Non-monotonic Operators in Strategic Games Krzysztof R. Apt CWI - - PowerPoint PPT Presentation

non monotonic operators in strategic games
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Non-monotonic Operators in Strategic Games Krzysztof R. Apt CWI - - PowerPoint PPT Presentation

Non-monotonic Operators in Strategic Games Krzysztof R. Apt CWI and University of Amsterdam Non-monotonic Operators in Strategic Games p. 1/2 A Pointer to Maurices Work M. Denecker, M. Bruynooghe and V. Marek, Logic Programming


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Non-monotonic Operators in Strategic Games

Krzysztof R. Apt CWI and University of Amsterdam

Non-monotonic Operators in Strategic Games – p. 1/2

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A Pointer to Maurice’s Work

  • M. Denecker, M. Bruynooghe and V. Marek,

Logic Programming Revisited: Logic Programs as Inductive Definitions, ACM Transactions on Computational Logic (TOCL) 2(4),

  • pp. 623 - 654, 2001.

Special issue devoted to Robert A. Kowalski. “The thesis is developed that logic programming can be understood as a natural and general logic of inductive

  • definitions. In particular, logic programs with negation

represent nonmonotone inductive definitions.”

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Strategic Games: 2 Examples

Prisoner’s Dilemma C D C 2, 2 0, 3 D 3, 0 1, 1 A 4 by 3 game X Y Z A 2, 1 0, 1 1, 0 B 0, 1 2, 1 1, 0 C 1, 1 1, 0 0, 0 D 1, 0 0, 1 0, 0

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Strategic Games: Definition

Strategic game for n ≥ 2 players: (possibly infinite) set Si of strategies, payoff function pi : S1 × . . . × Sn → R, for each player i. Basic assumptions: players choose their strategies simultaneously, each player is rational: his objective is to maximize his payoff, players have common knowledge of the game and of each others’ rationality.

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Dominance by a Pure Strategy

X Y A 2, − 1, − B 1, − 0, − C 2, − 0, − A strictly dominates B. A weakly dominates C.

Non-monotonic Operators in Strategic Games – p. 5/2

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Dominance by a Mixed Strategy

X Y A 2, − 0, − B 0, − 2, − C 0, − 0, − D 1, − 0, − 1/2A + 1/2B strictly dominates C. 1/2A + 1/2B weakly dominates D.

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Iterated Elimination: Example (1)

Consider weak dominance.

X Y Z A 2, 1 0, 1 1, 0 B 0, 1 2, 1 1, 0 C 1, 1 1, 0 0, 0 D 1, 0 0, 1 0, 0

We first get

X Y A 2, 1 0, 1 B 0, 1 2, 1 C 1, 1 1, 0

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Iterated Elimination: Example (2)

X Y A 2, 1 0, 1 B 0, 1 2, 1 C 1, 1 1, 0

Next, we get

X A 2, 1 B 0, 1 C 1, 1

and finally

X A 2, 1

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4 Operators

Given: initial finite strategic game

H := (H1, . . ., Hn, p1, . . ., pn).

4 operators on the lattice of all subgames of H: S(G): outcome of eliminating from G all strategies strictly dominated by a pure strategy (once). W(G): . . . weakly dominated by a pure strategy, SM(G): . . . strictly dominated by a mixed strategy, WM(G): . . . weakly dominated by a mixed strategy.

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4 Inclusions

Note For all G WM(G) ⊆ W(G) ⊆ S(G), WM(G) ⊆ SM(G) ⊆ S(G).

WM SM S W

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Iterated Elimination

Do these inclusions extend to the outcomes of iterated elimination? None of these operators is monotonic. Example In

X A 1, 0 B 0, 0 B is strictly dominated, but not in X B 0, 0

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Operators

T: operator on a finite lattice (D, ⊆ ). T 0 = D, T k: k-fold iteration of T, T ω := ∩k≥0T k. T is monotonic if G ⊆ G′ implies T(G) ⊆ T(G′).

Lemma T and U operators on a finite lattice (D, ⊆ ). For all G, T(G) ⊆ U(G), at least one of T and U is monotonic. Then T ω ⊆ Uω.

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Approach

Given: two strategy elimination operators Φ and Ψ such that for all G, Φ(G) ⊆ Ψ(G). To prove

Φω ⊆ Ψω

we define their ‘global’ versions Φg and Ψg and prove instead

Φω

g ⊆ Ψω g .

Need to show:

Φω

g = Φω and Ψω g = Ψω,

for all G, Φg(G) ⊆ Ψg(G), at least one of Φg and Ψg is monotonic.

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Global Operators

G: a subgame of H. si, s′

i: strategies of player i in H.

s′

i ≻G si: s′ i strictly dominates si over G:

Then S(G) := G′, where

G′

i := {si ∈ Gi | ¬∃s′ i ∈ Gi s′ i ≻G si}.

GS(G) := G′, where

G′

i := {si ∈ Gi | ¬∃s′ i ∈ Hi s′ i ≻G si}.

Similar definitions for GW, GSM, GWM.

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Inclusion 1: SMω ⊆ Sω

Lemma For all G SM(G) ⊆ S(G). GSω = Sω. GSMω = SMω. (Brandenburger, Friedenberg and Keisler ’06) For all G GSM(G) ⊆ GS(G). GSM and GS are monotonic. Conclusion: GSMω ⊆ GSω, so SMω ⊆ Sω.

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Other Inclusions

By the same approach the inclusions Wω ⊆ Sω, WMω ⊆ SMω hold.

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What about WMω ⊆ Wω?

Consider

X Y Z A 2, 1 0, 1 1, 0 B 0, 1 2, 1 1, 0 C 1, 1 1, 0 0, 0 D 1, 0 0, 1 0, 0

Applying WM we get

X Y A 2, 1 0, 1 B 0, 1 2, 1

Another application of WM yields no change.

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Conclusion

The inclusion WMω ⊆ Wω does not hold.

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Summary

WM SM S W WM Sω SMω Wω

Reference K.R. Apt, Relative Strength or Strategy Elimination Procedures, Economics Bulletin, 3(21), pp. 1-9, 2007.

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Happy Birthday Maurice

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Happy Birthday Maurice

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