Download Theory of Mechanism Design - Assignment 3 1. We consider the

Theory of Mechanism Design - Assignment 3
1. We consider the house allocation model with existing tenants. Consider the following
form of manipulation by a coalition of agents in the TTC mechanism. A coalition of
agents S exchange their houses before the start of the mechanism (i.e., they end up
with an endowment which is different from their actual endowment). Now, the TTC
mechanism is executed. Do you think each agent in S will now get a house which is
either the same house he gets if he had not done the manipulation or a house which is
higher ranked than the house he gets if he does not do manipulation?
Solution: Such a manipulation is possible. We give an example with four agents with
four houses. Let N = {1, 2, 3, 4} and the set of houses be {a1 , a2 , a3 , a4 }. The initial
endowments of houses are given by a∗ : a∗ (i) = ai for all i ∈ N. The preferences
of agents are shown in Table 1 - some of the preferences of agents are not shown
completely, implying that it can be anything in the parts not shown.
≻1
a2
a3
≻2
a1
≻3
a4
a2
a3
≻4
a4
Table 1: An example for housing model
If we run the TTC mechanism on this problem, the outcome will be a: a(1) = a2 , a(2) =
a1 , a(3) = a3 , a(4) = a4 .
Now, suppose agents 2 and 3 swap their endowments. So, the initial endowments of
agents look as a′ : a′ (1) = a1 , a′ (2) = a3 , a′ (3) = a2 , a′ (4) = a4 . If we run the TTC
mechanism on this problem, the outcome will be â: â(1) = a3 , â(2) = a1 , â(3) =
a2 , â(4) = a4 . Note that â(2) = a(2) and â(3) ≻3 a(3). Hence, agents 2 and 3
successfully manipulated their initial endowments.
2. Consider the marriage market model (with n men and n women) where every man m
ranks women as:
w1 ≻m w2 ≻m . . . ≻m wn
and every woman w ranks men as:
m1 ≻w m2 ≻w . . . ≻w mn .
What are the outcomes of the men-proposing and the women-proposing versions of the
deferred acceptance algorithm at this profile?
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Solution. The unique efficient matching in this profile is man mi is matched to woman
wi for all i ∈ {1, . . . , n}. Hence, both the versions of the deferred acceptance algorithm
must terminate at this matching.
3. Consider the house allocation model with three agents N = {1, 2, 3} and three objects
M = {a, b, c}. Let f be a social choice function defined as follows. At any preference
profile ≻≡ (≻1 , . . . , ≻n ), if ≻2 (1) = a, then agent 1 gets the best element in {b, c}
according to his preference ordering ≻1 , agent 2 gets a, and agent 3 gets the remaining
object (i.e., a serial dictatorship with the highest priority to agent 2, followed by agent
1, and finally to agent 3). In all other cases, agent 1 gets the best object in M, agent 2
gets the best remaining object according to ≻2 , and agent 3 gets the remaining object
(i.e., a serial dictatorship with the highest priority to agent 1, followed by agent 2, and
finally to agent 3).
Is f strategy-proof? Is f non-bossy, i.e., can an agent change the outcome at a profile
without changing the object assigned to him?
Answer: f is strategy-proof. Agents 1 and 3 cannot change the priority. So, they
have no incentive to manipulate. Agent 2 can change the priority. But he will not
manipulate if he gets the top priority. When agent 2 gets the second priority, he can
change the priority by saying that his top is a, and in this case he gets a. But a is
not his top according to his true preference. So he gets an object which is at least
his second preferred object. But that he could have got even if he did not change the
priority. So, he does not gain by manipulation.
f is also non-bossy. Note that if the serial dictatorship with a given priority is nonbossy. So, if an agent does not change his own allocation in f , it does not change the
priority in f . So, by the same reasoning, it is non-bossy - other agents will continue to
choose the best from same set of available objects to them.
4. Consider a two-sided matching model with men and women. Let ≻ be a profile of
preference orderings as shown in Table 2.
≻m1
w1
w3
w2
≻m2
w2
w1
w3
≻m3
w2
w1
w3
≻w1
m2
m1
m3
≻w2
m1
m2
m3
≻w3
m1
m2
m3
Table 2: Preference orderings of men and women
Suppose µ is the outcome of the women-proposing deferred acceptance algorithm for
the preference profile ≻. Let µ′ be the outcome of the fixed-priority TTC mechanism
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where the priorities of men are fixed according to their preference orderings in ≻ in
Table 2, and then each woman points to the woman with her favorite man in every
stage of the TTC.
(a) Verify that µ 6= µ′ .
(b) Verify that µ′ is not stable by identifying a blocking pair.
(c) Verify that µ′ women-dominates µ.
5. Suppose the set of colleges in Delhi is C, each with capacity 1. Let S be the set of
students applying to colleges in Delhi. Assume that |S| = |C|. There was a common
entrance test, which generated a strict ordering ≻ over the set of students S. The
preference of each college c ∈ C over the students S is given by this common preference
≻.
(a) If you are asked to design a mechanism (social choice function) that is stable and
strategy-proof for colleges, then which one will you recommend?
Answer. The college-proposing deferred acceptance algorithm will be strategyproof for colleges and stable.
(b) Consider a preference profile P ≡ (P1 , . . . , P|S| ), where Pi is a strict preference
ordering of student i ∈ S over the set of colleges. At this preference profile P of
students, what will be the outcome of the college-proposing deferred acceptance
algorithm?
Answer. The top ranked student according to ≻ gets his best college, the second
ranked student according to ≻ gets his best college among remaining colleges, and
so on.
(c) Can this outcome be also achieved using a version of the serial dictatorship social
choice function?
Answer. Yes, this is equivalent to running the serial dictatorship SCF where the
priority over students is given by ≻ and students choose their best college among
remaining colleges.
(d) Will the outcome change if we run the student-proposing deferred acceptance
algorithm?
Answer. No. The outcome will not change. To see this, all the students propose
to their top ranked colleges. The top student according to ≻ gets accepted at his
top college, say c1 . The second ranked student according to ≻ either applies at c1
and gets rejected, or applies at his top ranked college 6= c1 and gets accepted. If
he is rejected at c1 , then he applies to his next ranked college, and gets accepted.
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We can now use induction. Suppose, all students upto rank k in ≻ get exactly the
outcome as in college-proposing DA, then the kth ranked student either proposes
to all these colleges and gets rejected and then gets the best of the remaining
colleges. Hence, the outcome remains the same.
(e) Suppose you have to choose a mechanism that is stable. If you are a student and
you are asked to choose between student-proposing and college-proposing deferred
acceptance algorithm, which one will you choose in this case?
Answer. Since the student-proposing deferred acceptance algorithm is studentoptimal, you should choose it. But in this case, both the student-proposing and
the college-proposing DA give the same outcome. So, either one will work.
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