1 4.1. First- and Second-Price Auctions 2

Tutorat 5

These notes attempt to provide a discussion of the concepts and to explicate the exercises covered in the Micro 2 tutorial on Friday, 4 February 2011. The first two exercises and half of the third exercise are covered in detail. The final two exercises have yet to be revised; please continue to check the website for updated versions. Please contact [email protected] with any corrections or clarifications to these notes.

SHW (4 f´ ev 2011)

Contents

1 4.1. First- and Second-Price Auctions 2

1.1 Solution to Part (a) . . . . 2 1.2 Solution to Part (b) . . . . 3 1.3 Solving the Differential Equation . . . . 5

2 4.2 All-Pay Auctions 6

2.1 Solution . . . . 6

3 4.3 Optimal Auctions 8

3.1 Solution . . . . 8

4 4.4 The Revenue Equivalence Theorem 9

4.1 Solution . . . . 9

5 4.5.1 Private Values 10

5.1 Solution . . . . 10

This tutorial: Exercises 1-5. Next tutorial: Exercises 6-10.

Two methods to solve auctions problems.

1. “Traditional” method. Compute the probability P _i (v) that Bidder i wins the object when his valuation is v (and he bids b i (v)); then compute Bidder i’s expected payment T _i (v), his optimal bidding strategy b _i (v), etc.

2. Use the revelation principle.

1 4.1. First- and Second-Price Auctions

A seller has one good to sell to one of n possible buyers. Each potential buyer has a valuation v _i for the good; this valuation is known only by each buyer. Buyers’

valuations are independently drawn from a uniform distribution on [0, 1]; this is common knowledge.

(a) Consider the second-price auction: Each buyer submits a bid, and the object is allocated to the bidder that submits the highest bid, who then must pay the second-highest bid that was submitted. Show that in this auction, each buyer has a dominant strategy. Derive the expected revenue of the seller who uses the second-price auction to sell the object.

(b) Now consider the first-price auction: Each buyer submits a bid, and the object is allocated to the bidder with the highest bid, who then must pay the bid he submitted. Look for the optimal bidding strategy in the unique symmetric Bayesian-Nash equilibrium of this auction. Derive the expected revenue of a seller who uses this mechanism to sell the object. Compare this revenue with the revenue of the second-price auction derived in part (a).

1.1 Solution to Part (a)

In a second-price auction, bidding one’s true valuation is a (weakly) dominant strategy. We prove this by showing that there exists no profitable deviation from truthful bidding.

Overbidding, i.e. b _i (v _i ) > v _i , is weakly dominated by the truthful bidding strategy. If v _i ≥ max _j6=i b _j , or max _j6=i b _j ≥ b _i , the outcome is the same. If v i < max j6=i b j < b i , then Bidder i wins the object and must pay max j6=i b j > v i , an outcome that yields Bidder i negative utility.

Underbidding, i.e. b _i (v _i ) < v _i , is also weakly dominated by truthful bidding.

If max j6=i b j ≥ v _i or max j6=i b j ≤ b _i , then the outcome from underbidding is the same as the outcome of truthful bidding. If v i > max _j6=i b j > b i , Bidder i loses the auction and forfeits the positive payoff, instead getting a payoff of zero.

(Had Bidder i bid truthfully, he would have won the object and enjoyed positive surplus v i − max _j6=i b j .)

Thus, given that all other bidders are bidding truthfully, there is no strictly profitable deviation for any bidder i; the strategy profile in which all bidders bid truthfully is an equilibrium. This result is robust — it applies to bidders of all risk appetites.

In this truth-telling equilibrium, Bidder i’s expected payoff equals

Prob(v i ≥ v _j ) ⁿ⁻¹ (v i − E[v (2) | v ₍₂₎ ≤ v _i ]).

Under the assumption that v j ∼ U [0, 1], Prob(v _i ≥ v _j ) = v i . Prob(v ₍₂₎ ≤ ˆ v | v ₍₂₎ ≤ v _i ) = ˆ v ⁿ⁻¹

v ⁿ⁻¹ f (ˆ v | v ₍₂₎ ≤ v _i )

= (n − 1) v ˆ ⁿ⁻² v ⁿ⁻¹ .

The expected amount that the winner of the second-price auction will pay is also the expected revenue of the seller:

E[v (2) | v ₍₂₎ ≤ v _i ] = Z v

0

ˆ

v(n − 1) v ˆ ⁿ⁻² v ⁿ⁻¹ dˆ v

= n − 1 n v _i . 1.2 Solution to Part (b)

Denote Bidder i’s valuation by v _i and his bidding strategy by b _i (v _i ). Consider first a first-price auction. Bidder i’s expected payoff is



 Y

j6=i

Prob 

b _i (v _i ) ≥ b _j (v _j ) 



 (v _i − b _i (v _i )).

Assume a symmetric bidding strategy that is monotonic (i.e. assume all bidders use the same bidding strategy, b _i (·) = b(·) for all i, and this strategy is increasing with respect to the bidder’s valuation, b ⁰ (·) > 0). Since bidders’ valuations are independently drawn from a common distribution, the assumption of symmetric bidding strategies implies that Prob(b _i (v _i ) ≥ b _j (v _j )) is the same for all j 6= i; thus the product of probabilities in the above expression reduces to [Prob(b _i (v _i ) ≥ b j (v j ))] ⁿ⁻¹ . Bidder i’s utility maximization problem can thus be written as

max b

h Prob 

b(v _i ) ≥ b(v _j ) i n−1

(v _i − b(v _i )).

If the bidding strategy b(·) is strictly monotonic, then its inverse b ⁻¹ (·) exists;

and under the assumption that b(·) is strictly increasing, if Bidder i’s bid weakly exceeds Bidder j’s bid, then Bidder i’s valuation must weakly exceed Bidder j’s valuation. Hence we have

Prob(b(v _i ) ≥ b(v _j )) = Prob(b ⁻¹ (b(v _i )) ≥ v _j ).

From Bidder i’s point of view, v _i is known, and hence b ⁻¹ (b(v _i )) is a constant.

Since v _j ∼ U [0, 1], the probability that the random variable v _j is less than this constant b ⁻¹ (b(v i )) is

Prob(b ⁻¹ (b(v _i )) ≥ v _j ) = b ⁻¹ (b(v _i )).

Therefore we can rewrite Bidder i’s utility maximization problem as max b

 

b ⁻¹ (b(v _i ))  n−1

(v _i − b(v _i ))



.

The first-order condition ¹ of this problem is 0 = set (n − 1)



b ⁻¹ (b(v i ))

 n−2 1

b ⁰ (b ⁻¹ (b(v i ))) (v i − b(v _i )) − (b ⁻¹ (b(v i ))) ⁿ⁻¹ . Applying the relation b ⁻¹ (b(v _i )) = v _i and writing b in place of b(v _i ) for simplicity, we can rewrite this expression as

0 = (n − 1)v ⁿ⁻² _i 1

b ⁰ (v _i − b) − v _i ⁿ⁻¹ v i = (n − 1) 1

b ⁰ (v i − b) v _i b ⁰ = (n − 1)v _i − (n − 1)b b ⁰ + n − 1

v _i b = n − 1. (1)

This is a first-order differential equation; its solution (see the section Solving the Differential Equation below) is

b _i (v _i ) = e ⁻

R

vi

dv

Z

(n − 1)e ⁻

R

vi

dx

dv _i + c



= v _i ⁻⁽ⁿ⁻¹⁾



(n − 1) 1 n v _i ⁿ + c



= n − 1

n v i + cv _i ⁻⁽ⁿ⁻¹⁾ .

Applying the boundary condition b(0) = 0 (which simply says that a bidder with a valuation of zero bids zero for the object) yields c = 0, so we obtain that the optimal bidding strategy is

b _i (v _i ) = n − 1 n v _i .

In a first-price auction, the winner pays the amount he bid, so this expression also gives the seller’s expected revenue under a first-price auction. This is the same result we found in part (a) above: first- and second-price auctions yield the same expected revenue to the seller. Notice that this result follows immediately from the Revenue Equivalence Theorem: both the first- and second-price auctions are efficient, and they give zero payoff to the lowest-valuation bidders (because they win the auction with probability zero). Hence by the Revenue Equivalence Theorem, both auctions yield the same expected revenue to the seller.

1

In deriving the first-order condition, we make use of the fact that

f

(x) =

. To obtain this result, denote g(x) = f

(x). Then f (g(x)) = x. Differentiating both sides of this identity yields

d

dx f (g(x)) = d dx x f

(g(x))g

(x) = 1

g

(x) = 1 f

(g(x)) ,

so long as f

(g(x)) 6= 0. Writing f

(x) in place of g(x) yields the desired result.

In our current problem, take f

(x) = b

(b(v

)) and x = b(v

). Differentiating b

(b(v

)) then yields

d db

“

b

(b(v

))

”

= 1

b

(b

(b(v

))) .

Note that under the assumption that b(·) is strictly monotonic, b

(b

(·)) is never zero.

1.3 Solving the Differential Equation

Let us return to the differential equation (1) we encountered in the exercise:

b ⁰ + n − 1 v i

b = n − 1.

Recall that b(·) is a function of v _i . This is a non-homogenous, first-order differen- tial equation, which we can solve as follows. Any solution b(v i ) to this differential equation can be written as the sum of a particular solution (i.e. a function b that solves (1)), which we will denote b _p , and a solution to the homogenous counterpart to (1), namely

b ⁰ + n − 1

v _i b = 0, (2)

whose solution we will denote b _h . That is, any solution b can be written b = b _p +b _h (all functions of v _i ).

Solving the homogenous differential equation (2) is relatively easy, so let us start with that. We can rearrange the homogenous equation as

b ⁰

b = −(n − 1)v _i ⁻¹ . Notice that the left side of this equation is _dv ^d

i

(ln b). Making this replacement and then integrating both sides of the equation, we have

Z d dv i

(ln b) dv i = −(n − 1) Z

v _i ⁻¹ dv i

ln b = −(n − 1) ln v i + c 1 .

We can use the property of logarithms to make the −(n − 1) term into an exponent,

ln b = ln v _i ⁻⁽ⁿ⁻¹⁾ + c ₁ , then exponentiate both sides to obtain

b = e ^{ln v}

^+c

= e ^{ln v}

e ^c

= cv ⁻⁽ⁿ⁻¹⁾ _i .

This is our homogenous solution: b _h (v _i ) = cv ⁻⁽ⁿ⁻¹⁾ _i .

To compute a particular solution b p , we return to the non-homogenous dif- ferential equation (1):

b ⁰ + n − 1

v _i b = n − 1.

We guess a solution of the form b(v _i ) = kv _i , where k is a constant whose value we will have to compute. Plugging this proposed solution into the differential equation (1) and solving for k, we obtain

k + (n − 1)k = n − 1

k = n − 1

n .

Thus a particular solution to the non-homogenous equation is b p (v i ) = ⁿ⁻¹ _n v i . Combining this result with the solution b _h to the homogenous equation gives us the general solution to the differential equation (1):

b(v i ) = n − 1

n v i + cv ⁻⁽ⁿ⁻¹⁾ _i .

2 4.2 All-Pay Auctions

Consider a sealed-bid, all-pay auction with n ex-ante identical buyers. Each buyer’s valuation is independently drawn from [0, 1] according to the uniform distribution. The rules of the auction are as follows: Each buyer submits a bid, the highest bidder receives the good, and every bidder pays his bid regardless of whether he wins. We study the Bayesian equilibrium of the game generated by this auction.

(a) Denote Buyer i’s bidding strategy by b i (v i ). Write the equilibrium condi- tions for a Bayesian equilibrium

(b) Suppose all other buyers adopt an exponential bidding strategy of the form b _j (v _j ) = αv ^β _j , for all j 6= i, where α and β are two parameters. What is Buyer i’s best response?

(c) Show that there exists an equilibrium in which all buyers adopt a bid- ding strategy of the type described above, and compute the equilibrium strategies.

(d) Is the equilibrium in part (c) efficient? What is the seller’s expected rev- enue in this equilibrium?

(e) Compare the buyers’ strategies and the seller’s expected revenue in this equilibrium with the efficient equilibria of the first-price sealed-bid auction.

Discuss.

2.1 Solution

Note: The discussion in Friday’s tutorial addressed the optimal bidding strategy in an all-pay and first-price auction, and did not follow parts (a) through (e) of the exercise.

Assume a symmetric equilibrium in which bidders’ strategies are monoton- ically increasing in valuation. Bidder i seeks to maximize his expected utility, given by

max

 

Prob(b(v i ) ≥ b(v j ))

 n−1

v i − b _i (v i )

 .

Under the assumption that the bidding strategy is monotonically increasing, b(v _i ) ≥ b(v _j ) implies that b ⁻¹ (b(v _i )) ≥ v _j ; we can therefore write the utility- maximization problem as

max

 

Prob(b ⁻¹ (b(v _i )) ≥ v _j )  n−1

v _i − b _i (v _i )

 .

From Bidder i’s point of view, v _i is given, so b ⁻¹ (b(v _i )) is just a constant. Since

v j ∼ U [0, 1], the probability that the random variable v _j is weakly less than

some constant c ∈ [0, 1] is simply c. Hence we can further simplify the utility- maximization problem to

max

 

b ⁻¹ (b(v i ))

 n−1

v i − b _i (v i )

 . The first-order condition is

(n − 1) 1

b ⁰ (b ⁻¹ (b(v _i ))) (b ⁻¹ (b(v i ))) ⁿ⁻² v i − 1 =

set 0,

which we can simplify using the relation b ⁻¹ (b(v _i )) = v _i and the notation b = b(v i ) to obtain

(n − 1) 1

b ⁰ v ⁿ⁻² _i v i − 1 = 0.

Isolating b ⁰ reveals a particularly simple first-order differential equation, b ⁰ = (n − 1)v _i ⁿ⁻¹ ,

whose solution is

b = n − 1 n v ⁿ _i .

Let’s solve the same problem via the Revelation Principle. Bidder i’s ex- pected surplus π _b , given his true valuation v _i and his bid ˆ v _i , is

π _b = Prob(ˆ v _i ≥ v _j ) ⁿ⁻¹ v _i − T (ˆ v _i ).

Bidder i chooses his bid ˆ v i in order to maximize this expected surplus. The information rent R is simply this maximum expected surplus (???correct???):

R(v _i ) = max

ˆ v

π _b (ˆ v _i , v _i ) Using the envelope theorem,

∂R

∂v i

= ∂π _b

∂v i

   v ˆ

=v

= [Prob(v i ≥ v _j )] ⁿ⁻¹ = v _i ⁿ⁻¹ , (3) where in the last step we have used the fact that v j ∼ U [0, 1] to write Prob(v _i ≥ v _j ) = v _i . Assume that the lowest type gets zero rent: R(0) = 0. Then integration of (3) yields

R(v i ) = R(0) + Z v

0

s ⁿ⁻¹ ds = v _i ⁿ n .

In a first-price auction, the winning bidder pays what he bid. The probability that Bidder i wins the auction is Prob(v j ≤ v _i ) ⁿ⁻¹ = v ⁿ⁻¹ _i , and when he wins he enjoys a surplus of v _i − b _i (v _i ); thus his expected gain is v _i ⁿ⁻¹ (v _i − b _i (v _i )). Setting this equal to the information rent R(v _i ), we have

v _i ⁿ⁻¹ (v _i − b _i (v _i )) = v _i ⁿ

n ⇒ b _i (v _i ) = n − 1 n v _i .

In an all-pay auction, bidders pay their bid whether they win or not. Thus Bidder i’s expected gain is v _i ⁿ⁻¹ v _i − b _i (v _i ); setting this equal to the information rent and solving for b i (v i ), we find

v _i ⁿ⁻¹ v _i − b _i (v _i ) = v _i ⁿ

n ⇒ b _i (v _i ) = n − 1

n v ⁿ _i .

3 4.3 Optimal Auctions

A single seller with a private value v ₀ for the good faces n potential risk-neutral buyers, where Buyer i has reservation value v i . The reservation values are in- dependently and identically distributed, drawn from a common differentiable distribution F (·) on [v, ¯ v], with F (v) = 0 and F (¯ v) = 1.

(a) For a given reserve price v ^∗ , compute the total expected return to the seller.

(b) Compute the optimal reserve price. Compare this reserve price with the private value of the seller.

3.1 Solution

We use the Revelation Principle approach. Denote Bidder i’s true valuation by v _i and his bid by ˆ v _i . Bidder i’s expected gain equals the probability that he wins the object, (F (ˆ v i )) ⁿ⁻¹ , times his true valuation v i , minus his expected payment T (ˆ v _i ):

π _b = F ⁿ⁻¹ (ˆ v _i )v _i − T (ˆ v _i ).

Using the Envelope Theorem to find the information rent R(v _i ), we compute

∂R

∂v i

= ∂π b

∂v i

   ˆ v

=v

= F ⁿ⁻¹ (v i ).

The value of the information rent R(v _i ) is found by integrating this expression from the reserve price v ^∗ to Bidder i’s true valuation v i :

R(v _i ) = R(v ^∗ ) + Z v

v

F ⁿ⁻¹ (s) ds

The auction designer can use the reserve price to deter low-valuation bidders from participating in the auction. By setting R(v ^∗ ) equal to zero, the auction designer makes bidders with true valuation v ^∗ indifferent between participating in the auction and not participating; and if R(·) is non-decreasing, bidders with valuations below v ^∗ are deterred from participating as well. The auction de- signer’s goal is to find the reserve price v ^∗ that maximizes the seller’s expected surplus. ² That is, the auction designer seeks to solve

max X

i

Z ¯ v v



v _i P _i (v _i ) Z v

v

F ⁿ⁻¹ (s) ds − v ₀



f (v _i ) dv _i .

Performing an integration by parts, we have Z _{v ¯}

v

Z _v

v

F ⁿ⁻¹ (s) ds dF (v _i ) = F (v _i ) Z _v

v

F ⁿ⁻¹ (s) ds   

¯ v v

−

Z _{¯ v}

v

F ⁿ (v _i ) dv _i This last integral (? or whole expression ?) is

= Z _{¯ v}

v

1 − F (v _i )

f (v _i ) F ⁿ⁻¹ (v _i )f (v _i ) dv _i

2

Notice that in this problem, because the seller has a private value v

that is a priori non- zero, the seller’s revenue is not the same as the seller’s surplus. This is why we subtract v

from the expression.

Hence our maximization problem (i.e. maximizing the seller’s revenue) becomes max p

X

i

Z v ¯ v



F ⁿ⁻¹ (v _i )(v _i − v ₀ ) − 1 − F (v _i )

f (v i ) F ⁿ⁻¹ (v _i )



f (v _i ) dv _i

Factoring the common F ⁿ⁻¹ (v _i ) term out of the parentheses (since it depends on v i , we must keep it inside the integral), we have

X

i

Z ¯ v v



v _i − v ₀ − 1 − F (v _i ) f (v _i )



F ⁿ⁻¹ (v _i )f (v _i ) dv _i

The optimal reserve price v ^∗ solves

v ^∗ − v ₀ − 1 − F (v ^∗ ) f (v ^∗ ) = 0.

Rearranging this expression slightly, we observe that v ^∗ = v ₀ + 1 − F (v ^∗ )

f (v ^∗ ) ;

since F (·) ≤ 1 and f (·) ≥ 0, the second term on the right is always non-negative, and we conclude that v ^∗ ≥ v ₀ . That is, the optimal reserve price is (weakly) greater than the seller’s private value, suggesting that this auction will be inef- ficient.

4 4.4 The Revenue Equivalence Theorem

(The following exercise is taken from Mas-Colell, Whinston, & Green, Exercise 23.D.6.) Suppose that I symmetric individuals wish to acquire the single re- maining ticket to a concert. The ticket office opens at 9 a.m. on Monday. Each individual must decide what time to go to get in line: The first individual in line will get the ticket. An individual who waits t hours incurs a (monetary equiva- lent) disutility βt. Suppose also that an individual showing up after the first one can go home immediately and thus incurs no waiting cost. Individual i’s value of receiving the ticket is θ ⁱ , and each individual’s θ ⁱ is independently drawn from a uniform distribution on [0, 1]. What is the expected value of the number of hours that the first individual in line will wait? (Hint: Note the analogy to a first-price sealed-bid auction, and use the Revenue Equivalence Theorem.) How does this vary when β doubles? When I doubles?

4.1 Solution

We again use the revelation principle to solve this problem.

π _b = (θ − βt(ˆ θ))[Prob(ˆ θ _i ≥ θ _j )] ⁿ⁻¹ The information rent is

R = max

θ ˆ

π _b (ˆ θ, θ) By the envelope theorem,

∂R

∂θ = [Prob(θ i ≥ θ _j )] ⁿ⁻¹ = F ^I−1 (θ)

Since F (·) is uniform, F ^I−1 (θ) = θ ^I−1 . Hence the information rent is R =

Z θ 0

s ^I−1 ds = θ ^I

I = θ ^I−1 (θ − βt(θ)) Solving for t(θ), we get

t(θ) = I − 1 βI θ

The “seller’s expected revenue” (which is meaningless here — we compute it only because the seller’s expected revenue equals the buyer’s expected waiting cost) is

E[π S ] = I Z 1

0

 I − 1 βI θ



θ ^I−1 dθ

= I I − 1 βI

θ ^I+1 I + 1

1

0

= I − 1 β(I + 1)

5 4.5.1 Private Values

Consider an auction between two buyers, denoted 1 and 2, that give private and independent values to the object. Before the auction starts, the potential buyer i observes a random variable t _i ; the variables t ₁ , t ₂ are independent and identically distributed according to a uniform distribution on [0, 1]. The vale that the potential bidder gives to the object is v i = t i + ¹ ₂ .

(a) Consider a first-price sealed-bid auction. Each buyer i proposes a price b _i . The buyer that proposes the higher price gets the object; in case of equality, the winner is determined randomly (both buyers have an equal probability of winning).

5.1 Solution

max Prob(ˆ v _i ≥ v _j )(v _i − T (ˆ t _i ))

ˆ

v i = ˆ t i + 1

2 , Prob(ˆ v i ≥ v _j ) = Prob(ˆ t i ≥ t _j )

π b = Prob(ˆ t i ≥ t _j )(t i + 1

2 − T (ˆ t i ))

= ˆ t i

 t i + 1

2 − T (ˆ t i )



Use the envelope theorem to calculate the information rent

∂R

∂t i

= t i

solving gives

R = 1

2 t ² _i

1

2 t ² _i = t _i (t _i + 1

2 − T (t _i )) which yields the solution

T (t _i ) = t _i 1 + 1

2

E[π S ] = 2 Z 1

0



t _i (t _i + 1 2 ) − 1

2 t ² _i

 dt

= 5

6