We compare here the relative strength of four widely used procedures on
finite strategic games: iterated elimination of weakly/strictly dominated
strategies by a pure/mixed strategy. A complication is that none of these
procedures is based on a monotonic operator. To deal with this problem we use
'global' versions of these operators.Comment: 8 page

Apt, Krzysztof R.

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arXiv

Apt, K.R. (Krzysztof)

CWI's Institutional Repository

arXiv:0706.1617v1  [cs.GT]  12 Jun 2007Relative Strength of Strategy EliminationProceduresKrzysztof R. AptCWI, Amsterdam, the Netherlandsand University of AmsterdamAbstractWe compare here the relative strength of four widely used proce-dures on finite strategic games: iterated elimination of weakly/strictlydominated strategies by a pure/mixed strategy. A complication is thatnone of these procedures is based on a monotonic operator. To dealwith this problem we use ’global’ versions of these operators.1 IntroductionIn the literature four procedures of reducing finite strategic games have beenwidely studied: iterated elimination of weakly/strictly dominated strategies bya pure/mixed strategy. Denote the corresponding operators (the mnemonicsshould be clear, ’L’ refers to ’local’ the meaning of which will be clarified later)respectively by LW, MLW, LS and MLS. When these operators are applied toa specific game G we get the following obvious inclusions:MLW(G)⊆ LW(G)⊆ LS(G) and MLW(G)⊆MLS(G)⊆ LS(G).It is then natural to expect that these inclusions carry on to the outcomesof the iterations of these operators. It turns out that this is not completelytrue. Moreover, proofs of some of the apparently obvious implications are not,in our view, completely straightforward. One of the complications is that noneof these operators is monotonic. To reason about them we use their ‘global’versions.More precisely, given two strategy elimination operators Φl and Ψl suchthat for all games G, Φl(G)⊆Ψl(G) we prove the inclusion Φωl ⊆Ψωl betweenthe outcomes of their iterations by means of the following generic procedure:1(i) define the corresponding ‘global’ versions of these operators, Φg and Ψg,(ii) prove that Φωg = Φωl and Ψωg = Ψωl ,(iii) show that for all games G, Φg(G)⊆Ψg(G),(iv) show that at least one of Φg and Ψg is monotonic.The last two steps then imply Φωg ⊆Ψωg by a general lemma. The desiredinclusion Φωl ⊆Ψωl then follows by (ii). The main work is in proving (ii).2 Preliminaries2.1 Strategic gamesBy a strategic game (in short, a game) for n players (n > 1) we mean asequence(S1, . . ., Sn, p1, . . ., pn),where for each i ∈ [1..n]• Si is the non-empty, finite set of strategies available to player i,• pi is the payoff function for the player i, so pi : S1 × . . . × Sn →R,where R is the set of real numbers.Given a sequence of sets of strategies S1, . . ., Sn and s ∈ S1 × . . .× Sn wedenote the ith element of s by si, denote (s1, . . ., si−1, si+1, . . ., sn) by s−i andsimilarly with S−i, and write (s′i, s−i) for (s1, . . ., si−1, s′i, si+1, . . ., sn), wherewe assume that s′i ∈ Si. We denote the strategies of player i by si, possiblywith some superscripts.Given a finite non-empty set A we denote by ∆A the set of probabilitydistributions over A and call any element of ∆Si a mixed strategy of playeri. The payoff functions are extended in the standard way to mixed strategies.We say that G := (S1, . . ., Sn) is a restriction of a game H := (T1, . . ., Tn,p1, . . ., pn) if each Si is a (possibly empty) subset of Ti. We identify the restric-tion (T1, . . ., Tn) with H . A subgame of H is a restriction (S1, . . ., Sn) withall Si non-empty.To analyze various ways of iterated elimination of strategies from an initialgame H we view such procedures as operators on the set of subgames of H . Aminor complication is that this set together with the componentwise inclusionon the players’ strategy sets does not form a lattice. Consequently, we extend2these operators to the set of all restrictions of H , which together with thecomponentwise set inclusion does form a lattice.Given a restriction G := (S1, . . ., Sn) of H = (T1, . . ., Tn, p1, . . ., pn) and twostrategies si ∈ Ti and mi ∈ ∆Ti we write mi ≻G si as an abbreviation for∀s−i ∈ S−i pi(mi, s−i) > pi(si, s−i)and mi ≻wG si as an abbreviation for∀s−i ∈ S−i pi(mi, s−i) ≥ pi(si, s−i) ∧ ∃s−i ∈ S−i pi(mi, s−i) > pi(si, s−i).In the first case we say that si is strictly dominated on G by mi and inthe second one that si is weakly dominated on G by mi. In particular mican be a pure strategy, i.e. an element of Ti.Given an operator T on a finite lattice (D, ⊆ ) we denote by T k the k-folditeration of T , where T 0 = D (so the iterations start ’at the top’) and letT ω := ∩k≥0Tk. We call T monotonic if for all G,G′G⊆G′ implies T (G)⊆ T (G′).When comparing two ways of eliminating strategies from a strategic game,represented by the operators T and U on the lattice of all restrictions of H ,we would like to deduce T ω ⊆ Uω from the fact that for all G, T (G)⊆ U(G).Unfortunately, in general this implication does not hold; a revealing exampleis provided in Section 4. What does hold is the following simple lemma thatrelates to steps (iii) and (iv) of the generic procedure from the Introductionand reveals the importance of the monotonicity.Lemma 1 Consider two operators T and U on a finite lattice (D, ⊆ ), suchthat• for all G, T (G)⊆ U(G),• at least one of T and U is monotonic.Then T ω ⊆ Uω.Proof. We prove by induction that for all k ≥ 0 we have T k ⊆ Uk. The claimholds for k = 0. Suppose it holds for some k. Then by the assumptions and theinduction hypothesis we have the following string of inclusions and equalities:• if T is monotonic: T k+1 = T (T k)⊆ T (Uk)⊆ U(Uk) = Uk+1,• if U is monotonic: T k+1 = T (T k)⊆ U(T k)⊆ U(Uk) = Uk+1. ✷33 Strict dominanceFrom now we fix an initial game H = (T1, . . ., Tn, p1, . . ., pn). Given a restric-tion G := (S1, . . ., Sn) of H we denote Si by Gi. In particular we denote Ti byHi.First we focus on two operators on the restrictions of H :LS(G) := G′,where for all i ∈ [1..n]G′i := {si ∈ Gi | ¬∃s′i ∈ Gi s′i ≻G si},and MLS defined analogously but with G′i := {si ∈ Gi | ¬∃mi ∈ ∆Gi mi ≻Gsi}.Starting with Luce and Raiffa [1957], the iterated elimination of strictlydominated strategies is customarily defined as the outcome of the iteration ofthe MLS operator starting with the initial game H .To reason about the above two operators we introduce two related operatorsdefined by (’G’ stands for ‘global’):GS(G) := G′where for all i ∈ [1..n]G′i := {si ∈ Gi | ¬∃s′i ∈ Hi s′i ≻G si}.and MGS defined analogously but with G′i := {si ∈ Gi | ¬∃mi ∈ ∆Hi mi ≻Gsi}.So in the LS and MLS operators we limit our attention to strict dominanceby a pure/mixed strategy in the current game, G, while in the other twooperators we consider strict dominance by a pure/mixed strategy in the initialgame, H .The difference is crucial because the operators LS and MLS are not mono-tonic, while GS and MGS are. To see the former just take the following gameH :XA 1, 0B 0, 0Note that LS(H)=MLS(H)=({A}, {X}) and LS({B}, {X})=MLS({B}, {X})= ({B}, {X}). So ({B}, {X})⊆H , while neither LS({B}, {X})⊆ LS(H) norMLS({B}, {X})⊆MLS(H).4Monotonicity of GS and MGS follows directly from their definitions. TheMGS operator is studied in Brandenburger, Friedenberg and Keisler [2006b](it is their operator Φ) and in Apt [2007b]. For the case of strict dominanceby a pure strategy it was introduced for arbitrary games in Milgrom andRoberts [1990, pages 1264-1265], studied for compact games with continuouspayoffs in Ritzberger [2001, Section 5.1] and considered for arbitrary gamesin the presence of transfinite iterations in Chen, Long and Luo [2005] andApt [2007b].The fact that the MGS operator is monotonic has some mathematical ad-vantages. For example, by virtue of a general result established in Apt [2007b],it is automatically order independent. Moreover, thanks to monotonicity, asargued in Apt [2007a], it can be used in the epistemic framework of gametheory based on possibility correspondences as ‘a stand alone’ concept of ra-tionality.The following result, the second part of which was proved (as Proposition2.2 (ii)) in Brandenburger, Friedenberg and Keisler [2006b], relates the originaloperators to their global versions and corresponds to step (ii) of the genericprocedure from the Introduction.Lemma 2 GSω = LSω and MGSω = MLSω.Proof. See the appendix. ✷This allows us to establish the expected inclusion.Theorem 1 MLSω ⊆ LSω.Proof. This is a consequence of the mentioned generic procedure, since step(iii) holds: for all restrictions G, MGS(G)⊆GS(G), and step (iv) holds: bothMGS and GS are monotonic. ✷4 Weak dominanceNext we compare the LS and MLS operators with their weak dominance coun-terparts, LW and MLW, defined in the same way, but using the ≻wG relationinstead of ≻G. In the literature, starting with Luce and Raiffa [1957], the iter-ated elimination of weakly dominated strategies is customarily defined as theoutcome of the iteration of the MLW operator starting with the initial gameH .To reason about these two operators we use the corresponding ‘global’versions, GW and MGW, defined in the same way as for strict dominance. We5have following counterpart of Lemma 2, the second part of which was proved(as Lemma F.1) in Brandenburger, Friedenberg and Keisler [2006a].Lemma 3 GWω = LWω and MGWω = MLWω.Proof. See the appendix. ✷This allows us to establish the following result.Theorem 2 LWω ⊆ LSω and MLWω ⊆MLSω.Proof. Again, this is a consequence of the generic procedure from the Intro-duction. Indeed, step (iii) holds: for all restrictions G, GW(G)⊆GS(G) andMGW(G)⊆MGS(G), and step (iv) holds: both GS and MGS are monotonic.✷However, surprisingly, the inclusion MLWω ⊆ LWω does not hold.Example 1 Consider the following game H :X Y ZA 2, 1 0, 1 1, 0B 0, 1 2, 1 1, 0C 1, 1 1, 0 0, 0D 1, 0 0, 1 0, 0Applying to it the MLW operator we getX YA 2, 1 0, 1B 0, 1 2, 1Another application of MLW yields no change. In contrast, after three itera-tions of the LW operator to the initial game we reachXA 2, 1✷Since the conclusion of Lemma 1 does not apply here, while Lemma 3holds, we conclude that none of the operators LW,MLW,GW and MGW ismonotonic.6AcknowledgementWe acknowledge helpful email discussions with Adam Brandenburger.ReferencesK. R. Apt[2007a] Epistemic analysis of strategic games with arbitrary strategy sets, in: Proc. 11thConference on Theoretical Aspects of Reasoning about Knowledge (TARK07),ACM Press. To appear.[2007b] The many faces of rationalizability, The B.E. Journal of Theoretical Economics,7(1). (Topics), Article 18, 39 pages. Available from http://arxiv.org/abs/cs.GT/0608011.A. Brandenburger, A. Friedenberg, and H. Keisler[2006a] Admissibility in games. Working paper. Available from http://pages.stern.nyu.edu/~abranden.[2006b] Fixed points for strong and weak dominance. Working paper. Available fromhttp://pages.stern.nyu.edu/~abranden/.Y.-C. Chen, N. V. Long, and X. Luo[2005] Iterated strict dominance in general games. Available from http://www.sinica.edu.tw/~xluo/pa10.pdf.R. D. Luce and H. Raiffa[1957] Games and Decisions, John Wiley and Sons, New York.P. Milgrom and J. Roberts[1990] Rationalizability, learning, and equilibrium in games with strategic complemen-tarities, Econometrica, 58, pp. 1255–1278.K. Ritzberger[2001] Foundations of Non-cooperative Game Theory, Oxford University Press, Oxford.AppendixProof of Lemma 2.This result is a corollary to Theorem 5 of Apt [2007b]. We provide here adirect proof. Note first that for all restrictions G we have GS(G)⊆ LS(G) soby Lemma 1 we have GSω ⊆ LSω.To establish the first equality we prove by induction that for all k ≥ 0LSk ⊆GSk.7We only need to establish the induction step. Take si ∈ LSk+1i . By definitionsi ∈ LSki and ¬∃s′i ∈ LSki s′i ≻LSk si. By the induction hypothesis si ∈ GSki .LetA := {s′i ∈ Hi | s′i ≻LSk si}.Suppose by contradiction that A 6= ∅. Choose a maximal element s∗i ∈ Aw.r.t. the ≻LSk ordering on A. Then for all l ∈ [0..k], s∗i ∈ LSli, so in particulars∗i ∈ LSki , which contradicts the assumption about si. So A = ∅, which impliesby the induction hypothesis that ¬∃s′i ∈ Hi s′i ≻GSk si. So si ∈ GSk+1i . ✷Proof of Lemma 3.We prove by induction that for all k ≥ 0LWk = GWk.Again, we only need to establish the induction step. We prove first thatLWk+1 ⊆GWk+1. Take si ∈ LWk+1i . By the induction hypothesis LWk =GWk, so si ∈ GWki . LetA := {s′i ∈ Hi | s′i ≻LWk si}.Suppose by contradiction that A 6= ∅. Define the function index : A→ [0..k]byindex(s′i) :={k if s′i ∈ LWkij if s′i ∈ LWji \ LWj+1iLet B be the set of elements of A with the maximal index, i.e.,B = argmaxs′i∈Aindex(s′i).Let now j0 be the index of the elements in B and s∗i a maximal element of Bw.r.t. the ≻wLWj0ordering on B.Note that j0 = k. Indeed, otherwise s∗i 6∈ LWj0+1i . Then for some s′′i ∈ LWj0iwe have s′′i ≻wLWj0s∗i . But s∗i ≻LWk si and LWk ⊆ LWj0, so s′′i ≻wLWk si. Hences′′i ∈ B, which contradicts the choice of s∗i .So j0 = k, which means that s∗i ∈ LWki and s∗i ≻wLWk si. But this contradictsthe fact that si ∈ LWk+1i . So A = ∅, which implies by the induction hypothesisthat ¬∃s′i ∈ Hi s′i ≻wGWk si. So si ∈ GWk+1i .Next we prove that GWk+1 ⊆ LWk+1. Take si ∈ GWk+1i . Then si ∈ GWki =LWki . Also ¬∃s′i ∈ Hi s′i ≻wGWk si, so ¬∃s′i ∈ LWki s′i ≻wGWk si, so by theinduction hypothesis ¬∃s′i ∈ LWki s′i ≻wLWk si, which means that si ∈ LWk+1i .✷8

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