<p>It is not, only in intro problems.</p>

<p>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.)</p>

<p>Mixed strategies with full support are called completely mixed strategies.</p>

<p>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.</p>

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2018-07-04T20:41:24.247Z
2018-07-04T21:34:18.140Z
2018-07-04T21:34:18.140Z
Giskard
<p>While evaluating the mixed strategy equilibrium, in most of the cases, we consider that all actions of a player are in the <em>support</em> of the equilibrium.Is it always so? 
I was going through a problem (36.1, Chapter 3 in Ariel Rubinstein’s <em>A Course in Game Theory</em>) 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 <em>support</em> from Ariel Rubinstein’s <em>A Course in Game Theory.</em></p>

<game-theory>
Why is every action considered to be in the support while calculating mixed strategy equilibrium?
superhulk
# Why is every action considered to be in the support while calculating mixed strategy equilibrium?

(Asked by: username_0 on 2018-07-04)

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.*

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2018-07-04. Score: 2 | Accepted answer

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.


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