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Where I can find a simple proof for this fact?

For example, a trivial bimatrix game with alternating move has the following payoff matrix:

\begin{array}{|c|c|c|} \hline & 1 & 2 \\ \hline U & (0,0) & (0,0)\\ \hline L& (0,0)& (0,0)\\ \hline \end{array}

Then all of the pure and mixed strategies are trivially the equilibrium strategies.

I guess that, if the game structural is so complicated that it becomes impossible for the players to solve the game, then this complete information game becomes effectively like an incomplete information game. But I am not sure how to rigorously describe this.

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    $\begingroup$ You should probably clarify/define what you mean by: simple game and non-trivial mixed equilibrium. $\endgroup$ – Herr K. Sep 30 at 20:36
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    $\begingroup$ This question might be closed as unclear, but it doesn't look like homework to me. $\endgroup$ – Michael Greinecker Sep 30 at 21:01
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As is clear from the answer of VARulle, complete information is of no use. Every (finite) game in normal-form is the normal form of an extensive form game of complete information.

The situation is different for games of perfect information, and one can prove a result to the effect that "Almost all finite games of perfect information have equilibria that look like equilibria in pure strategies along the equilibrium path of play."

Making this precise requires a bit of work and gets us into fairly deep water. In the following, all games are assumed to be finite. The set equilibria in (potentially degenerate) mixed strategies of a game in normal form can be represented as a closed subset of a Euclidean space of suitable dimension and, by a result of Kohlberg and Mertens (1986), the set of equilibria has finitely many connected components, even when there are infinitely many equilibria. Moreover, by a result of Kreps and Wilson (1982) if you fix an extensive-form game of perfect recall apart from the assignment of payoffs to terminal nodes, then the set of payoff-assignments for which there are infinitely many Nash equilibrium paths of play is a manifold of lower dimension; almost all extensive form games have finitely many possible equilibrium plays. However, it is possible that there are still infinitely many equilibria, but these equilibria vary off the equilibrium paths. Taking these results together, for almost all extensive form games of perfect recall, the set of plays is constant on each of the finitely many connected components. We call extensive-form games of perfect recall with this property generic.

So far, we covered preliminaries. Now, these aspects of games have been examined in the setting of games of perfect information in [Demichelis, Stefano, Klaus Ritzberger, and Jeroen M. Swinkels. "The simple geometry of perfect information games." International Journal of Game Theory 32.3 (2004): 315-338.] A working paper version of the paper without a pay-wall can be found here. One of the results of the authors says that each connected component of Nash equilibria of a generic game of perfect information contains an equilibrium in pure strategies. So for generic games of perfect information, every Nash equilibrium induces the same equilibrium play as some Nash equilibrium in pure strategies. In particular, mixing can only play a role outside the equilibrium path, and the actual outcome is deterministic.

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This statement is wrong. Consider Alternating Matching Pennies with imperfect information (the follower doesn't observe the leader's move). The strategic form of this game is just the classical (simultaneous-move) Matching Pennies Game and the unique NE has both players mixing.

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  • $\begingroup$ Ok, I did not understand the difference between imperfect information and incomplete information. My bad may I update the question or start a new question? $\endgroup$ – High GPA Oct 1 at 19:11

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