3
$\begingroup$

Look at the following payoff matrix, there are two stage Nash equlibria, and we consider a infinitely repeated game with discount factor $\delta\in(0,1)$, can players sustain an average payoff of (2,2) under any strategy? (I am confused because the individually rational(i.e. minmax) payoff is (3,3)) If not, why?

enter image description here

$\endgroup$
  • $\begingroup$ How is the mimax payoff (3,3)? The minimum for H is 0. The minimum for L is 3. Which means that if both player play minimax, they both choose L, and the payoff is (5,5). $\endgroup$ – Acccumulation Mar 15 at 15:49
  • 2
    $\begingroup$ @Acccumulation: In the repeated games literature, it's customary to define the set of individually rational payoffs as $\{(u_1,\dots,u_n): u_i\ge\min_{a_{-i}}\max_{a_i}u_i(a_i,a_{-i})\,\forall i\}$. In this case, $(3,3)$ is the minimum of this set. $\endgroup$ – Herr K. Mar 15 at 16:09
3
$\begingroup$

I think the result you are looking for is Lemma 2 (p.7) of these lecture notes by Johannes Hörner. Also see the papers he references in the end. There is a formal proof, but the idea is exactly what Herr K. wrote in his post. In your repeated game, all players must have an average payoff of at least 3 in Nash equilibrium.

$\endgroup$
  • $\begingroup$ Wonderful! thanks! Just wondering how to convert this "obvious" idea into a formal proof. $\endgroup$ – The R Mar 16 at 20:45
3
$\begingroup$

I'm inclined to say $(2,2)$ is not sustainable, though I don't have a formal proof at the moment.

To achieve an average payoff of $(2,2)$, the players must play $(H,H)$ in some stages. But then a player can deviate to playing $L$ and avoid any subsequent punishments by ensuring that he can always get at least $3$, which is better than $2$.

$\endgroup$

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.