The Instability of Stable Matchings: The Influence of One Strategic Agent on The Matching Market

06/11/2018 ∙ by Ron Kupfer, et al. ∙ Hebrew University of Jerusalem 0

Consider a matching problem with n men and n women, with preferences drawn uniformly from the possible (n!)^2n full ranking options. We analyze the influence of one strategic agent on the quality of the other agents' matchings under the Gale--Shapley algorithm. We show that even though the Gale--Shapley algorithm is famous for being optimal for men, one small change in the reported preferences is enough for the women to get a near optimal match. In this case, the quality of the matching dramatically improves from the women's perspective. The expected women's rank is O(^4(n)) and almost surely the average women's rank is O(^2+ϵ(n)) rather than a rank of O(n/(n)) in both cases under a truthful regime.

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1 Introduction

The stable matching problem concerns a scenario where we must find a matching between two disjoint sets of agents that satisfies natural stability constraints. This problem has received an enormous amount of attention, starting with the seminal work of Gale and Shapley [7], and has been used as a paradigm in a host of applications ranging from matching doctors to hospitals [23] to matching kidney donors to recipients [25]. The basic formalism considers a matching between a set of men and women. Each man has a preference order over all women and an option of being unmatched, and the same goes for the women. A matching between the set of men and the set of women is called stable if there exists no “blocking pair”, i.e. a man and a woman who prefer each other to their current matching.

The Men-Proposing Deferred-Acceptance algorithm (also known as the Gale–Shapley algorithm [7]) is an algorithm for finding a stable matching and its proof of correctness shows that such a matching always exists. The algorithm works in an iterative way. In each round, every unmatched man proposes to his most preferred woman who has not previously rejected him. Each woman chooses her most preferred man out of those who proposed to her and releases all the other proposers to continue with their lists. The algorithm terminates when all men are matched or reached to the end of their list.

It is known [7] that the algorithm is optimal for men in the following sense: for each man , there is no other stable matching in which is matched with someone whom he prefers more. On the other hand, as shown in [17], the algorithm yields the worst stable matching for any woman. That is, for each woman , there is no other stable matching in which is matched with someone whom she prefers less.

Under the Gale–Shapley mechanism it is a dominant strategy for men to report their preferences truthfully [5]. This is not the case for women. The following simple example with two men (,) and two women (,) demonstrate this. Suppose that their preference rankings are as follows:

prefers prefers
prefers prefers

and for all of the agents the least preferred option is to stay single. The algorithm matches to . If falsely reports that she prefers staying single to being matched with , the algorithm matches her to , whom she prefers to .

Strategic behavior under stable matching algorithms has been a topic of vast research. In [22] it is shown that there is no algorithm for which reporting the true preferences is a dominant strategy for both men and women. A partial list of works on strategic behavior by women under the Gale–Shapley algorithm includes [5, 8, 11, 4, 28, 24, 16].

Notice that strategic behavior by a woman affects the outcome of the other women as well, e.g., in the example above, perhaps surprisingly, the strategic behavior of improves the matching of . It is known [9, 3] that in general, under the Gale–Shapley algorithm, strategic behavior by any set of women can only benefit the other women in the sense that if none of the former are worse off, then neither are any of the latter. A natural and well known way to manipulate the algorithm is to truncate the list of the reported preferences, i.e., to set a threshold so that only mates from some given rank and above are acceptable. This strategy is optimal for any woman when all the other agents’ preferences are known to her, assuming that all the other agents report truthfully [24]. In the same work it is shown that this is also true for a wide range of partial information structures.

In this paper, we analyze the effect of a selfish reporting strategy by one or more women, on the quality of the matching obtained by the other agents, both men and women.

Specifically, we explore the characteristics of such strategies in the commonly studied setting of a balanced market with men and women (e.g., [12, 13, 29]). The set of preferences is drawn uniformly at random from the set of all possible full ranking options (e.g., [19, 15, 4, 21]). A simple measure of the quality of a match from the perspective of a given agent is the rank of her or his match (e.g., [13]). We say that a person has a rank in some matching if they are matched with their th favorite mate (where rank denotes being matched to the most preferred mate and rank denotes being matched to the least preferred mate). In this model, under a truthful regime, it is known [20] that the expected rank of any woman is of order . The expected rank of the men is of order .

Our main theorem shows that one strategic agent is expected to dramatically affect the entire outcome of the matching. In particular, the average women’s rank is polylogarithmic in compared to an expected average rank of order of

under a truthful reporting regime. These results hold in expectation and with high probability. Formally:

Theorem 1.1

In a random uniformly ranked balanced matching market with men and women, where a single woman uses her optimal strategy and all the other agents report truthfully, we have that:

  1. Almost surely, a fraction of the women are matched with a man from their top men. Furthermore, the expected average women’s rank is of order of .

  2. Almost surely, the average men’s rank is no better than .

These results do not hinge on the strategic woman having full information of the preferences or taking the exact optimal truncation, but are in fact more robust, as we will show later.

Since the Gale–Shapley algorithm yields the worst stable matching for women, any upper bound for the women’s rank holds for any matching algorithm that outputs a stable matching for the true preferences. Assume that the strategic woman considers only truncation strategies. The optimal truncation strategy is indeed an optimal strategy for her among all the possible strategies. Since an optimal truncation strategy yields a stable matching in the true preferences under any matching algorithm that guarantees to output a stable matching for the reported preferences, our results hold for a large family of algorithms. Such algorithms are discussed in ,e.g., [26, 6, 14, 1, 13, 10].

We present simulation results in the setting described above, for different market sizes. These simulation results reinforce the theorem’s conclusions and suggest that the actual effect might be even stronger than the formally proven effect.

One interpretation of the results is that although the Gale–Shapley algorithm is optimal for the men, in fact strategic behavior by only one woman is sufficient to almost completely eliminate the men’s advantage. Notice that strategic behavior narrows the set of possible stable matchings attainable by any algorithm. Hence, one strategic agent may strongly influence the matching in favor of that agent’s sex, be it the men or the women.

2 Influence under Full Information

2.1 The Model

The Process

We analyze the effect of one strategic agent in a balanced market of men and women where the set of preferences are chosen uniformly from the possible full ranking options. In this section, we assume that one woman, , gets to look at all the other agents’ preference lists before the algorithm is run. Then acts according to a preference list chosen by her in a strategic way that maximizes her utility from the matching. This maximization is done with respect to her original preferences.

It is shown in [17] that the outcome of the Gale–Shapley algorithm is independent of the order of proposals made by the men. Using this fact and the principle of deferred decisions, we follow [15] by describing the algorithm as follows: a man needs to choose his th preference only after the first women have already rejected him. The fact that the preferences are chosen uniformly allows another simplification. When a man chooses his th woman he chooses with “amnesia”; i.e., the realization is done over all possibilities, allowing him to choose a woman who has already rejected him and make her a redundant proposal. In addition, each woman also reveals her preferences in an online manner and she accepts her th (nonredundant) proposal with probability . It is convenient to look at the running of the algorithm under ’s strategy as a process in which women keep getting proposals and keeps rejecting all of the offers she gets. At some point, decides to accept one of the offers and the process terminates. The stopping point is selected optimally given the preferences of all the agents. As we will show, the outcome is a stable matching for the original preferences. Combining these observations, we follow [15] by describing the process using the following algorithm. In order to account for the property that ’s strategy is optimal, we add an oracle to that tells her when to stop the rejection process. That is, the oracle is exposed to the realization of all the players’ preferences yet to come. This process is equivalent to the original Gale–Shapley algorithm with acting in a strategically optimal way.
sets representing the women proposed to so far by men to .
the number of men who have proposed so far.
the man who is currently proposing.
the woman who is currently being proposed to.
the men who have made the best offer so far to women to . if the woman has received no offer.
the number of proposals received by women to .

  1. Let , , and , for ; also let .

  2. If , increase by and let . Otherwise, if ’s oracle tells her to stop, the process terminates and this is the final matching. Otherwise, let reject and set .

  3. Let be a random number uniformly chosen between and . We say that man proposes to woman . If (i.e., if ’s proposal is redundant), repeat this step. Otherwise replace with and go on to step 4.

  4. Increase by one. With probability , return to step 3 (in this case we say that woman rejects the proposal). Otherwise interchange (that is, accepts the proposal and her former match is the next proposer). If the new value of is zero, or if , go back to step 2; otherwise continue with step 3.

Probabilistic Notations

Some asymptotic notations and probabilistic bounds are noted. We say that an event occurs almost surely (a.s.) or with high probability (w.h.p.), if the probability that it doesn’t happen is , i.e., if the probability of nonoccurrence approaches zero as goes to infinity.

Multiplicative Chernoff Bound Let

be independent random variables taking values in

. Let denote their sum and denote the sum’s expected value. Then,

2.2 Main Results

The main result in this section is that in the above scenario, one strategic agent is expected to dramatically affect the entire outcome of the matching. In particular, the expected rank for the women is polylogarithmic in compared to an expected rank of under a truthful regime. Similarly, with a probability that goes to as grows, the rank of almost all the women is polylogarithmic in .

Due to the proposing nature of the Gale–Shapley algorithm, any optimal strategy for a single woman outputs the same matching not only for her but to all other women as well.

Proposition 1

The process described above is equivalent to reporting in a strategically optimal way under the Gale–Shapley algorithm.

Another desired property would be that the process yields a stable matching and all women are matched.

Lemma 1

The process terminates with a stable matching where all the women are matched.

Proof

Let be the matching resulting from the process. It is stable for the reported preferences. For any possible pair not including the strategic woman , the stability for the reported preferences implies stability for the true preferences. For any pair that includes and a man , either rejected him and got a better match, or didn’t get an offer from . In the former case and are not a blocking pair due to ’s preferences, and in the latter case and not a blocking pair due to ’s preferences. Hence, the matching is stable. Since there is a matching where all agents are matched (for example, the women-optimal matching), by the lone wolf theorem (see for example [18]), in any stable matching all agents are matched. ∎

By the work in [20], it is known that ’s best stable match is w.h.p of polylogarithmic order. We denote the event where gets one of her top preferences by .

Lemma 2

For any function that goes to infinity and for any large enough , the following inequalities hold:

(1)
(2)
Proof

The first inequality is a corollary of Theorem 6.1 in [20]. The second term follows from Markov’s inequality and the fact that the expected rank of the best stable matching is of order . ∎

By the previous lemma, the probability of the matching process terminating before gets one of her top preferences goes to zero as grows. The next lemma shows that the number of proposals made to conditioned on this event is not very different from the distribution of the proposals without this condition.

Lemma 3

Let , and be the total number of distinct (i.e., nonredundant) proposals made to before the termination of the algorithm. Then, for large enough and any .

Proof

First, we examine without conditioning that the event occur. We start by considering a random order of proposers and look for the first time gets a proposal from a given man . Since the preference list of is determined online and independently of the proposals she gets, the probability that proposals are needed is exactly .111Item will be in position in exactly of all possible orders of proposals. Using the union bound, we deduce that when counting the proposals till gets a proposal from one of her most preferred men. By Bayes’ rule, and conditioning on the fact that we are in the event , we get

where the last inequality holds by Lemma 2. ∎

The next step is to show that all women get a similar number of (not necessarily distinct) proposals. This lemma is not directly connected to the termination of the process by but rather a general statement about the uniform nature of proposals.

Lemma 4

Let be the total number of proposals made by a men . If then with a probability of at least , all of the women get at least proposals.

Proof

First we estimate the probability of an arbitrary woman getting fewer than

proposals. Since in each proposal the woman who gets a proposal is chosen uniformly and independently of previous proposals, we may use the Chernoff bound with parameters and . Thus, the probability of such an event is less than . By the union bound, the probability of at least one woman not getting enough proposals is bounded by

So far, we know the number of proposals distributed in an approximately equal way among the women. But this is so only because we allow redundant proposals. For distinct proposals, however, this does not necessarily hold. In the next lemma we show that when conditioning on the event that the process terminates with all men matched, we can get a similar result, albeit a slightly less tight one. This conditioning is reasonable by Lemma 1.

Lemma 5

With probability greater than , no men propose to the same woman more than times.

Proof

Assume that some man, , proposes proposals in total and let be some woman.

We consider two cases:

We show that these events happen with negligible probability. The probability that some woman gets no proposal from is . For , this probability is no more than . Because only full matchings are considered, we know that no man reaches to the end of his preference list (up to at most one proposal to his least preferred mate). Therefore the probability of the event is at most .

In the case of , makes at most proposals. We use the Chernoff bound, with and . The probability that more than proposals are made to the same women is at most

Summing over all men and women and using the union bound, we get that with probability at least , no man proposes to the same woman more than times. ∎

Corollary 1

If a woman gets proposals then with probability at least she gets at least distinct proposals.

Lemma 6

Let be a constant, a woman, and the number of different proposals made to . Then, with probability at least , gets matched with a man who ranks among her most preferred men.

Proof

Since ’s preferences are independent of the proposals she gets, the probability of her not getting a proposals from any subset of men out of men is . ∎

Lemma 7

Denote by the random variable denoting woman ’s rank, given that she gets at least distinct proposals. Let be the average rank of the women. Then, and with probability , .

Proof

We start by estimating using the following equality:

where (and thus ) and, by Lemma 6,

where the last inequality is due to the convergence of the sum of geometric series. This term is bounded in the following way:

where the first inequality is due to the approximation and the second is true for , which is precisely our case. By the linearity property of the expectation, we get that .

Next we show that with high probability is small. We start by rearranging ’s summing order:

This can easily be bounded by

For , the number of women with rank less than is of course no more than . Therefore,

On the other hand, if is larger we can show that only a small number of women have a rank far from .

Using the Chernoff bound, with and , we get that the probability of more than of the women getting a worse rank than is at most .

For , this probability is less than and we can neglect the event that more than women have rank greater than . We assume that the rank of any of these women is .

For , using the same bound, we get that the number of women with rank or better is at least . If there are more than such women, we count the extra ones as if they are in the next segment (i.e., with rank or better), thus only enlarging our estimation of . Thus, the number of women in each segment is .Therefore,

which completes the lemma. ∎

By now, we have a good knowledge of ’s rank and see that all women get a similar rank with high probability. We are almost ready for the main result. Denote by
– the number of (distinct) proposals made to .
– the total number of (not necessarily distinct) proposals made to all women.
.

Given a parameter , we consider three disjoint events:

and estimate each event probability and what the average rank is when conditioned on each one. The parameter is adjusted to the requirements of Lemma 4. The event that doesn’t occur will be handled separately later on.

Lemma 8
Proof

By Lemma 3, the probability that exactly proposals are made to during the process is at most . Using the union bound we get that the probability of getting less than proposals is bounded by .

The outcome of the algorithm is a full matching where all women get at least one proposal. Hence, the worst possible average rank is . Thus, . ∎

Lemma 9
Proof

Since , it follows that, in particular, gets at least proposals, including the redundant ones. The probability that out of the proposals made in total, at least are made to is bounded using the Chernoff bound. Set . Then

The outcome of the algorithm is a full matching where all the women get at least one proposal. Hence, the worst possible average rank is . Thus, . ∎

The next lemma implies the first part of Theorem 1.1 and it will be used later to prove the bound the average men’s rank. The bound for the expected women’s rank will be proven separately.

Lemma 10

Let be any function that goes to infinity as a function of . Then, there exist constants such that w.h.p. the expected rank for women is less than , and any arbitrary women get a match with one of their top options.

Proof

First, we restrict the analysis to the case where gets matched with one of her top men. By Lemma 2 this event happens with probability approaching 1.

From Lemmas 8 and 9, the events and are negligible and only should be considered.

By Lemma 3 and the union bound, the probability of getting fewer than proposals is at most .

By Lemma 4, with probability at least , each woman gets at least proposals.

By Lemma 5, with probability at least , each woman gets at least distinct proposals.

By Lemma 7, with probability at least , the average rank of the women is

Hence, we bound the average rank with for some constant and large enough . The events in not included in the calculations are of order of . Even when we assume worst possible match in these cases, it will only add a constant factor to the average rank. Thus, getting the requested.

In order to show that almost all the women get a good matching, we start with the same arguments as before. Each woman gets at least distinct proposals. Let be any function that goes to infinity as a function of . By Lemma 6, the probability that a woman with that number of proposals getting a rank worse than is no greater than and goes to . may be chosen such that , which completes the proof.

In [20] it is shown that with probability that goes to any stable matching with an average women’s rank of order of has an average men’s rank that is of order . Thus, the first part of Lemma 10 gives a bound on the average men’s rank.

Corollary 2

For any , w.h.p. the average men’s rank is of order .

In the last part of this section, we bound the expected rank of the women.

Lemma 11

For there exist such that .

Proof

Let be the random variable counting the number of distinct proposals made to . By Lemma 4, for any realization of , all the women get at least proposals with probability at least .

By Lemma 5, all of the women get at least distinct proposals with probability at least .

For , Lemma 3 holds.

By Lemma 7 we conclude that with probability , the expected women’s rank is

for some constant and large enough . ∎

At this point, we can deduce that in expected women’s rank is , using and verifying that all “bad” scenarios happen with small enough probability.

Proof

Let be the random variable for the average women’s rank. The expected rank is given by -

for any . Setting and using Lemma 2 we get that

We conjecture that the actual expected rank is of order and we examine this conjecture in the next parts.

2.3 The Set of Stable Husbands

An alternative point of view about the effect of a single strategic woman on the entire market is given via observation on the set of stable husbands for each woman. A man is called a stable husband of if there is a stable matching in which they are matched together.

By the optimality of ’s strategy, gets a proposal from her best stable husband . Due to the lattice structure of the set of stable matchings, it is known that she is his worst stable wife (e.g., [27]). Thus, we know that at some point in the process proposed to his second worst woman, , and was rejected. Since all the women are eventually matched, we know that ends up with a proposal from her best stable husband. In the same manner, this event initiates a series of events that eventually prove some set of women are all guaranteed to be matched to their best stable husband at the end of the process.

Assume that the size of this set is distributed like a size of a cycle in a random permutation. Then, with constant probability, at least fraction of the women are matched to their best stable husband. In addition, as the experimental results in part 2.5 show, we conjecture that the expected number of women getting their best stable matching is .

Note that received a proposal not only from her second best stable husband but also from all of her possible stable husbands. Let ’s second best stable husband be . If we could reason that she is his second worst stable wife, we would have got a new set of women guaranteed to be matched to their second best stable possible husband. By Theorem 2 in [27], this is true when ordering the stable spouses with repetitions (one for each possible stable matching). Unfortunately, this may not be the case when counting each spouse only once. It will be interesting to see if we could show that w.h.p. this is still true even for the distinct case.

It seems that it would be possible to show that at least fraction of size of the women are expected to be matched to their th stable husband. Furthermore, it is known [20] that the expected rank of a woman who is matched to her best stable husband is . Looking at the process described in the previous section, we observe that each temporarily best proposer to the strategic woman in the rejecting part of the process (i.e., after the men-optimal match is reached) is a stable husband and is randomly located in the proposed-to woman’s preference list. Hence, the rank of the woman when she is matched to her th stable husband is expected to be twice as good compared to when she is matched to the th stable husband. Hence, the expected rank of a woman who is matched to her th stable husband is . Notice that some of the women have few stable husbands (see [21]) and it is unclear how to handle this in a rigorous way. Combining these two observations rigorously hopefully will yield an improved bound of order for the expected rank of the women.

2.4 Unbalanced Market Perspective

An unbalanced market of men and women can be described as a balanced market of size in which one woman rejects all of the proposals made to her. In this case, the men’s utility from being matched with her is irrelevant and may be ordered arbitrarily. In this market, assuming uniform and independent preferences, the average rank of the women is of order even in the women-pessimal matching [2]. On the other hand, a truthful balanced market can be viewed as a market in which this one woman decides not to trim her preference list at all. In this case, the average women’s rank is of order in the woman-pessimal match.

Adding the results of this paper, we get that a strategic woman can affect this range of ranks in an almost continuous way. To see this, observe that as long as all the men are matched at the end of the process, all of the lemmas in the previous section hold if we choose a suitable set . An interesting corollary is that if ’s list is of length of order and all the other lists are long enough, we get that the women’s rank is around up to a logarithmic factor. By the hyperbola matching rule, this match captures some of the properties of fairness between the sexes.

Another observation is for the case of lists that are uniformly ordered of different lengths. This scenario can be described as a market with full preferences where each woman decides how she trims her list. In this case, the expected rank is asymptotically determined by the woman with the shortest list. By using different truncations, the women may induce any possible stable matching (see Theorem 4 in [17]). It should be noted that when there are many strategic agents there are more profitable actions that a coalition of agents may achieve.

2.5 Experimental Results

The theoretical bounds of the previous section are quite impressive asymptomatically, but for real-life market, they are not of much use. For a better understanding of the behavior in typical market sizes, we simulated different markets. We tested our settings for market sizes for between to in leaps of . For any market size, the mean of average ranks over iterations was calculated.

Figure 1 shows the mean of the average ranks for men and women, in the truthful and the strategic scenarios. Figure 2 zooms in on the mean of the average ranks for women in the strategic scenario.

Figure 1: Mean average rank of men and women in both scenarios
Figure 2: Mean average rank of women with strategic agent compared to and

For the same settings, we counted the number of women who got their best stable husband in the strategic scenario. It would appear that in expectation half of the women get their best stable husband when another woman act strategically. For comparison, we also count the number of women who got their worst stable husband and the women who got matched to either of them. The results are shown in Figure 3. The simulation supports the intuition described in Section 2.3.

Figure 3: Number of women who gets best/worst/(best or worst) stable match

3 Discussion and Future Work

3.1 Actual Expected Rank

We’ve shown that the expected women’s rank is . A logarithmic factor was added since the number of distinct proposals to an arbitrary woman was bounded by a fraction of order of the distinct proposals made to . In fact, in most cases, we can show that those two are the same up to a constant multiplicative factor. Showing this rigorously will prove that the expected rank is of order . It will also help to conclude that the expected rank is . However, additional work is needed in this direction. An alternative direction is to formalize the observations stated in Section 2.3.

3.2 Bayesian Information

In the settings of this work, is assumed to have full information on the market and thus knows how to choose an optimal strategy. This assumption is later relaxed to having access to an oracle that hints when to terminate a process. Most of the Lemmas do not directly use ’s knowledge of the other preferences. Although it is tempting to deduce that this implies exactly the same results when knows only the other lists’ distributions, it is not the case. For example, assume that has an extremely high utility for not staying single and hardly distinguishes between all possible mates. In that case, it is likely that ’s optimal strategy is just to report truthfully and thus guarantees that she will be matched. But now, has no effect on the rest of the market and the expected rank for all agents is unchanged. Note that although ’s rank isn’t monotone in her truncation and she needs to be careful not to trim her list too much, the average rank of all the other women is indeed monotone in ’s decision and they can only benefit from being too picky.

The optimal strategy of given she have Bayesian information is discussed in [24, 4]. When the uniform independent assumption holds, truncation strategy is still optimal for her when she maximizes her expected utility, and the exact point of truncation is determined by her utility function. Furthermore, [4] shows that reduction in the risk aversion causes the length of ’s list after truncation to be negligible compared to the number of agents. If we assume truncation in we get similar results to those of the full knowledge model: either the strategy was successful and the expected rank is or the strategic woman is unmatched and we are in a -men -women case. In this case, it should be reasonable to say that the other women get rank due to [2]. A more rigorous analysis is needed since the fact that ’s rank when matched with her best stable husband is worse than her truncation location might imply something about the preference lists of the other agents.

3.2.1 Other Distributions

The independent uniform distribution of the preferences is widely common assumption in theoretical research, even though in real-life matching markets it is not always justified to assume such a distribution. The following distribution (presented in

[11]) preserves some of the properties that were used while adding correlation between the preferences. Men still choose the preferences online as in the process before but instead of choosing uniformly between the women, they use an arbitrary distribution over the set of women, thus making some of the women more popular than others while keeping variety in the men’s preferences. It seems the result of our work may be extended to this distribution with minor adjustments assuming doesn’t make no women extremely popular or unpopular. The main constraint is that no woman is much more popular than the others; i.e. constraints on or will hopefully be sufficient for us to get similar results to those received in the uniform case.

3.3 Many Strategic Women

As we have seen, the fact that one agent acts strategically affects in a nontrivial way the quality of the matching from the other women’s perspective. This raises some interesting questions about a scenario in which the number of women act strategically.

Full Information How many women need to act strategically in order to induce the women-optimal matching? The intuition in Section 2.3 suggests that this number should be close to the number of cycles in a random permutation, i.e., order of .

Bayesian Information Almost all of the women benefit from not reporting their preferences truthfully, given that the other women are truthful. On the other hand, as soon as at least one woman truncates her list, many of the other women will have already promised a better match and will not gain much from not being truthful (their probability of staying unmatched is insensitive to the truncation of the first woman). It is interesting to examine the dynamics of the strategies in such environments (with some suitable distributions of the preferences and cardinality over the matchings). The existence of equilibrium in truncation strategies under incomplete information is proved in [4]. Some considerable directions for further research:

  • In many real-life matchings, the players report their preferences to a centralized mechanism that runs the algorithm for them. Assume that the women report to the mechanism in an order that is known in advance. How would it affect the strategies of the women?

  • Assume that the men can truncate their lists as well. What would be the set of equilibria and how would a centralized mechanism affects it?

Acknowledgments

I would like to thank my supervisor Noam Nisan for his guidance. I would also like to thank Assaf Romm and Yannai Gonczarowski for their interest in this work and their thoughtful comments.

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