<p>It is true because of the discounting. If the discount parameter were $\delta = 1$ then players 1 and 2 could alternate playing TL and MR in equilibrium and then reach the average payoff vector $1,1,5$. But if $\delta &lt; 1$ someone will not play along. Suppose the players are supposed to start with TL. Then MR would follow in the next round then, then TL again, and so on. In this case
$$
U_1 = 0 + 2 \cdot \delta + 0 \cdot \delta^2 + 2 \cdot \delta^3 + ...
$$
Yet by always playing B player 1 would achieve the payoff
$$
U_1' = 1 + 1 \cdot \delta + 1 \cdot \delta^2 + 1 \cdot \delta^3 + ...
$$
which is larger then $U_1$. Similarly player 2 would not want to start with MR. So these are not equilibrium playoffs. (Not even Aumann's correlated-equilibrium would do the trick.)</p>

CC BY-SA 3.0
2015-11-10T14:05:33.190Z
This is much clearer than the explanation in the book. Thanks!
Erel Segal-Halevi
CC BY-SA 3.0
2015-11-11T05:43:08.720Z
But, you only proved that the players don't play TL and MR in certainty.  It should also be proved that they don't play TL and MR in any positive probability. I think the idea is as follows: if the players play TL with positive probability, then, since they don't play TR and ML, this can only happen if the column player plays L with certainty and the row player mixes T and B. But, such a mix gives the row player strictly less than 1 in the first step. Hence, the row player has an incentive to deviate to B with probability 1, as you said.
Erel Segal-Halevi
CC BY-SA 3.0
2015-11-10T13:17:12.583Z
2015-11-10T13:17:12.583Z
Giskard
<p>The textbook I am currently reading claims that, in an infinitely-repeated game with discount, there might be a payoff vector which is feasible and individually-rational, but it is not an equilibrium payoff vector in the repeated game. The example is the following basic game for three players:</p>

<pre><code>     L       R
T  0,2,5    0,0,0
M  0,1,0    2,0,5
B  1,1,0    1,1,0
</code></pre>

<p>The third player is a dummy player with only one possible action.</p>

<p>In this game:</p>

<ul>
<li>The minimax values of the players are 1,1,0.</li>
<li>The payoff vector 1,1,5 is both individually-rational and feasible (e.g. by mixing TL and MR with equal frequencies).</li>
</ul>

<p>The book claims that the only payoff vector in equilibrium is 1,1,0! Why?</p>

<p>Let $E$ be some equilibrium in the repeated game. The payoffs of the row and column players in $E$ must be 1, because:</p>

<ul>
<li>They must be at least 1 because these are the minimax values;</li>
<li>They must be at most 1 because the sum of utilities of these players in every outcome is at most 2.</li>
</ul>

<p>In the TR and ML cells, the sum of utilities of the row and column players is less than 2; hence, these cells are not played in equilibrium at all.</p>

<p>So in equilibrium, the only cells that may be played with positive frequency are: TL, MR, BL, BR.</p>

<p>Now, the authors claim that TL and MR are also not played at all in equilibrium. From this they conclude that the payoff of the dummy player is 0. I didn't understand this part. Is it true that TL and MR are never played in equilibrium? Why?</p>

<game-theory><repeated-games>
A feasible rational payoff that is not an equilibrium payoff in the repeated game
Erel Segal-Halevi
# A feasible rational payoff that is not an equilibrium payoff in the repeated game

(Asked by: username_0 on 2015-11-10)

The textbook I am currently reading claims that, in an infinitely-repeated game with discount, there might be a payoff vector which is feasible and individually-rational, but it is not an equilibrium payoff vector in the repeated game. The example is the following basic game for three players:
     L       R
T  0,2,5    0,0,0
M  0,1,0    2,0,5
B  1,1,0    1,1,0

The third player is a dummy player with only one possible action.
In this game:

The minimax values of the players are 1,1,0.
The payoff vector 1,1,5 is both individually-rational and feasible (e.g. by mixing TL and MR with equal frequencies).

The book claims that the only payoff vector in equilibrium is 1,1,0! Why?
Let $E$ be some equilibrium in the repeated game. The payoffs of the row and column players in $E$ must be 1, because:

They must be at least 1 because these are the minimax values;
They must be at most 1 because the sum of utilities of these players in every outcome is at most 2.

In the TR and ML cells, the sum of utilities of the row and column players is less than 2; hence, these cells are not played in equilibrium at all.
So in equilibrium, the only cells that may be played with positive frequency are: TL, MR, BL, BR.
Now, the authors claim that TL and MR are also not played at all in equilibrium. From this they conclude that the payoff of the dummy player is 0. I didn't understand this part. Is it true that TL and MR are never played in equilibrium? Why?

**Tags:** game-theory, repeated-games

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## 1 answers

### Answer by username_1 on 2015-11-10. Score: 5 | Accepted answer

It is true because of the discounting. If the discount parameter were $\delta = 1$ then players 1 and 2 could alternate playing TL and MR in equilibrium and then reach the average payoff vector $1,1,5$. But if $\delta < 1$ someone will not play along. Suppose the players are supposed to start with TL. Then MR would follow in the next round then, then TL again, and so on. In this case
$$
U\_1 = 0 + 2 \cdot \delta + 0 \cdot \delta^2 + 2 \cdot \delta^3 + ...
$$
Yet by always playing B player 1 would achieve the payoff
$$
U\_1' = 1 + 1 \cdot \delta + 1 \cdot \delta^2 + 1 \cdot \delta^3 + ...
$$
which is larger then $U\_1$. Similarly player 2 would not want to start with MR. So these are not equilibrium playoffs. (Not even Aumann's correlated-equilibrium would do the trick.)

> **Comments:**
> **username_0**: This is much clearer than the explanation in the book. Thanks!  
> **username_0**: But, you only proved that the players don't play TL and MR in certainty.  It should also be proved that they don't play TL and MR in any positive probability. I think the idea is as follows: if the players play TL with positive probability, then, since they don't play TR and ML, this can only happen if the column player plays L with certainty and the row player mixes T and B. But, such a mix gives the row player strictly less than 1 in the first step. Hence, the row player has an incentive to deviate to B with probability 1, as you said.  

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