Epistemic Game Theory Lecture 2 ESSLLI12, Opole Eric Pacuit - PowerPoint PPT Presentation

Dominance vs MEU For the converse direction, we sketch the proof for two player games and where X = S − i . 1 Let G = � S 1 , S 2 , u 1 , u 2 � be a two-player game. (Let U i : ∆( S 1 ) × ∆( S 2 ) → R be the expected utility for i ) Suppose that α ∈ ∆( S 1 ) is not a best response to any p ∈ ∆( S 2 ). ∀ p ∈ ∆( S 2 ) ∃ q ∈ ∆( S 1 ) , U 1 ( q , p ) > U 1 ( α, p ) 1 The proof of the more general statement uses the supporting hyperplane theorem from convex analysis. Eric Pacuit and Olivier Roy 7

Dominance vs MEU For the converse direction, we sketch the proof for two player games and where X = S − i . 1 Let G = � S 1 , S 2 , u 1 , u 2 � be a two-player game. (Let U i : ∆( S 1 ) × ∆( S 2 ) → R be the expected utility for i ) Suppose that α ∈ ∆( S 1 ) is not a best response to any p ∈ ∆( S 2 ). ∀ p ∈ ∆( S 2 ) ∃ q ∈ ∆( S 1 ) , U 1 ( q , p ) > U 1 ( α, p ) We can define a function b : ∆( S 2 ) → ∆( S 1 ) where, for each p ∈ ∆( S 2 ), U 1 ( b ( p ) , p ) > U 1 ( α, p ). 1 The proof of the more general statement uses the supporting hyperplane theorem from convex analysis. Eric Pacuit and Olivier Roy 7

Dominance vs MEU Consider the game G ′ = � S 1 , S 2 , u 1 , u 2 � where u 1 ( s 1 , s 2 ) = u 1 ( s 1 , s 2 ) − U 1 ( α, s 2 ) and u 2 ( s 1 , s 2 ) = − u 1 ( s 1 , s 2 ) Eric Pacuit and Olivier Roy 8

Dominance vs MEU Consider the game G ′ = � S 1 , S 2 , u 1 , u 2 � where u 1 ( s 1 , s 2 ) = u 1 ( s 1 , s 2 ) − U 1 ( α, s 2 ) and u 2 ( s 1 , s 2 ) = − u 1 ( s 1 , s 2 ) By the minimax theorem, there is a Nash equilibrium ( p ∗ 1 , p ∗ 2 ) such that for all m ∈ ∆( S 2 ), U ( p ∗ 1 , m ) ≥ U 1 ( p ∗ 1 , p ∗ 2 ) ≥ U 1 ( b ( p ∗ 2 ) , p ∗ 2 ) Eric Pacuit and Olivier Roy 8

Dominance vs MEU Consider the game G ′ = � S 1 , S 2 , u 1 , u 2 � where u 1 ( s 1 , s 2 ) = u 1 ( s 1 , s 2 ) − U 1 ( α, s 2 ) and u 2 ( s 1 , s 2 ) = − u 1 ( s 1 , s 2 ) By the minimax theorem, there is a Nash equilibrium ( p ∗ 1 , p ∗ 2 ) such that for all m ∈ ∆( S 2 ), U ( p ∗ 1 , m ) ≥ U 1 ( p ∗ 1 , p ∗ 2 ) ≥ U 1 ( b ( p ∗ 2 ) , p ∗ 2 ) We now prove that U 1 ( b ( p ∗ 2 ) , p ∗ 2 ) > 0: Eric Pacuit and Olivier Roy 8

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ > 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ 2 )( x ) U 1 ( α, p ∗ = 2 ) · � 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ > 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ 2 )( x ) U 1 ( α, p ∗ = 2 ) · � 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ > 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ 2 )( x ) U 1 ( α, p ∗ = 2 ) · � 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ > 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ 2 )( x ) U 1 ( α, p ∗ = 2 ) · � 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ > 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ 2 )( x ) U 1 ( α, p ∗ = 2 ) · � 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ y ∈ S 2 p ∗ = 2 ) − � 2 )( x ) � 2 ( y ) U 1 ( α, y ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ 2 )( x ) U 1 ( α, p ∗ = 2 ) · � 2 ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ y ∈ S 2 p ∗ = 2 ) − � 2 )( x ) � 2 ( y ) U 1 ( α, y ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ = 2 ) · � 2 )( x ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ = 2 ) · � 2 )( x ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU U 1 ( b ( p ∗ 2 ) , p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ 2 ) = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y )[ u 1 ( x , y ) − U 1 ( α, y )] x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ = � � 2 ( y ) u 1 ( x , y ) x ∈ S 1 y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( b ( p ∗ 2 ) , p ∗ = 2 ) y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ y ∈ S 2 b ( p ∗ 2 )( x ) p ∗ > 2 ) − � � 2 ( y ) U 1 ( α, y ) x ∈ S 1 U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ y ∈ S 2 p ∗ = 2 ) − � 2 )( x ) � 2 ( y ) U 1 ( α, y ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ x ∈ S 1 b ( p ∗ = 2 ) · � 2 )( x ) U 1 ( α, p ∗ 2 ) − U 1 ( α, p ∗ = 2 ) = 0 Eric Pacuit and Olivier Roy 9

Dominance vs MEU Hence, for all m ∈ ∆( S 2 ) we have U ( p ∗ 1 , m ) ≥ U 1 ( p ∗ 1 , p ∗ 2 ) ≥ U 1 ( b ( p ∗ 2 ) , p ∗ 2 ) > 0 Eric Pacuit and Olivier Roy 10

Dominance vs MEU Hence, for all m ∈ ∆( S 2 ) we have U ( p ∗ 1 , m ) ≥ U 1 ( p ∗ 1 , p ∗ 2 ) ≥ U 1 ( b ( p ∗ 2 ) , p ∗ 2 ) > 0 which implies for all m ∈ ∆( S 2 ), U 1 ( p ∗ 1 , m ) > U 1 ( α, m ), and so α is strictly dominated by p ∗ 1 . Eric Pacuit and Olivier Roy 10

Dominance vs MEU Important Issue: Correlated Beliefs x l r y l r z l r 1,1,3 1,0,3 1,1,2 1,0,0 1,1,0 1,0,0 u u u 0,1,0 0,0,0 0,1,0 1,1,2 0,1,3 0,0,3 d d d Eric Pacuit and Olivier Roy 11

Dominance vs MEU Important Issue: Correlated Beliefs x l r y l r z l r 1,1,3 1,0,3 1,1,2 1,0,0 1,1,0 1,0,0 u u u 0,1,0 0,0,0 0,1,0 1,1,2 0,1,3 0,0,3 d d d ◮ Note that y is not strictly dominated for Charles. Eric Pacuit and Olivier Roy 11

Dominance vs MEU Important Issue: Correlated Beliefs x l r y l r z l r 1,1,3 1,0,3 1,1,2 1,0,0 1,1,0 1,0,0 u u u 0,1,0 0,0,0 0,1,0 1,1,2 0,1,3 0,0,3 d d d ◮ Note that y is not strictly dominated for Charles. ◮ It is easy to find a probability measure p ∈ ∆( S A × S B ) such that y is a best response to p . Suppose that p ( u , l ) = p ( d , r ) = 1 2 . Then, EU ( x , p ) = EU ( z , p ) = 1 . 5 while EU ( y , p ) = 2. Eric Pacuit and Olivier Roy 11

Dominance vs MEU Important Issue: Correlated Beliefs x l r y l r z l r 1,1,3 1,0,3 1,1,2 1,0,0 1,1,0 1,0,0 u u u 0,1,0 0,0,0 0,1,0 1,1,2 0,1,3 0,0,3 d d d ◮ Note that y is not strictly dominated for Charles. ◮ It is easy to find a probability measure p ∈ ∆( S A × S B ) such that y is a best response to p . Suppose that p ( u , l ) = p ( d , r ) = 1 2 . Then, EU ( x , p ) = EU ( z , p ) = 1 . 5 while EU ( y , p ) = 2. ◮ However, there is no probability measure p ∈ ∆( S A × S B ) such that y is a best response to p and p ( u , l ) = p ( u ) · p ( l ). Eric Pacuit and Olivier Roy 11

Dominance vs MEU x l r y l r z l r u 1,1,3 1,0,3 u 1,1,2 1,0,0 u 1,1,0 1,0,0 d 0,1,0 0,0,0 d 0,1,0 1,1,2 d 0,1,3 0,0,3 ◮ To see this, suppose that a is the probability assigned to u and b is the probability assigned to l . Then, we have: • The expected utility of y is 2 ab + 2(1 − a )(1 − b ); • The expected utility of x is 3 ab + 3 a (1 − b ) = 3 a ( b + (1 − b )) = 3 a ; and • The expected utility of z is 3(1 − a ) b + 3(1 − a )(1 − b ) = 3(1 − a )( b + (1 − b )) = 3(1 − a ). Eric Pacuit and Olivier Roy 12

Dominance vs MEU Weak Dominance and MEU Fact . Suppose that G = � N , { S i } i ∈ N , { u i } i ∈ N � is a strategic game and X ⊆ S − i . A strategy s i ∈ S i is weakly dominated (possibly by a mixed strategy) with respect to X iff there is no full support probability measure p ∈ ∆ > 0 ( X ) such that s i is a best response to p . Eric Pacuit and Olivier Roy 13

Dominance vs MEU Some preliminary remarks Eric Pacuit and Olivier Roy 14

Dominance vs MEU Propositional Attitudes ◮ We will talk about so-called propositional attitudes . These are attitudes (like knowledge, beliefs, desires, intentions, etc...) that take propositions as objects. Eric Pacuit and Olivier Roy 15

Dominance vs MEU Propositional Attitudes ◮ We will talk about so-called propositional attitudes . These are attitudes (like knowledge, beliefs, desires, intentions, etc...) that take propositions as objects. ◮ Proposition will be taken to be element of a given algebra. I.e. measurable subsets of a state space (sigma- and/or power-set algebra), formulas in a given language (abstract Boolean algebra)... Eric Pacuit and Olivier Roy 15

Dominance vs MEU All-out vs graded attitudes ◮ A propositional attitude A is all-out when, for any proposition p , the agent can only be in three states of that attitude regarding p : 1. Ap : the agent “believes” that p . 2. A ¬ p : the agent “disbelieve” that p . 3. ¬ Ap ∧ ¬ A ¬ p : the agent “suspends judgment” about p . Eric Pacuit and Olivier Roy 16

Dominance vs MEU All-out vs graded attitudes ◮ A propositional attitude A is all-out when, for any proposition p , the agent can only be in three states of that attitude regarding p : 1. Ap : the agent “believes” that p . 2. A ¬ p : the agent “disbelieve” that p . 3. ¬ Ap ∧ ¬ A ¬ p : the agent “suspends judgment” about p . ◮ A propositional attitude A is graded when, for any proposition p , the states of that attitude that the agent be in w.r.t. a proposition p can be compared according to their strength on a given scale. P ¬ P p i A 1/8 3/8 Eric Pacuit and Olivier Roy 16

Knowledge and beliefs in games Hard and Soft Attitudes ◮ Hard attitudes: • Truthful. • Unrevisable. • Fully introspective. ◮ Soft attitudes: • Can be false / mistaken. • Revisable / can be reversed. • Not fully introspective. Eric Pacuit and Olivier Roy 17

Knowledge and beliefs in games Models of graded beliefs Eric Pacuit and Olivier Roy 18

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) The other players’ types Eric Pacuit and Olivier Roy 19

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: ◮ A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) The other players’ types Eric Pacuit and Olivier Roy 19

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: ◮ A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. ◮ Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) The other players’ types Eric Pacuit and Olivier Roy 19

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: ◮ A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. ◮ Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) Player i ’s types Eric Pacuit and Olivier Roy 19

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: ◮ A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. ◮ Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) The set of all probability distributions Eric Pacuit and Olivier Roy 19

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: ◮ A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. ◮ Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) The other players’ types Eric Pacuit and Olivier Roy 19

Models of graded beliefs Harsanyi Type Space Based on the work of John Harsanyi on games with incomplete information , game theorists have developed an elegant formalism that makes precise talk about beliefs, knowledge and rationality: ◮ A type is everything a player knows privately at the beginning of the game which could affect his beliefs about payoffs and about all other players’ possible types. ◮ Each type is assigned a joint probability over the space of types and actions λ i : T i → ∆( T − i × S − i ) The other players’ choices Eric Pacuit and Olivier Roy 19

Models of graded beliefs Returning to the Example: A Game Model Bob One type for Ann ( t A ) and two types U H M for Bob ( t B , u B ) A state is a tuple of choices and 3,3 0,0 H Ann types: ( M , t A , M , u B ) 0,0 1,1 M Calculate expected utility in the usual way... U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 t B u B t A Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ One type for Ann ( t A ) and two types U H M for Bob ( t B , u B ) A state is a tuple of choices and 3,3 0,0 H Ann types: ( M , t A , M , u B ) 0,0 1,1 M Calculate expected utility in the usual way... U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 t B u B t A Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ One type for Ann ( t A ) and two types U H M for Bob ( t B , u B ) ◮ A state is a tuple of choices and 3,3 0,0 H Ann types: ( M , M , t A , t B ) 0,0 1,1 M Calculate expected utility in the usual way... U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 t B u B t A Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ One type for Ann ( t A ) and two types U H M for Bob ( t B , u B ) ◮ A state is a tuple of choices and 3,3 0,0 H Ann types: ( M , t A , M , u B ) 0,0 1,1 M ◮ Calculate expected utility in the usual way... U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 t B u B t A Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob Ann ( t A ) is rational U H M 0 · 0 . 5 + 1 · 0 ≥ 3 · 0 . 5 + 0 · 0 . 2 Bob is rational 3,3 0,0 H Ann 0 · 0 . 5 + 1 · 0 ≥ 3 · 0 . 5 + 0 · 0 . 2 0,0 1,1 M Bob thinks Ann is irrational P B ( Irrat ( Ann )) = 0 . xx U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 u B t A t B Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ M is rational for Ann ( t A ) U H M 0 · 0 . 2 + 1 · 0 . 8 ≥ 3 · 0 . 2 + 0 · 0 . 8 Bob is rational 3,3 0,0 H Ann 0 · 0 . 5 + 1 · 0 ≥ 3 · 0 . 5 + 0 · 0 . 2 0,0 1,1 M Bob thinks Ann is irrational P B ( Irrat ( Ann )) = 0 . xx U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 u B t A t B Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ M is rational for Ann ( t A ) U H M 0 · 0 . 2 + 1 · 0 . 8 ≥ 3 · 0 . 2 + 0 · 0 . 8 ◮ M is rational for Bob ( t B ) 3,3 0,0 H Ann 0 · 0 + 1 · 1 ≥ 3 · 0 + 0 · 1 0,0 1,1 M Bob thinks Ann is irrational P B ( Irrat ( Ann )) = 0 . xx U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 u B t A t B Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ M is rational for Ann ( t A ) U H M 0 · 0 . 2 + 1 · 0 . 8 ≥ 3 · 0 . 2 + 0 · 0 . 8 ◮ M is rational for Bob ( t B ) 3,3 0,0 H Ann 0 · 0 + 1 · 1 ≥ 3 · 0 + 0 · 1 0,0 1,1 ◮ Ann thinks Bob may be irrational M P B ( Irrat ( Ann )) = 0 . xx U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 u B t A t B Eric Pacuit and Olivier Roy 20

Models of graded beliefs Returning to the Example: A Game Model Bob ◮ M is rational for Ann ( t A ) U H M 0 · 0 . 2 + 1 · 0 . 8 ≥ 3 · 0 . 2 + 0 · 0 . 8 ◮ M is rational for Bob ( t B ) 3,3 0,0 H Ann 0 · 0 + 1 · 1 ≥ 3 · 0 + 0 · 1 0,0 1,1 ◮ Ann thinks Bob may be irrational M P A ( Irrat [ B ]) = 0 . 3, P A ( Rat [ B ]) = 0 . 7 U H M U H M U H M t B 0 0.5 t A t A 0 1 0.4 0.6 u B 0.2 0.3 u B t A t B Eric Pacuit and Olivier Roy 20

Models of graded beliefs Rationality Let G = � N , { S i } i ∈ N , { u i } i ∈ N � be a strategic game and T = �{ T i } i ∈ N , { λ i } i ∈ N , S � a type space for G . Eric Pacuit and Olivier Roy 21

Models of graded beliefs Rationality Let G = � N , { S i } i ∈ N , { u i } i ∈ N � be a strategic game and T = �{ T i } i ∈ N , { λ i } i ∈ N , S � a type space for G . For each t i ∈ T i , we can define a probability measure p t i ∈ ∆( S − i ): � p t i ( s − i ) = λ i ( t i )( s − i , t − i ) t − i ∈ T − i Eric Pacuit and Olivier Roy 21

Models of graded beliefs Rationality Let G = � N , { S i } i ∈ N , { u i } i ∈ N � be a strategic game and T = �{ T i } i ∈ N , { λ i } i ∈ N , S � a type space for G . For each t i ∈ T i , we can define a probability measure p t i ∈ ∆( S − i ): � p t i ( s − i ) = λ i ( t i )( s − i , t − i ) t − i ∈ T − i The set of states (pairs of strategy profiles and type profiles) where player i chooses rationally is: Rat i := { ( s i , t i ) | s i is a best response to p t i } The event that all players are rational is Rat = { ( s , t ) | for all i , ( s i , t i ) ∈ Rat i } . Eric Pacuit and Olivier Roy 21

Models of graded beliefs Common “knowledge” of rationality In much of this literature, “full belief” or sometimes “knowledge” is identified with probability 1. (This is not a philosophical commitment, but rather a term of art!) Eric Pacuit and Olivier Roy 22

Models of graded beliefs Common knowledge of rationality Define R n i by induction on n : Eric Pacuit and Olivier Roy 23

Models of graded beliefs Common knowledge of rationality Define R n i by induction on n : Let R 1 i = Rat i . Eric Pacuit and Olivier Roy 23

Models of graded beliefs Common knowledge of rationality Define R n i by induction on n : Let R 1 i = Rat i . Suppose that for each i , R n i has been defined. Define R n − i as follows: R n − i = { ( s , t ) | s ∈ S − i , t ∈ T − j , and for each j � = i , ( s j , t j ) ∈ R n j } . Eric Pacuit and Olivier Roy 23

Models of graded beliefs Common knowledge of rationality Define R n i by induction on n : Let R 1 i = Rat i . Suppose that for each i , R n i has been defined. Define R n − i as follows: R n − i = { ( s , t ) | s ∈ S − i , t ∈ T − j , and for each j � = i , ( s j , t j ) ∈ R n j } . For each n > 1, R n +1 = { ( s , t ) | ( s , t ) ∈ R n i and λ i ( t ) assigns probability 1 to R n − i } i Eric Pacuit and Olivier Roy 23

Models of graded beliefs Common knowledge of rationality Define R n i by induction on n : Let R 1 i = Rat i . Suppose that for each i , R n i has been defined. Define R n − i as follows: R n − i = { ( s , t ) | s ∈ S − i , t ∈ T − j , and for each j � = i , ( s j , t j ) ∈ R n j } . For each n > 1, R n +1 = { ( s , t ) | ( s , t ) ∈ R n i and λ i ( t ) assigns probability 1 to R n − i } i Common knowledge of rationality is: � R n � R n � R n 1 × 2 × · · · × N n ≥ 1 n ≥ 1 n ≥ 1 Eric Pacuit and Olivier Roy 23

Models of graded beliefs B U L R ◮ Consider the state ( d , r , a 3 , b 3 ). Both a 3 2,2 0,0 U and b 3 correctly believe that (i.e., assign A probability 1 to) the outcome is ( d , r ) 0,0 1,1 D λ A ( a 1 ) λ A ( a 2 ) λ A ( a 3 ) L R L R L R 0 . 5 0 . 5 0 . 5 0 0 0 b 1 b 1 b 1 0 0 0 0 0 0 . 5 b 2 b 2 b 2 0 0 0 0 . 5 0 0 . 5 b 3 b 3 b 3 λ B ( b 1 ) λ B ( b 2 ) λ B ( b 3 ) U D U D U D a 1 0 . 5 0 a 1 0 . 5 0 a 1 0 0 a 2 0 0 . 5 a 2 0 0 a 2 0 0 . 5 a 3 0 0 a 3 0 0 . 5 a 3 0 0 . 5 Eric Pacuit and Olivier Roy 24

Models of graded beliefs B ◮ This fact is not common knowledge: a 3 U L R assigns a 0.5 probability to Bob being of type b 2 , and type b 2 assigns a 0.5 2,2 0,0 U probability to Ann playing l . Ann does not A 0,0 1,1 know that Bob knows that she is playing r D λ A ( a 1 ) L R λ A ( a 2 ) L R λ A ( a 3 ) L R b 1 0 . 5 0 . 5 b 1 0 . 5 0 b 1 0 0 0 0 0 0 0 0 . 5 b 2 b 2 b 2 0 0 0 0 . 5 0 0 . 5 b 3 b 3 b 3 λ B ( b 1 ) λ B ( b 2 ) λ B ( b 3 ) U D U D U D 0 . 5 0 0 . 5 0 0 0 a 1 a 1 a 1 0 0 . 5 0 0 0 0 . 5 a 2 a 2 a 2 0 0 0 0 . 5 0 0 . 5 a 3 a 3 a 3 Eric Pacuit and Olivier Roy 24

Models of graded beliefs B U L R ◮ Furthermore, while it is true that both 2,2 0,0 U Ann and Bob are rational, it is not A common knowledge that they are rational. 0,0 1,1 D λ A ( a 1 ) λ A ( a 2 ) λ A ( a 3 ) L R L R L R 0 . 5 0 . 5 0 . 5 0 0 0 b 1 b 1 b 1 0 0 0 0 0 0 . 5 b 2 b 2 b 2 0 0 0 0 . 5 0 0 . 5 b 3 b 3 b 3 λ B ( b 1 ) λ B ( b 2 ) λ B ( b 3 ) U D U D U D a 1 0 . 5 0 a 1 0 . 5 0 a 1 0 0 a 2 0 0 . 5 a 2 0 0 a 2 0 0 . 5 a 3 0 0 a 3 0 0 . 5 a 3 0 0 . 5 Eric Pacuit and Olivier Roy 24

Models of graded beliefs General Comments ◮ Suppressed mathematical details about probabilities ( σ -algebra, etc.) Eric Pacuit and Olivier Roy 25

Models of graded beliefs General Comments ◮ Suppressed mathematical details about probabilities ( σ -algebra, etc.) ◮ “Impossibility” is identified with probability 0, but it is an important distinction (especially for infinite games) Eric Pacuit and Olivier Roy 25

Models of graded beliefs General Comments ◮ Suppressed mathematical details about probabilities ( σ -algebra, etc.) ◮ “Impossibility” is identified with probability 0, but it is an important distinction (especially for infinite games) ◮ We can model “soft” information using conditional probability systems, lexicographic probabilities, nonstandard probabilities (more on this later). Eric Pacuit and Olivier Roy 25

Models of all-out attitudes. Eric Pacuit and Olivier Roy 26

Models of all-out attitudes Hard Information Eric Pacuit and Olivier Roy 27

Models of all-out attitudes Example Bob r l Ann u 1, -1 -1, 1 d -1,1 1, -1 Eric Pacuit and Olivier Roy 28

Models of all-out attitudes Example Bob r l Ann u 1, -1 -1, 1 d -1,1 1, -1 u , r w 1 w 3 u , l w 2 w 4 d , l d , r Eric Pacuit and Olivier Roy 28

Models of all-out attitudes Example Bob r l Ann u 1, -1 -1, 1 d -1,1 1, -1 u , r w 1 w 3 u , l w 2 w 4 d , l d , r Eric Pacuit and Olivier Roy 28

Models of all-out attitudes Example Bob r l Ann u 1, -1 -1, 1 d -1,1 1, -1 A u , r w 1 w 3 u , l A w 2 w 4 d , l d , r Eric Pacuit and Olivier Roy 28

Models of all-out attitudes Example Bob r l Ann u 1, -1 -1, 1 d -1,1 1, -1 A u , r w 1 w 3 u , l B B A w 2 w 4 d , l d , r Eric Pacuit and Olivier Roy 28

Models of all-out attitudes Epistemic Model Suppose that G is a strategic game, S is the set of strategy profiles of G , and Ag is the set of players. An epistemic model based on S and Ag is a triple � W , { Π i } i ∈ Ag , σ � , where W is a nonempty set, for each i ∈ Ag , Π i is a partition 2 over W and σ : W → S is a strategy function. 2 A partition of W is a pairwise disjoint collection of subsets of W whose union is all of W . Elements of a partition Π on W are called cells , and for w ∈ W , let Π( w ) denote the cell of Π containing w . Eric Pacuit and Olivier Roy 29

Models of all-out attitudes Epistemic Model Suppose that G is a strategic game, S is the set of strategy profiles of G , and Ag is the set of players. An epistemic model based on S and Ag is a triple � W , { Π i } i ∈ Ag , σ � , where W is a nonempty set, for each i ∈ Ag , Π i is a partition 2 over W and σ : W → S is a strategy function. 2 A partition of W is a pairwise disjoint collection of subsets of W whose union is all of W . Elements of a partition Π on W are called cells , and for w ∈ W , let Π( w ) denote the cell of Π containing w . Eric Pacuit and Olivier Roy 29

Models of all-out attitudes Game models Game G Eric Pacuit and Olivier Roy 30

Models of all-out attitudes Game models Strategy Space Game G b a Eric Pacuit and Olivier Roy 30

Models of all-out attitudes Game models Strategy Space Game G b a Game Model Eric Pacuit and Olivier Roy 30

Models of all-out attitudes Game models Strategy Space Game G b a Game Model Eric Pacuit and Olivier Roy 30

Models of all-out attitudes Game models Strategy Space Game G b a Game Model Eric Pacuit and Olivier Roy 30

Models of all-out attitudes Kripke Model for S5 Prop is a given set of atomic propositions and Ag is a set of agents. An epistemic model based on Prop and Ag is a triple � W , { Π i } i ∈ Ag , V � , where W is a nonempty set, for each i ∈ Ag , Π i is a partition over W and V : W → P ( Prop ) is a valuation function. Eric Pacuit and Olivier Roy 31

Models of all-out attitudes Example A r l S u 1, -1 -1, 1 d -1,1 1, -1 A u , r w 1 w 3 u , l S S A w 2 w 4 d , l d , r Eric Pacuit and Olivier Roy 32

Models of all-out attitudes Example A w 1 u , r w 3 u , l S S A w 2 w 4 d , l d , r Eric Pacuit and Olivier Roy 33

Models of all-out attitudes Example A w 1 u , r w 3 u , l S S A w 2 w 4 d , l d , r = K i ϕ iff for all w ′ ∈ π i ( w ), M w ′ | ◮ M , w | = ϕ . Eric Pacuit and Olivier Roy 33

Models of all-out attitudes Example A w 1 u , r w 3 u , l U S S A w 2 w 4 d , l d , r = K i ϕ iff for all w ′ ∈ π i ( w ), M w ′ | ◮ M , w | = ϕ . ◮ One assumption: Ex-interim condition. • If w ′ ∈ π i ( w ) then σ ( w ) i = σ ( w ′ ) i . Eric Pacuit and Olivier Roy 33

Models of all-out attitudes Hard Information, Axiomatically 1. Closed under known implication (K): K i ( ϕ → ψ ) → ( K i ϕ → K i ψ ) 2. Logical truth are known (NEC): If | = ϕ then | = K i ϕ 3. Truthful, (T): K i ϕ → ϕ 4. Positive introspection (4): K i ϕ → K i K i ϕ 5. Negative introspection (5): ¬ K i ϕ → K i ¬ K i ϕ Eric Pacuit and Olivier Roy 34