1
$\begingroup$

Consider a 2-player symmetric game given by a payoff matrix $A\in [0,1]^{n,n}$ for the row player (i.e. the column player matrix is $A^t$).

Define the social welfare as the sum of payoffs for both players, i.e. $$SW(i,j)=A(i,j)+A(j,i)$$

Define the social-welfare of a (possibly mixed) equilibrium in a straight forward manner: $$SW(s_1,s_2) = \sum _{i\in [n]}\sum_{j\in [m]}SW(i,j)\Pr_{s_1}(i)\Pr_{s_2}(j)$$

Is it true that for a symmetric equilibrium $s$ (one has to exist from Nash theorm) and a asymmetric equilibrium $a$, $SW(s)\leq SW(a)$?


For example, consider the following simple game:

$A= \left( \begin{array}{ccc} 1/3 & 2/3 \\ 1/3 & 1/6 \\ \end{array} \right) $

And the column player profit, given by $A^t$ is:

$A^t= \left( \begin{array}{ccc} 1/3 & 1/3 \\ 2/3 & 1/6 \\ \end{array} \right) $

There exist a (pure-strategies) symmetric equilibrium where both players play strategy 1. The social welfare of this equilibrium is 2/3.

A (pure-strategies) asymmetric equilibrium exist as well, where some player (say the rows player) plays strategy 1 while the other plays 2, and this gives a social-welfare of 1.


EDIT: Stef's answer made me think if the statement failed because the strategies in the symmetric equilibrium support were different than the ones in the asymmetric equilibrium. I've created a new question to discuss this possibility here.

$\endgroup$
5
  • $\begingroup$ But an asymmetric does not always exist. So you mean when an asymmetric exists, whether it yields in that case a greater welfare, or not? $\endgroup$
    – Jimmy R.
    Dec 3, 2014 at 10:41
  • $\begingroup$ @Stef - yes, thanks. $\endgroup$
    – R B
    Dec 3, 2014 at 10:46
  • $\begingroup$ If you're interested, you could consider posting questions like these on economics.stackexchange.com $\endgroup$
    – jmbejara
    Dec 3, 2014 at 19:55
  • $\begingroup$ @jmbejara - thanks for the suggestion ! While I still opted to keep the math handling of the new variant (see edit) in $MO$, I'm also interested in opinions of economists about what would happened in the real world, i.e. "does the fact that symmetric players act the same hurts the social-welfare". Do you think it's enough well-defined to be posted in EC.SE? $\endgroup$
    – R B
    Dec 4, 2014 at 11:48
  • $\begingroup$ Yeah, if the proposition turns out to be true, its definitely worth discussing. Also, don't hesitate to put the handling of the math on EC.SE if you want to. $\endgroup$
    – jmbejara
    Dec 5, 2014 at 2:32

1 Answer 1

1
$\begingroup$

No, it is not true. Consider the game with payoff matrices $$A=\begin{pmatrix}0&0&1/5\\0&1&0\\1/5&0&0\\\end{pmatrix} \quad \text{ vs } \quad B=A^T=\begin{pmatrix}0&0&1/5\\0&1&0\\1/5&0&0\\\end{pmatrix}$$ Then $e^2 \text{ vs } e^2$ is a symmetric Nash equilibrium with payoffs $(1,1)$ but $e^1 \text{ vs } e^3$ is an asymmetric Nash equilibrium with payoffs $(1/5,1/5)$.

$\endgroup$
9
  • 1
    $\begingroup$ This is true, thanks! I'm now wondering about the case where the asymmetric equilibrium support is a super set of the symmetric one (the intuition is that the utility of the players using only the strategies in the symmetric's support is unlikely, as otherwise they wouldn't play any of the strategies). I'll formalize it in a new question. $\endgroup$
    – R B
    Dec 3, 2014 at 12:09
  • 1
    $\begingroup$ @ R B Oh, I should also divide the entries of the matrices with 5 so that they are less or equal than 1. I also thought once that if a Nash equilibrium is contained in the support of another that it should then have always higher/or lower payoff than that other one but it is not true. I will try to follow your other question though because perhaps it holds for symmetric games! $\endgroup$
    – Jimmy R.
    Dec 3, 2014 at 13:00
  • 1
    $\begingroup$ not exactly what we discussed, but here's a question aiming towards it: mathoverflow.net/questions/188726/… $\endgroup$
    – R B
    Dec 3, 2014 at 13:02
  • 1
    $\begingroup$ @ R B. A game has always an odd number of equilibria or infinitely many. So it is not correct that the game in the other post has two equilibria. I will try to answer it, but if not good luck with it! $\endgroup$
    – Jimmy R.
    Dec 3, 2014 at 13:18
  • 1
    $\begingroup$ It is true that every asymmetric equilibrium $<s_1,s_2>$ has a counterpart equilibrium $<s_2,s_1>$, but for all I care they are the same equilibrium so I consider the game as having 2 equilibriums. I don't see how this implies an answer to what I asked there, so if you could answer the other question that willl be great, thanks ! $\endgroup$
    – R B
    Dec 3, 2014 at 13:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .