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Consider the extensive game \( \Gamma \) given in Figure 3.2.1. Let \( {x}_{1} \) and \( {x}_{2} \) be the unique nodes of \( {w}_{2}^{1} \) and \( {w}_{2}^{2} \), respectively. \( {L}_{2} \) is a Nash equilibrium of \( {\Gamma }_{{x}_{1}} \) and \( {l}_{2} \) is a Nash equilibrium of \( {\Gamma }_{{x}_{2}} \) . The ga... | By construction, \( \left( {{R}_{1},{L}_{2}{l}_{2}}\right) \) is not only a Nash equilibrium, but also a subgame perfect equilibrium. | Yes |
Theorem 3.4.2. Every extensive game with perfect recall has, at least, one sub-game perfect equilibrium. | Proof. We make the proof by induction on \( L\left( \Gamma \right) \) . If \( L\left( \Gamma \right) = 1 \), the result is straightforward. Assume that the result is true up to \( t - 1 \) and let \( \Gamma \) be a game with length \( t \) . If \( \Gamma \) has no proper subgame, then all its Nash equilibria are subgam... | Yes |
Theorem 3.4.3. Every extensive game with perfect information \( \Gamma \) has, at least, one pure strategy profile that is a subgame perfect equilibrium. | Proof. Again, we make the proof by induction on \( L\left( \Gamma \right) \) . If \( L\left( \Gamma \right) = 1 \), the result is straightforward. Assume that the result is true up to \( t - 1 \) and let \( \Gamma \) be a game with length \( t \) . Decompose \( \Gamma \) in a set of nodes \( {x}_{1},\ldots ,{x}_{m} \) ... | Yes |
Example 3.4.3. (The centipede game (Rosenthal 1981)). Two players are engaged in the following game. There are two pots with money: a small pot and a big pot. At the start of the game, the small pot contains one coin and the big pot contains four coins. When given the turn to play, a player has two options: i) stop the... | can be easily solved by backward induction. If node \( {x}_{2}^{3} \) is reached, then player 2 will surely stop, which gives him payoff 128, instead of the 64 he would get by passing. Now, if node \( {x}_{1}^{3} \) is reached, player 1, anticipating that player 2 will stop at \( {x}_{2}^{3} \), will surely stop as wel... | Yes |
Proposition 3.4.4 (One shot deviation principle). Let \( \Gamma \) be an extensive game with perfect information and let \( b \in B \) . Then, the two following statements are equivalent,\n\ni) \( b \) is a subgame perfect equilibrium of \( \Gamma \) ,\n\nii) no one shot deviation from \( b \) is profitable. | Proof. \( {}^{15} \) Clearly, i) \( \Rightarrow \) ii). Next we prove that ii) \( \Rightarrow \) i) also holds. Suppose that \( b \) satisfies ii) but it is not a subgame perfect equilibrium. Let \( {\Gamma }_{x} \) be a subgame where a player, namely \( i \), can profitably deviate. Among all such deviations, let \( {... | No |
Proposition 3.5.1. Let \( \Gamma \) be an extensive game. If \( \left( {b,\mu }\right) \) is a sequential equilibrium of \( \Gamma \), then \( b \) is a subgame perfect equilibrium. Moreover, if \( \Gamma \) is a game with perfect information, then every subgame perfect equilibrium is also a sequential equilibrium. | Proof. Let \( \Gamma \) be an extensive game and let \( \left( {b,\mu }\right) \) be a sequential equilibrium of \( \Gamma \) . Let \( x \in X \) be such that \( {\Gamma }_{x} \) is a proper subgame and let \( w \) be the information set containing \( x \) . Recall that, for \( {\Gamma }_{x} \) to be a subgame, \( w \)... | Yes |
Theorem 3.5.2. Let \( \Gamma \) be an extensive game with perfect recall. Then, \( \Gamma \) has, at least, one sequential equilibrium. | Proof. We do not provide a direct proof of this result. It is an immediate consequence of Proposition 3.6.1 and Corollary 3.6.4 in Section 3.6.1; there we define perfect equilibrium for extensive games and show that every perfect equilibrium is a sequential equilibrium and that every extensive game with perfect recall ... | No |
Consider the extensive game \( \Gamma \) of Figure 3.5.3. We claim that the pure strategy profile \( a = \left( {{R}_{1},{R}_{2}}\right) \) is a sequential equilibrium of \( \Gamma \) that is not sensible. | We now construct the sequential equilibrium assessment based on \( \left( {{R}_{1},{R}_{2}}\right) \) . For player 2 to be a best reply to play \( {R}_{2} \) he must believe that it is more likely to be at \( {\widehat{x}}_{2} \) than at \( {x}_{2} \) . Hence, a natural candidate for\n\n![9931c6b0-4777-405a-ae05-2404ea... | Yes |
Consider the extensive game \( {\Gamma }_{1} \) in Figure 3.5.4 (a). The strategy \( U \) can be supported as part of a sequential equilibrium. | Namely, let \( b \) be such that \( {b}_{1}\left( U\right) = 1 \) and \( {b}_{2}\left( {L}_{2}\right) = 1 \) ; and define the corresponding beliefs such that \( \mu \left( {x}_{2}\right) = 1 \) . So defined, the assessment \( \left( {b,\mu }\right) \) is sequentially rational. Moreover, it is also consistent. Just take... | Yes |
Proposition 3.6.1. Let \( \Gamma \) be an extensive game. Then, every perfect equilibrium of \( \Gamma \) is (part of) a sequential equilibrium. | Proof. Let \( \Gamma \) be an extensive game. Let \( b \) be a perfect equilibrium of \( \Gamma \) and let \( \left\{ {\eta }^{k}\right\} \subset T\left( \Gamma \right) \) and \( \left\{ {b}^{k}\right\} \subset {B}^{{\eta }^{k}} \) be two sequences taken according to Definition 3.6.4. For each \( k \in \mathbb{N},{b}^{... | Yes |
Consider the extensive game \( \Gamma \) depicted in Figure 3.6.2 (a). This game has two subgame perfect equilibria in pure strategies: \( \left( {{L}_{1}{l}_{1},{L}_{2}}\right) \) and \( \left( {{R}_{1}{l}_{1},{l}_{2}}\right) \) . Moreover, they are also perfect equilibria. Figure 3.6.2 (b) shows \( {G}_{\Gamma } \), ... | Note that the strategy \( {R}_{1}{l}_{1} \) is dominated by \( {L}_{1}{l}_{1} \) . Hence, this example also shows that a perfect equilibrium in extensive games can involve dominated strategies. This might be surprising since one of the advantages of the perfect equilibrium for strategic games was precisely to avoid thi... | Yes |
Lemma 3.6.2. Let \( \Gamma \) be an extensive game with perfect recall and let \( \eta \in T\left( \Gamma \right) \) . Then, each Nash equilibrium of \( \left( {\Gamma ,\eta }\right) \) induces a Nash equilibrium of \( \left( {{G}_{A\Gamma },\eta }\right) \) and vice versa. | Proof. Each strategy profile \( \left( {\Gamma ,\eta }\right) \) has the property that all the information sets are reached with positive probability. The result in Remark 3.3.1 is also true when we restrict attention to strategies in \( {B}^{\eta } \) instead of \( B \) . Therefore, a Nash equilibrium of \( \left( {\G... | Yes |
Proposition 3.6.3. Let \( \Gamma \) be an extensive game with perfect recall. Then, each perfect equilibrium of \( \Gamma \) induces a perfect equilibrium of \( {G}_{A\Gamma } \) and vice versa. | Proof. Let \( b \in B \) be a perfect equilibrium of \( \Gamma \) . Let \( \{ {\eta }^{k}\} \subset T\left( \Gamma \right) \) and let \( \{ {b}^{k}\} \) be such that each \( {b}^{k} \in {B}^{{\eta }^{k}} \) is an equilibrium of the perturbed game \( \left( {\Gamma ,{\eta }^{k}}\right) \) . By Lemma 3.6.2, the sequence ... | No |
Corollary 3.6.4. Let \( \Gamma \) be an extensive game with perfect recall. Then, \( \Gamma \) has, at least, one perfect equilibrium. | Proof. It follows from the combination of the existence result of perfect equilibrium for strategic games (Theorem 2.9.2) and the proposition above. | No |
We claim that the undesirable perfect equilibrium \( \left( {{R}_{1},{R}_{2}}\right) \) is not proper. | We argue informally why this is so (Exercise 3.7 asks to formalize our arguments). Suppose that both players think that \( \left( {{R}_{1},{R}_{2}}\right) \) is the equilibrium to be played, but there is some probability that player 2 makes the mistake \( {L}_{2} \) . Then, between the two errors of player \( 1,{L}_{1}... | No |
Consider the extensive game \( \Gamma \) of Figure 3.6.5 (a). Note that the strategy profile \( \left( {{R}_{1}{r}_{1},{R}_{2}}\right) \) is a perfect equilibrium of \( E\left( {G}_{A\Gamma }\right) \) (we only need that the agent playing at \( {w}_{1}^{1} \) makes the mistake \( {L}_{1} \) with a smaller probability t... | Player 1 has to decide whether to force the outcome \( \left( {2,2}\right) \) or pass the move to player 2 . In the latter case, player 2 does not know in which of his two nodes he is at. Yet, provided that player 1 is rational, only node \( {x}_{2} \) can be reached. Hence, player 2 should play \( {L}_{2} \) . Player ... | Yes |
Consider the one-player game in Figure 3.6.6, taken from van Damme (1984). Both \( {Ll} \) and \( {Lr} \) are proper equilibria of \( E\left( {G}_{\Gamma }\right) \). Yet, since the choice \( r \) is strictly dominated at \( {w}_{1}^{2},{Lr} \) seems quite unreasonable. Indeed, the behavior strategy induced by \( {Lr} ... | We use the one-player game in Example 3.6.6 to illustrate the idea of the previous definition. We now show that if a behavior strategy profile of \( \Gamma \) chooses \( L \) at \( {w}_{1}^{1} \) and assigns positive probability at the choice \( r \) at \( {w}_{1}^{2} \), then it is not a limit behavior strategy profil... | Yes |
A Stackelberg duopoly only differs from a Cournot duopoly by the fact that the two producers do not choose simultaneously the number of units they produce and bring to the market. On the contrary, one of the producers (the leader) makes his choice first. Then, the second producer (the follower), after observing the cho... | A subgame perfect equilibrium of this game is a pair \( \left( {{a}_{1}^{ * },{g}^{ * }}\right) \), with \( {a}_{1}^{ * } \in {A}_{1} \) and \( {g}^{ * } \in {A}_{2}^{{A}_{1}} \), satisfying that:\n\n\[ \text{- for each}{a}_{1} \in {A}_{1},{g}^{ * }\left( {a}_{1}\right) \in {\mathrm{{BR}}}_{2}\left( {a}_{1}\right) \tex... | Yes |
Proposition 3.7.1. Let \( G\left( {\delta, T}\right) \) be a repeated game. Assume that the stage game is the mixed extension of a finite game. Then, the strategy profile \( \sigma \) is a subgame perfect equilibrium if and only if it is (part of) a sequential equilibrium. | Proof. \( {}^{25} \) By Proposition 3.5.1, every sequential equilibrium induces a sub-game perfect equilibrium. Now, we prove that the converse is true in this context. Let \( \sigma \) be a subgame perfect equilibrium of \( G\left( {\delta, T}\right) \) . There is some system of beliefs \( \mu \) such that \( \left( {... | Yes |
Example 3.7.1. (The repeated prisoner's dilemma). We already know that the only Nash equilibrium of the prisoner's dilemma game (Example 2.1.1) is the action profile in which both players defect. Now, consider the repeated prisoner’s dilemma, \( G\left( {\delta, T}\right) \) . In the infinite-horizon case, i.e., \( T =... | Yet, if \( T \in \mathbb{N} \) there is no way in which an outcome different from \( \left( {-{10}, - {10}}\right) \) can be achieved in equilibrium. The idea follows a simple backward induction reasoning. For a strategy profile to be a Nash equilibrium of a finitely repeated game, the actions chosen in the last period... | Yes |
Example 3.7.2. (The chain store paradox (Selten 1978, Kreps and Wilson 1982a)) Consider the following variation of the chain store game (Example 3.3.5). Suppose that the chain store game is to be played a finite number of times (periods). Now, the monopolist faces one different entrant in each period and each entrant o... | It is natural to think that, in this repeated environment, the monopolist can decide to fight early entrants in order to build a reputation and deter later entries. In doing so, the short run loss experienced by fighting early entrants might be compensated by the lack of competitors in the long run. Yet, although the s... | Yes |
Theorem 3.7.2 (Infinite horizon Nash folk theorem). Let \( G\left( {\delta ,\infty }\right) \) be an infinitely repeated game. Then, for each \( v \in \bar{F} \), there is \( {\delta }^{0} \in \left( {0,1}\right) \) such that, for each \( \delta \in \left\lbrack {{\delta }^{0},1}\right) \), there is a Nash equilibrium ... | Proof. Let \( a \) be a (possibly correlated) action profile such that \( u\left( a\right) = v \) . Let \( \sigma \) be a strategy profile defined, for each \( i \in N \), as follows:\n\nA) Main path: Play \( {a}_{i} \) . If one and only one player \( j \) does not play \( {a}_{j} \) at some period, go to \( B \) ).\n\... | Yes |
Theorem 3.7.3 (Finite horizon Nash folk theorem; adapted from Benoît and Krishna \( {\left( {1987}\right) }^{27} \) ). Let \( G\left( {\delta, T}\right) \) be a finitely repeated game. Suppose that, for each player \( i \in N \), there is a Nash equilibrium \( {e}^{i} \) of \( G \) such that \( {u}_{i}\left( {e}^{i}\ri... | Proof. Let \( a \) be a (possibly correlated) action profile such that \( u\left( a\right) = v \) . We claim that there are \( {\delta }^{0} \in \left( {0,1}\right) \) and natural numbers \( {T}^{0},{T}^{1},\ldots ,{T}^{n} \) such that there is a Nash equilibrium of \( G\left( {{\delta }^{0},{T}^{0}}\right) \) with pat... | Yes |
Theorem 3.7.4 (Infinite horizon perfect folk theorem; adapted from Fuden-berg and Maskin \( {\left( {1986}\right) }^{29} \) ). Let \( G\left( {\delta ,\infty }\right) \) be an infinitely repeated game. If \( F \) is full dimensional, then, for each \( v \in \bar{F} \), there is \( {\delta }^{0} \in \left( {0,1}\right) ... | Proof. Let \( a \) be a (possibly correlated) action profile such that \( u\left( a\right) = v \) . The construction we present below relies heavily on the full dimensionality assumption. Let \( \widetilde{v} \) be a payoff in the interior of \( \bar{F} \) such that \( \widetilde{v} < v \) . For each \( i \in N \), let... | Yes |
Theorem 3.7.5 (Finite horizon perfect folk theorem; adapted from Benoît and Krishna \( {\left( {1985}\right) }^{30} \) ). Let \( G\left( {\delta, T}\right) \) be a finitely repeated game. Suppose that \( F \) is full dimensional and that, for each player \( i \in N \), there are two Nash equilibria \( {e}^{i} \) and \(... | Proof. \( {}^{32} \) Let \( a \) be a (possibly correlated) action profile such that \( u\left( a\right) = v \) . As in Theorem 3.7.4, let \( \widetilde{v} \) be a payoff vector in the interior of \( \bar{F} \) such that \( \widetilde{v} < v \) . For each \( i \in N \), let \( \lambda > 0 \) be such that\n\n\[\n\left( ... | Yes |
Proposition 4.2.1. Let \( {BG} \) be a Bayesian game.\n\ni) If a pure/mixed/behavior strategy profile in BG is a Bayesian Nash equilibrium, then the induced strategy profile in \( {\Gamma }^{BG} \) is a Nash equilibrium.\n\nii) Conversely, if a pure/mixed/behavior strategy profile in \( {\Gamma }^{BG} \) is a Nash equi... | Proof. Exercise 4.1. | No |
Corollary 4.2.2. Every Bayesian game has at least one Bayesian Nash equilibrium in behavior strategies. | Proof. It follows from the combination of Theorem 3.4.2, which states that every extensive game with perfect recall has, at least, a subgame perfect equilibrium, and Proposition 4.2.1 above. | No |
Player 1 has two information sets: \( {w}_{1}^{1} \) and \( {w}_{1}^{2} \). Since both of them are reached with positive probability, for a strategy profile to be a Nash equilibrium of \( {\Gamma }^{B\widehat{G}} \), player 1 should be best replying at both \( {w}_{1}^{1} \) and \( {w}_{1}^{2} \). At \( {w}_{1}^{1},{NI... | \[ {b}_{2}\left( {w}_{2}\right) = \left\{ \begin{array}{ll} {NE} & p < 1/2 \\ {s}_{2} \in \Delta {A}_{2} & p = 1/2 \\ E & p > 1/2 \end{array}\right. \] | Yes |
Lemma 4.3.1. Let \( t \in \{ 1,\ldots, T\} \) .\ni) If \( {q}_{t} < {\beta }^{T - t + 1} \), then, if the players follow the above strategies, the total expected payoff of a weak monopolist from period \( t \) onwards is 0 .\nii) If \( {q}_{t} = {\beta }^{T - t + 1} \), then, if the players follow the above strategies,... | Proof. The proof is made by (backward) induction on \( t \) . Suppose \( t = T \) and \( {q}_{t} < {\beta }^{T - t + 1} \) . Then, entrant \( T \) enters the market and the weak monopolist yields and gets payoff 0 . This establishes claim i) for \( t = T \) . Suppose \( t = T \) and \( {q}_{t} = {\beta }^{T - t + 1} \)... | No |
Proposition 4.4.1. Let \( \left( {\Theta ,\rho, A,{u}^{II}}\right) \) be a second-price auction. Then, for each \( i \in N \), the strategy defined by \( {\widehat{a}}_{i}^{II}\left( {\theta }_{i}\right) \mathrel{\text{:=}} {\theta }_{i} \) is a dominant strategy, i.e., for each \( {\widehat{a}}_{-i} \in {\widehat{A}}_... | Proof. Let \( {\widehat{a}}_{-i} \in {A}_{-i} \) and \( {\widehat{a}}_{i} \in {A}_{i} \) . Let \( \theta \in {\left\lbrack 0,\bar{v}\right\rbrack }^{n} \) . We now claim that \( {\theta }_{i} \) is a weakly dominant strategy for \( i \) at \( \theta \) . Let \( \widehat{b} \mathrel{\text{:=}} \mathop{\max }\limits_{{j ... | Yes |
Proposition 4.4.2. The strategy profile \( {\widehat{a}}^{II} \) is the unique symmetric, differentiable, and increasing Bayesian Nash equilibrium of the second-price auction. | Proof. Let \( \widehat{y} : \left\lbrack {0,\bar{v}}\right\rbrack \rightarrow \lbrack 0,\infty ) \) be a symmetric, differentiable, and increasing Bayesian Nash equilibrium of the second-price auction \( \left( {\Theta ,\rho, A,{u}^{II}}\right) \) . We show in two steps that, for each \( i \in N \) and each \( \theta \... | Yes |
We now solve for the simplest case of all, where \( \rho \) corresponds with the uniform distribution over \( \left\lbrack {0,1}\right\rbrack \) . In this case, for each \( v \in \left\lbrack {0,1}\right\rbrack \), we have \( F\left( v\right) = v \) and \( \widehat{F}\left( v\right) = {v}^{n - 1} \) . Bidding behavior ... | \[ {\widehat{a}}^{I}\left( \theta \right) = {\int }_{0}^{\theta }\frac{z\widehat{f}\left( z\right) }{\widehat{F}\left( \theta \right) }{dz} = \frac{1}{{\theta }^{n - 1}}{\int }_{0}^{\theta }z\left( {n - 1}\right) {z}^{n - 2}{dz} = \frac{n - 1}{n}\theta \] that is, whereas in the second-price auction a player always bid... | Yes |
Proposition 4.4.4 (Revenue equivalence principle). If the valuations of the players are private and independently and identically distributed, then the expected revenues for the seller in first-price and second-price auctions coincide. | Proof. Note that the expected revenue of the seller coincides with the sum of the expected payments of the bidders. We derive these expected payments below. Let \( v \in \left\lbrack {0,\bar{v}}\right\rbrack \) . Let \( \pi \left( v\right) \) be the expected payment of a bidder of type \( v \) .\n\nSecond-price auction... | Yes |
Theorem 4.5.1 (The revelation principle). Let \( \left( {\Theta ,\rho, M,\mathcal{S}, u}\right) \) be a mechanism and \( \widehat{m} \) a Bayesian equilibrium. Then, there is a direct mechanism in which there is a truthful equilibrium whose outcome coincides with the outcome of the original equilibrium, i.e., for each ... | Proof. Consider the direct mechanism \( \left( {\Theta ,\rho ,\bar{M},\overline{\mathcal{S}}, u}\right) \) defined as follows: \( \bar{M} : = \mathop{\prod }\limits_{{i \in N}}{\Theta }_{i} \cup \{ r\} \) and \( \bar{\mathcal{S}} \) is defined, for each \( \theta \in \Theta \), as \( \bar{\mathcal{S}}\left( \theta \rig... | Yes |
Proposition 4.6.1. Let \( \left( {\Theta ,\rho ,\mathcal{A}, u}\right) \) be a multistage game with incomplete information. If an assessment \( \left( {b,\mu }\right) \) is consistent, then it is reasonable. | Proof. Exercise 4.9. | No |
Corollary 4.6.2. Let \( \left( {\Theta ,\rho ,\mathcal{A}, u}\right) \) be a multistage game with incomplete information. If an assessment \( \left( {b,\mu }\right) \) is a sequential equilibrium then it is a perfect Bayesian equilibrium. | Proof. Follows from Proposition 4.6.1. | No |
Corollary 4.6.3. Let \( \left( {\Theta ,\rho ,\mathcal{A}, u}\right) \) be a multistage game with incomplete information. Then, \( \left( {\Theta ,\rho ,\mathcal{A}, u}\right) \) has at least one perfect Bayesian equilibrium in behavior strategies. | Proof. Follows from the combination of Theorem 3.5.2 and Corollary 4.6.2. | No |
Consider the fragment of a multistage game in Figure 4.6.1. It depicts the situation of a two player multistage game at stage \( t \) after a certain history \( h \), where player 1 has three possible types and player 2 has a unique possible type \( {\theta }_{2} \). Suppose that, using Bayes rule, player 2 infers that... | check that they can never be part of a consistent assessment. Informally, just note that, when justifying his beliefs after history \( {h}_{l} \), he has to come up with a \ | No |
Proposition 5.3.1. Let \( \\left( {F, d}\\right) \\in {B}^{N} \) . Then, there is a unique \( z \\in {F}_{d} \) that maximizes the function \( {g}^{d} \) over the set \( {F}_{d} \) . | Proof. Since \( {g}^{d} \) is continuous and \( {F}_{d} \) is compact, \( {g}^{d} \) has a maximum in \( {F}_{d} \) . Suppose that there are \( z,\\widehat{z} \\in {F}_{d} \), with \( z \\neq \\widehat{z} \), such that\n\n\[ \n\\mathop{\\max }\\limits_{{x \\in {F}_{d}}}{g}^{d}\\left( x\\right) = {g}^{d}\\left( z\\right... | Yes |
Lemma 5.3.2. Let \( \left( {F, d}\right) \in {B}^{N} \) and let \( z \mathrel{\text{:=}} \mathrm{{NA}}\left( {F, d}\right) \) . For each \( x \in {\mathbb{R}}^{N} \), let \( \begin{matrix} h\left( x\right) : = \mathop{\sum }\limits_{{i \in N}}\mathop{\prod }\limits_{{j \neq i}}\left( {{z}_{j} - {d}_{j}}\right) {x}_{i}.... | Proof. Suppose that there is \( x \in F \) with \( h\left( x\right) > h\left( z\right) \) . For each \( \varepsilon \in \left( {0,1}\right) \), let \( {x}^{\varepsilon } \mathrel{\text{:=}} {\varepsilon x} + \left( {1 - \varepsilon }\right) z \) . By the convexity of \( F,{x}^{\varepsilon } \in F \) . Since \( z \in {F... | Yes |
Proposition 5.3.4. None of the axioms used in the characterization of the Nash solution given by Theorem 5.3.3 is superfluous. | Proof. We show that, for each of the axioms in the characterization, there is an allocation rule different from the Nash solution that satisfies the remaining three.\n\nRemove EFF: The allocation rule \( \varphi \) defined, for each bargaining problem \( \left( {F, d}\right) \), by \( \varphi \left( {F, d}\right) \math... | No |
Consider the two-player bargaining problem \( \left( {F, d}\right) \), where \( d = \left( {0,0}\right) \) and \( F \) is the comprehensive hull of the set \( \left\{ {\left( {{x}_{1},{x}_{2}}\right) \in {\mathbb{R}}^{2}}\right. \) : \( \left. {{x}_{1}^{2} + {x}_{2}^{2} \leq 2}\right\} \) . Since NA satisfies SYM and E... | These two bargaining problems and the corresponding proposals made by the Nash solution are depicted in Figure 5.3.1. | No |
Theorem 5.3.5. The Kalai-Smorodinsky solution is the unique allocation rule for two-player bargaining problems that satisfies EFF, SYM, CAT, and IM. | Proof. It is easy to check that KS satisfies EFF, CAT, SYM, and IM. Let \( \varphi \) be an allocation rule for two-player bargaining problems that satisfies the four properties and let \( \left( {F, d}\right) \in {B}^{2} \) . We now show that \( \varphi \left( {F, d}\right) = \operatorname{KS}\left( {F, d}\right) \) .... | Yes |
Proposition 5.3.6. Let \( n \geq 3 \) . Then, there is no solution for \( n \) -player bargaining problems satisfying EFF, SYM, and IM. | Proof. Let \( n \geq 3 \) and suppose that \( \varphi \) is a solution for \( n \) -player bargaining problems satisfying EFF, SYM, and IM. Let \( d = \left( {0,\ldots ,0}\right) \) and\n\n\[ \widehat{F} \mathrel{\text{:=}} \{ x \in {\mathbb{R}}^{N} : \text{there is }y \in \mathrm{{conv}}\{ \left( {0,1,\ldots ,1}\right... | Yes |
Example 5.4.4. (The visiting professor). We now illustrate how TU-games can also model situations that involve costs instead of benefits. Three research groups, from the universities of Milano (group 1), Genova (group 2), and Santiago de Compostela (group 3), plan to invite a Japanese professor to give a course on game... | \[ v\left( S\right) = \mathop{\sum }\limits_{{i \in S}}c\left( i\right) - c\left( S\right) \] Thus, \( v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = 0, v\left( {12}\right) = {1500}, v\left( {13}\right) = {500}, v\left( {23}\right) = {500}, \) and \( v\left( N\right) = {2000} \) . | Yes |
Lemma 5.4.1. Let \( v \in {G}^{N} \) . Then, \( v \) is weakly superadditive if and only if it is zero-monotonic. | Proof. Exercise 5.5. | No |
Proposition 5.5.1. Let \( v \in {G}^{N} \) . Then,\ni) If \( x \in C\left( v\right), x \) is undominated.\nii) If \( v \in S{G}^{N}, C\left( v\right) = \{ x \in I\left( v\right) : x \) is undominated \( \} \) . | Proof. i) Let \( x \in C\left( v\right) \) and suppose there is \( y \in I\left( v\right) \) and \( S \subset N, S \neq \varnothing \) , such that \( y \) dominates \( x \) through \( S \) . Then, \( v\left( S\right) \geq \mathop{\sum }\limits_{{i \in S}}{y}_{i} > \mathop{\sum }\limits_{{i \in S}}{x}_{i} \geq v\left( S... | Yes |
Proposition 5.5.2. Let \( v \in {S}^{N} \) be a simple game. Then, \( C\left( v\right) \neq \varnothing \) if and only if there is at least one veto player in \( v \) . Moreover, if \( C\left( v\right) \neq \varnothing \), then\n\n\[ C\left( v\right) = \left\{ {x \in I\left( v\right) : \text{ for each nonve to player }... | Proof. Let \( v \in {S}^{N} \) . Let \( x \in C\left( v\right) \) and let \( A \) be the set of veto players. Suppose that \( A = \varnothing \) . Then, for each \( i \in N, v\left( {N\smallsetminus \{ i\} }\right) = 1 \) and, hence,\n\n\[ 0 = v\left( N\right) - v\left( {N\smallsetminus \{ i\} }\right) \geq \mathop{\su... | No |
Theorem 5.5.3 (Bondareva-Shapley theorem). Let \( v \in {G}^{N} \) . Then, \( C\left( v\right) \neq \varnothing \) if and only if \( v \) is balanced. | Proof. \( {}^{7} \) Let \( v \in {G}^{N} \) be such that \( C\left( v\right) \neq \varnothing \) . Let \( x \in C\left( v\right) \) and let \( \mathcal{F} \) be a balanced family with balancing coefficients \( \left\{ {{\alpha }_{S} : S \in \mathcal{F}}\right\} \) . Then,\n\n\[ \mathop{\sum }\limits_{{S \in \mathcal{F}... | Yes |
Lemma 5.5.4. Let \( v \in {G}^{N} \) be such that \( v\left( N\right) > \mathop{\sum }\limits_{{i \in N}}v\left( i\right) \) . Then, there is a unique \( 0 - 1 \) -normalized game \( \widehat{v} \) such that \( v \) and \( \widehat{v} \) are S-equivalent. | Proof. Exercise 5.6. | No |
Lemma 5.5.5. Let \( v,\widehat{v} \in {G}^{N} \) be S-equivalent and such that \( v\left( N\right) > \mathop{\sum }\limits_{{i \in N}}v\left( i\right) \) . Then, \( v \) is balanced if and only if \( \widehat{v} \) is balanced. | Proof. Let \( k > 0 \) and \( {a}_{1},\ldots ,{a}_{n} \in \mathbb{R} \) be such that, for each \( S \subset N,\widehat{v}\left( S\right) = \) \( {kv}\left( S\right) + \mathop{\sum }\limits_{{i \in S}}{a}_{i} \) . Suppose that \( v \) is balanced. Let \( \mathcal{F} \) be a balanced family of coalitions with balancing c... | Yes |
Lemma 5.5.6. Let \( v,\widehat{v} \in {G}^{N} \) be S-equivalent and such that \( v\left( N\right) > \mathop{\sum }\limits_{{i \in N}}v\left( i\right) \) . Then, \( C\left( v\right) \neq \varnothing \) if and only if \( C\left( \widehat{v}\right) \neq \varnothing \) . | Proof. It immediately follows from definitions of core and \( S \) -equivalence (definitions 5.5.2 and 5.5.9). | No |
Lemma 5.5.7. Let \( v,\widehat{v} \in {G}^{N} \). If \( v \) and \( \widehat{v} \) are totally balanced, then \( v \land \widehat{v} \) is totally balanced. | Proof. Let \( S \subset N \). Assume, without loss of generality, that \( v\left( S\right) \leq \widehat{v}\left( S\right) \). Since \( v \) is totally balanced, \( C\left( {v}_{S}\right) \neq \varnothing \). Let \( x \in C\left( {v}_{S}\right) \). Then, it is straightforward to see that \( x \in C\left( {\left( v \lan... | No |
Theorem 5.5.8. A nonnegative TU-game \( v \in {G}^{N} \) is totally balanced if and only if it is the minimum game of a finite collection of nonnegative additive games. | Proof. A nonnegative additive game is totally balanced and, hence, the \ | No |
Theorem 5.6.1. The Shapley value is the unique allocation rule in \( {G}^{N} \) that satisfies EFF, NPP, SYM, and ADD. | Proof. First, \( \Phi \) satisfies both NPP and ADD. Moreover, it should be clear at this point that it also satifies EFF and SYM. Each vector of marginal contributions is an efficient allocation and, hence, EFF follows from Eq. (5.6.2). Also, SYM can be easily derived from Eq. (5.6.2).\n\nNow, let \( \varphi \) be an ... | Yes |
The Shapley value for the glove game is \( \left( {2/3,1/6,1/6}\right) \) | Remember that the core of this game is \( \{ \left( {1,0,0}\right) \} \). The core and the Shapley value of this game are represented in Figure 5.6.1(a). Hence, even when the core is nonempty, the Shapley value may not be a core allocation. | No |
The Shapley value in the visiting professor game discussed in Example 5.4.4 is \( \Phi \left( v\right) = \left( {{5000}/6,{5000}/6,{2000}/6}\right) \) . These are the savings for the players. According to this allocation of the savings, the players have to pay \( \left( {{4000}/6,{4600}/6,{9400}/6}\right) \) . Note tha... | \[ \Phi _{i}\left( v\right) = \mathop{\sum }\limits_{{S \subset N\smallsetminus \{ i\} }}\frac{s!\left( {n - s - 1}\right) !}{n!}\left( {v\left( {S\cup \{ i\} }\right) - v\left( S\right) }\right) \] \[ = \mathop{\sum }\limits_{{S \subset N\smallsetminus \{ i\} }}\frac{s!\left( {n - s - 1}\right) !}{n!}\left( {c\left( S... | Yes |
Lemma 5.7.1. Let \( v \in {G}^{N} \) . Let \( x, y \in {\mathbb{R}}^{N} \) be such that \( x \neq y \) and \( \theta \left( x\right) = \theta \left( y\right) \) . Let \( \alpha \in \left( {0,1}\right) \) . Then, \( \theta \left( x\right) { \succ }_{L}\theta \left( {{\alpha x} + \left( {1 - \alpha }\right) y}\right) \) ... | Proof. Note that, for each \( S \subset N, e\left( {S,{\alpha x} + \left( {1 - \alpha }\right) y}\right) = \alpha \left( {v\left( S\right) - \mathop{\sum }\limits_{{i \in S}}{x}_{i}}\right) + \) \( \left( {1 - \alpha }\right) \left( {v\left( S\right) - \mathop{\sum }\limits_{{i \in S}}{y}_{i}}\right) = {\alpha e}\left(... | Yes |
Theorem 5.7.2. Let \( v \in {G}^{N} \) be such that \( I\left( v\right) \neq \varnothing \) . Then, the set \( \eta \left( v\right) \) contains a unique allocation. | Proof. Let \( {I}^{0} \mathrel{\text{:=}} I\left( v\right) \) . For each \( k \in \left\{ {1,\ldots ,{2}^{n}}\right\} \), let \( {I}^{k} \) be the set\n\n\[ \n{I}^{k} \mathrel{\text{:=}} \left\{ {x \in {I}^{k - 1} : \text{ for each }y \in {I}^{k - 1},{\theta }_{k}\left( x\right) \leq {\theta }_{k}\left( y\right) }\righ... | Yes |
Consider again the TU-game \( v \in {G}^{N}\; \) given by \( \;N = \{ 1,2,3\} \) and \( v\left( 1\right) = v\left( 2\right) = 0, v\left( 3\right) = 1, v\left( {12}\right) = 2, v\left( {13}\right) = v\left( {23}\right) = 1 \), and \( v\left( N\right) = 2 \) (introduced in Example 5.5.1 and already discussed in this sect... | \[ \text{Minimize}{\alpha }_{1} \] \[ \text{subject to}\;{x}_{i} + {\alpha }_{1} \geq 0,\;i \in \{ 1,2\} \] \[ {x}_{3} + {\alpha }_{1} \geq 1 \] \[ {x}_{1} + {x}_{2} + {\alpha }_{1} \geq 2 \] \[ {x}_{1} + {x}_{3} + {\alpha }_{1} \geq 1 \] \[ {x}_{2} + {x}_{3} + {\alpha }_{1} \geq 1 \] \[ {x}_{i} \geq 0,\;i \in \{ 1,2\}... | Yes |
Consider the visiting professor game introduced in Example 5.4.4, where \( N = \{ 1,2,3\}, v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = 0, v\left( {12}\right) = {1500} \) , \( v\left( {13}\right) = {500}, v\left( {23}\right) = {500} \), and \( v\left( N\right) = {2000} \) . This game is convex. Moreover, as... | <table><thead><tr><th>Permutation \( \pi \)</th><th>\( {m}_{1}^{\pi }\left( v\right) \)</th><th>\( {m}_{2}^{\pi }\left( v\right) \)</th><th>\( {m}_{3}^{\pi }\left( v\right) \)</th></tr></thead><tr><td>123</td><td>0</td><td>1500</td><td>500</td></tr><tr><td>132</td><td>0</td><td>1500</td><td>500</td></tr><tr><td>213</td... | Yes |
Theorem 5.8.1. Let \( v \in {G}^{N} \) . The following statements are equivalent:\n\ni) The game \( v \) is convex.\n\nii) For each \( \pi \in \Pi \left( N\right) ,{m}^{\pi }\left( v\right) \in C\left( v\right) \).\n\niii) \( C\left( v\right) = \operatorname{conv}\left\{ {{m}^{\pi }\left( v\right) : \pi \in \Pi \left( ... | Proof. \( {}^{13}\mathrm{i} \) ) \( \Rightarrow \mathrm{{ii}} \) ). Let \( \pi \in \Pi \left( N\right) \) . We have that \( \mathop{\sum }\limits_{{i \in N}}{m}_{i}^{\pi }\left( v\right) = v\left( N\right) \) . Let \( S \varsubsetneq N \) . Let \( i \in N \smallsetminus S \) be such that, for each \( j \in N \smallsetm... | Yes |
Corollary 5.8.2. Let \( v \in {G}^{N} \) be a convex game. Then, \( \Phi \left( v\right) \in C\left( v\right) \) . | Proof. It immediately follows from the formula of the Shapley value (given in Eq. (5.6.2)) and the statement iii) in Theorem 5.8.1 above. | No |
Consider the TU-game \( \left( {N, v}\right) \), where \( N = \{ 1,2,3\} \) and \( v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = 0, v\left( {12}\right) = v\left( {13}\right) = 2, v\left( {23}\right) = 6 \) and \( v\left( N\right) = 7 \) . This game is not convex because \( v\left( {23}\right) - v\left( 2\rig... | However, the reader can easily verify that \( \Phi \left( v\right) = \left( {1,3,3}\right) \in C\left( v\right) \) . | No |
Proposition 5.9.1. The set of pure Nash equilibrium payoff vectors of the negotiation game \( {G}^{\mathrm{{NA}}} \) consists of all Pareto efficient allocations of \( F \) along with the disagreement point. | Proof. Let \( a \in A \) . If \( a \) belongs to \( F \) but it is not Pareto efficient, then there \( \operatorname{are}i \in N \) and \( {\widehat{a}}_{i} \in {A}_{i} \) such that \( \left( {{a}_{-i},{\widehat{a}}_{i}}\right) \in F \) and \( {\widehat{a}}_{i} = {u}_{i}\left( {{a}_{-i},{\widehat{a}}_{i}}\right) > {u}_... | Yes |
Suppose that \( T = 0 \), that is, if the proposal of player 1 in period 0 is not accepted, then both players get 0 . It is easy to see that this game has an infinite number of Nash equilibria. We now show that all but one are based on incredible threats. More precisely, we show that there is a unique subgame perfect e... | We proceed backwards. In any subgame perfect equilibrium, when asking whether to accept or not, player 2 will accept any share \( \left( {{r}_{1},{r}_{2}}\right) \) with \( {r}_{2} > 0 \) . However, no such share can be part of a subgame perfect equilibrium since, given proposal \( \left( {{r}_{1},{r}_{2}}\right) \) wi... | Yes |
Example 5.9.2. (A simple bargaining game). Suppose that \( T = 1 \), i.e., the game has two periods, labeled 0 and 1 . Again, we proceed backwards to show that this game also has a unique subgame perfect equilibrium. If period 1 is reached, the corresponding subgame is an ultimatum game. Thus, in equilibrium, if period... | However, repeating the arguments above, no such share can be part of a subgame perfect equilibrium since, given proposal \( \left( {{r}_{1},{r}_{2}}\right) \) with \( {r}_{2} > \delta \), player 1 can profitably deviate by offering \( \left( {{r}_{1} + \frac{{r}_{2} - \delta }{2},{r}_{2} - \frac{{r}_{2} - \delta }{2}}\... | Yes |
Proposition 5.9.3. Consider the alternating offers game with perfectly patient players (i.e., \( \delta = 1 \) ). Then, an allocation \( x \in {\bar{F}}_{\left( 0,0\right) } \) is a subgame perfect equilibrium payoff if and only if \( x \) is Pareto efficient. | Proof. Let \( \left( {{r}_{1},{r}_{2}}\right) \in R \) be such that \( x = \left( {{u}_{1}\left( {r}_{1}\right) ,{u}_{2}\left( {r}_{2}}\right) }\right) \) is Pareto efficient in \( \bar{F} \), i.e., \( {r}_{1} + {r}_{2} = 1 \) . Consider the following strategy profile. Both players, when playing in the role of the prop... | Yes |
Corollary 5.9.5. The equilibrium allocation of the alternating offers game converges to the Nash solution of the bargaining problem \( \left( {\bar{F},\left( {0,0}\right) }\right) \) as \( \delta \) converges to 1 . | Proof. Given an alternating offers game with discount factor \( \delta \), let \( {x}^{\delta } \) be the equilibrium payoff vector when player 1 proposes first, and let \( {y}^{\delta } \) be the equilibrium payoff vector when player 2 proposes first. These payoff vectors satisfy that \( {x}_{1}^{\delta }{x}_{2}^{\del... | Yes |
Proposition 5.9.6. Let \( \left( {F, d}\right) \in {B}^{2} \) and let \( x \in {F}_{d} \) . Then, there is a Nash equilibrium of \( {\Gamma }^{\mathrm{{KS}}} \) whose payoffs equal \( x \) . | # Proof. Let \( x \in F \) . Consider the following strategy profile:\n\nStage 1: Both players bid 0.\n\nStage 2: The proposer chooses \( x \) . The player who is not the proposer\n\naccepts \( x \) and rejects any other proposal.\n\nStage 3: The proposer chooses the disagreement point.\n\nThis is a Nash equilibrium of... | No |
Consider the airport problem of Example 5.10.1. In the first step we get\n\n\[ \min \left\{ {\frac{30}{4},\min \left\{ {\frac{12}{2},\frac{28}{3}}\right\} }\right\} \] | and, hence, \( {\bar{\alpha }}_{1} = - 6 \) and \( {k}_{1} = 1 \) . Then, we have\n\n\[ \min \left\{ {\frac{{30} - 6}{4 - 1},\min \left\{ \frac{{28} - 6}{3}\right\} }\right\} \]\n\nwhich leads to \( {\bar{\alpha }}_{2} = - \frac{{28} - 6}{3} = - \frac{22}{3} \) and \( {k}_{2} = 2 \) . Finally, \( {\bar{\alpha }}_{3} = ... | Yes |
Consider the bankruptcy problem in Example 5.11.1 when the estate is \( E = {200} \) . The players are \( N = \{ 1,2,3\} \) and the demands are \( {d}_{1} = {100},{d}_{2} = {200} \), and \( {d}_{3} = {300} \) . The characteristic function assigns \( v\left( {23}\right) = \max \{ 0,{200} - {100}\} = {100} \) to coalitio... | The Shapley value proposes the division \( \Phi \left( v\right) \; = \;\left( {\frac{3}{3} + \frac{1}{3},\frac{8}{3} + \frac{1}{3},\frac{8}{3} + \frac{1}{3}}\right) \) and the nucleolus \( \eta \left( v\right) = \left( {{50},{75},{75}}\right) \) . In particular, the proposal of the nucleolus coincides with the proposal... | Yes |
The motivation for the random arrival rule we presented above resembles the interpretation of the Shapley value we made after its definition in Section 5.6. Actually, the division proposed by the random arrival rule coincides with the Shapley value of the associated bankruptcy game. | The proof of this result is quite straightforward and is left as an exercise. | No |
Proposition 5.11.1. Let \( \left( {E, d}\right) \) be a bankruptcy problem and let \( v \) be the associated bankruptcy game. Then, \( {f}^{RA}\left( {E, d}\right) = \Phi \left( v\right) \) . | Proof. Exercise 5.15. | No |
Theorem 5.11.2. Let \( \left( {E, d}\right) \) be a bankruptcy problem and let \( v \) be the associated bankruptcy game. Then, \( {f}^{TR}\left( {E, d}\right) = \eta \left( v\right) \) . | Proof. The case \( n = 1 \) is immediate and hence, we assume that \( n \geq 2 \) . We begin by computing the excess of a coalition at any imputation in the bankruptcy game. Let \( S \subset N \) and let \( x \in I\left( v\right) \) . Then, \( e\left( {S, x}\right) = v\left( S\right) - \mathop{\sum }\limits_{{i \in S}}... | Yes |
Theorem 5.11.3. A division rule \( f \) for banckruptcy problems can be represented through an allocation rule for bankruptcy games if and only if, for each bankruptcy problem \( \left( {E, d}\right), f\left( {E, d}\right) = f\left( {E,{d}^{T}}\right) \) . | Proof. The \ | No |
Let \( \left( {E, d}\right) \) be the bankruptcy problem given by \( E = {200} \) , \( {d}_{1} = {100},{d}_{2} = {200} \), and \( {d}_{3} = {400} \) . The proposal of the random arrival rule is \( \left( {{33} + 1/3,{83} + 1/3,{83} + 1/3}\right) \) and the one of the Talmud rule is \( \left( {{50},{75},{75}}\right) \) ... | Now, let \( \left( {E,\widehat{d}}\right) \) be the problem in which player 3 splits into two players with demands 200 and 200 . Then, \( {\widehat{d}}_{1} = {100},{\widehat{d}}_{2} = {200},{\widehat{d}}_{3} = {200} \), and \( {\widehat{d}}_{4} = {200} \) ; in particular, \( {d}_{3} = {\widehat{d}}_{3} + {\widehat{d}}_... | Yes |
Consider the simple game given by \( N = \{ 1,2,3,4,5\} \) and \( {W}^{m} = \{ \{ 1,2,3\} ,\{ 4,5\} \} \) . This simple game does not arise from any quota \( q \) and system of weights \( {p}_{1},\ldots ,{p}_{5} \) . | Suppose, on the contrary, that there is a quota \( q > 0 \) and a system of weights \( {p}_{1},\ldots ,{p}_{5} \) for this simple game. Then, \( {p}_{1} + {p}_{2} + {p}_{3} \geq q\;\mathrm{{and}}\;{p}_{4} + {p}_{5} \geq q.\;\mathrm{{Moreover}}, q > {p}_{1} + {p}_{2} + {p}_{4} \geq q - {p}_{3} + {p}_{4} \) and, hence, \... | Yes |
Theorem 5.12.1. The Shapley-Shubik index is the unique allocation rule in \( {S}^{N} \) that satisfies EFF, NPP, SYM, and TF. | Proof. From Theorem 5.6.1, the Shapley-Shubik index satisfies EFF, NPP, and SYM. Moreover, given two simple games \( v \) and \( \widehat{v}, v + \widehat{v} = v \vee \widehat{v} + v \land \widehat{v} \) and, hence, since the Shapley value satisfies ADD, the Shapley-Shubik index satisfies TF.\n\nWe prove the uniqueness... | Yes |
Theorem 5.12.3. The Banzhaf index is the unique allocation rule \( \varphi \) in \( {S}^{N} \) that satisfies NPP, SYM, TF, and that, for each \( v \in {S}^{N} \) ,\n\n\[ \mathop{\sum }\limits_{{i \in N}}{\varphi }_{i}\left( v\right) = \frac{\bar{\mu }\left( v\right) }{{2}^{n - 1}} \] | Proof. It follows similar lines to the proof of Theorem 5.12.1 and it is left to the reader (see Exercise 5.16). | No |
We now show that, in general, the power indices \( \beta \) and \( \Phi \) do not coincide. Consider the simple game given by \( N = \{ 1,2,3,4\} \) and \( {W}^{m} = \{ \{ 1,2,3\} ,\{ 1,2,4\} \} \) . | The swings of player 1 are \( \left( {\{ 1,2,3\} ,\{ 2,3\} }\right) \) , \( \left( {\{ 1,2,4\} ,\{ 2,4\} }\right) \), and \( \left( {\{ 1,2,3,4\} ,\{ 2,3,4\} }\right) \) and, hence, \( {\mu }_{1}\left( v\right) = 3 \) . Similarly, we obtain \( {\mu }_{2}\left( v\right) = 3,{\mu }_{3}\left( v\right) = 1 \), and \( {\mu ... | Yes |
Theorem 5.13.1. For each linear production process \( \left( {N,\mathcal{A}, b, c}\right) \), the associated linear production game is totally balanced. | Proof. Let \( \left( {N,\mathcal{A}, b, c}\right) \) be a linear production process and let \( v \) be the characteristic function of the associated linear production game. For each coalition \( S \), the restriction of the characteristic function to \( S \) is also a linear production game. When combined with Bondarev... | Yes |
Consider the linear production process \( \left( {N,\mathcal{A}, b, c}\right) \) with \( N = \{ 1,2,3\} \)\n\n\[ \mathcal{A} = \left( \begin{array}{lll} 1 & 0 & 2 \\ 1 & 1 & 3 \end{array}\right), b = \left( {3,1,8}\right) ,{c}_{1} = \left( {1,0}\right) ,{c}_{2} = \left( {2,2}\right) \text{, and }{c}_{3} = \left( {0,2}\... | The optimal value is 11 and the optimal solution set is \( \{ \left( {1,2}\right) \} \) . Thus, \( v\left( N\right) = {11} \) . The complete characteristic function of the linear production problem is given by\n\n\[ v\left( 1\right) = 0,\;v\left( 2\right) = 6,\;v\left( 3\right) = 2, \]\n\n\[ v\left( {12}\right) = 6,\;v... | Yes |
Theorem 5.13.2. Let \( v \in {G}^{N} \) be a nonnegative TU-game. If \( v \) is totally balanced, then \( v \) is a linear production game. | Proof. Let \( v \in {G}^{N} \) be a nonnegative totally balanced game. We now show that \( v \) can be seen as the game associated with a linear production process. The set of agents is \( N \) . There are \( n \) commodities and each agent \( i \in N \) only has a unit of commodity \( i \), that is, \( {c}_{i}^{i} \ma... | Yes |
Corollary 5.13.3. The class of nonnegative totally balanced TU-games coincides with the class of linear production games. | Proof. Follows from the combination of Theorems 5.13.1 and 5.13.2. | No |
Lemma 5.13.4. Every nonnegative additive game is a maximum flow game. | Proof. Let \( v \in {G}^{N} \) be a nonnegative additive game given by the nonnegative real numbers \( v\left( 1\right) ,\ldots, v\left( n\right) \) . Let \( X \mathrel{\text{:=}} \left\{ {{x}_{o},{x}_{a}}\right\} \) and, for each \( i \in N \) , introduce an edge from \( {x}_{o} \) to \( {x}_{a} \), owned by player \(... | Yes |
Lemma 5.13.5. Let \( {v}^{1} \) and \( {v}^{2} \) be two maximum flow games. Then, the minimum game \( {v}^{1} \land {v}^{2} \) is a maximum flow game. | Proof. Let \( \left( {{X}^{l},{E}^{l},{\operatorname{cap}}^{l}}\right) \) and \( {o}^{{E}^{l}} \) be the maximum flow problem and the ownership function, respectively, of game \( {v}^{l} \), with \( l \in \{ 1,2\} \) . Now, let \( \left( {X, E,\text{cap}}\right) \) be the maximum flow problem whose network is obtained ... | Yes |
Theorem 5.13.6. A nonnegative TU-game \( v \in {G}^{N} \) is totally balanced if and only if it is a maximum flow game. | Proof. From Theorem 5.5.8, Lemma 5.13.4, and Lemma 5.13.5, every nonnegative totally balanced game is a maximum flow game. It remains to prove the converse. Given a maximum flow game \( v \) its restriction to a coalition \( S \subset N,{v}_{S} \), is also a maximum flow game. Hence, to get the result it suffices to sh... | Yes |
Corollary 5.13.7. A nonnegative TU-game is a linear production game if and only if it is a maximum flow game. | Proof. Follows immediately from the combination of Corollary 5.13.3 and Theorem 5.13.6. | Yes |
Proposition 5.13.8. Given an inventory game \( \left( {N, c}\right) \), the SOC rule gives an allocation in the core of the game. | Proof. Exercise 5.20. | No |
Theorem 1 Every extenstive game \( \Gamma \) with perfect information has at least one pure strategy profile \( a \in A \) that is a subgame perfect equilibrium of \( \Gamma \) . | We focus on finite game only: finite game tree \( \rightarrow \) finite width; finite length (depth, horizon) O O O 定理一 證明概要 Sketch: Induction on the number of internal nodes in \( \Gamma \) .) Induction Step: 找一個最低的internal node \( x \) ,令 \( {a}_{x} \) 為 \( {\Gamma }_{x} \) 的任一個pure Nash。 根據inductive hypothesis, \( {... | Yes |
Theorem 2 Let \( \Gamma \) be an extensive game with perfect information. For any behavior strategy profile \( b \in B \) of \( \Gamma, b \) is a subgame perfect equilibrium of \( \Gamma \) if and only if no one-shot deviation from \( b \) is profitable. One-shot deviation: 对某个子博弈, 这个子博弈开始节点的玩家只在该节点改變他的choice,造成跟 \( b ... | SPE \( \rightarrow \) No profitable one-shot deviation: of course \( - \leftarrow ?? \) 证明概要: 反证法 - - 假设 \( b \) 不是SPE 则存在子博弈 \( {\Gamma }_{x} \) 和 \( i \) ,其中 \( {b}_{i} \) 不是 \( i \) 的最佳选择 \( \left( {\text{仍记}{b}_{x} = b}\right) \) O O O, 分） 清火掌 Game T&A 证明概要: 反证法例如 Among all such deviations, let \( {\widehat{b}}_{i}... | No |
Theorem 3 Every extenstive game \( \Gamma \) with behavior Nash perfect recall has at least one behavior strat- equilibrium egy profile \( b \in B \) that is a subgame perfect equilibrium of \( \Gamma \). | ## 定理三 證明概要\n\nSketch: Induction on the number of decomposable nodes in \( \Gamma \) .\n\nInductive basis: 如果都不能分割, 則任何每一個behavior Nash都是一個subgame perfect equilibrium。而根據perfect recall得知一定有behavior Nash, 所以OK。 Induction Step: 找一個可以分割的internal node \( x \) 。先替 \( {\Gamma }_{x} \) 找到一個subgame perfect equilibrium \( {b}_{... | No |
Theorem 3.7.5 (Finite horizon perfect folk theorem; adapted from Benoît and Krishna (1985)). Let \( G\left( {\delta, T}\right) \) be a finitely repeated game. Suppose that \( F \) is full dimensional and that, for each player \( i \in N \), there are two Nash equilibria \( {e}^{i} \) and \( {\widetilde{e}}^{t} \) of \(... | Proof. Omitted (see GSM-115) | No |
Theorem 4.5.1 (The revelation principle). Let \( \left( {\Theta ,\rho, M,\mathcal{S}, u}\right) \) be a mechanism and \( \widehat{m} \) a Bayesian equilibrium. Then, there is a direct mechanism in which there is a truthful equilibrium whose outcome coincides with the outcome of the original equilibrium, i.e., for each ... | Proof. Consider the direct mechanism \( \left( {\Theta ,\rho ,\bar{M},\overline{\mathcal{S}}, u}\right) \) defined as follows: \( \bar{M} \mathrel{\text{:=}} \mathop{\prod }\limits_{{i \in N}}{\Theta }_{i} \cup \{ r\} \) and \( \overline{\mathcal{S}} \) is defined, for each \( \theta \in \Theta \), as \( \overline{\mat... | Yes |
Lemma 5.3.2. Let \( \left( {F, d}\right) \in {B}^{N} \) and let \( z \mathrel{\text{:=}} \mathrm{{NA}}\left( {F, d}\right) \) . For each \( x \in {\mathbb{R}}^{N} \), let \( h\left( x\right) : = \mathop{\sum }\limits_{{i \in N}}\mathop{\prod }\limits_{{j \neq i}}\left( {{z}_{j} - {d}_{j}}\right) {x}_{i} \) . Then, for ... | Proof Suppose that there is \( x \in F \) with \( h\left( x\right) > h\left( z\right) \) for sufficiently small \( \varepsilon \in \left( {0,1}\right) ,{x}^{\varepsilon } \mathrel{\text{:=}} {\varepsilon x} + \left( {1 - \varepsilon }\right) z \in {F}_{d} \n\n\[ \n{g}^{d}\left( {x}^{\varepsilon }\right) = \mathop{\prod... | Yes |
Example 5.4.1. (Divide a million). A wealthy man dies and leaves one million euros to his three nephews, with the condition that at least two of them must agree on how to divide this amount among them; otherwise, the million euros will be burned. | \[ N = \{ 1,2,3\}, v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = 0, \] \[ v\left( {12}\right) = v\left( {13}\right) = v\left( {23}\right) = v\left( N\right) = 1. \] | No |
Example 5.4.2. (The glove game). Three players are willing to divide the benefits of selling a pair of gloves. Player 1 has a left glove and players 2 and 3 have one right glove each. A left-right pair of gloves can be sold for one euro. | \[ N = \{ 1,2,3\} \] \[ v\left( 1\right) = v\left( 2\right) = v\left( 3\right) = v\left( {23}\right) = 0, \] \[ v\left( {12}\right) = v\left( {13}\right) = v\left( N\right) = 1\text{.} \] | Yes |
Proposition 5.5.2. Let \( v \in {S}^{N} \) be a simple game. Then, \( C\left( v\right) \neq \varnothing \) if and only if there is at least one veto player in \( v \) . Moreover, if \( C\left( v\right) \neq \varnothing \), then \( C\left( v\right) = \left\{ {x \in I\left( v\right) : }\right. \) for each nonveto player ... | Proof. Easy (GSM-115, Page 220) | No |
Theorem 1. ELS games are balanced. | Proof. Let \( B \) be a balanced collection of \( {2}^{N} \) and let \( {\lambda }_{S} \in {\mathbb{Q}}^{ + }, S \in B \), be the corresponding weights. It is sufficient to prove that \( \;\mathop{\sum }\limits_{{S \in B}}{\lambda }_{S}c\left( S\right) \geq c\left( N\right) \) . | No |
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