While evaluating the mixed strategy equilibrium, in most of the cases, we consider that all actions of a player are in the support of the equilibrium.Is it always so? I was going through a problem (36.1, Chapter 3 in Ariel Rubinstein’s A Course in Game Theory) where the mixed strategy equilibrium was calculated by considering all the possible actions in the support of the probability distribution. I was able to reach to this solution using the basic theory, but I cannot prove that every action will be in the support. Here I am borrowing the definition of support from Ariel Rubinstein’s A Course in Game Theory.
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It is not, only in intro problems.
You have already seen counterexamples: The prisoner's dilemma has an equilibrium, yet it does not have full support. (But it is a pure strategy equilibrium! So? All pure strategies are just very particular mixed strategies.)
Mixed strategies with full support are called completely mixed strategies.
It is a good exercise to show that strictly dominated pure strategies are never in the support of mixed equilibrium strategies. Using this you yourself can construct a simple example where the only equilibrium is a mixed but not completely mixed strategy profile.
