Uniqueness in Symmetric First-Price Auctions

with Affiliation

David McAdams

MIT Sloan School of Management, Cambridge, MA 02142

Abstract

The first-price auction has a unique monotone pure strategy equilibrium when there

are n symmetric risk-averse bidders having affiliated types and interdependent val-

ues.

Key words: First-price auction, uniqueness, affiliation, interdependent values,

all-pay auction.

1 Introduction

A growing empirical literature studies symmetric first-price auctions in which

bidders do not have independent private values but rather affiliated types and

interdependent values (also known as ‘common values’). 1 The standard prac-

tice in this literature is to assume that bidders play the symmetric monotone

pure strategy equilibrium (MPSE) described by Milgrom and Weber [15].

Email address: mcadams@mit.edu (David McAdams).1 See e.g. Hendricks, Pinkse, and Porter [6]. For a survey of experimental work, see

Kagel [7].

Preprint submitted to Elsevier 12 July 2006

Unfortunately, theory has provided no justification for focusing on this sym-

metric equilibrium, even though the possibility of asymmetric equilibria in

symmetric auctions is a real concern. In models with independent private val-

ues, the symmetric war of attrition has multiple equilibria, including a contin-

uum of asymmetric equilibria [Nalebuff and Riley [16]]. Once the private values

assumption is relaxed, the second-price auction and the open ascending-price

auction also are well-known to have a continuum of asymmetric equilibria (in

undominated strategies) [Milgrom [13], Bikhchandani and Riley [4]]. 2 These

asymmetric equilibria have a natural ‘winner’s curse’ intuition. When some

bidders are more aggressive, this decreases other bidders’ expected value con-

ditional on winning with any given bid, leading these other bidders to bid less

aggressively, and vice versa. This intuition would seem to apply to the first-

price auction as well, suggesting that symmetric first-price auctions might also

have asymmetric equilibria.

This paper proves that symmetric first-price auctions with affiliated types

and interdependent values do not have asymmetric MPSE. More precisely, a

unique MPSE exists and this equilibrium is in symmetric strategies. Could

still other mixed strategy equilibria exist, not in monotone pure strategies?

This possibility has been ruled out in some important special cases, but not

in general. 3 See Athey and Haile [2], especially Theorem 2.1(ii), for a survey

2 The symmetric private value second-price auction is also well-known to have many

asymmetric equilibria in weakly dominated strategies. See Blume and Heidhues [5].3 If (asymmetric) bidders have private values or independent signals, McAdams

“Monotonicity in Asymmetric First-Price Auctions with Affiliation”, mimeo (2003),

proves that any mixed strategy equilibrium is outcome-equivalent to a MPSE, i.e.

bidding strategies are identical to those in a MPSE except possibly for subsets of

types whose equilibrium bids win with probability zero.

2

of the best available results on uniqueness of mixed strategy equilibrium.

Lebrun [9,10], Maskin and Riley [12], and Bajari [3] prove uniqueness of mixed

strategy equilibrium given independent private values and any number of

asymmetric bidders. 4 Given two asymmetric bidders having affiliated types

and interdependent values, Lizzeri and Persico [11] (LP) proves uniqueness

of MPSE. This paper differs from LP by allowing for more than two bidders

while requiring symmetry.

The rest of the paper is organized as follows. Section 2 lays out the model and

assumptions. Section 3 then proves the main result on uniqueness of MPSE

given symmetric bidders. Section 4 concludes with an extension to all-pay

auctions.

2 Model and preliminaries

Information: Bidder types are one-dimensional random variables having joint

density f(t) on the unit cube [0, 1]n. For each subset I ⊂ {1, ..., n}, the

conditional joint density will be denoted f(tI |t−I) where t ≡ (t1, ..., tn),

tI ≡ (ti : i ∈ I), and−I ≡ {1, ..., n}\I. (Bold notation will be used throughout

the paper to refer to vectors of types, bids, and strategies.)

(A1) Bidder types are affiliated, i.e. f(t′ ∨ t)f(t′ ∧ t) ≥ f(t′)f(t) for all type

profiles t′, t where t′ ∨ t, t′ ∧ t are their component-wise maximum and

minimum, respectively.

4 Lebrun [10] proves uniqueness under the weakest distributional assumptions.

3

Affiliation is a powerful form of positive correlation; see Milgrom and Weber

[15] for more detailed discussion.

(A2) There exists fhigh, flow > 0 such that f(t) ∈ [flow, fhigh] for all t.

(A3) f is continuously differentiable on [0, 1]n.

Bids and Payoffs: After learning its type, each bidder submits a bid bi ∈

OUT ∪ [r,∞) where r is the reserve price. Bidding is voluntary: a bidder

who chooses not to participate ‘bids’ OUT . If all bidders bid OUT , then the

auction is cancelled. Otherwise the highest bidder wins the object, with ties

broken by a coin-flip: if k bidders each submit the highest bid, then each wins

the object with probability 1/k. Bidder i’s utility upon losing is zero and upon

winning with bid b has form ui(ti; t−i; b). I make the following assumptions on

utility: for all i,

(A4) ui is twice continuously differentiable.

(A5) ui is strictly increasing in ti, non-decreasing in tj for all j 6= i, and strictly

decreasing in b

(A6) ∂ui

∂bis non-decreasing in t and non-increasing in b

(A7) ui(1;1; bh) < 0 for some bid level bh < ∞

An important special case arises when utility takes the form ui(vi(ti; t−i)− b),

where vi is strictly increasing in ti, non-decreasing in t−i, and so on. In this

case, (A6) is satisfied when u′′i ≤ 0. Thus, the model is consistent with any

sort of risk-aversion.

The model is a special case of Reny and Zamir [17]. The most important

additional restriction imposed here is symmetry.

4

(A8) (i) Density f is symmetric in all types. (ii) Each bidder’s utility is sym-

metric in others’ types. (iii) Different bidders’ utility depends on own

type and others’ types in the same way. 5

Strategies: In a monotone pure strategy, type ti bids bi(ti), where bi(t′i) ≥ bi(ti)

for all t′i > ti. (By definition, the non-participation ‘bid’ OUT < b for all

other bids b ≥ r.) A monotone pure strategy equilibrium (MPSE) is a Nash

equilibrium in monotone pure strategies, i.e. for all i, bi(ti) is a best response

for bidder i for a full measure set of types ti ∈ [0, 1]. 6

All assumptions presented in this section, including (A1)-(A8), are main-

tained throughout the entire analysis.

Preliminaries and useful shorthand

Let b(·) be a given MPSE. (bi(·) will always refer to an equilibrium strategy.)

Several basic results are gathered together here and proven in the Appendix.

First, some results derived from Milgrom and Weber [15] (‘MW’) Theorem 23,

based on our assumption that t are affiliated. (A is a decreasing subset of X

when x ∈ A, y ≤ x ∈ X implies y ∈ A.)

Lemma 1 (a) Let X be a sublattice of [0, 1]n and let A be a decreasing subset

5 Formally, for all b, i, j, k ∈ {1, ..., n}, t, t′ ∈ [0, 1], and t−ij , t−jk ∈ [0, 1]n−2, (i)

f(ti = t, tj = t′, t−ij) = f(ti = t′, tj = t, t−ij), (ii) ui(ti; tj = t, tk = t′, t−ijk; b) =

ui(ti; tj = t′, tk = t, t−ijk; b), and (iii) ui(ti = t; tj = t′, t−ij ; b) = uj(tj = t; ti =

t′, t−ij ; b).6 In principle, some bidder types might fail to have a best response. A side-

implication of the proof, however, is that every type of every bidder has a best

response in every MPSE.

5

of X. Then

E [ui(ti; t−i; b)|t ∈ A] ≤ E [ui(ti; t−i; b)|t ∈ X] ≤ E [ui(ti; t−i; b)|t ∈ X\A]

for all i. (b) Ui(ti, b) is strictly increasing in ti for all b ≥ r.

Lemma 2 (No ties) Suppose that Pr (b ≥ maxi bi(ti)) > 0 for some b ≥ r.

Then Pr (bi(ti) = bj(tj) = b) = 0 for all i, j.

Lemma 3 (No atoms) Suppose that Pr (b > maxi bi(ti)) > 0 for some b >

r. Then Pr(bi(ti) = b) = 0 for all i.

Limits of bids: Define bi(t−) ≡ limε→0 bi(t− ε) for all t ∈ (0, 1] and bi(0−) ≡

bi(0). Define bi(t+) ≡ limε→0 bi(t + ε) for all t ∈ [0, 1) and bi(1+) ≡ bi(1).

Lowest bids: bi ≡ bi(0+) is bidder i’s ‘lowest bid’. bI ≡ bI is the highest lowest

bid among bidders in I, with shorthand b ≡ b1,...,n.

Several useful facts about lowest bids follow immediately from the definition

(proof omitted): for all I ⊂ {1, ..., n}, (a) b > bI iff Pr(b > maxi∈I bi(ti)) > 0,

(b) b < bI implies Pr(b ≥ maxi∈I bi(ti)) = 0, and (c) b = bI implies Pr(b ≥

maxi∈I bi(ti)) = Pr(b = maxi∈I bi(ti)). By (a,b), bidder i will win with positive

probability whenever he bids more than b−i and win with zero probability

whenever he bids less than b−i. By (c), bidder i may win after bidding exactly

b−i, but only by tying with other(s).

Inverse bid function: For all b ≥ r, define φi(b) ≡ inf{ti : bi(ti) ≥ b} and

φi(b+) = limε→0 φi(b + ε). By Lemma 3, φi(b+) = φi(b) for all b > b.

Winning event: Define Wi(b) ≡ ×j 6=i[0, φj(b)] ≡ [0, φ−i(b)]. Wi(b) is the event

6

(up to a zero measure set) in which all bidders j 6= i bid strictly less than b.

As long as b > r, there are no atoms at bid-level b by Lemma 3, and Wi(b) is

the event in which bidder i would win with bid b.

Winning probability: Pi(ti, b) ≡ Prt−i|ti (Wi(b)|ti)

Winning expected utility: Ui(ti, b) ≡ E [ui(ti; t−i; b)|ti, t−i ∈ Wi(b)]

Interim expected payoff: Πi(ti, b) is bidder i’s expected utility (or ‘payoff’) from

bidding b conditional on own type ti and others’ equilibrium strategies. For

all b > max{b, r}, 7

Πi(ti, b) ≡ Pi(ti, b)Ui(ti, b) =∫t−i∈Wi(b)

ui(ti; t−i; b)f(t−i|ti)dt−i

Πi(ti, bi(ti)) = supb Πi(ti, b) for all types ti having a best response.

Lemma 4 (a) For all i, supb Πi(ti, b) is continuous in ti. (b) For all i, bi(ti−),

bi(ti+) is a best response for type ti when bi(ti−), bi(ti+) > max{b, r}, respec-

tively.

Lemma 5 Pi(ti, b(ti)) > 0 implies supb Πi(t′i, b) > 0 for all t′i > ti.

Highest bids: Let bi ≡ bi(1−) be bidder i’s ‘highest bid’, and b ≡ maxi bi.

Lemma 6 Either b = OUT or bi > max{b, r} and Πi(1, bi) = supb Πi(1, b)

for all i.

7 See equation (7) in the Appendix for a general formulation of payoffs allowing for

atoms.

7

3 Uniqueness

Reny and Zamir [17] guarantees existence of monotone pure strategy equilib-

rium (‘MPSE’) in a more general model allowing for bidder asymmetry.

Theorem 1 There is a unique MPSE in the symmetric first-price auction,

up to the bids made by a zero measure set of types.

The rest of the paper proves Theorem 1, via several intermediate ‘claims’.

More technical parts of the proof are relegated to the Appendix.

Symmetry at the highest types

Recall our notation for highest bids, bi(1−) = bi and b = maxi bi.

Claim 1 Either bi(1−) = OUT for all i or bi(1−) = b > max{b, r} for all i.

Proof of Claim 1. By Lemma 6, either bi = OUT for all i or bi > max{b, r}

for all i. We need to prove that bi = b for all i given that bi > max{b, r} for

all i.

Without loss, suppose for the sake of contradiction that b1 = b and b2 <

b. By Lemma 3, there are no atoms at bid-level bi for any i. By Lemma

6, Πi(1, bi) = supb Πi(1, b) for all i. In particular, Π1(1, b1) ≥ Π1(1, b2) and

Π2(1, b2) ≥ Π2(1, b1).

Since there are no atoms at b1, each bidder would win with probability one if

he were to bid b1. Thus, bidder 2 must get the same expected utility as bidder

8

1 given type t = 1 when bidding b1:

Π2(1, b1) = Pr(W2(1, b1)|t2 = 1)E[u2(1; t−2; b1)|t2 = 1, t−2 ∈ W2(1, b1)]

= Pr(W1(1, b1)|t1 = 1)E[u1(1; t−1; b1)|t1 = 1, t−1 ∈ W1(1, b1)]

= Π1(1, b1)

On the other hand, if bidder 1 were to bid b2, he would win more frequently

than bidder 2 does with the same bid, and have a weakly higher expected

utility when winning:

Π1(1, b2) = Pr(W1(b2)|t1 = 1

)E[u1(t1; t−1; b2)|t1 = 1, W1(b2)

]≥ Pr

(W1(b2)|t1 = 1

)E[u2(t2; t−2; b2)|t2 = 1, W2(b2)

](1)

> Pr(W2(b2)|t2 = 1

)E[u2(t2; t−2; b2)|t2 = 1, W2(b2)

]= Π2(1, b2)

(2)

(1) follows from symmetry of bidders 1,2 combined with Lemma 1(a). (Set X ≡

W1(b2) = [0, 1] × [0, φ−12(b2)] and A ≡ W2(b2) = [0, φ1(b2)] × [0, φ−12(b2)].)

(2) uses the fact that bidder 1 is strictly more likely to win with bid b2 than

bidder 2. (φ1(b2) < φ2(b2) = 1 by presumption since b1 > b2.) Thus, Π1(1, b2) >

Π2(1, b2) ≥ Π2(1, b1) = Π1(1, b1) = supb Π1(1, b), a contradiction.

Intuition for Claim 1. The crucial step was to show that b1 > b2 implies

Π1(1, b2) > Π2(1, b2). There are two reasons for this. First, bidder 2 bids less

than b2 with probability one while bidder 1 bids more than b2 with positive

probability. Consequently, conditional on bidding b2, bidder 1 is strictly more

likely to win the object than bidder 2. Second, conditional on bidding b2 and

winning, bidder 1 faces ‘winner’s curse’ to a lesser degree than bidder 2. This is

because bidder 1 wins the object regardless of bidder 2’s type, whereas bidder

2 only wins when bidder 1 has a relatively low type.

9

Continuous differentiability near where symmetric

Claim 2 Suppose bi(t+) = b > max{b, r} for all i. Then bi(t−) = b for all i.

Claim 3 Suppose bi(t+) = b for all i, where max{b, r} < b ≤ b. Then there

exists γ > 0 such that bi(φi(b)) = b for all i and all b ∈ (b− γ, min{b + γ, b}).

Claim 3 is the most technically challenging result in the paper.

Claim 4 Suppose bi(t+) = b for all i, where max{b, r} < b ≤ b. Then there

exists γ > 0 so that, for all i, φi(b) is continuously differentiable at all b ∈

(b− γ, min{b + γ, b}).

Local uniqueness near where symmetric

Claim 5 Suppose bi(t+) = b for all i, where max{b, r} < b ≤ b. Then there

exists a strictly increasing, continuously differentiable function φ(·) and γ > 0

such that φi(b) = φ(b) for all i and all b ∈ (b− γ, min{b + γ, b}).

Proof of Claim 5. Preliminaries.

Definition 1 (aij(b), ci(b)) For all i, j 6= i, b, and φ ∈ [0, 1]n, define

aij(b, φ) ≡∫ φ−ij

0u(φi; φj, t−ij; b)f(φj, t−ij|φi)dt−ij

ai,i(b, φ) ≡ 0

ci(b, φ) ≡ −∫ φ−i

0

∂u(φi; t−i; b)

∂bf(t−i|φi)dt−i

Similarly, define aij(b) ≡ aij(b, φ(b)) and ci(b) ≡ ci(b, φ(b)).

Given (A3,4), it is easy to check that aij(b, φ) and ci(b, φ) are continuously

differentiable (in all variables).

10

By assumption, b has the property that φi(b) = t for all i. This symmetry

implies that aij(b) ≡ a and ci(b) ≡ c for all i, j. Furthermore, b > b implies

a, c > 0. To see this, let tI denote a vector of types for bidders I all equal to

t. Then

a(b) =∫ t−ij

0u(t; t, t−ij; b)f(t, t−ij|ti = t)dt−ij

≥∫ t−i

0u(t; t−i; b)f(t−i|ti = t)dt−i > 0

The weak inequality follows from Lemma 1(a). (Set X ≡ {t : ti = t, tk ≤

t ∀k 6= i} and X\A ≡ {t : ti = t, tj = t, tk ≤ t ∀k 6= i, j}.) The strict

inequality follows from Lemma 5, since bi(t−) = b > b implies that type t

must get positive expected utility. (bi(t−) = b > b implies that there exists ti

such that bi(ti) > b and t > ti.) c > 0 by (A5) since ∂u∂b

< 0.

Definition 2 (A(b), C(b)) For all i, b, and φ ∈ [0, 1]n, define matrix A(b, φ) ≡

(aij(b, φ) : 1 ≤ i, j ≤ n) and row vector C(b, φ) ≡ (ci(b, φ) : 1 ≤ i ≤ n). Sim-

ilarly, define A(b) ≡ A(b, φ(b)) and C(b) ≡ C(b, φ(b)).

Using shorthand a defined above, the matrix A(b) = A(b, t, ..., t) takes the

special symmetric form

A(b) =

0 a ... a a

a 0 ... a a

... ... ... ... ...

a a ... 0 a

a a ... a 0

11

Since a > 0, this matrix is invertible (straightforward proof omitted). Thus,

its determinant is non-zero (Anton [1], Theorem 2.3.4). Since each aij(b, φ) is

continuously differentiable, so is the determinant of A(b, φ). In particular, the

inverse A−1(b, φ) exists for all (b, φ) in a neighborhood of (b, t, ..., t), and the

entries a−1ij (b, φ) of this inverse matrix are continuously differentiable as well.

Finally, by Lemma 3, b > b implies that φi(·) is continuous in a neighborhood

of b. Thus, the inverse A−1(b) = A−1(b, φ(b)) exists for all b in a neighborhood

of b.

First-order conditions on bidding. Consider any bid-level b ∈ (b − γ, b + γ) .

By Claim 4, derivatives φ′i(b) exist for all i. Thus,

∂Πi(ti, b)

∂b=∑j 6=i

{φj

′(b)∫ φ−ij(b)

0u(ti; φj(b), t−ij; b)f(φj(b), t−ij|ti)dt−ij

}

+∫ φ−i(b)

0

∂u(ti; t−i; b)

∂bf(t−i|ti)dt−i

Since type φi(b) finds bid b to be a best response, the following system of n

equations must be satisfied by φ(b) = (φ1(b), ..., φn(b)): for all i,

0 =∑j 6=i

{φj

′(b)∫ φ−ij(b)

0u(φi(b); φj(b), t−ij; b)f(φj(b), t−ij|φi(b))dt−ij

}(3)

+∫ φ−i(b)

0

∂u(φi(b); t−i; b)

∂bf(t−i|φi(b))dt−i

As long as A(b) is invertible, we may express system (3) as:

(φ′1(b), ..., φ′n(b)) ≡ (g1(b, φ(b)), ..., gn(b, φ(b))) = C(b) ∗ A−1(b)

Note that gi(b, φ) = C(b, φ) ∗ A−1(b, φ) is continuously differentiable in a

neighborhood of (b, t) for all i. This is more than enough to imply the Lipschitz

condition needed to apply the Fundamental Theorem of Differential Equations

12

(FTODE). 8 All bidders’ inverse bid functions φi(b) are uniquely determined

and continuously differentiable over a neighborhood (b− γ, min{b + γ, b}) for

some γ > 0. Uniqueness implies symmetry of this local solution since any

asymmetric solution would lead to another asymmetric solution after permut-

ing the identities of the bidders. So, φi(b) = φ(b) for all b ∈ (b−γ, min{b+γ, b})

where φ(·) is strictly increasing and continuously differentiable.

All MPSE are symmetric.

Recall our shorthand for the lowest winning bid b ≡ maxi bi(0+). For each

bidder i, define ti ≡ φi(max{b, r}+).

Claim 6 t ∈ [0, 1] and b(·) exist such that (i) ti = t for all i and (ii) bi(t) =

b(t) for all i and all t ∈ [0, t)∪ (t, 1), where (iii) b(·) is strictly increasing and

continuously differentiable over (t, 1) and (iv) b(t) = OUT for all t ∈ (0, t).

Proof of Claim 6. If b = OUT , (i)-(iv) are immediate: set t = 1 and

b(t) = OUT for all t < 1. Otherwise, bi(1−) = b > max{b, r} for all i by Claim

1. By Claim 5, furthermore, there exists strictly increasing, continuously dif-

ferentiable φ(·) and γ1 > 0 such that φi(b) = φ(b) for all b ∈ (b−γ1, b). Define

b∗ ≡ min{b ∈ [max{b, r}, b) : φi(b) = φ(b) for all i and all b ∈ (b, b), where φ(·)

is strictly increasing and continuously differentiable}. (So far, we have shown

8 The Lipschitz condition in our case requires that M < ∞ exists such that, for

all i, |gi(b,φ)−gi(b,φ)|max1≤i≤n |φi−φi|

≤ M for all b in a neighborhood of b and all φ, φ in a neigh-

borhood of t. (See Theorem 2′ from Kolmogorov and Fomin [8], p. 72.) Contin-

uous differentiability of each gi implies that |gi(b,φ)−gi(b,φ)|max1≤i≤n |φi−φi|

< 2∑

1≤j≤n∂gi

∂φj(b, t)

for all i when these neighborhoods are small enough. Thus, we may set M =

maxi

(2∑

1≤j≤n∂gi

∂φj(b, t)

)< ∞.

13

that b∗ ≤ b− γ1.)

I claim that b∗ = max{b, r}. Suppose otherwise that b∗ > max{b, r}. By

construction, bi(t) = b(t) for all t > φ(b∗), where b(·) is strictly increasing and

continuously differentiable over this range of types. In particular, bi(φ(b∗)+) =

b∗ for all i. Claim 5 then implies that we can (uniquely) extend φ(·) to the

wider range of bid-levels (b∗ − γ2, b) for some γ2 > 0. But then b∗ ≤ b∗ − γ2,

a contradiction. We conclude that

φi(b) = φ(b) for all b > max{b, r}

where φ(·) is strictly increasing and continuously differentiable over (max{b, r}, b).

Equivalently, bi(t) = b(t) for all t ∈ (t, 1), where t = φ(max{b, r}) and b(·) is

strictly increasing and continuously differentiable over (t, 1).

To complete the proof, we need to show that bi(ti) = OUT for all ti < t. There

are two cases to consider.

First, suppose that max{b, r} = b. By definition, there exists j∗ such that

b = bj∗(0+). This implies φj∗(b) = 0 so that t = 0. bi(ti) = OUT for all ti < t

is vacuous in this case.

Second, suppose that b = OUT so that max{b, r} = r. In this case, bi(ti) ∈

{OUT, r} for all i and all ti < t. By Lemma 2, at most one bidder (say bidder

1) can have an atom at r, so bi(ti) = OUT for all i 6= 1 and all ti < t. Thus,

supb Πi(ti, b) = 0 for all i 6= 1 and almost all ti < t. By symmetry of bidder

strategies above r, supb Πi(t, b) ≥ limδ→0 Πi(t, r + δ) = limδ→0 Π1(t, r + δ) for

all i and all t. Since no bidder i 6= 1 has an atom at r, limδ→0 Π1(t1, r +

δ) = Π1(t1, r) for all t1. Lastly, by (A3-4), Π1(t1, r) is continuous in t1. All

together, we conclude that Π1(t, r) ≤ 0. Since P1(t, r) > 0, Π1(t, r) ≤ 0

14

implies U1(t, r) ≡ E [u(t; t−1; r)|t1 = t, t−1 ≤ 1] ≤ 0. Since U1(t1, r) is strictly

increasing in t1 (Lemma 1(b)), however,

Π1(t1, r) = P1(t1, r)U1(t1, r) < 0 for all t1 < t. (4)

All types t1 < t strictly prefer not to participate rather than bid r, a contra-

diction.

Unique ‘minimal winning type’ and ‘minimal winning bid’.

Define h(t, b) ≡ E[u(t1; t−1; b)|t1 = t, t1 ≤ t for all i 6= 1]. h(t, b) is strictly

increasing in t and strictly decreasing in b. (The proof of this is very similar

to that of Lemma 1(b) and omitted to save space.)

Claim 7 Every MPSE has the same ‘minimal winning type’ t and the same

‘minimal winning bid’ b, where these depend on the environment. Case I: If

h(0, r) ≥ 0, then t = 0 and b ≥ r solves h(0, b) = 0. Case II: If h(0, r) < 0 and

h(1, r) > 0, then t solves h(t, r) = 0 and b = OUT . Case III: If h(1, r) ≤ 0,

then t = 1 and b = OUT .

Figures 1, 2 illustrate Cases I,II given risk-neutral bidders, i.e. u(ti; t−i; b) =

v(ti; t−i)− b. In this setting, h(0, r) = v(0;0)− r.

Proof of Claim 7. Each bidder gets zero payoff given type t, since he gets

zero payoff given any type less than t and payoffs are continuous in ti (Lemma

4(a)). Thus,

Pr(t−i < t|ti = t)E[ui(t; t−i; b(t+))|ti = t, t−i < t] = 0

and either t = 0 or E[ui(t; t−i; b(t+))|ti = t, t−i < t] = 0. When t = 0, further,

u(0;0; b(0+)) = 0 else each bidder i would prefer to deviate given types ti ≈ 0.

15

ti

bi

r

v(0;0)

OUT

b(·)

Fig. 1. If v(0;0) > r, then t = 0

and b(0+) = v(0;0).

ti

bi

r

OUTt

b(·)

Fig. 2. If v(0;0) ≤ r, then b(t+) = r and

E [v(ti; t−i)|ti = t, t−i < t] = r.

Thus, in either case, equilibrium requires that h(t, b(t+)) = 0.

Furthermore, from the proof of Claim 6, b(t+) = max{b, r} and b > r is only

possible when t = 0. Thus, equilibrium requires h(t, max{b, r}) = 0 and either

t = 0 or b = OUT .

Case I: h(0, r) ≥ 0. t = 0 in this case: otherwise, max{b, r} = r so that

h(t, max{b, r}) ≥ h(t, r) > h(0, r) ≥ 0. b ≥ r is then uniquely determined by

h(0, b) = 0.

Case II: h(0, r) < 0 and h(1, r) > 0. Now t ∈ (0, 1). Otherwise, h(t, max{b, r}) ≤

h(0, r) < 0. Thus, b = OUT and t is uniquely determined by h(t, r) = 0.

Case III: h(1, r) ≤ 0. Here t = 1 and b ≤ r. Otherwise, h(t, max{b, r}) <

h(1, r) ≤ 0. As discussed earlier, b ≤ r implies b = OUT . Thus, t = 1 and

b = OUT . This completes the proof.

Uniqueness of MPSE.

Consider Case I in which t = 0, i.e. the reserve price is not binding. (The proof

for Cases II, III proceeds in a similar way and is omitted.)

16

So far, we have shown that any MPSE must be symmetric: bi(t) = b(t) for

all i and all t ∈ (0, 1), where b(·) is strictly increasing and continuously dif-

ferentiable. More precisely, given only that maxi bi(1−) = b, we showed by

construction that there is at most one such bidding function that is consistent

with MPSE. Thus, if there are multiple MPSE b1(·) = (b1(·), ..., b1(·)) and

b2(·) = (b2(·), ..., b2(·)), then it must be that b1(1−) 6= b2(1−).

Further, all MPSE must be strictly ordered in the sense that b1(1−) > b2(1−)

implies b1(t) > b2(t) for all t ∈ (t, 1). Suppose to the contrary that b1(1−) >

b2(1−) but b = b1(t) = b2(t) for some t ∈ (t, 1). By Claim 4, γ > 0 exists so

that b1(t) = b2(t) for all t ∈ (t − γ, t + γ). Indeed, by this logic b1(t) = b2(t)

for all t ∈ (t− γ, 1) so that b1(1−) = b2(1−), 9 a contradiction. On the other

hand, b1(0+) = b2(0+) = b where b is defined implicitly by u(0;0; b) = 0.

Finally, suppose that (b1(·), ..., b1(·)) and (b2(·), ..., b2(·)) are two MPSE where

b1(t) > b2(t) for all t ∈ (0, 1) and b1(0+) = b2(0+) = b. Fix any type t ∈ (0, 1)

and let b1 ≡ b1(t) and b2 ≡ b2(t). By symmetry, each bidder’s first-order

condition (3) in each equilibrium can be re-arranged as:

φ1′(b1) =−∫ t−1

0∂u(t;t−1;b1)

∂bf(t−1|t1 = t)dt−1

(n− 1)∫ t−12

0 u(t; t, t−12; b1)f(t2 = t, t−12|t1 = t)dt−12

(5)

φ2′(b2) =−∫ t−1

0∂u(t;t−1;b2)

∂bf(t−1|t1 = t)dt−1

(n− 1)∫ t−12

0 u(t; t, t−12; b2)f(t2 = t, t−12|t1 = t)dt−12

(6)

where bolded notation tI = (ti = t : i ∈ I).

By (A5), u is strictly decreasing in b. Since b1 > b2, the denominator of (5)

is strictly less than the denominator of (6). (One can show that the denom-

9 Define t∗ ≡ min{t > t : b1(t) = b2(t) for all t ∈ (t, t)}. One shows t∗ = 1 by

repeating, with minor modifications, the argument in the proof of Claim 6 that

b∗ = max{b, r}.

17

t1/.../tn

b1/.../bn

V (t)

b2(t)

b1(t)

t

b2(·) b1(·)

Fig. 3. Graphical intuition why equilibria can not be ordered.

inators in (5,6) are positive, in the same way that we showed a > 0 in the

proof of Claim 5.) By (A6), ∂u/∂b is non-increasing in b and negative, so the

numerator of (5) is weakly greater than the numerator of (6) (and both are

positive). Thus, φ1′(b1) > φ2′(b2) so that b2′(t) > b1′(t) for all t > t. But this

implies b1(t) = b(t) +∫ tt b1′(t)dt < b(t) +

∫ tt b2′(t)dt = b2(t), a contradiction.

Graphical intuition why strictly ordered MPSE can not exist. Consider bidders’

first-order conditions in both equilibria given type t > 0. Figure 3 summarizes

bidder i’s trade-off associated with bidding slightly higher than b1(t) in equilib-

rium 1 and/or slightly higher than b2(t) in equilibrium 2. The extra expected

payment from bidding higher is the ‘area’ of a horizontal rectangle; the extra

expected surplus from the marginal winning event is the ‘area’ of a vertical

rectangle. (Rectangles corresponding to equilibrium 2 are filled. maxj 6=i tj = t

in the marginal winning event, so bidder i’s conditional expected value is

V (t) ≡ E[v(t; t−1)|t2 = t, t−12 ≤ t−12].) For the same small increase in the

bids, the two horizontal rectangles have the same area. Since b2(t) > b1(t), the

vertical rectangle for equilibrium 1 has more height. Since bidder 1 must be

indifferent to raising its bid in each equilibrium, the vertical and horizontal

18

areas must be the same in each equilibrium, implying that the vertical rectan-

gle for equilibrium 1 has less width. That is to say, b1′(t) > b2′(t) for all t > 0.

But this contradicts the presumption that b1(t) < b2(t) since b1(0) = b2(0)

and these bid functions are continuous.

4 Extension: All-pay auctions

The first-price auction has the property that bidders get the same utility from

losing as from not participating. This property is not essential to my analysis.

What is essential is that each bidder’s payoff does not depend on others’

bids (except insofar as others’ bids determine the winner). More precisely,

suppose that bidders have different utilities from winning and losing the object,

uWi (ti; t−i; bi;b−i) and uL

i (ti; t−i; bi;b−i). The analysis depends on: (i) uWi , uL

i

each do not depend on b−i, (ii) uWi is strictly decreasing in bi while uL

i is non-

increasing in bi, and (iii) uWi (ti; t−i; bi)− uL

i (ti; t−i; bi) is strictly increasing in

ti and non-decreasing in t−i.

All-pay auctions: In the all-pay auction, losers pay their own bid and one

can check that (i,ii,iii) are satisfied. Thus, the proof of Theorem 1 implies

that the all-pay auction has a unique monotone pure strategy equilibrium. 10

Yet whether the all-pay auction has mixed strategy or non-monotone pure

strategy equilibria is unknown, so the question of uniqueness remains partially

unresolved.

10 To apply to the all-pay auction, the equations for bidders’ first-order conditions

(see e.g. (3,20,27)) must be modified slightly to reflect the fact that each bidder

always pays its bid. This does not change the argument in any substantive way.

19

Appendix

Proof of Lemma 1

(a) follows immediately from MW Theorem 23, since ui is non-decreasing in

t for all i. For (b), fix t ∈ [0, 1] and b ≥ r. Define an intermediate function

gi(ti; t−i) ≡ ui(ti; t−i; b) that is non-decreasing in t and constant in ti. Consider

any t′i ≥ ti and b′ ≤ b such that (t′i, b′) 6= (ti, b).

Ui(t′i, b

′) = E[ui(t

′i; t−i; b

′)|ti = t′i, t−i ≤ φ−i(b′)]

> E[ui(ti; t−i; b)|ti = t′i, t−i ≤ φ−i(b

′)]

= E[gi(t

′i; t−i)|ti = t′i, t−i ≤ φ−i(b

′)]

≥ E[gi(t

′i; t−i)|ti = t′i, t−i ≤ φ−i(b)

]≥ E

[gi(ti; t−i)|ti = ti, t−i ≤ φ−i(b)

]= E

[ui(ti; t−i; b)|ti = ti, t−i ≤ φ−i(b)

]= Ui(ti, b)

The strict inequality holds since ui is strictly increasing in ti and strictly

decreasing in ti.11 The weak inequalities hold by successive applications of

MW Theorem 23. 12

Proof of Lemma 2

Preliminaries: By definition, bj(tj) < b when tj ∈ [0, φj(b)] and bj(tj) = b

when tj ∈ [φj(b), φj(b+)] (up to zero measure boundaries), where φj(b+) ≡

11 E[ui(ti; t−i; b)

∣∣ti = t′i, t−i ≤ φ−i(b′)]≡∫ φ−i(b

′)0

ui(ti;t−i;b)f(t−i|ti=t′i)dt−i∫ φ−i(b′)

0f(t−i|ti=t′i)dt−i

and like-

wise for similar conditional expectations.12 gi is non-decreasing in t. For the first inequality, consider sublattice X ≡ {t′i} ×

[0,φ−i(b′)] and decreasing subset A ≡ {t′i} × [0,φ−i(b)]. For the second inequality,

consider sublattice X ≡ {t′i, ti} × [0,φ−i(b)] and decreasing subset A ≡ {ti} ×

[0,φ−i(b)].

20

limδ→0 φj(b+δ). The assumption Pr (b ≥ maxj bj(tj)) > 0 implies that φj(b+) >

0 for all j.

Consider any b ≥ r. The result is immediate if every bidder bids b with zero

probability. Suppose without loss that bidder 1 bids b with positive probability,

so that φ1(b+) > φ1(b). All types in (φ1(b), φ1(b+)) bid b and almost all of

them find b to be a best response. Let t1 ∈ (φ1(b), φ1(b+)) be some type for

which b is a best response. Interim expected payoff for this type is

Π1(t1, b) ≡∫ φ−1(b+)

0

1

1 + # {j 6= 1 : bj(tj) = b}u1(t1; t−1; b)f(t−1|t1)dt−1 (7)

Define the following shorthand:

G(k) ≡{t−1 ≤ φ−1(b+) : # {j 6= 1 : bj(tj) = b} = k − 1

}.

A(k) ≡ Pr(G(k)

∣∣∣t1)E[u1(t1; t−1; b)

∣∣∣t1, t−1 ∈ G(k)]

G(k) is the event in which bidder 1 wins the object with probability 1/k

because he ties with k − 1 others at b. Let k∗ − 1 be the number of other

bidders who bid b with positive probability. Thus, A(k) = 0 for all k > k∗. In

terms of this shorthand, note that

Π1(t1, b) =k∗∑

k=1

A(k)

k, Π1(t1, OUT ) = 0, lim

δ→0Π1(t1, b + δ) =

k∗∑k=1

A(k) (8)

Step 1: Restrictions imposed by best response. Since b is a best response, type

t1 does not prefer to submit the null bid OUT nor to bid slightly more than

b. Thus,

k∗∑k=1

1

kA(k) ≥ 0,

k∗∑k=2

k − 1

kA(k) ≤ 0 (9)

Step 2: Either k∗ = 1 or∑k∗

k=1 A(k) = 0. In words, either no other bidder

21

bids b with positive probability or bidder 1 gets approximately zero expected

utility given type t1 when he bids slightly more than b. When k∗ = 1, the first

inequality (9) is satisfied when∑k∗

k=1 A(k) ≥ 0 while the second inequality is

vacuously satisfied.

Suppose instead that k∗ > 1. Note that[0, φ−1(b+)

]is a lattice and that, for

all m ∈ {1, ..., k∗}, ∪mk=1G(k) is a decreasing subset of

[0, φ−1(b+)

]. 13 Thus,

Lemma 1(a) implies that∑k∗

k=1 A(k) ≥ 0 since

k∗∑k=1

A(k) < 0 ⇒m∑

k=1

A(k) < 0 for all m = 1, ..., n

⇒k∗∑

k=1

1

kA(k) =

1

k∗

k∗∑k=1

A(k) +n−1∑m=1

((1

m− 1

m + 1

) m∑k=1

A(k)

)< 0

contradicting (9). Similarly,∑k∗

k=1 A(k) ≤ 0 since

k∗∑k=1

A(k) > 0 ⇒k∗∑

k=m

A(k) > 0 for all m = 1, ..., k∗

⇒k∗∑

k=2

k − 1

kA(k) =

1

2

k∗∑k=2

A(k) +k∗∑

m=3

((m− 1

m− m− 2

m− 1

) k∗∑k=m

A(k)

)> 0

contradicting (9). Finally, E[u1(t; t−1; b)

∣∣∣t1 = t]

=∑k∗

k=1 A(k) is strictly in-

creasing in t by Lemma 1(b). Thus, at most one bidder 1-type finds b to be

a best response. But almost all types in (φ1(b), φ1(b+)) find b to be a best

response, a contradiction.

Proof of Lemma 3

By assumption, Pr (b > maxi bi(ti)) > 0, so φi(b) > 0 for all i. Furthermore,

13 Each set in this union has the form, up to a zero measure set, of

G(k) =⋃

J⊂{2,...,n}:#(J)=k−1

∏j∈J

[φj(b), φj(b+)]∏

j∈{2,...,n}\J

[0, φj(b)]

So, if t−1 ∈ G(k) and t′−1 < t−1, then t′−1 ∈ G(k′) for some k′ ≤ k.

22

given no ties (Lemma 2) there is at most one bidder (say bidder 1) with an

atom at b.

Step 1: Someone else bids just below b. There can not be a gap in the

distribution of maxj 6=1 bj(tj) below b. If there were, bid b would be strictly

dominated for bidder 1, contradicting the presumption that almost all types

t1 ∈ (φ1(b), φ1(b+)) find b to be a best response. Thus, there exists a bid-

der j∗ 6= 1 with a convergent sequence of types {tj∗,k}k=1,2,... ↗ t∗j such that

b− bj∗(tj∗,k) ≡ δk ↘ 0. Without loss, we can select this sequence so that type

tj∗,k finds bj∗(tj∗,k) to be a best response for all k.

For sufficiently large K, Pj∗(tj∗,K , bj∗(tj∗,K)) > 0. Thus, Πj∗(tj∗,K , bj∗(tj∗,K)) >

0 by Lemma 5. (The proof of Lemma 5 does not depend on Lemma 3.) Indeed,

profits of slightly higher types are strictly bounded above zero: for all k > K,

Πj∗(tj∗,k, bj∗(tj∗,k)) ≥ Πj∗(tj∗,k, bj∗(tj∗,K)) = Uj∗(tj∗,k, bj∗(tj∗,K))Pj∗(tj∗,k, bj∗(tj∗,K))

> Uj∗(tj∗,K , bj∗(tj∗,K))Pj∗(tj∗,k, bj∗(tj∗,K))

≥ flow

fhigh

Uj∗(tj∗,K , bj∗(tj∗,K))Pj∗(tj∗,K , bj∗(tj∗,K)) =flow

fhigh

Πj∗(tj∗,K , bj∗(tj∗,K))

The first weak inequality is by revealed preference, the strict inequality is by

Lemma 1(b), and the last inequality follows from assumption (A2).

Step 2: No one else bids just below b. Each type tj∗,k must at least weakly

prefer to bid bj∗(tj∗,K) = b− δk rather than b+ δk. Since bidder 1 has an atom

at b and all other bidders bid strictly less than b with positive probability,

raising his bid by 2δk allows bidder j∗ to increase his probability of winning

by an amount that does not disappear as δk → 0.

For these deviations to be unprofitable, then, bidder j∗ must not strictly prefer

23

to win the object at price b conditional on tying (in the limit):

0 ≥ limk→∞

E[u (tj∗ ; t−j∗ ; b)

∣∣∣tj∗ = tj∗,k, t−j∗ : maxi6=j∗

bi(ti) ∈ (b− δk, b + δk)]

≥ limk→∞

E[u (tj∗ ; t−j∗ ; b)

∣∣∣tj∗ = tj∗,k, t−j∗ ∈ maxi6=j∗

bi(ti) ≤ b− δk

]= lim

k→∞Uj∗(tj∗,k, bj∗(tj∗,k))

The second inequality follows from Lemma 1(a). (Set X ≡ ×ni=1[0, φi(b + δk)]

and A ≡ ×ni=1[0, φi(b−δk)].) In this case, however, limk→∞ Πj∗(tj∗,k, bj∗(tj∗,k)) ≤

0, a contradiction.

Proof of Lemma 4

Proof of (a). By (A3-4), Πi(ti, b) =∫t−i≤φ−i(b)

ui(ti; t−i; b)f(t−i|ti)dt−i is abso-

lutely continuous in ti for each fixed b and

∂Πi(ti, b)

∂ti=

∂(∫

t−i≤φ−i(b)ui(ti; t−i; b)f(t−i|ti)dt−i

)∂ti

=∫t−i≤φ−i(b)

(∂ui(ti; t−i; b)

∂tif(t−i|ti) + ui(ti; t−i; b)

∂f(t−i|ti)∂ti

)dt−i

exists for each fixed b. Indeed, by (A3-4), ui(ti; t−i; b) and f(t−i|ti) are contin-

uously differentiable, so ∂ui(ti;t−i;b)∂ti

f(t−i|ti)+ui(ti; t−i; b)∂f(t−i|ti)

∂tiis continuous

and hence uniformly bounded above and below for all t ∈ [0, 1]n and all

b ≤ b. Thus, there exists an integrable function x : [0, 1] → R+ such that∣∣∣∂Πi(ti,b)∂ti

∣∣∣ ≤ x(ti) for all b ≤ b.

By Theorem 2 of Milgrom and Segal [14], we may therefore conclude that

supb≤b Πi(ti, b) is absolutely continuous in ti. Since all bids strictly greater

than b are strictly dominated, finally, supb Πi(ti, b) = supb≤b Πi(ti, b). Thus,

supb Πi(ti, b) is (absolutely) continuous in ti.

Proof of (b). Consider any type ti such that bi(ti−) > max{b, r} and any

24

increasing sequence {tk} ↗ ti such that bi(tk) is a best response for type

tk for all k. Πi(t, b) is continuous in t by assumptions (A3-4), supb Πi(t, b)

is continuous in t by Lemma 4(a), and Πi(t, b) is continuous in b at bid-

level bi(ti−) by Lemma 3. (Since bi(ti−) > max{b, r}, Lemma 3 implies that

there are no atoms at bi(ti−).) Thus, supb Πi(ti, b) = limk→∞ supb Πi(tk, b) =

limk→∞ Πi(tk, bi(t

k)) = Πi(ti, bi(ti−)) and bid bi(ti−) is a best response.

The proof that bi(ti+) is a best response when bi(ti+) > max{b, r} is very

similar and omitted to save space.

Proof of Lemma 5

Consider any type ti such that Pi(ti, bi(ti)) > 0, any t′i > ti, and any ti ∈

(ti, t′i) having a best response. f(t) ∈ [flow, fhigh] for all t by (A2), imply-

ing f(t−i|ti) ∈ [flow, fhigh] for all t. In particular, Pi(t′i, bi(ti)), Pi(ti, bi(ti)) ≥

flow

fhighPi(ti, bi(ti)) > 0. Further, bi(ti) ≥ bi(ti) implies Pi(t

′i, bi(ti)), Pi(ti, bi(ti)) >

0. By revealed preference, Πi(ti, bi(ti)) ≥ Πi(ti, OUT ) = 0 so Pi(ti, bi(ti)) > 0

implies Ui(ti, bi(ti)) ≥ 0. By Lemma 1(b), Ui(t, b) ≡ E[u(t; t−i; b)|ti = t, t−i ≤ φ−i(b))

]is strictly increasing in t for all b. Thus Ui(t

′i, bi(ti)) > 0. All together,

supb

Πi(t′i, b) ≥ Πi(t

′i, bi(ti)) = Ui(t

′i, bi(ti))Pi(t

′i, bi(ti)) > 0

Proof of Lemma 6

If b = OUT we are done, so suppose that b ≥ r for the rest of the proof.

Without loss, suppose that b1 = b.

Step I: b 6= b. Suppose otherwise. Since b ≥ b1(1−) ≥ b1(0+) = b = b, we con-

clude that b1(t1) = b for all t1 ∈ (0, 1). By Lemma 2, no other bidder can have

25

an atom at this bid-level. Thus, bi(ti) < b and supb Πi(ti, b) = Πi(ti, bi(ti)) = 0

for all i 6= 1 and almost all ti < 1, and bidder 1 wins the object with prob-

ability one when bidding b. Repeating the argument in the text leading to

(4), replacing type t with type 1 and bid r with bid b = b, we conclude that

Π1(t1, b) < 0 for all t1 < 1. Thus, all types t1 < 1 strictly prefer not to

participate rather than bid b, a contradiction.

Step II: b 6= r. Suppose otherwise. Since b1(1−) = r, bidder 1 must have an

atom at bid-level r while (by Lemma 2) bi(ti) = OUT for all i 6= 1 and all

ti < 1. This leads to a contradiction, as in Step I.

Step III: supb Πi(1, b) ≥ Π1(1, b1) = supb Π1(1, b) > 0 for all i. By Steps I-II,

b1 > max{b, r}. Thus, by Lemma 3 there are no atoms at bid-level b1 and

any bidder can win with probability one by bidding b1. Thus, supb Πi(1, b) ≥

Πi(1, b1) = Π1(1, b1) > 0. The equality Πi(1, b1) = Π1(1, b1) holds by sym-

metry while the inequality Π1(1, b1) > 0 follows from Lemma 5. (Since b1 >

max{b, r}, P1(1−ε, b1(1−ε)) > 0 for all small enough ε > 0.) Finally, consider

any sequence {tk} ↗ 1 such that, for all k, every player plays a best response

given type tk. Π1(1, b1) = limk→∞ Π1(tk, b1(t

k)) = limk→∞ supb Π1(tk, b) =

supb Π1(1, b). The first equality holds since there are no atoms at b1, the sec-

ond since each type tk plays a best response, and the third by Lemma 4.

Step IV: bi > max{b, r} for all i. By Step I-II, b > max{b, r}. Without

loss, suppose that b1 = b and b2 ≤ max{b, r}. By definition, bidder 2 never

wins the object and gets zero payoff when he bids less than max{b, r}. By

Lemma 4, supb Π2(1, b) = limk→∞ supb Π2(tk, b) = limk→∞ Π2(t

k, b2(tk)). If

b2 < max{b, r}, then all types t2 < 1 bid less than max{b, r}. Consequently,

26

Π2(tk, b2(t

k)) = 0 for all k and supb Π2(1, b) = 0, a contradiction of Step III.

Finally, suppose that b2 = max{b, r}. Again, Π2(tk, b2(t

k)) = 0 for all k and a

contradiction is reached unless b2(tk) = b2 for all large enough k, i.e. bidder 2

has an atom at max{b, r}. In this case, Lemma 2 implies that no other bidder

has an atom at b2. But then supb Π1(1, b) ≥ limδ→0 Π1(1, b2 + δ) > Π2(1, b2),

since

limδ→0

Π1(1, b2 − δ) = Pr(W1(b2)|t1 = 1

)E[u1(t1; t−1; b2)|t1 = 1, W1(b2)

]≥ Pr

(W1(b2)|t1 = 1

)E[u2(t2; t−2; b2)|t2 = 1, W2(b2)

](10)

> Pr(W2(b2)|t2 = 1

)E[u2(t2; t−2; b2)|t2 = 1, W2(b2)

]= Π2(1, b2) (11)

For the first equality, note that W1(b2) is by definition the event in which all

bidders i 6= 1 bid strictly less than b2; inequalities (10,11) are then identical

to (1,2) in the text. This again contradicts Step III.

Step V: Πi(1, bi) = supb Πi(1, b) for all i. Since bi > max{b, r}, we may sim-

ply repeat for all bidders i 6= 1 the argument used in Step III to show that

Π1(1, b1) = supb Π1(1, b). This completes the proof.

Proof of Claim 2

Without loss, assume that b1(t−) ≥ ... ≥ bn(t−).

Suppose that b1(t−) < b. Since b > r, there must be a gap in the distribution

of bids below b, so that b is strictly dominated for all bidders. Since b > b,

however, Lemma 4 implies that all bidders must find b = bi(t+) to be a best

response given type t. This is a contradiction, so b1(t−) = b.

Suppose that bn(t−) < b. Several steps establish a contradiction.

27

First, Π1(t, b1(t)) = Π1(t, b) = limε→0 Πn(t−ε, bn(t−ε)). As mentioned above,

bidder 1 finds b to be a best response given type t, so Π1(t, b1(t) = Π1(t, b).

By Lemma 4, limε→0 Πn(t − ε, bn(t − ε)) = limε→0 Πn(t + ε, bn(t + ε)). There

are no atoms at b by Lemma 3, so limε→∞ bn(t + ε) = b implies limε→0 Πn(t +

ε, bn(t + ε)) = Πn(t, b). Since φi(b) = t for all i, finally, symmetry implies

Πn(t, b) = Π1(t, b).

Second, Πn(t−ε, bn(t−ε)) > 0 for all small enough ε > 0. Since b1(t−) = b > b,

b1(t − ε) > b for small enough ε > 0. Hence bidder 1 wins the object with

positive probability given type t − ε. By Lemma 5, then, Π1(t, b) > 0. The

desired result follows now from the first point above.

Third, (i) φn(bn(t−)) = t, (ii) φ1(bn(t−)) < t, and (iii) φi(bn(t−)) > 0 for all

i 6= n. (i) holds since bn(tn) ≥ b > bn(t−) for all tn > t while bn(tn) ≤ bn(t−)

for all tn < t. (ii) is immediate from bn(t−) < b1(t−). (Bidder 1 has types less

than t that bid more than bn(t−).) To prove (iii), suppose that φi∗(bn(t−)) = 0

for some i∗, i.e. bi∗(ti∗) ≥ bn(t−) for all ti∗ > 0. Bidder n wins the object

with zero probability with any bid b < bn(t−). As we have seen, however,

Πn(t− ε, bn(t− ε)) > 0 for some ε > 0. Since bn(t− ε) ≤ bn(t−), we conclude

that bn(tn) = bn(t−) for all tn ∈ (t−ε, t), i.e. bidder n has an atom at bid-level

bn(t−). By Lemma 2, no other bidder can have an atom at bn(t−), including

bidder i∗. Thus, bi∗(ti∗) > bn(t−) for all ti∗ > 0. Hence, all bidder-n types in

(t − ε, t) win the object with probability zero and get zero profit. This is a

contradiction.

Fourth, no bidder i 6= n has an atom at bid-level bn(t−). There are two cases

to consider. (φi(bn(t−) > 0 for all i implies bn(t−) ≥ b.) (A) If bn(t−) > b, no

bidder has an atom at bn(t−) by Lemma 3. (B) If bn(t−) = b, it must be that

28

bn(t− ε) = b for all small enough ε, i.e. bidder n has an atom at b. But then

no other bidder can have an atom at b by Lemma 2.

Fifth, Π1(t, b1(t)) > limε→0 Πn(t − ε, bn(t − ε)), contradicting the first point.

By bidder 1’s revealed preference, Π1(t, b1(t)) ≥ Π1(t, b) for all b. Thus,

Π1(t, b1(t)) ≥ limε→0 Π1(t, bn(t−) + ε).

limε→0

Π1(t, bn(t−) + ε)

= Pr(t−1 ≤ φ−1(bn(t−)+)|t1 = t

)E[u1(t; t−1; bn(t−))|t1 = t, t−1 ≤ φ−1(bn(t−)+))

]≥ Pr

(t−1 ≤ φ−1(bn(t−))|t1 = t

)E[u1(t; t−1; bn(t−))|t1 = t, t−1 ≤ φ−1(bn(t−)))

]> Pr

(t−n ≤ φ−n(bn(t−))|tn = t

)E[u1(t; t−1; bn(t−))|t1 = t, t−1 ≤ φ−1(bn(t−)))

]≥ Pr

(t−n ≤ φ−n(bn(t−))|tn = t

)E[un(t; t−n; bn(t−))|tn = t, t−n ≤ φ−n(bn(t−)))

]= lim

ε→0Πn(t− ε, bn(t− ε))

The first equality holds by definition: φi(b) ≡ sup{ti : bi(ti) ≥ b}, so φi(b+) =

φi(b). The two inequalities are essentially identical to (1,2); see their discussion

in the text. (bn(t−) < b1(t−) implies that φn(bn(t−)) = t > φ1(bn(t−)).) The

last equality is valid since no bidder i 6= n has an atom at bn(t−). This

completes the proof.

Proof of Claim 3

Part I: preliminaries. bi(t+) = b for all i implies bi(t−) = b and hence bi(t) = b

for all i (Claim 2). Since b > b, there are no atoms at b by Lemma 3, so

bi(φi(b)) = b for all i. To complete the proof, it suffices to show that (A) γ > 0

exists such that bi(t−) = b implies bi(φi(b)) = b for all b ∈ (b − γ, b) and (B)

when b 6= b, γ > 0 exists such that bi(t+) = b for all i implies bi(φi(b)) = b

for all b ∈ (b, b + γ). (If b = b, only (A) is relevant.) The rest of the proof

29

establishes (A). The proof of (B) is symmetrical and omitted to save space.

First, I show that it suffices to establish (A’) γ > 0 exists such that bi(ti−) ∈

(b− γ, b) for some ti implies bi(ti−) = bi(ti+). When (A’) holds with respect

to γ > 0, I claim that (A) holds with respect to γ∗ ≡ b − maxi bi, where

bi ≡ bi(φi(b− γ)+) for all i. Before proceeding, observe that b− γ ≤ bi < b for

all i, so that (bi, b) ⊂ (b− γ, b). 14

Suppose for the sake of contradiction that bi(φi(b)) 6= b for some i and some

b ∈ (b − γ∗, b). This is only possible if bi(φi(b)−) < bi(φi(b)+). To reach a

contradiction, it suffices to show that bi(φi(b)−) ∈ (bi, b): if so, (A’) implies

that bi(φi(b)−) = bi(φi(b)+).

By definition of γ∗, b ∈ (b − γ∗, b) implies b ∈ (maxi bi, b) ⊂ (bi, b). Since

bi = bi(φi(b−γ)+), b > bi implies φi(b) > φi(b−γ). In particular, bi(φi(b)−) ≥

bi = bi(φi(b − γ)+). By the argument of footnote 14, further, bi(φi(b)−) 6= bi

since there can not be an atom at bid-level bi. Similarly, b = bi(t−) and b < b

implies that φi(b) < t. Thus, bi(φi(b)−) ≤ bi(t−) = b. Again by the argument

of footnote 14, bi(φi(b)−) 6= b since there can not be an atom at bid-level

b. All together, we conclude that bi(φi(b)−) ∈ (bi, b). This yields the desired

contradiction.

Before continuing, I collect important notation in a series of ‘definitions’.

Definition 3 (Bidder iγ and bid-level bγ) For any given γ > 0, let iγ

14 bi ≥ b − γ by definition of bi. Since b = bi(t−), φi(b′) < t for all b′ < b. In

particular, φi(b− γ) < t and bi = bi(φi(b− γ)+) ≤ bi(t−) = b. Thus, bi = b is only

possible if bi(ti) = b for all ti ∈ (φi(b− γ), t). Since b > b, however, Lemma 3 rules

out such atoms. We conclude that b− γ ≤ bi < b for all i.

30

denote any bidder for whom biγ (tiγ−) = bγ ∈ (b − γ, b) for some tiγ but

biγ (tiγ+) > bγ.

To complete the proof, we need to show that such a bidder iγ does not exist

for small enough γ > 0.

Recall from Definition 1 the shorthand notation aij(b) and ci(b). Recall from

the text that aij(b) and ci(b) are each continuous when b > b. Thus, ε(γ) is

well-defined for all γ > 0 small enough that b− γ > b:

Definition 4 (ε(γ)) Define ε(γ) > 0 so that limγ→0 ε(γ) = 0 and

∣∣∣∣∣ aij(b)

(n− 1)ci(b)− a(b)

(n− 1)c(b)

∣∣∣∣∣ , ∣∣∣aij(b)− a(b)∣∣∣ , ∣∣∣ci(b)− c(b)

∣∣∣ < ε(γ) (12)

for all i, j and all b ∈ (b− γ, b).

Definition 5 (Bidder i∗ and sequences {bki∗},{tki∗}) Let {bk

i∗} be a decreas-

ing sequence such that (i) limk→∞ bki∗ = bγ and there exists bidder i∗ and a de-

creasing sequence of types {tki∗} such that (ii) i∗ finds bki∗ to be a best response

given type tki∗ for all k and (iii) limk→∞φi∗ (bk

i∗ )−φi∗ (bγ)

bki∗−bγ

> c(bγ)(n−1)a(bγ)

− ε(γ).

Part II: constructing {bki∗}, {tki∗}, and i∗. We will show that such a bidder

i∗ exists. This is the hardest part of the proof and essential for reaching a

contradiction in Part III. Say that bidder i is ‘active above b’ if bi(ti+) = b

for some type ti. Observe that bγ > b implies that at least two bidders must

be active above bγ. (This is a standard argument, so some steps are sketched

without full details.) First, if no one is active above bγ, then there is a gap

in the distribution of maxi bi(ti) from bγ up to mini bi(φi(bγ)+). This leads a

contradiction, since bid mini bi(φi(bγ)+) is strictly dominated for all bidders,

but some bidder must find it to be a best response (Lemma 4). Second, suppose

31

that only bidder j is active above bγ. Now there is a gap in the distribution of

maxi6=j bj(tj) from bγ up to mini6=j bi(φi(bγ)+). This leads to a contradiction,

since all bids in between bγ and mini6=j bi(φi(bγ)+) are strictly dominated for

bidder j.

Without loss, suppose that bidder 1 is one of the bidders who is active above

bγ. For any decreasing sequence of types {tk1} ↘ φ1(bγ), define a sequence of

bids {bk1} by bk

1 = b1(tk1) for all k. {bk

1} ↘ bγ since bidder 1 is active above bγ.

Without loss, we may assume that limk→∞φi(b

k1)−φi(bγ)

bk1−bγ

exists for all i (possibly

infinite). (Otherwise, select a subsequence such that this limit exists for bidder

1, a further subsequence so that the limit exists for bidder 2, and so on.)

Revealed preference of bidder 1. By construction, bidder 1 finds bk1 to be a

best response for type tk1 for all k. Consequently, Π1(tk1, b

k1) − Π1(t

k1, b

k′1 ) ≥ 0

and Π1(tk′1 , bk

1) − Π1(tk′1 , bk′

1 ) ≤ 0 for all k and all k′ > k. (Note: bk1 > bk′

1 and

φi(bk1) ≥ φi(b

k′1 ) for all i.)

Π1(tk1, b

k1)− Π1(t

k1, b

k′1 ) ≥ 0 implies

0 ≤∫ φ−1(bk

1)

0u1(t

k1; t−1; b

k1)f(t−1|tk1)dt−1 −

∫ φ−1(bk′1 )

0u1(t

k1; t−1; b

k′

1 )f(t−1|tk1)dt−1

=∫

t−1∈[0,φ−1(bk1)]\[0,φ−1(bk′

1 )]

u1(tk1; ti, t−1i; b

k1)f(ti, t−1i|tk1)dt−1

+∫ φ−1(bk′

1 )

0

(u1(t

k1; t−1; b

k1)− u1(t

k1; t−1; b

k′

1 ))f(t−1|tk1)dt−1

Now [0, φ−1(bk1)]\[0, φ−1(b

k′1 )] ≡ ∪J({2,...,n}ZJ , where the sets ZJ ≡

(×i∈J [0, φi(b

k1)])×(

×i6∈J [φi(bk′1 ), φi(b

k1)])

have measure zero intersection. For each J ( {2, ..., n},

define shorthand WJ ≡∫t−1∈ZJ

u1(tk1; ti, t−1i; b

k1)f(ti, t−1i|tk1)dt−1. By Lemma

1(a),

WJ ≥∫ φ−1(bk′

1 )

0u1(t

k1; t−1; b

k′

1 )f(t−1|tk1)dt−1 = Π1(tk1, b

k′

1 )

32

(Observe that [0, φ−1(bk′1 )]∪ZJ is a lattice with decreasing subset [0, φ−1(b

k′1 )].)

Furthermore, Π1(tk1, b

k′1 ) > 0. (Recall that bγ > b so P1(t

k′1 , bk′

1 ) > 0. By the

proof of Lemma 5, we may then conclude that P1(tk1, b

k′1 ) > 0 and U1(t

k1, b

k′1 ) >

0.) We conclude that WJ > 0 for all J ( {2, ..., n}. Thus, in particular,∑J({2,...,n} WJ ≤

∑i6=1

∑J⊂{2,...,n}:i6∈J WJ .

Next, observe that [φi(bk′1 ), φi(b

k1)]× [0, φ−1i(b

k1)] = ∪J⊂{2,...,n}:i6∈JZJ . Thus

∫ φi(bk1)

φi(bk′1 )

∫ φ−1i(bk1)

0u1(t

k1; ti, t−1i; b

k1)f(ti, t−1i|tk1)dt−1idti =

∑J⊂{2,...,n}:i6∈J

WJ

for all i. All together, we conclude that

0 ≤∫ φ−1(bk

1)

0u1(t

k1; t−1; b

k1)f(t−1|tk1)dt−1 −

∫ φ−1(bk′1 )

0u1(t

k1; t−1; b

k′

1 )f(t−1|tk1)dt−1

(13)

≤∑i6=1

∫ φi(bk1)

φi(bk′1 )

∫ φ−1i(bk1)

0u1(t

k1; ti, t−1i; b

k1)f(ti, t−1i|tk1)dt−1idti (14)

+∫ φ−1(bk′

1 )

0

(u1(t

k1; t−1; b

k1)− u1(t

k1; t−1; b

k′

1 ))f(t−1|tk1)dt−1

≤∑i6=1

(φi(b

k1)− φi(b

k′

1 ))Y k,k′

1,i +∫ φ−1(bk′

1 )

0(u1(t

k1; t−1; b

k1)− u1(t

k1; t−1; b

k′

1 ))f(t−1|tk1)dt−1

(15)

for all k′ > k, where

Y k,k′

1,i ≡ maxti∈[φi(bk′

1 ),φi(bk1)]

∫ φ−1i(bk1)

0u1(t

k1; ti, t−1i; b

k1)f(ti, t−1i|tk1)dt−1i

Since these inequalities hold for all k, they must also hold when we divide

both sides by bk1 − bk′

1 > 0. In particular, for any sequence {(kl, k′l)} such that

33

kl →∞ and k′l > kl for all l (shorthand ‘k, k′ →∞’), we have

limk,k′→∞

∑i6=1

φi(bk1)− φi(b

k′1 )

bk1 − bk′

1

Y k,k′ +∫ φ−1(bk′

1 )

0

u1(tk1; t−1; b

k1)− u1(t

k1; t−1; b

k′1 )

bk1 − bk′

1

f(t−1|tk1)dt−1

=∑i6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

∫ φ−1i(bγ)

0u1(φ1(bγ); φi(bγ), t−1i; bγ)f(φi(bγ), t−1i|φ1(bγ))dt−1i

+∫ φ−1(bγ)

0

∂u1(φ1(bγ); t−1; bγ)

∂bf(t−1|φ1(bγ))dt−1

≡∑i6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

a1i(bγ)− c1(bγ) ≥ 0 (16)

The first equality relies on the fact that each bidder’s inverse bid function is

continuous (Lemma 3) and that limk→∞ bk1 = bγ and limk→∞ tk1 = φ1(bγ), as

well as our assumption that u is continuously differentiable and f continuous.

The second equality is by definition.

Similarly, Π1(tk′1 , bk

1)− Π1(tk′1 , bk′

1 ) ≤ 0 implies

0 ≥∫ φ−1(bk

1)

0u1(t

k′

1 ; t−1; bk1)f(t−1|tk

′

1 )dt−1 −∫ φ−1(bk′

1 )

0u1(t

k′

1 ; t−1; bk′

1 )f(t−1|tk′

1 )dt−1

(17)

≥∑i6=1

∫ φi(bk1)

φi(bk′1 )

∫ φ−1i(bk′1 )

0u1(t

k′

1 ; ti, t−1i; bk1)f(ti, t−1i|tk

′

1 )dt−1idti (18)

+∫ φ−1(bk′

1 )

0

(u1(t

k′

1 ; t−1; bk1)− u1(t

k′

1 ; t−1; bk′

1 ))f(t−1|tk

′

1 )dt−1

≥∑i6=1

(φi(b

k1)− φi(b

k′

1 ))

minti∈[φi(bk′

1 ),φi(bk1)]

∫ φ−1i(bk1)

0u1(t

k′

1 ; ti, t−1i; bk1)f(ti, t−1i|tk

′

1 )dt−1i

+∫ φ−1(bk′

1 )

0

(u1(t

k′

1 ; t−1; bk1)− u1(t

k′

1 ; t−1; bk′

1 ))f(t−1|tk

′

1 )dt−1

The sets [φi(bk′1 ), φi(b

k1)]× [0, φ−1i(b

k′1 )] are disjoint and their union is a strict

subset of [0, φ−1i(bk1)]\[0, φ−1i(b

k′1 )]. Thus, (18) follows from (17) since we are

34

now ‘under-counting’ types. 15 We conclude that

∑i6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

∫ φ−1i(bγ)

0u1(φ1(bγ); φi(bγ), t−1i; bγ)f(φ1(bγ), φi(bγ), t−1i)dt−1i

+∫ φ−1(bγ)

0

∂u1(φ1(bγ); t−1; bγ)

∂bf(φ1(bγ), t−1)dt−1 ≤ 0

≡∑i6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

a1i(bγ)− c1(bγ) ≤ 0 (19)

Combined with (16), we conclude

0 =∑i6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

a1i(bγ)− c1(bγ) (20)

In particular, since bγ ∈ (b− γ, b) by presumption,

maxi6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

≥ c1(bγ, φ(bγ))

(n− 1) maxi6=1 a1i(bγ)>

c(b)

(n− 1)a(b)− ε(γ)

The weak inequality follows from (20); the strict inequality is by definition of

ε(γ).

We are now ready to define bidder i∗ and the sequences {tki∗},{bki∗}:

i∗ ≡ arg maxi6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

Next, define tki∗ ≡ φi∗(bk1) and bk

i∗ = bi∗(tki∗−) for all k. By Lemma 4, bidder

i∗ finds bki∗ to be a best response given type tki∗ for all k. Furthermore, note

that φi∗(bki∗) = φi∗(b

k1) while bk

i∗ ≤ bk1 for all k. 16 Consequently, by our choice

of bidder i∗

limk→∞

φi∗(bki∗)− φi∗(bγ)

bki∗ − bγ

≥ maxi6=1

limk→∞

φi(bk1)− φi(bγ)

bk1 − bγ

>c(b)

(n− 1)a(b)− ε(γ) (21)

15 The formal derivation of inequality (18) is very similar to that of (14); (14) holds

since we ‘overcount types’.16 Since bγ > b, Lemma 3 implies that φi∗(bi∗(t−)) = t whenever bi∗(t) > bγ . In

particular, φi∗(bki∗) = φi∗(bi∗(φi∗(bk

1))−) = φi∗(bk1).

35

Revealed preference of bidder i∗. By construction, bidder i∗ finds bki∗ to be a

best response for type tki∗ for all k. Repeating the steps used to derive equation

(20), we conclude

0 =∑i6=i∗

limk→∞

φi(bki∗)− φi(bγ)

bki∗ − bγ

ai∗i(bγ)− ci∗(bγ) (22)

Part III: Profitable deviation for bidder iγ.

Revealed preference of bidder iγ. By construction, bγ = biγ (tiγ−) so bidder

iγ finds bγ to be a best response given type tiγ (Lemma 4). In particular,

Πiγ (tiγ , bk) − Πiγ (tiγ , bγ) ≤ 0 for all k. Repeating the steps used to derive

inequality (19), we conclude

0 ≥∑i6=iγ

limk→∞

φi(bki∗)− φi(bγ)

bki∗ − bγ

aiγ i(bγ)− ciγ (bγ) (23)

Bidder iγ strictly prefers bki∗ over bγ given type tiγ , for all large enough k. By

presumption, bidder iγ is not active above bγ, which implies φiγ (b) = φiγ (bγ)

for all b in a neighborhood above bγ. Thus, limk→∞φiγ (bk

i∗ )−φiγ (bγ)

bki∗−bγ

= 0 and (22)

becomes

0 =∑

i6=i∗,iγ

limk→∞

φi(bki∗)− φi(bγ)

bki∗ − bγ

ai∗i(bγ)− ci∗(bγ) (24)

Subtracting (24) from (23),

0 ≥ limk→∞

φi∗(bki∗)− φi∗(bγ)

bki∗ − bγ

aiγ i(bγ) +∑

i6=i∗,iγ

limk→∞

φi(bki∗)− φi(bγ)

bki∗ − bγ

(aiγ i(bγ)− ai∗i(bγ)

)

−(ciγ (bγ)− ci∗(bγ)

)(25)

>

(c(b)

(n− 1)a(b)− ε(γ)

)aiγ i(bγ)− ε(γ)

∑i6=i∗,iγ

limk→∞

φi(bki∗)− φi(bγ)

bki∗ − bγ

(26)

(26) follows from (25) by definition of ε(γ) and i∗, see (12,21). By (24), the

36

sum

∑i6=i∗,iγ

limk→∞

φi(bki∗)− φi(bγ)

bki∗ − bγ

≤ ci∗(bγ)

mini6=i∗,iγ ai∗i(bγ)<

c(b)

a(b)+ δ(γ)

where limγ→0 δ(γ) = 0. Since a, c > 0, the right-hand-side of (26) is positive

for all small enough γ > 0. This is a contradiction and completes the proof.

Proof of Claim 4

Let γ > 0 be that identified by Claim 3.

Right-derivatives φ′i(b+) well-defined. Consider any bid-level bγ ∈ (b−γ, max{b+

γ, b}) and any decreasing sequence {bk} ↘ bγ such that limk→∞φj(b

k)−φj(bγ)

bk−bγ

exists (possibly infinite) for all i. (This involves no loss of generality; see the

proof of Claim 3.) By Claim 3, bidder i bids bk and hence finds bk to be a

best response given type tki ≡ φi(bk) for all i and all k. (Without loss, we may

assume that bk ∈ (bγ, max{b + γ, b}) so that Claim 3 applies to bid-level bk.)

We can now repeat that part of the proof of Claim 3 labeled ‘Revealed pref-

erence for bidder 1’ for each bidder i. As in equation (20), we conclude that

0 =∑j 6=i

limk→∞

φj(bk)− φj(bγ)

bk − bγ

aij(bγ)− ci(bγ) for all i = 1, ..., n (27)

As long as A(bγ) is invertible, there is a unique solutionlimk→∞

φ1(bk)−φ1(bγ)bk−bγ

... limk→∞φn(bk)−φn(bγ)

bk−bγ

= C(bγ) ∗ A−1(bγ)

(Recall the meaning of C(bγ), A−1(bγ) from definition 2 in the text.) We

conclude infk→∞φj(b

k)−φj(bγ)

bk−bγ= supk→∞

φj(bk)−φj(bγ)

bk−bγand does not depend on

the chosen sequence {bk}. Hence, the right-derivatives

φ′i(bγ+) ≡ limε→0

φi(bγ + ε)− φi(bγ)

ε

37

exist for all i and all bγ ∈ (b− γ, max{b + γ, b}).

Derivatives φ′i(bγ) well-defined and continuous. Similarly, to prove that left-

derivatives φ′i(bγ−) are well-defined, consider any increasing sequence {bk} ↗

bγ such that limk→∞φj(b

k)−φj(bγ)

bk−bγexists (possibly infinite) for all i. Repeating

the same steps, the same solution is the unique solutionlimk→∞

φ1(bγ)−φ1(bk)bγ−bk ... limk→∞

φn(bγ)−φn(bk)bγ−bk

= C(bγ) ∗ A−1(bγ)

Thus, the left-derivative and derivative exists:

φ′i(bγ−) ≡ limε→0

φi(bγ)− φi(bγ − ε)

ε= φ′i(bγ+)

Finally, the derivative is continuous at bγ since C(bγ) ∗A−1(bγ) is continuous.

Thus, φi(·) is continuously differentiable over (b− γ, max{b + γ, b}) for all i.

Acknowledgments

I thank seminar participants at Arizona, Illinois, Pitt, Rutgers, WZB Berlin,

Susan Athey, and Nicola Persico for helpful comments. I especially thank the

editor Alessandro Lizzeri and two anonymous referees for suggestions that

strengthened the paper.

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