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The fee has to be paid before June 1st, Without the payment of the course fee, your registration is not complete. You can apply for a scholarship on the online registration form. You will only be eligible for a scholarship if:. Of these students, the first 15 students that have registered will receive a scholarship. Hence, the earlier you register, the higher the chance that you will obtain a scholarship. The payment of this scholarship will be after the course. More precisely, it explores how we reason about the minds of others before we make a choice. Hence, the reasoning of people like you and me is at the center stage of epistemic game theory.

Audience: The course is primarily meant for advanced bachelor students, master students and PhD students all over the world, but researchers are also very welcome. It is highly recommended to purchase this book for the course, as most lectures, and all exercise sessions, will be based on the book. Organization of the course: The course stretches over 11 days, of which the last day consists of a written exam. Typically, a day consists of two theory lectures, and one exercise session. In the exercise sessions we will discuss some problems from the book which you should prepare the day before.

On the last Friday, we will discuss the exam from last year. Your grade will be based on the exam. To pass the exam, you must score at least a 5. Since a1 is a best response to b3 , b3 is a best response to a3 , a3 is a best response to b1 , and b1 is a best response to a1 the actions a1 , a3 , b1 , and b3 are rationalizable.

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Comment That b4 is not rationalizable also follows from Lemma Since the game is symmetric, the set of rationalizable actions is the same for both players; denote it by Z. Hence using Definition By the argument in the solution to Exercise The result follows from Definition This belief must assign positive probability to both points in the set otherwise the best response is to locate at one of the points. In fact 12 is the only action that survives iterated elimination of strictly dominated actions: on the first round Out is strictly dominated by 21 , and in every subsequent round each of the remaining most extreme actions is strictly dominated by This game is dominance solvable, the only surviving outcome being T, L.

However, if B is deleted then neither of the remaining actions of player 2 is dominated, so that both T, L and T, R survive iterated elimination of dominated actions. At every subsequent round up to 50 one action is eliminated for each player: at the second round this action is , at the third round it is 99, and so on.

After round 50 the single action pair 51, 51 remains, with payoffs of 50, Since U is determined by a finite set, the set B is closed.

## Book:Martin J. Osborne/A Course in Game Theory

Comment This exercise is quite difficult. Knowledge and Equilibrium The proof of the second part follows the lines of the proof of Proposition Hence E is not common knowledge. Comment If this result were not true then a mutually profitable trade between individuals could be made. The existence of such a pair of beliefs is necessary for existence of a rational expectations equilibrium in which the individuals are aware of existing price, take it into consideration, and trade the lottery L even though they risk-neutral.

Then G is the strategic form of the extensive game with perfect information in Figure The other case is similar. Suppose that player 1 moves first. In this game L, AD is a subgame perfect equilibrium, with a payoff of 1, 0 , while the solution of the maximization problem is R, C , with a payoff of 2, 1. Extensive Games with Perfect Information The payoff to the single infinite history is 1.

This game has a finite horizon the length of the longest history is 1 but has no subgame perfect equilibrium since [0, 1 has no maximal element. In the infinite-horizon one-player game the beginning of which is shown in Figure The game has no subgame perfect equilibrium. Figure The payoff to the single infinite history is 0. Assume the hypothesis for all subgames with length at most k. We now show that the equilibria are interchangeable. For any subgame perfect equilibrium we can attach to every subgame the outcome according to the subgame perfect equilibrium if that subgame is reached.

Thus by the one deviation property Lemma Chapter 6. Extensive Games with Perfect Information 17 The first player to move is player 1. We do not specify the game formally. There are two cases. By the induction hypothesis army j does not launch a counterattack in any subgame perfect equilibrium, so that the attack is worthwhile. Extensive Games with Perfect Information In each subgame that follows a proposal x of player 1 there are two types of Nash equilibria. In one equilibrium, which we refer to as Y x , players 2 and 3 both accept x.

In all the remaining equilibria the proposal x is not implemented; we refer to the set of these equilibria as N x. Consequently the equilibria of the entire game are the following. In the subgame that follows the proposal x of player 1, the equilibrium is Y x. In the subgame that follows any proposal y of player 1 in which y1 0, the equilibrium is in N y.

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In the subgame that follows the history Continue there is a unique subgame perfect equilibrium, in which both players choose 0. Thus the game has a unique subgame perfect equilibrium, in which player 1 chooses Stop and, if she chooses Continue, both players choose 0. Note that if the set of actions of each player after player 1 chooses Continue were bounded by some number M then there would be an additional subgame perfect equilibrium in which player 1 chooses Continue and each player names M , with the payoff profile M 2 , M 2.

This game has no subgame perfect equilibrium. Extensive Games with Perfect Information 19 In every period before k both players choose C; in every subsequent period both players choose S. The outcome is that the game stops in period k. In period k player 2 is better off choosing S given the strategy of player 1 , and in subsequent periods the action that each player chooses has no effect on the outcome.

Weakly dominated actions can be iteratively eliminated as follows. D, S, S is weakly dominated for player 1 by 0, B, B Every strategy a, b of player 2 in which either a or b is D, B is weakly dominated by the strategy that differs only in that D, B is replaced by 0, S. Every strategy x, y, B of player 1 is weakly dominated by x, y, S since there is no remaining strategy of player 2 in which he chooses D, B.

Every strategy a, b of player 2 in which b is either 0, B or 0, S is weakly dominated by the strategy that differs only in that b is replaced by D, S since in every remaining strategy player 1 chooses S after player 2 chooses D. The game that remains is shown in Figure Weakly dominated actions can be eliminated iteratively as follows. DB is weakly dominated for player 1 by 0B 2. BA is weakly dominated for player 2 by AA 5. It is straightforward to check that the strategy pair is a subgame perfect equilibrium.

Let Mi Gi and mi Gi be as in the proof of Proposition By the argument for Then by the argument for It is clear that this is a subgame perfect equilibrium. Note that the definitions of M2 and m1 differ from those in the proof of Proposition A Model of Bargaining 23 the arguments for The following proof of uniqueness is taken from Osborne and Rubinstein Similarly, let m1 and m2 be the infima of these payoffs.

We proceed in a number of steps. The proof is the same as that for Step 2. The proof is analogous to those for Steps 1 and 2. Step 4. Step 5. Step 6. A Model of Bargaining Proof. These inequalities follow from the subgame perfect equilibrium described in the text as in Step 4. Step 7. Hence b 8 Repeated Games Since there is no cycle in this sequence, the strategy cannot be executed by a machine with finitely many states.

Repeated Games The game has a unique subgame perfect equilibrium, in which player 1 chooses D in every period and each of the other players chooses D.

## Osborne-Rubinstein books

Choose a sequence of outcomes C, C and D, D whose average payoff to player 1 is x. Let m be the minimum payoff of any player in a subgame perfect equilibria of the repeated game. The full dimensionality assumption in Proposition Chapter 8. Repeated Games 27 If a player can gain by a one-period deviation then the strategy profile is obviously not a subgame perfect equilibrium. Now assume that no player can gain by a one-period deviation from s after any history but there is a history h after which player i can gain by switching to the strategy s0i.

Thus we can assume that there exists some period T such that s0i differs from si only in the first T periods. To cover this case, the strategy in the proof of Proposition Suppose that player 2 uses a machine with a cycle of length 2, player 3 uses a machine with a cycle of length 3, and player 1 wants to coordinate with players 2 and 3. Then player 1 needs to have six states in her machine. Then M1 , M2 , M3 is a Nash equilibrium of the machine game.

To show that a player cannot obtain the same payoff in the repeated game by a less complex machine assume that player 1 uses a machine M1 with fewer than five states and player 2 uses the machine M. Thus M, M is a Nash equilibrium of the machine game. The new pair of machines is not a Nash equilibrium since a player can obtain the same payoff by omitting the state I3 and transiting from I2 to R2 if the other player chooses D.

Now suppose that there is an equilibrium that has no introductory phase. Denote the states in the cycle by q 1 ,. Suppose that in state q k the outcome is C, C. Complexity Considerations in Repeated Games that follows q K would be at least z, so that a deviation in state q k would be profitable. If we continue in the same manner then, since K is finite, we eventually return to the state q k that we began with. In this way we cover the cycle an integer number of times and thus conclude that the average payoff in the cycle q 1 ,. From Lemma The former case is impossible by the following argument.

The path in which the outcome in every period is B, A is not an equilibrium outcome since players 1 and 2 then use one-state machines that play B and A respectively, and player 1 can profitably gain by switching to the one-state machine that plays A. A Nash equilibrium can be constructed by having a long enough introductory phase in which B, B occurs in every period, with deviations in the cycling phase sending each machine back to its initial state. Any Nash equilibrium of the machine game for the repeated extensive game is a Nash equilibrium of the machine game for the repeated strategic game.

Thus by part a in all possible equilibria of the machine game for the repeated extensive game the outcome is either A, A or B, B in every period.

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But if there is any occurrence of A, A then player 2 can drop the state in which he chooses B and simply choose A in every period. Thus in the only possible equilibria the outcome is B, B in every period; it is clear that both players choosing a one-state machine that chooses B in every period is indeed a Nash equilibrium. Suppose that a game form G with outcome function g Nash-implements f.

That is, by choosing s1 , player 1 guarantees that the outcome is a. We conclude that f is not Nash-implementable. Now, by repeatedly applying inflationâ€”deflation we obtain a game of perfect information. Repeated applications of the principle of coalescing of moves yields a game with a single non-terminal history. The strategic form of this game is given in Figure Now, there is a unique Nash equilibrium, in which the mixed strategy of player 1 assigns probability 25 to HL and probability 53 to HH and player 2 concedes with probability In behavioral strategies this equilibrium is: player 1 chooses H when her card is H and chooses H with probability 35 and L with probability 52 when her card is L; player 2 concedes with probability Chapter First note that by the assumption of perfect recall, if the information set Ii0 of player i contains a history h, a1 ,.

To show the one deviation property, use backwards induction. Sequential Equilibrium 39 It is straightforward to check that each of these assessments satisfies sequential rationality and consistency. The first equilibrium has the following undesirable feature. If he derives such a belief from beliefs about the behavior of players 1 and 3 then he must believe that player 1 chooses c with positive probability and player 3 chooses e with positive probability. Assume that player 2 believes that players 1 and 3 attempted to adhere to their behavioral strategies but made errors in carrying out these strategies.

Then the fact that he believes that there is an equal probability that each of them made a mistake does not mean that he has to assign a positive probability to a mistake in the future. It is straightforward to check that the following assessment is a sequential equilibrium: type 0 always offers the price of 2 and type 3 always offers the price of 5. In both periods player 2 accepts any price at most equal to 2 and rejects all other prices regardless of the history.

If player 2 observes a price different from 5 in either period then he believes that he certainly faces type 0. Thus having rejected a price of 5 in the first period, which he believed certainly came from type 3, he concludes, in the event that he observes a price different from 5 in the second period, that he certainly faces type 0. Comment There are other sequential equilibria, in which both types offer a price between 3 and 3.

Sequential Equilibrium Consider the pooling equilibrium. A similar argument shows that the separating equilibrium is a sequential equilibrium.