12
$\begingroup$

In Osborne's An Introduction to Game Theory Nash equilibrium is described as follows (p. 21–22):

First, each player chooses her action according to the model of rational choice, given her beliefs about the other players' actions. Second, every player's belief about the other players' actions is correct.

It seems to me that this definition is not completely equivalent to the usual definition of the Nash equilibrium as a strategy profile where each player's strategy is a best response to the strategies of the others.

The usual definition says nothing about beliefs and therefore allows for the possibility that beliefs might be incorrect.

To take a trivial possibility, consider the Prisoner's Dilemma. Suppose each player believes that the other player will not confess. Since confessing is a dominant strategy each player would still confess. So the actions constitute a Nash equilibrium even though the players' beliefs are completely the opposite of the actual equilibrium actions.

Am I right in this understanding that Osborne's definition characterizes something other than Nash's equilibrium?

$\endgroup$
  • $\begingroup$ Doesn't "The usual definition says nothing about beliefs and therefore allows for the possibility that beliefs might be incorrect." Simply hinge on the fact that you always have an underlying rationality assumption in these definitions? $\endgroup$ – Thorst Nov 28 '14 at 10:17
9
$\begingroup$

Introducing the language of beliefs here is slightly strange, given that beliefs do have a very specific meaning in other parts of game theory.

Indeed, Osborne's description is reminiscent of a Bayes Nash Equilibrium. We could introduce the notion of beliefs into the normal form of a complete information game as follows: suppose that with probability $a_i$ each player, $i$, is a "strategic" type who will play according to (Nash) equilibrium, and with probability $1-a_i$ he will select some strategy uniformly at random (because, say, he is indifferent across all actions). We thus have a Bayesian game where thinking about beliefs is more natural.

The Bayes Nash solution concept then says that $i$'s strategy must be optimal given the expected play induced by the other players' strategies and the beliefs over their types implied by $\{a_j\}_{j\neq i}$. If we look at the limit as $a_i\rightarrow 1$ for all $i$ then the Bayes Nash equilibrium of this game will coincide with the solution concept described by Osborne.


I guess the reason Osborne wrote it like this is a pedagogical one, given that this is an introductory text. When we introduce students to static games, we tell them that player $i$ best responds to the actions of the other players. Students naturally want to know "how can they respond to a strategy chosen simultaneously without knowing what that strategy will be?" This is, in many senses, a philosophical question. Common answers are

  • If the game is one that is played often then (putting aside issues of other outcomes that can be sustained in repeated games) we can think of Nash as being an equilibrium in the sense that if we converge there we can develop a norm whereby people continue to play that equilibrium indefinitely (and expect others to do the same).
  • If the game is really one-shot then we usually invoke the idea that players are going to try to predict what others will do—and our equilibrium notion embeds the idea that these predictions must be correct.

It seems that the predictions in the second point correspond to the "beliefs" invoked by Osborne. However, it is important to stress that these predictions/"beliefs", are merely an informal/intuitive tool for helping us to conceptualise what is going on in an equilibrium and are not part of the definition of such an equilibrium. The concept of Nash equilibrium itself is completely agnostic on the notion of beliefs (as you note in a comment, it is defined only over actions), which is why, when Osborne goes on to formally define Nash equilibrium, he does so without invoking the idea of beliefs at all.

$\endgroup$
4
$\begingroup$

Introducing belief makes the concept of NE comparable to other refinement concepts such as PBE and sequential equilibrium, but the meaning of NE is not changed.

The graduate micro textbook by Mas-Colell, Whinston, and Green (MWG) has a result for this

Proposition 9.C.1. A strategy profile $\sigma$ is a Nash equilibrium of an extensive form game $\Gamma_E$ if and only if there exists a system of beliefs $\mu$ such that

  1. The strategy profile $\sigma$ is sequentially rational given belief system $\mu$ at all information sets $H$ such that $\Pr(H|\sigma)>0$.
  2. The system of beliefs $\mu$ is derived from strategy profile $\sigma$ through Bayes' rule whenever possible.

Thus, the Prisoner's Dilemma example you give where players have beliefs opposite to what the opponent's actual strategy fails the second condition, which requires beliefs to be derived from Bayes' rule whenever possible. In fact, this is the mathematical equivalent of the second requirement of Osborne's definition: that a player's belief about the other players' actions is correct.

$\endgroup$
  • $\begingroup$ I think there is a difference between MWG and Osborne. MWG are saying that for a Nash equilibrium "there exists" a system of beliefs which makes it sensible. We are silent on what beliefs, if any, players actually have. Osborne is saying that the players actually have beliefs and they are the right ones. I thought that the latter changes the conceptual meaning of NE since the usual definition does not mention beliefs at all and the Prisoner's Dilemma example shows that the strategies do not uniquely determine the beliefs. $\endgroup$ – Jyotirmoy Bhattacharya Nov 28 '14 at 1:56
  • $\begingroup$ @JyotirmoyBhattacharya: I don't think MWG is "silent on what beliefs, if any, players actually have". Condition 2 of the proposition actually requires that this belief be derived from the equilibrium strategy profile using Bayes' rule whenever possible. Thus, in the PD example, when one player chooses defect with probability 1, the belief of the other player must also put probability 1 on the action defect and best respond given such a belief (which leads to him also choosing defect). $\endgroup$ – Herr K. Nov 28 '14 at 18:36
  • $\begingroup$ @JyotirmoyBhattacharya: Belief for NE need not be unique, however. This is because if for a given equilibrium, a path on a game tree is taken with probability zero, then Bayes' rule does not apply, and so any belief on that path would be considered "correct" in a NE. This is also why refinements like sequential equilibrium are introduced, so as to rule out unreasonable beliefs off equilibrium paths. $\endgroup$ – Herr K. Nov 28 '14 at 18:41
  • $\begingroup$ @JyotirmoyBhattacharya: Also, because it is an undergrad textbook, Osborne might have chosen a language that is more intuitive than mathematically rigorous for pedagogical reasons. For me, the two conditions in Osborne's definition are exact counterparts in MWG's proposition. $\endgroup$ – Herr K. Nov 28 '14 at 18:47
3
$\begingroup$

Your prisoner's dilemma example only works because that is a game with dominant strategies. Osborne is correct.

To be best responding to another player's strategy, as in the definition you give, I must know their strategy. In other words, I must have beliefs about what they are doing, and those beliefs must be correct. This is a strengthening of the concept of rationalizability.

You make an interesting point about how you can get strange "equilibria" in games with dominant strategies. It amounts to outcome equivalent $(\sigma,\mu_1)$ and $(\sigma,\mu_2)$ where $\mu_2$ might be wrong and placing positive weight on nonrationalizable strategies. But, I have never seen a Nash equilibrium that included beliefs. The definitions I recall go, "a strategy profile $\sigma\in \Sigma$ is a Nash equilibrium if $\sigma_i\in B_i(\sigma_{-i})$..." I believe this to mean that defining the beliefs is unnecessary, because the beliefs are exactly a correct assessment of the strategy profile. Referencing, one of my books, it gives the usual definition with a Nash (1950) citation, and then goes on to discuss two underlying assumptions. One is correct beliefs and the other is rational play given those correct beliefs.

$\endgroup$
  • $\begingroup$ But to disprove something one countrexample is enough. If you take Osborne as claiming that his definition is equivalent to Nash's then how does one deal with the Prisoner's Dilemma counterexample. I understand that Osborne's definition is a strengthening of rationalizability, I submit that it is not Nash equilibrium for the simple reason that here equilibrium is defined over actions and beliefs whereas Nash equilibrium is completely silent about beliefs. $\endgroup$ – Jyotirmoy Bhattacharya Nov 22 '14 at 5:26
  • 1
    $\begingroup$ It is a definition, not a proof. $\endgroup$ – Pburg Nov 22 '14 at 5:28
  • $\begingroup$ Fair enough. But it is a definition of a concept that already has another well-accepted definition. So I expect that if the author does not mention otherwise he is claiming that the two definitions are equivalent. $\endgroup$ – Jyotirmoy Bhattacharya Nov 22 '14 at 5:30
  • $\begingroup$ to be clear, does he include those comments in the definition, or in the discussion? p.s. I edited my answer $\endgroup$ – Pburg Nov 22 '14 at 5:55
  • $\begingroup$ The part I quoted is discussion. Immediately after that he says (p.22) "These two components are embodied in the following definition" and then gives the standard definition in terms of best response to strategies which does not mention beliefs at all. So where are the beliefs embodied in the definition? And the problem is not just in games with dominant strategies. It is quite possible to build examples where there is no dominant strategies but a Nash equilibrium strategies are best responses to beliefs different from equilibrium play. $\endgroup$ – Jyotirmoy Bhattacharya Nov 22 '14 at 6:05
2
$\begingroup$

I might be repeating things which have been said before, but here is my take on this.

I think we face a usual problem when comparing two different models. What an "equivalence" means is not completely obvious because the two definitions lie in different worlds, or different models. However, if "equivalence" is properly define, I think one can make sense of Osborne definition and show that it is indeed "equivalent" to a NE.

The solution concept underlying the quoted section would be something like the following :

Belief equilibrium (BE) : > A strategy profile $s^*$ and a belief profile $b^*$ in which for all player $i$

$$ u_i ( s^*_i ~|~ s_{-i} = b^*_{i}) \geq u_i ( s' ~|~ s_{-i} = b^*_{i}) \text{ for all } s' \in S_i$$ and $$b^*_i = s^*_{-i}$$

Now the problem if we are to get to any "equivalence" statement is that on the one hand, we have the BE which "lives" in a world with... beliefs, and on the other the NE notion which lives in a world... exempt of beliefs. So what would an equivalence statement like "NE $\Leftrightarrow$ BE" possibly mean?

1) BE $\Rightarrow$ NE

This direction of the implication is probably uncontroversial, because we go from a more complex to a more simple model. "Every BE is a NE" should mean that if we look at the equilibrium strategy profile of a BE alone (that is without its supporting belief profile $p$), it should be a NE. One can check that this is the case.

2) NE $\Rightarrow$ BE

This is the tricky part. What does it mean that "Every NE is a BE"? Certainly not that "a NE plus any belief profile is a BE", as the OP showed with his counter-example. Yet, it is the case that "any NE can be made a BE for some belief profile". I think it is in this sense that one should understand Osborne's "equivalence" claim

Notice that we also have the following more "equivalence-like" statement : "An outcome of the game is a NE outcome if and only if it is a BE outcome".

$\endgroup$
  • $\begingroup$ But every BE is not a NE since the BE concept is a weakening of rationalizability which us strictly weaker than NE. $\endgroup$ – Jyotirmoy Bhattacharya Nov 25 '14 at 16:13
  • $\begingroup$ As I wrote, I have trouble understanding what it means for a BE "not to be" a NE because they seem to live in different models. Do you mean that some strategies played at a BE are not NE? I thought that was incorrect, but I might have missed something. If that's what you mean, could you point me toward a counter-example? That would be very helpful. $\endgroup$ – Martin Van der Linden Nov 25 '14 at 16:34
  • $\begingroup$ I am not sure that BE is a weakening of rationalizability. In my understanding, BE is rationalizability along with the condition that conjectures are correct, right? If that is correct, then wouldn't BE we stronger than rationalizability? (it might of course depend on what is called a "weaker/stronger" solution concept) $\endgroup$ – Martin Van der Linden Nov 25 '14 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.