<p>The definition of a NE is not that a player cannot increase his payoff by changing his action, its that he cannot increase his payoff by changing his <em>strategy</em>. A strategy in a repeated game is a &quot;plan of action&quot; and such a plan can be complicated and history-dependent, as Michael Greinecker remarked in his comment.</p>
<p>So (high, high) cannot be a NE of the repeated game, but it can be the <em>action pair induced by a NE</em> of the repeated game in each round (if the players don't discount future payoffs too strongly). One such NE is (trigger, trigger), but there are infinitely many others. And there are also infinitely many NE that induce (low, low) in each round.</p>
<p>Neither of these NE should be called an &quot;optimal choice&quot;, and the strategies in NE should not be called &quot;optimal strategies&quot;. These expressions are not well defined. The (trigger, trigger) NE is <em>Pareto-optimal</em>, but that's not the same as &quot;optimal&quot; in everyday language, and trigger is only &quot;optimal&quot; in the sense that it is a <em>best reply</em> against trigger.</p>

CC BY-SA 4.0
2023-11-03T05:00:07.330Z
To be fair, in NE a player also cannot increase their payoff by changing their action.
Giskard
CC BY-SA 4.0
2023-11-03T00:16:01.660Z
2023-11-03T00:16:01.660Z
VARulle
<p>See in one shot game ,(low, low) is the visible Nash Equilibrium. As we define Nash Equilibrium as a situation or payoff where no player can gain by changing his strategy. In other words if other player's best response is low so knowing this fact choosing low strategy by other player is the best move as he would be worse off if he choses anything else.
However in an infinitely repeated game, strategies like Grim Trigger and Tit for Tat are commonly used as we can see the past actions of former players. So we can play at high forcing other player to cooperate in a long run and can always go back to original Nash Equilibrium if other player deviates. So in this situation (high,high) becomes a potential Nash Equilibrium until we use Grim Trigger . So to answer your question that my action affects the opponent is true since we are using Trigger Strategy and as long as our action is known and any change is there in it, opponent can easily deviate to &quot;at least&quot; the least possible gain he can have.</p>

CC BY-SA 4.0
2024-05-22T12:37:58.843Z
2024-05-22T12:37:58.843Z
Lksonice
<p>Say I have the following pay-off matrix:</p>
<p><a href="https://i.sstatic.net/BTQVi.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BTQVi.png" alt="enter image description here" /></a></p>
<p>For a one-shot game, it is easy to see, that (low, low) is the only Nash Equillibrium in the payoff-matrix.</p>
<p>However, say we're playing an infinitely repeated game, and that both players have a trigger strategy in place.</p>
<p>Such that, the optimal strategy for both players is to stay on (high,high,) and under the assumption that the other player will go low.</p>
<p>Does this mean, that the Nash Equillibrium under the assumption is now (high,high) or is the Nash Equillibrium still (low,low) and the optimal choice something entirely different from the Nash Equillibrium?</p>
<p>The definition (as I've been taught it,) is that the Nash Equillibrium is the decision where a single player cannot increase his payoff by only changing his own actions. However, if we assume that changing my actions affects the opponent, then does that still constitute a Nash Equillibrium?</p>

CC BY-SA 4.0
2023-11-01T18:48:07.920Z
The strategies in the repeated games are complicated history-dependent objects. (high, high) is not a strategy profile in the repeated game.
Michael Greinecker
<game-theory><nash-equilibrium>
Nash Equillibrium - Depend On The Opponent's Strategy?
BurgerMan
# Nash Equillibrium - Depend On The Opponent's Strategy?

(Asked by: username_0 on 2023-11-01)

Say I have the following pay-off matrix:

[![enter image description here](https://i.sstatic.net/BTQVi.png)](https://i.sstatic.net/BTQVi.png)

For a one-shot game, it is easy to see, that (low, low) is the only Nash Equillibrium in the payoff-matrix.

However, say we're playing an infinitely repeated game, and that both players have a trigger strategy in place.

Such that, the optimal strategy for both players is to stay on (high,high,) and under the assumption that the other player will go low.

Does this mean, that the Nash Equillibrium under the assumption is now (high,high) or is the Nash Equillibrium still (low,low) and the optimal choice something entirely different from the Nash Equillibrium?

The definition (as I've been taught it,) is that the Nash Equillibrium is the decision where a single player cannot increase his payoff by only changing his own actions. However, if we assume that changing my actions affects the opponent, then does that still constitute a Nash Equillibrium?

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: The strategies in the repeated games are complicated history-dependent objects. (high, high) is not a strategy profile in the repeated game.  

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## 2 answers

### Answer by username_2 on 2023-11-03. Score: 1

The definition of a NE is not that a player cannot increase his payoff by changing his action, its that he cannot increase his payoff by changing his *strategy*. A strategy in a repeated game is a "plan of action" and such a plan can be complicated and history-dependent, as Michael Greinecker remarked in his comment.

So (high, high) cannot be a NE of the repeated game, but it can be the *action pair induced by a NE* of the repeated game in each round (if the players don't discount future payoffs too strongly). One such NE is (trigger, trigger), but there are infinitely many others. And there are also infinitely many NE that induce (low, low) in each round.

Neither of these NE should be called an "optimal choice", and the strategies in NE should not be called "optimal strategies". These expressions are not well defined. The (trigger, trigger) NE is *Pareto-optimal*, but that's not the same as "optimal" in everyday language, and trigger is only "optimal" in the sense that it is a *best reply* against trigger.

> **Comments:**
> **username_3**: To be fair, in NE a player also cannot increase their payoff by changing their action.  

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### Answer by username_4 on 2024-05-22. Score: 0

See in one shot game ,(low, low) is the visible Nash Equilibrium. As we define Nash Equilibrium as a situation or payoff where no player can gain by changing his strategy. In other words if other player's best response is low so knowing this fact choosing low strategy by other player is the best move as he would be worse off if he choses anything else.
However in an infinitely repeated game, strategies like Grim Trigger and Tit for Tat are commonly used as we can see the past actions of former players. So we can play at high forcing other player to cooperate in a long run and can always go back to original Nash Equilibrium if other player deviates. So in this situation (high,high) becomes a potential Nash Equilibrium until we use Grim Trigger . So to answer your question that my action affects the opponent is true since we are using Trigger Strategy and as long as our action is known and any change is there in it, opponent can easily deviate to "at least" the least possible gain he can have.


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