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Corollary 2 Suppose that Assumption 2 holds and that \( \left\{ {\alpha }_{k}\right\} ,\left\{ {\beta }_{k}\right\} \), and \( \left\{ {\lambda }_{k}\right\} \) in Algorithm 2 are set to (2.27) and (2.30). Also assume that an optimal solution \( {x}^{ * } \) exists for problem (1.3). Then for any \( N \geq 1 \) , we ha... | Proof. The results directly follow by plugging the value of \( {\Gamma }_{k} \) in (2.33), the value of \( {\lambda }_{1} \) in (2.30), and the bound (2.36) into (2.44) and (2.45), respectively. | Yes |
a) If \( \left\{ {\alpha }_{k}\right\} ,\left\{ {\beta }_{k}\right\} ,\left\{ {\lambda }_{k}\right\} \), and \( \left\{ {p}_{k}\right\} \) are chosen such that (2.7) holds and\n\n\[ \n{p}_{k} = \frac{{\lambda }_{k}{C}_{k}}{\mathop{\sum }\limits_{{k = 1}}^{N}{\lambda }_{k}{C}_{k}},\;k = 1,\ldots, N, \n\] \n\n(3.4) \n\nw... | Proof. We first show part a). Denote \( {\delta }_{k} \mathrel{\text{:=}} G\left( {{x}_{k}^{md},{\xi }_{k}}\right) - \nabla \Psi \left( {x}_{k}^{md}\right) \) and \( {\Delta }_{k} \mathrel{\text{:=}} \nabla \Psi \left( {x}_{k - 1}\right) - \nabla \Psi \left( {x}_{k}^{md}\right) \) . By (\n\n(2.1)\n\nand (3.2), we have\... | Yes |
If \( \left\{ {\alpha }_{k}\right\} \) and \( \left\{ {\lambda }_{k}\right\} \) in the RSAG method are set to (2.27) and (2.28), respectively, \( \left\{ {p}_{k}\right\} \) is set to (3.4), \( \left\{ {\beta }_{k}\right\} \) is set to \n\n\[ \n{\beta }_{k} = \min \left\{ {\frac{8}{{21}{L}_{\Psi }},\frac{\widetilde{D}}{... | Proof. We first show part a). It follows from (2.28), (2.35), and (3.12) that \n\n\[ \n{C}_{k} \geq 1 - \frac{21}{16}{L}_{\Psi }{\beta }_{k} \geq \frac{1}{2} > 0\text{ and }{\lambda }_{k}{C}_{k} \geq \frac{{\beta }_{k}}{2}. \n\] \n\nAlso by (2.28), (2.33), (2.34), and (3.12), we have \n\n\[ \n{\lambda }_{k}^{2}\left\lb... | Yes |
Theorem 4 Suppose that \( \left\{ {\alpha }_{k}\right\} ,\left\{ {\beta }_{k}\right\} ,\left\{ {\lambda }_{k}\right\} \), and \( \left\{ {p}_{k}\right\} \) in Algorithm 4 satisfy (2.9), (2.10), and (3.7). Then under Assumptions 1 and 2, we have\n\n\[ \mathbb{E}\left\lbrack {\begin{Vmatrix}\mathcal{G}\left( {x}_{R}^{md}... | Proof. Denoting \( {\bar{\delta }}_{k} \equiv {\bar{G}}_{k} - \nabla \Psi \left( {x}_{k}^{md}\right) \) and \( {\bar{\delta }}_{\left\lbrack k\right\rbrack } \equiv \left\{ {{\bar{\delta }}_{1},\ldots ,{\bar{\delta }}_{k}}\right\} \) for any \( k \geq 1 \), and using Lemma 2 of [11] for the solutions of subproblems (3.... | Yes |
Corollary 4 Suppose that the stepsizes \( \left\{ {\alpha }_{k}\right\} ,\left\{ {\beta }_{k}\right\} \), and \( \left\{ {\lambda }_{k}\right\} \) in Algorithm 4 are set to (2.27) and (2.30), respectively, and \( \left\{ {p}_{k}\right\} \) is set to (3.7). Also assume that an optimal solution \( {x}^{ * } \) exists for... | Proof. Similar to Corollary 1b), we can easily show that (2.9) and (2.10) hold. By (3.30), (2.27), (2.30), (2.33), and (2.36), we have\n\n\[ \mathbb{E}\left\lbrack {\begin{Vmatrix}\mathcal{G}\left( {x}_{R}^{md},\nabla \Psi \left( {x}_{R}^{md}\right) ,{\beta }_{R}\right) \end{Vmatrix}}^{2}\right\rbrack \leq \frac{{192}{... | Yes |
Corollary 5 Suppose that the stepsizes \( \left\{ {\alpha }_{k}\right\} ,\left\{ {\beta }_{k}\right\} \), and \( \left\{ {\lambda }_{k}\right\} \) in Algorithm 4 are set to (2.27) and (2.30), respectively, and \( \left\{ {p}_{k}\right\} \) is set to (3.7). Also assume that an optimal solution \( {x}^{ * } \) exists for... | Proof. By (3.38), we have \[ \frac{{\sigma }^{2}}{{L}_{\Psi }{N}^{3}}\mathop{\sum }\limits_{{k = 1}}^{N}\frac{{k}^{2}}{{m}_{k}} \leq \frac{{\widetilde{D}}^{2}}{{N}^{3}}\mathop{\sum }\limits_{{k = 1}}^{N}{k}^{2}\max \left\{ {\frac{{L}_{f}}{k},\frac{{L}_{\Psi }}{{k}^{2}N}}\right\} \leq \frac{{\widetilde{D}}^{2}}{{N}^{3}}... | Yes |
Corollary 6 Suppose that the stepsizes \( \left\{ {\alpha }_{k}\right\} ,\left\{ {\beta }_{k}\right\} \), and \( \left\{ {\lambda }_{k}\right\} \) in Algorithm 4 are set to (2.27) and (2.30), respectively, and \( \left\{ {p}_{k}\right\} \) is set to (3.7). Also assume that an optimal solution \( {x}^{ * } \) exists for... | Proof. Observe that by (3.44), we have\n\n\[ \n\frac{{\sigma }^{2}}{{L}_{\Psi }{N}^{3}}\mathop{\sum }\limits_{{k = 1}}^{N}\frac{{k}^{2}}{{m}_{k}} \leq \frac{{\widetilde{D}}^{2}}{{N}^{3}}\mathop{\sum }\limits_{{k = 1}}^{N}k \leq \frac{{\widetilde{D}}^{2}}{N}. \n\] \n\nUsing this observation and (3.36), we obtain (3.45). | Yes |
Proposition 1.1.1. Let \( \\left( {A, \\succcurlyeq }\\right) \) be a decision problem. Then,\ni) The strict preference, \( \\succ \), is asymmetric and transitive.\n\nii) The indifference, \( \\sim \), is an equivalence relation, i.e., it is reflexive, symmetric, and transitive.\n\nMoreover, for each triple \( a, b, c... | We now show that, to each decision problem \( \\left( {A, \\succcurlyeq }\\right) \), we can associate a new one whose weak preference is antisymmetric. For each \( a \\in A \), let \( I\\left( a\\right) \\mathrel{\\text{:=}} \\{ b \\in A : b \\sim a\\} \) . Take \( \\left( {A/ \\sim ,{ \\succcurlyeq }_{a}}\\right) \),... | No |
Example 1.2.1. Let \( \left( {A, \succcurlyeq }\right) \) be a decision problem where \( A = \{ a, b, c, d, e\} \) and \( a \succcurlyeq a, a \succcurlyeq c, a \succcurlyeq d, b \succcurlyeq a, b \succcurlyeq b, b \succcurlyeq c, b \succcurlyeq d, b \succcurlyeq e, c \succcurlyeq a \) , \( c \succcurlyeq c, c \succcurl... | In this case, for instance, it would be enough to say that \( A/ \sim = \{ \{ a, c\} ,\{ b, e\} ,\{ d\} \} \) and that \( \{ a, c\} { \succ }_{a}\{ d\} ,\{ b, e\} { \succ }_{a}\{ a, c\} \) , \( \{ b, e\} { \succ }_{a}\{ d\} \) . There are infinitely many utility functions representing \( \succcurlyeq \) . For instance,... | Yes |
Let \( \left( {{\mathbb{R}}^{n},{ \succcurlyeq }_{L}}\right) \) be a decision problem where \( { \succcurlyeq }_{L} \) is the lexicographic order, i.e., for each pair \( x, y \in {\mathbb{R}}^{n}, x{ \succcurlyeq }_{L}y \) if and only if either \( x = y \) or there is \( i \in \{ 1,\ldots, n\} \) such that, for each \(... | To see this, suppose that \( u \) is a utility function representing \( { \succcurlyeq }_{L} \). For each \( x \in \mathbb{R},\left( {x,\ldots, x,1}\right) { \succ }_{L}\left( {x,\ldots, x,0}\right) \). Hence, \( u\left( {x,\ldots, x,1}\right) > u\left( {x,\ldots, x,0}\right) \) and we can find \( f\left( x\right) \in ... | Yes |
Proposition 1.2.1. Let \( A \) be a countable set and \( \left( {A, \succcurlyeq }\right) \) a decision problem. Then, there is a utility function \( u \) representing \( \succcurlyeq \) . | Proof. Since \( A \) is a countable set, we have \( A = \left\{ {{a}_{1},{a}_{2},\ldots }\right\} \) . For each pair \( i, j \in \mathbb{N} \), define\n\n\[ \n{h}_{ij} \mathrel{\text{:=}} \left\{ \begin{array}{ll} 1 & {a}_{i},{a}_{j} \in A\text{ and }{a}_{i} \succ {a}_{j} \\ 0 & \text{ otherwise. } \end{array}\right.\n... | Yes |
Lemma 1.2.2. Let \( \\left( {A, \\succcurlyeq }\\right) \) be a decision problem and assume that \( \\succcurlyeq \) is antisymmetric.\ni) If there is a countable set \( B \\subset A \) that is order dense in \( A \), then \( {A}^{ * } \) is countable.\nii) If there is a utility function representing \( \\succcurlyeq \... | Proof. i) Let \( {A}_{1}^{ * } \) and \( {A}_{2}^{ * } \) be the sets of superior and inferior gap extremes, respectively, and let \( B \\subset A \) be a countable set order dense in \( A \) . If \( \\left( {{a}_{1},{a}_{2}}\\right) \) is a gap, since \( \\succcurlyeq \) is antisymmetric, then there is \( b \\in B \) ... | Yes |
Corollary 1.2.4. Let \( \left( {A, \succcurlyeq }\right) \) be a decision problem. Then, \( \succcurlyeq \) can be represented by a utility function if and only if there is a countable set \( B \subset A \) that is order dense in \( A \) . | Proof. Consider the decision problem \( \left( {A/ \sim ,{ \succcurlyeq }_{a}}\right) \), and recall that \( { \succcurlyeq }_{a} \) is antisymmetric. Then, \( \succcurlyeq \) can be represented by a utility function if and only if \( { \succcurlyeq }_{a} \) can be represented by a utility function. It is easy to check... | Yes |
Proposition 1.3.1. Let \( \\left( {X, \\succcurlyeq }\\right) \) be a convex decision problem. Assume that \( \\succcurlyeq \) is independent and assume that there are \( x, y \\in X \) such that \( y \\succ x \) . Let \( s, t \\in \\left\\lbrack {0,1}\\right\\rbrack \) , \( s > t \) . Then, \( {sy} + \\left( {1 - s}\\... | Proof. First, note that \( t < 1 \) . Since \( \\succcurlyeq \) is independent,\n\n\[ \n\\frac{s - t}{1 - t}y + \\frac{1 - s}{1 - t}x \\succ \\frac{s - t}{1 - t}x + \\frac{1 - s}{1 - t}x = x.\n\]\n\nOn the other hand,\n\n\[ \n{sy} + \\left( {1 - s}\\right) x = {ty} + \\left( {1 - t}\\right) \\left( {\\frac{s - t}{1 - t... | Yes |
Proposition 1.3.4. Let \( \left( {{\Delta A}, \succcurlyeq }\right) \) be a convex decision problem. Then, there is a von Neumann and Morgenstern utility function representing \( \succcurlyeq \) if and only if there is a linear utility function representing \( \succcurlyeq \) . | Proof. Suppose that \( u \) is a von Neumann and Morgenstern utility function representing \( \succcurlyeq \) and define \( \bar{u} : {\Delta A} \rightarrow \mathbb{R} \), for each \( x \in {\Delta A} \), by \( \bar{u}\left( x\right) \mathrel{\text{:=}} \) \( \mathop{\sum }\limits_{{a \in A}}u\left( a\right) x\left( a\... | Yes |
Suppose that a decision maker has to choose between the following two options: i) a sure 1 million dollars and ii) 2 million dollars with probability 0.5 and 0 dollars with probability 0.5 . Expected utility does not imply that the decision maker will be indifferent toward the two lotteries, that is, the decision maker... | A decision maker whose utility for a monetary amount is exactly this monetary amount \( \left( {u\left( x\right) = x}\right) \) is said to be risk neutral; he would be indifferent regarding i) and ii). A risk averse decision maker \( \left( {u\left( x\right) = \sqrt{x}}\right) \) would go for option i) and a risk prone... | Yes |
Example 1.3.2. (Allais paradox (Allais 1953)). The objective of this paradox was to show an inconsistency of actual observed behavior with the independence property. The paradox is as follows. Consider that a decision maker is asked to choose between option \( {O}_{1} \) (a sure win of 1 million euro) and option \( {P}... | Allais paradox has been one of the motivations for the development of alternative utility theories for decision making under risk. However, von Neumann and Morgenstern utility theory is widely used nowadays, since Allais paradox can be interpreted as an example of the appearance of irrational behavior in singular situa... | Yes |
Example 2.1.4. (A second-price auction). In a second-price auction, the rules are the same as in a first-price auction, except that the player who gets the object pays the highest of the bids of the other players. The strategic game that corresponds with this situation is the same as in Example 2.1.3, but now | \[ {u}_{i}\left( a\right) = \left\{ \begin{array}{ll} {v}_{i} - \max \{ {a}_{j} : j \in N, j \neq i\} & i = \min \{ j \in N : {a}_{j} = \mathop{\max }\limits_{{l \in N}}{a}_{l}\} \\ 0 & \text{ otherwise. } \end{array}\right. \] | Yes |
The only Nash equilibrium of the prisoner's dilemma is \( {a}^{ * } = \left( {D, D}\right) \) . | Moreover, as we have already argued, this is the rational behavior in a noncooperative environment. | No |
Example 2.2.2. We now study the Nash equilibria in a Cournot model like the one in Example 2.1.2. We do it under the following assumptions:\n\n- We deal with a duopoly, i.e., \( n = 2 \) .\n\n- For each \( i \in \{ 1,2\} ,{c}_{i}\left( {a}_{i}\right) = c{a}_{i} \), where \( c > 0 \) .\n\n- Let \( d \) be a fixed number... | Now, we compute a Nash equilibrium of this game. For each \( i \in \{ 1,2\} \) and each \( a \in A \), let \( {f}_{i}\left( a\right) \mathrel{\text{:=}} {a}_{i}\left( {d - {a}_{1} - {a}_{2} - c}\right) \) . Then,\n\n\[ \frac{\partial {f}_{1}}{\partial {a}_{1}}\left( a\right) = - 2{a}_{1} + d - {a}_{2} - c\;\text{ and }... | Yes |
It is easy to check that in the second-price auction described in Example 2.1.3, the strategy profile \( {a}^{ * } = \left( {{v}_{1},\ldots ,{v}_{n}}\right) \) satisfies the following condition: for each \( i \in N \), each \( {\widehat{a}}_{i} \in {A}_{i} \), and each \( {a}_{-i} \in {A}_{-i} \) , \( {u}_{i}\left( {{a... | This condition implies that \( {a}^{ * } \) is a Nash equilibrium of this game. Note that, if \( {a}^{ * } \) is played, player 1 gets the object. However, there are Nash equilibria of the second-price auction in which player 1 is not the winner. Take, for instance, \( a = \left( {0,{v}_{1} + 1,0,\ldots ,0}\right) \) . | No |
Consider the strategic game in Example 2.1.5. The strategy profile \( \left( {{L}_{1}{l}_{1},{R}_{2}{r}_{2}}\right) \) is the unique Nash equilibrium of that game. | Observe that, in order to find a Nash equilibrium in a game in which players have finite strategy sets, given the \ | No |
Theorem 2.2.1 (Kakutani fixed-point theorem). Let \( X \subset {\mathbb{R}}^{n} \) be a nonempty, convex, and compact set. Let \( F : X \rightarrow X \) be an upper hemicontinuous, nonempty-valued, closed-valued, and convex-valued correspondence. Then, there is \( \bar{x} \in X \) such that \( \bar{x} \in F\left( \bar{... | Proof. Refer to Section 2.13. | No |
Proposition 2.2.2. Let \( G = \left( {A, u}\right) \) be a strategic game such that, for each \( i \in N \) ,\n\ni) \( {A}_{i} \) is a nonempty and compact subset of \( {\mathbb{R}}^{{m}_{i}} \) .\n\nii) \( {u}_{i} \) is continuous.\n\niii) For each \( {a}_{-i},{u}_{i}\left( {{a}_{-i}, \cdot }\right) \) is quasi-concav... | Proof. Let \( i \in N \) .\n\nNonempty-valuedness: Obvious, since every continuous function defined on a compact set reaches a maximum.\n\nClosed-valuedness: Also straightforward by the continuity of the payoff functions and the compactness of the sets of strategies.\n\nConvex-valuedness: Let \( {a}_{-i} \in {A}_{-i} \... | Yes |
Example 2.3.1. (A finite two-player zero-sum game that is strictly determined). Consider the two-player zero-sum game in Figure 2.3.1. Clearly,  | \[ \Delta \left( {L}_{1}\right) = 2\text{and}\Delta \left( {R}_{1}\right) = 1\text{, so}\lambda = 2\text{. Besides,}\bar{\Lambda }\left( {L}_{2}\right) = 2\text{and}\bar{\Lambda }\left( {R}_{2}\right) = 3\text{, so} \]\( \bar{\lambda } = 2 \) . Hence, the value of this game is 2 and \( {L}_{1} \) and \( {L}_{2} \) are ... | Yes |
Example 2.3.2. (The matching pennies revisited). In this game, introduced in Example 2.2.6, \( \lambda = - 1 \) and \( \bar{\lambda } = 1 \) . Hence, it is not strictly determined. However, consider the following extension of the game. Each player, instead of selecting either \( E \) or \( O \), can randomize over the ... | For each \( {a}_{1} \in \left\lbrack {0,1}\right\rbrack \) ,\n\n\[ \n\Delta \left( {a}_{1}\right) = \mathop{\inf }\limits_{{{a}_{2} \in \left\lbrack {0,1}\right\rbrack }}\left( {\left( {4{a}_{1} - 2}\right) {a}_{2} - 2{a}_{1} + 1}\right) = \left\{ \begin{array}{ll} 2{a}_{1} - 1 & {a}_{1} < 1/2 \\ 0 & {a}_{1} = 1/2 \\ 1... | Yes |
Example 2.3.3. (An infinite two-player zero-sum game that is not strictly determined). Take the two-player zero-sum game \( \left( {\left\lbrack {0,1}\right\rbrack ,\left\lbrack {0,1}\right\rbrack ,{u}_{1}}\right) \), where, for each \( \left( {{a}_{1},{a}_{2}}\right) \in \left\lbrack {0,1}\right\rbrack \times \left\lb... | For each \( {a}_{1} \in \left\lbrack {0,1}\right\rbrack \) ,\n\n\[ \n\Delta \left( {a}_{1}\right) = \mathop{\inf }\limits_{{{a}_{2} \in \left\lbrack {0,1}\right\rbrack }}\frac{1}{1 + {\left( {a}_{1} - {a}_{2}\right) }^{2}} = \left\{ \begin{array}{ll} \frac{1}{1 + {\left( {a}_{1} - 1\right) }^{2}} & {a}_{1} \leq 1/2 \\ ... | Yes |
Example 2.3.4. (An infinite two-player zero-sum game with a value and with optimal strategies only for one player). Consider the two-player zero-sum game \( \left( {\left( {0,1}\right) ,\left( {0,1}\right) ,{u}_{1}}\right) \), where, for each pair \( \left( {{a}_{1},{a}_{2}}\right) \in \left( {0,1}\right) \times \left(... | It is easy to check that, for each \( {a}_{1} \in \left( {0,1}\right) ,\Delta \left( {a}_{1}\right) = 0 \) . Hence, \( \lambda = 0 \) and, for each \( {a}_{2} \in \left( {0,1}\right) ,\bar{\Lambda }\left( {a}_{2}\right) = {a}_{2} \) . Thus, \( \bar{\lambda } = 0 \) . Therefore, the game is strictly determined, its valu... | Yes |
Consider the two-player zero-sum game \( \left( {\left( {0,1}\right) ,\left( {1,2}\right) ,{u}_{1}}\right) \), where, for each pair \( \left( {{a}_{1},{a}_{2}}\right) \in \left( {0,1}\right) \times \left( {1,2}\right) ,{u}_{1}\left( {{a}_{1},{a}_{2}}\right) = {a}_{1}{a}_{2} \) . | It is easy to check that, for each \( {a}_{1} \in \left( {0,1}\right) ,\Delta \left( {a}_{1}\right) = {a}_{1} \). Hence, \( \lambda = 1 \) and, for each \( {a}_{2} \in \left( {1,2}\right) ,\bar{\Lambda }\left( {a}_{2}\right) = {a}_{2} \). Thus, \( \bar{\lambda } = 1 \). Therefore, the game is strictly determined with v... | Yes |
Proposition 2.3.1. Let \( G = \left( {{A}_{1},{A}_{2},{u}_{1}}\right) \) be a two-player zero-sum game and let \( \left( {{a}_{1}^{ * },{a}_{2}^{ * }}\right) \in {A}_{1} \times {A}_{2} \) be a Nash equilibrium of \( G \) . Then:\n\ni) \( G \) is strictly determined.\n\nii) \( {a}_{1}^{ * } \) is optimal for player 1 an... | Proof. By Eq. (2.3.1) we have:\n\n\[ \text{-}\underline{\lambda } = \mathop{\sup }\limits_{{{\widehat{a}}_{1} \in {A}_{1}}}\underline{\Delta }\left( {\widehat{a}}_{1}\right) \geq \underline{\Delta }\left( {a}_{1}^{ * }\right) = \mathop{\inf }\limits_{{{\widehat{a}}_{2} \in {A}_{2}}}{u}_{1}\left( {{a}_{1}^{ * },{\wideha... | Yes |
Proposition 2.3.2. Let \( G = \left( {{A}_{1},{A}_{2},{u}_{1}}\right) \) be a two-player zero-sum game. Let \( G \) be strictly determined and let \( {a}_{1} \in {A}_{1} \) and \( {a}_{2} \in {A}_{2} \) be optimal strategies of players 1 and 2, respectively. Then \( \left( {{a}_{1},{a}_{2}}\right) \) is a Nash equilibr... | Proof. Since \( {a}_{1} \) and \( {a}_{2} \) are optimal strategies we have that, for each \( {\widehat{a}}_{1} \in {A}_{1} \) and each \( {\widehat{a}}_{2} \in {A}_{2} \) ,\n\n\[ \n{u}_{1}\left( {{\widehat{a}}_{1},{a}_{2}}\right) \leq \bar{\Lambda }\left( {a}_{2}\right) = V = \Delta \left( {a}_{1}\right) \leq {u}_{1}\... | Yes |
Consider the matching pennies game (see Example 2.2.6). Suppose that the players, besides choosing \( E \) or \( O \), can choose a lottery \( L \) that selects \( E \) with probability \( 1/2 \) and \( O \) with probability \( 1/2 \) (think, for instance, of a coin toss). The players have von Neumann and Morgenstern u... | Observe that this game has a Nash equilibrium: \( \left( {L, L}\right) \) . The mixed extension of the matching pennies is a new strategic game in which players can choose not only \( L \), but also any other lottery over \( \{ E, O\} \) . It is easy to check that the only Nash equilibrium of the mixed extension of the... | Yes |
Theorem 2.4.1. Let \( G = \left( {A, u}\right) \) be a finite strategic game. Then, the mixed extension of \( G, E\left( G\right) \), has, at least, one Nash equilibrium. | Proof. Easily follows from Nash theorem. | No |
Proposition 2.4.2. Let \( G \) be a finite game and \( E\left( G\right) \) its mixed extension. Then, for each \( i \in N \), each \( {s}_{i} \in {S}_{i} \), and each \( s \in S \), the following properties hold:\n\ni) \( {s}_{i} \in {\mathrm{{BR}}}_{i}\left( {s}_{-i}\right) \) if and only if \( \mathcal{S}\left( {s}_{... | Proof. It is clear that \( {u}_{i}\left( s\right) = \mathop{\sum }\limits_{{{a}_{i} \in {A}_{i}}}{u}_{i}\left( {{s}_{-i},{a}_{i}}\right) {s}_{i}\left( {a}_{i}\right) \) . From this fact, the proposition immediately follows. | No |
Example 2.5.1. (The battle of the sexes). A couple is deciding where to go for the evening. He (player 1) prefers to go to the cinema, but she (player 2) would like to go to the theater. Finally, both prefer to go together to the least preferred show than being alone at their first options. Figure 2.5.1 shows the strat... | is a coordination problem: there are two Nash equilibria in pure strategies, \( \left( {C, C}\right) \) and \( \left( {T, T}\right) \), but player 1 prefers \( \left( {C, C}\right) \) while player 2 prefers \( \left( {T, T}\right) \) . Just to illustrate the procedure we described above, we compute all the Nash equilib... | Yes |
A thief is deciding whether or not to steal from a warehouse tonight. A guard who works for the owner of the warehouse, guards it during the night. The guard is deciding whether to sleep or not tonight while he is supposed to be guarding the warehouse. If the thief steals and the guard sleeps, the thief will get a very... | identify a strategy of player 1 (the thief) with the probability that he steals \( \left( {x \in \left\lbrack {0,1}\right\rbrack }\right) \), and a strategy of player 2 (the guard) with the probability that he sleeps \( \left( {y \in \left\lbrack {0,1}\right\rbrack }\right) \) . Then, it is easy to check that:\n\n\[ \b... | Yes |
Example 2.5.3. This example is taken from Deutsch (1958, 1960) and Straf-fin (1993). It illustrates an application of the prisoner's dilemma game in experimental psychology. In the 1950s, the F-scale was designed by a group of psychologists to test susceptibility to authoritarian ideologies. In the two works mentioned ... | play this game twice: once as player 1, and once as player 2 . The students never knew who the other player was in either case. In fact, every time a student played the role of player 2, player 1 was fictitious: the experimenter was player 1 and chose \( {ND} \) . Deutsch used the following definitions for his analysis... | Yes |
Proposition 2.6.1. Take an \( l \times m \) matrix game given by matrix \( \mathcal{A} \) . Then, for each \( x \in {S}_{l} \) and each \( y \in {S}_{m} \) , \n\n\[ \n\Delta \left( x\right) = \mathop{\min }\limits_{{j \in M}}x{a}_{\cdot j}\;\text{ and }\;\bar{\Lambda }\left( y\right) = \mathop{\max }\limits_{{i \in L}}... | Proof. We only prove the first equality, since the second one is analogous. Let \( x \in {S}_{l} \) . Recall that, for each \( i \in M,{e}_{i} \in {S}_{m} \) denotes the mixed strategy that selects \( i \) with probability 1; hence, according to the notation in this section, \( {e}_{i} \in {\mathbb{R}}^{m} \) and \( {\... | Yes |
This example explores the use of mixed strategies in penalty kicks in soccer. A penalty kick involves two players, the kicker (player 1) and the goalkeeper (player 2). A goalkeeper, due to the speed of a typical kick and his own reaction time, cannot wait until the kicker has kicked the ball to start his movement. Thus... | Given these probabilities, the unique Nash equilibrium is given by the mixed strategy \( \left( {{0.38},{0.62}}\right) \) for the kicker and \( \left( {{0.42},{0.58}}\right) \) for the goalkeeper. \( {}^{15} \) One of the main findings in Palacios-Huerta (2003) is that the observed frequencies chosen by the professiona... | Yes |
In this work, Davenport illustrates with an example that social behavior patterns can sometimes be functional responses to problems that the society must solve. Davenport studied a small village in Jamaica, where about two hundred inhabitants made their living by fishing. The fishing grounds can be divided into inside ... | Then, the captains of the canoes faced a decision problem that can be treated as a \( 3 \times 2 \) matrix game. If we solve this game by any of the two methods described in Section 2.7 below, the result is that the first player has a unique optimal strategy \( \left( {{0.67},0,{0.33}}\right) \) . Surprisingly enough, ... | No |
Take the matrix game given by\n\n\[ \mathcal{A} = \left( \begin{array}{lll} 2 & 2 & 3 \\ 1 & 4 & 3 \end{array}\right) \] | In Figure 2.7.2, we show the segments corresponding to the columns of \( \mathcal{A} \) and their lower envelope. We have that \( V = 2,{O}_{1}\left( \mathcal{A}\right) = \{ \left( {1,0}\right) \} \), and \( {O}_{2}\left( \mathcal{A}\right) = \operatorname{conv}\left( {\{ \left( {1,0,0}\right) ,\left( {2/3,1/3,0}\right... | No |
Now we solve the matrix game given by\n\n\[ \mathcal{A} = \left( \begin{array}{ll} 0 & 3 \\ 2 & 2 \\ 3 & 0 \end{array}\right) \] | Note first that, although this is not a \( 2 \times m \) matrix game, it can be solved with our method by interchanging the names of the players. \( {}^{16} \) The resulting\n\n--- \n\n\( {}^{16} \) Another possibility would be to design the method for \( l \times 2 \) matrix games; and this design would be very simila... | Yes |
Proposition 2.7.1. Let \( \mathcal{A} \) be an \( l \times m \) matrix game. Let \( x \in {S}_{l} \) and \( y \in {S}_{m} \) . Then,\ni) \( x \in {O}_{1}\left( \mathcal{A}\right) \) if and only if, for each \( j \in M, x{a}_{\cdot j} \geq V \) .\nii) \( y \in {O}_{2}\left( \mathcal{A}\right) \) if and only if, for each... | Proof. We only prove the first equivalence, the second one being analogous. Suppose that \( x \in {O}_{1}\left( \mathcal{A}\right) \) . Then, by Proposition 2.6.1, \( V = \Delta \left( x\right) = \) \( \mathop{\min }\limits_{{j \in M}}x{a}_{\cdot j} \) and, hence, for each \( j \in M, x{a}_{\cdot j} \geq V \) . Convers... | Yes |
Proposition 2.7.2. Let \( \mathcal{A} \) be an \( l \times m \) matrix game. Then, \( {O}_{1}\left( \mathcal{A}\right) \) and \( {O}_{2}\left( \mathcal{A}\right) \) are convex and compact sets. | Proof. In view of the last proposition, \( {O}_{1}\left( \mathcal{A}\right) \) and \( {O}_{2}\left( \mathcal{A}\right) \) are convex sets. Since the sets of strategies are bounded sets, \( {O}_{1}\left( \mathcal{A}\right) \) and \( {O}_{2}\left( \mathcal{A}\right) \) are also bounded sets. Since \( \Delta \) and \( \ba... | Yes |
Proposition 2.7.3. Let \( \mathcal{A} \) be an \( l \times m \) matrix game. Let \( x \in {S}_{l} \) and \( y \in {S}_{m} \) . Then, \( x \in {O}_{1}\left( \mathcal{A}\right) \) and \( y \in {O}_{2}\left( \mathcal{A}\right) \) if and only if, for each \( i \in L \) and each \( j \in M \) , \( x{a}_{\cdot j} \geq {a}_{i... | Proof. The \ | No |
Proposition 2.7.4. Let \( K \in \mathbb{R} \) and let \( \mathcal{A} \) and \( \mathcal{B} \) be two \( l \times m \) matrix games such that, for each \( i \in L \) and each \( j \in M,{b}_{ij} = {a}_{ij} + K \) . Then, \( {V}_{\mathcal{B}} = {V}_{\mathcal{A}} + K \) , \( {O}_{1}\left( \mathcal{A}\right) = {O}_{1}\left... | Proof. Since, for each \( x \in {S}_{l} \) and each \( y \in {S}_{m},{\Delta }_{\mathcal{B}}\left( x\right) - {\Delta }_{\mathcal{A}}\left( x\right) = {\bar{\Lambda }}_{\mathcal{B}}\left( y\right) - {\bar{\Lambda }}_{\mathcal{A}}\left( y\right) = K \), the result is straightforward. | Yes |
Theorem 2.7.5 (Krein-Milman theorem). Let \( S \subset {\mathbb{R}}^{m} \) be a nonempty, convex, and compact set. Then, \( i)\operatorname{ext}\left( S\right) \neq \varnothing \) and ii) \( \operatorname{conv}\left( {\operatorname{ext}\left( S\right) }\right) = S \) . | Proof. Refer to Section 2.14. | No |
Lemma 2.8.1 (Farkas lemma). Let \( \mathcal{A} \) be an \( l \times m \) matrix and let \( c \) be an 1 \times \( m \) vector. Then, one and only one of the following systems of inequalities has a solution:\ni) \( x\mathcal{A} \leq 0 \) and \( c{x}^{t} > 0 \).\nii) \( \mathcal{A}{y}^{t} = {c}^{t} \) and \( y \geq 0 \). | Proof. Suppose first that ii) has a solution \( y \in {\mathbb{R}}^{m} \) and let \( x \in {\mathbb{R}}^{l} \) be such that \( x\mathcal{A} \leq 0 \) . Then, since \( y \geq 0 \) and \( x\mathcal{A} \leq 0, c{x}^{t} = y{\mathcal{A}}^{t}{x}^{t} = x\mathcal{A}{y}^{t} \leq 0 \) . Thus, i) has no solution.\n\nConversely, s... | Yes |
Theorem 2.8.3 (Duality theorem). Let (P) and (D) be a pair of dual linear programming problems. Then,\ni) (P) has an optimal solution if and only if (D) has an optimal solution.\n\nii) Let \( x \) and \( y \) be feasible solutions of \( \left( \mathrm{P}\right) \) and \( \left( \mathrm{D}\right) \), respectively. Then,... | Proof. We prove both statements together. Let \( \widehat{x} \) and \( \widehat{y} \) be feasible solutions of (P) and (D), respectively. Then,\n\n(2.8.6)\n\n\[ c{\widehat{x}}^{t} \geq \widehat{y}{\mathcal{A}}^{t}{\widehat{x}}^{t} = \widehat{y}{\left( \widehat{x}\mathcal{A}\right) }^{t} \geq \widehat{y}{b}^{t} \]\n\nwh... | Yes |
Proposition 2.8.4. The following two statements hold:\n\ni) Let \( \\left( {x,\\Delta }\\right) \) be an optimal solution of (2.8.8.a). Let \( v \\in {\\mathbb{R}}^{l} \) be defined, for each \( i \\in \\{ 1,\\ldots, l\\} \), by \( {v}_{i} \\mathrel{\\text{:=}} {x}_{i}/\\Delta \) . Then, \( v \) is an optimal solution ... | ## Proof. Exercise 2.12. | No |
Proposition 2.8.5. The following two statements hold:\n\ni) Let \( \left( {y,\bar{\Lambda }}\right) \) be an optimal solution of (2.8.10.a). Let \( w \in {\mathbb{R}}^{m} \) be defined, for each \( j \in \{ 1,\ldots, m\} \), by \( {w}_{j} \mathrel{\text{:=}} {y}_{j}/\bar{\Lambda } \) . Then, \( w \) is an optimal solut... | Proof. Exercise 2.12. | No |
Proposition 2.8.6. Let \( \mathcal{A} \) be an \( m \times m \) symmetric matrix game. Then, \( {O}_{1}\left( \mathcal{A}\right) = \) \( {O}_{2}\left( \mathcal{A}\right) \) and \( V = 0 \) . | Proof. Let \( x, y \in {S}_{m} \) be such that \( x \in {O}_{1}\left( \mathcal{A}\right) \) and \( y \in {O}_{2}\left( \mathcal{A}\right) \) . Then, for each pair \( \widehat{x},\widehat{y} \in {S}_{m},\widehat{x}\mathcal{A}{y}^{t} \leq x\mathcal{A}{y}^{t} \leq x\mathcal{A}{\widehat{y}}^{t} \) . Equivalently, after tak... | Yes |
Proposition 2.8.7. Let \( \mathcal{A} \) be an \( l \times m \) matrix game.\n\ni) \( {a}_{\cdot j} \) is a relevant column if and only if, for each \( x \in {O}_{1}\left( \mathcal{A}\right), x{a}_{\cdot j} = V \) .\n\nii) \( {a}_{i} \) . is a relevant row if and only if, for each \( y \in {O}_{2}\left( \mathcal{A}\rig... | Proof. We prove only the first statement; the second one being analogous. Let \( {a}_{\cdot j} \) be a relevant column. Recall that, for each \( x \in {O}_{1}\left( \mathcal{A}\right) \) and each \( k \in M, x{a}_{\cdot k} \geq V \) . Suppose there is \( x \in {O}_{1}\left( \mathcal{A}\right) \) such that \( x{a}_{\cdo... | Yes |
Theorem 2.8.8. A pair of dual linear programming problems (P) and (D) have optimal solutions if and only if the associated matrix game \( \mathcal{B} \) has an optimal strategy whose last component is positive. | Proof. First, let \( x \in {\mathbb{R}}^{l}, y \in {\mathbb{R}}^{m} \), and \( \alpha > 0 \) be such that \( \left( {x, y,\alpha }\right) \) is an optimal strategy of \( \mathcal{B} \) . Then,\n\n\[ \left( \begin{matrix} 0 & \mathcal{A} & - {c}^{t} \\ - {\mathcal{A}}^{t} & 0 & {b}^{t} \\ c & - b & 0 \end{matrix}\right)... | Yes |
Consider the bimatrix game in Figure 2.9.1. This game has two Nash equilibria in pure strategies: \( \left( {{L}_{1},{L}_{2}}\right) \) and \( \left( {{R}_{1},{R}_{2}}\right) \) . However, \( \left( {{R}_{1},{R}_{2}}\right) \) is not really self-enforcing. | Suppose that the players have informally agreed to play \( \left( {{R}_{1},{R}_{2}}\right) \) and take, for instance, player 1 . He does not lose if he plays \( {L}_{1} \) instead of \( {R}_{1} \) and, in addition, he might gain by doing this (if player 2 also deviates from \( {R}_{2} \) ). So, he will probably deviate... | Yes |
Theorem 2.9.1. If \( s \in S \) is a perfect equilibrium of \( E\left( G\right) \), then it is a Nash equilibrium of \( E\left( G\right) \) . | Proof. Let \( s \in S \) be a perfect equilibrium of \( E\left( G\right) \). Let \( \left\{ {\eta }^{k}\right\} \) and \( \left\{ {s}^{k}\right\} \) be in the conditions of Definition 2.9.3. Note that, for each \( {s}^{k} \in S \) and each \( i \in N \), \[ {u}_{i}\left( {s}^{k}\right) = \mathop{\sum }\limits_{{{a}_{i}... | Yes |
Theorem 2.9.2. The mixed extension of a finite game \( G \) has, at least, one perfect equilibrium. | Proof. Let \( \left\{ {\eta }^{k}\right\} \subset T\left( G\right) \), with \( \left\{ {\eta }^{k}\right\} \rightarrow 0 \) . For each \( k \in \mathbb{N} \), let \( {s}^{k} \) be a Nash equilibrium of \( \left( {G,{\eta }^{k}}\right) \) . Since \( S \) is a compact set, the sequence \( \left\{ {s}^{k}\right\} \subset ... | Yes |
Proposition 2.9.3. Let \( E\left( G\right) \) be the mixed extension of a finite game \( G \) and let \( s \in S \) . The following three statements are equivalent.\n\ni) \( s \) is a perfect equilibrium of \( E\left( G\right) \) .\n\nii) There are two sequences \( \left\{ {\varepsilon }^{k}\right\} \subset \left( {0,\... | Proof. i) \( \Rightarrow \) ii). Since \( s \) is perfect, there are two sequences \( \left\{ {\eta }^{k}\right\} \) and \( \left\{ {s}^{k}\right\} \) in the conditions of Definition 2.9.3. Recall that if \( {s}^{k} \) is a Nash equilibrium of \( \left( {G,{\eta }^{k}}\right) \), then only best replies are chosen with ... | Yes |
Lemma 2.9.4. Let \( G = \left( {\left\{ {{S}_{l},{S}_{m}}\right\}, u}\right) \) be a bimatrix game and let \( {\bar{s}}_{1} \in {S}_{l} \) . Then, \( {\bar{s}}_{1} \) is undominated if and only if \( V\left( {G}_{{\bar{s}}_{1}}\right) = 0 \) and each \( j \in M \) belongs to the support of some optimal strategy of play... | Proof. First, note that \( V\left( {G}_{{\bar{s}}_{1}}\right) \geq 0\left( {\bar{s}}_{1}\right. \) ensures payoff 0 to player 1). The strategy \( {\bar{s}}_{1} \) is dominated by \( {s}_{1} \) in \( G \) if and only if, for each \( j \in M,{u}_{1}\left( {{s}_{1}, j}\right) \geq \) \( {u}_{1}\left( {{\bar{s}}_{1}, j}\ri... | Yes |
Theorem 2.9.5. Let \( E\left( G\right) \) be the mixed extension of a finite game \( G \) . Then:\n\ni) Every perfect equilibrium of \( E\left( G\right) \) is an undominated Nash equilibrium of \( E\left( G\right) \) .\n\nii) If \( G \) is a two-player game, i.e., if \( E\left( G\right) \) is a bimatrix game, then a st... | Proof. Statement i) is quite straightforward by Theorem 2.9.1 and the fact that every strategy profile that satisfies statement iii) in Proposition 2.9.3 is undominated. Now, because of i), only one implication has to be proved to prove ii). Let \( G = \left( {\left\{ {{S}_{l},{S}_{m}}\right\}, u}\right) \) be a bimatr... | Yes |
Corollary 2.9.6. The mixed extension of a finite game \( G \) has, at least, one undominated Nash equilibrium. | Proof. Follows from the combination of Theorem 2.9.2 and the first statement in Theorem 2.9.5. | No |
Example 2.9.4. Consider the mixed extension of the finite two-player game in Figure 2.9.4. This game is a modification of the game in Example 2.9.1 | after the addition of a dominated strategy for each player. By statement ii) in Theorem 2.9.5, since we have a bimatrix game and \( \left( {{R}_{1},{R}_{2}}\right) \) is an undominated Nash equilibrium, it is also a perfect equilibrium. Alternatively, the perfection of \( \left( {{R}_{1},{R}_{2}}\right) \) can be check... | Yes |
Proposition 2.9.7. Let \( G \) be a finite game in which all players have two strategies. Then, \( s \) is a proper equilibrium of \( E\left( G\right) \) if and only if it is perfect. | Proof. It is sufficient to prove that, under the above conditions, perfect implies proper. Let \( s \in S \) a perfect equilibrium of \( E\left( G\right) \) and let \( \left\{ {\varepsilon }^{k}\right\} \) and \( \left\{ {s}^{k}\right\} \) be two sequences as in statement ii) in Proposition 2.9.3. For each \( k \in \ma... | Yes |
Theorem 2.9.8. The mixed extension of a finite game \( G \) has, at least, one proper equilibrium. | Proof. We show first that, for each \( k \in \mathbb{N} \smallsetminus \{ 1\} \), there is a \( \frac{1}{k} \) -proper equilibrium of \( E\left( G\right) \) . Let \( k \in \mathbb{N} \smallsetminus \{ 1\} \) . Let \( m \mathrel{\text{:=}} \mathop{\max }\limits_{{i \in N}}\left| {A}_{i}\right| \) and \( {\delta }_{k} \m... | Yes |
Take the mixed extension of the finite three-player game in Figure 2.9.5. This game can be seen as a modification of the game in Example 2.9.1 after adding a dominated strategy for player three. Note that \( \left( {{R}_{1},{R}_{2},{L}_{3}}\right) \) is now a proper equilibrium. | To check it, simply take the sequences \( \left\{ {\varepsilon }^{k}\right\} \) and \( \left\{ {s}^{k}\right\} \) given, for each \( k \in \mathbb{N} \), by \( {\varepsilon }^{k} \mathrel{\text{:=}} \frac{1}{k + {10}},{s}_{1}^{k} \mathrel{\text{:=}} \) \( \left( {\frac{1}{k + {100}},1 - \frac{1}{k + {100}}}\right) ,{s}... | Yes |
Consider the mixed extension of the finite two-player game in Figure 2.9.6. This game has two pure Nash equilibria that, moreover, are proper: \( \left( {{L}_{1},{L}_{2}}\right) \) and \( \left( {{R}_{1},{R}_{2}}\right) \). However, the strategy profile \( \left( {{R}_{1},{R}_{2}}\right) \) is not really self-enforcing... | Suppose that the players have informally agreed to play \( \left( {{R}_{1},{R}_{2}}\right) \), and take player 2 . If player 1 sticks to the agreement, player 2 is indifferent between playing \( {L}_{2} \) or \( {R}_{2} \) . Yet, if player 2 deviates, he might gain (if player 1 also deviates from \( {R}_{1} \) and choo... | Yes |
Proposition 2.9.9. Let \( G \) be a finite game and \( E\left( G\right) \) its mixed extension. Let \( s \in S \) be a strict equilibrium of \( E\left( G\right) \) . Then, \( s \) is a strictly perfect equilibrium. | Proof. Let \( s \in S \) be a strict equilibrium of \( E\left( G\right) \) . Recall that \( s \) has to be a pure strategy profile, i.e., there is \( a \in A \) such that \( s = a \) . Let \( \left\{ {\eta }^{k}\right\} \subset T\left( G\right) \) with \( \left\{ {\eta }^{k}\right\} \rightarrow 0 \) . For each \( k \in... | Yes |
Every (pure) Nash equilibrium \( a \in A \) of \( G \) induces a correlated equilibrium of \( G,\left( {I,{\tau }^{ * }}\right) \) | \[ \text{-}I = \left( {\Omega ,\rho ,{\left\{ {\mathcal{P}}_{i}\right\} }_{i \in N}}\right) \text{, where}\Omega = \{ a\} ,\rho \left( a\right) = 1\text{, and, for each}i \in N\text{,}\]\n\[{\mathcal{P}}_{i} = \{ \{ a\} \}\]\n\[ \text{- For each}i \in N,{\tau }_{i}^{ * }\left( a\right) = {a}_{i}\text{.} \] | Yes |
In the battle of the sexes (Example 2.5.1), the following specification of \( \\left( {I,{\\tau }^{ * }}\\right) \) is a correlated equilibrium: | \[ \text{-}I = \\left( {\\Omega ,\\rho ,{\\left\\{ {\\mathcal{P}}_{i}\\right\\} }_{i \\in N}}\\right) \\text{, where}\\Omega = \\{ C, T\\} ,\\rho \\left( C\\right) = \\rho \\left( T\\right) = 1/2\\text{, and} \] \[ {\\mathcal{P}}_{1} = {\\mathcal{P}}_{2} = \\{ \\{ C\\} ,\\{ T\\} \\} \] \[ \\text{-}{\\tau }_{1}^{ * }\\l... | Yes |
Theorem 2.11.3. Let \( G \) be a finite strategic game. Then, the set of correlated equilibrium payoff vectors of \( G \) is convex. | Proof. Let \( {v}^{1},{v}^{2} \in {\mathbb{R}}^{n} \) be a pair of correlated equilibrium payoff vectors of \( G \) . Let \( {\alpha }_{1},{\alpha }_{2} \in \mathbb{R} \) be such that \( {\alpha }_{1} \geq 0,{\alpha }_{2} \geq 0 \), and \( {\alpha }_{1} + {\alpha }_{2} = 1 \) . We now show that there is a correlated eq... | Yes |
Consider the finite two-player game \( G \) in Figure 2.11.1. Different variations of this game are referred to as the chicken game or the hawk-dove game. The chicken game represents a situation in which two drivers drive towards each other; one must swerve (play \( L \) ) to avoid the crash, but if only one swerves, h... | Let \( \left( {I,{\tau }^{ * }}\right) \) be given by \[ \text{-}I = \left( {\Omega ,\rho ,{\left\{ {\mathcal{P}}_{i}\right\} }_{i \in N}}\right) \text{, where}\Omega = \left\{ {{\omega }_{1},{\omega }_{2},{\omega }_{3}}\right\} ,\rho \left( {\omega }_{1}\right) = \rho \left( {\omega }_{2}\right) = \] \[ \rho \left( {\... | Yes |
Theorem 2.12.1. Let \( G \) be a finite game and \( I \in {\mathcal{I}}^{G} \) an information model with common prior. If each player is rational at each state of the world, then the pair \( \left( {I, a\left( \cdot \right) }\right) \) is a correlated equilibrium. | Proof. Let \( I \mathrel{\text{:=}} \left( {\Omega ,\rho ,{\left\{ {\mathcal{P}}_{i}\right\} }_{i \in N}}\right) \in {\mathcal{I}}^{G} \), where \( \rho \) is the common prior. First, note that \( a\left( \cdot \right) \) is an \( I \) -consistent correlated strategy profile. Since all the players are rational at each ... | Yes |
Corollary 2.12.2. Let \( G \) be a finite game and \( I \in {\mathcal{I}}^{G} \) an information model with common prior. Assume that for each pair \( i, j \in N \), each \( {P}_{i} \in {\mathcal{P}}_{i} \), and each \( {P}_{j} \in {\mathcal{P}}_{j} \) the events \( {P}_{i} \) and \( {P}_{j} \) are independent according... | Proof. Follows from Theorem 2.12.1 and the fact that for any such information model a correlated strategy profile corresponds with a mixed strategy profile. | No |
Lemma 2.12.3. Let \( G \) be a finite game and let \( {\bar{a}}_{i} \in {A}_{i} \) . Then, the following two statements are equivalent:\n\ni) There is \( \alpha \in {\mathcal{C}}_{-i}\left( {A}_{-i}\right) \) such that \( {\bar{a}}_{i} \in {\mathrm{{BR}}}_{i}\left( \alpha \right) \).\n\nii) The strategy \( {\bar{a}}_{i... | Proof. \( {}^{39} \)\n\ni) \( \Rightarrow \) ii) Suppose that there is \( {s}_{i} \in {S}_{i} \) that strictly dominates \( {\bar{a}}_{i} \) . Then, for each \( \beta \in {\mathcal{C}}_{-i}\left( {\widehat{A}}_{-i}\right) ,{u}_{i}\left( {\beta ,{s}_{i}}\right) > {u}_{i}\left( {\beta ,{\bar{a}}_{i}}\right) \) . Hence, \... | Yes |
Theorem 2.12.4. Let \( G \) be a finite game and \( I \in {\mathcal{I}}^{G} \) be an information model. If there is common knowledge of rationality at each state of the world, then, for each \( i \in N \) and each \( \omega \in \Omega \), the strategy \( {a}_{i}\left( \omega \right) \in {A}_{i} \) is rationalizable. | Proof. The first part of the statement essentially follows from the discussion preceding the definition of rationalizable strategy (Definition 2.12.5). Let \( I \in {\mathcal{I}}^{G} \) be an information model with common knowledge of rationality. Let \( i \in N \) and \( \omega \in \Omega \) . By common knowledge of r... | Yes |
Proposition 2.12.5. Let \( \left( {{I}^{ * },{\tau }^{ * }}\right) \) be a correlated equilibrium of \( G \), with \( {I}^{ * } \in {\mathcal{I}}^{G} \) . Let \( {a}_{i} \in {A}_{i} \) be a strategy that is played with positive probability according to the common prior. Then, \( {a}_{i} \) is rationalizable, i.e., \( {... | Proof. For each \( i \in N \), let \( {\bar{A}}_{i} \) be the set of strategies of player \( i \) that are played with positive probability according to \( \left( {{I}^{ * },{\tau }^{ * }}\right) \) . Let \( \widetilde{A} \mathrel{\text{:=}} \mathop{\prod }\limits_{{i \in N}}{\bar{A}}_{i} \) . To prove the result it su... | Yes |
Theorem 2.13.2 (Knaster-Kuratowski-Mazurkiewicz theorem). Let \( S \) be the simplex in \( {\mathbb{R}}^{n} \) defined by \( S \mathrel{\text{:=}} \operatorname{conv}\left( {\left\{ {x}^{i}\right\} }_{i \in I}\right) \) . Let \( {\left\{ {A}^{i}\right\} }_{i \in I} \) be such that \( i \) ) for each \( i \in I,{A}^{i} ... | Proof. Let \( {\left\{ {\mathcal{S}}^{m}\right\} }_{m \in \mathbb{N}} \) be a sequence of dissections of \( S \) such that \( \left\{ {d}_{{\mathcal{S}}^{m}}\right\} \rightarrow 0 \) . For each \( m \in \mathbb{N} \), take the Sperner labelling \( {f}_{l}^{m} \) associated with \( S \) and \( {\mathcal{S}}^{m} \) defin... | Yes |
Theorem 2.14.1 (Separation of a point and a convex set). Let \( S \subset {\mathbb{R}}^{n} \) be a nonempty and convex set and let \( y \in {\mathbb{R}}^{n} \smallsetminus S \) . Then, there is a hyperplane \( H\left( {v, r}\right) \) that separates \( \{ y\} \) from \( S \) . If \( S \) is also closed, the separation ... | Proof. First of all, let \( \bar{x} \in {\mathbb{R}}^{n} \) be such that \( \parallel \bar{x} - y\parallel = \mathop{\inf }\limits_{{x \in S}}\parallel x - y\parallel \) .\n\nWe begin with the strict separation of a point from a nonempty, closed, and convex set \( S \) . Since \( S \) is closed, \( \bar{x} \in S \) and... | Yes |
Theorem 2.14.2 (Supporting hyperplane theorem). Let \( S \subset {\mathbb{R}}^{n} \) be a nonempty and convex \( n \) -dimensional set and let \( y \in \partial S \) . Then, there is a supporting hyperplane for \( S \) at \( y \) . | Proof. The set \( \widehat{S} \) is nonempty and convex and, moreover, \( y \notin \widehat{S} \) . Hence, by Theorem 2.14.1, there are \( v \in {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) and \( r \in \mathbb{R} \) such that \( H\left( {v, r}\right) \) separates \( \{ y\} \) and \( S \) . Thus, \( H\left( {v, r}\right) ... | Yes |
Theorem 2.14.3 (Separating hyperplane theorem). Let \( S \) and \( \widehat{S} \) be two disjoint, nonempty, and convex subsets of \( {\mathbb{R}}^{n} \). Then, there is a separating hyperplane for \( S \) and \( \widehat{S} \). | Proof. Let \( S - \widehat{S} \mathrel{\text{:=}} \left\{ {z \in {\mathbb{R}}^{n} : z = x - y, x \in S}\right. \) and \( \left. {y \in \widehat{S}}\right\} \). The convexity of \( S \) and \( \widehat{S} \) implies the convexity of \( S - \widehat{S} \). Since, \( S \cap \widehat{S} = \varnothing ,0 \notin S - \dot{\wi... | Yes |
Consider the two 1-player games depicted in Figure 3.2.2. | We begin by discussing the game \( {\Gamma }^{1} \), which does not have perfect recall since, at \( w \), the player does not know what he played at \( r \) . Let \( s \) be the mixed strategy that selects the pure strategy \( \left( {{L}_{1},{L}_{2}}\right) \) with probability \( 1/2 \) and \( \left( {{R}_{1},{R}_{2}... | Yes |
Proposition 3.3.1. Let \( \Gamma \) be an extensive game and let \( {a}^{ * } \in A \) . Then, \( {a}^{ * } \) is a Nash equilibrium of \( \Gamma \) in behavior strategies if and only if, for each \( i \in N \) and each \( {\widehat{a}}_{i} \in {A}_{i} \)\n\n\[ \n{u}_{i}\left( {a}^{ * }\right) \geq {u}_{i}\left( {{a}_{... | Proof. Since \( A \subset B \), if \( {a}^{ * } \) is a Nash equilibrium of \( \Gamma \) in behavior strategies, then Definition 3.3.2 implies that, for each \( i \in N \) and each \( {a}_{i} \in {A}_{i},{u}_{i}\left( {a}^{ * }\right) \geq \) \( {u}_{i}\left( {{a}_{-i}^{ * },{a}_{i}}\right) \) . Conversely, let \( {a}^... | Yes |
Theorem 3.3.2. Let \( \Gamma \) be an extensive game with perfect recall. Then, \( \Gamma \) has, at least, one Nash equilibrium. | Proof. By Nash theorem, the strategic game \( E\left( {G}_{\Gamma }\right) \) has, at least, one Nash equilibrium. Let \( {s}^{ * } \in S \) be one of such equilibria. By Kuhn theorem, there is \( {b}^{ * } \in B \) such that \( {s}^{ * } \) and \( {b}^{ * } \) are realization equivalent and, hence, also payoff equival... | Yes |
Theorem 3.3.3. Let \( \Gamma \) be an extensive game with perfect recall. If \( {b}^{ * } \) is a Nash equilibrium of \( \Gamma \), then \( {b}^{ * } \) is also a Nash equilibrium of \( E\left( {G}_{\Gamma }\right) \) . | Proof. First, recall that \( B \) can be considered as a subset of \( S \) . If \( {b}^{ * } \) is not a Nash equilibrium of \( E\left( {G}_{\Gamma }\right) \), then there are \( i \in N \) and \( {\widehat{s}}_{i} \in {S}_{i} \) such that \( {u}_{i}\left( {b}^{ * }\right) < {u}_{i}\left( {{b}_{-i}^{ * },{\widehat{s}}_... | Yes |
Consider again the interactive situation described in Example 3.1.2 and the corresponding extensive game \( {\Gamma }_{1} \) in Figure 3.1.3. The strategic game associated with \( {\Gamma }_{1} \) is the two-player zero-sum game \( {G}_{{\Gamma }_{1}} \) depicted in Figure 3.3.1. It is easy to check that the unique Nas... | \[ \text{-}{B}_{1} = \left\{ {{b}_{1} = \left( {{\alpha \beta },\alpha \left( {1 - \beta }\right) ,\left( {1 - \alpha }\right) \beta ,\left( {1 - \alpha }\right) \left( {1 - \beta }\right) }\right) : \alpha ,\beta \in \left\lbrack {0,1}\right\rbrack }\right\} \text{.} \] \[ \text{-}{B}_{2} = \left\{ {{b}_{2} = \left( {... | Yes |
Consider again the interactive situation of Example 3.1.2 but, this time, modeled as a three-player game. The corresponding extensive game is \( {\Gamma }_{2} \) (see Figure 3.1.4). Figure 3.3.2 shows its associated strategic game, \( {G}_{{\Gamma }_{2}} \) . | It is easy to see that this game has two pure Nash equilibria: \( \left( {K, K, N}\right) \) and \( \left( {S, S, E}\right) \) . | Yes |
Consider the extensive game \( \Gamma \) in Example 3.1.1. Its corresponding strategic game \( {G}_{\Gamma } \) is shown in Figure 3.3.3. The strategies of player 2 mean the following:\n\n- \( {MM} \) : Player 2 plays \( M \) regardless of the choice of player 1 .\n- \( {MH} \) : Player 2 plays the same as player 1 .\n... | Note that this is a constant-sum two-player game. Analyzing it is the same as analyzing the zero-sum game corresponding to player 1 's payoff function. It is an easy exercise to check that, in constant-sum two-player games, the payoff for a player is the same in all the Nash equilibria of the game (as in the zero-sum c... | No |
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,\ni) \( b \) is a subgame perfect equilibrium of \( \Gamma \) ,\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 \( {... | Yes |
Consider the extensive game \( \Gamma \) of Figure 3.4.4. Since \( \Gamma \) does not have a subgame different from itself, all its Nash equilibria are subgame perfect. Thus \( \left( {{R}_{1},{R}_{2}}\right) \) is a subgame perfect equilibrium of \( \Gamma \) . | However, it is not really self-enforcing because if player \( {2}^{\prime }\mathrm{s} \) information set is reached, he will play \( {L}_{2} \) ; since player 1 is able to realize this, he will play \( {L}_{1} \) . Note that the problem with \( \left( {{R}_{1},{R}_{2}}\right) \) in this example is the same one we had r... | Yes |
Consider the extensive game \( \Gamma \) of Figure 3.5.1 and the strategy profile \( b = \left( {{R}_{1},{L}_{2},{R}_{3}}\right) \). To associate a system of beliefs with \( b \), it suffices to say what are the beliefs at the information set of player 3, i.e., at \( {w}_{3} \). Since \( {w}_{3} \) is not in the path o... | By doing so, we have that \( \left( {b,\mu }\right) \) is a weakly perfect Bayesian equilibrium. However, \( b \) is not subgame perfect. | Yes |
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![758908bd-9288-4780-8b49-f39ec7... | 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}^{... | No |
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 |
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