Almost common value auctions and discontinuous equilibria

Abstract

In almost common value auctions, even a small private payoff advantage is usually supposed to have an explosive effect on the outcomes in a second-price sealed-bid common value auction. According to (Bikhchandani in J. Econ. Theory 46:97–119, 1988) and (Klemperer in Eur. Econ. Rev. 42:757–769, 1998) the large set of equilibria obtained for common value auction games drastically shrinks, so that the advantaged player always wins the auction, at a price that sharply decreases the seller’s payoff. Yet this result has not been observed experimentally. In this paper, we show that Bikhchandani’s equilibria are not the only equilibria of the game. By introducing discontinuities in the bids, we establish a new family of perfect equilibria with interesting properties, among them: (i) the advantaged bidder does no longer win the auction regardless of her private information, (ii) she may pay a much higher price than in Bikhchandani’s equilibria, (iii) there is no ex-post regret, (iv) the intersection with level-k reasoning is not empty. We also show that a private advantage limits the number of possible discontinuities: one can introduce any number of discontinuities in the common value auction, but this is not possible in presence of a private advantage.

Appendices

Appendix A

Proof of Proposition 2

Let us fix, w.l.o.g. i=1 and j=2.

We first establish that b ₁(x ₁) a best response:

• If x ₁>c _N , player 1 wins the auction because she bids more than any type of player 2, given that player 2’s highest bid is 1+c _N ; and she is best off winning the auction because she gets at least x ₁−c _N >0.

• If c _N−k−1<x ₁≤c _N−k with k from 0 to N−2, she wins against any type of player 2 with x ₂<d _N−k because she bids more than c _N−k+1+d _N−k whereas player 2 bids less, and winning is her best strategy because she gets at least x ₁−c _N−k−1>0.

  But she loses the auction if x ₂≥d _N−k and x ₁<c _N−k because she bids less than c _N−k +d _N−k , whereas player 2 bids at least this amount, and she is best off losing the auction because winning would at best lead to the payoff x ₁−c _N−k <0. If x ₁=c _N−k and x ₂>d _N−k , she loses the auction because player 2’s bid is higher than her bid c _N−k +d _N−k and she is best off losing because winning would at best lead to a null payoff. If x ₁=c _N−k and x ₂=d _N−k , then she makes the same bid as player 2. Proposition 2 does not specify how the wallet is shared in case of a tie. But this is of no importance because player 1 gets 0, whether getting or not getting the wallet.

• If x ₁≤c ₁, she wins against any type of player 2 such that x ₂<d ₁ (because she bids at least d ₁ and player 2 bids less) and winning is the best strategy because she gets x ₁≥0. She loses against any type of player 2 such that x ₂>d ₁ because player 2 bids more than c ₁+d ₁, and losing is the best strategy because winning leads to the payoff x ₁−c ₁≤0. For similar reasons, if x ₂=d ₁ and x ₁<c ₁, she loses the auction and is happy to lose (she bids less than player 2’s bid c ₁+d ₁ and would get x ₁−c ₁<0 upon winning). And if x ₁=c ₁ and x ₂=d ₁, both players bid c ₁+d ₁, and player 1 gets 0, whether she gets or not the wallet (Proposition 2 does not specify how to share the wallet in case of a tie).

We now establish that b ₂(x ₂) is a best response:

• If x ₂≥d _N , player 2 bids less than any type of player 1 such that x ₁>c _N , because he bids at most 1+c _N , and player 1 bids more. Losing the auction is his best strategy because winning would lead to the payoff x ₂−1≤0. And player 2 bids more than any type of player 1 such that x ₁<c _N because he bids at least c _N +d _N and player 1 bids less. And winning is the best strategy because he gets at least x ₂−d _N ≥0. If x ₁=c _N , and x ₂>d _N , he wins the auction because he bids more than player 1 and winning is the best strategy because he gets x ₂−d _N >0. If x ₁=c _N and x ₂=d _N , both players make the same bid and player 2 gets 0, whether getting or not getting the wallet (given that Proposition 2 does not specify how to share the wallet in case of a tie).

• If d _N−k−1≤x ₂<d _N−k with k from 0 to N−2, player 2 bids less that any type of player 1 such as c _N−k−1<x ₁, because he bids less than d _N−k +c _N−k−1 whereas player 1 bids more. And losing the auction is the best strategy because winning would at best lead to the payoff x ₂−d _N−k <0.

  And player 2 bids more than any type of player 1 such as x ₁<c _N−k−1, because he bids at least d _N−k−1+c _N−k−1 (and player 1 bids less) and winning is his best strategy because he gets at least x ₂−d _N−k−1≥0.

  If x ₁=c _N−k−1 and x ₂>d _N−k−1, player 2 wins the auction because his bid is higher than player 1’s, d _N−k−1+c _N−k−1, and winning is the best strategy because he gets x ₂−d _N−k−1>0. And if x ₁=c _N−k−1 and x ₂=d _N−k−1, then both players bid d _N−k−1+c _N−k−1 and player 2 gets 0, whether he gets or not the wallet (given that Proposition 2 does not specify how to share the wallet in case of a tie).

• If x ₂<d ₁, player 2 bids less than any type of player 1 and losing the auction is his best strategy because winning would at best lead to the payoff x ₂−d ₁<0.

And no strategy is dominated, given that each player of type x bids an amount between x and x+1. □

Appendix B

Proof of Proposition 4

We first establish that b ₁(x ₁) is a best response:

If x ₁>c−K player 1 best replies against any type of player 2 because:

• First she wins against any type of player 2, because x ₁+1+K>c+1 and player 2 never bids more than c+1.

• Second, she earns a positive payoff by winning, because she gets either x ₁+x ₂+K−x ₂−c or x ₁+x ₂+K−x ₂ depending on player 2’s type, which are both positive.

If x ₁≤c−K:

• She loses against any type of player 2 such that x ₂≥d, because x ₁+d≤c−K+d and x ₂+c≥d+c. And losing gives her the best payoff, 0, because winning would lead to the payoff x ₁+x ₂+K−x ₂−c≤0.

• She wins against any type of player 2 such that x ₂<d, because x ₂<d≤x ₁+d, and she gets a positive payoff by winning, given that x ₁+x ₂+K−x ₂>0.

We now establish that b ₂(x ₂) is a best response:

If x ₂≥d:

• Player 2 loses against any type of player 1 such that x ₁>c−K (see above). And he is best off losing, because winning would lead to the payoff x ₂+x ₁−x ₁−1−K<0.

• He wins against player 1 if x ₁≤c−K (see above); and he is best off winning because he gets x ₂+x ₁−x ₁−d≥0.

If x ₂<d, he loses against any type of player 1 (see above), and he is best off losing because, by winning, he would get x ₂+x ₁−x ₁−d<0, if facing a player 1 type x ₁≤c−K, and he would get x ₂+x ₁−x ₁−1−K<0 if facing a player 1 type x ₁>c−K.

The equilibrium contains no weakly dominated strategy given that player 1’s bid is between x ₁+K and x ₁+1+K, and player 2’s bid is between x ₂ and x ₂+1. □

Appendix C

Proof of Proposition 5

Let us generalise the bidding strategies of Fig. 1 to bidding strategies with any number of discontinuities. The case with two discontinuities is illustrated in Fig. 5.

Fig. 5

• means that the point is excluded, a bracket means that it is included. The full lines are player 1’s bids, the dashed lines are player 2’s bids

The key feature of Fig. 5 is the shaded area, which includes the segments [FG] and [HI] and the points E and J, and which doesn’t exist in Fig. 1. It is immediate that similar shaded areas exist for any number of discontinuities higher than one (n−1 shaded areas for n discontinuities).

In what follows, attention will be focused on Fig. 5’s shaded area to show that the bidding strategies in this area are incompatible with no ex-post regret Nash equilibrium. So, given that n discontinuities lead to n−1 similar areas, we prove the nonexistence of equilibria with n discontinuities simply by focusing on the case with 2 discontinuities.

It actually comes out that we only need to prove that the bidding strategies associated with player 1’s type C ₁ and player 2’s type D ₂ are not compatible with a Nash equilibrium with no ex-post regret.

Let us precise the bidding behaviour:

Player 1 bids:

$$\begin{array}{l@{\quad}l@{\quad}l} {b}_{13}({x}_{1})& \mbox{if } {x}_{1} > {C}_{2}\\\noalign{\vspace{3pt}} {b}_{12}({x}_{1})& \mbox{if } {C}_{1} < {x}_{1} \leq{C}_{2}& \mbox{with } {b}_{13}({C}_{2} ) > {b}_{12}(C_{2})\\\noalign{\vspace{3pt}} {b}_{11}({x}_{1})& \mbox{if } {x}_{1} \leq{C}_{1}&\mbox{with } {b}_{12}({C}_{1} ) > {b}_{11}(C_{1}) \end{array}$$

Player 2 bids:

$$\begin{array}{l@{\quad}l@{\quad}l} {b}_{23}({x}_{2})& \mbox{if } {x}_{2} \geq{D}_{2}\\\noalign{\vspace{5pt}} {b}_{22}({x}_{2})& \mbox{if } {D}_{1} \leq{x}_{2} < {D}_{2}& \mbox{with } {b}_{23}({D}_{2} ) > {b}_{22}(D_{2})\\\noalign{\vspace{5pt}} {b}_{21}({x}_{2})& \mbox{if } {x}_{2} < {D}_{1}&\mbox{with } {b}_{22}({D}_{1} ) > {b}_{21}(D_{1}) \end{array}$$

with b _ij (x _i ), i=1,2, j=1,2,3 increasing continuous functions in x _i .

Given that we try to generalise the equilibria of Proposition 4 in order to come close to equilibria of Proposition 2, we have b ₁₂(C ₁+)≥b ₂₂(D ₂−), with C ₁+ close to C ₁ (and C ₁+>C ₁) and D ₂− close to D ₂ (and D ₂−<D ₂). By continuity of the functions b _ij (x _j ), i=1,2, j=1,2,3, it follows that b ₁₂(C ₁)≥b ₂₂(D ₂).

Now let us focus on player 1 of type C ₁ and player 2 of type D ₂ (the types associated to the points E and J).

Player 1 of type C ₁+ bids more than any type of player 2 in [D ₁ D ₂); so, given the no ex-post regret property, C ₁+ has to get a positive or null payoff when she faces D ₂−, i.e. C ₁+K+D ₂−b ₂₂(D ₂)≥0 given the continuity of the bid functions. And player 1 of type C ₁ has to get a negative or null payoff when she faces any type of player 2 in [D ₁ D ₂) (because she bids less and loses the auction). So she has to get a negative or null payoff when she faces D ₂−, hence C ₁+K+D ₂−b ₂₂(D ₂)≤0 given the continuity of the bid functions.

It thus follows that

$$ {C}_{1} +{K}+ {D}_{2}-b_{22}(D_{2} ) = 0 $$

(1)

Player 2 of type D ₂ bids more than any type of player 1 in (C ₁,C ₂], so, given the no ex-post regret property, D ₂ has to get a positive or null payoff when he faces C ₁+, i.e. C ₁+D ₂−b ₁₂(C ₁)≥0 given the continuity of the bid functions. And player 2 of type D ₂− has to get a negative or null payoff when he faces any type of player 1 in (C ₁,C ₂]. So he has to get a negative or null payoff when he faces C ₁+; hence C ₁+D ₂−b ₁₂(C ₁)≤0 given the continuity of the bid functions.

It thus follows that

$$ {C}_{1} + {D}_{2}-b_{12}(C_{1} ) = 0 $$

(2)

Given that b ₁₂(C ₁)≥b ₂₂(D ₂), from (1) and (2) we get:

$${b}_{12}(C_{1} ) = {C}_{1} +{D}_{2} \geq{b}_{22}(D_{2} )= {C}_{1} +{K}+ {D}_{2}$$

which is clearly impossible, given K>0.

Let us add that, to get this result, we do not need the linear bid functions illustrated in Fig. 5: the proof works with any increasing continuous bid functions b(x). □