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Tutorat 6

These notes attempt to provide a discussion of the concepts and to explicate the exercises covered in the Micro 2 tutorial on Friday, 11 February 2011.

SHW (11 f´ev 2011)

Contents

1 4.5.2 Common Values 2

1.1 Solution . . . 2

2 4.6 Asymmetric Auction 1 3

2.1 Solution . . . 3

3 4.7 6

4 4.8 Matching Auction 6

4.1 Solution . . . 7

5 4.9 7

6 4.10 7

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1 4.5.2 Common Values

Remark: In a common-value auction, bidders may not know the exact value of the object being auctioned. Bidders may then use the valuations of the other bidders to update their own valuation.

Suppose that the two bidders have the same value for the object. As in Exercise 4.5.1, before the auction starts, each player i observes a random variable ti, where the variables t1 and t2 are independent and identically distributed according to the uniform distribution on [0, 1]. The difference is that now the value of the object is V = t1+ t2. (Each bidders i observes his “type” ti, but does not observe the type of the other bidder.) The auction then happens as in 4.5.1.

(a) Determine the conditional expected value given to the object by Buyer i given his type ti. Compare this answer to the answer you obtained in part (a) of 4.5.1.

(b) Show that there exists a Bayesian equilibrium in which Buyer i’s bid, bi, is of the form

1.1 Solution

πi = Prob(ˆti ≥ tj)(ti+ tj) − T (ˆti).

By the Revelation Principle, ti ∈ arg maxˆt

iπi. According to the envelope theorem,

∂R

∂ti

= ∂πi(ti, ˆti)

∂ti

ˆ

ti=ti

= ti

since (?) Prob ˆti ≥ tj = F (ˆti) = ˆti. Thus the information rent is given by R = R(0) +

Z ti

0

s ds = 1 2t2i. In equilibrium, R = maxˆt

iπi(ˆti, ti) = ti(ti + tj) − T (ti) = 12t2i. The optimal bidding strategy is

bi= T (ti) Prob(ti ≥ tj) =

1

2t2i + titj

ti = 1

2ti+ tj. The following is NOT correct:

E[bi] = Z 1

0

s ds + 1 2ti= 1

2 +1 2ti.

It is not correct because of the winner’s curse: if you win the object, you know your type is bigger than the other bidder’s type. (In particular, if Bidder i wins the object, then the value of the object is ti+ tj ≤ ti+ ti = 2ti.) Thus we should not integrate from 0 to 1, but instead from 0 to ti (remembering to renormalize the distribution, since if Bidder i wins the object, then tj = s is known to lie in [0, ti] and not [0, 1]):

E[bi] = Z ti

0

s ti

ds +1 2ti= 1

2ti+1

2ti = ti.

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2 4.6 Asymmetric Auction 1

A seller wishes to sell an object to one of two buyers who have independent values for the object. Buyer 1’s valuation is drawn from a uniform distribution over [0, 1], whereas Buyer 2’s valuation is drawn from a uniform distribution over [1, 2]. The seller has no opportunity cost and would therefore gain from selling the object at any positive price.

We consider the following mechanism: First, both bidders announce a value, ˆ

v1 and ˆv2 respectively. Then, each bidder gets the object with probability pi(ˆvi, ˆvj) ≥ 0 (where the probabilities satisfy p1(ˆv1, ˆv2) + p2(ˆv2, ˆv1) = 1) and pays ti(ˆvi, ˆvj), whether he gets the object or not.

Buyer i thus gets an expected utility Eˆvj[pi(ˆvi, ˆvj)vi− ti(ˆvi, ˆvj)]. We further restrict attention to mechanisms where each buyer i is induced to announce his true valuation vi.

(a) Is there a loss of generality in restricting attention to such a mechanism?

Assuming that Buyer j always tells the truth, write Buyer i’s expected utility when she has a true value vi and announces ˆvi. Show that this expected utility depends only on vi, Pi(ˆvi) = Eˆvj[pi(ˆvi, ˆvj)], and Ti(ˆvi) = Eˆvj[ti(ˆvi, ˆvj)].

(b) Write Buyer i’s truth-telling incentive-compatibility condition. Show that, necessarily, Buyer i’s expected utility, Ui(vi), is non-decreasing in vi. (c) Write Buyer i’s participation constraint, assuming that buyers’ reservation

levels of utility are zero. For which value of vi will this constraint be binding? Determine the relation linking Buyer i’s expected utility to Pi(·).

(d) Assuming that the seller wants to implement given probabilities p1(v1, v2) and p2(v1, v2), write the seller’s expected revenue (1) as a function of P1(·) and P2(·), and (2) as a function of p1(·, ·) and p2(·, ·).

(e) What are the probabilities p1(v1, v2) and p2(v1, v2) that maximize the seller’s expected revenue? Is the trade resulting from these probabilities efficient?

(f ) Show that those optimal probabilities are implementable through a biased first-price auction (where Buyer 2 wins only if she outbids Buyer 1 by an amount ∆) with asymmetric entry fees f1 and f2. In this auction, what are the equilibrium bidding strategies? Are these optimal probabilities implementable through an asymmetric second-price auction?

2.1 Solution

Tradeoff between extracting some information rent from Bidder 2 and giving the good to the low-valuation bidder, Bidder 1. Auction designer gives some probability of winning to the low-valuation bidder in order to induce the high- valuation bidder to bid above 1.

(a) No loss of generality.

(b) The expected gain for Bidder i is given by πi(vi) = Pi(vi)vi− Ti(vi)

(4)

The incentive-compatibility condition is simply that each bidder tell the truth:

vi∈ arg max

ˆ vi

Pi(ˆvi)vi− Ti(ˆvi).

Given a true valuation vi,

Pi(vi)vi− Ti(vi) ≥ Pi(ˆvi)vi− Ti(ˆvi) Similarly, given a true valuation ˆvi,

Pi(ˆvi)ˆvi− Ti(ˆvi) ≥ Pi(vi)ˆvi− Ti(vi) Adding these two equations

Pi(vi)(vi− ˆvi) ≥ Pi(ˆvi)(vi− ˆvi) (Pi(vi) − Pi(ˆvi))(vi− ˆvi) ≥ 0

Thus vi ≥ ˆvi ⇒ Pi(vi) ≥ Pi(ˆvi); that is, P is non-decreasing in vi. Ui(vi) = Pi(vi)vi− Ti(vi) ≥ Pi(ˆvi)vi− Ti(ˆvi) By assumption, vi≥ ˆvi,

Pi(ˆvi)vi− Ti(ˆvi) ≥ Pi(ˆvi)ˆvi− Ti(ˆvi) = Ui(ˆvi) Hence vi≥ ˆvi implies that

Ui(vi) ≥ Ui(ˆvi)

which is simply the statement that Ui(·) is non-decreasing in vi. Ui(vi) ≥ 0 implies Ui(v) = 0; that is, give the least information rent to the lowest-type buyer.

(c) Ui(ˆvi; vi) = Pi(ˆvi) − T (ˆvi). The information rent is R = maxˆviUi(ˆvi, vi).

From the envelope theorem,

∂R

∂vi

= ∂Ui(ˆvi; vi)

∂vi

vˆi=vi

= Pi(vi) The seller’s profit is then

πS = Z 1

0



P1(v1)v1− Z v1

0

P1(s) ds

 1 dv1+

Z 2 1



P2(v2)v2− Z v2

1

P2(s2) ds2

 1 dv2

where ??? follows from v1 ∼ U [0, 1] so f (v1) = 1 and v2 ∼ U [1, 2] so f (v2) = 1.

Look at each piece:

Z 1 0

Z v1

0

P1(s1) ds1



dv1 = v1 Z v1

0

P (s1) ds1

1 0

− Z 1

0

v1P1(v1) dv1

= Z 1

0

(1 − v1)P1(v1) dv1 Next

Z 2

1

Z v2

1

P2(s2) ds2



dv2 = (v2− 1) Z v2

1

P2(s2) ds2

2 1

Z 2

1

(v2− 1)p2(v2) dv2

= Z 2

1

(2 − v2)P2(v2) dv2

(5)

Substituting these integration by parts results into the profit function of the sellers, we get

πS= Z 1

0

(2v1− 1)p1(v1) dv1+ Z 2

1

(2v2− 2)p2(v2) dv2

The seller can only control the form of the probability functions p1(·) and p2(·).

If 2v1− 1 ≤ 0, i.e. if v112, then a profit-maximizing seller will set P (v1) = 0.

If 2v1 − 1 ≥ 2v2 − 2, then the seller should give the object to Buyer 1. If instead 2v2− 1 ≤ 2v2− 2, then give the object to Buyer 2. These two situations correspond to v1 ≥ v212 and v2 ≥ v1+12, respectively. Hence,

P1(v1) =

0 if v112 v1−1

2 if 12 ≤ v1 ≤ 1

From the seller’s perspective, give the object to Buyer 1 when v1 + 12 ≥ v2, Prob(v1+12 ≥ v2) = F2(v1+12) = F2∼ U [1, 2] = v1+12 − 1 = v112.

p2(v2) =

F1(v2−1

2) = v2−1

2 if 1 ≤ v232

1 if v232

(e) They are

p1(v1, v2) =

(1 if v1≥ v212 0 if v1< v212 and

p2(v1, v2) =

(1 if v2≥ v1+ 12 0 if v2≤ v1+ 12

Trade is not efficient, since the low-valuation bidder wins the good with non-zero probability.

(f) The information rent of Bidder 1 is R1 =

Z v1 1 2

(s − 1

2) ds = 1 2s2−1

2s

v1 1 2

= 1 2v12− 1

2v1+1 8 Also

R1 = max U1 = P1(v1)(v1− b1(v1)) Hence

v1− b1(v1) = 1 2

(v112)2 v112 Solving, we find

b1(v1) = 1 2v1+ 1

4

We can use the information rent trick only for Bidder 1 — not Bidder 2! Be- cause the probability distribution of the two bidders is not the same! Bidder 1’s probability of winning is

[0,1 2]

| {z }

Prob=0

and [1 2, 1]

| {z }

Prob=v112

(6)

The bidders with valuation in [0,12] are out of the game, and we need not leave them any information rent. This is not the case with Bidder 2:

[1,3 2]

| {z }

However, the Revenue Equivalence Theorem still holds. b2(v2) = av2 + k2. Introduce a bias (note that this makes the bidding asymmetric): If b2(v2) − ∆ ≥ b1(v1), then give the object to Bidder 2. In this asymmetric auction, Buyer 2 seeks to maximize

max(v2− b2) Prob(b1(v1) + ∆ ≤ b2(v2) b1(v1) + ∆ ≤ b2(v2)

1

2v1+14 + ∆ ≤ b2

v1b2

1 4−∆

1 2

Prob = F

b

21

4−∆

1 2

 v2− 2b2+ ∆ + 14 = 0 b2 = 12v2+2 +18 b2 = av2+ k2

Hence a = 1/2 and k2 = 2 +18 v212 ≥ v1

b2− ∆ ≥ b1

These last two conditions must have the same probability. Thus 1

2v2+1 8 −∆

2 ≥ 1 2v1+ 1

4 which yields the solution ∆ = 14 and k2 = 14 = 12v1+ 14.

Let’s see how to solve this using the traditional method.

b1 = a1v1+ k1, b2 = a2v2+ k2.

max(v1− b1) Prob(a2v2+ k2≤ b1+ ∆) maxb1(v1− b1)

b1+∆−k2

a2 − 1 You should obtain the result b1 = 12v1+a2+k22−∆ = a1v1+ k1

which yields the result that a1= 12. Similarly for Bidder 2 max(v2− b2) Prob(b2− ∆ ≥ b1)

maxb2(v2− b2)b2−∆−ka 1

1

Thus we have three equations in three unknowns:

1

2v2∆+k2

2 = a2v2+ k2 1

2v1a2+k22−∆ = a1v1+ k1 b1+ ∆ ≥ b2⇔ v1+12 ≥ v2

You should obtain the results ∆ = 14 and k1 = k2 = 14.

3 4.7

4 4.8 Matching Auction

Suppose there are only two bidders. The seller (with no private value) uses the following auction rule.

(7)

Round 1: There is a single round of bidding. Bidder 1 is given the oppor- tunity to quote a price b1 > v, where v is the reservation price set by the seller.

Round 2:

4.1 Solution

(a) The optimal strategy for Bidder 2 is simple: If b1 ≤ v2, then bid b2 = b1. (Recall that in the case of a tie, Bidder 2 wins the good.

(b)

max(v1− b1) Prob(b1 ≥ v2) giving the solution b1= v21, v1∼ [0, 1] so max b1 = 12.

b1 =

 1

2 if v112 0 if v112.

(c) The matching auction does not always lead to an efficient allocation. For example, consider v1 = 1 and v2 = 0.6. Bidder 1 would bid b1 = 1/2, and then Bidder 2 would submit a matching bid b2 = b1 = 1/2 and win the object, even though v2 = 0.6 < 1 = v1.

We can restore efficiency by making the matching auction asymmetric. For example, to win the object Bidder 2 must bid at least two times as high as Bidder 1.

Let us carry the analysis of this auction a bit further. Denote by πS1 the seller’s expected revenue in a standard first-price auction, πS2 a standard auction with reservation price, πS3 is a matching auction. The standard auction with reservation price is an optimal auction, therefore it is the best. We have the following ordering:

πS1 ≤ πS3 ≤ πS2 8

24 ≤ 9 24 ≤ 10

24

5 4.9

6 4.10

References

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