In this Appendix, we consider the situation in which being selected is a “bad” thing so the worst alternative for every agent is for himself to be chosen while the enemy being chosen is his best alternative. In this case, the class of admissible preference function for an agent is defined as follows.
For the mechanism that we are going to propose below to work, the following property must be fulfilled:
Proof
Let
\(N=\{1,...,n\}\) be the a numbering of the agents as defined in Mechanism
3. For each pair
\(i,j\in N\), let
\(k_{i,j}\in N\backslash \{i,j\}\) be as defined in Mechanism
3. Given any state,
\((\omega ^{d},R)\in S(e)\), consider a profile of strategies that is a backward induction equilibrium of Mechanism
3 at
\((\omega ^{d},R)\). For each
\(k\in N\) , let
\(x_{k}\) denote the agent who would be ultimately chosen by Mechanism
3 in case Stage
k is reached, given the previous profile of equilibrium strategies. Similarly, for each
\(k\in N\backslash \{n\}\) and each
\(m_{k}\in N\backslash \{k,e_{k}\}\), let
\(x_{k.m_{k}}\) denote the agent who would be ultimately chosen by Mechanism
3 in case Stage
\(k.m_{k}\) is reached, given the previous profile of equilibrium strategies.
Claim 1. Let \(k\in N\backslash \{n\}\) and \(m_{k}\in N\backslash \{k,e_{k}\}\).
(1.1) If \(\omega ^{d}=k\), then \(x_{k.m_{k}}=k.\)
(1.2) If \(\omega ^{d}=m_{k}\), then \(x_{k.m_{k}}=m_{k}\) .
(1.3) If \(\omega ^{d}\notin \{k,m_{k}\}\), then \( x_{k.m_{k}}\in \{k,m_{k}\}.\)
In this case, Stage \(k.m_{k}\) is reached, agent \(k_{m_{k},k}\) chooses between \(m_{k}\) and k, the choice of agent \(k_{m_{k},k}\) will be selected as the winner, and the mechanism will stop. By definition, agent \( k_{m_{k},k} \) is such that \(m_{k}\) \(P_{k_{m_{k},k}}(m_{k})\) k and k \( P_{k_{m_{k},k}}(k)\) \(m_{k}\). Therefore, if \(\omega ^{d}=k\), agent \( k_{m_{k},k}\) will announce k, and if \(\omega ^{d}=m_{k}\), he will announce \(m_{k}\). If \(\omega ^{d}\notin \{k,m_{k}\}\), it can happen either \(m_{k}\) \( R_{k_{m_{k},k}}(\omega ^{d})\) k or k \(P_{k_{m_{k},k}}(\omega ^{d})\) \( m_{k}\), so agent \(k_{m_{k},k}\) can announce either \(m_{k}\) or k.
Claim 2. Let \(k\in N\backslash \{n\}\).
(2.1) If \(\omega ^{d}=k\), we have \(x_{k}=x_{k+1}.\)
(2.2) If \(\omega ^{d}=e_{k}\), then:
(2.2.1) if \(x_{k+1}\ne e_{k}\), we have \(x_{k}\ne e_{k}\), and
(2.2.2) if \(x_{k+1}=e_{k}\), we have \(x_{k}=e_{k}\) .
(2.3) If \(\omega ^{d}\notin \{k,e_{k}\}\), then:
(2.3.1) if \(x_{k+1}\ne e_{k}\), we have \( x_{k}=\omega ^{d}\), and
(2.3.2) if \(x_{k+1}=e_{k}\), we have \(x_{k}=e_{k}\) .
Suppose that
\(\omega ^{d}=k\). Then, by definition of Mechanism
3 and by Claim 1, at Stage
k, agent
k has to decide between selecting
k at that stage or selecting
\(x_{k+1}\) at some later stage: if
\(m_{k}=k\), then
k is chosen at Stage
k; if
\(m_{k}\notin \{k,e_{k}\}\), then
k is chosen at Stake
\(k.m_{k}\); if
\(m_{k}=e_{k}\), then
\(x_{k+1}\) is ultimately chosen at some later stage. Because
k is the worst alternative for agent
k, he will announce
\(m_{k}=e_{k}\), and agent
\(x_{k+1}\) will be ultimately chosen (if
\(x_{k+1}=k\), he is indifferent between all announcements).
Suppose that \(\omega ^{d}=e_{k}\) and \(x_{k+1}\ne e_{k}\). In this case, agent k cannot do anything to ensure that \(e_{k}\) gets selected neither at Stage k nor at any subsequent stage: if his announcement is \(m_{k}=k\), then k is chosen at Stage k; if his announcement is \(m_{k}\notin \{k,e_{k}\}\), then the mechanism advances to Stage \(k.m_{k}\), in which either \(m_{k}\) or k will be chosen (depending on the preferences of agent \( k_{m_{k},k}\)); if his announcement is \(m_{k}=e_{k}\), then the mechanism advances to Stage \(k+1\), and agent \(x_{k+1}\ne e_{k}\) is ultimately selected. Among the previous agents that k can make to be chosen once Stage k is reached (all different from \(e_{k}\)), \(x_{k}\) will be the one he prefers the most.
Suppose that \(\omega ^{d}=e_{k}\) and \(x_{k+1}=e_{k}\). Then, because \(e_{k}\) is the best alternative for agent k, he will announce \(m_{k}=e_{k}\) and agent \(x_{k+1}=e_{k}\) will be chosen at some subsequent stage.
Suppose that \(\omega ^{d}\notin \{k,e_{k}\}\) and \(x_{k+1}\ne e_{k}\). In this case, using the same argument as when \(\omega ^{d}=e_{k}\) and \( x_{k+1}\ne e_{k}\), we conclude that agent k cannot do anything to ensure that \(e_{k}\) is selected neither at Stage k nor at any subsequent stage. However, by Claim 1, now agent k can ensure that \(\omega ^{d}\) is chosen at Stage k by announcing \(m_{k}=\omega ^{d}\). Since \(\omega ^{d}\notin \{k,e_{k}\}\), then \(\omega ^{d}\) is the second most preferred alternative for agent k. Therefore, agent k will announce \(m_{k}=\omega ^{d}\), and \( \omega ^{d}\) will be selected at Stage k.
Suppose that \(\omega ^{d}\notin \{k,e_{k}\}\) and \(x_{k+1}=e_{k}\). Because \( e_{k}\) is the best alternative for agent k, he will announce \(m_{k}=e_{k}\) , and agent \(x_{k+1}=e_{k}\) will be ultimately chosen at some subsequent stage.
Claim 3. Suppose that Stage n is reached. Then, \( x_{n}=e_{n}.\)
It follows from the fact that \(e_{n}\) is the most preferred alternative for agent n.
Claim 4. Suppose that Stage \(n-1\) is reached.
(4.1) If \(\omega ^{d}=n-1\), then \(x_{n-1}=e_{n}.\)
(4.2) If \(\omega ^{d}=e_{n-1}\), then \(x_{n-1}\ne e_{n-1}\) .
(4.3) If \(\omega ^{d}\notin \{n-1,e_{n-1}\}\), then \( x_{n-1}=\omega ^{d}.\)
It follows from Claims 2 and 3 and the fact that Mechanism
3 is such that
\(e_{n-1}\ne e_{n}\).
Claim 5. Suppose that Stage \(n-2\) is reached.
(5.1) If \(\omega ^{d}=e_{n-1}\) and \(x_{n-1}=e_{n-2}\) , then \(x_{n-2}=e_{n-2}.\)
(5.2) Otherwise, \(x_{n-2}=\omega ^{d}.\)
Suppose that
\(\omega ^{d}=e_{n-2}\). Since Mechanism
3 is such that
\(n-1\ne e_{n-2}\) and
\(e_{n-1}\ne e_{n-2}\), then
\(\omega ^{d}\notin \{n-1,e_{n-1}\}\) . Therefore, by point 4.3 of Claim 4,
\(x_{n-1}=\omega ^{d}\). Then, because
\( \omega ^{d}=e_{n-2}\), by point 2.2.2 of Claim 2, we have
\( x_{n-2}=e_{n-2}=\omega ^{d}\).
Suppose that
\(\omega ^{d}=n-2\). Since Mechanism
3 is such that
\(n-2\ne e_{n-1}\) and
\(n-2\ne n-1\), then
\(\omega ^{d}\notin \{n-1,e_{n-1}\}\). Therefore, by point 4.3 of Claim 4,
\(x_{n-1}=\omega ^{d}\). Then, because
\( \omega ^{d}=n-2\), by point 2.1 of Claim 2, we have
\(x_{n-2}=x_{n-1}=\omega ^{d}\).
Suppose that
\(\omega ^{d}\notin \{n-2,e_{n-2}\}\). Then we have three possibilities:
(i)
If \(\omega ^{d}=e_{n-1}\), by point 4.2 of Claim 4, \(x_{n-1}\ne e_{n-1}\) . If, in addition, \(x_{n-1}\ne e_{n-2}\), by point 2.3.1 of Claim 2, we have \(x_{n-2}=\omega ^{d}\). If, on the contrary, \(x_{n-1}=e_{n-2}\), by point 2.3.2 of Claim 2, we have \(x_{n-2}=e_{n-2}\).
(ii)
If
\(\omega ^{d}=n-1\), by point 4.1 of Claim 4,
\(x_{n-1}=e_{n}\). Then, because
\(\omega ^{d}\notin \{n-2,e_{n-2}\}\) and
\(x_{n-1}=e_{n}\), and since Mechanism
3 is such that
\(e_{n}\ne e_{n-2}\), by point 2.3.1 of Claim 2, we have
\(x_{n-2}=\omega ^{d}\).
(iii)
If \(\omega ^{d}\notin \{n-1,e_{n-1}\}\), by point 4.3 of Claim 4, \( x_{n-1}=\omega ^{d}\). Then, because \(\omega ^{d}\notin \{n-2,e_{n-2}\}\) and \( x_{n-1}=\omega ^{d}\ne e_{n-2}\), by point 2.3.1 of Claim 2, we have \( x_{n-2}=\omega ^{d}\).
Claim 6. Suppose that Stage \(n-3\) is reached. Then, \( x_{n-3}=\omega ^{d}.\)
Suppose that
\(\omega ^{d}=e_{n-3}\). By definition of Mechanism
3, we have
\( e_{n-3}\ne e_{n-1}\), and then
\(\omega ^{d}\ne e_{n-1}\). Therefore, by point 5.2 of Claim 5,
\(x_{n-2}=\omega ^{d}\). Hence, by point 2.2 2 of Claim 2,
\(x_{n-3}=e_{n-3}=\omega ^{d}\).
Suppose that
\(\omega ^{d}=n-3\). By definition of Mechanism
3, we have
\( n-3\ne e_{n-1}\), and then
\(\omega ^{d}\ne e_{n-1}\). Therefore, by point 5.2 of Claim 5,
\(x_{n-2}=\omega ^{d}\). Hence, by point 2.1 of Claim 2,
\( x_{n-3}=x_{n-2}=\omega ^{d}\).
Suppose that \(\omega ^{d}\notin \{n-3,e_{n-3}\}\). Then we have two possibilities:
(i) Suppose that
\(\omega ^{d}=e_{n-1}\). If, in addition,
\(x_{n-1}=e_{n-2}\), by point 5.1 of Claim 5, we have
\(x_{n-2}=e_{n-2}\). By definition of Mechanism
3, we have
\(e_{n-2}\ne e_{n-3}\). Then, by point 2.3.1 of Claim 2,
\(x_{n-3}=\omega ^{d}\). If, on the contrary,
\(x_{n-1}\ne e_{n-2}\), by point 5.2 of Claim 5, we have
\(x_{n-2}=\omega ^{d}\). By definition of Mechanism
3, we have
\(e_{n-1}\ne e_{n-3}\). Then,
\(x_{n-2}=\omega ^{d}=e_{n-1}\ne e_{n-3}\), and by point 2.3.1 of Claim 2,
\(x_{n-3}=\omega ^{d}\).
(ii) Suppose that \(\omega ^{d}\ne e_{n-1}\). Then, by point 5.2 of Claim 5, \( x_{n-2}=\omega ^{d}\). Because \(\omega ^{d}\ne e_{n-3}\), then \(x_{n-2}\ne e_{n-3}\). Hence, by point 2.3.1 of Claim 2, \(x_{n-3}=\omega ^{d}\).
Claim 7. Suppose that Stage \(n-4\) is reached. Then, \( x_{n-4}=\omega ^{d}.\)
Suppose that \(\omega ^{d}=n-4\). Then, by point 2.1 of Claim 2 and by Claim 6, \(x_{n-4}=x_{n-3}=\omega ^{d}\).
Suppose that \(\omega ^{d}=e_{n-4}\). By Claim 6, \(x_{n-3}=\omega ^{d}\). Then, since \(\omega ^{d}=e_{n-4}\) and \(x_{n-3}=e_{n-4}\), by point 2.2.2 of Claim 2, \(x_{n-4}=e_{n-4}=\omega ^{d}\).
Suppose that \(\omega ^{d}\notin \{n-4,e_{n-4}\}\). By Claim 6, \( x_{n-3}=\omega ^{d}\). Then, since \(\omega ^{d}\ne e_{n-4}\), by point 2.3.1 of Claim 2, \(x_{n-4}=\omega ^{d}\).
Claim 8. Suppose that \(n\ge 6.\) Suppose that Stage \(n-t\) is reached for some \(t\ge 5.\) Then, \(x_{n-t}=\omega ^{d} .\)
The demonstration of this claim follows from repeatedly applying the same argument used in the proof of Claim 7.
From Claims 6, 7, and 8, we have that, given any state
\((\omega ^{d},R)\in S(e)\), every profile of backward induction equilibrium strategies of Mechanism
1 at
\((\omega ^{d},R)\) is such that
\(\omega ^{d}\) is ultimately chosen as the winner.
\(\square \)