Is it true that for two player zero sum game, Perfect Bayesian Nash equilibrium is simply Nash Equilibrium?
I am learning game theory and our lecturer does not explicitly cover it.
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1 Answer
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The quick answer is Yes.
A Bayesian Nash equilibrium (BNE) is Nash equilibrium (NE) generalized to situations where there is uncertainty over players' payoffs; that is, games with incomplete information. For example, in the following zero-sum game, $t$ is a "state variable" that may take values $\epsilon>0$ or $-\epsilon$, each with probability $1/2$, say. Then the equilibrium you solve for would be a BNE.
On the other hand, if $t$ is not a random variable (or a degenerate random variable, as the jargons would have it), e.g. $t=0$ with probability $1$, then the equilibrium would simply be a NE.
In other words, a NE (of a complete information game) is a special case of a BNE (of a game of incomplete information, where there is only one payoff-relevant state, so to speak).
$$ \begin{array}{|c|c|c|}\hline & H & T \\\hline H & 1+t,-1+t & -1+t, 1+t \\\hline T & -1,1 & 1,-1\\\hline \end{array} $$
Adding perfection to BNE puts more restrictions on the players' off equilibrium beliefs about their opponents' strategies. In the case of matching pennies (the game above with $t=0$), which I assume you have in mind when saying "zero-sum games", this consideration wouldn't matter.
