microqual 2013 spring solutions

8/9/2019 MicroQual 2013 Spring Solutions

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Department of EconomicsRice University

Micro Qual Solutions16 May 2013

Pareto-optimal and individually rational allocation (z j) j∈N , each agent is indifferent betweenher bundle z j and her initial endowment. Is this conclusion preserved if preferences are not

necessarily convex?

1.1 Solution to part (a)

Since the agents’ ideal outcomes ai are real numbers and R is well-ordered, we can rank the agents’ideal outcomes. Let a[i] denote the ith-lowest ideal, so that a[1] ≤ . . . ≤ a[n]. Similarly, let [i] denotethe agent with ideal outcome a[i].

Proposition 1.1. The set of Pareto-optimal outcomes is given by the interval [a[1], a[n]] ∩ [0, 1].

The intersection with [0, 1] is simply to ensure feasibility. If ai ∈ [0, 1] for all agents i ∈ N , thenthe set of Pareto-optimal outcomes is simply [a[1], a[2]]. We make this assumption in the proof thatfollows; the general proof is identical, just more notationally cumbersome.

Proof. First, consider any outcome x < a[1]. Intuitively, if we move the outcome x closer to a[1], theoutcome becomes closer to al l agents’ ideals, making all of them strictly better off. More precisely,denote ε := a[1] − x. Agent i’s utility under the public outcome x is

ui(x) = −(x − ai)2 = −(a[1] − ai − ε)2.

Now consider the public outcome x := a[1]. Agent i’s utility under x is

ui(x) = −(x − ai)2 = −(a[1] − ai)

2

Since a[1] ≤ ai for all i ∈ N , the quantity a[1] − ai ≤ 0; since x < a[1] by assumption, it follows thatε >

0, so that a

[1] −ai −

ε < a[1] −

ai ≤ 0, and therefore

ui(

x)

< ui(

x

). Since agent i

∈ N

wasarbitrary, we conclude that all agents are strictly better off under the public outcome x comparedto x, and therefore x is not Pareto-optimal. The proof that any x > a[n] is not Pareto-optimal isanalogous.

Now consider any outcome x ∈ [a[1], a[n]]; we wish to show that any such outcome is Pareto-optimal. Intuitively, if we move the outcome to the left, we make agent [n] strictly worse off;whereas if we move the outcome to the right, we make agent [1] strictly worse off. More precisely,consider any other outcome x, and denote ε := x − x. We consider three cases.

Case 1: ε = 0. This is the trivial case in which x = x; clearly x cannot Pareto dominate x.

Case 2: ε < 0. This is the case in which x < x. Consider agent [n]; we have

u[n](x) = −(x − a[n])2 = −(x + ε − a[n])2.

Since x ≤ a[n], we have x − a[n] ≤ 0; moreover, since ε < 0, it follows that x + ε − a[n] <

x − a[n] ≤ 0, and therefore

u[n](x) = −(x + ε − a[n])2 < −(x − a[n])

2 = u[n](x),

so that agent [n] is strictly worse off under x compared to x.1 Thus x does not Paretodominate x.

1In fact, this argument shows that any agent i ∈ N such that a[i] > x will be made strictly worse off by the changeto x .

2





Case 3: ε > 0. This is the case in which x > x. The proof that x does not Pareto dominate x isanalogous to Case 2, with agent [1] in place of agent [n].

In all cases, the new outcome x does not Pareto dominate the original outcome x, so we concludethat x is Pareto-optimal.

1.2 Solution to part (b)

Let (z j) j∈N be a Pareto-optimal allocation. We prove each subpart in turn.

Proof of (i). Assume for the sake of contradiction that the statement does not hold. That is, foreach agent i ∈ N there exists some other agent j ∈ N such that z j i zi. It follows from theproperties of directed graphs that there exists a cycle of agents j1, j2, . . . , jm (with 1 < m ≤ n)such that

z j1 jm z jm jm−1 z jm−1 jm−2 . . . j2 z j2 j1 z j1;

in words, each agent strictly prefers the allocation of the next agent in the cycle to her ownallocation. Thus, by permuting the allocations within this cycle (namely, give agent j1’s allocationto agent jm, agent jm’s allocation to agent jm−1, etc.) and keeping all other allocations outsidethe cycle the same, we obtain a feasible allocation that Pareto dominates the original allocation(z j) j∈N , contradicting the hypothesis that (z j) j∈N is Pareto-optimal.

Proof of (ii). Assume for the sake of contradiction that the statement does not hold. That is, foreach agent k ∈ N there exists some other agent j ∈ N such that zk j z j. The argument nowproceeds analogously to our proof of (i). In particular, there exists a cycle

z j1 j2 z j2 j3 . . . m z jm j1 z j1 ,

and we can permute the allocations within this cycle to obtain a Pareto improvement, a contradic-tion.

Notice that our proofs in part (b) do not require any assumptions on individual preferencesother than completeness.

1.3 Solution to part (c)

To be completed.

2 Exercise 2

Consider the following economy, consisting of three goods, two consumers, and two firms. Good3 is used as an input in the production process and provides no utility to the consumers. Firm1, which is owned entirely by Alice, has a technology that allows good 3 to be made into good 1according to the production function

f 1(x3) = 3x3.

3





Firm 2, owned entirely by Bob, uses good 3 to make good 2 according to the production function

f 2(x3) = 4x3.Consumers’ initial endowments of goods are ωA = ωB = (0, 0, 5), and their utility functions are

uA(x1, x2, x3) = .4ln(x1) + .6ln(x2), uB(x1, x2, x3) = .5ln(x1) + .5ln(x2).

Compute the Walrasian equilibria of this economy.

2.1 Solution

We begin by reminding ourselves of the following definition (see Mas-Colell et al. (1995), p 579).

Definition 2.1. Given a private-ownership economy

{(X i, ui)}i∈I , {Y j} j∈J , {(ωi, θi)}i∈I ,

the allocation (x, y) and price vector p constitute a Walrasian equilibrium if the following threeconditions hold.

1. Each firm j ∈ J is solving its profit-maximization problem, given p:

p · y j ≥ p · y j, ∀y j ∈ Y j .

2. Each consumer i ∈ I is solving her utility-maximization problem, given p:

xi ∈ arg maxxi∈X i

ui(xi) | p · xi ≤ p · ωi + j∈J

θi j p · y j

.

3. Markets clear, given p: i∈I

xi =i∈I

ωi + j∈J

y j.

Remark 2.1. We make the following two observations, which will simplify the subsequent analysis.First, note that both firms have a constant returns to scale (CRS) production function and no sunkcosts. It follows that profit-maximizing firms will either earn zero or infinite profit, so that inparticular both firms must earn zero profit in any Walrasian equilibrium. Second, note that (i)both consumers have utility functions that are strongly monotone in goods 1 and 2; (ii) neitherconsumer obtains any utility from good 3; and (iii) if either x1 = 0 or x2 = 0 for a consumer,then her utility is −∞. From (i) it follows that in any Walrasian equilibrium, p1, p2 > 0 and thebudget constraint of each consumer holds with equality. From (ii) it follows that xi

3 = 0 for bothconsumers in any Walrasian equilibrium.2 From (iii) it follows that xi

> 0 for goods = 1, 2, forboth consumers i = A, B, in any Walrasian equilibrium, provided that their budget set containssuch a point.3 As a corollary, note that (iii) implies that both firms have strictly positive output,and therefore demand strictly positive amounts of good 3, in any Walrasian equilibrium.

2If xi3 > 0 for consumer i, she could strictly increase her utility by selling xi

3 at the market price p3 and using theproceeds p3xi3 to purchase a strictly positive amount of goods 1 or 2 (or both).

3Let Bi( p) = {x ∈ X i | p · x ≤ p · ωi + θi · π} denote agent i’s budget set. Provided there exists some bundlex ∈ X i such that x1, x2 > 0, since ui(x) > −∞ = ui(x) for any x such that x

1 = 0 or x

2 = 0, agent i always strictlyprefers such an “interior” bundle x to any boundary bundle x .

4





We begin with the firms. Firm 1’s profit-maximization problem is

maxy3≥0 [ p1f 1(y3) − p3y3] = maxy3≥0 [(3 p1 − p3)y3] .

Solving the first-order condition for p1, we find

p1 ≤ 1

3 p3. (2.1)

Similarly, firm 2’s profit-maximization problem is

maxy3≥0

[ p2f 2(y3) − p3y3] = maxy3≥0

[(4 p2 − p3)y3] .

Solving the first-order condition for p2, we find

p2 ≤ 1

4 p3. (2.2)

In Remark 2.1 we argued that in equilibrium, both firms are demanding a strictly positive amountof good 3; therefore (2.1) and (2.2) hold with equality. Normalizing the price of good 3 to p3 = 1,we obtain the equilibrium price vector (unique up to normalization) of

p = (1

3, 1

4, 1). (2.3)

Next we turn to the consumers. In Remark 2.1 we argued that in equilibrium, (i) all budgetconstraints hold with equality, (ii) xi

3 = 0 for both consumers i = A, B, and (iii) xi > 0 for = 1, 2

for both consumers i = A, B (i.e., the solution is interior in (x1, x2) space); we also argued thatthe profit of both firms must be 0. Using these results and the equilibrium prices found in (2.3),consumer A’s utility-maximization problem writes as

maxx1,x2≥0

[.4 ln x1 + .6 ln x2]

s.t. 1

3x1 +

1

4x2 = 5.

Solving the budget constraint for x2 in terms of x1 and substituting the resulting expression intothe objective function, we obtain

maxx1≥0

.4 ln x1 + .6 ln20 − 4

3x1 .

Taking the first-order condition (which is satisfied with equality, since the solution is interior), wehave

2

5xA1

− 4

3

3

5

1

20 − 43xA

1

=set

0,

which solving for xA1 yields

xA1 = 6.

5





Substituting this solution into consumer A’s budget constraint and solving for xA2 gives xA

2 = 12.Similarly, consumer B’s utility-maximization problem writes as

maxx1,x2≥0

[.5 ln x1 + .5 ln x2]

s.t. 1

3x1 +

1

4x2 = 5.

Proceeding in the same manner as above, we can simplify the maximization problem to

maxx1≥0

.5 ln x1 + .5 ln

20 −

4

3x1

.

The first-order condition (satisfied with equality) is

1

2xB

1

− 4

3

1

2

1

20 −

4

3xB

1

=set

0,

which solving for xB1 yields

xB1 =

15

2 .

Substituting this solution into consumer B ’s budget constraint gives xB2 = 10.

We can infer the firms’ input-output vectors from the aggregate consumption, which is x =xA + xB = (13.5, 22, 0), and the production functions. Since only firm 1 produces good 1 and onlyfirm 2 produces good 2, we have

x1 = f 1(x13) ⇔ 13.5 = 3x1

3 ⇔ x13 = 4.5

x2 = f 2(x2

3

) ⇔ 22 = 4x2

3

⇔ x2

3

= 5.5.

Note that the total input of good 3 is x3 = 4.5+5.5 = 10, and aggregate endowment is ω = (0, 0, 10).To summarize, the Walrasian equilibrium of this economy (unique up to normalization of prices)

is

xA = (6, 12, 0), xB = (7.5, 10, 0), y1 = (13.5, 0, −4.5), y2 = (0, 22, −5.5), p =

1

3, 1

4, 1

.

3 Exercise 3

A pure exchange economy has 1000 people and one consumption good. There are two possiblestates of the word for the future (date 1). If state A occurs, then each person will have an initial

endowment of 10 units of the consumption good. If state B occurs, then each person will havean initial endowment of 20 units of the consumption good. Each person is an expected utilitymaximizer, with von Neumann–Morgenstern utility

ui(cA, cB) = πi ln(cA) + (1 − πi)ln(cB),

where π i is person i’s subjective probability that event A happens, and cA, cB denote consumptionin states A and B, respectively.

Before the future state is realized (i.e., at date 0), people can trade in a full set of Arrow (state-contingent) securities. Let pA (respectively, pB) denote the (date-0) price of the Arrow securitythat pays 1 in state A (state B).

6





(a) Let the Arrow security for state B be the numeraire. Find the demand of person i for theArrow security corresponding to state A as a function of pA and πi.

(b) What is the competitive equilibrium price for the Arrow security corresponding to state A?


Agent i’s utility-maximization problem (UMP) is

max

πi ln cA + (1 − πi) ln cB

s.t. pAcA + pBcB ≤ pAωA + pBωB

cA, cB ≥ 0.

Note that, provided πi ∈ (0, 1), agent i’s utility is strongly monotone in both state-contingent

consumption goods. Thus, when πi ∈ (0, 1), the budget constraint holds with equality. More-over, if cA = 0 or cB = 0, then agent i’s utility is −∞. Thus, provided her budget set has anonempty interior, agent i will choose cA, cB > 0. Normalizing pB := 1 as suggested, solving thebudget constraint for cB and substituting the result into the objective function, and dropping thenonnegativity constraints on consumption, agent i’s UMP becomes

max

πi ln cA + (1 − πi) l n ( pAωA + ωB − pAcA)

.

The first-order condition (with respect to cA) is

πi

cA−

(1 − πi) pA

pAωA + ωB − pAcA=set

0,

where the equality follows since agent i’s budget set has nonempty interior,4 and therefore cA > 0.Solving for cA, we find

cA = πi

pA( pAωA + ωB); (3.1)

plugging in the given endowment ωi = (ωA, ωB) = (10, 20) gives

cA = πi

pA(10 pA + 20).

Remark 3.1. As a quick check, let’s analyze expression (3.1) for agent i’s consumption of thestate-contingent good A. The expression shows that each agent i spends a fixed fraction πi

pAof her

total wealth pAωA + ωB on good A. Mathematically,

∂cA

∂π i =

1

pA( pAωA + ωB),

4In particular, each agent i is assumed to have endowment ωi = (10, 20), and therefore her budget set is the closedtriangle in R2defined by the points (0, 0), (10pA+20pB

pA, 0), and (0, 10pA+20pB

pB).

7





which is strictly greater than 0 provided that ωB > 0, as assumed, and pA is bounded above, aswe show in part (b). This makes sense: Agent i’s consumption of good A increases as she assigns

more probability πi to state A occurring. Second,

∂cA

∂pA= −

πiωB

p2A,

which is strictly less than 0 under the same assumptions as previously stated. This also makessense: As the price of good A increases, agent i’s consumption of the good decreases.


Let I denote the set of consumers. By definition, in any competitive equilibrium all markets clear.In particular, the market for the state-contingent consumption good A clears, i.e.

i∈I

ciA =

i∈I

ωiA. (3.2)

By assumption, ωiA = ωA for all i ∈ I . Consumer i’s demand ci

A for good A is given by (3.1).Substituting these results into (3.2), we have

i∈I

πi

pA( pAωA + ωB) =

i∈I

ωA.

Solving this expression for pA, we obtain

pA = ωB

i∈I πi

ωA

i∈I (1 − πi), (3.3)

or with the given values (recall that |I | = 1000 by assumption),

pA = 20

i∈I π

i

10(1000 −

i∈I πi)

Remark 3.2. Returning to expression (3.3), observe that pA equals the ratio of the (common)endowment in state B to state A, times the ratio of the sum of probabilities assigned to stateA to the sum of probabilities assigned to state B. Note that pA increases (i) as relatively moreaggregate weight is assigned to state A than state B (agents place a greater premium on the good as

state A becomes more likely in aggregate), and (ii) as the relative endowment in state A decreases(relatively less good A is available compared to good B, so agents bid up its price).

As a final remark, notice the role of the common-endowment assumption in our analysis in part(b). If ωi

A or ω iB varied among agents, then we would have obtained

pA =

i∈I π

iωiB

i∈I (1 − πi)ωiA

.

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4 Exercise 4

One dollar is placed in a “pot” in period 0; its value will diminish by a discount factor δ eachperiod. (Thus, after k periods, the pot is worth δ k to both players.) Starting with player 1, the twoplayers alternate moves. When it is player i’s move, she has two choices: she can stop the game, inwhich case she collects 40% of the pot and the remaining 60% goes to her opponent; or she can lether opponent have the next move. The game continues until someone stops; if neither player everstops, both players receive a payoff of 0.

(a) Show that if δ is small enough, the only Nash equilibrium of the game is that player 1 choosesto stop the game immediately. Specify what is meant by “small enough”.

(b) Is there any value of δ such that, in some Nash equilibrium, someone stops the game aftereach player has declined to do so at least once?

(c) Show that if δ is large enough, there is a subgame perfect Nash equilibrium in which player 1does not stop the game, and player 2 does on the next turn. Specify what is meant by “largeenough”.


Player 1 stopping the game immediately (i.e., in period 0) is the unique Nash equilibrium outcomeif and only if, no matter what strategy player 2 uses, stopping the game immediately is player 1’sunique best response. Among all of player 2’s possible strategies, the ones that yield player 1 thegreatest utility are those in which player 2 stops the game on his first move (i.e., in period 1). Thusplayer 1 stopping the game immediately is the unique Nash equilibrium if and only if

.4 > .6δ ⇔ δ < 2

3.

Remark 4.1. When δ = 23 , both players are indifferent between stopping the game in a given

period and letting their opponent stop the game in the next period. It follows that the strategyprofile in which player 1 “passes” in each period she moves and player 2 stops the game in period1 (and takes either action in subsequent periods) is a Nash equilibrium, as is the strategy profile inwhich player 1 stops the game in period 0 and threatens to “pass” in each subsequent period andplayer 2 plans to stop the game in period 1 (and again takes either action in subsequent periods).Thus when δ = 2

3 , there are Nash equilibrium outcomes in which player 1 stops and does not stopthe game immediately.


No. Assume for the sake of contradiction that there did exist a Nash equilibrium in which someonestopped the game after each player has declined to do so at least once. Intuitively, the player whostops the game can strictly increase her payoff by stopping the game in the earlier period whenshe declined to stop the game. Since a strictly profitable unilateral deviation exists, the originalstrategy profile cannot be a Nash equilibrium.

More precisely, let t1 and t2 denote the first period in which player 1 and 2, respectively, declineto stop the game; note that in any profile in which both players have declined to stop the game at

9





least once, we must have t1 = 0 and t2 = 1. Suppose that player 1 stops the game in period t > 1.Her payoff is then .4δ t. However, if she deviates and stops the game in period 0, then her payoff

is .4. Provided that δ < 1, this deviation is strictly profitable, and therefore the original strategyprofile cannot be a Nash equilibrium. The argument if player 2 stops the game is analogous.


We remind ourselves of the definition of subgame-perfect Nash equilibrium.Consider the strategy profile in which player 1 “passes” after every history and player 2 stops the

game after every history. By the one-shot deviation principle, this strategy profile is a subgame-perfect Nash equilibrium if (and only if) there are no strictly profitable one-shot deviations foreither player. If player 1 one-shot deviates after any history ht, she will stop the game in thatperiod t and obtain payoff .4δ t (if that history realizes); if she does not deviate, then player 2’sstrategy specifies that he stops the game in the next period, yielding player 1 a payoff of .6δ t+1

(again, if that history realizes). Thus player 1’s deviation is not strictly profitable provided that

.4δ t ≤ .6δ t+1 ⇔ δ ≥ 2

3.

If player 2 one-shot deviates after any history ht, he will not stop the game in that period t, player1’s strategy specifies that she passes in period t + 1 (regardless of player 2’s actions), and player 2returns to his original strategy in period t + 2, so he stops the game then and receives payoff .4δ t+2

(if that history realizes); if he does not deviate, then he stops the game in period t and receivespayoff .4δ t (again, if that history realizes). Provided that δ < 1, not deviating always strictlydominates deviating for player 2.

We conclude that there are no profitable one-shot deviations, and hence the proposed strategyprofile is a subgame-perfect Nash equilibrium, if δ ≥ 2

3 .

5 Exercise 5

Bob has just insulted Mob in a bar, and Mob will challenge Bob to a fight. Mob must decidewhether to challenge Bob immediately or to leave and challenge him in a couple of hours in a quietplace. Then Bob must decide whether to run away or accept the fight. On a day when he usesdrugs, Mob is strong and beats Bob; on a drug-free day, Mob is weak and loses to Bob.

The payoffs are as follows. For Mob, 5 if Bob runs (either in the bar or outside); 10 (respectively,−10) if they fight outside and he wins (respectively, loses); and 20 (respectively, −30) if they fightin the bar and he wins (respectively, loses). For Bob, −10 if he runs (either in the bar or outside);3 (respectively, −6) if they fight outside and he wins (respectively, loses); and 5 (respectively, −15)if they fight in the bar and he wins (respectively, loses).

Mob knows whether he took drugs today. Bob does not know, but he was told by the bar ownerthat Mob uses drugs on average one day out of three.

(a) Describe this game in extensive form, then in matrix form. Eliminate dominated strategies,if any.

(b) Find all Bayesian Nash equilibria of this game.

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nd ( 2

3)d ( 1

3) P0

Od

Bd

[1.1] M

Ond

Bnd

M [1.2]

B [2.1]

B [2.2]

F b

,−1

Rb

,−1

F b

− ,

Rb

,−1

F o

10,−6

Ro

5,−10

F o

−10, 3

Ro

5,−10

Figure 5.1: Extensive form of the game in Exercise 5. Player labels: P0 denotes Nature, Mdenotes Mob, and B denotes Bob. Payoffs are listed as (Mob’s payoffs, Bob’s payoffs). Numbers inparentheses after Nature’s actions denote probabilities; numbers in square brackets denote labelsfor information sets.

RbRo RbF o F bRo F bF oBdBnd 5, −10 5, −10 −40

3 , −53

∗−40

3 , −53

∗

BdOnd 5, −10 −5, −43∗ ∗10, −35

3∗0, −3

OdBnd 5, −10 ∗ 203 , −26

3 −553 , 0 −50

3 , 43∗

OdOnd 5, −10 −103 , 0∗ 5, −10 −10

3 , 0∗

Figure 5.2: Normal form of the game in Exercise 5; payoffs listed in the form (Mob’s payoff, Bob’spayoff). Best responses are indicated by an asterisk (∗) next to the corresponding payoff.


Remark 5.1. Note that a strategy for Mob specifies two actions: whether to fight in the bar ( B)or outside (O) on days when he has taken drugs, and whether to fight in the bar or outside ondays when he has not taken drugs. Denote the conditioning events “drugs” and “no drugs” bysubscripts d and nd, respectively, on Mob’s actions. Similarly, a strategy for Bob specifies twoactions: whether to fight (F ) or run (R) when the fight occurs in the bar, and whether to fight

or run when the fight occurs outside. Denote the conditioning events “bar” and “outside” bysubscripts b and o, respectively, on Bob’s actions.

Throughout this exercise we will present payoffs in the order (Mob’s payoffs, Bob’s payoffs).We will assume that Mob must set his strategy before knowing whether he will take drugs that day(this assumption was implicit when we remarked that a strategy for Mob specifies two actions),and that the probabilities given by the bar owner are accurate.

The extensive form of the game is presented in Figure 5.1. The corresponding normal form,obtained from a weighted sum of the payoffs of the respective strategy profiles (with weight 1

3 on“drug” payoffs and weight 2

3 on “no drug” payoffs), is given in Figure 5.2.

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Analyzing the normal form in Figure 5.2, we observe that Bob’s pure strategy RbRo is strictlydominated by F bF o; also, F bRo is weakly dominated by F bF o. Less obviously, Mob’s pure strategy

OdOnd is dominated by the mixed strategy σOOM ( p) := p · BdOnd + (1 − p) · OdBnd for any p ∈ [45 , 67 ];the dominance is weak in the full 4 × 4 game, and strict for p ∈ (45 , 67) in the 4 × 3 reducedgame obtained by eliminating Bob’s strictly dominated pure strategy RbRo. The computations arepresented in Appendix A.


Remark 5.2. Since Bob’s pure strategy RbRo is strictly dominated, it will never be used in any(Bayesian) Nash equilibrium, and so we may restrict our attention to the 4 × 3 reduced gameobtained by eliminating RbRo. Moreover, in part (a) we found that Bob’s pure strategy F bRo isweakly dominated by F bF o; since Bob’s payoff from F bF o is strictly better than his payoff from F bRo

against all of Mob’s strategies except Bd

Bnd

(for which Bob’s payoffs from F b

F o

and F b

Ro

are equal),it follows that Bob will strictly prefer F bF o to F bRo whenever Mob plays BdBnd with probabilityless than 1. That is, provided Mob mixes nontrivially in equilibrium, we may further eliminateBob’s pure strategy F bRo and analyze the 4 × 2 reduced game without any loss of generality.

We present two methods for finding Bayesian Nash equilibria: analysis of the payoff matrix toobtain BNE in mixed strategies (from which the realization-equivalent behavior strategies can beinferred), and analysis of the extensive form to obtain BNE in behavior strategies directly.

Mixed-Strategy Approach We begin by looking for pure-strategy Bayesian Nash equilibria(BNE). Analyzing the pattern of best responses in Figure 5.2,5 we find no pure-strategy BNE.Better yet, we can show that Mob cannot use a pure strategy in any BNE.

In any BNE, given Bob’s strategy, Mob is playing a best response, corresponding to choosingthe pure strategy (or mixture of pure strategies) that maximizes his own payoff. Graphing Mob’spayoff as a function of the probability α that Bob places on RbF o as shown in Figure 5.3, Mob’sbest response corresponds to the upper envelope. We find two points where two or more of Mob’spure strategies intersect along this upper envelope: (i) α = 4

7 , where BdBnd and BdOnd yield thesame payoff; and (ii) α = 2

3 , where BdBnd and OdBnd yield the same payoff.First consider intersection (i) between Mob’s pure strategies BdBnd and BdOnd. In order for

Bob to be indifferent between his two pure strategies RbF o and F bF o, Mob’s mixture σ(i)M := β ◦

BdBnd + (1 − β ) ◦ BdOnd must satisfy

uB(σ(i)M , RbF o) = uB(σ

(i)M , F bF o)

β (−10) + (1 − β )

− 43

= β

− 5

3

+ (1 − β )(−3);

solving for β , we obtain β = 16 . Thus the strategy profile σ(i) :=(σ

(i)M , σ

(i)B ), where

σ(i)M :=

1

6 ◦ BdBnd +

5

6 ◦ BdOnd, σ

(i)B :=

4

7 ◦ RbF o +

3

7 ◦ F bF o,

5Note that the payoff line corresponding to Mob’s pure strategy OdOnd lies everywhere below the upper envelope.This illustrates our assertion in part (a) that OdOnd is a dominated strategy.

12





Figure 5.3: Mob’s expected payoff from each of his four pure strategies as a function of α, whenBob uses the mixed strategy α ◦ RbF o + (1 − α) ◦ F bF o.

is a BNE. The realization-equivalent behavior strategy profile is

σ(i)M :=

Bd,

1

6 ◦ Bnd +

5

6 ◦ Ond

, σ

(i)B :=

4

7 ◦ Rb +

3

7 ◦ F b, F o

,

and the associated payoffs are (−20

7 , −25

9 ).Next consider intersection (ii) between Mob’s pure strategies BdBnd and OdBnd. From Figure5.2, we observe that for both of these pure strategies, Bob’s pure strategy F bF o strictly dominatesRbF o (in fact, F bF o is Bob’s unique best response to either of these two pure strategies). Inparticular, Bob will never want to put positive probability on RbF o, so 2

3 ◦ RbF o + 13 ◦ F bF o cannot

be an equilibrium response to any valid mixture of BdBnd and OdBnd.6

We conclude that the unique BNE of this game is σ(i), detailed above.

Behavior-Strategy Approach As in the mixed-strategy approach, we begin by checking forpure-strategy Bayesian Nash equilibria (BNE). Analyzing the pattern of best responses in thepayoff matrix in Figure 5.2, we find that no pure-strategy BNE exist.

Thus we look for BNE in behavior strategies involving strict mixing in at least one informationset for each player. To do this, we assign general probabilities to the actions in each information set,

6If we were to follow the analysis performed for intersection (i), we would proceed as follows: In order for Bob to

be indifferent between his two pure strategies RbF o and F bF o, Mob’s mixture σ(ii)M := γ ◦ BdBnd + (1 − γ ) ◦ OdBnd

must satisfy

uB(σ(ii)M , RbF o) = uB(σ

(ii)M , F bF o)

γ (−10) + (1 − γ )

„−

26

3

« = γ

„−

5

3

« + (1 − γ )

„4

3

«;

solving for γ , we obtain γ = 6. Since this is not a valid probability, we conclude that Mob cannot mix over his twopure strategies BdBnd and OdBnd in such a way that Bob is indifferent between RbF o and F bF o.

13





and derive conditions on these probabilities such that the opponent is indifferent between certainof his pure strategies (and hence willing to mix).

Prior to assigning general probabilities, however, let us see if we can simplify our analysis byeliminating dominated actions in a given information set. Consider information set [2.2] in Figure5.1. Given that Bob finds himself in information set [2.2], F o is strictly dominant to Ro given hisbeliefs about the probabilities on Nature’s actions (d,nd) (indeed, given any valid probabilities forthese beliefs). Thus Bob will play F o with probability 1 at information set [2.2]. We can check thatthere are no other dominated actions.

Let β d and β nd denote the probability that Mob plays Bd and Bnd, respectively; let ρb denotethe probability that Bob plays Rb. For Bob to be willing to play a strictly mixed strategy atinformation set [2.1], it must be the case that the expected payoffs from the actions over whichhe mixes are equal, given Mob’s strategy (i.e., the probabilities β d and β nd) and his beliefs aboutthe probabilities on Nature’s actions (d,nd). Let 2.1.d and 2.1.nd denote the left and right node,

respectively, in information set [2.1]. Then the condition that Bob’s expected payoffs from his twopure actions Rb and F b are equal, conditional on being in information set [2.1], writes as

E[uB((β d, β nd), Rb) | [2.1]] = E[uB((β d, β nd), F b) | [2.1]]

−10 = P(2.1.d)uB(Bd, F b) + P(2.1.nd)uB(Bnd, F b)

The second line follows from the fact that, conditional on being in information set [2 .1], Mob musthave played Bd or Bnd, so the payoffs are those corresponding to actions in information set [2.1];moreover, if Bob plays Rb, then his payoff is −10 with probability 1. Using Bayes’ rule to computethe probability of being at each node in information set [2.1], conditional on being in informationset [2.1], we obtain

−10 =

1

3

β d13β d + 2

3β bd (−15) +

2

3

β nd

13β d + 2

3β bd (5).

Moving the denominator to the left and multiplying both sides by 3, we obtain

−10β d − 20β nd = −15β d + 10β nd,

which simplifies to

β d = 6β nd. (5.1)

We next find conditions on ρb such that Mob is willing to mix at information sets (nodes) [1.1]and [1.2]. For Mob to be willing to mix at information set [1.1], we must have

E[uM (Bd, (ρb, 1)) | [1.1]] = E[uM (Od, (ρb, 1)) | [1.1]]

ρb(5) + (1 − ρb)(20) = 10

ρb = 2

3. (5.2)

For Mob to be willing to mix at information set [1.2], we must have

E[uM (Bnd, (ρb, 1)) | [1.1]] = E[uM (Ond, (ρb, 1)) | [1.1]]

ρb(5) + (1 − ρb)(−30) = −10

ρb = 4

7. (5.3)

14





2

3

1

3 P0

0

1

[1.1] M

5

6

1

6

M [1.2]

B [2.1]

B [2.2]

37

1

7

47

4

21

37

1

21

47

4

63

1

0

0

0

1

5

9

0

0

Figure 5.4: Extensive form of the game in Exercise 5, depicting the probabilities over outcomesinduced by the unique BNE σ. Player labels: P0 denotes Nature, M denotes Mob, and B denotesBob. Numbers on branches indicate the probability with which those branches are played; numbersat terminal nodes indicate the probability that outcome is realized; numbers in square bracketsdenote labels for information sets.

Note that (5.2) and (5.3) are mutually exclusive. This does not mean that one (or both) of ourcomputations are wrong, but rather that Mob is willing to strictly mix at at most one of his twoinformation sets (exactly one when one of (5.2) and (5.3) hold).

In particular, if ρb = 23 , so that Mob is willing to mix at information set [1.1], then the expected

payoff from his actions in information set [1.2] satisfy

E[uM (Bnd, (ρb, 1)) | [1.2]] = −20

3 > −10 = E[uM (Ond, (ρb, 1)) | [1.2]],

so that Mob strictly prefers Bnd to Ond. That is, ρb = 23 implies β nd = 1. However, result (5.1)

then implies that β d = 6, which is not a valid probability. It follows that ρb = 23 does not yield a

valid BNE.If ρb = 4

7 , so that Mob is willing to mix at information set [1.2], the the expected payoff fromhis actions in information set [1.1] satisfy

E[uM (Bd, (ρb, 1)) | [1.1]] = 80

7 > 10 = E[uM (Od, (ρb, 1)) | [1.1]],

so that Mob strictly prefers Bd to Od. That is, ρb = 47 implies β d = 1. Result (5.1) then implies

that β nd = 16 . Combining these results, we conclude that the unique BNE is σ = (σM , σB), where

σM :=

Bd,

1

6 ◦ Bnd +

5

6 ◦ Ond

, σB :=

4

7 ◦ Rb +

3

7 ◦ F b, F o

.

A game tree depicting the probabilities over outcomes induced by the strategy profile σ is presentedin Figure 5.4. The payoffs associated with this BNE are (−20

7 , −259 ).

15





6 Exercise 6

Consider the following interaction between a seller and a buyer. There is an object worth a to theseller and b to the buyer. Both a and b are i.i.d. on the interval [0, 300], with uniform distribution.The players play a sealed-bid double auction game: they independently and simultaneously sendan ask price x (seller) and an offer price y (buyer). If x > y, then no trade takes place; if x ≤ y ,then trade takes place at the price p = x+y

2 .

(a) Given α ∈ [0, 300], consider the following pair of strategies:

x(a) =

α if a ∈ [0, α],

300 if a ∈ (α, 300];y(b) =

0 if b ∈ [0, α),

α if b ∈ [α, 300].

Show that the strategy pair (x, y) constitutes a Bayesian Nash equilibrium. Show that the

expected gain is α(300−α)2

2(300)2 for the buyer and α2(300−α)

2(300)2 for the seller. Compute the total welfare

loss (compared to the first-best) and show that it is never less than 25%.

(b) Consider the following take-it-or-leave-it mechanism. The seller chooses a price x ∈ [0, 300],which the buyer than accepts or rejects. Compute the unique Bayesian Nash equilibrium,and compare its welfare loss to that found in part (a).

(c) Consider the following mechanism. The seller and buyer independently submit bids x and y ,respectively. Then

• trade occurs at price y2 if y ≥ 3x and x + y ≤ 300;

• trade occurs at price x2 + 150 if y ≥

x3 + 200 and x + y > 300;

• no trade occurs, and no money changes hands, in all other cases.

Show that sincerely reporting one’s valuation (i.e., the strategies x(a) = a and y(b) = b for alla, b ∈ [0, 300]) is a Bayesian Nash equilibrium. Compare the welfare loss of this mechanismto those found in parts (a) and (b).


Remark 6.1. Before we begin the mathematical analysis, let’s ask ourselves what is happening withthe economics. The mechanism in part (a) operates as follows: If the seller’s reported valuationis higher than the buyer’s reported valuation, then no trade takes place; if the seller’s reported

valuation is less than or equal to the buyer’s reported valuation, then trade takes place at theaverage of the reported values. The given strategy profile has the seller report a fixed valuation α

whenever her true valuation does not exceed α, and has her report the maximum possible valuation300 if her true valuation exceeds α; similarly, it has the buyer report the lowest possible valuation 0if his true valuation is less than α, and has him report α when his true valuation is at least α. Notethat in this strategy profile, the seller never reports a valuation less than α, and the seller neverreports a valuation greater than α. Given the mechanism and the other player’s strategy, it seemsreasonable that each player’s strategy is a best response. Our job is to prove this mathematically.

16





To show that the given strategy pair (x, y) is a Bayesian Nash equilibrium, we must prove thatx is a best response to y given the seller’s belief about the buyer’s valuation, and vice versa. Here

we perform a relatively simple checking procedure; a more general approach is detailed in AppendixB.

First, fix the buyer’s strategy y, and consider the seller. If trade takes place, then the seller’spayoff is p − a; if trade does not take place, then the seller’s payoff is 0.7 If the buyer realizesa valuation b ∈ [0, α), then he submits an offer price of y(b) = 0, so that by the rules of the

mechanism, trade takes place if and only if x(a) ≤ 0, in which case the price is p = x(a)+02 ≤ 0 and

the seller’s payoff is p − a. Since a ≥ 0, we have p − a ≤ 0, so trade is not optimal for the seller inthis case;8 the seller can preclude trade from occurring by submitting any ask price x(a) > 0.

If the buyer realizes a valuation b ∈ [α, 300], then he submits an offer price of y(b) = α. By therules of the mechanism, trade takes place if and only if the seller submits any ask price x(a) ≤ α,

in which case the price is p = x(a)+α

2 and the seller’s payoff is p − a. If trade takes place, then the

seller wants to maximize the price she receives, which corresponds to maximizing the ask price shesubmits subject to the condition that the mechanism specifies trade, given the buyer’s offer price.As we noted, trade takes place if x(a) ≤ α, so the seller maximizes p by setting x(a) = α (yielding p = α) when trade occurs. Given that the buyer submits an offer price of α, trade is only profitableto the seller if p − a ≥ 0, i.e. if a ≤ p = α. In particular, when a > α, the seller never wantstrade to take place.9 Given y, she can preclude trade from occurring by submitting any ask pricex(a) > α whenever a > α.

The results of the previous two paragraphs show that the following strategy is a best responseof the seller to the buyer’s strategy y:

x(a) = α if a ∈ [0, α],

pS (a) if a ∈ (α, 300].In particular, setting pS (a) = 300 for all a ∈ (α, 300] works.

Next, fix the seller’s strategy x, and consider the buyer. The analysis follows the same line of reasoning that we used for the seller. If trade takes place, then the buyer’s payoff is b − p; if tradedoes not take place, then the buyer’s payoff is 0. If the seller realizes a valuation a ∈ (α, 300], thenshe submits an ask price of x(a) = 300, so that by the rules of the mechanism trade takes place if

and only if y(b) ≥ 300, in which case the price is p = 300+y(b)

2 ≥ 300 and the buyer’s payoff is b − p.Since b ≤ 300, we have b − p ≤ 0, so trade is not optimal for the buyer in this case; the buyer canpreclude trade from occurring by submitting any offer price y(b) < 300.

If the seller realizes a valuation a ∈ [0, α], then she submits an ask price of x(a) = α, so thatby the rules of the mechanism, trade takes place if and only if y(b) ≥ α, in which case the price is

p = α+y(b)2 and the buyer’s payoff is b − p. If trade takes place, then the buyer wants to minimize the

7These payoffs may seem surprising — for example, why isn’t the seller’s payoff a when she keeps the object?The answer lies in our point of reference. If we assume the seller possesses the object before entering the mechanism,then when trade takes place she gains p and loses a, so her payoff is p − a; and when trade does not take place,she gains nothing and loses nothing, so her payoff is 0. (These are the payoffs given in the text.) If we assume themechanism designer possesses the object prior to the mechanism, then when trade takes place the seller gains p andloses nothing, so her payoff is p; and when trade does not take place, she receives the object, so her payoff is a. Ineither case, we obtain the expected utility given in (B.1), up to a constant (namely, a).

8When a = 0 (which occurs with probability 0), the seller is indifferent between trade at price p = 0 and no trade.9Suppose a > α. Given the buyer’s strategy y, the maximum offer price is α, so for trade to take place the seller

must submit an ask price x(a) ≤ α, in which case the price is p ≤ x(a)+α2 ≤ α < a, so that the seller’s payoff p−a < 0.

17





price he pays, which corresponds to minimizing the offer price he submits subject to the conditionthat the mechanism specifies trade, given the seller’s ask price. The buyer minimizes p by setting

y(b) = α (yielding p = α) when trade occurs. Given that the seller submits an ask price of α, tradeis only profitable to the seller if b − p ≥ 0, i.e. if b ≥ p = α. In particular, when b < α, the buyernever wants trade to take place.10 Given x, he can preclude trade from occurring by submittingany offer price y(b) < α whenever b < α.

Thus, a best response of the buyer to the seller’s strategy x is

y(b) =

0 if b ∈ [0, α),

pB(b) if b ∈ [α, 300].

In particular, setting pB(b) = 0 for all b ∈ [0, α) works.We have shown that the strategies x and y are mutual best responses; thus, (x, y) is a Bayesian

Nash equilibrium. Let A = B = [0, 300] denote the set of valuations of the seller and buyer,respectively. The set T :={(a, b) ∈ A × B | x(a) ≤ y(b)} in which trade takes place is shown inFigure 6.1(a). The seller’s expected surplus from the strategy profile (x, y) under this mechanismis

E[uS (x, y)] =

A

B

T (x(a), y(b))( p − a) dF B(b) dF A(a)

=

α0

300α

(α − a)

1

300 db

1

300 da

= 300 − α

3002

α0

(α − a) da

= α

2

(300 − α)2(300)2 . (6.1)

Similarly, the buyer’s expected surplus from the strategy profile (x, y) under this mechanism is

E[uB(x, y)] =

B

A

T (x(a), y(b))(b − p) dF A(a) dF B(b)

=

300α

α0

(b − α)

1

300 da

1

300 db

= α

3002

300α

(b − α) db

= α

(300 −α

)

2

2(300)2 . (6.2)

Whenever trade occurs, the total surplus generated equals the sum of the buyer’s and the seller’ssurplus; for this mechanism, this sum is

(b − p) + ( p − a) = b − a.

10Suppose b < α. Given the seller’s strategy x, the minimum ask price is α, so for trade to take place the buyermust submit an ask price y(b) ≥ α, in which case the price is p ≥ α+y(b)

2 ≥ α > b, so that the buyer’s payoff is

b − p < 0.

18





(a) (b) (c)

Figure 6.1: Regions of trade (shaded) for the three mechanisms described in Exercise 6.

From an efficiency standpoint, trade should take place whenever the buyer’s valuation (weakly)exceeds the seller’s valuation, i.e. whenever b ≥ a. Graphically, this condition is satisfied by allpoints in the closed triangular region above (and including) the 45-degree line. Welfare loss ariseswhen there are regions in which trade should take place but does not. For the mechanism in part(a), welfare loss arises from the two smaller triangular regions above the 45-degree line in Figure6.1(a).

We can compute the expected welfare loss of the mechanism Γ(a) under the strategy profile(x, y) with parameter α — let us denote it L(Γ(a), (x, y), α) — in two ways. One is by directlycomputing the integrals over the regions in which welfare loss arises:

L(Γ(a), (x, y), α) =

α0

αa

(b − a) dF B(b) dF A(a) +

300α

300a

(b − a) dF B(b) dF A(a)

= 1

3002

1

6α3 +

1

6(300 − α)3

=

α2 − 300α + 30000

600 .

Alternatively, we can compute the total expected surplus from the first-best (in which trade takesplace if and only if it is efficient),

S fb =

3000

300a

(b − a) dF B(b) dF A(a) = 50,

and subtract off the total expected surplus generated by the mechanism Γ (a) under the strategyprofile (x, y) with parameter α, found by summing (6.1) and (6.2),

S (Γ(a), (x, y), α) = α(300 − α)

600 ;

the difference is the expected welfare loss,

L(Γ(a), (x, y), α) = S fb − S (Γ(a), (x, y), α) = 50 − α(300 − α)

600 =

α2 − 300α + 30000

600 .

19





Note that the expected welfare loss is a convex quadratic in α; the minimizing value of α is foundby setting the first derivative equal to zero and solving for α:

1

600(2α − 300) =

set0 ⇔ α = 150,

corresponding to the minimum value

L(Γ(a), (x, y), 150) = 25

2 .

As a fraction of the total possible (i.e. first-best) expected welfare, this minimum expected welfare

loss is 25/250 = 1

4 , as we wished to show.


Given a price x(a), the buyer’s best response is to accept if b ≥ x(a),11 and to reject otherwise;denote this strategy by y . Knowing this, the seller’s expected utility from naming the price x(a) ∈[0, 300] is

E[uS (x, y) | a] =

B

T (x(a), y(b))(x(a) − a) dF (b),

where T :={(a, b) | x(a) ≤ y(b)}, as in part (a); note that here, x(a) is a constant for each givena ∈ A. The integral becomes

E[uS (x, y) | a] =

300x(a)

(x(a) − a)

1

300 db

=

1

300(x(a) − a)(300 − x(a)).

This is a concave quadratic in x(a); it is maximized when

1

300(−2x(a) + 300 + a) =

set0 ⇔ x(a) = 150 +

a

2.

Thus the unique Bayesian Nash equilibrium is

x(a) = 150 + a

2, y(b) =

accept if b < x(a),

reject if b ≥ x(a).

The total expected welfare (see Figure 6.1(b)) is

S (Γ(b)

, (x, y)) = 3000

300150+a(b − a) dF B(b) dF A(a) =

25

2 .

Equivalently, the expected welfare loss of this mechanism is

L(Γ(b), (x, y)) = S fb − S (Γ(b), (x, y)) = 75

2 ,

three times greater than the minimum expected welfare loss of the mechanism Γ (a) under thestrategy profile (x, y) and the expected-welfare-maximizing parameter value α = 150.

11Since, given a cdf with no mass points, the indifference case b = x occurs with probability zero, we need notconcern ourselves with indifference.

20






The utilities are described in Figure 7.2. Note that the player’s payoffs agree everywhere except inthe state (θ1 = 1, θ2 = 1).

Agent 1’s payoffs:

θ2 = 0 θ2 = 1

θ1 = 0 1 0θ1 = 1 0 1

a = 0

θ2 = 0 θ2 = 1

θ1 = 0 0 1θ1 = 1 1 0

a = 1

Agent 2’s payoffs:

θ2 = 0 θ2 = 1

θ1 = 0 1 0θ1 = 1 0 0

a = 0

θ2 = 0 θ2 = 1

θ1 = 0 0 1θ1 = 1 1 0

a = 1

Figure 7.2: Payoffs for the game in Exercise 7.


There are two commonly used notions of incentive compatibility. The stronger notion12 requiresthat, for each agent, truth-telling be a dominant strategy, i.e. a best response regardless of thestrategy used by the other agent(s). We use this stronger notion in what follows.

Using the notation described in the statement of Exercise 7, the four mechanisms we have toconsider (respecting the decision imposed in states (0, 0) and (1, 1)) can be described as

(0, 0, 0, 1), (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 1). (7.1)

Denote these mechanisms by Γ1, . . . , Γ4, respectively. Let ui(Γk(θ1, θ2); (θ1, θ2)) denote the payoff to agent i under mechanism Γk when the reported type profile is (θ1, θ2) and the true type profile is(θ1, θ2). Let σs

i denote agent i’s “sincere” strategy in which she always truthfully reports her type,and let σm

i denote agent i’s “misreport” strategy in which she always lies about her type.Consider the scenario under Γ1 in which agent 1’s true type is θ1 = 0 and agent 2 uses his

sincere strategy. Agent 1’s expected payoff when she truthfully reports her type is

E[u1(σs1, σs

2) | θ1 = 0] = P(θ2 = 0)u1

Γ1(0, 0);(0, 0)

+ P(θ2 = 1)u1

Γ1(0, 1);(0, 1)

=

1

2 u1(0;(0, 0)) +

1

2 u1(0;(0, 1)) =

1

2 (1) +

1

2 (0) =

1

2 ,

whereas her expected payoff if she misreports her type is

E[u1(σm1 , σs

2) | θ1 = 0] = P(θ2 = 0)u1

Γ1(1, 0);(0, 0)

+ P(θ2 = 1)u1

Γ1(1, 1);(0, 1)

=

1

2u1(0;(0, 0)) +

1

2u1(1;(0, 1)) =

1

2(1) +

1

2(1) = 1.

12The weaker notion of incentive compatibility is a Bayesian Nash equilibrium notion of incentive compatibility; amechanism is incentive compatible if truthfully reporting one’s type is a best response, given that the other agent(s)is truthfully reporting her type.

22





We see that agent 1 has a strict incentive to misreport her type when she is type θ1 = 0 and agent2 uses his sincere strategy. Thus truth-telling is not a dominant strategy for agent 1 in mechanism

Γ1. An analogous argument shows that truth-telling is not a dominant strategy for agent 1 inmechanism Γ4.13

To be completed.


The short answer is that each agent’s expected utility depends on the strategy used by her opponent.We consider two approaches to providing a response to which mechanism is “most favorable”: wefind each agent’s maximum payoff under each mechanism provided that (i) her opponent alwaystruthfully reports her type, and (ii) that the agents play a Bayesian Nash equilibrium.

Mutual truth-telling is a Bayesian Nash equilibrium of mechanism Γ4 = (0, 1, 1, 1), guaranteeingboth players an ex ante expected payoff of

E[ui(σs1, σs

2)] =θ∈Θ

P(θ)ui(Γ4(θ); θ)

= 1

4ui(0;(0, 0)) +

1

4ui(1;(0, 1)) +

1

4ui(1;(1, 0)) +

1

4ui(1;(1, 1))

= 3

4.

To be completed.

References

Mas-Colell, Andreu, Whinston, Michael D., & Green, Jerry R. 1995. Microeconomic Theory . NewYork: Oxford University Press.

A Computations for Mixed-Strategy Dominance (Exercise 5)

To show that Mob’s pure strategy OdOnd is dominated by σOOM ( p) := p · BdOnd + (1 − p) · OdBnd, we

solve the four inequalities implied by the requirement that E[uM (σOOM ( p), sB)] ≥ E[uM (OdOnd, sB)]

13More precisely, consider the scenario under Γ4 in which agent 1’s true type is θ1 = 1 and agent 2 uses hismisreporting strategy. Agent 1’s expected payoff when she truthfully reports her type is

E[u1(σs1, σ

m2 ) | θ1 = 1] = P(θ2 = 0)u1

`Γ4(1, 1); (1, 0)

´ + P(θ2 = 1)u1

`Γ4(1, 0); (1, 1)

´=

1

2u1(1; (1, 0)) +

1

2u1(1; (1, 1)) =

1

2(1) +

1

2(0) =

1

2,

whereas her expected payoff if she misreports her type is

E[u1(σm1 , σ

m2 ) | θ1 = 1] = P(θ2 = 0)u1

`Γ1(0, 1); (1, 0)

´ + P(θ2 = 1)u1

`Γ4(0, 0); (1, 1)

´=

1

2u1(1; (1, 0)) +

1

2u1(0; (1, 1)) =

1

2(1) +

1

2(1) = 1.

23





for all of Bob’s pure strategies sB . In particular, the three nontrivial conditions are

p(−5) + (1 − p)

203

≥ −10

3 ⇔ p ≤ 6

7

p(10) + (1 − p)

−

55

3

≥ 5 ⇔ p ≥

8

17

p(0) + (1 − p)

−

50

3

≥ −

10

3 ⇔ p ≥

4

5;

the final condition, for sB = RbRo, is always trivially satisfied. The conditions are simultaneouslysatisfied for all p ∈ [45 , 67 ].

Since Mob’s pure strategy BdBnd is also not a best response to any of Bob’s pure strategies,we might suspect that BdBnd is dominated by σBB

M (q ) := q · BdOnd + (1 − q ) ◦ OdBnd for some

q ∈ Q ⊂ [0, 1]. Solving the four inequalities implied by the requirement that E

[uM (σBB

M ( p), sB)] ≥E[uM (BdBnd, sB)] for all of Bob’s pure strategies sB , we obtain the three nontrivial conditions

q (−5) + (1 − q )

20

3

≥ 5 ⇔ q ≤

1

7

q (10) + (1 − q )

−

55

3

≥ −

40

3 ⇔ q ≥

3

17

q (0) + (1 − q )

−

50

3

≥ −

40

3 ⇔ q ≥

1

5;

the final condition, for sB = RbRo, is always trivially satisfied. Since 17 < 1

5 , the conditions arenever simultaneously satisfied.

B General Approach to Mechanism Design (Exercise 6)

Let A = B = [0, 300] denote the set of valuations of the seller and buyer, respectively; let x : A → A

and y : B → B denote generic strategies for the seller and buyer, respectively; and let T ⊆ A × B

denote the set of valuation pairs (a, b) for which the mechanism specifies that trade takes place.First consider the seller. If trade takes place, then the seller’s payoff is p − a; if trade does not

take place, then the seller’s payoff is 0.14 Thus, given her valuation a, the expected payoff to theseller from the strategy x when the buyer uses strategy y is

E[uS (x, y) | a] =

B [

T (x(a), y(b))( p − a) + (1 −

T (x(a), y(b)))0]dF B(b)

=

B

T (x(a), y(b))( p − a) dF B(b). (B.1)

14These payoffs may seem surprising — for example, why isn’t the seller’s payoff a when she keeps the object?The answer lies in our point of reference. If we assume the seller possesses the object before entering the mechanism,then when trade takes place she gains p and loses a, so her payoff is p − a; and when trade does not take place,she gains nothing and loses nothing, so her payoff is 0. (These are the payoffs given in the text.) If we assume themechanism designer possesses the object prior to the mechanism, then when trade takes place the seller gains p andloses nothing, so her payoff is p; and when trade does not take place, she receives the object, so her payoff is a. Ineither case, we obtain the expected utility given in (B.1), up to a constant (namely, a).

24





For the mechanism in part (a), T = {(a, b) | x(a) ≤ y(b)} and p = x(a)+y(b)

2 . Also, b ∼ U [0, 300],independent of the seller’s valuation, so we have dF B(b) = 1

300 db. The seller’s problem is to choose

a strategy x that maximizes (B.1), given the buyer’s strategy y . Our approach will be to maximizethis integral pointwise, i.e. for each value of the buyer’s valuation b ∈ B. Essentially, the sellerwants to set T (x(a), y(b)) = 0 if the integrand p − a is nonpositive for every possible choice of x(a); and set T (x(a), y(b)) = 1, and choose x(a) to maximize the integrand p − a, otherwise.

Consider the particular buyer’s strategy y specified in (a). If b ∈ [0, α), then y(b) = 0; for thesevalues of b, trade takes place if and only if x(a) = 0, in which case p = 0. Given any a > 0, theseller strictly prefers not to trade (corresponding to setting the integrand of (B.1) equal to 0) thanto trade (setting the integrand equal to p − a < 0). Given a = 0, which occurs with probability0, the seller is indifferent between not trading and trading. Given the buyer’s particular strategyy, trade will never take place for b ∈ [0, α) provided the seller always reports a strictly positivevaluation. That is, if x(a) > 0 for all a ∈ A, then the integrand in (B.1) is 0 for all b ∈ [0, α). If

b ∈ [α, 300], then y(b) = α, and therefore p = x(a)+α

2 . The seller’s problem is therefore to maximizethe expected payoff

E[uS (x, y) | a] =

300α

T (x(a), α)

x(a) + α

2 − a

1

300 db.

The integrand is nonnegative when

x(a) + α

2 − a ≥ 0 ⇔ x(a) ≥ 2a − α.

The seller will want to set T (x(a), α) = 1, and maximize x(a) to maximize the price she receives,whenever this inequality is satisfied. However, remember that the mechanism specifies that tradetakes place if and only if the seller’s reported valuation does not exceed the buyer’s reportedvaluation. The buyer’s reported valuation is y(b) = α for all b ∈ [α, 300], which gives us a secondinequality,

x(a) ≤ α.

Combining these two results, we have that the seller’s best response is to trade whenever 2a − α ≤x(a) ≤ α. It follows that the seller’s best response is to report the maximum value at which tradestill occurs, i.e. x(a) = α, for all a ≤ α. When a > α, the seller’s best response is to report anyvaluation that guarantees trade does not occur; x(a) = 300 is one such report.

The analysis for the buyer is similar. The buyer’s expected utility, given his valuation b, is

E[uB(x, y) | b] =

A

T (x(a), y(b))(b − p) dF S (a).

Given the seller’s particular strategy x specified in part (a), this expression becomes

E[uB(x, y) | b] =

α0

T (α, y(b))

b −

α + y(b)

2

1

300 da

+

300α

T (300, y(b))

b −

300 + y(b)

2

1

300 da. (B.2)

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In the first integral, trade takes place if and only if y(b) ≥ x(a) = α; and the integrand is nonnegative

if and only if b − α+y(b)2 ≥ 0, which rearranges to y(b) ≤ 2b − α. Combining these two results, we

have

α ≤ y(b) ≤ 2b − α. (B.3)

The buyer maximizes his payoff when the price is as low as possible, which corresponds to choosingthe minimum reported valuation y(b) such that trade takes place, namely, y(b) = α. This reportedvaluation satisfies the upper bound of (B.3) provided b ≥ α. Turning our attention to the secondintegral in (B.2), we obtain the inequalities

300 ≤ y(b) ≤ 2b − 300,

which is satisfied for the minimizing reported valuation y(b) = 300 if and only if b ≥ 300. It follows

that a best response for the buyer to the particular seller strategy y is

x(b) =

0 if b ∈ [0, α),

α if b ∈ [α, 300].

To summarize, we have found that x is a best response to y, and that y is a best response tox, given the beliefs about the distribution of types. That is, (x, y) is a Bayesian Nash equilibrium,as we wished to show.

microqual 2013 spring solutions

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