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Uninformed Bidding in Sequential Auctions

  • Emmanuel Lorenzon ORCID logo EMAIL logo
Published/Copyright: February 28, 2022

Abstract

We consider a private-value auction with one-sided incomplete information in which two objects are sold in a sequence of two second-price auctions. The buyers have multiunit demand and are asymmetrically informed at the ex ante stage of the game. One buyer perfectly knows his type, and the other buyer is uninformed about her own type. We consider interim information acquisition by the uninformed buyer and derive an asymmetric equilibrium that is shown to produce a declining price sequence across both sales. The supermartingale property of the price sequence stems from the uninformed buyer’s incentives to gather private information, which leads to aggressive bidding in the first-stage auction.

JEL Classification: D44; D82; L86; M37

Corresponding author: Emmanuel Lorenzon, CNRS, INRAE, Grenoble INP, GAEL, Université Grenoble Alpes, Grenoble, 38000, France; and Governance and Regulation Chair, University Paris-Dauphine, PSL Research University, Paris, F-75016, France, E-mail:

Acknowledgments

I am grateful to two anonymous referees and to Petros Sekeris, Wilfried Sand-Zantman, David Ettinger, Noemi Navaro, Nicolas Carayol, David Malueg and Laurent Bergé for their helpful comments and suggestions. All errors are mine.

Appendix A

A.1 Proof of Lemma 1

We look for a Bayesian equilibrium in which player I with valuation x I chooses a bid b I 1 = β I 1 x I to maximize his expected payoff. From his viewpoint, buyer U acts as a dummy player who submits a quantity b 0 drawn from some distribution F . on 0 , x ̄ . Assume that for the first round, he chooses to act as type zx I and bids b = β I 1 z . His overall expected payoff becomes:

Π I + z ; x I = 0 β I 1 z x I b 0 + x I E x U + d F b 0 + 1 F β I 1 z 0 x I x I x U d G x U .

The first integral corresponds to the event in which he wins both periods against b U 1 = b 0 in the first round and b U 2 = E x U in the second round. The second integral corresponds to the event in which he loses the first round against b U 1 = b 0 and wins the second auction against β U 2 x U = x U by bidding truthfully β I 2 x I = x I .

If he acts as type z < x I and bids b = β I 1 z , then his payoff becomes:

Π I z ; x I = 0 β I 1 z x I b 0 + x I E x U + d F b 0 + 1 F x I 0 x I x I x U d G x U + β I 1 z x I 0 x I x I x U d G x U d F t .

The first integral is identical to the previous case. The second line expresses the situation in which he loses the first round against bid b 0 > x I > b, and the third line denotes the situation in which buyer I loses the first round against bid b < b 0x I .

If for the informed player, it is optimal to act as type z = x I , then it should be the case that:

Π I + z ; x I z z = x I 0 and Π I z ; x I z z = x I 0 .

Both FOCs are given by:

Π I + z ; x I z z = x I : β I 1 x I f β I 1 x I x I β I 1 x I + x I E x U + 0 x I x I x U d G x U 0 Π I z ; x I z z = x I : β I 1 x I f β I 1 x I x I β I 1 x I + x I E x U + 0 x I x I x U d G x U 0 ,

which leads to the following equilibrium strategy at z = x I :

(1) β I 1 x I = x I 1 G x I + x I E x U + + E X U | X U x I G x I = x I + x I E x U + 0 x I G x U d x U .

To ensure that the proposed bid function constitutes an equilibrium strategy candidate, substitute the expression of β I 1 x I at z in the first-order condition. Thus, we obtain the following:

Π I z ; x I z β I 1 z f β I 1 z = Ψ z = x I β I 1 z + x I + E x U + 0 x I x I x U d G x U = x I z z E x U + + x I E x U + 0 x I d G x U + 0 z d G x U .

Suppose that z > x I > Ex U ; then, we obtain:

Ψ z = x I β I 1 z + x I + E x U + 0 x I x I x U d G x U = 2 x I z + x I z d G x U 0 .

If Ex U > z > x I , then:

Ψ z = x I z + x I z d G x U = z x I G z 1 0 .

For any type z > x I , player I can increase his payoff by acting as his true type. It is not optimal to mimic any higher type. Consider now the situation Ex U < z < x; then, we also obtain:

Ψ z = 2 x I z z x I d G x U 0 .

If z < x I < Ex U , then:

Ψ z = x I z z x I d G x U 0 .

Finally, if the following situation z < Ex U < x I is realized, we obtain the following:

Ψ z = x I z + x i E x U + z x I d G x U 0 .

As a result, mimicking any type z lower than his true type is also not optimal. He can again increase his payoff by acting as his true type, which shows that σ I * = b I 1 , b I 2 forms an equilibrium strategy profile.

To realize now that β I 1 x I x I , note that if x I < Ex U , we have that β I 1 x I = x I 0 x I G x U d x U x I . Then, if x I Ex U , we have that:

β I 1 x I = 2 x I E x U 0 x I G x U d x U = 2 x I x ̄ + x I x ̄ G x U d x U 2 x I x ̄ + x ̄ x I = x I ,

and the assertion that the informed player underbids in the first auction follows.

A.2 Proof of Lemma 2

This proof consists of three parts. First, we prove that for some range of first-period bids b U 1 = b 0 , the uninformed player’s expected utility payoff function is strictly increasing so that there is no best response for this range. Second, we prove that her expected utility payoff is a concave function for the opposite range of bid values and that a unique maximand exists within this set. Finally, we show that the optimal first-period bid b 0 * involves an equilibrium quantity greater than Ex U . As a matter of technical simplification, consider the following property:

Property 2

x 0 0 , x ̄ : b 0 x 0 and b 0 β I 1 x I , ϕ b 0 E x U X I [ E x U , x ̄ ] .

Proof

Denote by β I x I and β I + x I , respectively, player I’s bid functions x I 0 , E x U and x I E x U , 0 . We have for x ̃ = E x U that β I x ̃ = β I + x ̃ = x 0 γ x 0 = ϕ x 0 = x ̃ = E x U . By Property 1, ϕ b 0 > ϕ x 0 = E x U > x 0 ; hence, β I 1 x I b 0 x 0 x I ϕ b 0 ϕ x 0 = E x U . The explicit value of x 0 is simply equal to ϕ x 0 = E x U x 0 = β I 1 x ̃ x 0 = x ̃ + x ̃ E x U + 0 x ̃ x ̃ x U d G x U = x ̃ 0 x ̃ G x U d x U , in which the last equality follows from integration by parts. □

This property indicates that there exists some threshold x 0 > Ex U so that any first-period losing bid b 0 higher than x 0 implies that player I’s valuation is at least equal to the expected value of X U . Therefore, any truthful bid b U 2 = E x U in the second period after losing the first unit is a losing bid. Under Property 2, we can easily split buyer U’s decision. The bid is considered to be the expected utility payoff function Π U b 0 from playing any quantity b 0x 0 and the expected utility payoff function Π U + b 0 from playing any quantity b 0 > x 0.

Any bid b 0 in the range 0 , x 0 leads to the following expected utility payoff function:

Π U b 0 = 0 γ b 0 x I x ̄ 2 x U β I 1 x I x I d G x U d G x I + 0 γ b 0 0 x I x U β I 1 x I d G x U d G x I + γ b 0 E x U γ b 0 E x U 0 x ̄ x U x I d G x U d G x I d G t + E x U x ̄ 0 d G t .

Considering the first two lines, buyer U wins both units if in the second auction x U > X I and wins only the first unit otherwise. The third line addresses the case in which she loses the first unit but still wins the second with bid Ex U , which is the conjunction of two events: buyer U can lose the first auction if:

i E x U > x 0 > β I 1 x I > b 0 ii E x U > β I 1 x I > x 0 > b 0 .

The situation in which β I 1 x I > E x U > x 0 > b 0 is realized implies a certain loss in the second period.

  1. No best response in the range of Π U b 0 . We first prove that there is no best response for the uninformed bidder within the range 0 , x 0 by showing that Π U b 0 is strictly increasing in b 0. Differentiating Π U over b 0, we obtain:

    Π U b 0 = g γ b 0 γ b 0 b 0 γ b 0 x ̄ 2 x U b 0 γ b 0 d G x U + 0 γ b 0 x U b 0 d G x U γ b 0 E x U E x U x I d G x I + γ b 0 γ b 0 E x U g t d t E x U γ b 0 E x U g t d t .

    The product g γ b 0 γ b 0 b 0 is strictly positive, and upon rearranging the term in brackets and replacing E x U = γ x 0 , we obtain relation T b 0 = γ x 0 b 0 + A b 0 B b 0 , in which:

    A b 0 = γ b 0 x ̄ t γ b 0 d G t B b 0 = γ b 0 γ x 0 2 γ x 0 t γ b 0 d G t .

    The problem is nested to show that this relation is positive; thus, if it is positive at the lower boundary and at the upper boundary, the relation is monotonic so that it is positive everywhere. Let us first compute the value of this function at both b 0 = 0 and b 0 = x 0. For b 0 = 0, we obtain:

    T 0 = E x U + 0 x ̄ t d G t 0 E x U 2 E x U t d G t

    since γ 0 = 0 . Rearranging the relation, we obtain:

    T 0 = 2 E x U 2 E x U 0 E x U d G t + 0 E x U t d G t > 0 .

    Now, for b 0 = x 0, that is, γ x 0 = E x U , we obtain:

    T x 0 = E x U x 0 + E x U x ̄ t E x U d G t > 0 .

    Thus, the relation is positive at its two boundaries. Next, since γ x 0 E x U , we obtain that:

    (2) A ( b 0 ) B ( b 0 ) = 2 γ b 0 E x U ( t γ x 0 ) d G ( t ) + E x U x ̄ ( t γ b 0 ) d G ( t ) .

    Given that:

    0 x ̄ t E x U d G ( t ) = 0 ,

    we can write:

    (3) 0 γ b 0 ( t E x U ) d G ( t ) + γ b 0 E x U ( t E x U ) d G ( t ) + E x U x ̄ ( t E x U ) d G ( t ) = 0 .

    Substituting (3) into expression (2) yields:

    A ( b 0 ) B ( b 0 ) = E x U x ̄ ( E x U t ) d G ( t ) + E x U x ̄ ( E x U γ b 0 ) d G ( t ) 2 0 γ b 0 ( t E x U ) d G ( t ) .

    Hence:

    T ( b 0 ) = E x U b 0 E x U x ̄ ( t E x U ) d G ( t ) + E x U x ̄ E x U γ b 0 d G ( t ) + 2 0 γ b 0 ( E x U t ) d G ( t ) ,

    which, using relation (3) again, yields:

    T ( b 0 ) = E x U b 0 + 0 γ b 0 ( E x U t ) d G ( t ) + E x U x ̄ ( E x U γ b 0 ) d G ( t ) γ b 0 E x U ( E x U t ) d G ( t ) .

    Since Ex U equally splits the interval [ 0 , x ̄ ] and since γ b 0 [ 0 , E x U ] , we obtain that:

    E x U x ̄ ( E x U γ b 0 ) d G ( t ) γ b 0 E x U ( E x U γ b 0 ) d G ( t ) .

    We also have that:

    ( E x U γ b 0 ) ( E x U t ) , t γ b 0 , E x U .

    As a result, the following is true t γ b 0 , E x U :

    γ b 0 E x U ( E x U γ b 0 ) d G ( t ) γ b 0 E x U ( E x U t ) d G ( t ) ,

    so that t γ b 0 , E x U :

    E x U x ̄ ( E x U γ b 0 ) d G ( t ) γ b 0 E x U ( E x U γ b 0 ) d G ( t ) γ b 0 E x U ( E x U t ) d G ( t ) .

    Therefore:

    E x U x ̄ ( E x U γ b 0 ) d G ( t ) γ b 0 E x U ( E x U t ) d G ( t ) 0 ,

    and so we obtain that:

    T ( b 0 ) = γ x 0 b 0 + A ( b 0 ) B ( b 0 ) 0 .

    Therefore, Π b 0 is monotonically increasing for all b 0 0 , x 0 , and there is no best response in this interval for the uninformed bidder.

  2. Existence of a maximand in the range of Π U + b 0 . Consider now all bids b 0x 0. Then, according to Property 2, any losing bid in the first auction implies a strict loss in the second auction, which leads to the following overall expected payoff:

    Π U + b 0 = ϕ x 0 ϕ b 0 x I x ̄ 2 x U β I 1 x I x I d G x U d G x I + ϕ x 0 ϕ b 0 0 x I x U β I 1 x I d G x U d G x I + 0 ϕ x 0 x I x ̄ 2 x U β I 1 x I x I d G x U d G x I + 0 ϕ x 0 0 x I x U β I 1 x I d G x U d G x I + ϕ b 0 x ̄ 0 d G x I .

    Differentiating Π U + b 0 over b 0, we obtain the following FOC:

    b 0 0 ϕ b 0 x U g x U d x U 2 ϕ b 0 x ̄ x U g x U d x U + ϕ b 0 1 G ϕ b 0 = 0 .

    Rearranging the formulation, we obtain the relation of the proposition:

    (4) b 0 * = E x U + Γ b 0 ,

    with Γ b 0 = 1 G ϕ b 0 E X U | X U ϕ b 0 ϕ b 0 .

  3. Overbidding is optimal. It is now straightforward to note that in relation (4), Γ b 0 0 . Therefore, b 0 * E x U , and the assertion that it is optimal for the uninformed player to overbid with respect to her expected valuation in the first auction stage follows.

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Received: 2021-01-15
Revised: 2021-09-13
Accepted: 2021-12-19
Published Online: 2022-02-28

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