uniqueness in symmetric first-price auctions with affiliationdm121/papers/fpaunique_final.pdfthe...
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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: [email protected] (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.
References
[1] H. Anton, Elementary Linear Algebra, 6th edition, John Wiley and Sons, 1991.
[2] S. Athey, P. Haile, Nonparametric approaches to auctions, in: J. Heckman,
E. Leamer (Eds.), Handbook of Econometrics, Volume 6, North Holland,
38
Amersterdam, forthcoming, 2006
[3] P. Bajari, Comparing competition and collusion: a numerical approach, Econ.
Theory 18 (2001) 187-205.
[4] S. Bikhchandani, J. Riley, Equilibria in open common value auctions, J. Econ.
Theory 53 (1991) 101-130.
[5] A. Blume, P. Heidhues, All equilibria of the Vickrey auction, J. Econ. Theory
114 (2004) 170-177.
[6] K. Hendricks, J. Pinkse, R. Porter, Empirical implications of equilibrium
bidding in first-price, symmetric, common value auctions, Rev. Econ. Stud.
70 (2003) 115-146.
[7] J. Kagel, Auctions: A survey of experimental research, in: J. Kagel, A. Roth,
Handbook of Experimental Economics, Princeton Univ. Press, Princeton, NJ,
1995.
[8] A. N. Kolmogorov, S. V. Fomin, Introductory Real Analysis, Dover, New York,
1975.
[9] B. Lebrun, First price auctions in the asymmetric N bidder case, Int. Econ.
Rev. 40 (1999) 125-142.
[10] B. Lebrun, Uniqueness of the equilibrium in first-price auctions, Games Econ.
Behav. 55 (2006) 131-151.
[11] A. Lizzeri, N. Persico, Uniqueness and existence of equilibrium in auctions with
a reserve price, Games Econ. Behav. 30 (2000) 83-114.
[12] E. Maskin, J. Riley, Equilibrium in sealed high-bid auctions, Rev. Econ. Stud.
67 (2000) 439-454.
[13] P. Milgrom, Rational expectations, information acquisition, and competitive
bidding, Econometrica 49 (1981) 921-943.
39
[14] P. Milgrom, I. Segal, Envelope theorems for arbitrary choice sets, Econometrica
70 (2002) 583-601.
[15] P. Milgrom, R. Weber, A theory of auctions and competitive bidding,
Econometrica 50 (1982) 1089-1122.
[16] B. Nalebuff, J. Riley, Asymmetric equilibria in the war of attrition, J.
Theoretical Biology 113 (1985) 517-527.
[17] P. Reny, S. Zamir, On the existence of pure strategy monotone equilibria in
asymmetric first-price auctions, Econometrica 72 (2004) 1105-1125.
40