Informazioni sul documento
- Università
- Politecnico di Milano
- Corso di laurea
- Management Engineering
- Materia
- GAME THEORY
- Classificazione
- Esame · Esame completo
- Contenuto
- Testo d’esame
- Formato originale
- Testo
- Testo ricercabile
Esame completo di GAME THEORY per il corso di Management Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.
Esame completo di GAME THEORY per il corso di Management Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.
Qualità dell’importazione: il testo è stato estratto direttamente dal documento originale.
Passaggi rappresentativi riconosciuti nelle diverse parti del materiale. Il testo completo resta presente nella pagina per la ricerca, mentre l’anteprima compatta rende più semplice la lettura.
GAME THEORY June 11, 2021 Surname: Name: Matricola: Exercise 1 Given the following bimatrix game, depending on the parameters a,b∈ R, (12, 6) (4 , 4) (3 , 5) (3, 6) (8 , 5) (2 , 7) (a, 3) (6 , 7) (3 ,b ) , 1. find the pure Nash equilibria for every a,b ; 2. find the best reaction of the first player to ( 1 3, 1 3, 1 3 ); 3. find a,b such that the game is a potential game (if any); 4. find a,b such that there is a Nash equilibrium with second player playing ( 1 3, 1 3, 1 3 ); 5. show that for every a,b there is a Nash equilibrium with the first player playing a pure strategy and the second one a mixed strategy where exactly two pure strategies are played with strictly positive probability. Answer of exercise 1 1. The pure Nash equilibria outcomes are (12, 6) for every a,b such that a≤ 12 and (3,b ) if b≥ 7; 2. The expected utilities form the three rows (multiplied by 3) are respectively 19, 13, 9 + a. Thus the best reaction of the first player to ( 1 3, 1 3, 1 3 ) is (1, 0, 0) if a< 10, (0, 0, 1) if a> 10, and (p, 0, 1−p), 0≤p≤ 1 if a = 10. 3. A candidate potential (up to a constant) should look like 6 4 5 −3 8 . However, looking at the second player, with second row fixed for the first one, we see that 6− 5̸=−3− 8 thus for no a,b the game is a potential game; 4. For no pure strategy of the first player the best reaction of the second one is ( 1 3, 1 3, 1 3 ). Thus, if such a strategy exists, it must be a = 10 and such a strategy should be of the form (p, 0, 1−p). If player one plays (p, 0, 1−p), in order that the second mixes among the three columns, all of them must be optimal, thus providing the same expected value. Thus 6p + 3− 3p = 4p + 7− 7p = 5p +b−bp providingp = 2 3,b = 5; 5. the only possible case is when b = 7 and first player playing the last row,…
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