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

3 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. The Optimality of Auctions An example: ?

4 First best 4 – p 2 Posting price 3 3 Posting price 4 4(1 – p 2 ) 1 st price auction 2 nd price auction 3+(1 – p) 2 Modified 2 nd price auction 4 – p 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 in Binmore p.564

8 First best 4 – p 2 Posting price 3 3 Posting price 4 4(1 – p 2 ) 1 st price auction 2 nd price auction 3+(1 – p) 2 Modified 2 nd price auction 4 – p 3+(1 – p) 2 10 ¼ p 4 – p4(1 – p 2 ) max Modified 2 nd price auction 4 – p Posting price 4 4(1 – p 2 ) Posting price 4 Modified 2 nd price auction Is there a better mechanism, one that yields a higher payoff to the seller?

9 The Optimality of Auctions A general mechanism G Player 2 types: H, L Player 1 H types: L 4, 3 An Equilibrium: $ tHtH tLtL sHsH sLsL

10 Direct A Direct mechanism Player 2 types: H, L Player 1 H types: L 4, 3 The strategy set of player 1: { H,L } $ tHtH tLtL sHsH sLsL The strategy set of player 2: { H,L } When the players choose the strategies X,Y The outcome is G( s X, t Y ) L H sLsL tHtH

11 Direct A Direct mechanism Player 2 types: H, L Player 1 H types: L 4, 3 Truth telling is an equilibrium of the direct mechanism. tHtH tLtL sHsH sLsL The payoff in this equilibrium is the same as the payoff of the quilibrium of G

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

13 A general direct mechanism D Each player is allowed to announce H o r L. The mechanism can be described by 4 numbers: h, l, H, L Assuming the other player tells the truth: A player who announces H wins the object with probability h, and expects to pay H. 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 Assuming the other player tells the truth: A player who announces L wins the object with probability l, and expects to pay L.

14 Incentive Compatibility constraints The 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 L 4h - H ≥ 4l - L IC H 3l - L ≥ 3h -H IC L subject to: Assume we are at the optimal h,l,H,L, which of the constraints are equalities?

15 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 L s.t. If I R L is a strict inequality 3l – L > 0 Then by I C H : 4h - H ≥ 4l - L ≥ 3l –L > 0 4h –H > 0 Now, increasing both H and L by the (same, small) constant will keep I R H, IR L and will not change I C 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 payoff IR L must be an equality.

16 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 L s.t. If I C H is a strict inequality 4h – H > 4l - L Then by IC H and IR L : 4h - H > 4l - L ≥ 3l –L = 0 4h –H > 0 Now, increasing H by a small constant will keep I C H, IR H and will not change I R 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 payoff IC H must be an equality. IR H is a strict inequality.

17 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 L s.t. Moreover, If I C H and I R L are equalities, then I R H is satisfied. And if h ≥ l then also I C L is satisfied. 4h - H = 4l - L ≥ 3l –L = 0 4h –H ≥ 0 IR H is satisfied 4h - H = 4l - L 3h - H = 4l – L - h = 3l – L + l - h ≤ 3l – L IC L is satisfied Note, that if IC L is satisfied then h ≥ l

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

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

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

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

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

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

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

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

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

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

28 (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 – p 2 ) For p < ¼ the maximum is at l = 0, h = ½(1 + p) L never wins. H wins against L, a nd against H with prob. ½ : p + ½ (1 - p) = ½ (1 + p) and pays 4 when he wins (expected payoff 2 (1 + p) )