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strategy m of G , when we say that m is a Nash equilibrium of G we implicitly
stipulate that each strategy used (with positive probability) in m is a strategy
in G .
Theorem 1.16 (IESDMS) Suppose that G is a finite strategic game.
(i) If G is an outcome of IESDMS from G, then m is a Nash equilibrium
of G iff it is a Nash equilibrium of G .
(ii) If G is solved by IESDMS, then the resulting joint strategy is a unique
Nash equilibrium of G (in, possibly, mixed strategies).
Exercise 1.6
Provide the proof.
To illustrate the use of this result let us return to the Beauty Contest
game discussed in Example 1.8. We explained there why (1 ,..., 1) is a Nash
equilibrium. Now we can draw a stronger conclusion.
Example 1.17 One can show that the Beauty Contest game is solved by
IESDMS in 99 rounds. In each round the highest strategy of each player is
removed and eventually each player is left with the strategy 1. On account
of the above theorem we now conclude that (1 ,..., 1) is a unique Nash
Exercise 1.7 Show that the Beauty Contest game is indeed solved by
IESDMS in 99 rounds.
As in the case of strict dominance by a pure strategy we now address the
question of whether the outcome of IESDMS is unique. The answer, as before,
is positive. The following result was established by Osborne and Rubinstein
Theorem 1.18 (Order independence III) All iterated eliminations of strat-
egies strictly dominated by a mixed strategy yield the same outcome.
1.5.2 Elimination of weakly dominated strategies
Next, we consider iterated elimination of pure strategies that are weakly
dominated by a mixed strategy.
As already noticed in Subsection 1.3.2 an elimination by means of weakly
dominated strategies can result in a loss of Nash equilibria. Clearly, the
same observation applies here. We also have the following counterpart of the
IEWDS Theorem 1.7, where we refer to Nash equilibria in mixed strategies.
Instead of 'the iterated elimination of strategies weakly dominated by a
mixed strategy' we write IEWDMS .
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