<p>This depends on the kinds of beliefs players can have about their opponents’ play. If opponents' strategies can be correlated, then the strategies surviving IESDS are exactly the rationalizable ones. But if you only allow opponents to choose independently (this is the standard approach in most textbooks), then these two sets are identical for 2-player games but need not be identical for games with more than 2 players.</p>
<p><a href="https://i.sstatic.net/wA9aG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wA9aG.png" alt="" /></a></p>
<p>Here is an example: Player 3's strategy Z is undominated, but not rationalizable (assuming independent mixtures of players 1 and 2).</p>

CC BY-SA 4.0
2020-11-06T11:29:59.377Z
Okay, so it depends on whether the decisions are correlated, if so , both procedures are equivalent, if not, it depends on the number of players. If the the number of players is 2 in thise last case, then IESDS ==rationalization and not otherwise. For the first, in Hotelling's standard linear model, the IESDS yields the strategy "Middle", as such, "Middle" for both players is the rationalizabel set of strategies, right?
ceins
CC BY-SA 4.0
2020-11-09T11:25:13.197Z
@ceins, this equivalence only holds for finite games. If I'm not mistaken, it's not true for Hotelling's location game. Only 1/2 is rationalizable there, since all other strategies are never-best-replies, but only 0 and 1 are iteratively strictly dominated. But for any finite discretization of Hotelling's game the equivalence holds again.
VARulle
CC BY-SA 4.0
2020-11-06T10:43:11.723Z
2020-11-06T10:43:11.723Z
VARulle
<blockquote>
<p>the set of rationalizable strategies can comprehend the result not only the IESDS but include the weakly dominant strategies and as such, be wider than the set remaining after IESDS.</p>
</blockquote>
<p>Your wording suggests that the result of iterated eliminations of strictly dominated strategies cannot include a weakly dominant strategy. However, this is not true. Consider the following two-player game.</p>
<p><span class="math-container">\begin{array} {|c|c|c|}
\hline
&amp; L &amp; R \\
\hline
T &amp; 1,0 &amp; 0,0 \\
\hline
B &amp; 0,0 &amp; 0,0 \\
\hline
\end{array}</span></p>
<p>The action <span class="math-container">$T$</span> for the row player weakly dominates <span class="math-container">$B$</span>, but there are no strictly dominated strategies in this game, so no strategies are eliminated by IESDS.</p>

CC BY-SA 4.0
2020-11-06T09:45:44.400Z
2020-11-06T09:45:44.400Z
jnanin
<p>When is the set of rationalizable strategies not equal to the set that is left after IESDS?
My thoughts are:
I know that a rationalizable strategy is that which is a best response for Player i given what Player -i does and thus the set of rationalizable strategies would include all the strategies of player i such that each of them is a BR to any other strategy of Player-i. As such, for sure we will include in the set all of the strictly dominant strategies so if there are only strictly dominant and strictly dominated strategies the set after IESDS will result in the same as the set of rationalizable strategies.</p>
<p>Nevertheless, if there exists also some weakly dominant strategies, the set of rationalizable strategies will also contain this because a weakly dominant strategy is a best response to some strategy -i as it pays more in at least some case. In this sense, the set of rationalizable strategies can comprehend the result not only the IESDS but include the weakly dominant strategies and as such, be wider than the set remaining after IESDS.</p>
<p>I know the last is incorrect because the set of rationalizable strategies cannot be greater than the set surviving IESDS. Can someone clarify the matter?</p>

<game-theory>
Rationalizabilitiy and Strict Dominance
ceins
# Rationalizabilitiy and Strict Dominance

(Asked by: username_0 on 2020-11-06)

When is the set of rationalizable strategies not equal to the set that is left after IESDS?
My thoughts are:
I know that a rationalizable strategy is that which is a best response for Player i given what Player -i does and thus the set of rationalizable strategies would include all the strategies of player i such that each of them is a BR to any other strategy of Player-i. As such, for sure we will include in the set all of the strictly dominant strategies so if there are only strictly dominant and strictly dominated strategies the set after IESDS will result in the same as the set of rationalizable strategies.

Nevertheless, if there exists also some weakly dominant strategies, the set of rationalizable strategies will also contain this because a weakly dominant strategy is a best response to some strategy -i as it pays more in at least some case. In this sense, the set of rationalizable strategies can comprehend the result not only the IESDS but include the weakly dominant strategies and as such, be wider than the set remaining after IESDS.

I know the last is incorrect because the set of rationalizable strategies cannot be greater than the set surviving IESDS. Can someone clarify the matter?

**Tags:** game-theory

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## 2 answers

### Answer by username_1 on 2020-11-06. Score: 2

This depends on the kinds of beliefs players can have about their opponents’ play. If opponents' strategies can be correlated, then the strategies surviving IESDS are exactly the rationalizable ones. But if you only allow opponents to choose independently (this is the standard approach in most textbooks), then these two sets are identical for 2-player games but need not be identical for games with more than 2 players.

[![](https://i.sstatic.net/wA9aG.png)](https://i.sstatic.net/wA9aG.png)

Here is an example: Player 3's strategy Z is undominated, but not rationalizable (assuming independent mixtures of players 1 and 2).

> **Comments:**
> **username_0**: Okay, so it depends on whether the decisions are correlated, if so , both procedures are equivalent, if not, it depends on the number of players. If the the number of players is 2 in thise last case, then IESDS ==rationalization and not otherwise. For the first, in Hotelling's standard linear model, the IESDS yields the strategy "Middle", as such, "Middle" for both players is the rationalizabel set of strategies, right?  
> **username_1**: @ceins, this equivalence only holds for finite games. If I'm not mistaken, it's not true for Hotelling's location game. Only 1/2 is rationalizable there, since all other strategies are never-best-replies, but only 0 and 1 are iteratively strictly dominated. But for any finite discretization of Hotelling's game the equivalence holds again.  

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### Answer by username_2 on 2020-11-06. Score: 0

> the set of rationalizable strategies can comprehend the result not only the IESDS but include the weakly dominant strategies and as such, be wider than the set remaining after IESDS.

Your wording suggests that the result of iterated eliminations of strictly dominated strategies cannot include a weakly dominant strategy. However, this is not true. Consider the following two-player game.

\begin{array} {|c|c|c|}
\hline
& L & R \\
\hline
T & 1,0 & 0,0 \\
\hline
B & 0,0 & 0,0 \\
\hline
\end{array}

The action $T$ for the row player weakly dominates $B$, but there are no strictly dominated strategies in this game, so no strategies are eliminated by IESDS.


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