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I'm asked to prove that there is, at best, one equilibrium in a two-player zero sum game where every payoff is different from one another in the matrix representation.

My idea is to suppose there are two equilibria and arrive at some contradiction, but I am having trouble formulating my proof. I was able to prove that these two equilibria would not happen in the same row or column and now I'm trying to develop my proof using the fact that, if an equilibrium happens, one of the players gets $a$ and the other gets $-a$ but, since there is another equilibrium with payoff $b \neq a$, one of them would be more interested in switching strategies (because $b>a$ or $-b > -a$). But this doesn't seem enough to be the contradiction.

Any form of help is appreciated.

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As stated the claim is false. Firstly, the "at best" doesn't make sense, since every game has a Nash equilibrium. Secondly, here is a counter example:

3 x 2 Payoff matrix A:

    1  -2
   -4   3
  -10   9



3 x 2 Payoff matrix B:

  -1   2
   4  -3
  10  -9

EE = Extreme Equilibrium, EP = Expected Payoff

Decimal Output

  EE  1  P1:  (1)  0.700000  0.300000  0.000000  EP=  -0.5  P2:  (1)  0.500000  0.500000  EP=  0.5
  EE  2  P1:  (2)  0.863636  0.000000  0.136364  EP=  -0.5  P2:  (1)  0.500000  0.500000  EP=  0.5

Rational Output

  EE  1  P1:  (1)   7/10  3/10     0  EP=  -1/2  P2:  (1)  1/2  1/2  EP=  1/2
  EE  2  P1:  (2)  19/22     0  3/22  EP=  -1/2  P2:  (1)  1/2  1/2  EP=  1/2

Connected component 1:
{1, 2}  x  {1}

I suspect you have been asked to prove that if every payoff is distinct then there is at most one pure Nash equilibrium.

For proving that, you can do a proof by contraction.

Hint, assume there are two distinct pure Nash equilibria:

  • row $i$, column $j$
  • row $k$, column $l$

such that $A_{ij} = a \ne b = A_{kl}$. Now consider the cells $A_{il}=c$ and $A_{kj}=d$ and reason about $a, b, c$, and $d$ using that facts that $(i,j)$ and $(k,l)$ are equilibria, and by distinctness of payoffs each pure strategy in these pure equilibria are unique pure best responses.

Here's a picture of a 2 by 2 game, which is sufficient for the intuition of what's happening: $$ \left( \begin{matrix} a & c\\ d & b\\ \end{matrix} \right)\ . $$ Hopefully you can now complete the proof yourself.

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