On Sustainable Equilibria...Myerson ConjectureHofbauer FormulationHofbauer-Myerson...

Transcript of On Sustainable Equilibria...Myerson ConjectureHofbauer FormulationHofbauer-Myerson...

Myerson Conjecture Hofbauer Formulation Hofbauer-Myerson Conjecture Proof Extending Sustainability

On Sustainable Equilibria

Hari Govindan, Rida Laraki & Lucas Pahl

Game Theory World SeminarAugust, 10, 2020

Govindan, Laraki, & Pahl

The Beginning: Essays in Honor of Selten 65th Birthday


Myerson Essay (1995)


Desirable Criteria for Refinement


Sustainable Equilibria in the Battle of the Sexes

Myerson considers, and then dismisses existing refinements thatyield the same prediction in the Battle-of-Sexes game:

Persistent Equilibria (Kalai-Samet): fail invariance

ESS (Maynard-Smith): fail existence


Intuition from Fixed Point Theory


Illustration: odd number of crosses, sum of indices = +1


Link with Fixed Point and Equilibria


Myerson’s Conjecture

Myerson was seemingly unaware of the existence at thatmoment of an index theory for Nash equilibria:

Gül, Pearce & Stacchetti (1993) Ritzberger (1994)


Myerson’s Conjecture

Myerson was seemingly unaware of the existence at thatmoment of an index theory for Nash equilibria:

Gül, Pearce & Stacchetti (1993) Ritzberger (1994)


Index of equilibria for economists: THE BOOK


Myerson’s Conclusion

Hofbauer (2000) formulated a precise definition ofsustainability for generic games and some conjectures.


Myerson’s Conclusion

Hofbauer (2000) formulated a precise definition ofsustainability for generic games and some conjectures.


Hofbauer Formulation of Myerson’s Criteria

I (A1) Strict Nash equilibria are sustainable.

I (A2) Battle of sexes: only strict equilibria are sustainable.

I (A3) If a game has a unique equilibrium, it is sustainable.

I (A4) Every generic game has a sustainable equilibrium.

I (A5) Sustainable equilibria are invariant under addition orremoval of inferior replies.

A5 is a form of independence of irrelevant alternative.

It is a very strong axiom as, combined with A3, they imply A1.


Hofbauer Definition of Sustainability

I Hofbauer defines an equivalence relation among pairs(G, σ) where G is a game and σ is an equilibrium of G.

I (G, σ) ∼ (Ĝ, σ̂) if σ = σ̂ (up to a relabelling) and therestriction of G and Ĝ to the best replies to σ and σ̂,resp., are the same game (up to a relabelling).

I By A5 (IIA) and A3 (uniqueness => sustainability):If an equilibrium σ of a game G is unique in anequivalent pair (Ĝ, σ̂), it must be sustainable.

Hofbauber cleverly combined A5 & A3 with minimality:

I An equilibrium of a game G is sustainable iff it isthe unique equilibrium in an equivalent pair.


Battle of the Sexes3 Nash equilibria: 2 strict σ = (t, l) and θ = (b, r), and 1 mixed.

G =l r

t (3, 2) (0, 0)b (0, 0) (2, 3)

By adding two strategies, σ is the unique equilibrium of Ĝ:

Ĝ =

l r yt (3, 2) (0, 0) (0, 1)b (0, 0) (2, 3) (−2, 4)x (1, 0) (4,−2) (−1,−1)

I Hence, the strict equilibrium σ is sustainable in G.

I The mixed equilibrium is not sustainable (prove it?).

I This is in line with Myerson requirements A2.



G =l r

t (3, 2) (0, 0)b (0, 0) (2, 3)


Ĝ =

l r yt (3, 2) (0, 0) (0, 1)b (0, 0) (2, 3) (−2, 4)x (1, 0) (4,−2) (−1,−1)






G =l r

t (3, 2) (0, 0)b (0, 0) (2, 3)


Ĝ =

l r yt (3, 2) (0, 0) (0, 1)b (0, 0) (2, 3) (−2, 4)x (1, 0) (4,−2) (−1,−1)






G =l r

t (3, 2) (0, 0)b (0, 0) (2, 3)


Ĝ =

l r yt (3, 2) (0, 0) (0, 1)b (0, 0) (2, 3) (−2, 4)x (1, 0) (4,−2) (−1,−1)






G =l r

t (3, 2) (0, 0)b (0, 0) (2, 3)


Ĝ =

l r yt (3, 2) (0, 0) (0, 1)b (0, 0) (2, 3) (−2, 4)x (1, 0) (4,−2) (−1,−1)





A Sustainable Mixed EquilibriumG has 7 equilibria, all symmetric: 3 strict, 3 mixed (playersrandomise between 2 strategies), and 1 completely mixed.

G1 =

l m rt (10, 10) (0, 0) (0, 0)m (0, 0) (10, 10) (0, 0)b (0, 0) (0, 0) (10, 10)

The 3 strict equilibria and the completely mixed aresustainables, as Ĝ shows (von Schemde & von Stengel).

Ĝ1 =

l m r x y zt (10, 10) (0, 0) (0, 0) (0, 11) (10, 5) (0,−10)m (0, 0) (10, 10) (0, 0) (0,−10) (0, 11) (10, 5)b (0, 0) (0, 0) (10, 10) (10, 5) (0,−10) (0, 11)

3 remaining mixed equilibria are not sustainable (prove it?)



G1 =

l m rt (10, 10) (0, 0) (0, 0)m (0, 0) (10, 10) (0, 0)b (0, 0) (0, 0) (10, 10)


Ĝ1 =

l m r x y zt (10, 10) (0, 0) (0, 0) (0, 11) (10, 5) (0,−10)m (0, 0) (10, 10) (0, 0) (0,−10) (0, 11) (10, 5)b (0, 0) (0, 0) (10, 10) (10, 5) (0,−10) (0, 11)




G1 =

l m rt (10, 10) (0, 0) (0, 0)m (0, 0) (10, 10) (0, 0)b (0, 0) (0, 0) (10, 10)


Ĝ1 =

l m r x y zt (10, 10) (0, 0) (0, 0) (0, 11) (10, 5) (0,−10)m (0, 0) (10, 10) (0, 0) (0,−10) (0, 11) (10, 5)b (0, 0) (0, 0) (10, 10) (10, 5) (0,−10) (0, 11)




G1 =

l m rt (10, 10) (0, 0) (0, 0)m (0, 0) (10, 10) (0, 0)b (0, 0) (0, 0) (10, 10)


Ĝ1 =

l m r x y zt (10, 10) (0, 0) (0, 0) (0, 11) (10, 5) (0,−10)m (0, 0) (10, 10) (0, 0) (0,−10) (0, 11) (10, 5)b (0, 0) (0, 0) (10, 10) (10, 5) (0,−10) (0, 11)

3 remaining mixed equilibria are not sustainable (prove it?)Govindan, Laraki, & Pahl

Notations

I Set of players is N = { 1, . . . , N }.

I ∀n, a finite set Sn of pure strategies.

I The set of pure strategy profiles S ≡∏

n∈N Sn.

I Σn is player n’s set of mixed strategies.

I The set of mixed strategy profiles Σ ≡∏

n∈N Σn.

I For each n, let S−n =∏

m6=n Sm and Σ−n =∏

m 6=n Σm.

I G : S → RN is the payoff vector function.

I Γ ≡ RN×S : is the space of games as payoffs vary.

I E = { (G, σ) ∈ Γ× Σ | σ is a Nash equilibrium ofG }is the Nash equilibrium correspondence.


Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Notations




n∈N Sn.



n∈N Σn.



m 6=n Σm.





Index of Equilibria

I Let G be a game and U be a neighbourhood of Σ in

Index of Equilibria


Index of Equilibria


Index of Equilibria


Index of Equilibria


Index of Equilibria


Index of Equilibria


Hofbauer-Myerson conjectureHofbauer-Myerson conjecture: A regular equilibrium issustainable if and only if it has index +1.

I von Schemde & von Stengel (2008) proved the conjecturefor 2-player games using polytopal geometry.

I We prove the Hofbauer-Myerson conjecture for all N -playergames using algebraic topology.

I Corollary 1: since the sum of the indices of equilibria is+1, any regular game has a sustainable equilibrium.

I Corollary 2: Since the set of regular games is open anddense, almost every game has a sustainable equilibrium.

This implies Myerson requirements A1, A2, A4 and A5.

As our proof extends to isolated equilibria, we obtain A3(because a unique equilibrium is isolated and has index +1).


Proof: the Easy Direction

Let σ be a regular equilibrium of game G.

I Let (G, σ) ∼ (Ĝ, σ) and σ = unique equilibrium of Ĝ.

I Let G∗ be obtained from G by deleting inferior replies to σ.

I It follows from a property of the index that:the index of σ in G = the index of σ in G∗.

I G∗ is also obtained from Ĝ by deleting inferior replies.

I Therefore, the index of σ in G∗ = index of σ in Ĝ.

I As σ is the unique equilibrium of Ĝ:the index of σ in Ĝ = +1.


Difficult Direction 1: Index=DegreeThe projection map Π projects a pair (G, σ) in the equilibriumcorrespondence E to the game G ∈ G.

Definition: The degree of an equilibrium σ of a game G= the local orientation of projection map Π at (G, σ).

Theorem Govindan & Wilson (2005): Degree = Index




Theorem Govindan & Wilson (2005): Degree = IndexGovindan, Laraki, & Pahl

Difficult Direction 2: Use Hopf Extension TheoremLet G be a regular game and σ a +1 equilibrium.

I Let f be the better reply map f defined in Nash’s PhDwhose fixed points are the equilibria of G.

I Since the sum of degrees over all equilibria is +1, the sumof degrees over all equilibria other than σ is zero.

I This implies that we can alter f outside a neighbourhoodU of σ so that the new map f0 has only one fixed point: σ.

I The possibility of such a construction follows from a deepresult in differential topology: Hopf Extension Theorem.

Hopf Theorem Let W be a compact, connected, oriented k+ 1dimensional manifold with boundary, and let f : ∂W → Sk be asmooth map. Then f extends to a globally defined mapF : W → Sk, with F = f , if and only if the degree of f is zero.


Illustrative Example

L RL (1, 1) (0, 0)R (0, 0) (1, 1)

I For simplicity, focus on symmetric strategies.

I A symmetric profile is represented by a number x ∈ [0, 1].

I Two strict equilibria x = 1 and x = 0 (index=degree= +1);one mixed x = 1/2 (index=degree −1).

I The Nash map f of this game is:

f(x) ≡

{x

1−2x2+x if x ∈ [0, 1/2]x−2x2+3x−11−2x2+3x−1 if x ∈ (1/2, 1]



L RL (1, 1) (0, 0)R (0, 0) (1, 1)





f(x) ≡

{x

1−2x2+x if x ∈ [0, 1/2]x−2x2+3x−11−2x2+3x−1 if x ∈ (1/2, 1]



L RL (1, 1) (0, 0)R (0, 0) (1, 1)





f(x) ≡

{x

1−2x2+x if x ∈ [0, 1/2]x−2x2+3x−11−2x2+3x−1 if x ∈ (1/2, 1]



L RL (1, 1) (0, 0)R (0, 0) (1, 1)





f(x) ≡

{x

1−2x2+x if x ∈ [0, 1/2]x−2x2+3x−11−2x2+3x−1 if x ∈ (1/2, 1]



L RL (1, 1) (0, 0)R (0, 0) (1, 1)





f(x) ≡

{x

1−2x2+x if x ∈ [0, 1/2]x−2x2+3x−11−2x2+3x−1 if x ∈ (1/2, 1]



L RL (1, 1) (0, 0)R (0, 0) (1, 1)





f(x) ≡

{x

1−2x2+x if x ∈ [0, 1/2]x−2x2+3x−11−2x2+3x−1 if x ∈ (1/2, 1]


Illustration of Hopf Extension Theorem

Figure: Graphs of f (black) and f0 (green)


Illustration of Hopf Extension Theorem

Figure: Graphs of f (black) and f0 (green)


Difficult Direction 3: add payoff-irrelevant strategies

We have a map f0 with no fixed points other than σ.

I The map f0 is meant to be a “better-reply” function.

I But f0n is defined over Σ and not just Σ−n.

I Construct an equivalent game G̃ that incorporates this.

I Strategy set of player n is S̃n ≡ Sn × Sn+1, where the2nd-coordinate is payoff-irrelevant.

I Construct a map f̃0n : Σ̃−n → Σ̃n which uses player n− 1’s2nd-coordinate choice to determine nth component in f0.


The proof needs a Delaunay triangulation of Σ̃n

Figure: Horizontal axis represents the duplicated strategy set in theequivalent game. Vertical axis represents the original strategy set.


It is NOT a Delaunay triangulation


It is NOT a Delaunay triangulation


Delaunay triangulation for points in general position

I Let C = co{x0, x1, . . . , xk } in


I Let C = co{x0, x1, . . . , xk } in


I Let C = co{x0, x1, . . . , xk } in


I Let C = co{x0, x1, . . . , xk } in


I Let C = co{x0, x1, . . . , xk } in


I Let C = co{x0, x1, . . . , xk } in


I Let C = co{x0, x1, . . . , xk } in

VERY Difficult Direction 5: Add Bonus

I Consider a sufficiently fine Delaunay triangulation of Σ̃n.

I Add vertices of the triangulation as pure strategies to G.

I Give a bonus to the new strategies using f̃0 (followingseveral tricks, as in Govindan & Wilson 2005)

I The ρ in Delaunay triangulation will be part of the bonus.

I Conclude using that f0 has only one fixed point σ.


Importance of not adding/removing best replies in IIA

Consider the following three-player game G, where player 3 hasa unique action (is a dummy player).

G =l r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)

Three Nash equilibria

I Two strict, (t, l) and (b, r) (index +1)

I One completely mixed σ, with index −1.

I Adding strategies leads to larger game where σ is theunique equilibrium.

I This cannot be done for a two player game!




G =l r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)









G =l r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)









G =l r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)









G =l r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)







−1 equilibrium can become unique by new strategies!

L R

Tl r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)

(3, 3, 0)

B (3, 0, 1) (0, 3, 1)

L R

Tl r

t (−3, 0, 4) (1, 4, 0)b (1, 4, 0) (1, 4, 0)

(1, 0, 1)

B (3, 0, 0) (0, 3, 0)

L R

Tl r

t (1, 4, 0) (1, 4, 0)b (1, 4, 0) (−3, 0, 4)

(3, 0, 1)

B (3, 0, 0) (0, 3, 0)

Something wrong?

NO: (G, σ) is not equivalent to (Ĝ, σ) because some ofstrategies we added are best replies to the equilibrium!

Open problem: Can any equilibrium be made unique byadding (non necessarily inferior replies) strategies and players?



L R

Tl r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)

(3, 3, 0)

B (3, 0, 1) (0, 3, 1)

L R

Tl r

t (−3, 0, 4) (1, 4, 0)b (1, 4, 0) (1, 4, 0)

(1, 0, 1)

B (3, 0, 0) (0, 3, 0)

L R

Tl r

t (1, 4, 0) (1, 4, 0)b (1, 4, 0) (−3, 0, 4)

(3, 0, 1)

B (3, 0, 0) (0, 3, 0)

Something wrong?





L R

Tl r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)

(3, 3, 0)

B (3, 0, 1) (0, 3, 1)

L R

Tl r

t (−3, 0, 4) (1, 4, 0)b (1, 4, 0) (1, 4, 0)

(1, 0, 1)

B (3, 0, 0) (0, 3, 0)

L R

Tl r

t (1, 4, 0) (1, 4, 0)b (1, 4, 0) (−3, 0, 4)

(3, 0, 1)

B (3, 0, 0) (0, 3, 0)

Something wrong?





L R

Tl r

t (6, 6, 1) (0, 0, 1)b (0, 0, 1) (6, 6, 1)

(3, 3, 0)

B (3, 0, 1) (0, 3, 1)

L R

Tl r

t (−3, 0, 4) (1, 4, 0)b (1, 4, 0) (1, 4, 0)

(1, 0, 1)

B (3, 0, 0) (0, 3, 0)

L R

Tl r

t (1, 4, 0) (1, 4, 0)b (1, 4, 0) (−3, 0, 4)

(3, 0, 1)

B (3, 0, 0) (0, 3, 0)

Something wrong?




Extension to non generic games: a first attempt

I Fact 1: A non isolated equilibrium σ of a game G cannotbe made unique in a larger game Ĝ obtained from G byadding inferior replies (our result is sharp).

Proof: By contradiction, let σn → σ be a sequence ofequilibria of G. Since σ is the unique equilibrium of Ĝ,there is a (up to a subsequence) a fixed player i and a fixedprofitable best reply deviation θi to σn in Ĝ. By continuity,θi is a best reply to σ in Ĝ: a contradiction.

I Fact 2: A strict subset U of an equilibrium component Cin G can never be made the unique component in a largergame Ĝ obtained from G by adding inferior replies.

Proof: By contradiction, U must be closed and since C isconnected, there is σ ∈ U , and σn ∈ C but not in U suchthat σn → σ, then we use same argument as above.


First attempt does not extend to all games!

I Corollary: Only an (entire) equilibrium component canbe made unique by adding inferior replies.

I BIG PROBLEM: There are games where no equilibriumcomponent can be made unique by adding inferior replies.

Proof: In Hauk & Hurkens example, no equilibriumcomponent has index +1.

I An open problem:

Can a +1 index equilibrium component be made unique byadding inferior replies?






I An open problem:







I An open problem:







I An open problem:



Second attempt: an axiomatic approachWe want to construct a correspondence Φ that selects for eachfinite game G, a set of subsets of Nash equilibria of G (calledsustainable sets for G) satisfying the following axioms:

I A1: Existence: Every game has a sustainable set.I A2: IIA: If a set is sustainable for a game, it remains

sustainable after adding or removing inferior replies.

I A3: Invariance: Equivalent games (obtained by additivepositive transformation of payoffs, or addition/deletion ofduplicate of mixte strategies) have equivalent solutions.

I A4: Robustness: If a set is sustainable for G, then anynearby game has a nearby sustainable set.

I A5: Minimality: If Ψ satisfies A1 to A4 then Φ ⊂ Ψ.

Theorem Φ satisfies A1 to A5 iff it associates to each game:its set of Nash equilibrium components with positive index.



I A1: Existence: Every game has a sustainable set.

I A2: IIA: If a set is sustainable for a game, it remainssustainable after adding or removing inferior replies.












Theorem Φ satisfies A1 to A5 iff it associates to each game:

its set of Nash equilibrium components with positive index.


Robustness in Myerson’s Essai



Myerson ConjectureHofbauer FormulationHofbauer-Myerson ConjectureProofExtending Sustainability

On Sustainable Equilibria...Myerson ConjectureHofbauer FormulationHofbauer-Myerson...

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Transcript of On Sustainable Equilibria...Myerson ConjectureHofbauer FormulationHofbauer-Myerson...