《博弈论》(英文版) lect 8.5 More on Auctions

Eco514-Game Theory Lecture 8.5: More on Auctions; PS#1 Marciano siniscalchi October 14. 1999 Introduction These notes essentially tie up a few loose ends in Lecture 8; in particular, I exhibit examples of inefficiencies in first- and second-price auctions. I would also like to briefly comment on Questions 1 and 2 in Problem Set 2 The first-price auction may be inefficient even with private values Both examples I am going to show are due to Eric Maskin(to the best of my knowledge) The first point I wish to make is that, even in a private-values setting, asymmetries may render a first-price auction inefficient Consider a two-bidder situation in which the value of the first bidder is uniformly dis- tributed on 0, 1, and the value of the second bidder is uniformly distributed on 0, 2. We look for a(necessarily asymmetric)equilibrium in increasing, continuous bid fund (using a convenient shorthand notation) maps a: [0, i-R+ First, note that ai(0)=0 in any equilibrium. For suppose w l.o.g. that a1(0)=a>0 Then it must be the case that in equilibrium a wins with probability zero, so that also a2(0)=a. But since we are assuming that bid functions are increasing and continuous types e>0 with e a expect to win with positive probability and pay at least a 1. Note that the previous argument, together with the assumptions that bid functions are increasing and continuous implies that positive bids win with positive probability. Hence bidder 1 will never bid above 1, her maximum value. But then, from the point of view of bidder 2, any bid a> 1 may be improved upon by bidding Hence, bidder 2 never bids more than 1. But this means that for some realization of the signals close to(1, 2), by continuity we will have a2(u2)U1; the first-price auction assigns the object to bidder 1
Eco514—Game Theory Lecture 8.5: More on Auctions; PS#1 Marciano Siniscalchi October 14, 1999 Introduction These notes essentially tie up a few loose ends in Lecture 8; in particular, I exhibit examples of inefficiencies in first- and second-price auctions. I would also like to briefly comment on Questions 1 and 2 in Problem Set 2. The first-price auction may be inefficient even with private values Both examples I am going to show are due to Eric Maskin (to the best of my knowledge). The first point I wish to make is that, even in a private-values setting, asymmetries may render a first-price auction inefficient. Consider a two-bidder situation in which the value of the first bidder is uniformly distributed on [0, 1], and the value of the second bidder is uniformly distributed on [0, 2]. We look for a (necessarily asymmetric) equilibrium in increasing, continuous bid functions, i.e. (using a convenient shorthand notation) maps ai : [0, i] → R+. First, note that ai(0) = 0 in any equilibrium. For suppose w.l.o.g. that a1(0) = a > 0. Then it must be the case that in equilibrium a wins with probability zero, so that also a2(0) = a. But since we are assuming that bid functions are increasing and continuous, types > 0 with 1. Note that the previous argument, together with the assumptions that bid functions are increasing and continuous, implies that positive bids win with positive probability. Hence bidder 1 will never bid above 1, her maximum value. But then, from the point of view of bidder 2, any bid a > 1 may be improved upon by bidding a+1 2 . Hence, bidder 2 never bids more than 1. But this means that, for some realization of the signals close to (1,2), by continuity we will have a2(v2) v1; the first-price auction assigns the object to bidder 1. 1

[Incidentally, this argument does not rule out the possibility that equilibria in non- increasing or discontinuous bid functions might exist. Such equilibria can be ruled out under certain conditions: see Lizzeri and Persico. Games and Economic behavior. 1999 for details The second-price auction can be inefficient with interdependent values and more than two bidders This example is really clever. Consider a three-bidder, interdependent-values environment with signals distributed according to some law on a support which includes the point (1, 1, 1) The actual distribution of signals does not matter(you can rig things so you get the same conclusion even if signals are correlated), but to keep things simple, assume that signals are independen Suppose valuations are as follows 1 S1+-S2+ t2(s1,s2,S3)=82+71+s3 Consider(s1, S2, s3)in a(small)neighborhood of(1, 1, 1). For ex-post efficiency to obtain the object should be assigned to either bidder l or 2, depending on whether s3>l or ss <1 That is: efficient assignment depends on s3 But in a second-price auction, a bid depends only on a buyers own signal; thus, whether bidder 1 or bidder 2 receives the object depends on the relative magnitude of s1 and $2 Hence, for some realization of the signals in a neighborhood of (1, 1, 1), the outcome of the auction will be inefficient In case you were wondering, Dasgupta and Maskin offer a simple solution to this problem allow players to submit bids contingent on the signals of all the other players. That is, an action for bidder 1 with signal s1 is a function f(s1)(s2, s3)to be interpreted as follows: "If I, bidder 1, knew that my opponents' signal were S2 and S3 respectively, then I would bid f(s1(s2, S3). Every bidder is thus asked to report, not just a single bid, but a full bid function The seller then computes a fixpoint of these bid functions, and takes the fixpoint to be the actual vector of signals; the allocation is determined with respect to this vector. It turns out that it is an equilibrium to report truthfully: that is, to report f(s1)($2, S3)=U1(s1, $2, $3). On Problem set 2
[Incidentally, this argument does not rule out the possibility that equilibria in nonincreasing or discontinuous bid functions might exist. Such equilibria can be ruled out under certain conditions: see Lizzeri and Persico, Games and Economic Behavior, 1999 for details.] The second-price auction can be inefficient with interdependent values and more than two bidders This example is really clever. Consider a three-bidder, interdependent-values environment, with signals distributed according to some law on a support which includes the point (1,1,1). The actual distribution of signals does not matter (you can rig things so you get the same conclusion even if signals are correlated), but to keep things simple, assume that signals are independent. Suppose valuations are as follows: v1(s1, s2, s3) = s1 + 1 2 s2 + 1 4 s3 v2(s1, s2, s3) = s2 + 1 4 s1 + 1 2 s3 v3(s1, s2, s3) = s3 Consider (s1, s2, s3) in a (small) neighborhood of (1, 1, 1). For ex-post efficiency to obtain, the object should be assigned to either bidder 1 or 2, depending on whether s3 > 1 or s3 < 1. That is: efficient assignment depends on s3. But in a second-price auction, a bid depends only on a buyer’s own signal; thus, whether bidder 1 or bidder 2 receives the object depends on the relative magnitude of s1 and s2. Hence, for some realization of the signals in a neighborhood of (1,1,1), the outcome of the auction will be inefficient. [In case you were wondering, Dasgupta and Maskin offer a simple solution to this problem: allow players to submit bids contingent on the signals of all the other players. That is, an action for bidder 1 with signal s1 is a function f(s1)(s2, s3) to be interpreted as follows: “If I, bidder 1, knew that my opponents’ signal were s2 and s3 respectively, then I would bid f(s1)(s2, s3).” Every bidder is thus asked to report, not just a single bid, but a full bid function. The seller then computes a fixpoint of these bid functions, and takes the fixpoint to be the actual vector of signals; the allocation is determined with respect to this vector. It turns out that it is an equilibrium to report truthfully: that is, to report f(s1)(s2, s3) = v1(s1, s2, s3).] On Problem Set 2 2

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 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

and similarly for states wBxk. This inductively completes the definition of the posteriors, hence of the priors of the players Again, you should verify(by induction) that this yields the required result
and similarly for states ωθxk.This inductively completes the definition of the posteriors, hence of the priors of the players. Again, you should verify (by induction) that this yields the required result. 4
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