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Game Theory

Game Theory. Lecture 12. problem set 12. from Binmore’s Fun and Games. p.564 Ex. 41 p.565 Ex. 42. The Optimality of Auctions. An example:. 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.

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Game Theory

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  1. Game Theory Lecture 12

  2. problem set 12 from Binmore’s Fun and Games. p.564 Ex. 41 p.565 Ex. 42

  3. The Optimality of Auctions An example: • 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 3 or 4 with probability p, 1-p (respc.) • The seller wishes to design a mechanism that will yield the highest possible expected payoff. ?

  4. The results so far ???? ????

  5. The Optimality of Auctions First price auction

  6. The Optimality of Auctions First price auction

  7. The Optimality of Auctions First price auction see ex. 11.10.39 in Binmore p.564

  8. Is there a better mechanism, one that yields a higher payoff to the seller? 3+(1 – p)2 max 4(1 – p2) 4 – p p ¼ 0 1

  9. Player 2 types: H , L The Optimality of AuctionsA general mechanism G An Equilibrium: 4 , 3 tH tL sH Player 1 H types: L G sL $

  10. Player 2 types: H , L ADirectmechanism H The strategy set of player 1:{H,L} The strategy set of player 2:{H,L} 4 , 3 tH tL tH When the players choose the strategies X,Y sH The outcome is G(sX, tY) Player 1 H types: L L sL G sL $

  11. Player 2 types: H , L ADirectmechanism Truth telling is an equilibrium of the direct mechanism. 4 , 3 The payoff in this equilibrium is the same as the payoff of the quilibrium of G tH tL sH Player 1 H types: L G sL

  12. Allan Gibbard 1942 - The Revelation Principle For every mechanism G, and an equilibrium E of it, there is a direct mechanism D * in which truth telling is an equilibrium. The truth telling equilibrium implements the outcome of E. * (the strategy set of each player is his set of types)

  13. A general direct mechanism D Each player is allowed to announce HorL. D( , ): D(H ,H), D(H ,L), D(L ,H) ,D(L ,L)describes the probability of obtaining the object and the expected payoff We are interested in describing a truth telling equilibrium The mechanism can be described by 4 numbers: h , l , H , L Assuming the other player tells the truth: A player who announcesH wins the object with probability h, and expects to payH. Assuming the other player tells the truth: A player who announcesL wins the object with probability l, and expects to payL.

  14. The seller chooses h,l,H,L to maximize his expected profits: 2[(1-p)H + pL] subject to: Individual rationality (participation) constraints: 4h - H ≥ 0 IRH 3l - L ≥ 0 IRL Incentive Compatibility constraints 4h - H ≥ 4l - L ICH 3l - L ≥ 3h -H ICL Assume we are at the optimal h,l,H,L, whichof the constraints are equalities?

  15. max 2[(1-p)H + pL] s.t. 4h - H ≥ 0 IRH 3l - L ≥ 0 IRL 4h - H ≥ 4l - L ICH 3l - L ≥ 3h -H ICL If IRLis a strict inequality 3l – L > 0 Then by ICH : 4h - H ≥ 4l - L ≥ 3l –L > 0 4h –H > 0 Now, increasing both H and L by the (same, small) constant will keep IRH , IRL and will not change ICH , ICL. 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 payoff IRL must be an equality.

  16. max 2[(1-p)H + pL] s.t. 4h - H ≥ 0 IRH 3l - L = 0 IRL 4h - H ≥ 4l - L ICH 3l - L ≥ 3h -H ICL If ICHis a strict inequality 4h – H > 4l - L Then by ICH andIRL : 4h - H > 4l - L ≥ 3l –L = 0 4h –H > 0 IRH is a strict inequality. Now, increasing H by a small constant will keep ICH , IRH and will not change IRL , ICL. However, this means that we could not have been at a maximum of the objective function. An increase in H improves the seller’s payoff ICH must be an equality.

  17. max 2[(1-p)H + pL] s.t. 4h - H ≥ 0 IRH 3l - L = 0 IRL 4h - H = 4l - L ICH 3l - L ≥ 3h -H ICL Moreover, If ICH and IRL are equalities, then IRHis satisfied. And if h ≥ l then also ICL is satisfied. 4h - H = 4l - L ≥ 3l –L = 0 4h –H ≥ 0 IRH is satisfied 4h - H = 4l - L 3h - H = 4l – L - h =3l – L + l - h ≤ 3l – L ICL is satisfied Note, that if ICL is satisfied then h ≥l

  18. max 2[(1-p)H + pL] s.t. 4h - H ≥ 0 IRH 3l - L = 0 IRL max 2[(1-p)H + pL] 4h - H = 4l - L ICH s.t. 3l - L ≥ 3h -H ICL 3l - L = 0 IRL 4h - H = 4l - L ICH We therefore need to consider only ICH and IRL (and ensure that the outcome satisfies h ≥ l ) L = 3l H = 4h – l equivalent to maximizing:

  19. But symmetry imposes additional constraints The probability that a given player wins is≤ ½ (1-p)h + pl ≤½ The winning probabilities h,l cannot be too large h ≤ p+ ½(1-p) =½(1+p) l ≤ 1 - p+ ½p =1 -½p L wins againstH and wins with prob. ½againstL H wins againstL and wins with prob. ½againstH

  20. (1-p)h + pl ≤½ h ≤½(1 + p) l l ≤ 1 -½p slope: -(1-p)/p 1 -½p (1-p)h + pl =½ h ½(1 + p)

  21. For p <¼ the maximum is at l = 0, h = ½(1 + p) For p <¼the slope of h(1-p) + l(p- ¼)= Const is(1-p)/(¼-p) (1-p)h + pl ≤½ > 0 h ≤½(1 + p) l ≤ 1 -½p l slope: -(1-p)/p 1 -½p (1-p)h + pl =½ h ½(1 + p)

  22. For p >¼ the maximum is at l = ½p, h = ½(1 + p) For p >¼the slope of h(1-p) + l(p- ¼)= Const is -(1 - p)/(p - ¼) (1-p)h + pl ≤½ < -(1 - p)/p h ≤½(1 + p) l ≤ 1 -½p l slope: -(1-p)/p 1 -½p (1-p)h + pl =½ ½p h ½(1 + p)

  23. For p >¼ the maximum is at l = ½p, h = ½(1 + p) (1-p)h + pl ≤½ h ≤½(1 + p) For p <¼ the maximum is at l = 0, h = ½(1 + p) l ≤ 1 -½p l 1 -½p (1-p)h + pl =½ ½p h ½(1 + p)

  24. For p >¼ the maximum is at l = ½p, h = ½(1 + p) (1-p)h + pl ≤½ h ≤½(1 + p) For p <¼ the maximum is at l = 0, h = ½(1 + p) l ≤ 1 -½p l 1 -½p (1-p)h + pl =½ ½p h ½(1 + p)

  25. For p >¼ the maximum is at l = ½p, h = ½(1 + p) (1-p)h + pl ≤½ h ≤½(1 + p) l ≤ 1 -½p What is the seller’s payoff at this point ?? H = 4h – l = 2 + 3p/2 L = 3 l = 3p/2 = 4 - p Interpretation

  26. For p >¼ the maximum is at l = ½p, h = ½(1 + p) Modified 2nd Price Auction (1-p)h + pl ≤½ h ≤½(1 + p) H = 2 + 3p/2 L= 3p/2 l ≤ 1 -½p = 4 - p L wins against L with prob. ½:½p and pays3 when he wins (expected payoff 3p/2) H wins against L, and against H with prob. ½: p + ½(1 - p) = ½(1 + p) He pays4 when he wins against H, and 3½ against L (expected payoff: 3½·p +4·½(1 - p) =2+3p/2 )

  27. For p >¼ the maximum is at l = ½p, h = ½(1 + p) (1-p)h + pl ≤½ h ≤½(1 + p) H = 2 + 3p/2 L= 3p/2 l ≤ 1 -½p = 4 - p Indeed, he could payx when he wins against H, and y against L, with px +½(1 - p)y =2+3p/2.

  28. For p <¼ the maximum is at l = 0, h = ½(1 + p) Posting Price 4 (1-p)h + pl ≤½ h ≤½(1 + p) l ≤ 1 -½p What is the seller’s payoff at this point ?? H = 4h – l = 2(1 + p) L = 3 l = 0 = 4(1 – p2) L never wins. H wins against L, and against H with prob. ½:p+ ½(1 - p) =½(1 + p) and pays4 when he wins (expected payoff 2(1 + p))

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