波恩大学博弈论 讲义 GameT-12

The Optimality of Auctions• A seller sells an object whose value to him is zero, he faces two buyers.•The seller does not know the value of the object to the buyers.•Each of the buyers has the valuation 3or 4with probability p , 1-p (respc.)•The seller wishes to design amechanism that will yield the highest possible expected payoff .An example:?First best 4 –p2 Posting price 33 Posting price 44(1 –p2) 1st price auction2nd price auction3+(1 –p)2Modified 2nd priceauction 4 –pTheresultssofar ????[]We look for an equilibrium in which mixes with probability distribution on .⋅H F() 3,K Player will bid .L 3()()()When bidding he expects to earn :⎡⎤⎣⎦b 4-b p +1-p F b It is optimal for to play a pure strategy.not H ()()This should be constant over the support, with F 3=0, F K =1()()()When bidding he expects to earn :⎡⎤⎣⎦b 4-b p +1-p F b ()()This should be constant over the support, with F 3=0, F K =1()()().Const ⎡⎤=⎣⎦4-b p +1-p F b ()()().Const ⎡⎤==⎣⎦4-3p +1-p F 3p ()p b -3F b =1-p 4-bK =4-p()ndThe expected payoff to the seller is : (same as 2 price auction)23+1-p see ex. 11.10.39 in Binmore p.564First best 4 –p 2Posting price 331st price auction 2nd price auction3+(1 –p)2auction–3+(1 –p)2p4 –p4(1 –p 2)maxModified 2nd price4 pPosting price 44(1 –p 2)Posting price 4Is there a better mechanism, one that yields a higher payoff to the seller?The Optimality of Auctions A general mechanism G Player 2 types: H, LPlayer 1 H types: L 4 , 3An Equilibrium:$t H t L s Hs LH, LPlayer 1 H types: 4 , 3The strategy set of player 1:{H,L}$t t L s Hs LThe strategy set of player 2:{H,L} When the players choosethe strategies X,YThe outcome is G(s X, t Y)L HLHH , LPlayer 1 H types: L4 , 3Truth telling is an equilibrium of the direct mechanism.t Ht Ls Hs LThe payoff in this equilibrium is the same as the payoff ofthe quilibrium of GRevelation Principle(the strategy set of each player is his set of types)in which truth telling is an equilibrium.The truth telling equilibrium implements theoutcome of E .For every mechanism G,and an equilibriumE of it, there is a direct mechanismD **TheA general direct mechanism DEach player is allowed to announce H or L .The mechanism can be described by 4numbers:h , l , H , LAssuming the other player tells the truth: HD (,): D (H ,H ), D (H ,L ), D (L ,H ) ,D (L ,L )describes the probability of obtaining the object and the expected payoffWe are interested in describing a truth telling equilibrium A player who announces LIncentive Compatibility constraintsThe seller chooses h,l,H,L to maximize his expected profits:2[(1-p )H + pL ]Individual rationality (participation) constraints:4h -H ≥ 0 IR H 3l -L ≥ 0 IR L4h -H ≥ 4l -L IC H 3l -L ≥ 3h -H IC Lsubject to:Assume we are at the optimal h,l,H,L,4h -H ≥ 0 IR H 3l -L ≥ 0 IR L 4h -H ≥ 4l -L IC H 3l -L ≥ 3h -H IC LThen by IC H :4h -H ≥ 4l -L≥ 3l –L >04h –H >0Now, increasing both H and L by the (same, small) constant will keep IR H , IR L and will not change IC H , IC L.However, this means that we could not have been at a maximum of the objective function.An increase in H and L improves the seller’s payoffIR must be an equality.4h -H ≥ 0 IR H 3l -L =0 IR L 4h -H ≥ 4l -L IC H 3l -L ≥ 3h -H IC LThen by IC H and IR L :4h -H >4l -L ≥ 3l –L =04h –H >0Now, increasing H by a small constantwill keep IC H , IR H and will not change IR L , IC L.However, this means that we could not have been at a maximum of the objective function.An increase in H improves the seller’s payoffIR H is a strict inequality.max 2[(1-p )H + pL ]4h -H ≥ 0 IR H 3l -L =0 IR L 4h -H =4l -L IC H 3l -L ≥ 3h -H IC Ls.t.Moreover, If IC H and IR L are equalities, then IR H is satisfied. And if h ≥ l then also IC L is satisfied.4h -H =4l -L ≥ 3l –L =4h –H ≥IR H is satisfied4h -H =4l -L3h -H =4l –L -h =3l –L + l -h ≤3l –L4h -H ≥ 0 IR H 3l -L =0 IR L 4h -H =4l -L IC H3l -L ≥ 3h -H IC LWe therefore need to consideronly IC H and IR L (and ensurethat the outcome satisfies h ≥ l )max 2[(1-p )H + pL ]3l -L =0 IR L 4h -H =4l -L ICHs.t.L = 3l H = 4h –l()()()max ⎡⎤⎣⎦h,l21-p 4h -l +p 3l ()()max ⎡⎤⎣⎦h,l24h 1-p +l 4p -1equivalent to maximizing:()⎛⎫ ⎪1h 1-p +l p -But symmetry imposes additional constraintsh ≤ p+ ½(1-p)(1-p)h + pl ≤ ½()max ⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦h,l 1h 1-p +l p -4The probability that a given player wins is ≤ ½The winning probabilities h,lcannot be too large=½(1+p)l ≤ 1 -p+ ½p =1 -½p(1-p)h + pl ≤ ½()max ⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦h,l 1h 1-p +l p -4h ≤ ½(1 + p)l ≤ 1 -½phl1 -½p(1-p)h + pl = ½slope:-(1-p)/p(1-p)h + pl ≤ ½()max ⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦h,l 1h 1-p +l p -4h ≤ ½(1 + p)l ≤ 1 -½phl1 -½p(1-p)h + pl = ½slope:-(1-p)/p For p <¼the slope of h(1-p) + l(p-¼)= Const is (1-p)/(¼-p)>0For p <¼the maximum is at l = 0, h = ½(1 + p)(1-p)h + pl ≤ ½()max ⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦h,l 1h 1-p +l p -4h ≤ ½(1 + p)l ≤ 1 -½phl1 -½p (1-p)h + pl = ½slope:-(1-p)/p For p >¼the slope of h(1-p) + l(p-¼)= Const is -(1 -p)/(p -¼)<-(1 -p)/pFor p >¼the maximum is at l = ½p, h = ½(1 + p)½p(1-p)h + pl ≤ ½⎝⎭⎣⎦4h ≤ ½(1 + p)l ≤ 1 -½pl1 -½p(1-p)h + pl = ½½pFor p <¼the maximum is atl = 0, h = ½(1 + p)l = ½p, h = ½(1 + p)(1-p)h + pl ≤ ½⎝⎭⎣⎦4h ≤ ½(1 + p)l ≤ 1 -½pl1 -½p(1-p)h + pl = ½½pFor p <¼the maximum is atl = 0, h = ½(1 + p)l = ½p, h = ½(1 + p)(1-p)h + pl ≤ ½⎝⎭⎣⎦4h ≤ ½(1 + p)l ≤ 1 -½pl = ½p, h = ½(1 + p)What is the seller’s payoff at this point ??()max ⎡⎤⎣⎦2H 1-p +Lp H = 4h –l = 2 + 3p/2L = 3 l = 3p/2= 4 -pInterpretation(1-p)h + pl ≤ ½⎝⎭⎣⎦4h ≤ ½(1 + p)l ≤ 1 -½pl = ½p, h = ½(1 + p)()max ⎡⎤⎣⎦2H 1-p +Lp = 4 -pL wins against L with prob. ½:½pand pays 3when he wins (expected payoff 3p/2)H = 2 + 3p/2 L= 3p/2H wins against L,and against H with prob. ½:p + ½(1 -p) = ½(1 + p)He pays 4when he wins against H,and 3½ against L(1-p)h + pl ≤ ½⎝⎭⎣⎦4h ≤ ½(1 + p)l ≤ 1 -½pl = ½p, h = ½(1 + p)()max ⎡⎤⎣⎦2H 1-p +Lp = 4 -pH = 2 + 3p/2 L= 3p/2Indeed, he could pay x when he wins against H,and y against L, with p x +½(1 -p)y =2+3p/2.(1-p)h + pl ≤ ½()max ⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦h,l 1h 1-p +l p -4h ≤ ½(1 + p)l ≤ 1 -½pWhat is the seller’s payoff at this point ??()max ⎡⎤⎣⎦2H 1-p +Lp H = 4h –l = 2(1 + p) L = 3 l = 0= 4(1 –p 2)For p <¼the maximum is atl = 0, h = ½(1 + p)L never wins. H wins against L, and against Hwith prob. ½:p+ ½(1 -p) =½(1 + p)。