Q uestion In light of the proof of Proposition 4 in Lecture 7, choosing Q= A, ti(a)=ai A-i should come natural. The tricky part is how to define beliefs However, inspecting the iterative construction suggests the following. Of course, for each player i E N, we can specify conditional probabilities pi ((ti(a)) directly (a prior which generates them is easy to construct) Let K be such that AK+l= Ak. For each ai E Ai, let ki(ai)= maxk E 10, ...,KI a1∈A}.Ifk(a1)>0, there exists a∈△(A-) such that a∈r(a1)anda(4)=1 Then let pillai, a_i)lti(ai,a-iD)=a_(ali) Va-i, a_ E A-i Complete the description of the model by assigning pi (ti(ai, a-i)) arbitrarily for ai g Ai The rest of the problem is now merely a matter of verifying the claims Question 2 This one was a bit more tricky. As we did in Lecture 5, I will define "belief states' LA,0, 1, 2, ..,K with the following interpretation: in belief state A, a player is certain t 0=3, so that A is strictly dominant. In belief state 0, a player's beliefs are consistent with E="the probability that 0=2 is T", where T < 2. In belief state k>0, a player's beliefs are consistent with E, plus she is certain that her opponent's belief state is k-1 A state w E S comprises a specification of the actual value of 0 as well as of both player belief state. We use subscripts as follows: in state wBrv, 0=8, Player 1's belief state is a, and Player 2s belief state is y Finally, each player learns only her belief state: thus, ti(wAxy)=wery: c'=a) and similarly for Player 2 With this convention, define conditional probabilities as follows. First, for states of the form weAy and werA, let 3. P(=lt1(c4y)=1 and P1(6=]t1(4y)=1 Next, for states weou, let p(=2)1-(02u-1() and similarly for states wer0 Finally, assuming that probabilities for all states wery with e, y s k-1 have been assigned let P1(B=司]∩[y=k-1]t1(oky) P1 1∩y=k-1]t-1(aoky)Question 1 In light of the proof of Proposition 4 in Lecture 7, choosing Ω = A, ti(a) = ai × A−i should come natural. The tricky part is how to define beliefs. However, inspecting the iterative construction suggests the following. Of course, for each player i ∈ N, we can specify conditional probabilities pi(·|ti(a)) directly (a prior which generates them is easy to construct). Let K be such that AK+1 = AK. For each ai ∈ Ai , let ki(ai) = max{k ∈ {0, . . . , K} : ai ∈ Ak i }. If ki(ai) > 0, there exists α ai −i ∈ ∆(A−i) such that ai ∈ ri(α ai −i ) and α ai −i (A k−1 −i ) = 1. Then let pi((ai , a0 −i )|ti(ai , a−i)) = α ai −i (a 0 −i ) ∀a−i , a0 −i ∈ A−i Complete the description of the model by assigning pi(·|ti(ai , a−i)) arbitrarily for ai 6∈ A1 i . The rest of the problem is now merely a matter of verifying the claims. Question 2 This one was a bit more tricky. As we did in Lecture 5, I will define “belief states” {A, 0, 1, 2, . . . , K} with the following interpretation: in belief state A, a player is certain that θ = 3 2 , so that A is strictly dominant. In belief state 0, a player’s beliefs are consistent with E=“the probability that θ = 1 2 is π”, where π ≤ 1 2 . In belief state k > 0, a player’s beliefs are consistent with E, plus she is certain that her opponent’s belief state is k − 1. A state ω ∈ Ω comprises a specification of the actual value of θ as well as of both players’ belief state. We use subscripts as follows: in state ωθxy ¯ , θ = ¯θ, Player 1’s belief state is x, and Player 2’s belief state is y. Finally, each player learns only her belief state: thus, t1(ωθxy) = {ωθ 0x0y 0 : x 0 = x} and similarly for Player 2. With this convention, define conditional probabilities as follows. First, for states of the form ωθAy and ωθxA, let p1([θ = 3 2 ]|t1(ωθAy)) = 1 and p1([θ = 3 2 ]|t1(ωθAy)) = 1. Next, for states ωθ0y, let p1([θ = 1 2 ]|t1(ωθ0y)) = π = 1 − p1([θ = 3 2 ]|t − 1(ωθ0y)) and similarly for states ωθx0. Finally, assuming that probabilities for all states ωθxy with x, y ≤ k−1 have been assigned, let p1([θ = 1 2 ] ∩ [y = k − 1]|t1(ωθky)) = π = 1 − p1([θ = 3 2 ] ∩ [y = k − 1]|t − 1(ωθky)) 3