Strategy Elimination in Strategic Gameshomepages.cwi.nl/~apt/slides/krakow09-sli.pdf · Strategy Elimination and Nash Equilibrium Theorem If G is solved by IEWD(M)S, then the resulting

Strategy Elimination in StrategicGames

Krzysztof R. Apt

CWI and University of Amsterdam

Strategy Elimination in Strategic Games – p. 1/39

Executive Summary

1. We compare the relative strength of 4 procedures onfinite strategic games:

iterated elimination of strategies that are

weakly/strictly

dominated by a

pure/mixed strategy.

2. Discuss order independence of such procedures.

3. Discuss the case of infinite games.


Preliminaries


Example: Prisoner’s Dilemma

C DC 2, 2 0, 3D 3, 0 1, 1

Nash equilibrium:No player gains by unilateral deviation.

Here a unique Nash equilibrium: (D,D).

Pareto optimal outcome: (C,C).


Prisoner’s Dilemma in Practice


Dominance by a Pure Strategy

C2 D2

C1 2, 3 0, 3D1 3, 0 1, 1

D1 strictly dominates C1.

D2 weakly dominates C2.


Dominance by a Mixed Strategy

X YA 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.


Iterated Elimination: Example (1)

Consider weak dominance.X Y Z

A 2, 1 0, 1 1, 0B 0, 1 2, 1 1, 0C 1, 1 1, 0 0, 0D 1, 0 0, 1 0, 0

We first get

X YA 2, 1 0, 1B 0, 1 2, 1C 1, 1 1, 0


Iterated Elimination: Example (2)X Y

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

Next, we get

XA 2, 1B 0, 1C 1, 1

and finally

XA 2, 1

Conclusion: We solved the game by IEWDS.


Strategy Elimination and Nash Equilibrium

Theorem

If G is solved by IEWD(M)S, then the resulting jointstrategy is a Nash equilibrium of G.

If G is solved by IESD(M)S, then the resulting jointstrategy is a unique Nash equilibrium of G.


Beauty-contest Game

Example: The 2nd Maldives Mr & Miss Beauty Contest.


Beauty-contest Game (1)

[Moulin, ’86]

each set of strategies = {1, . . ., 100},

payoff to each player:1 is split equally between the players whose submittednumber is closest to 2

3of the average.

Examplesubmissions: 29, 32, 29; average: 30,payoffs: 1

2, 0, 1

2.

This game is solved by IEWDS.

Hence it has a Nash equilibrium, namely (1, . . ., 1).


Beauty-contest Game (2)

Also

This game is solved by IESDMS, in 99 steps.

Hence it has a unique Nash equilibrium, (1, . . ., 1).


Part IFinite Games


4 Operators

Given: initial finite strategic game

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

with each pi : H1 × · · · × Hn →R.

G: a restriction of H (Gi ⊆ Hi).

S(G): outcome of eliminating (once) from Gall strategies strictly dominated by a pure strategy.

W(G): . . . weakly dominated by a pure strategy,

SM(G): . . . strictly dominated by a mixed strategy,

WM(G): . . . weakly dominated by a mixed strategy.


4 Inclusions

Note For all games G

WM(G) ⊆ W(G) ⊆ S(G),

WM(G) ⊆ SM(G) ⊆ S(G).

WM

SM

S

W


Iterated Elimination

Do these inclusions extend to the outcomes of iteratedelimination?

None of these operators is monotonic.

Example

XA 1, 0B 0, 0

Then

S({A,B}, {X}) = ({A}, {X}),

S({B}, {X}) = ({B}, {X}).

So ({B}, {X}) ⊆ ({A,B}, {X}), but notS({B}, {X}) ⊆ S({A,B}, {X}).


Operators

(D, ⊆ ): a finite lattice with the largest element ⊤.

T : D → D is monotonic if

G ⊆ G′ implies T (G) ⊆ T (G′),

T 0 = ⊤,T k: k-fold iteration of T ,Tω := ∩k≥0T

k.

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 Tω ⊆ Uω.


Approach

Given: two strategy elimination operators Φ and Ψ such thatfor all G

Φ(G) ⊆ Ψ(G).

To proveΦω ⊆ Ψω

we define their ‘global’ versions Φg and Ψg,

prove Φωg = Φω and Ψω

g = Ψω,

show that for all G

Φg(G) ⊆ Ψg(G),

show that at least one of Φg and Ψg is monotonic,

Use the Lemma.


Global Operators

G: a restriction of H.si, s

′i: strategies of player i in H.

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

∀s−i ∈ G−i pi(s′i, s−i) > pi(si, s−i)

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}.

GS proposed in [Chen, Long and Luo, ’07].

Similar definitions for GW, GSM, GWM.


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: SMω ⊆ Sω.


Inclusion 2: Wω ⊆ Sω

Lemma

For all G

W(G) ⊆ S(G).

GSω = Sω.

GWω = Wω.

For all G

GW(G) ⊆ GS(G).

GS is monotonic.

Conclusion: Wω ⊆ Sω.


Inclusion 3: WMω ⊆ SMω

Lemma

For all G

WM(G) ⊆ SM(G).

GSMω = SMω.

GWMω = WMω.([Brandenburger, Friedenberg and Keisler ’06])

For all G

GWM(G) ⊆ GSM(G).

GSM is monotonic.

Conclusion: WMω ⊆ SMω.


What about WMω ⊆ Wω?

ReconsiderX Y Z

A 2, 1 0, 1 1, 0B 0, 1 2, 1 1, 0C 1, 1 1, 0 0, 0D 1, 0 0, 1 0, 0

Applying WM we get

X YA 2, 1 0, 1B 0, 1 2, 1

Another application of WM yields no change.


Conclusion

The inclusionWMω ⊆ Wω

does not hold.


Summary

WM

SM

S

W

WM

Sω

SMω

Wω


Part IIOrder Independence


Weak Confluence

A a set, → a binary (reduction) relation on A.→∗ : the transitive reflexive closure of → .

b is a → -normal form of a ifa →∗ b,no c exists such that b → c.

→ is weakly confluent if ∀a, b, c ∈ A

aւ ցb c

implies that for some d ∈ A

b cց∗ ∗ ւ

d


Unique Normal Form Property

If each a ∈ A has a unique normal form, then(A, → ) satisfies the unique normal form property.

Newman’s Lemma (’42)Consider (A, → ) such that

no infinite → sequences exist,

→ is weakly confluent.

Then → satisfies the unique normal form property.


Order Independence

Order independence of a dominance notion:no matter in what order the strategies are removed theoutcome is the same.


All roads lead to Rome


Approach

Treat the dominance notion as a reduction relation onthe set of restrictions of an initial game.

Then order independence = unique normal formproperty.

Try to use Newman’s Lemma.

Note It suffices to apply Newman’s Lemma to a ‘simpler’relation →1 such that

no infinite →1 sequences exist,

→1 is weakly confluent,

→+

1= →+.


Results: Glossary

– S: strict dominance,

– W : weak dominance,

– NW : nice weak dominance ([Marx and Swinkels ’97]),

– PE : payoff equivalence,

– RM: the ‘mixed strategy’ version of the dominancerelation R,

– inh-R: the ‘inherent’ version of the (mixed) dominancerelation R([Börgers ’90]),

– OI: order independence,

– ∼-OI: order independence up to strategy renaming.


Summary of Results

Notion Property Result originally due to

S OI [Gilboa, Kalai and Zemel, ’90]

[Stegeman ’90]

inh-W OI [Börgers ’90]

inh-NW OI

SM OI [Osborne and Rubinstein ’94]

inh-WM OI [Börgers ’90]: equal to SM

inh-NWM OI

PE ∼-OI

S ∪ PE ∼-OI

NW ∪ PE ∼-OI [Marx and Swinkels ’97]

PEM ∼-OI

SM ∪ PEM ∼-OI

NWM ∪ PEM ∼-OI [Marx and Swinkels ’97]


Part IIIInfinite Games


Strict Dominance

Note [Dufwenberg and Stegeman ’02]Strict dominance is not order independent for infinitegames.

Example Consider a two-players game G with

S1 = S2 = N ,

p1(k, l) := k,

p2(k, l) := l.

ThenG

ւ ց∅ G′

where G′ := ({0}, {0}).


Transfinite Reductions

Note [Chen, Long and Luo, ’07]In general transfinite sequences of reductions are neededto solve an infinite game.

Note [Zvesper ’09]For every ordinal α there is a game that can be solved inexactly α steps.


. . .

Wiecej nastepnym razem

Dziekuje za uwage


References

[1] K. R. Apt. Uniform proofs of order independence forvarious strategy elimination procedures. The B.E. Journalof Theoretical Economics, 4(1), 2004. (Contributions),Article 5, 48 pages.

[2] K. R. Apt. Order independence and rationalizability. InProceedings 10th Conference on Theoretical Aspects ofReasoning about Knowledge (TARK ’05), pages 22–38.The ACM Digital Library, 2005.

[3] K. R. Apt. The many faces of rationalizability. The B.E.Journal of Theoretical Economics, 7(1), 2007. (Topics),Article 18, 39 pages.

[4] K. R. Apt. Relative strength of strategy eliminationprocedures. Economics Bulletin, 3, pp. 1–9, 2007.


Documents

Strategy Elimination in Strategic Gameshomepages.cwi.nl/~apt/slides/krakow09-sli.pdf · Strategy Elimination and Nash Equilibrium Theorem If G is solved by IEWD(M)S, then the resulting