<p>The quick answer is Yes.</p>

<p>A Bayesian Nash equilibrium (BNE) is Nash equilibrium (NE) generalized to situations where there is uncertainty over players' payoffs; that is, games with <strong>incomplete information</strong>. For example, in the following zero-sum game, $t$ is a "state variable" that may take values $\epsilon&gt;0$ or $-\epsilon$, each with probability $1/2$, say. Then the equilibrium you solve for would be a BNE.</p>

<p>On the other hand, if $t$ is not a random variable (or a <em>degenerate random variable</em>, as the jargons would have it), e.g. $t=0$ with probability $1$, then the equilibrium would simply be a NE.</p>

<p>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).</p>

<p>$$
\begin{array}{|c|c|c|}\hline
  &amp; H &amp; T \\\hline
H &amp; 1+t,-1+t &amp; -1+t, 1+t \\\hline
T &amp; -1,1 &amp; 1,-1\\\hline
\end{array}
$$</p>

<p>Adding <em>perfection</em> 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. </p>

CC BY-SA 3.0
2015-12-11T06:27:16.447Z
2015-12-11T06:33:39.710Z
2015-12-11T06:33:39.710Z
Herr K.
<p>Is it true that for two player zero sum game, Perfect Bayesian Nash equilibrium is simply Nash Equilibrium?</p>

<p>I am learning game theory and our lecturer does not explicitly cover it.</p>

<nash-equilibrium>
perfect bayesian nash equilibrium is simply nash equilibrium
user40780
# perfect bayesian nash equilibrium is simply nash equilibrium

(Asked by: username_0 on 2015-12-11)

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.

**Tags:** nash-equilibrium


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## 1 answers

### Answer by username_1 on 2015-12-11. Score: 1

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.


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