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Epistemic Game Theory Lecture 4 ESSLLI12, Opole Eric Pacuit Olivier Roy TiLPS, Tilburg University MCMP, LMU Munich ai.stanford.edu/~epacuit http://olivier.amonbofis.net August 9, 2012 Eric Pacuit and Olivier Roy 1 Plan for the week 1.

  1. CKR in the tree A B A a a a (3 , 3) v 2 s 1 = ( da , d ), s 2 = ( aa , d ), d d d s 3 = ( ad , d ), s 4 = ( aa , a ), s 5 = ( ad , a ) (2 , 2) (1 , 1) (0 , 0) w 1 w 2 w 3 w 4 w 5 Bob is rational at v 2 in w 2 add asdf a def add fa sdf asdfa adds asdf asdf add fa sdf asdf adds f asfd Eric Pacuit and Olivier Roy 12

  2. CKR in the tree A B A a a a (3 , 3) v 2 v 3 s 1 = ( da , d ), s 2 = ( aa , d ), d d d s 3 = ( ad , d ), s 4 = ( aa , a ), s 5 = ( ad , a ) (2 , 2) (1 , 1) (0 , 0) w 1 w 2 w 3 w 4 w 5 Note that f ( w 1 , v 2 ) = w 2 and f ( w 1 , v 3 ) = w 4 , so there is common knowledge of S-rationality at w 1 . Eric Pacuit and Olivier Roy 12

  3. CKR in the tree Aumann’s Theorem : If Γ is a non-degenerate game of perfect information, then in all models of Γ, we have C ( A − Rat ) ⊆ BI Stalnaker’s Theorem : There exists a non-degenerate game Γ of perfect information and an extended model of Γ in which the selection function satisfies F1-F3 such that C ( S − Rat ) �⊆ BI . Eric Pacuit and Olivier Roy 13

  4. CKR in the tree Aumann’s Theorem : If Γ is a non-degenerate game of perfect information, then in all models of Γ, we have C ( A − Rat ) ⊆ BI Stalnaker’s Theorem : There exists a non-degenerate game Γ of perfect information and an extended model of Γ in which the selection function satisfies F1-F3 such that C ( S − Rat ) �⊆ BI . Revising beliefs during play: “Although it is common knowledge that Ann would play across if v 3 were reached, if Ann were to play across at v 1 , Bob would consider it possible that Ann would play down at v 3 ” Eric Pacuit and Olivier Roy 13

  5. CKR in the tree F4. For all players i and vertices v , if w ′ ∈ [ f ( w , v )] i then there exists a state w ′′ ∈ [ w ] i such that σ ( w ′ ) and σ ( w ′′ ) agree on the subtree of Γ below v . Theorem (Halpern). If Γ is a non-degenerate game of perfect information, then for every extended model of Γ in which the selection function satisfies F1-F4, we have C ( S − Rat ) ⊆ BI . Moreover, there is an extend model of Γ in which the selection function satisfies F1-F4. J. Halpern. Substantive Rationality and Backward Induction . Games and Eco- nomic Behavior, 37, pp. 425-435, 1998. Eric Pacuit and Olivier Roy 14

  6. CKR in the tree Proof of Halpern’s Theorem A B B (1 , 0) (2 , 3) (1 , 5) A (3 , 1) (4 , 4) ◮ Suppose w ∈ C ( S − Rat ). We show by induction on k that for all w ′ reachable from w by a finite path along the union of the relations ∼ i , if v is at most k moves away from a leaf, then σ i ( w ) is i ’s backward induction move at w ′ . Eric Pacuit and Olivier Roy 15

  7. CKR in the tree Proof of Halpern’s Theorem A B B (1 , 0) (2 , 3) (1 , 5) A (3 , 1) (4 , 4) ◮ Base case: we are at most 1 move away from a leaf. Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Eric Pacuit and Olivier Roy 16

  8. CKR in the tree Proof of Halpern’s Theorem A B B (1 , 0) (2 , 3) (1 , 5) A (3 , 1) (4 , 4) ◮ Base case: we are at most 1 move away from a leaf. Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Since w ∈ C ( S − Rat ), we know that w ′ ∈ C ( S − Rat ). Eric Pacuit and Olivier Roy 16

  9. CKR in the tree Proof of Halpern’s Theorem A B B (1 , 0) (2 , 3) (1 , 5) A (3 , 1) (4 , 4) ◮ Base case: we are at most 1 move away from a leaf. Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Since w ∈ C ( S − Rat ), we know that w ′ ∈ C ( S − Rat ). So i must play her BI move at f ( w ′ , v ). Eric Pacuit and Olivier Roy 16

  10. CKR in the tree Proof of Halpern’s Theorem A B B (1 , 0) (2 , 3) (1 , 5) A (3 , 1) (4 , 4) ◮ Base case: we are at most 1 move away from a leaf. Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Since w ∈ C ( S − Rat ), we know that w ′ ∈ C ( S − Rat ). So i must play her BI move at f ( w ′ , v ). But then by F3 this must also be the case at ( w ′ , v ). Eric Pacuit and Olivier Roy 16

  11. CKR in the tree Proof of Halpern’s Theorem A B B (1 , 0) (2 , 3) (1 , 5) A (3 , 1) (4 , 4) ◮ Base case: we are at most 1 move away from a leaf. Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Since w ∈ C ( S − Rat ), we know that w ′ ∈ C ( S − Rat ). So i must play her BI move at f ( w ′ , v ). But then by F3 this must also be the case at ( w ′ , v ). Eric Pacuit and Olivier Roy 17

  12. CKR in the tree Proof of Halpern’s Theorem A B B w ′′ (1 , 0) (2 , 3) (1 , 5) A w ’ ∗ w ′ (3 , 1) (4 , 4) ◮ Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Assume, towards contradiction, that σ ( w ) i ( v ) = a is not the BI move for player i . By the same argument as before, i must be rational at w ′′ = f ( w ′ , v ). Furthermore, by F3 all players play according to the BI solution after v at ( w ′ , v ). Furthermore, by IH, at all vertices below v the players must play their BI moves. Eric Pacuit and Olivier Roy 18

  13. CKR in the tree Proof of Halpern’s Theorem w 3 A B B w ′′ (1 , 0) (2 , 3) (1 , 5) A f w ’ ∗ w ′ (3 , 1) (4 , 4) ◮ Induction step. Suppose w ∈ C ( S − Rat ). Take any w ′ reachable from w . Assume, towards contradiction, that σ ( w ) i ( v ) = a is not the BI move for player i . Since w is also in C ( S − Rat ), we know by definition i must be rational at w ′′ = f ( w ′ , v ). But then, by F3 and our IH, all players play according to the BI solution after v at w ′′ . Eric Pacuit and Olivier Roy 19

  14. CKR in the tree Proof of Halpern’s Theorem w 3 A i B B w ′′ (1 , 0) (2 , 3) (1 , 5) A f w ’ ∗ w ′ (3 , 1) (4 , 4) ◮ i ’s rationality at w ′′ means, in particular, that there is a w 3 ∈ [ w ′′ ] i such that h v i ( σ i ( w ′′ ) , σ − i ( w 3 )) ≥ h v i (( bi i , σ − i ( w 3 ))) for bi i i ’s backward induction strategy. Eric Pacuit and Olivier Roy 20

  15. CKR in the tree Proof of Halpern’s Theorem w 3 A i B B w 4 w ′′ (1 , 0) (2 , 3) (1 , 5) A i f w ’ ∗ w ′ (3 , 1) (4 , 4) ◮ But then by F4 there must exists w 4 ∈ [ w ] i such that σ ( w 4 ) σ ( w 3 ) at the same in the sub-tree starting at v . Eric Pacuit and Olivier Roy 21

  16. CKR in the tree Proof of Halpern’s Theorem w 3 A i B B w 4 w ′′ (1 , 0) (2 , 3) (1 , 5) A i f w ’ ∗ w ′ (3 , 1) (4 , 4) ◮ But then by F4 there must exists w 4 ∈ [ w ] i such that σ ( w 4 ) σ ( w 3 ) at the same in the sub-tree starting at v . Since w 4 is reachable from w , in that state all players play according to the backward induction after v , and so this is also true of w 3 . Eric Pacuit and Olivier Roy 21

  17. CKR in the tree Proof of Halpern’s Theorem w 3 A i B B w 4 w ′′ (1 , 0) (2 , 3) (1 , 5) A i f w ’ ∗ w ′ (3 , 1) (4 , 4) ◮ But then by F4 there must exists w 4 ∈ [ w ] i such that σ ( w 4 ) σ ( w 3 ) at the same in the sub-tree starting at v . Since w 4 is reachable from w , in that state all players play according to the backward induction after v , and so this is also true of w 3 . But then since the game is non-degenerate, playing something else than bi i must make i strictly worst off at that state, a contradiction. Eric Pacuit and Olivier Roy 21

  18. CKR in the tree Some remarks Aumann has proved that common knowledge of substantive rationality implies the backward induction solution in games of perfect information. Joseph Halpern Eric Pacuit and Olivier Roy 22

  19. CKR in the tree Some remarks Aumann has proved that common knowledge of substantive rationality implies the backward induction solution in games of perfect information. Joseph Halpern Aumann’s theorem is a special case of Halpern’s, where the converse of (F4) also holds. Beliefs (in fact, knowledge) are fixed. Eric Pacuit and Olivier Roy 22

  20. CKR in the tree Some remarks Aumann has proved that common knowledge of substantive rationality implies the backward induction solution in games of perfect information. Stalnaker has proved that it does not. Joseph Halpern Aumann’s theorem is a special case of Halpern’s, where the converse of (F4) also holds. Beliefs (in fact, knowledge) are fixed. Stalnaker’s theorem, as we saw, uses a more liberal belief revision policy. Eric Pacuit and Olivier Roy 22

  21. CKR in the tree Some remarks Aumann has proved that common knowledge of substantive rationality implies the backward induction solution in games of perfect information. Stalnaker has proved that it does not. Joseph Halpern Aumann’s theorem is a special case of Halpern’s, where the converse of (F4) also holds. Beliefs (in fact, knowledge) are fixed. Stalnaker’s theorem, as we saw, uses a more liberal belief revision policy. Belief revision is key in extensive games. You might observe things you didn’t expect, revise your beliefs on that, and make your decision for the next move. Eric Pacuit and Olivier Roy 22

  22. CKR in the tree Some remarks Eric Pacuit and Olivier Roy 23

  23. CKR in the tree Some remarks Belief revision is key in extensive games. Eric Pacuit and Olivier Roy 23

  24. CKR in the tree Some remarks Belief revision is key in extensive games. Are there, then, epistemic conditions using more liberal belief revision policies that still imply BI? Eric Pacuit and Olivier Roy 23

  25. CKR in the tree Some remarks Belief revision is key in extensive games. Are there, then, epistemic conditions using more liberal belief revision policies that still imply BI? Yes. Eric Pacuit and Olivier Roy 23

  26. CKR in the tree Some remarks Belief revision is key in extensive games. Are there, then, epistemic conditions using more liberal belief revision policies that still imply BI? Yes. We just saw one... But by now dominant view on epistemic conditions for BI is: ◮ Rationality and common strong belief in rationality implies BI. Strong belief in rationality := a belief that you keep as long as you don’t receive information that contradicts it. Battigalli, P. and Siniscalchi, M. ”Strong belief and forward induction reasoning”. Journal of Economic Theory . 106(2), 2002. Keep ’hoping’ for rationality: a solution to the backward induction paradox. Synthese . 169(2), 2009. Eric Pacuit and Olivier Roy 23

  27. CKR in the tree From backward induction to weak dominance in the matrix (3, 3) A Bob Hi Ann (0, 0) B Hi Lo Bob (1, 1) Ann Lo (2, 2) Eric Pacuit and Olivier Roy 24

  28. CKR in the tree From backward induction to weak dominance in the matrix (3, 3) A Bob Hi Ann (0, 0) B Hi Lo Bob (1, 1) Ann Lo (2, 2) Hi, A Hi, B Lo, A Lo, B Hi 3, 3 0, 0 2, 2 2, 2 Lo 1, 1 1, 1 2, 2 2, 2 Eric Pacuit and Olivier Roy 24

  29. CKR in the tree From backward induction to weak dominance in the matrix (3, 3) A Bob Hi Ann (0, 0) B Hi Lo Bob (1, 1) Ann Lo (2, 2) Hi, A Hi, B Lo, A Lo, B Hi 3, 3 0, 0 2, 2 2, 2 Lo 1, 1 1, 1 2, 2 2, 2 Eric Pacuit and Olivier Roy 24

  30. Weak Dominance A B Eric Pacuit and Olivier Roy 25

  31. Weak Dominance A B Eric Pacuit and Olivier Roy 25

  32. Weak Dominance A = = = > > B Eric Pacuit and Olivier Roy 25

  33. Weak Dominance A = = = > > B ◮ All strictly dominated strategies are weakly dominated. Eric Pacuit and Olivier Roy 25

  34. Weak Dominance Suppose that G = � N , { S i } i ∈ N , { u i } i ∈ N � is a strategic game. A strategy s i ∈ S i is weakly dominated (possibly by a mixed strategy) with respect to X ⊆ S − i iff there is no full support probability measure p ∈ ∆ > 0 ( X ) such that s i is a best response with respect to p . Eric Pacuit and Olivier Roy 25

  35. Strategic Reasoning and Admissibility L R U 1,1 0,1 D 0,2 1,0 Eric Pacuit and Olivier Roy 26

  36. Strategic Reasoning and Admissibility L R U 1,1 0,1 D 0,2 1,0 Suppose rationality incorporates admissibility (or cautiousness ). Eric Pacuit and Olivier Roy 26

  37. Strategic Reasoning and Admissibility L R U 1,1 0,1 D 0,2 1,0 Suppose rationality incorporates admissibility (or cautiousness ). 1. Both Row and Column should use a full-support probability measure Eric Pacuit and Olivier Roy 26

  38. Strategic Reasoning and Admissibility L R U 1,1 0,1 D 0,2 1,0 Suppose rationality incorporates admissibility (or cautiousness ). 1. Both Row and Column should use a full-support probability measure 2. But, if Row thinks that Column is rational then should she not assign probability 1 to L ? Eric Pacuit and Olivier Roy 26

  39. Strategic Reasoning and Admissibility “The argument for deletion of a weakly dominated strategy for player i is that he contemplates the possibility that every strategy combination of his rivals occurs with positive probability. However, this hypothesis clashes with the logic of iterated deletion, which assumes, precisely, that eliminated strategies are not expected to occur.” Mas-Colell, Whinston and Green. Introduction to Microeconomics . 1995. Eric Pacuit and Olivier Roy 27

  40. Strategic Reasoning and Admissibility The condition that the players incorporate admissibility into their rationality calculations seems to conflict with the condition that the players think the other players are rational (there is a tension between admissibility and strategic reasoning) Eric Pacuit and Olivier Roy 28

  41. Strategic Reasoning and Admissibility The condition that the players incorporate admissibility into their rationality calculations seems to conflict with the condition that the players think the other players are rational (there is a tension between admissibility and strategic reasoning) Does assuming that it is commonly known that players play only admissible strategies lead to a process of iterated removal of weakly dominated strategies? Eric Pacuit and Olivier Roy 28

  42. Strategic Reasoning and Admissibility The condition that the players incorporate admissibility into their rationality calculations seems to conflict with the condition that the players think the other players are rational (there is a tension between admissibility and strategic reasoning) Does assuming that it is commonly known that players play only admissible strategies lead to a process of iterated removal of weakly dominated strategies? No! L. Samuelson. Dominated Strategies and Common Knowledge . Games and Economic Behavior (1992). Eric Pacuit and Olivier Roy 28

  43. Iterated Admissibility Bob T L R 1,1 1,0 T U Ann 1,0 0,1 B U T weakly dominates B Eric Pacuit and Olivier Roy 29

  44. Iterated Admissibility Bob T L R 1,1 1,0 T U Ann 1,0 0,1 B U T weakly dominates B Eric Pacuit and Olivier Roy 29

  45. Iterated Admissibility Bob T L R 1,1 1,0 T U Ann 1,0 0,1 B U Then L strictly dominates R . Eric Pacuit and Olivier Roy 29

  46. Iterated Admissibility Bob T L R 1,1 1,0 T U Ann 1,0 0,1 B U The IA set Eric Pacuit and Olivier Roy 29

  47. Iterated Admissibility Bob T L R 1,1 1,0 T U Ann 1,0 0,1 B U But, now what is the reason for not playing B? Eric Pacuit and Olivier Roy 29

  48. Iterated Admissibility Bob T L R 1,1 1,0 T U Ann 1,0 0,1 B U Theorem (Samuelson). There is no model of this game satisfying common knowledge of rationality (where “rationality” incorporates admissibility) Eric Pacuit and Olivier Roy 29

  49. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } There is no model of this game with common knowledge of admissibility. Eric Pacuit and Olivier Roy 30

  50. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } The ”full” model of the game: B is not admissible given Ann’s information Eric Pacuit and Olivier Roy 30

  51. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } The ”full” model of the game: B is not admissible given Ann’s information Eric Pacuit and Olivier Roy 30

  52. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } What is wrong with this model? asdf ad fa sdf a fsd asdf adsf adfs Eric Pacuit and Olivier Roy 30

  53. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } Privacy of Tie-Breaking/No Extraneous Beliefs : If a strategy is rational for an opponent, then it cannot be “ruled out”. Eric Pacuit and Olivier Roy 30

  54. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } Moving to choice sets . asdf ad fa sdf a fsd asdf adsf adfs Eric Pacuit and Olivier Roy 30

  55. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } Moving to choice sets . asdf ad fa sdf a fsd asdf adsf adfs Eric Pacuit and Olivier Roy 30

  56. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } Ann thinks: Bob has a reason to play L OR Bob has a reason to play R OR Bob has not yet settled on a choice Eric Pacuit and Olivier Roy 30

  57. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } Still there is no model with common knowledge that players have admissibility-based reasons asdf ad fa sdf a fsd asdf adsf adfs Eric Pacuit and Olivier Roy 30

  58. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } there is a reason to play T provided Ann considers it possible that Bob might play R (actually three cases to consider here) Eric Pacuit and Olivier Roy 30

  59. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } But there is a reason to play R provided it is possible that Ann has a reason to play B Eric Pacuit and Olivier Roy 30

  60. Common Knowledge of Admissibility Bob T , L T , R T , { L , R } T L R 1,1 1,0 T U B , L B , R B , { L , R } Ann 1,0 0,1 B U { T , B } , L { T , B } , R { T , B } , { L , R } But, there is no reason to play B if there is a reason for Bob to play R . ada dad asd a ds asd ad d Eric Pacuit and Olivier Roy 30

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