<p>I'm inclined to say <span class="math-container">$(2,2)$</span> is not sustainable, though I don't have a formal proof at the moment. </p>

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

CC BY-SA 4.0
2019-03-15T16:00:25.377Z
2019-03-15T16:00:25.377Z
Herr K.
<p>I think the result you are looking for is Lemma 2 (p.7) of these <a href="https://drive.google.com/file/d/0Bz9XuTKbxavEZGY3ODE4MGMtYmJhZi00NGM4LTkwZDYtZWU3NzU0ZWEyMWFi/view" rel="nofollow noreferrer">lecture notes</a> 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.</p>

CC BY-SA 4.0
2019-03-16T20:45:21.893Z
Wonderful! thanks! Just wondering how to convert this "obvious" idea into a formal proof.
The R
CC BY-SA 4.0
2019-03-16T13:53:17.950Z
2019-03-16T13:53:17.950Z
Bayesian
<p>Look at the following payoff matrix, there are two stage Nash equlibria, and we consider a infinitely repeated game with discount factor <span class="math-container">$\delta\in(0,1)$</span>, 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?</p>

<p><a href="https://i.sstatic.net/ERLim.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ERLim.png" alt="enter image description here"></a></p>

CC BY-SA 4.0
2019-03-15T15:49:32.623Z
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).
Acccumulation
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2019-03-15T16:09:32.857Z
@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.
Herr K.
<game-theory>
Can repeated game sustain a discounted average payoff that is not individually rational?
The R
# Can repeated game sustain a discounted average payoff that is not individually rational?

(Asked by: username_0 on 2019-03-15)

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](https://i.sstatic.net/ERLim.png)](https://i.sstatic.net/ERLim.png)

**Tags:** game-theory

> **Comments:**
> **username_1**: 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).  
> **username_2**: @username_1: 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.  

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## 2 answers

### Answer by username_3 on 2019-03-16. Score: 3 | Accepted answer

I think the result you are looking for is Lemma 2 (p.7) of these [lecture notes](https://drive.google.com/file/d/0Bz9XuTKbxavEZGY3ODE4MGMtYmJhZi00NGM4LTkwZDYtZWU3NzU0ZWEyMWFi/view) by Johannes Hörner. Also see the papers he references in the end.
There is a formal proof, but the idea is exactly what username_2 wrote in his post.
In your repeated game, all players must have an average payoff of at least 3 in Nash equilibrium.

> **Comments:**
> **username_0**: Wonderful! thanks! Just wondering how to convert this "obvious" idea into a formal proof.  

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### Answer by username_2 on 2019-03-15. Score: 3

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 $2$.


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