<p>A usual refinement concept used to deal with weakly dominated strategies is the <a href="http://core.ac.uk/download/pdf/6805957.pdf" rel="nofollow">trembling hand perfect equilibrium</a>. (I do not know others but this one works quite well.</p>

<p>The strategy in question is indeed weakly dominated by the following strategy</p>

<ol>
<li>In the first period, player plays $B$.</li>
<li>In all other periods, player plays $A$.</li>
</ol>

CC BY-SA 3.0
2015-06-22T09:18:12.510Z
TH is a refinement of NE, not dominance (and the question is about dominance, not NE).
The Almighty Bob
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2015-06-22T09:58:58.920Z
@TheAlmightyBob There are two questions. One is whether the specified strategy is dominated (in the infinitely repeated game). The other one is "What kind of ideas exist to rule out the kind of strategy described above?" TH is indeed a refinement of NE with just this purpose.
Giskard
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2015-06-20T20:35:38.510Z
2015-06-20T20:35:38.510Z
Giskard
<p>The problem you mentioned is why dominance is not really used frequently. It is a very weak concept in the sense that is usually has no "grip", i.e. not many strategies are eliminated. That is why we use Nash equilibria or even other concepts (e.g., the trembling hand perfect equilibrium which was mentioned by @desnesp, or, some form of subgame perfection).</p>

<p>And this has nothing to do with finitely or infinitely repeated games, this is true for every repeated game (or every game in general).</p>

<p>In <em>finitely repeated games</em> the only strategies that are weakly dominated (in your example) are strategies that play $B$ in the very last round, given some history.</p>

<p>Let me show you:
It is clear that it is dominated (by just switching to $A$ in the last round for this history).</p>

<p>Now the other direction: Assume there is a dominated strategy $S$ that plays $A$ in the last round, then there is a strategy which plays $B$ everywhere, but if you played $S$, then plays $A$ in the last round. To this strategy $S$ is strictly better than any other strategy and therefore not dominated.</p>

<p>How does that work for <em>infinitely repeated games</em>? In the same manner: You think $S$ is dominated and in $S$ there is always a chance that $A$ is played later? Then there is a strategy which plays $A$ then, iff $S$ was played. (I know this is a little bit sloppy, but I hope you still get the idea.)</p>

<p><strong>So, in short:</strong></p>

<blockquote>
  <p>Is the strategy in question itself weakly dominated?</p>
</blockquote>

<p>Yes, for example by switching to always play $A$ in the last round.</p>

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2015-06-22T07:16:29.877Z
In infinitely repeated games (and the question is about those) there is no last round.
Giskard
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2015-06-22T09:15:08.497Z
I know. That is why I write in finitely repeated games. I just say the problem arises in finitely repeated games, too. It is not a problem of infinitely repeated games but of dominance. And, in addition the construction of the counter strategy works in the same way.
The Almighty Bob
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2015-06-20T21:46:45.827Z
2015-06-20T21:53:39.933Z
2015-06-20T21:53:39.933Z
The Almighty Bob
<p>Repeated games are a special class of extensive form games. And to the extent that such games admit a strategic form representation, the usual dominance notions apply:</p>

<blockquote>
  <p><strong>Definition 1.</strong> Pure strategy $s_i\in S_i$ is <em>weakly dominated</em> for player $i$ if there exists a (mixed) strategy $\sigma_i'\in\Delta(S_i)$ such that $$u_i(\sigma_i',s_{-i})\ge u_i(s_i,s_{-i})\tag{1}$$ for all $s_{-i}\in S_{-i}$, and <strong>the above inequality is strict for at least one $s_{-i}$</strong>.    </p>
</blockquote>

<p>and</p>

<blockquote>
  <p><strong>Definition 2.</strong> Pure strategy $s_i^*\in S_i$ is <em>weakly dominant</em> for player $i$ if every other strategy $s_i\in S_i\setminus\{s_i^*\}$ is weakly dominated by $s_i^*$. </p>
</blockquote>

<p>Let's apply this definition to your example. Call the two players $i$ and $j$,  denote the grim trigger strategy you propose by $s_i^G$ (if used by player $i$, and $s_j^G$ if used by $j$) and denote by $s_i^A$ player $i$'s strategy that plays $A$ regardless of history (and similarly for $j$).</p>

<p>As you recognize, $s_i^A$ is not weakly dominant: $s_i^G$ is not dominated by $s_i^A$ in that there exists a strategy $s_j^G$ such that $u_i(s_i^A,s_j^G)&lt;u_i(s_i^G,s_j^G)$. But this doesn't imply that $s_i^A$ is weakly dominated. In particular, it's not dominated by $s_i^G$. </p>

<p>In their game theory textbook, Fudenberg and Tirole commented that </p>

<blockquote>
  <p>The notion of iterated strict dominance extends to extensive-form games as well; however, as we mentioned above, this concepts turns out to have little force in most extensive forms. The point is that a player cannot strictly prefer one action over another at an information set that is not reached given his opponents' play.</p>
</blockquote>

CC BY-SA 3.0
2015-06-28T22:00:05.553Z
2015-06-28T22:00:05.553Z
Herr K.
<p>How does the concept of weak dominance work with infinite games? The abundance of concepts seems to muddy things. </p>

<p>In particular, suppose two players play the following game an infinite number of times.</p>

<p>$$\begin{array}{l*{2}{c}r}
             &amp; A &amp; B  \\
\hline
A &amp; 1,1 &amp; 0,0 \\
B &amp; 0,0 &amp; 0,0
\end{array}$$</p>

<p>In the one shot game, it is clear that playing $A$ weakly dominates the action $B$. </p>

<p>However, what if we consider a bizarre kind of grim trigger:</p>

<ol>
<li>In the first period, a player plays $B$.</li>
<li>If the other played $A$ in the first period, play $B$ forever.</li>
<li>Otherwise, play $A$ forever.</li>
</ol>

<p>For a patient enough player, it is weakly best to respond to this strategy with the same one. Any strategy that begins with $B$ and then $A$ afterwards is a best response (the off-path details can be different, so that a best response may not stipulate $B$ forever if the other chooses $A$ at the beginning). </p>

<p>The pareto efficient equilibria involve always playing $A$ on the equilibrium path, and it's tempting to say that playing $A$ no matter what should be weakly dominant, but that doesn't seem to be the case. Has much been written on this kind of thing? What kind of ideas exist to rule out the kind of strategy described above? </p>

<p>Is the strategy in question itself weakly dominated? I would guess that we could concoct an equally sinister trigger strategy $s'$ so that the strategy in question is not dominated by any other $s''$.</p>

CC BY-SA 3.0
2015-06-20T20:22:54.690Z
What exactly is the payoff function of the super game? Discounted sum of the payoff of the one shot games, average payoff of the one shot games?
Giskard
CC BY-SA 3.0
2015-06-22T14:56:21.390Z
As it stands, no strategy can be better than simply playing A every time. This remains true regardless of how many times the game is played. To clarify, your point is that other strategies are *no worse*, so you'd like to know how to eliminate them?
RegressForward
CC BY-SA 3.0
2015-06-22T17:27:12.133Z
@RegressForward : The total payoff is somehow derived from the payoff of all the one-shot games. Assuming it is a discounted sum of the one shot payoffs and a discount factor of $\delta$: If in the repeated game if the other player plays the above strategy I get a payoff of $$ 0 + \delta + \delta^2 + \delta^3 ... =  \frac{\delta}{1 - \delta} $$ by playing the same strategy and a payoff of 0 by playing $A$ every time.
Giskard
CC BY-SA 3.0
2015-06-22T21:27:39.283Z
Ah, missed the must-play-B clause in the trigger, ty.
RegressForward
CC BY-SA 3.0
2015-06-23T12:31:45.087Z
@denesp Discounted sum.
Pburg
<game-theory><repeated-games>
Dominated Strategies in an Infinitely vs Finitely Repeated Game
Pburg
# Dominated Strategies in an Infinitely vs Finitely Repeated Game

(Asked by: username_0 on 2015-06-20)

How does the concept of weak dominance work with infinite games? The abundance of concepts seems to muddy things.

In particular, suppose two players play the following game an infinite number of times.

$$\begin{array}{l\*{2}{c}r}
& A & B \\
\hline
A & 1,1 & 0,0 \\
B & 0,0 & 0,0
\end{array}$$

In the one shot game, it is clear that playing $A$ weakly dominates the action $B$.

However, what if we consider a bizarre kind of grim trigger:

1. In the first period, a player plays $B$.
2. If the other played $A$ in the first period, play $B$ forever.
3. Otherwise, play $A$ forever.

For a patient enough player, it is weakly best to respond to this strategy with the same one. Any strategy that begins with $B$ and then $A$ afterwards is a best response (the off-path details can be different, so that a best response may not stipulate $B$ forever if the other chooses $A$ at the beginning).

The pareto efficient equilibria involve always playing $A$ on the equilibrium path, and it's tempting to say that playing $A$ no matter what should be weakly dominant, but that doesn't seem to be the case. Has much been written on this kind of thing? What kind of ideas exist to rule out the kind of strategy described above?

Is the strategy in question itself weakly dominated? I would guess that we could concoct an equally sinister trigger strategy $s'$ so that the strategy in question is not dominated by any other $s''$.

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

> **Comments:**
> **username_1**: What exactly is the payoff function of the super game? Discounted sum of the payoff of the one shot games, average payoff of the one shot games?  
> **username_2**: As it stands, no strategy can be better than simply playing A every time. This remains true regardless of how many times the game is played. To clarify, your point is that other strategies are *no worse*, so you'd like to know how to eliminate them?  
> **username_1**: @username_2 : The total payoff is somehow derived from the payoff of all the one-shot games. Assuming it is a discounted sum of the one shot payoffs and a discount factor of $\delta$: If in the repeated game if the other player plays the above strategy I get a payoff of $$ 0 + \delta + \delta^2 + \delta^3 ... =  \frac{\delta}{1 - \delta} $$ by playing the same strategy and a payoff of 0 by playing $A$ every time.  
> **username_2**: Ah, missed the must-play-B clause in the trigger, ty.  
> **username_0**: @denesp Discounted sum.  

---

## 3 answers

### Answer by username_1 on 2015-06-20. Score: 3

A usual refinement concept used to deal with weakly dominated strategies is the [trembling hand perfect equilibrium](http://core.ac.uk/download/pdf/6805957.pdf). (I do not know others but this one works quite well.

The strategy in question is indeed weakly dominated by the following strategy

1. In the first period, player plays $B$.
2. In all other periods, player plays $A$.

> **Comments:**
> **username_3**: TH is a refinement of NE, not dominance (and the question is about dominance, not NE).  
> **username_1**: @username_3 There are two questions. One is whether the specified strategy is dominated (in the infinitely repeated game). The other one is "What kind of ideas exist to rule out the kind of strategy described above?" TH is indeed a refinement of NE with just this purpose.  

---

### Answer by username_3 on 2015-06-20. Score: 3

The problem you mentioned is why dominance is not really used frequently. It is a very weak concept in the sense that is usually has no "grip", i.e. not many strategies are eliminated. That is why we use Nash equilibria or even other concepts (e.g., the trembling hand perfect equilibrium which was mentioned by @desnesp, or, some form of subgame perfection).

And this has nothing to do with finitely or infinitely repeated games, this is true for every repeated game (or every game in general).

In *finitely repeated games* the only strategies that are weakly dominated (in your example) are strategies that play $B$ in the very last round, given some history.

Let me show you:
It is clear that it is dominated (by just switching to $A$ in the last round for this history).

Now the other direction: Assume there is a dominated strategy $S$ that plays $A$ in the last round, then there is a strategy which plays $B$ everywhere, but if you played $S$, then plays $A$ in the last round. To this strategy $S$ is strictly better than any other strategy and therefore not dominated.

How does that work for *infinitely repeated games*? In the same manner: You think $S$ is dominated and in $S$ there is always a chance that $A$ is played later? Then there is a strategy which plays $A$ then, iff $S$ was played. (I know this is a little bit sloppy, but I hope you still get the idea.)

**So, in short:**

> Is the strategy in question itself weakly dominated?

Yes, for example by switching to always play $A$ in the last round.

> **Comments:**
> **username_1**: In infinitely repeated games (and the question is about those) there is no last round.  
> **username_3**: I know. That is why I write in finitely repeated games. I just say the problem arises in finitely repeated games, too. It is not a problem of infinitely repeated games but of dominance. And, in addition the construction of the counter strategy works in the same way.  

---

### Answer by username_4 on 2015-06-28. Score: 0

Repeated games are a special class of extensive form games. And to the extent that such games admit a strategic form representation, the usual dominance notions apply:

> **Definition 1.** Pure strategy $s\_i\in S\_i$ is *weakly dominated* for player $i$ if there exists a (mixed) strategy $\sigma\_i'\in\Delta(S\_i)$ such that $$u\_i(\sigma\_i',s\_{-i})\ge u\_i(s\_i,s\_{-i})\tag{1}$$ for all $s\_{-i}\in S\_{-i}$, and **the above inequality is strict for at least one $s\_{-i}$**.

and

> **Definition 2.** Pure strategy $s\_i^\*\in S\_i$ is *weakly dominant* for player $i$ if every other strategy $s\_i\in S\_i\setminus\{s\_i^\*\}$ is weakly dominated by $s\_i^\*$.

Let's apply this definition to your example. Call the two players $i$ and $j$, denote the grim trigger strategy you propose by $s\_i^G$ (if used by player $i$, and $s\_j^G$ if used by $j$) and denote by $s\_i^A$ player $i$'s strategy that plays $A$ regardless of history (and similarly for $j$).

As you recognize, $s\_i^A$ is not weakly dominant: $s\_i^G$ is not dominated by $s\_i^A$ in that there exists a strategy $s\_j^G$ such that $u\_i(s\_i^A,s\_j^G)<u\_i(s\_i^G,s\_j^G)$. But this doesn't imply that $s\_i^A$ is weakly dominated. In particular, it's not dominated by $s\_i^G$.

In their game theory textbook, Fudenberg and Tirole commented that

> The notion of iterated strict dominance extends to extensive-form games as well; however, as we mentioned above, this concepts turns out to have little force in most extensive forms. The point is that a player cannot strictly prefer one action over another at an information set that is not reached given his opponents' play.


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