If I have an specific extensive game, with only a finite set of strategies, how can I demonstrate that the game always have a subgame-perfect equilibrium in pure strategies? My first intuition was to show that in every subgame a pure strategy it's a solution to the problem (maybe not unique) and then there is always a pure strategies equilibrium. But I'm not sure.
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1$\begingroup$ I'm not sure the statement is true in general. The only SPE in matching pennies is in mixed strategies. $\endgroup$– Herr K.Commented Mar 29, 2017 at 22:23
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2$\begingroup$ Perhaps the OP has in mind finite extensive form games with perfect information, for which I believe the claim is true. $\endgroup$– Theoretical EconomistCommented Mar 30, 2017 at 2:13
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$\begingroup$ This is true in mixed strategies, I think. $\endgroup$– 123Commented Mar 30, 2017 at 15:05
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$\begingroup$ I'm not saying that it's true for all games, sorry if I did not explain myself good. Its if I had an specific extensive game. $\endgroup$– hllspwnCommented Mar 31, 2017 at 15:07
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$\begingroup$ Then you should tell us what this specific game is, and show that you've made some effort on trying to prove the result. $\endgroup$– Herr K.Commented Mar 31, 2017 at 15:39
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1 Answer
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You could represent the game in simultaneous form (matrix) to see all possible equilibria. If there is only one, then that's your subgame-perfect Nash equilibrium (since there are no other equilibria). If there are multiple equilibrias in the simultaneous representation, the extensive form allows you to see if some paths would never be reached given the structure of the game.
Essentially, the extensive form "proves" which paths would never be reached through ruling them out. The one path that is reached is your SPNE. An SPNE has a more strict definition. There is really no shortcut around the arithmetic in the extensive form since the definition of SPNE forces you to check for the one viable path.
