Are all Nash equilibrium pure strategies also Nash equilibrium mixed strategies.

Question

while going over wiki page on Battle of the Sexes game I found something funny.

This game has two pure strategy Nash equilibria, one where both go to the opera and another where both go to the football game. For the first game, there is also a Nash equilibrium in mixed strategies, where the players go to their preferred event more often than the other. For the payoffs listed above, each player attends their preferred event with probability 3/5.

http://en.wikipedia.org/wiki/Battle_of_the_sexes_%28game_theory%29#Equilibrium_analysis Also from wiki :

A mixed strategy is an assignment of a probability to each pure strategy. This allows for a player to randomly select a pure strategy. Since probabilities are continuous, there are infinitely many mixed strategies available to a player, even if their strategy set is finite.

Of course, one can regard a pure strategy as a degenerate case of a mixed strategy, in which that particular pure strategy is selected with probability 1 and every other strategy with probability 0.

So my Q is why arent pure strategies also considered Nash equilibrium for the BotS game? I mean if players play lets say uper left corner of the matrix(always, p1=1.0,p2=1.0) then either players expected utility will drop if it lowers its px without other player chaning his px. So shouldnt this be Nash eq also?

Answer 1

Yes. All pure Nash equilibria is a subset of mixed Nash equilibria. Pure strategy equilibrium is just a special case of mixed ones. However, some professors (or wikipedia entries) use "mixed equilibrium" to refer completely mixed equilibrium. Some professors distinguish between those cases : 1) completely mixed 2) semi mixed 3) pure equilibrium.

1. completely mixed: all pure strategies should be played with a probability >0
2. semi mixed: some pure strategies are not used (used with probability 0)
3. pure: a pure strategy is used with probability 1.

Answer 2

It's like tonic, which isn't considered a mixed drink ($0$ parts gin and $1$ part tonic). That's degenerate for you: "In mathematics [as in mixology], a degenerate case is a limiting case in which a class of object changes its nature so as to belong to another, usually simpler, class."