Auction Theory - Solutions Manual [2 ed.] 0123745071, 9780123745071

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Solutions Manual for AUCTION THEORY Alexey Kushnir and Jun Xiao August 2009

Contents 2 3 4 5 6 8 9 10 11 13 15 16 17

Private Value Auctions: A First Look . . . . . The Revenue Equivalence Principle . . . . . . . Quali…cations and Extensions . . . . . . . . . . Mechanism Design . . . . . . . . . . . . . . . . Auctions with Interdependent Values . . . . . . Asymmetries and Other Complications . . . . . E¢ ciency and the English Auction . . . . . . . Mechanism Design with Interdependent Values Bidding Rings . . . . . . . . . . . . . . . . . . . Equilibrium and E¢ ciency with Private Values Sequential Sales . . . . . . . . . . . . . . . . . . Nonidential Objects . . . . . . . . . . . . . . . Packages and Positions . . . . . . . . . . . . . .

V. Krishna, Auction Theory (2nd. Ed.), Elsevier, 2009.

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2 8 11 17 25 34 40 43 48 52 55 60 62

2

Private Value Auctions: A First Look

Problem 2.1 (Power distribution) Suppose there are two bidders with private values that are distributed independently according to the distribution F (x) = xa over [0; 1] where a > 0: Find symmetric equilibrium bidding strategies in a …rst-price auction. Solution. Since N = 2, G(x) = F (x) = xa : Thus, using the formula on page 16 of the text, Z x Z x a G (y) y a I (x) = x dy = dy = x x a 1+a 0 G (x) 0 x Problem 2.2 (Pareto distribution) Suppose there are two bidders with private values that are distributed independently according to a Pareto distribution F (x) = 1 (x + 1) 2 over [0; 1). Find symmetric equilibrium bidding strategies in a …rst-price auction. Show by direct computation that the expected revenues in a …rst- and secondprice auction are the same. Solution. Again, since N = 2, G (x) = F (x) = 1 (x + 1) 2 . Thus, Z x G (y) I (x) = x dy G (x) 0 Z x 1 (y + 1) 2 = x dy (x + 1) 2 0 1 x = x+2 In the …rst-price auction, the expected revenue of the seller is E RI

= 2E mI (x) I = 2E G (x) (x) Z 1 = 2 1 (x + 1)

x 2 (x + 1) x+2

2

0

3

dx

= 1=3

Let Y2 be the second highest value, and its density is f2 (y) = 2 (1 (see Appendix C). In a second-price auction, the expected revenue of the seller is E RII

= E [Y2 ] Z 1 = y2 (y + 1)

2

2 (y + 1)

3

dy

0

= 1=3

Therefore, the expected revenues in the two auctions are the same. 2

F (y)) g (y)

Problem 2.3 (Stochastic dominance) Consider an N -bidder …rst-price auction with independent private values. Let be the symmetric equilibrium bidding strategy when which each bidder’s value is distributed according to F on [0; !] : Similarly, let be the equilibrium strategy when each bidder’s value distribution is F on [0; ! ] : a. Show that if F dominates F in termsof the reverse hazard rate (see Appendix B for a de…nition) then for all x 2 [0; !] ; (x) (x) : p 1 2 b. By considering F (x) = 3x x on [0; 2 (3 5)] and F (x) = 3x 2x2 on 1 0; 2 , show that the condition that F …rst-order stochastically dominates F is not su¢ cient to guarantee that (x) (x) : Solution. Part a. Because G (x) = F (x)N 1 and g (x) = (N 1) F (x)N symmetric equilibrium in Proposition 2.2 could be rewritten as follows Z x 1 (x) = yg (y) dy G (x) 0 Z x 1 = y (N 1) F (x)N 2 f (x) dy [F (x)]N 1 0 Z x f (y) = (N 1) y dy F (y) Z0 x y (y) dy = (N 1)

2

f (x) ; the

0

where

(x) is the reverse hazard rate. Similarly, we have Z x (x) = (N 1) y (y) dy 0

So it is easy to see that if F dominates F in terms of reverse hazard rate, then (y) (y) for all y 2 [0; !] : Therefore (x) (x) for all x 2 [0; !]. Part b. Obviously, F (x) F (x), so F stochastically dominates F . The distributions F and F are illustrated in Figure S2.1, where the solid line represents F and the dashed line represents F .

3

1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0

0.1

0.2

0.3

0.4

0.5

x

Figure S2.1 Suppose there are two bidders, then Z

x

G (y) dy 0 G (x) Z x 3y y 2 = x dy x2 0 3x 1 2x 9 = x 6 x 3

(x) = x

for x 2 0; 12 3

p

5

: Similarly, (x) = x

Z

0

=

x

3y 3x

1 x (8x 6 2x 3

2y 2 dy 2x2 9)

p for x 2 0; 12 : It is easy to see that (x) < (x) for x 2 (0; 21 3 5 ]: The bidding strategies and are plotted in Figure S2.2, where is the solid line and is the dashed line.

4

0.20

0.15

0.10

0.05

0.00 0.0

0.1

0.2

0.3

0.4

0.5

x

Figure S2.2

Problem 2.4 (Mixed auction) Consider an N -bidder auction which is a “mixture” of a …rst- and second-price auction in the sense that the highest bidder wins and pays a convex combination of his own bid and the second-highest bid. Precisely, there is a …xed 2 (0; 1) such that upon winning, bidder i pays bi + (1 ) (maxj6=i bj ) : Find a symmetric equilibrium bidding strategy in such an auction when all bidders’values are distributed according to F: Solution. Suppose all bidders other than 1 follow the strategy payo¤ of bidder i from bidding b when his value is x is (b; x)) = G

1

= G

1

= G

1

(b) [x "

b

x

b

(1

(b) (x

b)

(1

(b)

(1

: The expected

)E [ (Y1 ) j (Y1 ) b1 ]] # R 1 (b) (y)g(y)dy ) 0 G( 1 (b)) Z 1 (b) (y)g(y)dy ) 0

Maximizing this with respect to b yields the …rst-order condition: 0 =

g

1

(b)

0

1

(b)

(1

0

(x) +

b) 1

) bg(

At a symmetric equilibrium, b = G (x)

(x

(b))

1

G

(b)

1 0

1

(b)

(x), so the …rst-order condition becomes 1

g (x) (x) = 5

1

xg (x)

Using G (x)(1= ) 1 as the integrating factor, the solution to the above di¤erential equation is easily seen to be Z x 1 yg (y) dy (x) = G (x) 0 where G

G1= and g = G0 .

Problem 2.5 (Resale) Consider a two-bidder …rst-price auction in which bidders’ values are distributed according to F: Let be the symmetric equilibrium (as derived in Proposition 2.2). Now suppose that after the auction is over, both the losing and winning bids are publicly announced. In addition, there is the possibility of postauction resale: The winner of the auction may, if he so wishes, o¤ er the object to the other bidder at a …xed “take-it-or-leave-it” price of p: If the other bidder agrees, then the object changes hands and the losing bidder pays the winning bidder p. Otherwise, the object stays with the winning bidder and no money changes hands. The possibility of post-auction resale in this manner is commonly known to both bidders prior to participating in the auction. Show that remains an equilibrium even if resale is allowed. In particular, show that a bidder with value x cannot gain by bidding an amount b > (x) even when he has the option of reselling the object to the other bidder. Solution. First, let us consider the resale stage. Suppose bidder 1 wins the auction and the announced bids are b1 and b2 . Hence bidder 1 can recover bidder 2’s private value by x2 = 2 1 (b2 ):Therefore bidder 1 suggests the price x2 which extracts all the surplus from bidder 2 if x2 x1 , and does not o¤er otherwise. Then bidder 1’s payo¤ is max(x1 b1 ; x2 b1 ). If bidder 1 loses the auction, he gets zero payo¤ because bidder 2 o¤ers price x1 to him and extracts all the surplus. Rx 1 Second, now we move to the auction stage. Let (x) = F (x) 0 yf (y)dy be the symmetric equilibrium without resale. We are going to show that any deviation of bidder 1 from (x) is not pro…table. Suppose bidder 1 deviates by bidding (z) when his private value is x, while bidder 2 still plays (x2 ). Bidder 1’s ex ante expected payo¤ is 1 (z; x)

=

(x (x

(z)) F (z) if x Rz z (z)) F (x) + x (y

If x2 < z x there is no resale. If z > x 2 and his payo¤ remains the same. If z x2

6

(z)) f (y)dy if x < z

x2 ; bidder 1 does not o¤er to bidder x, bidder 1 sells to bidder 2 and the

payo¤ after resale is x2

(z): Note that Z

= = = =

z

(y (z)) f (y)dy (z)) F (x) + x Z z yf (y)dy (z)F (z) xF (x) + x Z z Z z xF (x) + yf (y)dy yf (y)dy x 0 Z x 1 xF (x) F (x) yf (y)dy F (x) 0 (x (x)) F (x) (x

so we have 1 (z; x)

=

(x (x

(z)) F (z) if x z (x)) F (x) if x < z

which is not more than 1 (x; x). So no deviation strictly increases a bidder’s payo¤ and (x) is still an equilibrium in the presence of resale.

7

3

The Revenue Equivalence Principle

Problem 3.1 (War of attrition) Consider a two-bidder war of attrition in which the bidder with the highest bid wins the object but both bidders pay the losing bid. Bidders’ values independently and identically distributed according to F . a. Use the revenue equivalence principle to derive a symmetric equilibrium bidding strategy in the war of attrition. b. Directly compute the symmetric equilibrium bidding strategy and the sellers’ revenue when the bidders’ values are uniformly distributed on [0; 1]. Solution. Part a. Suppose that there is a symmetric, increasing equilibrium of the war of attrition, , such that the expected payment of a bidder with value 0 is 0. Since the assumptions of Proposition 3.1 are satis…ed, we must have that for all x, the expected payment is Z x

m (x) =

yf (y) dy

0

On the other hand, we also have

m (x) = E [ (Y1 ) j Y1 < x] Z x 1 = (y) f (y) dy F (x) 0 where Y1 is the bid from the other bidder. Combining the two equations, we have Z x Z x 1 yf (y) dy = (y) f (y) dy F (x) 0 0 Di¤erentiating both sides with respect to x and rearranging this, we get Z x (x) = F (x) x + yf (y) dy 0

Part b. Suppose that bidder 1 has valuation x: He chooses b to maximize his expected payo¤ Z 1 (b) 1 1 (y) f (y) dy F (b) x 1 F (b) 0 where the …rst term is the product of his probability of winning and his valuation, and the second term is his expected payment. Because F (x) = x and f (x) = 1, the expected payo¤ becomes 1

1

(b) x

1

(b)

Z

1

(b)

(y) dy

0

Maximizing with respect to b yields the …rst-order condition: 1

0=

1 0

1

(b)

x+

(b)

0

1

1

(

1

(b))

(b) + 1

8

(b)

0

2

1

(

1

(b))

R

1

0

(b)

(y) dy

In equilibrium, b =

(x) ; and thus the …rst-order condition becomes Z x 3 x (x) x + (y) dy = 0 0

Di¤erentiating both sides with respect to x, we obtain 0

(x) = 3x

Combining with the initial condition (0) = 0, we can solve the equilibrium bidding strategy as (x) = 23 x2 : Therefore the seller’s revenue is Z Z 2 x3 2 2 x (y) dy = 2E [m (x)] = y dy = 1=3 x 0 x 0 2 Problem 3.2 (Losers-pay auction) Consider a N -bidder losers-pay auction in which the bidder with the highest bid wins the object and pays nothing while all losing bidders pays their own bids. Bidders’ valuations independently and identically distributed according to F . a. Use the revenue equivalence principle to derive a symmetric equilibrium bidding strategy in the losers-pay auction. b. Directly compute the symmetric equilibrium bidding strategy for the case when the bidders’ values are distributed according to F (x) = 1 e ax over [0; 1). Solution. Part a. Suppose that there is a symmetric, increasing equilibrium of the losers-pay auction, , such that the expected payment of a bidder with value 0 is 0. Since the assumptions of Proposition 3.1 are satis…ed, we must have that for all x, the expected payment is Z x

m (x) =

yg (y) dy

0

On the other hand, we also have

m (x) = (1

G (x)) (x)

which is the product of probability of losing the auction and his own bid. If we combine the two equations, we have Z x yg (y) dy = (1 G (x)) (x) 0

and therefore (x) = =

Rx

yg (y) dy 1 G (x) Rx (N 1) 0 yF N 1 FN 0

9

2 (y) f 1 (x)

(y) dy

Part b. Suppose that bidder 1 has valuation x, then he chooses b to maximize his expected payo¤ 1 1 G (b) x 1 G (b) b where the …rst term is the product of the probability of winning and his valuation, and the second term is his expected payment. Maximizing with respect to b yields the …rst-order condition: ! 1 1 g (b) g (b) 1 x 1 G (b) b=0 0 1 0 1 (b) (b) In equilibrium, b =

(x), and thus the …rst-order condition becomes 0

(x)

g (x) g (x) x (x) = 1 G (x) 1 G (x)

R x g(y) Using exp 0 1 G(y) dy as the integrating factor and the initial condition 0, it is easily seen that R y g(z) Rx g(y)y exp 0 1 G(z) dz 1 G(y) dy 0 (x) = R x g(y) exp 0 1 G(y) dy Rx g (y) ydy = 0 1 G (x)

(0) =

(1)

where the last equality comes from the fact that Z x Z x g (y) 1 exp dy = exp dG (y) G (y) G (y) 0 1 0 1 Z x 1 = exp d [1 G (y)] G (y) 0 1 = exp (ln (1 G (x))) = 1

G (x)

Note that (1) is the same as the equilibrium strategy derived in Part a. When the bidders’values are distributed according to F (x) = 1 e G (x) = 1 g (x) = (N

1) 1

e e

ax N 1 ax N 2

e

ax

a

Therefore the equilibrium bidding strategy in (1) becomes Rx g (y) ydy (x) = 0 1 G (x) Rx N 2 (N 1) 0 [1 e ay ] e ay aydy = 1 [1 e ax ]N 1 R ax N 2 (N 1) a1 0 [1 e y ] e y ydy = 1 [1 e ax ]N 1

10

ax ,

we have

4

Quali…cations and Extensions

Problem 4.1 (Risk-averse bidders) There are two bidders with private values which are distributed independently according to the uniform distribution F (x) = x over p [0; 1] : Both bidders are risk-averse and have utility functions u (z) = z: Find symmetric equilibrium bidding strategies in a …rst-price auction. Solution. This is a special case of Example 4.1 with a = 1=2 and F (x) = x, so Z x Z x 1 2 1 yf (y) dy = 2 y2ydy = x (x) = F (x) 0 x 0 3

Problem 4.2 (Increase in risk aversion) Consider an N -bidder …rst-price auction where each bidder’s value is distributed according to F: All bidders are risk averse with a utility function u that satis…es u(0) = 0; u0 > 0; u00 < 0: Show that if one changed the utility function of all bidders from u (z) to (u (z)), where is an increasing and concave function satisfying (0) = 0; then this would lead to a higher symmetric equilibrium bidding strategy. Solution. Let be the equilibrium bidding strategy for utility u while ^ be the equilibrium bidding strategy for (u). From equation (4.2), we have

^ 0 (x) =

0

(x) =

0

(u (x

u (x u0 (x

(x)) (x))

(u (x ^ (x))) ^ (x))) u0 (x ^ (x))

Notice that is strictly concave and Using this fact, for x > 0 we have ^ 0 (x) = >

g (x) G (x) g (x) G (x)

(0) = 0, for all y > 0,

(u (x ^ (x))) (u (x ^ (x))) u0 (x ^ (x)) g (x) u (x (x)) = 0 (x) u0 (x (x)) G (x) 0

(y) =

0

(y) > y.

g (x) G (x)

or equivalently ^ 0 (x) >

0

(x) for x > 0

(2)

It is also easy to check that ^ (0) =

(0)

Equations (2) and (3) imply that for all x > 0; ^ (x) >

11

(x)

(3)

Problem 4.3 (Asymmetric …rst-price auction) Suppose that bidder 1’s value X1 is distributed according to F1 (x) = 14 (x 1)2 over [1; 3] and bidder 2’s value is distributed according to F2 (x) = exp 23 x 2 over [0; 3] : a. Show that 1 (x) = x 1 and 2 (x) = 23 x constitute equilibrium bidding strategies in a …rst-price auction. b. Compare the expected revenues in a …rst- and second-price auction. Solution. Part a. Given bidder 2’s strategy x chooses b to maximize his expected payo¤ 1

F2

2

(b) (x

2 (x)

b) = exp (b

= 23 x, bidder 1 with valuation

2) (x

b)

The …rst-order condition is exp (b In equilibrium, b =

1 (x),

2) (x

b)

exp (b

2) = 0

and thus the …rst-order condition yields

exp (

1 (x)

2) [x

1

1 (x)]

=0

It is easy to see that 1 (x) = x 1 satis…es the equation above, and it is easy to see that the marginal payo¤ is negative if 1 (x) > x 1 and positive if 1 (x) < x 1; therefore it is the best response to bid 1 (x) : On the other hand, given bidder 1’s strategy 1 (x) = x 1 , bidder 2 with valuation x chooses b to maximize his expected payo¤ 1

F1

1

1 b) = b2 (x 4

(b) (x

b)

The …rst-order condition is

1 1 2 b (x b) b =0 2 4 In equilibrium, b = 2 (x), and thus the …rst-order condition yields 3 4

2 (x)

2 x 3

2 (x)

=0

It obvious that 2 (x) = 32 x satis…es the equation above, and it is easy to see that the marginal payo¤ is negative if 2 (x) > 32 x and positive if 2 (x) < 23 x; therefore it is the best response to bid 1 (x) : In sum, given bidder 1’s strategy 1 (x) ; bidder 2’s best response is 2 (x), and given bidder 2’s strategy 2 (x) bidder 1’s best response is 1 (x), therefore 1 (x) and 2 (x) constitute an equilibrium. Part b. In the …rst-price auction, the expected revenue is E F2

1 2

(

1 (x1 ))

1 (x1 )

+ E F1

12

1 1

(

2 (x2 ))

2 (x2 )

where the …rst term is the expected payment from bidder 1 and the second term is the expected payment from bidder 2. Given the distributions and bidding strategies, the expected revenue could be rewritten as Z 3 I E R = F2 2 1 ( 1 (x1 )) 1 (x1 ) dF1 (x1 ) 1 Z 3 F1 1 1 ( 2 (x2 )) 2 (x2 ) dF2 (x2 ) + 0 Z 1 3 = exp (x1 3) (x1 1)2 dx1 2 1 Z 3 2 4 exp + x2 2 x32 dx2 81 0 3 = 1: 567 7 In the second-price auction, it is a dominant strategy to bid one’s value. We can either use the same method for the …rst price auction or calculate the expected revenue using price distribution, and here we use the latter. The distribution of price is 8 E RII . Problem 4.4 (Equilibrium with reserve price) Suppose that bidder 1’s value X1 is distributed uniformly on [0; 2] while bidder 2’s value X2 is distributed uniformly on 13

[ 32 ; 52 ]. The object is sold via a …rst-price auction with a reserve price r = 1. Verify that 1 (x) = x2 + 21 and 2 (x) = x2 + 14 constitute equilibrium strategies. Solution. F1 (x) = x2 for x 2 [0; 2] and F2 (x) = x 23 for x 2 32 ; 52 . Given bidder 2’s strategy 2 (x) = x2 + 41 , bidder 1 with valuation x chooses b to maximize his expected payo¤. If b r, the expected payo¤ is F2

1 2

= (2b

(b) (x

2) (x

b)

b) =2

and the corresponding marginal payo¤ is x

2b + 1

The …rst-order condition is x At equilibrium, b =

1 (x),

2b + 1 = 0

and thus the …rst-order condition yields x

2

1 (x)

+1=0

It is easy to see that 1 (x) = x+1 2 satis…es the equation above. Moreover, it is easy to see that the marginal payo¤ is negative if b > x+1 2 and the marginal payo¤ is positive x+1 x+1 if b < 2 : Therefore 1 (x) = 2 is the best response to 2 (x) : On the other hand, given bidder 1’s strategy 1 (x) = x2 + 21 , bidder 2 with valuation x chooses b to maximize his expected payo¤. If b r, The expected payo¤ is F1 1 1 (b) (x b) = (2b 1) (x b) =2 The …rst-order condition is 2 (x At equilibrium, b =

2 (x),

b)

(2b

1) = 0

and thus the …rst-order condition yields 2x

4

2 (x)

+1=0

It obvious that 2 (x) = x2 + 14 satis…es the equation above, and it is bidder 2’s best response to 1 (x) by the same reasoning in the case of bidder 1. We have shown that given bidder 1’s strategy 1 (x) bidder 2’s best response is 2 (x), and given bidder 2’s strategy 2 (x) bidder 1’s best response is 1 (x). Therefore 1 and 2 constitute an equilibrium. Problem 4.5 (Discrete values) Suppose that there is no uncertainty about bidder 1’s value and X1 = 2 always. Bidder 2’s value, X2 , is equally likely to be 0 or 2. a. Find equilibrium bidding strategies in a …rst-price auction. (Note that since values are discrete, the equilibrium will be in mixed strategies.) b. Compare the revenues in a …rst- and second-price auction. 14

Solution. Part a. Note that when bidder 2 has value 0, it is his weakly dominant strategy to bid zero. Let us assume bidder 1’s mixed-strategy has cumulative distribution M1 (b) and bidder 2 with valuation 2 has mixed-strategy with cumulative distribution M2 (b). We only consider the case that M1 and M2 are di¤erentiable. If bidder 1 bid b, his expected payo¤ is 1 1 + M2 (b) (2 2 2

b)

which is the product between the probability of winning and the payo¤ when he wins. Because bidder 1 plays mixed-strategy, his expected payo¤ should be the same for any b in the support of M1 ; therefore 1 1 + M2 (b) (2 2 2

b) = C

where C is a constant. So we have M2 (b) =

2C 2 b

1

(4)

It should also be true that M2 (0) = 0

(5)

(4) and (5) imply that b ; for b 2 [0; 1] 2 b If bidder 2 with valuation 2 bids b, this expected payo¤ is M2 (b) =

M1 (b) (2

b)

which is the product between the probability of winning and the payo¤ when he wins. Similar to the analysis for bidder 1, we have M1 (b) =

C0 2

b

(6)

for some constant C 0 . Because every bidder does not bid higher than his opponent, the highest bid is the same for both bidders. Note that the highest bid for bidder 2 is 1, so we have M1 (1) = 1 (6) and (7) imply that M1 (b) =

1

, for b 2 [0; 1] 2 b where 0 is a mass point of the distribution of M1 :

15

(7)

Part b. When X2 = 0, the cumulative distribution for price is M1 ; and when X2 = 2, the cumulative distribution for price is M1 M2 , so the expected revenue in the …rst-price auction is E RI

= Pr (X2 = 0) E [P j X2 = 0] + Pr (X2 = 2) E [P j X2 =] Z Z 1 1 1 1 = bdM1 (b) + bd [M1 (b) M2 (b)] 2 0 2 0 Z 1 Z 1 1 b 1 bd + bd = 2 0 2 b (2 b)2 0 Z 1 Z 1 1 b 1 1 db = db + 1 2 2 b b)2 0 0 (2 1 = 2

where the fourth equality comes from integration by parts. The expected revenue in the second-price auction is E RII =

1 2

0+

So we have E RI < E RII :

16

1 2

2=1

5

Mechanism Design

Problem 5.1 (Surplus extraction) Show that if buyers’values are independently distributed, then the seller cannot design an incentive compatible and individually rational mechanism that extracts the whole surplus from buyers. (In doing this problem, use only the results of Section 5.1 and not those from Section 5.2.) Solution. Suppose that there exists a mechanism extracting all the surplus from the buyers. We also assume that the supports of the buyer value distributions are …nite [ai ; bi ]. Using revelation principle we may say then there should be a direct mechanism which extracts the whole surplus from the buyers. Using the notation of Section 5.1 the fact that the seller extracts the whole surplus from the buyers means that for all i and for all xi , Ui (xi ) = 0: Because the buyers’values are independently distributed, when the buyer reports zi when his true value is xi , his expected payo¤ is (see Section 5.1) qi (zi )xi

mi (zi )

and Ui (xi ) = qi (xi )xi

mi (xi )

Incentive compatibility implies that Ui (xi )

qi (zi )xi = qi (zi )zi

mi (zi ) mi (zi ) + qi (zi ) (xi

= Ui (zi ) + qi (zi ) (xi

zi )

zi )

Because the seller extracts all the surplus from all the buyers, Ui (xi ) = Ui (zi ) = 0 and for all i and for all xi and zi ; qi (zi ) (xi

zi )

0

The inequality above implies qi (zi ) (bi zi ) 0; therefore qi (zi ) = 0 if zi < bi . So qi (zi ) = 1 only if zi is the highest value. Note that the mechanism is incentive compatible, so in equilibrium xi = zi ; so the seller allocates the object only when one of the buyers has the highest value. Then the seller does not extract the whole surplus when every buyer’s value is below the highest one. This contradicts our initial assumption. Problem 5.2 (Optimal auction) There is a single object for sale and there are 2 potential buyers. The value assigned by buyer 1 to the object X1 is uniformly drawn from the interval [0; 1 + k] whereas the value assigned by buyer 2 to the object X2 is uniformly drawn from the interval [0; 1 k] ; where k is a parameter satisfying 0 k < 1. The two values X1 and X2 are independently distributed. a. Suppose the seller decides to sell the object using a second-price auction with a reserve price r: What is the optimal value of r and what is the expected revenue of the seller? b. What is the optimal auction associated with this problem? 17

x1 and Solution. Part a. Let the cumulative distribution for X1 be F1 (x1 ) = 1+k x2 the cumulative distribution for X2 be F2 (x2 ) = 1 k . The corresponding density 1 functions are f1 (x1 ) = 1+k and f2 (x2 ) = 1 1 k . So the expected payment from buyer 1 is Z 1+k Z 1+k Z x1 1 [m1 (x1 ; r)] f1 (x1 ) dx1 = rF2 (r) + yf2 (y) dy dx1 1+k r r r Z 1+k Z x1 r 1 1 r = y + dy dx1 1 k 1 k 1+k r r

where the …rst equality comes from (2.9) and G = F2 . The payment from buyer 2 is Z

Z

1 k

[m2 (x2 ; r)] f2 (x2 ) dx2 =

1 k

rF1 (r) +

r

r

Z

=

1 k

r + r 1+k

r

Z

Z

x2

yf1 (y) dy

r x2

y

r

1 1

k

dx2

1 1 dy dx2 1+k 1 k

The optimal reserve price r maximizes the expected revenue below Z

E [m1 (X1 ; r)] + E [m2 (X2 ; r)] =

1+k

r

r

+

Z

r 1

1 k

r

r

=

k

x1

y

r

r + 1+k

1

3 (k 2

+

Z

3k 2

1)

Z

1 1

x2

y

r

k

dy

1 dy 1+k

1 dx1 1+k 1 1

4r3 + 3r2 + 1

Di¤erentiating this with respect to r, we obtain 12r2 + 6r = 0 So we have r =

1 2

and the corresponding expected revenue for the seller is 1 k2 )

3 (1

3k 2 +

5 4

Part b. The virtual valuation for buyer 1 with value x1 is 1 (x1 )

= x1

1

F1 (x1 ) = 2x1 f1 (x1 )

(1 + k)

The virtual valuation for buyer 1 with value x1 is 2 (x2 )

= x2

1

F2 (x2 ) = 2x2 f2 (x2 )

18

(1

k)

k

dx2

Therefore the smallest value for bidder 1 that wins against x2 is y1 (x2 ) = inf fz1 :

1 (z1 )

y2 (x1 ) = inf fz2 :

2 (z2 )

0 and

= inf fz1 : 2z1 (1 + k) 1+k = max ; x2 + k 2

1 (z1 )

0 and 2z1

2 (x2 )g

(1 + k)

2x2

(1

k)g

2x1

(1 + k)g

and 0 and

= inf fz2 : 2z2 (1 k) 1 k ; x1 k = max 2

2 (z2 )

0 and 2z2

1 (x1 )g

(1

k)

Because 1 and 2 are increasing functions, the design problem is regular and by Proposition 5.3, the optimal mechanism (Q; M ) is given as Q1 (x1 ; x2 ) =

1 if x1 0

x2 > k and x1 otherwise

1+k 2

Q2 (x1 ; x2 ) =

1 if x1 0

x2 < k and x2 otherwise

1 k 2

For i = 1; 2, Mi (x1 ; x2 ) =

yi (x i ) if Qi (x1 ; x2 ) = 1 0 if Qi (x1 ; x2 ) = 0

Problem 5.3 (Dissolving a partnership) Two agents jointly own a …rm and each has an equal share. The value of the whole …rm to each is a random variable Xi which is independently and uniformly distributed on [0; 1] : Thus in the current situation, agent 1 derives a value 21 X1 from the …rm and agent 2 derives a value 12 X2 from the …rm. Suppose that the two agents wish to dissolve their partnership and since the …rm cannot be subdivided, ownership of the whole …rm would have go to one of the two agents. a. Consider the following procedure for reallocating the …rm. Both agents bid amounts b1 and b2 and if bi > bj , then i gets ownership of the whole …rm and pays the other agent j the amount bi . Find a symmetric equilibrium bidding strategy in this auction. b. Is the procedure outlined above e¢ cient? Is it individually rational? c. Calculate each agent’s payments in the VCG mechanism associated with this problem. Solution. Part a. Let the symmetric equilibrium bidding strategy be (x). Given 1 the uniform distribution, (bi ) is the probability of winning for bidder i when he

19

bids bi ; and his expected payo¤ is 1 xi 2 1 1 (bi ) xi 2 1

=

1

bi + 1

(bi )

(bi ) E [ (X i ) j bi ; (X i ) > bi ] Z 1 (y) 1 (bi ) dy 1 1 (bi ) (bi ) 1

bi + 1

Maximizing this respect to bi yields the …rst-order condition 1 xi 2

1 0

1

(bi )

1

bi

At symmetric equilibrium bi = 0

1

(bi )

1

(bi )

0

1

(bi )

=0

(xi ) so the …rst-order condition becomes

1 1 xi (xi ) 2

(xi ) 0

(xi ) +

xi

(xi )

0

1 =0 (xi )

1 2 (xi ) = xi 2

Using the integrating factor x2i , we solve the equation above and have the solution (x) = 16 x which is the symmetric equilibrium bidding strategy. Part b. Because the bidding function is strictly increasing, the buyer with a higher valuation wins the …rm, therefore the procedure above is e¢ cient. The expected payment for buyer i with value xi is Z 1 (y) mi (xi ) = xi (xi ) (1 xi ) dy xi xi 1 Z 1 1 1 2 = xi ydy 6 xi 6 1 2 11 = x 1 x2i 6 i 62 1 1 + x2i = 12 4 so mi (0) < 0 therefore the procedure is individually rational. Part c. Using formula (5.22), the payments in VCG mechanism are MiV (xi ; x i ) = W ( i ; x i ) W i (xi ; x i ) 1 = x i W i (xi ; x i ) 2 1 2 x i if xi > x i = 0 if xi < x i where the second equality comes from W

i (xi ; x i )

=

0 1 2x i

20

if xi > x if xi x

i i

Problem 5.4 (Negative externality) The holder of a patent on a cost reducing process is considering the possibility of licensing it to one of two …rms. The two …rms are competitors in the same industry and so if …rm 1 obtains the license, its pro…ts will increase by X1 while those of …rm 2 will decrease by X2 , where is a known parameter satisfying 0 < < 1: This is because if …rm 1 gets the license, …rm 2 will have a cost disadvantage relative to …rm 1. Similarly, if …rm 2 obtains the license, its pro…ts will increase by X2 while those of …rm 1 will decrease by X1 . The variables X1 and X2 are uniformly and independently distributed on [0; 1] : Firm 1 knows the realized value x1 of X1 and only that X2 is uniformly distributed, and similarly for …rm 2. a. Suppose that the license will be awarded on the basis of a …rst-price auction. What are the equilibrium bidding strategies? What is the expected revenue of the seller, that is, the holder of the patent? b. Find the payments in the VCG mechanism associated with this problem. Are the expected payments the same as in a …rst-price auction? c. Suppose that the patent holder is a government laboratory and it wants to ensure that the license is allocated e¢ ciently and that the net payments of the buyers add up to zero, that is, the “budget” is balanced. What is the associated “expected externality” mechanism for this problem? Is it individually rational? Does there exist an e¢ cient, incentive compatible, and individually rational mechanism that also balances the budget? Solution. Part a. Let the symmetric equilibrium bidding strategy be …rm i with value xi bids bi , his expected payo¤ is 1

(bi ) [xi

1

bi ] + 1

(bi ) (

xi )

Maximizing this respect to bi yields the …rst-order condition 1 0

1

(bi )

[xi

At symmetric equilibrium bi = 0

1

bi ]

(bi ) +

0

xi =0 1 (bi )

(xi ) so the …rst-order condition becomes

1 [xi (xi )

(xi )]

xi +

0

xi =0 (xi )

which could be rewritten as 0

(xi ) +

1 (xi ) = 1 + xi

Therefore the equilibrium bidding strategy is (x) =

1+ x 2

The payment from bidder i with valuation xi is mi (xi ) = xi (xi ) = 21

1+ x2 2

(x) : If the

so the seller’s revenue is E R

I

= 2E [mi (Xi )] = 2

Z

1

0

1+ 1+ x2i dxi = 2 3

Part b. The VCG payments in this problem are MiV (xi ; x i ) = W (0; x i ) = x =

W

i (xi ; x i )

W i (xi ; x i ) x i if xi > x i 0 if xi < x i

i

where the last equality comes from W

i (xi ; x i )

0 if xi > x x i if xi x

=

i i

So the seller’s revenue from the VCG mechanism is Z 1 Z E RV = 2E [Xi E [Xj j Xi ; Xj < Xi ]] = 2 xi 0

xi

0

y 1 dydxi = xi 3

Therefore, we have E RI > E RV Part c. As in Section 5.3, the “expected externality”mechanism for this problem Q ; M A is de…ned by MiA (xi ; x i ) = EXi [Wi (Xi ; x i )] EX i [W Z 1 Z 1 xi dxi x i dx i = x

=

i (xi ; X i )]

xi

i

1 2 x 2 i

x2 i

where the second equality comes from Xi if Xi > x 0 if Xi < x

Wi (Xi ; x i ) = W

i (xi ; X i )

0 X

=

i

if xi > X if xi < X

i i i i

Note that mi (0) =

Z

0

1

MiA (0; x i ) dx i

=

Z

0

1

1 0 2

x2 i dx

i

=

1 6

0

so it is individual rational. Because VCG mechanism derived in Part b results an expected surplus, Proposition 5.6 guarantees the existence of an e¢ cient, incentive compatible, and individually rational mechanism that balances the budget. 22

Problem 5.5 (A non-standard selling method) There is a single object for sale and there are two interested buyers. The values assigned by the buyers to the object are independently and uniformly distributed on [0; 1] : As always, each buyer knows the value he or she assigns to the object but the seller knows only that each is independently and uniformly distributed. The seller assigns a value of 0 to the object. Suppose that the seller adopts the following selling strategy. She approaches one of the buyers (chosen at random) and makes a “take-it or leave-it” o¤ er at a …xed price p1 . If the …rst buyer accepts the o¤ er, the object is sold to him at the o¤ ered price. If the …rst buyer declines the o¤ er, the seller then approaches the other buyer with a “take-it or leave-it” o¤ er at a …xed price p2 . If the second buyer accepts the o¤ er, the object is sold to him at the o¤ ered price. If neither buyer accepts then the seller keeps the object. a. What are the optimal values of p1 and p2 ? b. What is the expected revenue to the seller if he adopts this selling scheme? c.How does it compare to a standard auction? In particular, does the revenue equivalence principle apply?. Solution. Part a. If the seller choose p1 and p2 , his expected revenue is (1

p1 ) p1 + p1 (1

p 2 ) p2

Maximizing this respect to p1 and p2 yields the …rst-order conditions 1

2p1

p22

p2 p1 (1

= 0

2p2 ) = 0

If p1 = 0, the expected revenue is 0, so it can not be a solution. Therefore p1 = p2 =

3 8 1 2

Part b. The expected revenue is (1

p1 ) p1 + p1 (1

p 2 ) p2 =

1

=

21 64

3 8

3 3 + 8 8

1

1 2

1 2

Part c. From Example 3.1, we know that the standard auctions has expected revenue 13 which is more than 21 64 . The selling scheme could also be viewed as a direct mechanism (Q; M ) where Q1 (x1 ; x2 ) = Q2 (x1 ; x2 ) =

1 if x1 p1 0 if x1 < p1

1 if x1 < p1 and x2 p2 0 if x1 p1 or x2 < p2 23

For i = 1; 2 ; Mi (x1 ; x2 ) =

pi if Qi (x1 ; x2 ) = 1 0 if Qi (x1 ; x2 ) = 0

However the mechanism does not have the same allocation rule as standard auctions. For example, If x2 > x1 > p1 ; buyer 1 gets the object, while in standard auction buyer 2 should get the object. As a result, the revenue equivalence principle does not apply here.

24

6

Auctions with Interdependent Values

Problem 6.1 (A¢ liation) Suppose there are two bidders who receive private signals X1 and X2 which are jointly distributed over the set n o p S = (x1 ; x2 ) 2 [0; 1]2 : x1 x2 (x1 )2

with a uniform density. The bidders attach a common value V = 21 (X1 + X2 ) to the object. a. Find symmetric equilibrium bidding strategies in both a …rst-price and a secondprice auction? b. Calculate the expected revenues from both auctions and show that the revenue in a second-price auction is greater than that in a …rst-price auction. Solution. Part a. First, we apply Proposition 6.3 to …nd the equilibrium strategy for the …rst-price auction. The area of the set S is Z 1 p 1 x1 x21 dx1 = 3 0 so the density function of X1 and X2 is f (x1 ; x2 ) = 3. Let G ( j x) be the distribution of X2 conditional on X1 = x and let g ( j x) be the associated conditional density function, then we have Ry y x2 x2 f (s; x)ds p G (y j x) = R p = x x x2 2 f (s; x)ds x

g (y j x) =

Hence,

Therefore,

f (y; x) 1 =p R px x x2 f (s; x)ds x2

g(x j x) 1 = G(x j x) x x2

L(y j x) = exp = exp

Z

x

y

Z

x

t

y

= exp

Z

x

y

1

g(t j t) dt G(t j t) t2

dt

1 1 + dt t 1 t

= exp ln t jxy + ln(1 1 x y = x 1 y 25

t) jxy

Proposition 6.3 implies that the symmetric equilibrium strategies for the …rst-price auction are given by Z x I (x) = v(y; y)dL(y j x) 0 Z x 1 x y yd = x 1 y 0 Z x 1 x x2 1 = 1 dy x 1 x 1 y 0 1 x 1 x x2 ( ln (1 x) x) = x 1 x x 1 x = 1+ ln(1 x) x Second, Proposition 6.1 implies the symmetric equilibrium strategies in the secondprice auction are given by II

(x) = v(x; x) =

x+x =x 2

Part b. Let us …rst consider the …rst-price auction. The expected payment from bidder i with value x is the product of his bid and his probability of winning mI (x) = =

I

(x) G (x j x) 1 x ln(1 1+ x

x x2 x) p x x2

So the expected revenue for the seller is E RI

= 2E mI (Xi ) Z 1 = 2 mI (x) fi (x) dx 0 Z 1 p = 6 mI (x) ( x x2 )dx 0 Z 1 Z 1 2 = 6 (x x )dx + (1 0

=

x)2 ln(1

x)dx

0

1 3

where the third equality comes from the fact that the density of the marginal distribution of Xi is Z px Z px p fi (x) = f (t; x) dt = 3dy = 3( x x2 ) x2

x2

In the second-price auction, the expected payment from the bidder with value x is mII (x) =

II

x x2 (x)G(x j x) = x p x x2 26

So the corresponding expected revenue for the seller is E[RII ] = 2E mII (Xi ) Z 1 mII (x)fi (x)dx = 2 0 Z 1 = 6 x(x x2 )dx 0

=

1 2

Thus, E RI < E RII . Problem 6.2 (Lack of a¢ liation) Suppose there are two bidders whose private values X1 and X2 which are jointly distributed over the set n o S = (x1 ; x2 ) 2 [0; 1]2 : x1 + x2 1 with a uniform density. Show that a …rst-price auction with this information structure does not have a monotonic pure strategy equilibrium.

Solution. This proof is taken from Section 6 in Reny and Zamir (2004). Suppose to the contrary that such an equilibrium exists, and denote the equilibrium bidding strategies as 1 ( ) and 2 ( ). Let us …rst show that x2 2 (x2 ) for all x2 2 [0; 1). Suppose not and there exists some x ^2 2 [0; 1) such that 2 (^ x2 ) > x ^2 . Since 2 is nondecreasing, 2 (x2 ) > x2 for all x2 2 (^ x2 ; 2 (^ x2 )). Because bidders do not bid above their highest value, x2 ) 1: Consider the event fx1 < x2 ; x2 2 (^ x2 ; 2 (^ x2 ))g. First, this event has 2 (^ positive probability. Second, if bidder 1 wins the object in this event, his bid 1 (x1 ) satis…es 1 (x1 ) > 2 (x2 ) > x2 > x1 therefore bidder 1 has negative payo¤. Third, if bidder 2 wins the object, he also has negative payo¤ because 2 (x2 ) > x2 . So there is a positive probability that at least one of the players has negative payo¤, which contradicts equilibrium. As a result, xi i (xi ) for i = 1; 2: For any " 2 (0; 1), suppose bidder i has value 1 ". He knows that the other bidder’s value is below ", so are the other bidder’s bids. Thus i’s bid should be no more than ", so i (1 ") " for any " 2 (0; 1). Therefore, the only possible nondecreasing strategy equilibrium is 1 (x1 ) = 2 (x2 ) = 0 for all (x1 ; x2 ) in S, which is clearly not an equilibrium. Problem 6.3 (Bidding gap) Consider a common value second-price auction with two bidders. The bidders receive signals X1 and X2 , respectively, and these are independently and uniformly distributed on [0; 1]. Each bidder’s value for the object is V = 21 (X1 + X2 ). a. If the seller sets a reserve price r > 0; show that there is no bid in the neighborhood of the reserve price. b. Find the optimal reserve price. 27

Solution. Part a. Let the symmetric equilibrium bidding function be bidder 1 with value x bids b r , his expected payo¤ is (b; x) =

Z

0

=

Z

1

Z

(b)

v (x; y) g (y j x) dy 1

1 1

(r) j x

(b)

(r)

(b)

0

1

rG

( ). If

(y) g (y j x) dy

x+y dy 2

1

r

Z

(r)

1

(b)

(y) dy 1

(r)

Maximizing it with respect to b yields the …rst-order condition 1

x+

(b)

1

2 At symmetric equilibrium, b =

1

(b)

0

1

=0

(b)

(x) so the …rst-order condition becomes

x+x 2

(x)

0

1 =0 (x)

So the equilibrium bidding strategy is (x) = x The corresponding payo¤ is ( (x) ; x) =

Z

0

=

x

x+y dy 2

1 2 x 4

r

2

Z

x

ydy

r

2r2

p which means 2r. Therefore only bidders with p that ( (x) ; x) 0 only when x values in 2r; 1 will participate in the auction, and there is a gap between reserve p price r and the lowest bid 2r. Part b. Suppose the object has no value to the seller. The expected revenue for the seller from setting a reserve price r > 0 is Z 1 E [R] = 2 p xdy = 1 2r2 2r

So the optimal reserve price is 0. Problem 6.4 (Common value auction with Pareto distribution) Suppose that two bidders are competing in a …rst-price auction for an object with common value V; which is distributed according to a Pareto distribution FV (v) = 1 v 2 over [1; 1): Prior to bidding, each bidder receives a signal Xi whose distribution, conditional on the realized common value V = v, is uniform over [0; v] ; that is, FXi jV (xi j v) = 28

(xi =v) : Conditional on V = v; the signals X1 and X2 are independently distributed. Verify that the following strategy constitutes a symmetric equilibrium of the …rst-price auction (x) = 32 max fx; 1g (1 + max fx; 1g 2 ) Solution. Part a. Let the symmetric equilibrium bidding function be bidder 1 with value x bids b r , his expected payo¤ is (b; x) =

Z

0

=

Z

1

Z

(b)

v (x; y) g (y j x) dy 1

1 1

(r) j x

(b)

(y) g (y j x) dy

(r)

(b)

0

1

rG

( ). If

x+y dy 2

1

r

Z

(r)

1

(b)

(y) dy 1

(r)

Maximizing with respect to b yields the …rst-order condition 1

x+

(b)

1

2 At a symmetric equilibrium, b =

1

(b)

0

1

=0

(b)

(x) so the …rst-order condition becomes

x+x 2

(x)

0

1 =0 (x)

So the equilibrium bidding strategy is (x) = x The corresponding payo¤ is ( (x) ; x) =

Z

0

=

x

x+y dy 2

1 2 x 4

r

2

Z

x

ydy

r

2r2

p which means 2r. Therefore only bidders with p that ( (x) ; x) 0 only when x values in 2r; 1 will participate in the auction, and there is a gap between reserve p price r and the lowest bid 2r. Part b. Suppose the object has no value to the seller. The expected revenue for the seller from setting a reserve price r > 0 is Z 1 E [R] = 2 p xdy = 1 2r2 2r

So the optimal reserve price is 0. 29

The density function of joint distribution of X1 ; X2 ; V is h (x1 ; x2 ; v) = fX1 jV (x1 j v) fX2 jV (x2 j v) fV (v) 5

= 2v

for 0

x1 ; x2

v.

so the the density of joint distribution of X1 and X2 is Z +1 h (x1 ; x2 ; v) dv h (x1 ; x2 ) = maxfx1 ;x2 ;1g +1

Z

=

2v

5

dv

maxfx1 ;x2 ;1g

1 max fx1 ; x2 ; 1g 4 2 and the density function of V conditional on X1 = x1 and X2 = x2 is =

h (x1 ; x2 ; v) 4v 5 = h (x1 ; x2 ) max fx1 ; x2 ; 1g max fx1 ; x2 ; 1g

h (v j x1 ; x2 ) = for v

The distribution of X2 given X1 = x1 is Z x2 Z G (x2 j x1 ) = h (x1 ; t) dt = 0

=

x2

0

2 3

1 max fx1 ; t; 1g 2

4 1 2 x2 max fx1 ; 1g 3 3 1 max fx1 ; 1g 6 x2

4

4

dt

if x2 max fx1 ; 1g if x2 > max fx1 ; 1g

and its density function is g (x2 j x1 ) =

1 2

max fx1 ; 1g 4 1 2 x2

4

if x2 max fx1 ; 1g if x2 > max fx1 ; 1g

and we also have v (x; y)

E [V j X1 = x; X2 = y] Z +1 = vh (v j x; y) dv maxfx;y;1g +1

=

Z

v

maxfx;y;1g

=

4v 5 max fx; y; 1g

4 dv

4 max fx; y; 1g 3

Note that v (y; y) is no longer strictly increasing in y and (x) is not strictly increasing neither, so the proof of Proposition 6.3 does not apply to this question. As a result, we are not going to use Proposition 6.3 to derive the equilibrium, but it is easy to check that Proposition 6.3 also gives the same equilibrium, so it is still true in this context. v (y; y) is illustrated in Figure S6.1 where the horizontal axis represents y and the vertical axis represents v (y; y). (x) is illustrated similarly in Figure S6.2. 30

4

3

2

1

0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

2.5

3.0

3.5

4.0

y

Figure S6.1 4

3

2

1

0 0.0

0.5

1.0

1.5

2.0

Figure S6.2

31

4.0

x

First, if a bidder with value x > 1 bids (z) where z > x; his expected payo¤ is Z z [v (x; y) (z)] g (y j x) dy (z; x) = 0 Z z 4 2z = max fx; yg 1 + z 2 g (y j x) dy 3 3 Z0 x 2z 1 4 4 x 1+z 2 x dy = 3 3 2 0 Z z 4 1 4 2z + 1+z 2 y y dy 3 3 2 x 1 2 1 1 4 3 + + 4 x z+ = x 2 2 9 z 9 z z It is easy to see that 4 @ (z; x) = @z 9

1

1 z2

1 z3

1 x3

x > 1

value x > 1 bids

(8)

(z) where 1 < z < x, his expected

4 2z max fx; yg 1 + z 2 g (y j x) dy 3 3 4 2z 1 4 x 1+z 2 x dy 3 3 2 0 z2 1 2 3 x z 1+ 2 x 4 3 3 z @ (z; x) 2 = x @z 3

4

(x

z) > 0

Then we have (z; x)
1 bids (z) where 0 z < 1, he can only win the object in a tie and his expected payo¤ is Z 4 1 4 1 1 4 (z; x) = x x dy 2 0 3 3 2 1 1 1 1 = 3 3x 3 x4 1 1 1 1 < 2 3x 3 x4 = (x; x) 32

so (z; x)
0: Show that the following constitutes a symmetric 34

equilibrium. First, bidders with value x < r do not participate. Second, bidders with value x r bid Z x

(x; r) = x

r

L (y j x) dy

(8.4)

where L (y j x) is de…ned in (8.4). Solution. First, any bidder with value strictly less than r does not participate the auction because he can not win the object with a price lower than his value. Second, let us consider a bidder with value x r. He does not bid less than r because then he has no chance of winning. Suppose that he bids (z) r, his expected payo¤ is (z; x) = G (z j x) [x (z)] Taking the derivative with respect to z yields @ = G (z j x) (x @z

(z))

g (z j x) G (z j x)

0

(z)

When z = x we have @ @z

x=z

g (z j z) 0 (z) G (z j z) = (z (z)) g (z j z) G (z j z) 0 (z) Z z Z z g (t j t) = g (z j z) z z + exp dt dy G (t j t) r y 3 2 R z g(tjt) 1 exp dt z G(tjt) 5 G (z j z) 4 R z R z g(tjt) g(zjz) exp dy dt y G(tjt) r G(zjz) = G (z j z) (z

(z))

= 0

Similar as in Proposition 6.3, when z < x we have @ @z

> G (z j x) (z =

@ @z

(z))

g (z j z) G (z j z)

0

(z)

=0 x=z

where the inequality comes from the fact that g (z j x) g (z j z) > G (z j x) G (z j z) because of a¢ liation. It can also be veri…ed similarly that maximized by choosing z = x:

@ @z

35

< 0 when z > x. Thus

(z; x) is

Problem 8.2 (Increase in number of bidders) Consider a …rst-price auction with N bidders who have private values X1 ; : : : ; XN : The values X1 ; : : : ; XN are symmetrically distributed over [0; 1]N and are a¢ liated. Speci…cally, the joint distribution 1 of values is determined as follows. First, a random variable Z 2 10 ; 2 is drawn and each value of Z is equally likely. Next, each bidder’s value Xi is drawn from the distribution FXjZ (x j z) = exp z 1 x 1 over (0; 1]: As in the previous problem, suppose that the seller sets a reserve price r = 0:5: Let (n) (x; r) denote the symmetric equilibrium bidding strategy (as in (8.4)) when there are n bidders. Now suppose that the number of bidders increases to n + 1 and let (n+1) (x; r) denote the symmetric equilibrium bidding strategy when there are n + 1 bidders (again as in (8.4)). Show that for some x > r, (n) (x; r) > (n+1) (x; r) ; that is, an increase in the number of bidders may cause bids to decrease. (Note: This problem is computationally intensive. A symbolic computation program will be of great help.) Solution. The density of joint distribution of X1 ; : : : ; XN conditional on Z = z is h (x1 ; : : : ; xN j z) = fXjZ (x1 j z) : : : fXjZ (xN j z) and the joint density of X1 ; : : : ; XN is 1 1 h (x1 ; : : : ; xN j z = 0:1) + h (x1 ; : : : ; xN j z = 2) 2 2 1 = f (x1 j 0:1) : : : fXjZ (xN j 0:1) 2 XjZ 1 + fXjZ (x1 j 2) : : : fXjZ (xN j 2) 2

h (x1 ; : : : ; xN ) =

Conditional on bidder 1’s value X1 = x ; the cumulative distribution function of the highest bid of the other bidders is

G (y j x) =

=

=

=

Ry 0 : : : 0 h (x; t2 ; : : : ; tN ) dt2 : : : dtN R1 R1 0 : : : 0 h (x; t2 ; : : : ; tN ) dt2 : : : dtN Ry Ry : : 0 12RfXjZ (x j 0:1) fXjZ (t2 j 0:1) : : : fXjZ (tN 0 :R y y + 0 : : : 0 12 fXjZ (x j 2) fXjZ (t2 j 2) : : : fXjZ (tN

Ry

j 0:1) dt2 : : : dtN j 2) dt2 : : : dtN

! R1 1 : : : f (x j 0:1) f (t j 0:1) : : : f (t j 0:1) dt : : : dt 2 N XjZ 2 XjZ N 0 R 0 2R XjZ 1 1 + 0 : : : 0 12 fXjZ (x j 2) fXjZ (t2 j 2) : : : fXjZ (tN j 2) dt2 : : : dtN Ry Ry 1 2 fXjZ (x j 0:1) 0R fXjZ (t2 j 0:1) dt2 : : : R0 fXjZ (tN j 0:1) dtN y y + 12 fXjZ (x j 2) 0 fXjZ (t2 j 2) dt2 : : : 0 fXjZ (tN j 2) dtN ! R1 R1 1 f (x j 0:1) f (t j 0:1) dt : : : f (t j 0:1) dt 2 N 2 XjZ 0R XjZ 2 R01 XjZ N 1 + 21 fXjZ (x j 2) 0 fXjZ (t2 j 2) dt2 : : : 0 fXjZ (tN j 2) dtN

R1

1 2 fXjZ

(x j 0:1) FXjZ (y j 0:1) 1 2 fXjZ

N 1

+ 12 fXjZ (x j 2) FXjZ (y j 2)

(x j 0:1) + 12 fXjZ (x j 2) 36

N 1

and the corresponding density function is N 2 1 1) FXjZ (y j 0:1) fXjZ 2 fXjZ (x j 0:1) (N N 2 1 fXjZ + 2 fXjZ (x j 2) (N 1) FXjZ (y j 2) 1 1 2 fXjZ (x j 0:1) + 2 fXjZ (x j 2)

g (y j x) =

(y j 0:1) (y j 2)

The expression of G (y j x) and g (y j x) give us the ratio g (t j t) G (t j t)

fXjZ (x j 0:1) (N 1) FXjZ (y j 0:1) +fXjZ (x j 2) (N 1) FXjZ (y j 2)

=

fXjZ (x j 0:1) FXjZ (y j 0:1) =

0:01t

2 (N

N 1

N 2

fXjZ (y j 0:1) 2 fXjZ (y j 2)

N

+ fXjZ (x j 2) FXjZ (y j 2) N

1) exp 0:1 1

t

1

0:1 [exp (0:1 (1

t

1 ))]N

2 (N

+ 4t

!

! N 1

1) exp 2 1

+ 2 [exp (2 (1

t

1

t

1 ))]N

where the second equality comes from the substitution of the function form of FXjZ . According to the de…nition in (6:7), we have Z x g (t j t) L (y j x) = exp dt y G (t j t) ! 0 1 N 0:01t 2 (N 1) exp 0:1 1 t 1 B Z x C N B C +4t 2 (N 1) exp 2 1 t 1 B = exp B dtC C N N 1 1 t ))) + 2 (exp (2 (1 t ))) y 0:1 (exp (0:1 (1 @ A = exp 0

+ 0:19(N 0:1N

1)

y

+ 0:19(N 0:1N

0:1N t y 0:1 exp(0:1N (1 t

= exp @

1) + 19(N N

exp(0:1N (1 y

2 (N

exp 0:1 (N

1) t 2 dt 2 exp 0:1N 1 ( (

2 (N

1 19

ln s

1)

1 x

ln

1 y

1 ))

1)

s19

1 x

))

+

1 y 1 20

19

+

1

t

1 ))

dt

A

exp(0:1N (1 x 1 )) exp(0:1N (1 y 1 ))

= 1 x

1

1 ds 0:1s+2s20

exp 0:1N 1

exp 0:1N 1

t

1 ))+2 exp(2N (1

2 (N 1) x1 y1 1 1) R exp(0:1N (1 x ))

= exp @ 0

Rx

Rx

1 20

1 y

19

+

1 20

1 A

N 1 N

N 1 N

where the fourth equality comes from changing variable s = exp 0:1N 1

37

!

t

1

.

N

Now applying formula (8.4) yields the bidding function with N bidders Z x L (y j x) dy N (x) = x 0:5

= x

Z

x

exp 0:1 (N

0:5

1)

1 x

1 y

exp 0:1N 1 1 x

exp 0:1N 1

19

+

1 20

1 y

19

+

1 20

N 1 N

dy

N 1 N

The following …gure plots the bidding functions with number of bidders N = 2 and N = 6, where the solid line represents the bidding function with 2 bidders, while the dash line is the bidding function with 6 bidders. Note that the dash line is below the solid line for small values, which means for those values the bids are lower with more bidders.

bid

1.0

0.9

0.8

0.7

0.6

0.5 0.5

0.6

0.7

0.8

0.9

1.0

x

Figure S8.1

Problem 8.3 (A second-price auction without a monotone equilibrium) Consider a second-price auction with three bidders. Each bidder i receives a private signal Xi . Bidder 1’s value V1 = X1 +X2 is the sum of his own signal and bidder 2’s signal, while bidder 2’s and 3’s values are private and equal their own signals— that is, V2 = X2 and V3 = X3 . Each bidder receives either a “high” or a “low” signal. Speci…cally, X1 2 f1; 2g, X2 2 f0; 4g, and X3 2 f0; 3g. Bidders 1 and 3 have perfectly correlated signals, both being low with probability 1=2 or both being high with probability 1=2. Bidder 2’s signal is independent of the others’ signals, being low with probability p and high with probability 1 p, where p 2 (2=3; 1). a. Verify that the bidders’ signals are a¢ liated. 38

b. Show that with the speci…ed information structure, there does not exist a purestrategy equilibrium of the second-price auction such that the equilibrium strategies are increasing and undominated. Solution. Part a. By the de…nition of a¢ liation in Appendix D, we need to show Pr x0 _ x00 Pr x0 ^ x00

Pr x0 Pr x00

for all x0 and x00 2 X

(13)

If bidder 2’s values are both 0 in x0 and x00 ; (13) becomes 1 2 p 4

1 2 p 4

If bidder 2’s values are both 1 in x0 and x00 ; (13) becomes 1 (1 4

1 (1 4

p)2

p)2

If bidder 2’s values are di¤erent in x0 and x00 ; (13) becomes 1 (1 4

1 (1 4

p) p

p) p

So the signals are a¢ liated. Part b. It is easy to see that bidding their values is a undominated strategy for bidders 2 and 3. Therefore bidder 2 and 3 each employ a pure-strategy which is increasing. It remains only to compute an undominated strategy for bidder 1. When X1 = 1, bidder 1 knows that 3 (X3 ) = X3 = 0 . Since 1’s value is v1 = 1 + X2 , his value is 1 if 2 (X2 ) = X2 = 0 ; while it is 5 is 2 (X2 ) = X2 = 4. Consequently, 1’s value is always strictly above the high bid of the others and so it is optimal for him to bid so that he wins with probability one. Because p < 1, this requires a bid more than 4. However, when X1 = 1 his value is never more than 5, and so in any undominated equilibrium we must have 1 (1) 2 (4; 5]: When X1 = 2; bidder 1 knows that 3 (X3 ) = X3 = 3 . Consequently, a bid above 4 still wins with probability one, but bidder 1 now pays 3 when 2 (X2 ) = X2 = 0 and v1 = 2, and pays 4 when 2 (X2 ) = X2 = 4 and v1 = 6. Hence, a bid above 4 yields bidder 1 a payo¤ of p (2 3) + (1 p) (6 4), which is negative because p > 2=3. Similarly, any bid weakly between 3 and 4 yields a negative payo¤. Consequently, we must have 1 (2) < 3. Hence, an undominated pure-strategy equilibrium exists, but in no such equilibrium is bidder 1’s strategy nondecreasing.

39

9

E¢ ciency and the English Auction

Problem 9.1 (Two-bidder auctions) Suppose that there are two bidders with valuations v1 (x1 ; x2 ) = 32 x1 + 13 x2 1 3 x1

v2 (x1 ; x2 ) =

+ 23 x2

and all signals lie in [0; 1]. a. Using the break-even conditions, …nd an e¢ cient ex post equilibrium of the English (in this case, also the second-price) auction. b. Show that the equilibrium strategies so determined survive the iterated elimination of weakly dominated strategies and are the only strategies to do so. (Note: Iterated elimination of dominated strategies is carried out stepwise as follows. In step 1, discard all weakly dominated strategies for both bidders. In step 2, discard all weakly dominated strategies in the reduced game obtained after step 1. In step 3, discard all weakly dominated strategies in the reduced game obtained after step 2. Continue in this fashion.) Solution. Part a. Let 1 (x) and 2 (x) be the prices the bidders drop out at and 1 and 2 be the corresponding inverse functions. According to equation (9.3) in section 9.2, the break-even conditions are

which can be rewritten as

v1 (

1 (p) ;

2 (p))

=p

v2 (

1 (p) ;

2 (p))

=p

2 (p)

=p

2 (p)

=p

2 3 1 3

1 3 2 1 (p) + 3 1 (p)

+

so 1 (p)

=

2 (p)

=p

Hence the bidding functions are 1 (x) = 2 (x) = x. Part b. Assume the conditional density functions fX1 jX2 fX2 jX1 are strictly positive, and 1 ( ) and 1 ( ) are strictly increasing. If bidder 1 with signal x1 bids b1 , his expected payo¤ is 1 (b1 ; x1 ) =

Z

1 2

(b1 )

2 1 x1 + y 3 3

0

2 (y)

fX2 jX1 (y j x1 ) dy

and the marginal expected payo¤ is 2 1 @ 1 = x1 + @b1 3 3

1 2

(b1 )

b1

40

fX2 jX1 0

1 2 1

(b1 ) j x1

(b1 )

1

Step 1. Because x2 2 [0; 1], we have the marginal expected payo¤ is @ 1 @b1

2 1 x1 + 3 3

=

1

(b1 )

2

2 1 x1 + 3 3

2

(b1 ) 2 [0; 1] : Therefore if b1 > 32 x1 + 13 , fX2 jX1

b1

0

1

fX2 jX1

b1

1 2

2

0

1

1

(b1 ) j x1

(b1 )

(b1 ) j x1

(b1 )

< 0 where the …rst inequality comes from 2 1 (b1 ) 1. Hence bidder 1 could bene…t by reducing the bid, so he does not bid more than 23 x1 + 13 : Similarly, if b1 < 23 x1 ; the marginal expected payo¤ is @ 1 @b1

2 1 x1 + 3 3

=

1

(b1 )

2

2 1 x1 + 3 3

fX2 jX1

b1

0

1

fX2 jX1

b1

2

0

1

1 2 1

(b1 ) j x1

(b1 )

(b1 ) j x1

(b1 )

> 0 1

where the …rst inequality comes from than 32 x1 : As a result, we have 1 (x1 )

2

2

(b1 )

0; hence bidder 1 does not bid less

2 2 1 x1 ; x1 + 3 3 3

Applying the same analysis to bidder 2 yields 2 (x2 )

3 2

2

2 1 2 x2 ; x2 + 3 3 3

Step 2. From step 1, we have 1 (x1 ) 2 23 x1 ; 23 x1 + 13 , therefore b1 13 ; 23 b1 : Repeating the same analysis in step 1 implies 1 (x1 )

2

1 2

(b1 ) 2

1 4 ; x1 3 3

4 x1 3

4 1 4 x2 ; x2 3 3 3 Step n: The strategies that survive the n steps of iterated eliminations are 2 (x2 )

i (xi )

2[

n xi ;

i (xi )

2[

n xi

2

n xi

n

n

+ 1;

1

=

2 3

n+1

=

+ 1] if n is odd

(14)

n xi ]

(15)

where

3 41

2 1=

n

if n is even

Taking limit of both sides of the equation above yields lim

n!1

lim

n!1

n

n

=

2 3 1= lim

n!1

n

; therefore

= 1. Therefore (14) and (15) imply that the set of bidder i’s strategies that

survives nth iterated elimination shrinks to fxi g as n goes to in…nity. Hence the equilibrium strategy derived in Part a. is the only strategy that survives the iterated elimination of weakly dominated strategies. Problem 9.2 (Ordinality) Suppose is an increasing and di¤ erentiable function such that (0) = 0. If the valuations v satisfy the average crossing condition, and for all i, wi ( ) = (vi ( )) then the valuations w also satisfy the average crossing condition. Solution. First, note that vi is maximal at x if and only if wi is maximal at x. This means that for all A N , the set of signals such that the values of all bidders in A are maximal is the same for both v and w. Then, for any such vector of signals x which satis…es that for all l 2 A, vl (x) = max vk (x) k2N

vx

(16)

and for all i; j 2 A, i 6= j, @ wA (x) = @xj = = = > = =

1 X @wi (x) i2A ]A @xj 1 X @vi 0 (vi (x)) (x) i2A ]A @xj 1 X @vi 0 (vx ) (x) i2A ]A @xj @ 0 vA (x) (vx ) @xj @vi 0 (vx ) (x) @xj @vi 0 (vi (x)) (x) @xj @wi (x) @xj

where the third equality comes from (16) and the inequality comes from the average crossing condition for v. Thus, the average crossing condition is also satis…ed by w.

42

10

Mechanism Design with Interdependent Values

All of the problems below concern the following environment. Suppose that there are two potential buyers for one indivisible object. Each buyer’s private value Xi for the object is drawn at random from the set X = f10; 20g : Buyers’ values are jointly distributed as follows: Pr [10; 10] = Pr [20; 20] = 0:2 Pr [10; 20] = Pr [20; 10] = 0:3 where Pr [x1 ; x2 ] denotes Pr [X1 = x1 ; X2 = x2 ]. Problem 10.1 (Generalized VCG mechanism) What is the generalized VCG mechanism (Q ; M ) for the environment speci…ed above? Determine the expected revenue from this mechanism. Solution. The buyers’ values are symmetric and private but correlated. Consider a direct mechanism (QA ; MA ). Recall that Qi (x) is i’s probability of winning the object when signals x are reported. And Mi (x) includes i’s payment upon “winning” and “losing” separately. In what follows, these components are written as Miw (x) and Mil (x); respectively.1 Buyer i’s beliefs about buyer j’s signal conditional on his own are given by Bayes’ rule: i (10 j 10) = i (20 j 20) = 0:4 and i (20 j 10) = i (10 j 20) = 0:6 for both buyers i = 1; 2. De…ne UiA (zi ; xi ) =

X

i (xj

xj

j xi ) QA i (zi ; xj ) xi

QA i (zi ; xj ))

+(1

MilA (zi ; xj )

A Miw (zi ; xj )

(17)

to be i’s expected payo¤ when his value is xi , he reports a value of zi and the other buyer j reports truthfully. The seller’s expected revenue in the truth-telling equilibrium of this mechanism (if it exists) is the weighted sum of the payments taken from each buyer i for each pair of signals x, where weights come from the probability with which each bidder wins Qi (x) and the probability of the signals P r[x]: X X A A E[RA ] = Pr[x] QA QA (18) i (x)Miw (x) + (1 i (x))Mil (x) x

i

The generalized Vickrey-Clarke-Groves (VCG) mechanism (Q ; M ) is shown in the table below. x Q (x) Miw (x) Mil (x) 1 1 10; 10 10 0 2; 2 10; 20 0; 1 10 0 20; 10 1; 0 10 0 1 1 20; 20 ; 20 0 2 2 1

In Chapter 5, Mi (x) was i’s expected payment and did not depend on who won the object.

43

The buyers have private values so the discrete version of the single crossing condition (10.4) clearly holds and Proposition 10.1 says that truth-telling is an equilibrium. The seller’s expected revenue in the truth-telling equilibrium is computed from (18), above: 1 1 1 1 10 + 0) + ( 10 + 0) 2 2 2 2 + Pr[10; 20]f(0 10 + 1 0) + (1 10 + 0 0)g

E[R ] = Pr[10; 10] (

+ Pr[20; 10]f(1 10 + 0 0) + (1 0 + 0 10)g 1 1 1 1 20 + 0) + ( 20 + 0) 2 2 2 2 = 0:2 10 + 0:3 10 + 0:3 10 + 0:2 20

+ Pr[20; 20] (

= 12

Problem 10.2 Consider the following mechanism. If both buyers report values z1 = z2 = 20, then pick a buyer randomly with probability 21 , say this is buyer i, and give him the object for a price of Mi = 20. The other buyer j pays nothing. If both report values z1 = z2 = 10, again pick a buyer randomly with probability 12 , say i, and give him the object for a price of Mi = 19: The other buyer, say j, pays Mj = 9 without, of course, getting the object. If one buyer reports zi = 20 and the other xj = 10, then give the object to buyer i for a price Mi = 20; The other buyer j receives a transfer of 6; that is, Mj = 6: a. Show that the mechanism described above is incentive compatible and individually rational. b. What is the expected revenue in the truthful equilibrium of this mechanism? c. Does the mechanism have other (non-truthful) equilibria? Solution. Call this mechanism (Q; M). Then x 10; 10 10; 20 20; 10 20; 20

Q(x) 1 1 2; 2 0; 1 1; 0 1 1 2; 2

Miw (x) 19 20 20 20

Mil (x) 9 6 6 0

Part a. Truth-telling is incentive compatible (IC) for buyer i if for all xi ; zi 2 Xi Ui (xi ; xi )

Ui (zi ; xi )

(IC)

where Ui was de…ned in (17), above. To verify this condition, it is su¢ cient to consider only buyer 1 and only z1 6= x1 . That is, we need to show U1 (10; 10) U1 (20; 10) and

44

U1 (20; 20)

U1 (10; 20).

U1 (10; 10) = 0:6f0 (10

20) + 1 6g + 0:4

1 (10 2

19) +

1 ( 9) 2

=0 U1 (20; 10) = 0:6 =

1 (10 2

20) +

1 0 + 0:4f1 (10 2

20) + 0 6g

1 (20 2

20) +

1 0 + 0:6f1 (20 2

20) + 0 6g

1 (20 2

19) +

1 ( 9) + 0:6f0 (20 2

7

U1 (20; 20) = 0:4 =0 U1 (10; 20) = 0:4

20) + 1 6g

=0 So, (IC) is satis…ed. And individual rationality (IR), Ui (xi ; xi )

Ui (xi )

0;

(IR)

is clearly also satis…ed. Note that this mechanism is e¢ cient and it extracts the entire surplus. Part b. The mechanism is IC and IR, so truth-telling is an equilibrium. The seller’s expected revenue in the truth-telling equilibrium is computed from (18), above: E[R] = 0:2(19 + 9) + 0:3(20

6) + 0:3(20

6) + 0:2(20 + 0) = 18

Part c. No, there are no other equilibria. First note that a low-valued buyer will never report a value of 20 because his payo¤ from doing so is Pr[opponent reports 10]( 9) + Pr[opponent reports 20]( 5) < 0: So each buyer must report 10 when his value is 10. Next suppose one buyer reports 10 with probability > 0 when his value is 20. Then the other buyer’s payo¤ when his value is 10 is (0:4 + 0:6 )( 9) + 0:6(1 )(6) < 0 which violates individual rationality. So each buyer must report 20 when his value is 20. Problem 10.3 (Crémer-McLean mechanism) What is the Crémer-McLean mechanism Q ; MC for the environment speci…ed above?

45

Solution. The construction of the Crémer-McLean (CM) mechanism is explained in the proof of Proposition 10.2. By symmetry of the value structure, we can restrict attention to buyer 1. The vector of buyer 1’s expected payo¤s in the truth-telling equilibrium of the VCG mechanism is U1 (10; 10) U1 (20; 20)

u1

0:4 0 + 0:6 0 0:4 0 + 0:6 10

=

0 6

=

The matrix of buyer 1’s conditional beliefs is 1 (10

1

1 (10

j 10) j 20)

1 (20 i (20

j 10) j 20)

=

0:4 0:6 0:6 0:4

:

This matrix has full row rank, so there is a unique solution to 1 c1

= u1

namely, c1 =

c1 (10) c1 (20)

=

18 12

The CM mechanism is (Q ; MC ) where M1C (x1 ; x2 ) = M (x1 ; x2 ) + c1 (x2 ): Buyer 2’s payments are analagous. The mechanism is summarized in the table below. The payments made in zero-probability events may be anything; they are written as in the table. x 10; 10 10; 20 20; 10 20; 20

Q (x) 1 1 2; 2 0; 1 1; 0 1 1 2; 2

C (x) M1w 28

C (x) M1l 18 12

28 8

12

C (x) M2w 28 28

C (x) M2l 12

8

12 12

Problem 10.4 (Non-negativity payo¤ constraint) For the environment speci…ed above, does there exist a mechanism that (a) is incentive compatible and individually rational; (b) gives each buyer a non-negative payo¤ for every realization of the values; and (c) extracts all the surplus from the buyers? Solution. No, there is no such mechanism. In the last two mechanisms, the buyers sometimes made negative payments (that is, the seller paid money to them). Suppose this is not allowed: Mik (x) 0 i = 1; 2 k = w; l: (NN) A revenue-maximizing mechanism will solve the problem max E[R] subject to (IC), (IR) and (NN): Q;M

46

(P)

This is a linear programming problem. After working it out, you will see that the best mechanism is a posted price of 20, as given in the table below. The expected revenue is easily calculated as Pr[at least one player has a value of 20] 20 = 16: x 10; 10 10; 20 20; 10 20; 20

Q(x) 0; 0 0; 1 1; 0 p; 1 p

Miw (x) x 20 20 20

47

Mil (x) 0 0 0 0

11

Bidding Rings

Problem 11.1 (Maximal loss from collusion) Consider a second-price auction with N 2 bidders. Each bidder’s private value Xi is independently and uniformly distributed according on [0; 1] : a. First, suppose bidders bid individually— that is, there is no bidding ring. As a benchmark, …nd the expected revenue of the seller if he sets an optimal reserve price r > 0 (as in Chapter 2). b. Now suppose that the N bidders form a perfectly functioning bidding ring. Find the expected revenue of the seller if he sets an optimal reserve price r > 0 in the face of such collusion. c. Show that for all n; the optimal revenue with collusion is at least one-half of the optimal revenue without collusion; that is, > 12 . Solution. Part a. Each player’s expected payment under a reserve price r can be found using (2.9) and G(y) = F n 1 (y): Z x II N m (x; r) = r + (N 1)y N 1 dy r

N 1 N 1 x = rN + N N

The seller’s expected revenue is the sum of the bidders’expected payments: Z 1 E[R] = N mII (y; r)dy r

N 1 (1 rN +1 ) (N + 1) 2N N +1 N 1 r + rN + N +1 N +1

N

= r (1 =

r) +

The seller maximizes this to …nd the optimal reserve price r = 21 . We could also …nd this by noting that the seller values the object at x0 = 0 and the hazard rate (x) = 1=(1 x) is increasing in x, so (2.12) is a su¢ cient condition for an optimal reserve price r . Plugging this into the expression above, we …nd the seller’s expected revenue under the optimal reserve price: =

1N 2

+N 1 N +1

Part b. Because the ring consists of all bidders, it does not need to protect itself from outside bids. If no bidder values the object at least at r, the ring will not bid. If any ring member has a value at or above r, the ring will submit a bid of r. Hence, the seller’s expected revenue from setting a reserve price of r is E[R] = r Pr max xi i

48

r = r(1

rN )

This is maximized at r = (N + 1) 1=N for expected revenue of = N=(N + N +1=N 1) . Part c. Because r is optimal, the revenue from setting r = 12 instead is less than or equal to (with equality at N = 1). The …rst line below uses this fact. 1 2

1 2

1

= 2

1 2N N +2 2N

1 2

1N 2

+N 1 N +1

2(N + 1)

>0

Problem 11.2 (Collusion in …rst-price auctions) Consider a …rst-price auction with three bidders. Bidder 1’s value X1 = 34 with probability 34 and X1 = 12 with probability 1 4 : Bidders 2 and 3 have …xed and commonly known values. Speci…cally, x2 = 1 and x3 = 14 : a. Find an equilibrium of the …rst-price auction when the three bidders act independently. (Note: Since values are discrete, this will be in mixed strategies). b. Now suppose that bidders 1 and 2 form a cartel. While the cartel cannot control the bids submitted by its members, it can arrange transfers and recommend bids .Further, suppose that the values of its members become commonly known among the cartel once it is formed. i. Find an equilibrium with the cartel, assuming that bidder 3 acts independently. ii. Is it possible for the cartel to ensure that only one member submits a bid? Solution. Part a. Suppose bidder 2 always bids b2 = 43 and wins, for a payo¤ of 0.25. Bidder 2 is already winning, so bidding more is not better. The strategies of bidders 1 and 3 must ensure that bidding less is also a bad idea. Let bidder 3 always bid zero. Suppose that bidding b < 43 gives 2 an expected payo¤ of b=3 < 0:25. Then bidder 1’s mixed strategy F1 (b) can be solved for b 2 [0; 3=4] by setting (1

b)F1 (b) = b=3

and so for b 2 [0; 3=4]

b=3 1 b It is straightforward to verify that F1 is indeed a distribution function and that bidders 1 and 3 cannot do better than follow the strategies given above. There are many other equilibria leading to the same outcome (allocation of the object and payments). For example, bidder 1’s strategy may depend on his value. Part b.i. Suppose the cartel makes no transfers and always lets bidder 2 have the object. The following argument follows that of part a. Bidder 2 bids b2 = x1 for F1 (b) =

49

a payo¤ of 1 x1 ; and bidder 1 uses the mixed strategies F1 (b j x1 ) so that 2 gets payo¤ b(1 x1 )=x1 < 1 x1 from bidding b < x1 : 1 2 3 b j x1 = 4

F1 b j x1 = F1

b 1 b b=3 = 1 b =

for b 2 [0; 1=2] for b 2 [0; 3=4]

Again, let bidder 3 drop out: b3 = 0. It remains to verify that bidders 1 and 3 are behaving optimally and that F1 ( j x1 ) is a distribution function for each x1 . Part b.ii. No, the cartel cannot enforce a one-bidder policy. Suppose bidder 1 has value x1 and does not submit a bid. If the support of bidder 3’s strategy extends to y > 41 , then bidder 1 will bid b1 < y, trading o¤ a lower payment and a lower probability of winning. This would make 3’s payo¤ negative, so the support of bidder 3’s strategy must lie inside [0; 14 ]. Bidder 1’s best response is some b1 < 14 . Bidder 2 would like to enter and bid some b2 < 41 for positive payo¤ (as opposed to no payo¤ from staying out). No pre-auction transfers can convince him to stay out. Similarly if only bidder 2 submitted a bid and that bid was a best response to bidder 3, 1 would be also choose to enter and bid. Problem 11.3 (PAKT) A single object is to be sold via a second-price auction to two bidders whose private values Xi are drawn independently from the uniform distribution on [0; 1]. Suppose that the bidders form a cartel. Find the equilibrium bidding strategies in the preauction knockout (PAKT). Solution. First, consider the second-price PAKT. As argued in the book, this is an incentive-compatible and individually rational direct mechanism. Hence truthful reporting is an equilibrium. Next, suppose the bidders use a …rst-price PAKT. Equilibrium bidding strategies are given in Proposition 11.3: 1 (x) = x 4 Problem 11.4 (Collusion-proof mechanism) A single object is to be sold to two bidders with private values drawn independently from the uniform distribution on [0; 1]. The following mechanism is used to sell the object. Each bidder i submits a bid bi : Suppose that bi > bj : Then the loser, bidder j, is asked to pay a …xed amount 13 to the seller. The winner, bidder i, is awarded the object and asked to pay bi to the losing bidder j. (If there is a tie, either bidder is assigned the role of a winner.) a. Find a symmetric equilibrium of this mechanism assuming that the bidders act noncooperatively. b. Can the two bidders gain by forming a cartel and colluding against the seller? Solution. Part a. Suppose is a symmetric equilibrium. If bidder 1 bids when his value is x, his expected payo¤ in this equilibrium is Z z Z 1 1 [x (z)] dv + (v) dv: 3 0 z 50

(z)

The …rst-order condition evaluated at z = x is x

2 (x)

x 0 (x) +

1 = 0: 3

This can be solved using the method of integrating factors. First, rearrange so that the ’s are on one side: 2 1 0 (x) + (x) = 1 + x 3x R Next, multiply by x2 = exp x2 dxg. x2 0 (x) + 2x (x) = x2 + x=3 d 2 [x (x)] = x2 + x=3 dx

Integration gives the solution: 1 1 (x) = x + 3 6 Part b. No. Roughly speaking, the bidders cannot take the seller’s piece of the pie; nor can they make the pie larger. Therefore they have nothing to gain from collusion. The seller’s pro…t is 31 no matter how the bidders try to collude. Each bidder’s expected payo¤ in the symmetric equilibrium above is 16 , for a total expected surplus of 23 . This is also the expectation of the highest value, so no more surplus is available.

51

13

Equilibrium and E¢ ciency with Private Values

Problem 13.1 (Uniform price auction) Consider a three-unit uniform-price auction with two bidders. Each bidder’s value vector Xi = X1i ; X2i ; X3i is independently and identically distributed on the set X = fx 2 [0; 1]3 : x1 x2 x3 g according to a density function f such that the marginal distributions are: F1 (x1 ) = (x1 )2 F2 (x2 ) = (2

x2 ) x2

F3 is left unspeci…ed. Show that the bidding strategy (x1 ; x2 ; x3 ) = (x1 ; (x2 )2 ; 0) constitutes a symmetric equilibrium of the uniform-price auction. Solution. Suppose one bidder with values (x1 ; x2 ; x3 ) bids (b1 ; b2 ; b3 ) where 1 b1 b2 b3 0: Let (y1 ; y2 ; y3 ) be the values for the other bidder. The expected payo¤ of the bidder is Z = [x1 + x2 + x3 3y1 ] dF (y1 ; y2 ) fy1 (y2 )2 ;b3 0;b2 X2 > Y1 > Y2 ; P1 = p1 ] 3 1 + E[P2 j X1 > Y1 > X2 > Y2 ; P1 = p1 ] 3 1 E[P2 j X1 > Y1 > Y2 > X2 ; P1 = p1 ] 3 In the …rst case P1 = (Y1 ) = 12 Y1 and P2 = Y1 . Because is strictly increasing, knowing P1 = p1 means knowing that the price tomorrow will be ; 1 (p1 ) = 2p1 . For the second case, the prices are (Y1 ) and X2 . X2 is the higher of two random draws from the uniform distribution on [0; 2p1 ], so its expectation is 32 2p1 . In the …nal case, the prices are also (Y1 ) and X2 . This time X2 is the lower of two random draws from the uniform distribution on [0; 2p1 ], so its expectation is 13 2p1 . Plugging these into the equation above, we have E[P2 j P1 = p1 ] =

1 2 1 2p1 1 + + 3 3 3

4 = p1 > p1 : 3

Problem 15.3 (Multiunit demand) Consider the same environment as in the previous problem. a. Show that the following strategy also constitutes a symmetric equilibrium of the second-price format: i. in the …rst auction, bid 1 xi1 ; xi2 = xi2 ; and ii. in the second auction, bid truthfully. b. What can you say about the resulting sequence of equilibrium prices (P1 ; P2 )? Solution. Part a. The following argument is similar to that found in Problem 15.2. We will show that 1 (x1 ; x2 ) = x2 is the unique strictly increasing strategy that only depends on the second-period value. If the second player is bidding according to a strictly increasing function (y2 ) in the …rst round and truthfully in the second, then it is still optimal for bidder 1 to bid truthfully in the second round. Bidder 1’s payo¤ from bidding as type t 2 (x2 ; x1 ) in the …rst round is Z tZ 1 + (t; x) = 2 (x1 (y2 ))dF (y1 )dF (y2 ) 0 y2 Z x2 Z x2 +2 (x2 y1 )dF (y1 )dF (y2 ) 0

+2

Z

t

x1

y2 1

Z

(x1

y2

58

y2 )dF (y1 )dF (y2 )

The …rst two lines are the payo¤ from winning the …rst round; and the last is the payo¤ from losing it. The payo¤ from bidding t < x2 is Z tZ 1 (t; x) = 2 (x1 (y2 ))dF (y1 )dF (y2 ) 0

+2

y2

Z tZ

+2

t

(x2

y1 )dF (y1 )dF (y2 )

y2

0

Z

x2

x1

Z

1

(x1

y2 )dF (y1 )dF (y2 )

y2

The …rst two lines are the payo¤ from winning the …rst round; and the last is the payo¤ from losing it. The …rst-order conditions lead to a characterization of : Z

1

x2

(x1

(x2 ))dF (y1 )

Z

1

(x1

x2 )dF (y1 ) = 0:

x2

This is clearly satis…ed by (x2 ) = x2 . We still need to show that the payo¤ from bidding t x1 is no greater than the payo¤ from this strategy. After writing out the payo¤ ++ from such a strategy and showing it is decreasing in t, we could show that the payo¤ at t = x1 is the same as at t = x2 , so it is weakly optimal to bid x2 . Part b. Consider the three possible orderings identi…ed in Problem 15.2.a, above: if X1 > X2 > Y1 > Y2 , the price sequence is Y2 < Y1 ; if X1 > Y1 > Y2 > X2 , the price sequence is X2 < Y2 ; and if X1 > Y1 > X2 > Y2 , the price sequence is Y2 < X1 . So the price sequence in this equilibrium is increasing.

59

16

Nonidential Objects

Problem 16.1 (Low revenue) Consider the problem of allocating a set of two objects in K = fa; bg to three buyers with values as follows: x1 x2 x2

a 0 10 10

b 0 10 10

ab 10 + " 10 10

where 0 " < 1: a. Find an e¢ cient allocation and the corresponding payments in the VCG mechanism. b. What is the total revenue accruing to the seller? Solution. Part a. It would be e¢ cient to give a to buyer 2 and b to buyer 3. With buyer 2, the other two buyers have total welfare of W 2 (x) = 10. Without buyer 2 (or equivalently if buyer 2 reported x2 = 0), it would be e¢ cient to give both items to buyer 1 for a total welfare of W 2 (0; x 2 ) = 10 + ". So 2’s payment in the VCG mechanism is W 2 (0; x 2 ) W 2 (x) = ". Similarly, buyer 3 would pay ". Part b. The seller collects 2" in revenue. Problem 16.2 (Complements) Consider the problem of allocating a set of four objects in K = fa1 ; a2 ; b1 ; b2 g to …ve buyers. Buyer 1 has use only for objects a1 and b1 ; buyers 2 and 3 have use only for objects a2 and b2 ; buyer 4 has use only for b1 and b2 ; and buyer 5 has use only for objects a1 and a2 . Speci…cally, the values attached by the buyers to these bundles are x1 (a1 b1 ) = 10 x2 (a2 b2 ) = 20 x3 (a2 b2 ) = 25 x4 (b1 b2 ) = 10 x5 (a1 a2 ) = 10 All other combinations (or packages) are valued at zero. a. Find an e¢ cient allocation and the corresponding payments in the VCG mechanism. Solution. It is e¢ cient to give a1 b1 to buyer 1 and a2 b2 to buyer 3. With buyer 1, the total welfare of the other buyers is W 1 (x) = 25. If buyer 1 reported a value vector of zeros, it would be e¢ cient to give a2 b2 to buyer 3 and a1 and b1 to no one, for total welfare of W 1 (0; x 1 ) = 25. Hence buyer 1 pays 0 in the VCG mechanism. With buyer 3 present, the other buyers have total welfare W 3 (x 3 ) = 10. If buyer 3 were not present, it would be e¢ cient to give a1 b1 to buyer 1 and a2 b2 to buyer 2 for total welfare W 3 (0; x 3 ) = 30. So buyer 3 pays 20. 60

Problem 16.3 Show that the conditions (16.6) and (16.7) are equivalent. Solution. (16:6) ) (16:7). Suppose (16.6) holds and take two sets S; T K. Because K is …nite, we can list T n S = a1 ; :::; an . De…ne T 0 T; S 0 T \ S and T m = T m 1 [ fam g and S m = S m 1 [ fam g for m = 1; :::; n. Applying (16.6) n times to S m T m 63 am+1 gives the following sequence of inequalities: x(S 1 )

X(S 0 )

x(T 1 )

x(T 0 )

x(S 2 )

X(S 1 )

x(T 2 )

x(T 1 )

::: x(S

n 1 n

)

x(S )

X(S

n 2

)

x(T n

X(S

n 1

)

n

x(T )

x(T

x(S 0 )

x(T n )

x(T 0 )

1

)

x(T n n 1

2

)

)

Adding these inequalities up gives x(S n )

And substituting S n = (T \ S) [ (T n S) = S, S 0 = T \ S, T n = T [ (T n S) = T [ S and T 0 = T gives x(S) x(T \ S) x(T [ S) x(T ) which can be rearranged to make (16.7). (16:7) ) (16:6). Suppose (16.7) holds and take a 2 K; a 2 = T; and S S 0 = S [ fAg. Applying (16.7) to the sets S 0 ; T gives x(S [ fag) + x(T )

x(T [ fag) + x(S)

which can be rearranged as (16.6).

61

T . De…ne

17

Packages and Positions

Problem 17.1 (Ine¢ ciency without package bidding) Suppose that there are two objects, a and b, for sale and two bidders with the following values x1 x2

a y 2

b z 2

ab 2 2

where y and z are parameters that lie between 0 and 1: Argue that an ascending auction format in which bidders can only bid on a and b individually, and not on the package ab; cannot allocate e¢ ciently. (Note: Without package bidding, the price of the package ab is necessarily the sum of the prices of the individual objects a and b.) Solution. Consider the ascending auction format described in Section 17.1 with increments ". Without loss of generality, let 0 < z y < 1 so that an e¢ cient allocation gives a to bidder 1 and b to bidder 2 (or also the reverse when y = z). Suppose that there is some ascending auction that leads to this outcome and payments (pa ; pb ). So that bidder 2 does not bid on a we need pb pa . So that bidder 1 does not attempt to win both goods, we need pa + pb 2. Together, these inequalities imply that pa = pb = 1. Moreover, y = 1 since 1’s payo¤ y pa must be nonnegative. This contradicts y < 1 so for any y and z strictly between zero and one, no ascending auction without package bidding can lead to an e¢ cient outcome. Problem 17.2 (Gross Substitutes) Show that if bidder i with value vector xi satis…es the gross substitutes condition (de…ned on page 241) then xi satis…es the substitutes condition (de…ned in (16.6)). Equivalently, show that the gross substitutes condition implies that xi (S) is submodular. Solution. This argument is taken from (1999).2 Take A 2 K, T 63 A and S T are too expensive. De…ne a price vector ( 0 pa = M

the proof of Lemma 5 in Gul and Stacchetti and let M > x(K) be a price at which goods

a 2 T [ fAg : a 62 T [ fAg

Recall that values are monotone: if A B then x(A) x(B). As a result, T [ fAg 2 D(p). For each " 0, de…ne a price vector 8 > a=A

: M a 62 T [ fAg 2

Gul, F. and E. Stacchetti (1999), “Walrasian Equilibrium with Gross Substitutes,” Journal of Economic Theory, 87, 95–124.

62

De…ne a critical price for good A by " = maxf" : T [ fAg 2 D(qT ("))g: By construction we have T [ fAg 2 D(qT (" )):

(20)

T 2 D(qT (" )):

(21)

Also for " > " we have qT (") p and T [ fAg 2 D(p) so by gross substitutes there must be a set C 2 D(qT (")) with T C. Goods outside T [ fAg are too expensive so we must also have C T [ fAg. However, by the de…nition of " we know that T [ fAg 62 D(qT (")). Hence T [ fAg 2 D(qT (")). This holds for all " > " , so by upper semi-continuity of D it also holds for " = " (see Gul and Stachetti for details):

Together (20) and (21) give x(T [ fAg) De…ne a price vector

8 > 2 > 3 : Each of the four bidders have “perclick” values of x1 > x2 > x3 > x4 : These values are commonly known and so we are considering a situation with complete information. a. Find an equilibrium of the sealed-bid generalized second-price auction (GSP). b. Is the equilibrium unique? If not, characterize all equilibria of the GSP. Solution. Part a. Consider the following strategies: b1 = x1 1

b2 = M (x)=

1

=

1

[(

1

2 2 )x

+(

[(

2

3 3 )x

+

2

1 3

2

b = M (x)=

2

=

1 2

3

b4 = M (x)=

3

=

1

4 3x

= x4

3

66

4 3x ]

3 3 )x

+

4 3x ]

Each bidder makes the same payment as in the VCG mechanism. Bidder 1 will not deviate to slot 2 for value v 1 (2) because v 1 (1) =

1 1x 1 2x

[(

1

[(

2

2 2 )x 3 3 )x

+(

3 3 )x

2

4 3x ]

+

4 3 x ]:

+

Similarly, none of the other bidders wish to win di¤erent slots at the associated prices. Part b. Here we follow Varian (2007) closely. Let I = f1; 2; 3; 4g be the set of bidders and K = f1; 2; 3; 4g the set of slots, with a rate of 4 = 0 clicks for the fourth slot. An outcome of the auction is a mapping : K ! I. An outcome is supported by some equilibrium bids if the following hold: 8k 2 K

k (x

(k)

b (k+1) )

8k; ` 2 K; ` < k

k (x

(k)

b (k+1) )

` (x

(k)

b (`+1) )

8k; ` 2 K; ` > k

k (x

(k)

b (k+1) )

` (x

(k)

b (`) )

b (1)

0

b (2)

b (3)

b (4)

b5

0

(28)

The …rst line is the individual rationality condition; the next two lines are incentive compatibility conditions; and the last must hold for good k to be assigned to bidder (k) by the auction. These inequalities de…ne the set of bids that can support an outcome . Some outcomes may only be supported for certain parameter values x; . With some abuse of notation, we will refer to such an equilibrium by the allocation alone. It is clear that the equilibrium is not unique. Suppose is an equilibrium outcome supported by bids b . Then bidder (1) can bid !b (1) for any ! > 1 and it is still an equilibrium. A subset of “symmetric”equilibria may be characterized as follows (see Varian, 2007): for all k; ` 2 K k (x

(k)

pk )

` (x

`)

k pk

(`)

p` )

(29)

or equivalently x (k) (

k

` p` ;

where pk bk+1 for k = 1; 2; 3; 4. To show that this condition characterizes a subset of equilibria, we must show that it implies the equilibrium conditions above. Claim. Condition (29) implies the conditions (28) and that is e¢ cient (that is, it is the identity mapping). Proof of claim. To show the …rst line of the equilibrium conditions, note that 4 = 0 so (k) (4) pk ) p4 ) = 0: 4 (x k (x The second line is a special case of (29). To show that is the identity mapping, it is su¢ cient to show that x (k) x (k+1) . For any k; ` the condition gives x (k) (

k

`)

k pk

` p`

x (`) (

`

k)

` p`

k pk :

67

Adding these gives (x (k)

x (`) )(

`)

k

0

so (k) (`) if and only if k `. So the slots are assigned e¢ ciently and we can write k instead of (k). To show that fourth line holds (decreasing prices), note that k k (x

k k 1 (x

pk )

pk

1)

can be rearranged for pk

1 k 1

pk

k

+ xk (

k 1

k)

pk

k

+ pk (

k 1

k)

= pk

k 1:

The second inequality uses the …rst line of the equilibrium conditions: xk Finally, to show that the third line holds: k k (x

k ` (x

pk )

k ` (x

p` )

p`

pk .

1 ):

This completes the proof. Second claim. If the condition (29) holds between k and k 1 and between k and k + 1 for all k, then it holds for all pairs. Sketch of proof. Suppose it holds between 1 and 3 and between 2 and 3. We will only show that it also holds between 2 and 3. Adding x1 (

2)

1

p1

p2

1

2

and x2 (

3)

2

p2

p3

2

3

) x1 (

3)

2

p2

2

p3

3

gives x1 (

3)

1

p1

p3

1

3

which is the condition for k = 1; ` = 3. For the other direction, add x3 (

2)

3

2

x (

p3

1)

2

3

p2

p2

2

x3 (

,

2

p1

1

p1

1

p3

3

)

x (

2

3)

p2

2

p3

3

1

2)

p1

1

p2

2

3

p1

1

to get x3 (

3)

1

3

, x3 (

3

1)

p3

which completes the sketch. Finally, we can use this claim to characterize prices in symmetric equilibria. The conditions between k and k + 1 are k k (x

xk (

k k+1 (x

pk )

k+1 )

k k

x (1

+ pk+1 k)

+

k+1 k+1 (x

pk+1 ) and k+1

pk

k pk+1

pk

k

68

k+1 pk ) k (x xk+1 ( k k+1 ) + pk+1 k+1 k+1 x (1 k ) + k pk+1

pk+1 )

where each of the lines above is equivalent and k k+1 = equilibrium bids is given by the following inequalities:

x2 (1 1

x (1

3)

+ b4

2)

3

+b

x3

b4

x4

3

b3

x3 (1

2

2

b

2

b

x (1

1

1

x (1

k.

So, the set of symmetric

3)

+ b4

2)

+b

3

+b

2

1)

3 2 1

This completes the characterization of symmetric equilibria of the GSP auction. Random-outcome equilibria. There are also equilibria with random outcomes, either because some winning bids are tied or because players are randomizing. Suppose each bidder i bids independently according to a random variable B i with no mass points. Then i’s payo¤ to bidding b is v i (b; B i ) =

3 X k=1

P rfb

Yk g

i k [x

E(Yk j b > Yk )]:

where Yk is the k th highest order statistic of the other bidders’bids. These strategies form a Nash equilibrium if, for all i 2 I, all b; b0 2 suppB i and all b00 62 suppB i we have v i (b; B i ) = v i (b0 ; B i ) and v i (b; B i )

69

v i (b00 ; B i ):