forum.jpg (4424 bytes)     "Inside  every small problem is a large problem struggling to get out."

Rules Forum Contributors [For contributors only]


Experimental Economics
General Equilibrium
Other Topics
Prisoners Dilemma
Zero Sum Games


Thread and Full Text View

Ask a question about: Other Topics
Respond to the question: min max operation in rubinsteins proof?

06/17/2012 07:52 AM by eyec; min max operation in rubinstein's proof of the folk theorem
hi. some of you do probably know rubinsteins 1979 paper "A non-cooperative equilibrium for supergames" on page 4 he defines v=min_i max_j (Payoffs for all combinations of i and j) as the smallest possible outcome opponents may force on an agent ... where j are the strategies of the agent and i those of her opponents. (pretty much everyone discussing the proof of the folk theorem continues to quote this equation.) now the usual convention is that nested operations like min, max etc are evaluated from right to left, i.e. first max them min. which does not seem to make any sense (and is furthermore not what rubinstein did in his example p.4/5). the smallest possible outcome other agents may force on me is (to my understanding) the maximin payoff, i.e. exactly the other way around v=max_j min_i (arguments) take rubinsteins example (payoffs of the agent in question only): opponent's strategy 1 opponent's strategy 2 agent's strategy 1 2 0 agent's strategy 1 3 1 thanks ... minimax (v=max_j min_i (arguments)) starts with minimizing over the rows which gives a vector (0, 1) [transposed] - maximizing leads to 1. that matches rubinstein's figures to this example on p.5. rubinsteins formula (v= min_i max_j (arguments)) starts with maximizing over the columns which gives a vector (2, 3) - minimizing gives 2. which does not make sense. am I somehow mistaken or uses rubinstein an order of operation deviating from the usual one? (i.e. then we have a notation "operation1 operation2 arguments" and do operation2(operation1(arguments)).) [Manage messages]