Graduate Microeconomics IILecture 11 - Surplus Extraction

Patrick Legros

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Outline

Surplus extraction

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Surplus extractionNature of the result

Full rent extraction result: Cremer-Mc Lean (1985), (1988) showthat as long as agents have correlated information a mechanismdesigner can extract this information without giving agents a rent.

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Surplus extractionCaveats

I Robert (“Continuity in Auction Design,” JET 2001) showsthat in auctions when agents are risk averse and have limitedliability, the relation between the common prior and theseller’s profit is continuous in the prior: hence for “nearlyindependent” types, full rent extraction fails with risk averseagents and limited liability.

I Constraints on communication

I Collusion; Laffont-Martimort (2000)I Interference: Legros-Newman (2003)

I Residual uncertainty

I Auriol-Laffont (1992)I Neeman (2003) shows that Cremer-Mc Lean result depends on

the assumption that beliefs pin down precisely the agents’preferences

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Two bidders

I T 1 = T 2 = {0, 1}I Suppose types are i.i.d., let πi = π

(t i = 1

). Suppose

πi ∈ (0, 1) for all i = 1, 2.

I Consider a Bayesian auction (p, x) where p (i , j) is theprobability that 1 gets the object when types are (i , j) .

I Full extraction requires that the auction is efficient. Hence,p (1, 0) = 1, p (0, 1) = 0.

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IR and full extraction require that

π2[p (1, 1)− x1 (1, 1)

]+(1− π2

) [1− x1 (1, 0)

]= 0 (1)

π2[−x1 (0, 1)

]+(1− π2

) [−x1 (0, 0)

]= 0 (2)

π1[1− p (1, 1)− x2 (1, 1)

]+(1− π1

) [1− x1 (0, 1)

]= 0 (3)

π1[−x2 (1, 0)

]+(1− π1

) [−x2 (1, 0)

]= 0 (4)

IC requires for t1 = 1

π2[p (1, 1)− x1 (1, 1)

]+(1− π2

) [1− x1 (1, 0)

]≥ π2

[−x1 (0, 1)

]+(1− π2

) [p (0, 0)− x1 (0, 0)

]hence from (1)-(2)

0 ≥(1− π2

)p (0, 0)

⇒ p (0, 0) = 0

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IC requires for t2 = 1

π1[1− p (1, 1)− x2 (1, 1)

]+(1− π1

) [1− x1 (0, 1)

]≥ π1

[−x2 (1, 0)

]+(1− π1

) [1− p (0, 0)− x2 (1, 0)

]or using (3) and (4)

0 ≥(1− π1

)[1− p (0, 0)]

⇒ p (0, 0) = 1

a contradiction.

Hence full extraction of the surplus is impossible with i.i.d.

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Consider now the same type space, same valuations but aninformation structure with correlation.

For instance, assume that

π (1, 1) = π (0, 0) = 1/3, π (1, 0) = π (0, 1) = 1/6.

Hence the conditional probabilities are

π (1|1) = π (0|0) = 2/3,

π (0|1) = π (1|0) = 1/3.

Total surplus is

S = [π (1, 1) + π (1, 0) + π (0, 1)]× 1 =2

3.

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The matrix of conditional probabilities is

[2/3 1/31/3 2/3

]has full

rank.

Therefore it is possible to find z i (0) , z i (1) such that[2/3 1/31/3 2/3

] [z1 (0)z1 (1)

]=

[0

23p (1, 1) + 1

3

].

and [2/3 1/31/3 2/3

] [z2 (0)z2 (1)

]=

[0

23 (1− p (1, 1)) + 1

3

]

Note that each system corresponds to binding IR constraints andp (1, 0) = p (0, 1) = 1. Therefore, there is full extraction of thesurplus.

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Surplus extractionCremer-Mc Lean (Econometrica 1988)

Auction

I seller has valuation 0I type of agent i : t i ∈ T i =

{t i1, ..., t

imi

}finite set

I buyer i of type ti has valuation v i(t i)

I if buyer i gets the object with probability pi and pays x i , hasutility piv i

(t i)− x i

Notation

I T = ×i∈NT i

I T−i = ×j∈N\{i}Tj , typical element is t−i

I t =(t i , t−i

)I t\t j =

(t j , t−j

)Beliefs

I (Common) prior belief π on TI buyer learns (interim stage) t i and has belief π

(t−i |t i

)I marginal π

(t i)> 0 for all i , t i ∈ T i (hence Bayes law for

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Auction is then(pi , x i

)where pi : T → ∆ (n) and x i : T → R

Basic constraint: interim IR (use revelation principle)∑t−i∈T−i

π(t−i |t i

) (pi (t) v

(t i)− x i (t)

)≥ 0.

Assume v i > 0 all i ∈ N

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Dominant strategy IC: for all t1 (“true type”), for all t1 ∈ T 1, allt ∈ T

p1 (t) v1(t1)− x1 (t) ≥ p1

(t\t1

)v1(t1)− x1

(t\t1

)[p1 (t)− p1

(t\t1

)]v1(t1)≥ x1 (t)− x1

(t\t1

)

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Bayesian Nash IC: for all t1 (“true type”), for all t1 ∈ T 1,

∑t−i∈T−i

π(t−i |t1

) [p1 (t) v1

(t1)− x1 (t)

]≥

∑t−i∈T−i

π(t−i |t1

) [p1(t\t1

)v1(t1)− x1

(t\t1

)]or ∑

t−i∈T−i

π(t−i |t1

) [p1 (t)− p1

(t\t1

)]v1(t1)

≥∑

t−i∈T−i

π(t−i |t1

) [x1 (t)− x1

(t\t1

)]

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Seller’s surplus is∑

t∈T

∑i∈N x i (t) . There is full extraction of

the surplus at (T , π) if for any v∑t∈T

∑i∈N

x i (t) =∑t∈T

π (t) max{v i(t i)

; i ∈ N}

(5)

Since v i > 0, full extraction requires that for all t ∈ T ,∑i∈N

pi(t i)

= 1. (6)

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Surplus extractionDominant Strategy

Remember Vickrey auction: dominant strategy because own payoffis independent of one’s announcement: for each i , t i 6= t i ,x i(t\t i

)= x i

(t\t i

)Use separation theorem

Proposition: An information structure guarantees full extraction ofthe surplus by a dominant strategy auction if and only if for alli ∈ N there do not exist

(αi(t i)

; t i ∈ T i)6= 0 such that∑

t i∈T i

αi(t i)π(t−i |t i

)= 0, all t−i ∈ T−i .

Equivalent to saying that the matrix of conditional probabilitiesπ(t−i |t i

)has full rank (equal to the dimension of T i ).

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Consider a Vickrey auction (p, x) satisfying IR and letui(t i)

=∑

t−i π(t−i |t i

) [pi (t) v i

(t i)− x i (t)

]. Know that

Vickrey auction satisfies (6).

Since the matrix[π(t−i |t i

);]

has full rank, there exists anon-zero solution z i : T−i → R to the system∑

t−i

π(t−i |t i

)z i(t−i)

= ui(t i), t i ∈ T i

(sufficiency) Let x i (t) = x i (t) + z i(t−i)

: then since (p, x) isVickrey, (p, x) is also Vickrey (if a lump sum payment is added, ICstill hold). Now, in the new auction,ui(t i)

= ui(t i)−∑

t−i π(t−i |t i

)z i(t−i)

= 0. Hence, there is fullextraction.

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(Necessity - for n = 2) Remember that (T , π) must allow fullextraction for any valuation profile v . Choose a profile wherev1 > 0 and v2 = 0. Since v1 > 0, needs p1 = 1.

For full extraction need that for all t,∑t2

(x1 (t) + x2

(t2))

= v1(t1). IC requires that for all t, t

v1(t1)− x1 (t) ≥ v1

(t1)− x1

(t\t1

)v1(t1)− x1

(t\t1

)≥ v1

(t1)− x1 (t)

hence

x1 (t) = x1(t\t1

)= z1

(t2).

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Since there is full extraction and IR, need that for all t1,and forall v1, there exists a solution z1 to the system.∑

t−i

π(t−i |t1

)z1(t−i)

= v1(t1).

it follows that[π(t2|t1

)]has full rank.

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Surplus extractionBayesian Nash

Proposition: An information structure (T , π) guarantees fullextraction of the surplus by a Bayeisan auction if and only if for alli ∈ N, there does not exist t i ∈ T i and a family(αi(t i)

; t i ∈ T i\{t i})

such that(i) αi

(t i)≥ 0 for all t i ∈ T i

(ii) π(t−i |t i

)=∑

t i 6=t i αi(t i)π(t−i |t i

)(Sufficiency) a direct consequence of (i) Gale “theorem of thealternative”, itself a consequence of Farkas lemma and (ii) theobservation that an auction satisfies a system of linear inequalities.

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(Necessity) Suppose (i) and (ii) hold for 1 ∈ N and some t1 ∈ T 1.Take v i = 0 if i ≥ 2, v1 > 0 such that v1

(t1)< v1

(t1), all

t1 ∈ T 1\{t1}. (IR) and full extraction require∑t2

π(t2|t1

) [v1(t1)− x1

(t\t1

)]= 0

(BIC1) requires that at any t∑t2

π(t2|t1

) [v1(t1)− x1

(t\t1

)]≥∑t2

π(t2|t1

) [v1(t1)− x1 (t)

]

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hence, ∑t2

π(t2|t1

) [v1(t1)− x1 (t)

]≤ 0

⇒∑t2

π(t2|t1

) [v1(t1)− x1 (t)

]< 0

since v1(t1)< v1

(t1). Using

π(t2|t1

)=∑

t1 6=t1 α1(t1)π(t−i |t1

), we have

α1(t1)∑

t2

π(t2|t1

) [v1(t1)− x1 (t)

]< 0

⇒∑t1 6=t1

α1(t1)∑

t2

π(t2|t1

) [v1(t1)− x1 (t)

]< 0

⇒∑t2

π(t2|t1

) [v1(t1)− x1 (t)

]< 0,

violating (IR) for type t1.

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Surplus extraction2 agents

Let T 1 = {1, 2, ...,m1} , T 2 = {1, 2, ....,m2} where order on typeindicates order on valuations.

Full rank: for each i , there does not exists λ (j) , j 6= i , such thatfor all k

π (k |i) =∑j 6=i

λ (j)π (k |j)

For Bayesian: requires that λ ≥ 0.

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If m1 > m2 : cannot have full extraction.

If m1 = m2 = 2, the two conditions coincide: indeed (cond. prob.for agent 1), π (2|j) = 1− π (1|j) for all j . Hence, if full rank,cannot find λ such that π (1|0) = 1− π (0|0) = λπ (0|0) and1− π (0|1) = λπ (0|1) ; clearly since π ≤ 1, need to consider onlyλ ≥ 0.

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If m1 = m2 = 3, the two conditions also coincide. Clearly enoughto show that a violation of full rank implies a violation of thecondition for the Bayesian case.

Suppose that there exists λ, µ such that (wlog) for allj = 1, 2, 3, π (j |1) = λπ (j |2) + µπ (j |3). Sinceπ (3|j) = 1−π (1|j)−π (2|j) , we have 1 = λ+µ. If µ < 0, λ mustbe strictly positive. Hence we have π (j |2) = 1

λπ (j |1)− µλπ (j |3)

and since −µλ > 0, the condition for the Bayesian case is violated.

With two agents, need more than 4 types to have a differencebetween full rank and the condition for the Bayesian case. Anexample is provided in their paper.

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Surplus extractionBeliefs and Valuations

Agents may have correlated information on some dimensions of thestate space but not on others.

I for instance, firms may have correlated information about themarket demand but i.i.d. shocks on their cost functions.

I In Cremer-Mc Lean, the type of agent bundles (i) his beliefsπ(t−i |t i

)and his valuation v i

(t i)

; however there is noreason to bundle the two. For instance, two types could havesame beliefs

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For instance consider two agents, types {L,H, LH} ;

Valuations: v (L) = v (LH) = 0, v (H) = 1

Beliefs of agent 1 about types (L, LH,H) of agent 2:

b (·|L) = b0 =

(2

3 + a,

a

3 + a,

1

3 + a

)b (·|LH) = b (·|H) = b1 =

(1

3 + a,

2a

3 + a,

2

3 + a

)

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As a = 0, as if types L and H and b (L) =(

23 ,

13

), b (H) =

(13 ,

23

),

correlation of types and of valuations. Full rank and full extractionwhen C < 1 (cost of public good).

As a > 0, likelihood of LH and H is a, but beliefs don’t pin downvaluations: agent with belief b0 has valuation 0 but agent withbelief b1 has valuation 0 with probability a

1+a of valuation 1 with

probability 11+a .

Since when beliefs are (b1, b1) , agents have valuations for thepublic good that are i.i.d., cannot get full extraction.

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