Economic and Game Theory
Forum

20 5 10 20 50 100 all
Ask a question about: Other Topics
Respond to the question: Prove it?

01/15/2001 03:19 PM by Rodrigo; About the game you suggested

Regarding part (a) of your question, player 1 has m1 strategies and player 2 has m2 strategies. Regarding part (b), player 1 has m1 strategies, but now player 2 can observe player 1's move before taking her action, so player 2 has (m2)^(m1) (that is, m2 to the power m1). This is because for any action taken by player 1 in the first period, she can take any of her m2 actions. We may prove part (c) by contradiction. (Recall that whenever we want to prove the existence of "something", it always pays to try first a proof by contradiction). Thus, suppose, by contradiction, that for all pairs of moves (m1,m2) and (m1',m2') with either [m1 not equal to 1'] or [m2 mot equal to m2'], we have [u1(m1,m2) not equal to u1(m1',m2')] and [u2(m1,m2) not equal to u2(m1',m2')]. Since, by way of contradiction, we assumed that player 2 is not indifferent between the pairs (m1,m2) and (m1',m2'), the she has a best response at each node (try to visualize the tree of this game). Moreover, her best response at each node is unique. If it were not unique, say there are two best responses at some node, then she would be indifferent between those responses, but we assumed away this possibility. Player 1 plays by backward induction. He also is not indifferent between any two pairs, so he will certainly have a unique best reponse to the startegy adopted by player 2. However this contradicts the multiplicity of SPNE. Regarding part (d) of your question, we know that player 2 is never indifferent between any two pairs of moves. Then she has a unique best response, by the same reasoning above. Now, suppose that (m1",m2") is a Nash equilibrium of the simultaneous game in part (a) of your question. This is also a Nash equilibrium of the extensive game in part (b), but not necessarily a SPNE. This is because player 2 is not necessarily playing a best response at each node. Player 1 can guarantee a least payoff of p1 when he plays m1", since player 2 will paly m2". Given 2's unique SPNE startegy in the sequential game, 1 can do at least as well as the Nash eq (m1",m2") in any SPNE. So his payoff is at least p1. [Manage messages]

12/23/2000 03:07 AM by Pavel; Prove it !!

I've an interesting (though an easy) problem set for you to try to solve ... e-mail your answers to me and you will have a chance to win J45 ... 1. Consider a game with two players, player 1 and player 2, in which each player i [View full text and thread]