64849r + 82r 1/2 + 41r 1/2 + 1r 3/2 + 2880r + r 1/2 z5
def
· rr + 1−1 0 First, investigating the case of z = 1, we obtain gr 1 = 2 −
√
√ 2
3 r + or 4
+ 874818r 11/2 + 45r 5 + 36r 9/2 − 2r 4 − 24r 7/2 −16r 3 −4r 5/2−43416r 2 −346682r 3/2 +rz4 √ + 626244−r 5 − r 9/2 + 4r 4 + 8r 7/2 +324323r 3 −165r 5/2 −246r 2 −82r 3/2 z3 + 218727r 6 + 90r 11/2 + 81r 5 − 36r 9/2 − 103r 4 √
− 70r 7/2 − 33r 3 − 16r 5/2 − 4r 2 z2
+ 626244r 1/2 +15 r 5/2 z−19683r 1/2 +16 r 3 We must show that for large r, qr z = 0 for 1 z r. Using a computer program like Maple, this can be done
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according to the following procedure: • Compute the fourth derivative qr4 z = a4 rz4 + a3 rz3 + a2 rz2 + a1 rz + a0 r. • Using Sturm sequences, verify that for large r, the polynomial qr4 z has no real roots in the interval 1 z r. • Check that qr4 1 > 0 for large r, so that qr4 z > 0 everywhere in the interval 1 z r. • Check that qr 1, qr 1, qr 1, and qr 1 are positive for large r. This shows that for sufficiently large r, qr z > 0 in the interval 1 z r. Lemma 9. Let r be the function defined in (8). For sufficiently large N , the inequality √ √ 3/2r 3/2r − 1 2 2 √ − − r √ N −r +1+1 3 N −r +1 3 √ holds for all N r N − 1. √ Proof. After the substitution z = N − r, our inequality transforms to √ √ 3/2r 2 3/2r − 1 2 − − √ (A2) r z+1 3 3 z2 + 1 + 1 √ and the interval N r N − 1 transforms to 1 z √ r 2 − r, which implies 1 z < r Evaluating √ r in (A2) and multiplying by 364rr + 1 · z + 13 z2 + 1 + 1 > 0 yields the equivalent inequality √ def pz r = 81 6 −2z3 − 4z2 − 2z − 1 + 2z2 + 4z + 1 z2 + 1 r 3 + 324z + 1 z2 + 1 + 1 r 2 √ +18 6 9z3 +23z2 +19z+5−4z+12 z2 +1 r − 184z + 13 z2 + 1 + 1 0 where pz r is clearly a cubic polynomial in r. Let us first show that the leading coefficient of pz r is positive, i.e., 2z2 + 4z + 1 z2 + 1 > 2z3 + 4z2 + 2z + 1 After squaring, expanding, and collecting, we obtain 4z3 + 9z2 + 4z > 0, which is true for all z 1. Using Sturm sequences, we show that there is a zˆ such that for all z > z, ˆ the polynomial pz r has no real roots greater than z, (see Appendix C). Because the leading coefficient of pz r is positive, we conclude that pz r > 0 whenever zˆ < z < r. It remains to prove positivity of pz r for 1 z z. ˆ For each z, let r + z be the largest absolute value of the three complex roots of the polynomial pz r. Because r + z is a continuous function of z, it attains a supremum on every compact interval, so we can define rˆ = ˆ (see Figure A.2). sup%r + z& 1 z z'
Figure A.2.
The graph of r + z. (We know that pz r > 0 above the curve.)
r
1 ≤ z ≤ r > rˆ
rˆ
1
zˆ
z
Consequently, for any r > r, ˆ it is true that pz r > 0 whenever 1 z < r. A large N implies a large r because √ n r, so the lemma follows.
Appendix B. Lemmas for Theorem 1 Here we prove the two lemmas that are needed in the proof of Theorem 1. Lemma 10. Let Tr be the function defined in (13), and let f N be any real function with limN → f N = . Then, for sufficiently large N , the inequality √ √ r+ r r −1+ r −1 Tr √ √ N −r +3 N −r +4 holds for all f N r N − 1. Proof. After the substitution z = ity transforms to √ √ r+ r r −1+ r −1 Tr √ z z2 + 1
√ N − r + 3, our inequal-
(B3)
and the condition r N − 1 transforms to z 2. We will prove that inequality (B3) holds for all z 1 and sufficiently large r. A large N implies a large r because f N r, so the lemma will follow. Evaluating Tr in (B3) yields √ def r −1+ r −1 grz = √ z2 +1 √ √ √ r + r3 2r + r2 2r 2 r + r − + + z3 z2 z − > 0 2rr +1 We see that gr 1 =
√ 1 2 − 1 r 3 + r 5/2 2rr + 1 √ + 2r − 1 + 1 r 2 + r − 3r 3/2 − 2 + 2 r
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which is positive for large r. Because gr z is continuous, it suffices to show that if r is large, gr z = 0 holds for all z 1. The zeros of gr z are the same as the zeros of √ r −1+ r −1 2 √ z2 +1 √ √ √
r + r3 2r + r2 2r 2 r + r 2 1 − − + + 2rr +1 2 z3 z2 z
Evaluating r in (B4) and multiplying by 2rz3 z − 2 + 1 + > 0
We multiply this by 2rr + 12 z2 + 1z6 and obtain a polynomial of degree 6 in z:
a quadratic polynomial in r. Let us first show that the leading coefficient of pz r is positive, i.e., 2z2 − 1 z − 2 + 1 > 2z3 + 1 − 2z2
√ def qr z = 4 2r 5 + r 4 − r 3 − r 2 r − 1 − 2r 11/2 − r 4 − r 3 z6 − 8r 5 + 3r 9/2 + 3r 4 + r 7/2 z5 +42r
11/2
5
+5r +4r
9/2
−4r
7/2
3
−6r −4r
5/2
2
−r z
4
+ 4−r 5 − r 9/2 + 4r 4 + 8r 7/2 + 5r 3 + r 5/2 z3 + 3r 6 + 10r 11/2 + 9r 5 − 4r 9/2 − 15r 4 − 22r 7/2 − 25r 3 − 16r 5/2 − 4r 2 z2 + 4r 5 + 5r 9/2 + 10r 4 + 10r 7/2 + 5r 3 + r 5/2 z −r 6 +6r 11/2 +15r 5 +20r 9/2 +15r 4 +6r 7/2 +r 3 We must show that if r is large, qr z = 0 holds for all z 1. Using a computer program like Maple, this is just a matter of verification: • Compute the fourth derivative qr4 z = a2 rz2 + a1 rz + a0 r. • Verify that a1 r/2a2 r2 − a0 r/a2 r < 0 for large r. This implies that qr4 z has no real roots. • Check that qr4 0 > 0 for large r so that qr4 z > 0 everywhere. This is immediately evident from the expression for qr z above. • Check that qr 1, qr 1, qr 1, and qr 1 are positive for large r. This shows that for sufficiently large r, the inequality qr z > 0 holds for all z 1. Lemma 11. Let r be the function defined in (14). For sufficiently large N , the inequality r +1 r r √ √ N − r + 3 + 0148 N − r + 4 + 0148 √ holds for all r in the interval N + 1 r N − 22. Proof. Let = 0148. After the substitution z = √ N − r + 3 + , our inequality transforms to r +1 r r (B4) z z − 2 + 1 + √ The interval N + 1 r N − 22 transforms to 5 z − r − 12 − r + 3
(B5)
yields
def pz r = 2z2 − 1 z − 2 + 1 − 2z3 + 1 − 2z2 r 2 + z − 2 z − 2 + 1 + r − z2 − z + 1 z − 2 + 1 + 0
After squaring, expanding, and collecting, we get 4z3 − 3z2 − 2z + 1 > 0. The zeros of this cubic polynomial in z are z1 ≈ −0592, z2 ≈ 0559, and z3 ≈ 5100, all of which are less than 5 + . There are always real roots to pz r because its constant term is negative for z 5 + . For each z, let r + z be the greater of the two roots of pz r. We can, of course, write down an explicit formula for r + z, and with a computer program like Maple, it is easy to check that limz→ r + z − z = 1 − . Thus, for any ) > 0, there is a zˆ) such that pz r > 0 when r > z+1−+) and z > zˆ) . We choose ) = 1/4. Inthe interesting interval (B5), we have z + 1 − + 1/4 r − 12 − r + 3 + 5/4 < r, where the last inequality follows from r − 12 − r + 3 < r − 5/42 , which is true if r 5. In the interval zˆ1/4 < z + r − 12 − r + 3, we know that pz r > 0, so we only have to worry about the interval 5 + z zˆ1/4 . Because r + z is a continuous function, we can define rˆ = sup%r + z& 5+ z zˆ1/4 '. Then, pz r > 0 in the interval ˆ A large N implies a large √ (B5) for all r > r. r because N + 1 r, so the lemma follows.
Appendix C. Sturm Sequences Given an n-degree polynomial P x with real coefficients, the Sturm sequence of P x is a finite sequence of polynomials P0 x P1 x Pn x defined recursively by P0 x &= P x P1 x &= P x Pk x &= −remPk−2 x Pk−1 x for 2 k n Here remP x Qx is the remainder when P x are divided by Qx. Let V a0 a1 an be the number of sign variations in the list a0 a1 an , i.e., V a0 a1 an = +%i& ai ai+1 < 0' The following famous theorem counts the real roots of P x. For a proof, we refer to Prestel and Delzell (2001). Theorem 3 (Sturm’s Theorem). If a < b are real numbers which are not roots of P x, then the number of real
Eriksson, Sjöstrand, and Strimling: Optimal Expected Rank in a Two-Sided Secretary Problem Operations Research 55(5), pp. 921–931, © 2007 INFORMS
roots of P x in the interval a b is
Straightforward computation in Maple yields
V P0 a Pn a − V P0 b Pn b
a0 a1 a2 a3 a4 = · + − − + +
A general cubic polynomial P x = bx3 + cx2 + dx + e has the Sturm sequence
b0 b1 b2 b3 b4 = · + + + − +
P0 x = bx3 + cx2 + dx + e P1 x = 3bx2 + 2cx + d 1 P2 x = 2c 2 − 6bdx + cd − 9be 9b 9b−4bd 3 + c 2 d 2 + 18bcde − 4c 3 e − 27b 2 e2 P3 x = 43bd − c 2 2 provided that the denominators do not vanish. Now let P x be the polynomial pz r in the proof of Lemma 9, identifying x with r. Straightforward computation in Maple yields lim Pi z = +
z→
and
lim lim Pi b = +
931
Therefore, lim V P0 1Pn 1−V P0 rPn r = 2−2 = 0
r→
and Sturm’s theorem proves that for sufficiently large r, qr z has no real roots in the interval 1 r .
Acknowledgments This research was partially supported by the Swedish Research Council and the European Commission’s IHRP Programme, Algebraic Combinatorics in Europe (grant HPRNCT-2001-00272).
z→ b→
References
for 0 i 3. Therefore,
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lim lim V P0 z Pn z − V P0 b Pn b
z→ b→
= 0 − 0 = 0 and Sturm’s theorem proves that there is a zˆ such that if z > z, ˆ then pz r has no real roots greater than z. Now consider a general fourth-degree polynomial P x = ax4 + bx3 + cx2 + dx + e with Sturm sequence P0 x P4 x. With Maple, we can compute these polynomials explicitly, but there is no point in displaying them here. Let P x be the polynomial qr z in the proof of Lemma 8, identifying x with z. For 0 i 4, define ai &= lim Pi 1 r→
bi &= lim Pi r r→
