1 Nash Implementation 2

Tutorat 4

These notes attempt to provide a discussion of the concepts and to explicate the exercises covered in the Micro 2 tutorial on Friday, 28 January 2011. Please contact [email protected] with any corrections or clarifications.

SHW (29 jan 2011)

Contents

1 Nash Implementation 2

2 Exercise 3.1: The Voting Game 3

2.1 Condorcet Winner . . . . 3

2.2 Solution to Part (a) . . . . 4

2.3 Solution to Part (b) . . . . 5

2.4 Solution to Part (c) . . . . 5

3 Question 9 6 3.1 Solution to Part (a) . . . . 6

3.2 Solution to Part (b) . . . . 7

3.3 Solution to Part (c) . . . . 7

3.4 Solution to Part (d) . . . . 7

3.5 Solution to Part (e) . . . . 9

1 Nash Implementation

The Nash equilibrium concept assumes complete information: all agents know everything about everyone else. By contrast, dominant-strategy equilibrium and Bayesian equilibrium can encompass asymmetric information and incomplete information settings.

How can a mechanism designer extract agents’ information? The basic idea in dominant-strategy and Bayesian implementation: Make each agent internalize the externalities, thus turning each agent into a de facto “social planner”. In the dominant-strategy and Bayesian settings, the mechanism designer has an informational advantage, since he can extract information from all the agents, whereas the agents all have uncertainty about the other agents. In the Nash setting, the mechanism designer is the only one who doesn’t know everything about the types. The object of the mechanism designer Nash implementation is to extract a full profile of agent types from each agent (and then to use this information to implement the target social choice rule (SCR)).

Nash implementation takes as the equilibrium concept the notion of Nash equi- librium, an equilibrium in which each agent i plays a best response against the actions of the other agents. Mathematically, θ = (θ ₁ , . . . , θ _n ) is a Nash equilib- rium if for all players i, the utility Player i gets from playing θ _i (i.e. playing according to the Nash equilibrium profile θ) is weakly greater than the utility she gets from playing any other possible action ˜ θ _i , holding fixed the actions of the other players:

u i (f (θ i , θ −i )) ≥ u i (f (˜ θ i , θ −i )).

The notion of Nash equilibrium is equivalent to “no profitable unilateral devi- ations” from the equilibrium profile. In Nash implementation, we seek a Nash equilibrium that results in the same alternative(s) being chosen under the mech- anism as would be chosen under the social choice rule.

Strong vs. Weak Nash Implementation

Strong Nash Implementation — Every Nash equilibrium of the mechanism leads to the same outcome(s) as the social choice rule f (θ).

The SCR f (·) can be strongly Nash implemented only if it can be imple- mented in dominant strategies.

Weak Nash Implementation — Easy to achieve (recall the “fatal conse- quences” game in which the mechanism designer kills everyone unless all reported profiles agree exactly).

Theorem (Maskin, 1977): (1) If the SCR f is strongly Nash-implementable, then it is monotonic. (2) If the number of agents n ≥ 3, and the SCR f is monotonic, and the no-veto-power condition holds, then f can be strongly Nash implemented.

The mechanism that implements f is designed as follows. All agents report the full type profile.

1. If all reports (u, a) coincide and a ∈ f (a), then the decision rule selects

g(u) = a.

2. If Agent i reports (˜ u, b) while all other agents report (u, a), then the decision rule selects

g(m) =

( b if u _i (a) ≥ u _i (b) a otherwise.

That is, the decision rule will select the alternative b proposed by the single dissenting agent only if b does not make that agent better off than a under the type profile u reported by all the other agents. Otherwise, the decision rule listens to the majority and selects alternative a.

3. In all other cases, the decision rule selects anything you want.

2 Exercise 3.1: The Voting Game

Let X be a finite set with at least three alternatives. Let N = {1, 2, . . . , n} be a finite set of voters, with n ≥ 2. For each voter, any complete and transitive preference ordering is considered a priori possible. Each possible profile R = (R 1 , . . . , R n ) of preference orderings represents a possible state of the world (we let R _i denote weak preference; the corresponding strict preference we denote P _i ).

By definition, a Condorcet winner in state R is an alternative a in X such that, for any other alternative b 6= a in X, #{i : aP i b} > #{i : bP i a}. (In words, for any alternative b 6= a, more people strictly prefer a to b than strictly prefer b to a.) Let x ^∗ be a fixed alternative in X. Define a social choice function f as follows. If there exists a Condorcet winner a in state R, then f (R) = a.

Otherwise, f (R) = x ^∗ . (a) Is f monotonic?

(b) Is f strategy-proof?

(c) Is f Nash-implementable?

2.1 Condorcet Winner

To illustrate the concept of Condorcet winner, consider the following game with three agents, I = {1, 2, 3}, and three alternatives, A = {P, W, X}. (To give the game a story line, let agents 1, 2, and 3 constitute a trio of flatmates trying to decide on which game console to buy, a PlayStation, a Wii, or an Xbox.) Each agent’s preferences are strictly ranked as follows.

1 2 3

First choice: X X P Second choice: W P X Third choice: P W W

Given this state of preferences, we can compute the number of agents who prefer alternative X to alternative b = P, W :

#{i : X  _i P } = 2 > 1 = #{i : P  _i X} → X beats P

#{i : X  i W } = 3 > 0 = #{i : W  i X} → X beats W ,

where a  i b denotes that Agent i strictly prefers a to b. Hence the Condorcet winner is X.

Remark 1: A Condorcet winner does not necessarily exist. Under the fol- lowing state of preferences,

1 2 3

First choice: P W X Second choice: W X P Third choice: X P W

there is no Condorcet winner, as the following computations show. For P ,

#{i : P  i W } = 2 > 1 = #{i : W  i P } → P beats W

#{i : P  _i X} = 1 < 2 = #{i : X  _i P } → X beats P . For W ,

#{i : W  i P } = 1 < 2 = #{i : P  i W } → P beats W

#{i : W  _i X} = 2 > 1 = #{i : X  _i W } → W beats X.

And for X,

#{i : X  i P } = 2 > 1 = #{i : P  i X} → X beats P

#{i : X  _i W } = 1 < 2 = #{i : W  _i X} → W beats X.

Remark 2: A Condorcet winner, if one exists, is unique. To prove this, suppose a game has two distinct alternatives a and b that are both Condorcet winners. Then by definition of Condorcet winner, since a is a Condorcet winner we must have #{i : a  i b} > #{i : b  i a}. Similarly, b is a Condorcet winner, so we must also have #{i : b  i a} > #{i : a  i b}, directly contradicting the previous statement.

2.2 Solution to Part (a)

No, the given SCR f cannot be monotonic. Recall that a social choice rule f is monotonic if ∀u, ˜ u ∈ U and ∀a ∈ f (u), then ∀i ∈ I and ∀b ∈ A,



u _i (a) ≥ u _i (b) ⇒ ˜ u _i (a) ≥ ˜ u _i (b) 

⇒ a ∈ f (˜ u).

That is, if the social choice rule f is monotonic, then if the alternative a is chosen when the preference profile is u, and if whenever an agent i preferred a to some other alternative b under u she continues to prefer a to this alternative b under

˜

u, then a will also be chosen when the preference profile is ˜ u.

The social choice function f given in this exercise fails to be monotonic due to the arbitrariness with which it must select an outcome when no Condorcet winner exists. Abstractly, we might imagine a state R ₁ in which no Condorcet winner exists and thus f chooses the fixed alternative x ^∗ ∈ X. If we move to a new state R 2 in which x ^∗ has the same rank as under R 1 for all agents but some of the other alternatives preferred to x ^∗ in R 1 have changed rank, it is conceivable that this new state R ₂ will have a Condorcet winner a 6= x ^∗ , so that monotonicity fails.

As a concrete example, take the same three-agent, three-alternative game

discussed above, and consider the initial state R ₁ given below.

State R 1

1 2 3

First choice: P X W Second choice: X W P Third choice: W P X

In this state, no Condorcet winner exists, and so the social choice function f chooses some fixed alternative, say W . Now consider moving to state R 2 given below.

State R ₂

1 2 3

First choice: X X W Second choice: P W P Third choice: W P X

In the move from state R 1 to state R 2 , all agents who preferred alternative W over some other alternatives in state R ₁ still prefer W over these same alterna- tives in state R 2 . (If W ranked above another alternative b in column i in state R 1 , then W still ranks above this alternative in column i in state R 2 .) Thus if f were monotonic, since W was chosen in state R ₁ , W should also be chosen in state R 2 . However, we can easily verify that a Condorcet winner exists in state R 2 , which is alternative X. Thus f (R 2 ) = X 6= W = f (R 1 ), and f is not monotonic.

2.3 Solution to Part (b)

No, f is not strategy-proof (i.e incentive compatible). Consider the state R 1

given in part (a) above. No Condorcet winner exists in this state, and so the social choice function f chooses some fixed alternative, which we set to be W . Given that the other agents truthfully report their types, Agent 1 has a profitable deviation: She can lie and claim that her preferences are those given in state R ₂ . When she does so, the alternative chosen by the social choice rule is f (R ₂ ) = X, which Agent 1 strictly prefers to W (remember that her true preferences are given by state R ₁ ). ¹

2.4 Solution to Part (c)

Yes and no. The social choice function f is weakly Nash-implementable; we can weakly Nash implement and equilibrium we want. However, since f is not mono- tonic, it cannot be strongly Nash-implementable. (Recall that monotonicity is a necessary condition for strong Nash implementation.)

1

The tutor provided a discussion on the order in which pair-wise votes are conducted, historically an important result. Consider again the state R

given in part (a), and suppose that agents are asked to reveal their preferences via a sequence of pair-wise votes (in which two alternatives a versus b are voted upon, and the winning alternative is then paired with the remaining alternative c and the winning alternative is chosen as the outcome). Which two alternatives agents vote on first affects the ultimate outcome. For example, if agents first vote on X versus W , followed by the winning alternative versus P , the outcome is P : (1) X vs.

W → X, then (2) X vs. P → P . If instead agents vote on X versus P first, followed by the

winning alternative versus W , the outcome is W : (1) X vs. P → P , then (2) W vs. P → W .

3 Question 9

One object needs to be allocated between two buyers, who arrive sequentially (the second buyer arrives only after the first buyer leaves). The buyers’ payoffs are quasilinear in money. The buyers’ values for the object are their private infor- mation and are drawn independently from the same distribution F (·). However, F is not known: ² with probability 1/2 it is uniform on [0, 1], and with probability 1/2 it is uniform on [1, 2]. (In the case the selected F (·) is U [0, 1], both buyers’

valuations are drawn from this distribution: v ₁ , v ₂ ∼ U [0, 1]. Similarly, if the selected F (·) is U [1, 2], then v 1 , v 2 ∼ U [1, 2].) These probabilities are common knowledge. Assume the seller has a valuation c = 0.

(a) Calculate the efficient allocation plan, i.e. one that maximizes the expected total surplus.

(b) Is it possible to implement the efficient allocation plan in a Bayesian incen- tive-compatible mechanism?

(c) Do the answers to parts (a) and (b) change if the first buyer can leave his credit card number, which could be charged after the second buyer arrives?

(d) Now consider instead the seller’s problem of designing an expected-revenue- maximizing, Bayesian, incentive-compatible, and interim individually ra- tional mechanism. Assume that the first buyer can leave his credit card number and that the first-period message and allocation can be kept secret from the second buyer. Derive an optimal condition.

(e) Design the expected-revenue-maximizing, incentive-compatible, interim in- dividually rational Bayesian mechanism.

3.1 Solution to Part (a)

In this incomplete information setting, an efficient allocation plan is one that maximizes expected total surplus. Ideally we would like to to design a mechanism that gives the object to the buyer with the higher valuation.

To find the ex ante efficient allocation, we use backward induction. In final stage, efficiency is achieved by giving the object to Buyer 2 at zero price. (At any positive stage-2 price p 2 > 0, there is some non-zero ex ante probability that v ₂ < p ₂ , and hence Buyer 2 would not agree to trade, and hence the positive surplus v ₂ would not be realized. ³

Now move to stage 1. Ask Buyer 1 what his valuation is. If he responds that his valuation lies in [ ¹ ₂ , 1] ∪ [ ³ ₂ , 2], i.e. in the upper half of either possible distribution, then give Buyer 1 the object; otherwise, give the object to Buyer 2.

This allocation is ex ante efficient because, given that v 1 ∈ [ ¹ ₂ , 1] and v 2 ∼ U [0, 1], Prob(v ₂ > v ₁ ) = 1 − v ₁ ≤ ¹ ₂ ; Buyer 2 has less than ⁴ a 50% chance of having

2

What is the difference between a

-

lottery over U [0, 1] and U [1, 2], and a uniform dis- tribution over [0, 2]? With one buyer, there is no difference. However, with multiple buyers whose distributions are all decided by a single

-

lottery, these two are very different. In the

1

2

-

lottery, buyers either both have valuations from U [0, 1] or they both have valuations from U [1, 2]. In contrast, in the U [0, 2] with probability 1 case, one buyer could have a valuation in [0, 1] while the other buyer has a valuation in [1, 2].

3

Notice that Buyer 2’s valuation is v

= 0 with probability zero.

4

The case in which Prob(v

> v

) =

corresponds to v

=

precisely, which occurs with

probability zero.

a higher valuation than Buyer 1, so we should give the object to Buyer 1. If instead v ₁ ∈ [0, ¹ ₂ ), then Prob(v ₂ > v ₁ ) = 1 − v ₁ > ¹ ₂ ; Buyer 2 has more than 50% chance of having a higher valuation, so we should give the object to him.

Analogous reasoning yields identical results for v 1 ∈ [ ³ ₂ , 2] and v 1 ∈ [1, ³ ₂ ).

3.2 Solution to Part (b)

No, there is no incentive-compatible Bayesian mechanism that can implement the efficient allocation plan detailed in part (a) above. A Buyer 1 with true type v ₁ ∈ [ ¹ ₂ , 1] will lie and report a valuation ˜ v ₁ = ¹ ₂ , the lowest reported valuation for which he still gets the object. By reporting ˜ v 1 < v 1 , Buyer 1 gets the object at a lower price. Similarly, a Buyer 1 with true type v ₁ ∈ [ ³ ₂ , 2] will lie and report a valuation ˜ v ₁ = ³ ₂ .

3.3 Solution to Part (c)

Yes. By enabling Buyer 1 to leave his credit card information, the allocation problem essentially becomes simultaneous. The efficient allocation plan becomes:

(1) Ask Buyer 1 for his valuation and credit card number in stage 1; (2) Ask Buyer 2 for his valuation in stage 2. If v ₂ > v ₁ , then give the object to Buyer 2.

If v ₂ ≤ v ₁ , then give the object to Buyer 1.

The question of whether the efficient allocation plan is Bayesian-implement- able via an incentive-compatible mechanism is left as an exercise. (Read: I haven’t come up with a convincing proof one way or the other yet. Be the first!

Then please let me know.) 3.4 Solution to Part (d)

To design an expected-revenue-maximizing, incentive-compatible, interim-indivi- dually rational Bayesian mechanism, we again use backward induction. In the final stage (stage 2), efficiency implies we must set p ₂ = 0. What about the price in stage 1? If we want any chance at distinguishing which of the two distributions U [0, 1] and U [1, 2] the buyers come from, we must have p ₁ > 1. Since the two distributions are chosen with equal probability and we have no prior information on Buyer 1, we can compute the probability that Buyer 1 accepts an offered price p 1 is obtained by treating him as if he drew his type from U [0, 2]; hence,

Prob(B1 accepts p ₁ ) = 2 − p ₁ 2 .

If Buyer 1 accepts the offered price p ₁ , then his valuation must equal or exceed this price: v 1 ≥ p ₁ . Thus the expected consumer surplus, given that Buyer 1 accepts p 1 , is

E[CS | B1 accepts p 1 ] = Z 2

p

(v 1 − p ₁ ) dF (v 1 | B1 accepts p ₁ )

= Z 2

p

(v ₁ − p ₁ ) 1 2 − p 1

dv ₁

= 1

2 − p ₁

 1

2 v ₁ ² − p ₁ v ₁

 v

=2 v

=p

= 2 − p ₁

2 .

The seller’s surplus, if Buyer 1 accepts price p 1 , is simply p 1 ; therefore the total surplus given that Buyer 1 accepts p ₁ equals

T S | B1 accepts p ₁ =  2 − p 1

2



+ p ₁ = 2 + p 1

2 .

The probability that Buyer 1 rejects an offered price of p 1 is given by the complementary probability:

Prob(B1 rejects p 1 ) = 1 − Prob(B1 accepts p 1 )

= 1 −  2 − p ₁ 2



= p 1

2 .

We can use the techniques of Bayesian inference to obtain more accurate probabilities on the buyers types, given Buyer 1’s response to the offered price p ₁ . Suppose that Buyer 1 rejects an offered price p ₁ . The fact that he rejected this offer reveals that his valuation v 1 is less than p 1 . In particular, if p 1 > 1, then a Buyer 1 who rejects p 1 can come from the high distribution only if v 1 ∈ [1, p ₁ ).

The probability that the buyers come from the low distribution U [0, 1], given that Buyer 1 rejects p 1 > 1, equals the probability that Buyer 1 rejects p 1 given that he is from the low distribution (1) times the prior probability that he is from the low distribution ( ¹ ₂ ), divided by the probability that Buyer 1 rejects p ₁ given that he is from the low distribution (1) times the prior probability that he is from the low distribution ( ¹ ₂ ) plus the probability that Buyer 1 rejects p 1

given that he is from the high distribution (this happens for a v ₁ ∈ [1, 2] only when v 1 < p 1 , which for a uniform distribution occurs with probability p 1 − 1) times the prior probability that he is from the high distribution ( ¹ ₂ ):

Prob (B ∈ [0, 1] | B1 rejects p 1 ) =

1 2 1

2 + ¹ ₂ (p 1 − 1) = 1 p 1

.

Notice that since p 1 ≤ 2, this probability is greater than or equal to ¹ ₂ , which is what we should expect: Given that Buyer 1 rejects p 1 > 1, we expect that the buyers are more likely to be low type than high type. (Notice that in the extreme case p 1 = 1, if Buyer 1 rejects this offered price, then we know with certainty that the buyers come from the low distribution, which is exactly what the above expression reveals: when p ₁ = 1, this updated probability equals 1.) The probability that the buyers come from the high distribution U [1, 2], given that Buyer 1 rejects p 1 , is the complementary probability ⁵

Prob (B ∈ [1, 2] | B1 rejects p 1 ) = 1 − 1 p 1

.

Recall that efficiency dictates that, if we find ourselves in the final round, we offer the object to Buyer 2 at price p 2 = 0. Thus the expected consumer surplus, given that Buyer 1 rejects p ₁ , equals the probability that Buyer 2 is a high-type

5

We could compute the probability that the buyers come from the high interval using Bayesian updating, as we did for the low interval:

Prob (B ∈ [1, 2] | B1 rejects p

) =

1 2

(p

− 1)

1

2

+

(p

− 1) = p

− 1 p

= 1 − 1 p

.

buyer times his expected utility from getting the object at zero price, plus the probability that Buyer 2 is a low-type buyer times his expected utility from getting the object at zero price:

E(CS | B1 rejects p 1 ) = (1 − 1

p ₁ )E[v 2 (high)] + 1

p ₁ E[v 2 (low)]

= (1 − 1

p ₁ )1.5 + 1

p ₁ 0.5 = 1.5 − 1 p ₁ .

Since the seller offers the good at zero price in stage 2, seller’s surplus is zero, and total surplus equals consumer’s surplus in this case.

To maximize the expected utility delivered by this mechanism, we therefore maximize, with respect to p 1 , the sum of the utility when Buyer 1 accepts p 1 (and therefore gets the good) and the utility when Buyer 1 rejects p ₁ (and therefore Buyer 2 gets the good at zero cost), weighted by their respective probabilities:

max p

  2 − p ₁ 2

  2 + p ₁ 2



| {z }

Buyer 1 accepts

+



1 − 2 − p ₁ 2

 

1.5 − 1 p 1



| {z }

Buyer 1 rejects



= max

p

 1

4 (4 − p ² ₁ ) + p 1

2

 3 2 − 1

p ₁



= max

p

 1 − 1

4 p ² ₁ + 3 4 p ₁ − 1

2



= max

p

 1 2 + 3

4 p ₁ − 1 4 p ² ₁

 .

Taking the first-order condition ⁶ with respect to p ₁ , we obtain 3

4 − 1 2 p ^∗ ₁ =

set 0 ⇒ p ^∗ ₁ = 3

2 .

Combining this result with the efficient stage-2 price p ^∗ ₂ = 0 gives us the optimal (totoal-surplus-maximizing) mechanism: In stage 1, offer the object to Buyer 1 at price p ^∗ ₁ = ³ ₂ ; if he rejects, then in stage 2, offer the object to Buyer 2 at price p ^∗ ₂ = 0.

3.5 Solution to Part (e)

Begin by remarking that, in contrast to the total-surplus-maximizing mechanism we analyzed in part (d), the expected-revenue-maximizing mechanism will never set the stage-2 price equal to zero: it is always better for the seller to charge a non-zero price and thus obtain positive revenue with some non-zero probability.

To design this mechanism, we again use backward induction. Start from stage 2. As we saw in part (d) above, if we are in stage 2, then Buyer 1 has rejected price p 1 ; from our Bayesian updating, this implies that Buyer 2 has a valuation in [0, 1] with probability _p ¹

1

, and a valuation in [1, 2] with probability 1 − _p ¹

1

. If we assume that p 2 ≤ 1 (an assumption we will have to check at the end of our analysis, once we have obtained a value for p ₂ ), then a Buyer 2 from the high distribution always accepts the offer and a Buyer 2 from the low distribution

6

Notice that the objective function is quadratic in p

with a negative square term; thus the

first-order condition indeed gives us a maximum.

accepts the offer with probability (1 − p 2 ). Thus the seller’s expected stage-2 profit is

E[π 2 ] = [Prob(B2 accepts p 2 )]p 2

=



1 − 1 p ₁

 1 + 1

p ₁ (1 − p 2 )

 p 2

=

 1 − p 2

p ₁

 p ₂

= p ₂ − 1

p ₁ p ² ₂ . (1)

Maximizing this with respect to the stage-2 price p ₂ via the first-order condition, ⁷ we obtain

1 − 2 p 1

p ^∗∗ ₂ =

set 0 ⇒ p ^∗∗ ₂ = p 1

2 .

(From this expression, we can remark immediately that since p ₁ ≤ 2, it follows that p 1 ≤ 1, so our earlier assumption on p ₁ is a safe one.) For the analysis of stage 1, it will be useful to have an expression for the expected payoff in stage 2 as a function of p ₁ , which we now compute. Plugging the expression for p ^∗∗ ₂ into (1), we get

E[π 2 (p ^∗∗ ₂ )] =  p ₁ 2

 − 1 p ₁

 p ₁ 2

 2

= 1 4 p ₁ .

Moving now to stage 1, if the seller offers Buyer 1 a price p ₁ ∈ [0, 2], then the probability that Buyer 1 will accept is ^2−p ₂

, in which case the seller gets profit p 1 . (Recall that we can treat Buyer 1’s valuation as coming from a uni- form distribution on [0, 2].) The probability that Buyer 1 rejects p ₁ is therefore 1 − ( ^2−p ₂

) = ¹ ₂ p ₁ , in which case the game moves to stage 2, the seller offers Buyer 2 the price p ^∗∗ ₂ = ¹ ₂ p 1 found above, and the seller makes expected profit E[π 2 (p ^∗∗ ₂ )] = ¹ ₄ p 1 . The seller’s expected profit over both stages is thus

E[π 1 ] =  2 − p ₁ 2



p ₁ +  p ₁ 2

 1 4 p ₁

= p ₁ − 1 2 p ² ₁ + 1

8 p ² ₁

= p ₁ − 3 8 p ² ₁ .

Maximizing this expression ⁸ with respect to p ₁ , we obtain 1 − 3

4 p ^∗∗ ₁ =

set 0 ⇒ p ^∗∗ ₁ = 4

3 .

The expected-profit-maximizing mechanism is thus p ^∗∗ ₁ = 4/3, p ^∗∗ ₂ = 2/3. (No- tice that our earlier assumption that p ₂ ≤ 1 is justified by this result.)

Let’s briefly compare these results to the prices of the optimal (total-surplus- maximizing) mechanism. The expected-profit-maximizing stage-1 price is slightly lower, whereas the stage-2 price is higher. We have already commented on why

7

The expression for E[π

] is quadratic in p

with the p

term negative, so the first-order condition yields the maximum.

8

This expression is yet another concave quadratic; the first-order condition yields the max-

imum.

the stage-2 price should be higher under the goal of profit maximization; can

we give some intuition for the stage-1 price? Essentially, the tradeoff between a

lower stage-1 price and a higher stage-1 sale probability leads the seller to set

her stage-1 price below that of the optimal mechanism.