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All-pay auctions with ties

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Abstract

We study the two-player, complete information all-pay auction in which a tie ensues if neither player outbids the other by more than a given amount. In the event of a tie, each player receives an identical fraction of the winning prize. Players thus engage in two margins of competition: losing versus tying, and tying versus winning. Two pertinent parameters are the margin required for victory and the value of tying relative to winning. We fully characterize the set of Nash equilibria for the entire parameter space. For much of the parameter space, there is a unique Nash equilibrium which is also symmetric. Equilibria typically involve randomizing over multiple disjoint intervals, so that in essence players randomize between attempting to tie and attempting to win. In equilibrium, expected bids and payoffs are non-monotonic in both the margin required for victory and the relative value of tying.

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Notes

  1. For example, in the first great engine war of the mid-1980s, the U.S. Air Force split the award of a $10 billion engine contract between General Electric and Pratt & Whitney. A key feature of that split-award decision was the strategic uncertainty as to whether a single proposal would sufficiently dominate the competition and win the contract outright or whether the proposals would be relatively close and result in a split contract. More recently, the second great engine war over the contract to supply engines for the F-35 joint strike fighter currently features a single winner (or supplier), Pratt & Whitney. See Drewes (1987) and Amick (2005) for further details.

  2. The list of applications is widespread and includes lobbying, litigation, R&D competitions, college admissions, election campaigns, warfare, etc. Konrad (2009) and Dechenaux et al. (2015), respectively, survey the theoretical and experimental literature.

  3. Again, the notion of a tie here is that player’s expenditures are close enough for an intermediate prize outcome to occur (such as political gridlock or a battlefield stalemate). Just as expenditures may differ in reaching such an outcome, players’ net payoffs (the value of the intermediate prize less the expenditure) may likewise differ.

  4. The use of split-awards is part of the broader question of how to best design procurement and innovation contests. Examples of this broader literature include Che and Gale (2003), Schöttner (2008), and Postl (2013).

  5. Ties have also been studied in the logit-type contest of Tullock (1980), as well as in the rank-order (difference-form) tournament of Lazear and Rosen (1981). Examples of the former include Blavatskyy (2010), Jia (2012), and Yildizparlak (2017), while examples of the latter include Nalebuff and Stiglitz (1983), Eden (2006), and Imhof and Kräkel (2014, 2015). The possibility of ties under these contest success functions is discussed in more detail in Gelder et al. (2015).

  6. Che and Gale show that if bidders in an all-pay auction have asymmetric valuations for the winning prize, then an auction-designer can increase expected revenue by introducing a bidding cap that levels the playing field by reducing the stronger player’s ability to outbid the weaker player. Szech (2015) then goes on to show that the auction-designer can do even better, with regard to equilibrium expected expenditure, by introducing an asymmetric tie-breaking rule that favors the weaker player (instead of using a symmetric tie-breaking rule where each player wins the prize with equal probability in the event of a tie).

  7. Stong focuses on equilibria where the upper bound of the support is no more than twice the size of the tie margin. (In our map of the parameter space, this corresponds to region II of Fig. 5.) His model has incomplete information in that players have privately known valuations of the winning prize and privately known bidding costs. The ratio of the tie prize to the winning prize is, however, common across players.

  8. A bid of precisely \(\delta \) is an exception. Tying instead of beating \(\alpha (0)>0\) leads to a strictly lower payoff at \(\delta \) compared to bids arbitrarily close to \(\delta \) from above. However, since distributional supports are necessarily closed sets, a bid of \(\delta \) is permitted to be in the support as the endpoint of an interval. Even still, equilibrium behavior is invariant to imposing a special tie-breaking rule where a bid of \(\delta \) beats a bid of zero—a bid of \(\delta \) occurs with zero probability either way. Other examples of select points within equilibrium supports having lower payoffs due to the presence of mass points include Osborne and Pitchik (1986) and Deneckere and Kovenock (1996).

  9. Although the exact number of interval pairs k is implicitly defined, there is a finite set of possible values it can take. The upper bound of that set \(\left\lfloor v/\delta \right\rfloor \), where \(\left\lfloor \cdot \right\rfloor \) is the floor function, is based on the fact that intervals within a pair have lower bounds that are spaced \(\delta \) apart, and a bid of zero strictly dominates bids greater than v.

  10. For instance, if \(\delta =v/5\), there is a mass point at zero of 1 / 5. Another 1 / 5 is distributed over \([\delta ,\ 2\delta ]\); 2 / 5 is distributed over \([2\delta ,\ 3\delta ]\); and 1 / 5 is distributed over \([3\delta ,\ 4\delta ]\). Note that there is no unique upper bound in this case as a multitude of distributions over \([\delta ,\ 2\delta ]\) and \([3\delta ,\ 4\delta ]\) can satisfy \({\mathscr {P}}\). When \(\delta = v/(4p+2)\), such as when \(\delta = v/10\), there is no mass in \([\delta ,\ 2\delta ]\), so \({\mathscr {P}}\) can only be satisfied when \([2\delta ,\ 4\delta ]\) has a density rate of 2 / v. This is a rare case of uniqueness. See Lemma 12 in the Appendix for further details.

  11. The intervals span \([0,\ v/2 - \delta (2p+1)]\) and \([\delta ,\ v/2 - 2\delta p]\). The lower bound of the bottommost length-\(4\delta \) pattern of \({\mathscr {P}}\) begins promptly at \(v/2 - 2\delta p\). This implies that the distribution has a unique upper bound.

  12. In terms of rent seeking, a tie prize of \(\beta > 1/2\) leads to rent creation in the event of a tie, whereas \(\beta < 1/2\) leads to rent destruction. The full rent of the game is endogenous and varies based on whether there is a tie or a clear winner.

  13. Expected payoffs monotonically decrease over \((\phi _{j} + 2\delta ,\ \phi _{j+1}-\delta ]\) since there is no additional mass to tie, and beating mass that was previously tied cannot compensate for the cost of bidding (i.e., \((1-\beta )/\beta < 1)\). Expected payoffs then rise over \([\phi _{j+1}-\delta ,\ \phi _{j+1})\) since tying mass at a rate of \(1/[(1-\beta )v]\) more than covers the bidding cost (i.e., \(\beta /(1-\beta ) > 1\)). Similar arguments rule out deviations in \([\phi _{1}-\delta ,\ \phi _{1})\) or \((\phi _{p}+2\delta ,\ \phi _{p}+3\delta ]\), where p is the total number of length-\(2\delta \) intervals. The high value of tying also precludes the incentive to win with certainty by outbidding the entire distribution by \(\delta \).

  14. Expected utilities at 0 and \(\phi _{1}\) are contingent on the presence or absence of an interval pair at zero and \(\delta \). When absent, \(u_{i}(0,\ G_{-i}) = \alpha (0)\beta v\) and \(u_{i}(\phi _{1},\ G_{-i}) = \alpha (0)v + \left[ \delta /(1-\beta )v\right] \beta v - \phi _{1}\), so \(\phi _{1} = \alpha (0) (1-\beta )v + [\delta \beta / (1-\beta )]\). The size of the mass point at zero is then simply the remainder after each of the length-\(2\delta \) intervals. Specifically, for p intervals of length-\(2\delta \), \(\alpha (0) = 1 - [\delta p/(1-\beta )\beta v]\).

  15. A bid of \(\delta \) technically ties a bid of zero, but a deviating player would want to bid on the extreme low end of \((\delta ,\ \phi _{1}-\delta )\) to reduce bidding costs. The notation \(u_{i}({\overline{\delta }},\ G_{-i})\) is repeatedly used in the proofs in the Appendix.

  16. Since a bid of \(\psi +\delta \) does not tie \(\phi _{1}\), this is equivalent to Eq. 4 with \(y_{1}\) replaced by \(\psi \).

  17. The existence of this equilibrium requires a slight alteration of Eq. 1: \(u_{i}(x_{i},\, x_{-i}) = v - x_{i}\) if \(x_{i} - x_{-i}=\delta \). Otherwise, there is no minimum distance greater than \(\delta \) to space the mass points by. A similar equilibrium with mass points spaced at regular intervals arises in the difference-form contest of Che and Gale (2000). There, when players’ bids are relatively close, there is some probability that the player with the lower bid may win. (The remaining probability is made up by luck.) Winning is only assured when the bidding difference is above a given threshold. In equilibrium, the mass points are spaced according to that threshold, just as they are here. Ewerhart (2015) identifies a related pattern of staggered mass points (although unevenly spaced) in Tullock contests with intermediate values of the decisiveness parameter (\(2< R < \infty \)). He explains that “intuitively, the suboptimality of bids placed [strictly between equilibrium mass points] captures a cost of being ‘halfhearted’ in the sense that such positive bids are too low to be effective against a decisive action by the opponent, but at the same time too high as a measured defense against speculative underbidding” (p. 66–67).

  18. If \(v/\delta = \left\lfloor v/\delta \right\rfloor \), the topmost mass point is at \(\left( \left\lfloor v/\delta \right\rfloor - 1\right) \delta \) and has a mass of \((\delta /v)\) like the other mass points. Otherwise, it is at \(\left\lfloor v/\delta \right\rfloor \delta \) and has the remaining mass of \(1 - \left\lfloor v/\delta \right\rfloor (\delta /v)\).

  19. The top red line at \(\delta = 33.33\) (or v / 3) involves one pair of intervals: \([0,\ 17.78]\) and \([33.33,\ 51.11]\), with respective density rates of 1 / 60 and 1 / 40 (alternatively, \(1/[(1-\beta )v]\) and \(1/\beta v\)). The second red line at \(\delta =25\) (or v / 4) has two pairs of intervals, the first pair in gray (\([0,\ 8.67]\) and \([25,\ 33.67]\)), and the second in blue (\([33.67,\ 45.67]\) and \([58.67,\ 70.67]\)). Then at \(\delta =20\) (or v / 5) there are three interval pairs, marked gray, blue, and gray again (\([0,\ 4.44]\) and \([20,\ 24.44]\); \([24.44,\ 33.14]\) and \([44.44,\ 53.14]\); \([53.14,\ 60.10]\) and \([73.14,\ 80.10]\)).

  20. To illustrate this, and to show the nature of the equilibrium support throughout the parameter space, the online appendix contains eight panels similar to Fig. 6 that show the equilibrium support for \(\delta \in (0,\, (1-\beta )v)\) when \(\beta \in \{0.1,\, 0.3,\, 0.45,\, 0.49999,\, 0.50001,\, 0.55,\, 0.7,\, 0.9\}\) and \(v=100\).

  21. For \(\beta = 1/2\), asymmetric equilibria occur along parts of the line where \(\delta \in (0,\, v/4]\). Such equilibria are governed by Property \({\mathscr {P}}\). Our focus in this section is \(\beta < 1/2\).

  22. The notation \(u_{i}({\overline{\delta }},\; G_{-i})\) is defined shortly before Eq. 9 on p. 11 as the limit of player i’s expected utility as \(\delta \) is approached from above.

  23. Recall that Eq. 14 is written in terms of player i’s indifference: \(u_{i}(x,\; G_{-i}) = u_{i}(y,\; G_{-i})\). So the corresponding version of Eq. 14 for player \(-i\) replaces each \(G_{-i}\) with \(G_{i}\).

  24. This pattern is easier to see by writing \(2\beta -1\) as \(\beta - (1-\beta )\) for \(\ell =1\); and then \(\beta ^{2} - \beta (1-\beta ) + (1-\beta )^{2}\) for \(\ell =2\); etc.

  25. For the case of \({\overline{y}}_{1}={\overline{w}}_{1} - \delta \), it might not be the case that \({\overline{y}}_{1} - \delta \in \text {supp}(G_{w})\), but we still have the desired result that \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\overline{w}}_{1}-4\delta )\). To see that, note that Eq. 29 can only be satisfied if \(G_{y}\) has the maximum density rate of 2 / v over \([{\overline{w}}_{1}-3\delta , {\overline{w}}_{1}-\delta ]\). If \(G_{y}\) had any mass in \(({\overline{w}}_{1}-4\delta ,\ {\overline{w}}_{1}-3\delta )\), then Eq. 14 could not be satisfied for any elements in the support of \(G_{w}\) over \(({\overline{w}}_{1}-3\delta , {\overline{w}}_{1}-2\delta )\) (since \(G_{y}\) already has a density rate of 2 / v over \([{\overline{w}}_{1}-3\delta , {\overline{w}}_{1}-\delta ]\)). If \(({\overline{w}}_{1}-3\delta , {\overline{w}}_{1}-2\delta )\) is not in the support of \(G_{w}\), then it must be that at least one point (possibly an endpoint) in \([{\overline{w}}_{1}-2\delta ,\ {\overline{w}}_{1}-\delta ]\) is in the support of \(G_{w}\). (This follows because at the maximal density rate of 2 / v, a mass of \(4\delta /v\) has to spread over at least \(2\delta \), and by Lemma 10, since \({\overline{w}}_{1}-\delta \in \text {supp}(G_{y})\), \([G_{w}({\overline{w}}_{1}) - G_{w}({\overline{w}}_{1}-4\delta )]=4\delta /v\).) For any \(x \in \text {supp}(G_{w})\cap [{\overline{w}}_{1}-2\delta ,\ {\overline{w}}_{1}-\delta ]\), we have \([G_{y}({\overline{y}}_{1}) - G_{y}(x-3\delta )]=4\delta /v\) by Lemma 10. Since \(x-3\delta \le {\overline{w}}_{1}-4\delta \) and Eq. 29 must hold, \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\overline{w}}_{1}-4\delta )\).

  26. As demonstrated previously, any bid in \([m_{z}-3\delta ,\ m_{z}-4\delta ]\) has the expected equilibrium payoff. For the density rates at \(x \in [m_{z}-3\delta ,\ m_{z}-2\delta ]\) and \(x-2\delta \) to sum to 2 / v, it must be that x, \(x-2\delta \), or both are in the support. If \(x-2\delta \) is in the support, it must have the expected equilibrium payoff. Otherwise, x must be in the support, and \(x-2\delta \) has the expected equilibrium payoff by Lemma 10.

  27. The claim that \(G_{w}\) has no mass in \([{\overline{y}}_{p}-5\delta ,\ \delta )\) is contingent on \({\overline{y}}_{p} \in (5\delta , {\overline{w}}_{p}]\). We can quickly rule out the possibility of \({\overline{y}}_{p} \in ({\overline{w}}_{p}-\delta ,\ 5\delta ]\): The lack of mass in \(G_{y}\) over \([{\overline{y}}_{p}-4\delta , {\overline{y}}_{p}-3\delta ]\) would preclude mass in \(G_{w}\) over \((0, {\overline{y}}_{p}-4\delta )\) (see Lemma 2). Since there is also no mass in \(G_{w}\) over \([2\delta , {\overline{w}}_{p}-3\delta ]\), there would be no way for player y to recover the bidding cost between \(\delta \) and \({\overline{w}}_{p}-4\delta \) (i.e., the next element in \(G_{y}\) above \({\overline{w}}_{p}-4\delta \) would need to be compensated for by more than the maximal density rate of 2 / v). Hence, \(u_{y}({\overline{\delta }},\ G_{w})\) would be strictly greater than all bids in \(G_{y}\) above \({\overline{w}}_{p}-4\delta \).

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Correspondence to Alan Gelder.

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We gratefully acknowledge helpful comments from David Malueg, Roman Sheremeta, Christian Ewerhart, Jaehong Kim, the faculty of the Economic Science Institute, two anonymous referees, and the co-editor. This work has also benefited from the comments of participants at many seminar presentations and conferences, including the 15th SAET (Cambridge, England), the 85th Southern Econ. Assoc. (New Orleans, LA), and the 2017 American Econ. Assoc. (Chicago, IL) conferences.

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Appendices

Appendices

General necessary conditions

Here we present the necessary conditions for equilibrium. Our first result limits the occurrence of a mass point to a bid of zero. If a mass point were to occur elsewhere, the opponent could profit by either barely tying it from below or slightly beating it from above, which in turn provides an incentive for moving the mass point. A mass point at zero is different, however, because it cannot be undercut.

Lemma 1

Let \(\delta >0\) and \(\beta > 0\). A positive mass point may only occur at zero in any equilibrium.

Proof

Suppose that \(G_{-i}\) includes a mass point at \(x^{*} > 0\) of size \(\alpha _{-i}(x^{*}) \in (0,1]\). For any \(k \in (0,\delta )\), unless \(G_{i}\) either has some mass in \([x^{*}+\delta -k,\; x^{*}+\delta ]\) which \(x^{*}\) can tie or some mass in \([x^{*}-\delta -k,\; x^{*}-\delta )\) which \(x^{*}\) can beat, then it is profitable for player \(-i\) to move \(x^{*}\) down since this reduces the cost of bidding but maintains the probability of a tie and a win. Suppose that \(G_{i}\) does indeed contain mass in \([x^{*}+\delta -k,\; x^{*}+\delta ]\) or \([x^{*}-\delta -k,\; x^{*}-\delta )\) for any arbitrarily small k. We will show that there exists a k such that placing mass in either interval is not optimal for player i, in which case a mass point at \(x^{*}\) cannot be a best response for player \(-i\). We begin by showing that there exists a sufficiently small \(\lambda > 0\) such that player i strictly prefers a bid of \(x^{*} + \delta + \lambda \) to a bid of \(x^{*}+\delta -\mu \) for any \(\mu \) such that \(\lambda > \mu \ge 0\). From Eq. 2, we have:

$$\begin{aligned}&u_{i}(x^{*}\!+\!\delta \!+\!\lambda ,\; G_{-i}) - u_{i}(x^{*}\!+\!\delta \!-\!\mu ,\; G_{-i}) \\&\quad = [G_{-i}(x^{*}\!+\!2\delta \!+\!\lambda ) - G_{-i}(x^{*}\!+\!2\delta \!-\!\mu )]\beta v\\&\qquad + [G_{-i}(x^{*}\!+\!\lambda ) - G_{-i}(x^{*}\!-\!\mu ) - \alpha _{-i}(x^{*}\!+\!\lambda ) \\&\qquad + \alpha _{-i}(x^{*}\!-\!\mu )](1-\beta )v - \lambda - \mu \end{aligned}$$

The mass in \(G_{-i}\) between \(x^{*}\!-\!\mu \) and \(x^{*}\!+\!\lambda \) must be at least as large as the mass point at \(x^{*}\): \(G_{-i}(x^{*}\!+\!\lambda ) - G_{-i}(x^{*}\!-\!\mu ) \ge \alpha _{-i}(x^{*})\). Since there are at most countably many mass points in \(G_{i}\), there exists such a \(\lambda > 0\) sufficiently small such that \(\alpha _{-i}(x^{*}+\lambda )=0\). Hence, for such a sufficiently small \(\lambda \):

$$\begin{aligned} u_{i}(x^{*}\!+\!\delta \!+\!\lambda ,\; G_{-i}) - u_{i}(x^{*}\!+\!\delta \!-\!\mu ,\; G_{-i})\ \ge \ \alpha _{-i}(x^{*})(1-\beta )v - \lambda - \mu \end{aligned}$$

Since \(\lambda > \mu \ge 0\), then \(u_{i}(x^{*}\!+\!\delta \!+\!\lambda ,\; G_{-i}) - u_{i}(x^{*}\!+\!\delta \!-\!\mu ,\; G_{-i})>0\) holds whenever \(\lambda + \mu < \alpha _{-i}(x^{*})(1-\beta )v\), or rather, whenever \(\lambda < \alpha _{-i}(x^{*})(1-\beta )(v/2)\). So for any k that is less than such a \(\lambda \), player i strictly prefers a bid of \(x^{*} + \delta + \lambda \) to any bid in \([x^{*}+\delta -k,\ x^{*}+\delta ]\). Since bids below zero are not possible, this completes the proof for \(x^{*} \in (0,\ \delta ]\). For \(x^{*} > \delta \), we must also rule out the possibility of mass in \([x^{*}-\delta -k,\; x^{*}-\delta )\). Similar to the previous argument, we will show that player i strictly prefers a bid of \(x^{*} - \delta \) to a bid of \(x^{*} - \delta - \gamma \) where \(\gamma > 0\):

$$\begin{aligned}&u_{i}(x^{*}\!-\!\delta \!,\; G_{-i}) - u_{i}(x^{*}\!-\!\delta \!-\!\gamma ,\; G_{-i}) \\&\quad = [G_{-i}(x^{*}) - G_{-i}(x^{*}\!-\!\gamma )]\beta v\\&\qquad +\, [G_{-i}(x^{*}\!-\!2\delta ) - G_{-i}(x^{*}\!-\!2\delta \!-\!\gamma ) - \alpha _{-i}(x^{*}\!-\!2\delta ) \\&\qquad +\, \alpha _{-i}(x^{*}\!-\!2\delta \!-\!\gamma )](1-\beta )v - \gamma \\&\quad \ge \alpha _{-i}(x^{*})\beta v - \alpha _{-i}(x^{*}-2\delta )(1-\beta )v - \gamma \end{aligned}$$

The inequality follows from observing that \(G_{-i}(x^{*}) - G_{-i}(x^{*}\!-\!\gamma ) \ge \alpha _{-i}(x^{*})\) and that the omitted terms are weakly positive. Then if \(\gamma < \alpha _{-i}(x^{*})\beta v - \alpha _{-i}(x^{*}-2\delta )(1-\beta )v\), player i profits from moving any mass in \([x^{*}-\delta -\gamma ,\; x^{*}-\delta )\) up to \(x^{*}-\delta \). Moreover, if \(\alpha _{-i}(x^{*})\beta > \alpha _{-i}(x^{*}-2\delta )(1-\beta )\), then such a \(\gamma \) exists, which in turn implies that there is a k that meets our requirements. (In particular, any \(k < \min \{\gamma ,\ \lambda \}\) where \(\gamma \) and \(\lambda \) satisfy the bounds specified above.) Since \(\alpha _{-i}(x^{*}-2\delta )\) necessarily equals zero for \(x^{*} \in (0,\ 2\delta )\), mass points in \((0,\ 2\delta )\) are not a best response. Now suppose that \(\alpha _{-i}(x^{*})\beta \le \alpha _{-i}(x^{*}-2\delta )(1-\beta )\) where \(x^{*} \ge 2\delta \). Following the above argument, \(x^{*}-2\delta \) can only be sustained as a mass point in equilibrium if \(\alpha _{-i}(x^{*}-2\delta )\beta \le \alpha _{-i}(x^{*}-4\delta )(1-\beta )\). More generally, a mass point in equilibrium at \(x^{*}-2q\delta \) requires that \(\alpha _{-i}(x^{*}-2q\delta )\beta \le \alpha _{-i}(x^{*}-2[q+1]\delta )(1-\beta )\) for \(q \in \{0,\ldots , \left\lfloor x^{*}/2\delta \right\rfloor - 1 \}\). However, since there are no mass points in \((0,\ 2\delta )\), then provided that \(x^{*}\) is not evenly divisible by \(2\delta \), there cannot be a mass point at \(x^{*}-2q\delta \). The final case to consider is a sequence of mass points at 0, \(2\delta \), \(4\delta \), etc. It suffices to show that a mass point at \(2\delta \) is not a best response; any successive mass points would then fail to hold in equilibrium. For \(\alpha _{-i}(2\delta )>0\) to be sustained in equilibrium, \(G_{i}\) must contain mass in a neighborhood immediately below \(\delta \). This, however, cannot be. Since \(\alpha _{-i}(2\delta )>0\) requires that \(\alpha _{-i}(0)>0\), player i strictly prefers a bid slightly above \(\delta \) to a bid of \(\delta -c\) where \(c < \alpha _{-i}(0)(1-\beta )v\). Specifically, \(u_{i}({\overline{\delta }},\; G_{-i})-u_{i}(\delta -c,\; G_{-i})=[G_{-i}(2\delta )-G_{-i}(2\delta -c)]\beta v + \alpha _{-i}(0)(1-\beta )v - c > 0\) for \(c < \alpha _{-i}(0)(1-\beta )v\).Footnote 22 Therefore, \(k<\min \{c,\ \lambda \}\) with c and \(\lambda \) meeting their respective bounds satisfies our requirements, so a mass point at \(2\delta \) is not optimal. \(\square \)

With mass points limited to zero, the next result stems from the indifference condition that must hold when players are randomizing between multiple bids. That is, \(u_{i}(x,\; G_{-i}) = u_{i}(y,\; G_{-i})\) for \(x, y \in \text {supp}(G_{i})\) where \(x > y\). Using Lemma 1 and Eq. 2, we can restate this indifference condition as:

$$\begin{aligned} \left[ G_{-i}(x+\delta ) - G_{-i}(y+\delta )\right] \beta v + \left[ G_{-i}(x-\delta ) - G_{-i}(y-\delta )\right] (1-\beta )v = x-y\nonumber \\ \end{aligned}$$
(14)

The added cost of the higher bid must either be compensated by tying mass in \([y+\delta ,\; x+\delta ]\) or beating mass in \([y-\delta ,\; x-\delta ]\). Notably, the absence of mass in either of these intervals pins down the necessary distribution over the other. This principle is formalized in the following lemma.

Lemma 2

For \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta > 0\), let \(G_{i}\) and \(G_{-i}\) be equilibrium distributions for players i and \(-i\).

A.:

Let \(b \ge 0\) satisfy \(G_{-i}(b) - \alpha _{-i}(b) = G_{-i}(b-c) - \alpha _{-i}(b-c)\) for some \(c \in (0,\; \delta ]\). If the subset \(({\underline{a}},\; {\overline{a}}] \subseteq (b+\delta -c,\; b+\delta ]\) is in the support of \(G_{i}\), then \(({\underline{a}} + \delta ,\; {\overline{a}} + \delta ]\) is in the support of \(G_{-i}\). Moreover, the distribution over \(({\underline{a}} + \delta ,\; {\overline{a}} + \delta ]\) in \(G_{-i}\) is uniform at the rate of \(1/(\beta v)\).

B.:

Let \(b>\delta \) satisfy \(G_{-i}(b) = G_{-i}(b+c)\) for some \(c \in (0,\; \delta ]\). If the subset \(({\underline{a}},\; {\overline{a}}] \subseteq (b-\delta ,\; b-\delta +c]\) is in the support of \(G_{i}\), then \(({\underline{a}}-\delta ,\; {\overline{a}}-\delta ]\) is in the support of \(G_{-i}\). Additionally, the distribution over \(({\underline{a}}-\delta ,\; {\overline{a}}-\delta ]\) in \(G_{-i}\) is uniform at the rate of \(1/[(1-\beta )v]\).

Proof

Part A. To show the first claim, suppose that the subset \(({\underline{a}} + \delta ,\; {\overline{a}} + \delta ]\) is not in the support of \(G_{-i}\). Since \(G_{-i}\) also has no mass in \([b-c,\; b)\), player i strictly prefers a bid of \({\underline{a}}\) to any \(k \in ({\underline{a}},\; {\overline{a}}]\). This is because the probability of winning or tying remains unchanged, but the cost of bidding is higher (i.e., \(u_{i}({\underline{a}},\; G_{-i}) - u_{i}(k,\; G_{-i}) = k - {\underline{a}} > 0\)). Hence, \(({\underline{a}},\; {\overline{a}}]\) cannot be in the support of \(G_{i}\). For the second claim, suppose that \(({\underline{a}},\; {\overline{a}}]\) is in the support of \(G_{i}\). Consequently, \(({\underline{a}} + \delta ,\; {\overline{a}} + \delta ]\) is then in the support of \(G_{-i}\). Let \(x,y \in ({\underline{a}},\; {\overline{a}}]\) where \(x > y\). By payoff equivalence, \(u_{i}(x,\; G_{-i}) = u_{i}(y,\; G_{-i})\), which then implies that \(G_{-i}(x+\delta ) - G_{-i}(y+\delta ) = (1/\beta v)(x-y)\). Since this equation holds for any x and y, including values which are arbitrarily close, the result then follows.

Part B. If \(({\underline{a}}-\delta ,\; {\overline{a}}-\delta ]\) is not in the support of \(G_{-i}\), then since \(G_{-i}\) has no mass in \((b,\; b+c]\), player i strictly prefers a bid of \({\underline{a}}\) to any \(k \in ({\underline{a}},\; {\overline{a}}]\) (i.e., the probability of a win and a tie remains the same, but k has a higher bidding cost than \({\underline{a}}\)). So \(({\underline{a}},\; {\overline{a}}]\) is not in the support of \(G_{i}\). For the next part, suppose that \(({\underline{a}},\; {\overline{a}}]\) is in the support of \(G_{i}\). By payoff equivalence, we have \(u_{i}(x,\; G_{-i}) = u_{i}(y,\; G_{-i})\) for any \(x, y \in ({\underline{a}},\; {\overline{a}}]\) such that \(x>y\). From this equality, \(G_{-i}(x-\delta ) - G_{-i}(y-\delta ) - \alpha _{-i}(x-\delta ) + \alpha _{-i}(y-\delta ) = (x-y)/[(1-\beta )v]\). Since mass points may only occur at zero in equilibrium (see Lemma 1), and since this equation holds for any x and y which are arbitrarily close, the result then follows. \(\square \)

Lemma 2 is particularly applicable at the upper and lower bounds of a distribution. If \({\overline{x}}_{i}\) is the upper bound of \(G_{i}\), then any mass in \(G_{-i}\) over \([{\overline{x}}_{i} - \delta ,\; {\overline{x}}_{i}]\) must be balanced by mass in \(G_{i}\), shifted down by \(\delta \) in \([{\overline{x}}_{i} - 2\delta ,\; {\overline{x}}_{i}-\delta ]\), with a density rate of \(1/[(1-\beta )v]\). Equation 14 can again be used to identify the necessary density rate for mass in \(G_{i}\) over \([{\overline{x}}_{i} - 4\delta ,\; {\overline{x}}_{i}-3\delta ]\). This process continues—iteratively moving down the distribution—and a similar process holds for moving up the distribution. We therefore obtain the following result.

Lemma 3

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta > 0\). If \(\beta \ne 1/2\), in any equilibrium, all continuously distributed mass must be uniform at a rate of either \(1/(\beta v)\) or \(1/[(1-\beta )v]\).

Proof

Let \([{\underline{s}}_{0},\, {\overline{s}}_{0}] \subseteq \text {supp}(G_{i})\) be given such that \(G_{-i}({\underline{s}}_{0}-\delta ) = G_{-i}({\overline{s}}_{0}-\delta )\). Since bids must be nonnegative, and since the initial naming of players i and \(-i\) is without loss of generality, there exists such an interval if either player has any mass in \((0,\; \delta )\) (mass points may only occur at zero in equilibrium by Lemma 1), and if neither player has any mass in \((0,\; \delta )\), then such an interval must necessarily exist if there is any mass above \(\delta \) in \(G_{i}\). By Lemma 2.A, \([{\underline{s}}_{0}+\delta ,\; {\overline{s}}_{0}+\delta ] \subseteq \text {supp}(G_{-i})\) with uniformly distributed mass at the rate of \(1/(\beta v)\). Likewise, by Lemma 2.B, if \(G_{i}({\underline{s}}_{0}+2\delta ) = G_{i}({\overline{s}}_{0}+2\delta )\), then the distribution over \([{\underline{s}}_{0},\, {\overline{s}}_{0}]\) in \(G_{i}\) is uniform at the rate of \(1/[(1-\beta )v]\). Suppose instead by way of contradiction that \(G_{i}({\underline{s}}_{0}+2\delta ) < G_{i}({\overline{s}}_{0}+2\delta )\). Let \([{\underline{s}}_{1},\, {\overline{s}}_{1}] \subseteq [{\underline{s}}_{0}+2\delta ,\; {\overline{s}}_{0}+2\delta ]\) be such that \([{\underline{s}}_{1},\, {\overline{s}}_{1}] \subseteq \text {supp}(G_{i})\). (By Lemma 1, such an interval must exist since mass points may only occur at zero.) We first show for \(\beta < 1/2\) that we obtain the contradiction that \([{\underline{s}}_{1},\, {\overline{s}}_{1}] \nsubseteq \text {supp}(G_{i})\). Recall that \([{\underline{s}}_{0}+\delta ,\; {\overline{s}}_{0}+\delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{-i}\). So for any \(x, y \in [{\underline{s}}_{1},\, {\overline{s}}_{1}]\), we have \(G_{-i}(x-\delta )-G_{-i}(y-\delta ) = (x-y)/\beta v\). By Eq. 14, player i’s indifference between bids of x and y implies that:

$$\begin{aligned} G_{-i}(x+\delta ) - G_{-i}(y+\delta ) = (x-y)\left( \frac{2\beta - 1}{\beta ^{2} v}\right) \end{aligned}$$
(15)

Since this holds for x and y which are arbitrarily close, the distribution over \([{\underline{s}}_{1}+\delta ,\; {\overline{s}}_{1}+\delta ]\) must be uniform at a rate of \((2\beta - 1)/(\beta ^{2} v)\). For \(\beta < 1/2\), this contradicts the monotonicity of \(G_{-i}\) by implying a negative density rate over \([{\underline{s}}_{1}+\delta ,\; {\overline{s}}_{1}+\delta ]\). So \([{\underline{s}}_{1},\, {\overline{s}}_{1}] \nsubseteq \text {supp}(G_{i})\). Working from the bottom of the distributions upward, this argument holds for any \([{\underline{s}}_{0},\, {\overline{s}}_{0}]\) that is in the bottom \(\delta \) of the continuously distributed mass in the two supports. The argument can also be applied to any \([{\underline{s}}_{0},\, {\overline{s}}_{0}]\) thereafter, since all earlier portions of the two distributions have density rates of \(1/(\beta v)\), \(1/[(1-\beta )v]\), or zero. This completes the proof for \(\beta < 1/2\). We therefore turn to the case where \(\beta > 1/2\), where the density rate of \((2\beta - 1)/(\beta ^{2} v)\) over \([{\underline{s}}_{1}+\delta ,\; {\overline{s}}_{1}+\delta ]\) in \(G_{-i}\) still holds. We next show that \([{\underline{s}}_{1}+2\delta ,\; {\overline{s}}_{1}+2\delta ]\) would also be in the support of \(G_{i}\).

Suppose by way of contradiction that \(G_{i}({\underline{s}}_{1}+2\delta ) = G_{i}({\overline{s}}_{1}+2\delta )\). By Lemma 2.B, mass over the interval \([{\underline{s}}_{1},\, {\overline{s}}_{1}]\) in \(G_{i}\) must be uniform with a density rate of \(1/[(1-\beta )v]\). Using Eq. 14, the indifference condition \(u_{-i}(x,\; G_{i}) = u_{-i}(y,\; G_{i})\), for \(x,y \in [{\underline{s}}_{1}-\delta ,\; {\overline{s}}_{1}-\delta ] \subseteq [{\underline{s}}_{0}+\delta ,\; {\overline{s}}_{0}+\delta ]\) with \(x>y\) implies:Footnote 23

$$\begin{aligned} G_{i}(x-\delta ) - G_{i}(y-\delta ) = (x-y)\left( \frac{1-2\beta }{(1-\beta )^{2}v}\right) \end{aligned}$$

For \(\beta > 1/2\), this implies a negative density rate in \(G_{i}\) over \([y-\delta ,\ x-\delta ] \subseteq [{\underline{s}}_{0},\ {\overline{s}}_{0}]\), contradicting the initial assumption that \([{\underline{s}}_{0},\, {\overline{s}}_{0}] \subseteq \text {supp}(G_{i})\). This contradiction is reached if any subset of \([{\underline{s}}_{1}+2\delta ,\; {\overline{s}}_{1}+2\delta ]\) is not in the support of \(G_{i}\). Thus, if \([{\underline{s}}_{1}+\delta ,\; {\overline{s}}_{1}+\delta ] \subseteq \text {supp}(G_{-i})\), it must be that \([{\underline{s}}_{1}+2\delta ,\; {\overline{s}}_{1}+2\delta ] \subseteq \text {supp}(G_{i})\). This argument holds more generally. Letting \([{\underline{s}}_{1}+(2\ell -3)\delta ,\; {\overline{s}}_{1}+(2\ell -3)\delta ]\) and \([{\underline{s}}_{1}+(2\ell -1)\delta ,\; {\overline{s}}_{1}+(2\ell -1)\delta ] \subseteq \text {supp}(G_{-i})\) for any \(\ell \in {\mathbb {N}}\), and supposing that \([{\underline{s}}_{1}+2\ell \delta ,\; {\overline{s}}_{1}+2\ell \delta ] \nsubseteq \text {supp}(G_{i})\), then the density rates over \([{\underline{s}}_{1}+(2\ell -2)\delta ,\; {\overline{s}}_{1}+(2\ell -2)\delta ]\) and \([{\underline{s}}_{1}+(2\ell -4)\delta ,\; {\overline{s}}_{1}+(2\ell -4)\delta ]\) in \(G_{i}\) are \(1/[(1-\beta )v]\) and \((1-2\beta )/[(1-\beta )^{2}v]\). A similar statement holds if \([{\underline{c}},\, {\overline{c}}]\nsubseteq \text {supp}(G_{i})\) for any \([{\underline{c}},\, {\overline{c}}] \subseteq [{\underline{s}}_{1}+2\ell \delta ,\; {\overline{s}}_{1}+2\ell \delta ]\). Hence, by contradiction, \([{\underline{s}}_{1}+2\ell \delta ,\; {\overline{s}}_{1}+2\ell \delta ] \subseteq \text {supp}(G_{i})\). Mass over intervals that are \(2\delta \) apart in \(G_{-i}\) requires that \(G_{i}\) has mass in an even higher interval. However, as we will show next, this in turn requires that \(G_{-i}\) has mass in yet a higher interval.

Again drawing on Eq. 14, in order for player i to be indifferent between \(x,y \in [{\underline{s}}_{1}+(2\ell -2)\delta ,\; {\overline{s}}_{1}+(2\ell -2)\delta ]\), where \(x>y\), then:

$$\begin{aligned} G_{-i}(x+\delta ) - G_{-i}(y+\delta ) = \left( \frac{x-y}{\beta v}\right) - [G_{-i}(x-\delta ) - G_{-i}(y-\delta )]\left( \frac{1-\beta }{\beta }\right) \end{aligned}$$

For \(\ell = 1\), the specific value of \(G_{-i}(x+\delta ) - G_{-i}(y+\delta )\) is given by Eq. 15. This value can then be directly inserted for \(G_{-i}(x-\delta ) - G_{-i}(y-\delta )\) when \(\ell = 2\). Iterating, we obtain the following general form for \(\ell \ge 1\):Footnote 24

$$\begin{aligned} G_{-i}(x+\delta ) - G_{-i}(y+\delta ) = (x - y) \left( \frac{1}{\beta ^{\ell +1}v}\right) \sum _{j=0}^{\ell } (-1)^{\ell -j}\beta ^{j}(1-\beta )^{\ell -j} \end{aligned}$$
(16)

The positivity of Eq. 16 for \(\beta > 1/2\) can be seen by doing a pairwise summation of right-hand side terms (i.e., sum \(j=\ell \) with \(j=\ell -1\); \(j=\ell -2\) with \(j=\ell -3\); etc.). Thus, we have:

$$\begin{aligned} \sum _{j=0}^{\ell } (-1)^{\ell -j}\beta ^{j}(1-\beta )^{\ell -j} \!=\! (1\!-\beta )^{\ell }\,{{I}}\,(\ell ) \!+\! (2\beta - 1)\sum _{j=0}^{\left\lfloor (\ell -1)/2 \right\rfloor } \beta ^{\ell -1-2j}(1\!-\beta )^{2j} > 0 \end{aligned}$$

where \({{I}}(\ell )\) is an indicator function equal to 1 if \(\ell \) is even and 0 otherwise, and \(\left\lfloor \cdot \right\rfloor \) is the floor function. Since Eq. 16 is strictly positive for any \(x, y \in [{\underline{s}}_{1}+(2\ell -2)\delta ,\; {\overline{s}}_{1}+(2\ell -2)\delta ]\), then \([{\underline{s}}_{1}+(2\ell -1)\delta ,\; {\overline{s}}_{1}+(2\ell -1)\delta ] \subseteq \text {supp}(G_{-i})\).

The escalating supports of \(G_{i}\) and \(G_{-i}\) ultimately rise above v where bids are strictly dominated, contradicting the initial supposition that \(G_{-i}\) has mass over intervals that are \(2\delta \) apart (i.e., there is no pair of intervals \([{\underline{s}}_{1}+(2\ell -3)\delta ,\; {\overline{s}}_{1}+(2\ell -3)\delta ]\) and \([{\underline{s}}_{1}+(2\ell -1)\delta ,\; {\overline{s}}_{1}+(2\ell -1)\delta ]\) that are both in the support of \(G_{-i}\) for any \(\ell \in {\mathbb {N}}\)). In particular, \([{\underline{s}}_{1}+\delta ,\; {\overline{s}}_{1}+\delta ] \nsubseteq \text {supp}(G_{-i})\). With no mass for player i to tie in \(G_{-i}\) over \([{\underline{s}}_{1}+\delta ,\; {\overline{s}}_{1}+\delta ]\), then \(u_{i}({\underline{s}}_{1},\; G_{-i}) > u_{i}(x,\; G_{-i})\) for any \(x \in ({\underline{s}}_{1},\; {\overline{s}}_{1}]\). Hence, \(G_{i}({\underline{s}}_{0}+2\delta ) = G_{i}({\overline{s}}_{0}+2\delta )\) and \([{\underline{s}}_{0},\, {\overline{s}}_{0}]\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\). Like the case for \(\beta < 1/2\), this argument holds for any \([{\underline{s}}_{0},\, {\overline{s}}_{0}]\) at the bottom of the distribution, and the argument can then be reapplied working up the distribution. \(\square \)

The density rates of \(1/(\beta v)\) and \(1/[(1-\beta )v]\) have intuitive appeal since \(\beta v\) is the marginal value of tying relative to losing and \((1-\beta )v\) is the marginal value of winning relative to tying. In isolating these density rates, we also derive the following corollary:

Corollary 1

Let \(\delta \in (0,\; (1-\beta )v\,)\), \(\beta > 0\), and \(\beta \ne 1/2\). For any \({\overline{z}} > {\underline{z}} \ge 0\) such that \({\overline{z}} - {\underline{z}} \le 2\delta \), in equilibrium, the interval \([{\underline{z}},\, {\overline{z}}]\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\) if and only if \([{\underline{z}}+\delta ,\, {\overline{z}}+\delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{-i}\). In this case, \(G_{-i}({\underline{z}}-\delta )=G_{-i}({\overline{z}}-\delta )\) and \(G_{i}({\underline{z}}+2\delta )=G_{i}({\overline{z}}+2\delta )\). Moreover, an interval with a density rate of \(1/[(1-\beta )v]\) cannot be more than \(2\delta \) in length.

Proofs specific to \(\beta < 1/2\)

Lemma 4

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta < 1/2\). In any equilibrium, any continuously distributed mass within \([0,\, \delta ]\) must be connected, have a lower bound of zero, and have a density rate of \(1/[(1-\beta )v]\). Similarly, if \(p, q \in \text {{supp}}(G_{-i})\) such that \(p<q\) and \(G_{-i}(p) = G_{-i}(q)\), then any continuously distributed mass in \(G_{i}\) within \([p+\delta ,\; q+\delta ]\) must also be connected, have a lower bound of \(p+\delta \), and have a density rate of \(1/[(1-\beta )v]\).

Proof

The density rate of \(1/[(1-\beta )v]\) follows from Lemma 2.B. By way of contradiction to the statement that mass within \([0,\, \delta ]\) must be connected with a lower bound of zero, for some \(z \in (0,\,\delta )\), suppose that there exists \([z,\; z+b] \in \text {supp}(G_{i})\) such that \(G_{i}(z) = G_{i}(z-c)\), where \(b>0\) and \(c \in (0,\, z]\). Since \(z \in \text {supp}(G_{i})\), then in equilibrium, \(u_{i}(z,\; G_{-i}) \ge u_{i}(z-c,\; G_{-i})\), and so:

$$\begin{aligned}{}[G_{-i}(z+\delta ) - G_{-i}(z + \delta - c)]\beta v \ge c \end{aligned}$$
(17)

By Lemma 3, any mass in \(G_{-i}\) over \([z + \delta - c,\; z+\delta ]\) must be uniform at a density rate of \(1/(\beta v)\) or \(1/[(1-\beta )v]\). Since \(G_{i}(z) = G_{i}(z-c)\), Corollary 1 clarifies that mass in \(G_{-i}\) over \([z + \delta - c,\; z+\delta ]\) must have a density rate of \(1/[(1-\beta )v]\). So for some \(s \in [0,\, c]\), Eq. 17 becomes:

$$\begin{aligned}{}[G_{-i}(z+\delta ) - G_{-i}(z + \delta - c)]\beta v = s \beta /(1-\beta ) \ge c \end{aligned}$$

However, this cannot hold since \(s\beta /(1-\beta ) < c\) for \(\beta < 1/2\). The same argument holds for \(z \in (p+\delta ,\; q+\delta ]\) and \(c \in (p+\delta ,\; z]\), where p and q are defined in the statement of the lemma. \(\square \)

Lemma 5

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta < 1/2\). In equilibrium, there does not exist a \({\underline{z}} \ge 0\) and a \({\overline{z}} > {\underline{z}}\) such that \(G_{i}({\underline{z}}-\delta ) = G_{i}({\underline{z}})\) and \([{\underline{z}},\; {\overline{z}}]\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\).

Proof

Suppose to the contrary that in equilibrium there exists a \({\underline{z}} \ge 0\) and a \({\overline{z}} > {\underline{z}}\) such that \(G_{i}({\underline{z}}-\delta )=G_{i}({\underline{z}})\) and \([{\underline{z}},\, {\overline{z}}]\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\). By Corollary 1, \([{\underline{z}}+\delta ,\, {\overline{z}}+\delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{-i}\) (and \({\overline{z}} - {\underline{z}}\) must be no greater than \(2\delta \)). In equilibrium, \(u_{-i}({\underline{z}}+\delta ,\; G_{i}) \ge u_{-i}({\underline{z}},\; G_{i})\). Drawing on Eq. 2 and recalling that \(G_{i}({\underline{z}}-\delta )=G_{i}({\underline{z}})\), this is equivalent to \([G_{i}({\underline{z}}+2\delta ) - G_{i}({\underline{z}}+\delta )]\beta v \ge \delta \). By Lemma 3, this can only be satisfied for \(\beta < 1/2\) if \([{\underline{z}}+\delta ,\, {\underline{z}}+2\delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{i}\) (in which case \({\overline{z}} - {\underline{z}}\) is further restricted to be no greater than \(\delta \)). So now \({\underline{z}}+2\delta \in \text {supp}(G_{i})\). We reach a contradiction in that equilibrium requires \(u_{i}({\underline{z}}+2\delta ,\; G_{-i}) \ge u_{i}({\overline{z}}+2\delta ,\; G_{-i})\). From Eq. 2, \(u_{i}({\underline{z}}+2\delta ,\; G_{-i}) - u_{i}({\overline{z}}+2\delta ,\; G_{-i}) \le ({\overline{z}} - {\underline{z}}) - [G_{i}({\overline{z}}+\delta ) - G_{i}({\underline{z}}+\delta )](1-\beta )v\). Given the density rate established above, this is equal to \(({\overline{z}} - {\underline{z}}) - ({\overline{z}} - {\underline{z}})[(1-\beta )v/\beta v]\) which is negative for \(\beta < 1/2\). \(\square \)

Since every segment with density \(1/[(1-\beta )v]\) must be preceded by some other mass (Lemma 5), equilibrium requires that at least one player must have a mass point at zero. (Recall that Lemma 1 restricts mass points to bids of zero, and Lemma 3 and Corollary 1 restrict the density rates and how they are placed.) The same player that has a mass point at zero must also have some mass in \([0,\; \delta ]\) that is connected, with a density rate of \(1/[(1-\beta )v]\), and a lower bound of zero (i.e., the properties in Lemma 4). If neither player’s distribution began this way, Corollary 1 and Lemmata 4 and 5 would prohibit the placing of any continuously distributed mass in either player’s distribution. Assuming that at least one player’s distribution complies, these results dictate the pattern for placing any further mass.

Labeling the distributions \(G_{w}\) and \(G_{y}\), suppose that \(G_{w}\) has a mass point \(\alpha _{w}(0) \in (0,\; 1)\) and a density rate of \(1/[(1-\beta )v]\) over \([0,\ w_{1}]\) where \(w_{1} > 0\). We can constrain \(w_{1}\) further. From Corollary 1, \(w_{1} \in (0,\ 2\delta ]\), and if \(G_{y}\) has a mass point \(\alpha _{y}(0) > 0\), then \( w_{1}\in (0,\ \delta )\) because player w would prefer to move any mass slightly below \(\delta \) to slightly above it to beat \(\alpha _{y}(0)\). Either way, \(G_{y}\) has a density rate of \(1/(\beta v)\) over \([\delta ,\ \delta +w_{1}]\) (Corollary 1). \(G_{y}\) may only have mass below \(\delta \) if it has a mass point at zero \(\alpha _{y}(0) > 0\), which may only be followed by a density rate of \(1/[(1-\beta )v]\) over \([0,\ y_{1}]\) where \(y_{1} \in [0, \delta )\) (Lemmata 4 and 5). If \(\alpha _{y}(0) > 0\) and \(y_{1} >0\), then \(G_{w}\) has a density rate of \(1/(\beta v)\) over \([\delta ,\ \delta +y_{1}]\) (Corollary 1). If there is any mass in \(G_{y}\) above \(\delta +w_{1}\) or in \(G_{w}\) above \(\delta +y_{1}\) (if \(\alpha _{y}(0) > 0\) and \(y_{1} >0\)), then either \(G_{y}\) must have a density rate of \(1/[(1-\beta )v]\) over \([\delta +w_{1},\ \delta +w_{1}+y_{2}]\) where \(y_{2}>0\), or \(G_{w}\) must have a density rate of \(1/[(1-\beta )v]\) over \([\delta +y_{1},\ \delta +y_{1}+w_{2}]\) where \(w_{2}>0\) (Lemmata 4 and 5).

Every segment with a density rate of \(1/[(1-\beta )v]\) must follow in the immediate wake of a mass point at zero or a segment with a density rate of \(1/(\beta v)\) (Lemmata 4 and 5), and the occurrence of segments with density \(1/(\beta v)\) is wholly determined by the occurrence of segments with density \(1/[(1-\beta )v]\) (Corollary 1). Any equilibrium must therefore be of the form depicted in Fig. 8. Mass points \(\alpha _{w}(0)\) and \(\alpha _{y}(0) \in [0, 1)\) are followed by alternating segments with densities \(1/[(1-\beta )v]\) and \(1/(\beta v)\). We label the length of the successive segments with density \(1/[(1-\beta )v]\) for player w as \(w_{1},\, w_{2}, \ldots ,\, w_{k} \ge 0\), and likewise for player y as \(y_{1},\, y_{2}, \ldots ,\, y_{k} \ge 0\). If indeed some \(w_{j}=0\) then \(y_{j+1}=0\), and like dominoes, \(w_{j+2}=0\), \(y_{j+3}=0\), etc. Symmetric equilibria are therefore restricted to the case where all \(y_{j}\) and \(w_{j}\) are strictly positive (and, by implication, \(\alpha _{w}(0)\), \(\alpha _{y}(0) > 0\)).

Proposition 1

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta < 1/2\). There exists a unique symmetric Nash equilibrium in mixed strategies. That equilibrium is the unique equilibrium satisfying the constraint that for some \(k \in {\mathbb {N}}\), \(y_{1},\,w_{1},\,y_{2},\,w_{2},\,\ldots ,\, y_{k},\, w_{k}\) are all strictly positive. Moreover, for that k, \(y_{1}=w_{1},\ y_{2}=w_{2},\, \ldots ,\, y_{k}=w_{k},\) and \(\alpha _{y}(0)=\alpha _{w}(0)\).

Proof

In equilibrium, all points within the support must have the same expected payoff. This property must particularly hold at each break in each player’s support. Given the constraint that \(y_{1},\,w_{1},\,\ldots ,\, y_{k},\, w_{k}\) are all strictly positive, each player’s support has k breaks. For player w:

$$\begin{aligned}&u_{w}({\overline{\delta }},\ G_{y}) = u_{w}(w_{1},\ G_{y})\\&u_{w}(2\delta + w_{1},\ G_{y}) = u_{w}(\delta + y_{1} + w_{2},\ G_{y})\\&u_{w}(3\delta + y_{1} + w_{2},\ G_{y}) = u_{w}(2\delta + w_{1} + y_{2} + w_{3},\ G_{y})\\&u_{w}(4\delta + w_{1} + y_{2} + w_{3},\ G_{y}) = u_{w}(3\delta + y_{1} + w_{2} + y_{3} + w_{4},\ G_{y}) \end{aligned}$$

In general, for an even integer q:

$$\begin{aligned}&u_{w}\left( q \delta + \sum _{j\ge 1,\, odd}^{q-1} \! w_{j} + \sum _{j\ge 2,\, even}^{q-2} \!\!\! y_{j},\ \ G_{y} \right) \\&\qquad = u_{w}\left( (q\!-\!1)\delta + \sum _{j\ge 1,\, odd}^{q-1} \! y_{j} + \sum _{j\ge 2,\, even}^{q} \!\!\! w_{j},\ \ G_{y}\right) \\&u_{w}\left( (q\!+\!1) \delta + \sum _{j\ge 1,\, odd}^{q-1} \! y_{j} + \sum _{j\ge 2,\, even}^{q} \!\!\! w_{j},\ \ G_{y} \right) \\&\qquad = u_{w}\left( q \delta + \sum _{j\ge 1,\, odd}^{q+1} \! w_{j} + \sum _{j\ge 2,\, even}^{q} \!\!\! y_{j},\ \ G_{y}\right) \end{aligned}$$

Corresponding equations for player y merely reverse the roles of all \(w_{j}\) and \(y_{j}\). Across players, this system has a total of 2k equations with \(2k + 2\) unknowns (i.e., \(\alpha _{y}(0),\, \alpha _{w}(0),\, y_{1},\, w_{1},\,\ldots ,\, y_{k},\, w_{k}\)). We close the system by requiring the total mass in each distribution to sum to one:

$$\begin{aligned}&1 = \alpha _{w}(0) + \sum _{j=1}^{k} \frac{w_{j}}{(1-\beta )v} + \frac{y_{j}}{\beta v}\\&1 = \alpha _{y}(0) + \sum _{j=1}^{k} \frac{y_{j}}{(1-\beta )v} + \frac{w_{j}}{\beta v} \end{aligned}$$

These mass constraints and the system of indifference conditions can be written in matrix form, where each row in the matrix represents an equation. The matrix for the case of \(k = 4\) is as follows:

$$\begin{aligned} { \left( \begin{array}{ccccccccccc} -1\; &{} 1 &{} 0 &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} \\ -1\; &{} 0 &{} 1 &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} \\ -\delta \; &{} (1\!-\!\beta )v &{} 0 &{} 0 &{} 1 &{} \frac{\beta }{1-\beta } &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 \\ -\delta \; &{} 0 &{} (1\!-\!\beta )v &{} 1 &{} 0 &{} 0 &{} \frac{\beta }{1-\beta } &{} 0 &{} 0 &{} 0 &{} 0 \\ -\delta \; &{} 0 &{} 0 &{} 1 &{} \frac{1-2\beta }{\beta } &{} 0 &{} 1 &{} \frac{\beta }{1-\beta } &{} 0 &{} 0 &{} 0 \\ -\delta \; &{} 0 &{} 0 &{} \frac{1-2\beta }{\beta } &{} 1 &{} 1 &{} 0 &{} 0 &{} \frac{\beta }{1-\beta } &{} 0 &{} 0 \\ -\delta \; &{} 0 &{} 0 &{} -1 &{} 1 &{} 1 &{} \frac{1-2\beta }{\beta } &{} 0 &{} 1 &{} \frac{\beta }{1-\beta } &{} 0 \\ -\delta \; &{} 0 &{} 0 &{} 1 &{} -1 &{} \frac{1-2\beta }{\beta } &{} 1 &{} 1 &{} 0 &{} 0 &{} \frac{\beta }{1-\beta } \\ -\delta \; &{} 0 &{} 0 &{} 1 &{} -1 &{} -1 &{} 1 &{} 1 &{} \frac{1-2\beta }{\beta } &{} 0 &{} 1 \\ -\delta \; &{} 0 &{} 0 &{} -1 &{} 1 &{} 1 &{} -1 &{} \frac{1-2\beta }{\beta } &{} 1 &{} 1 &{} 0 \\ \end{array} \right) \left( \begin{array}{c} 1\\ \alpha _{y}(0)\\ \alpha _{w}(0)\\ y_{1}\\ w_{1}\\ y_{2}\\ w_{2}\\ y_{3}\\ w_{3}\\ y_{4}\\ w_{4}\\ \end{array} \right) = \varvec{0} } \end{aligned}$$

The top two rows are the mass constraints for players y and w. The third and fourth rows correspond to \(u_{w}({\overline{\delta }},\ G_{y}) = u_{w}(w_{1},\ G_{y})\) and \(u_{y}({\overline{\delta }},\ G_{w}) = u_{y}(y_{1},\ G_{w})\); the fifth and sixth to \(u_{w}(2\delta + w_{1},\ G_{y}) = u_{w}(\delta + y_{1} + w_{2},\ G_{y})\) and \(u_{y}(2\delta + y_{1},\ G_{w}) = u_{y}(\delta + w_{1} + y_{2},\ G_{w})\). Each successive pair of rows corresponds to the indifference conditions over the next jump in each player’s support. The above matrix with \(k=4\) is also useful for visualizing the general form for an arbitrary k. We denote the system for a given k by \(\varvec{A_{_{k}}x\!=\!0}\). The matrices \(\varvec{A_{_{1}}}\), \(\varvec{A_{_{2}}}\), and \(\varvec{A_{_{3}}}\) are all partitions of the above matrix \(\varvec{A_{_{4}}}\). For \(\varvec{A_{_{1}}}\), it is the partition formed by the first four rows and five columns of \(\varvec{A_{_{4}}}\). The matrix \(\varvec{A_{_{2}}}\) comprises the first six rows and seven columns, and the first eight rows and nine columns form the matrix \(\varvec{A_{_{3}}}\). In general, \(\varvec{A_{_{k}}}\) has \(r_{k} = 2k + 2\) rows and \(c_{k} = 2k+3\) columns. The matrix \(\varvec{A_{_{k-1}}}\) constitutes the first \(r_{k}-2\) rows and \(c_{k}-2\) columns of \(\varvec{A_{_{k}}}\). Elements in the last two rows and columns of \(\varvec{A_{_{k}}}\) fit a clearly defined pattern. For \(k\ge 2\), the first three elements of rows \(r_{k}-1\) and \(r_{k}\) are \(\{-\delta ,\, 0,\, 0\}\). The fourth and fifth elements are shown in the matrix for \(k \le 4\). For \(k \ge 4\), the fourth and fifth elements of row \(r_{k}-1\) are the fourth and fifth elements of row \(r_{k}-2\), and for row \(r_{k}\) they are the fourth and fifth elements of row \(r_{k}-3\). Elements six through \(c_{k}\) of rows \(r_{k}-1\) and \(r_{k}\) are the same as the elements four through \(c_{k}-2\) of rows \(r_{k}-3\) and \(r_{k}-2\). The only nonzero elements of columns \(c_{k}-1\) and \(c_{k}\) are in rows 1, 2, and \(r_{k}-3\) through \(r_{k}\). The last two elements of rows 1 and 2 follow the established pattern for the mass constraints: \(\{1/[(1-\beta )v],\ 1/(\beta v)\}\) and \(\{1/(\beta v),\ 1/[(1-\beta )v]\}\). For rows \(r_{k}-3\) and \(r_{k}-2\), the last two elements are \(\{\beta /(1-\beta ),\ 0\}\) and \(\{0,\ \beta /(1-\beta )\}\).

We obtain our uniqueness result by showing that the \(r_{k}\) rows of \(\varvec{A_{_{k}}}\) are linearly independent for any \(k \in {\mathbb {N}}\). The \(r_{k}\) rows are linearly independent if and only if the system \(\varvec{A_{_{k}}x\!=\!0}\) has a unique solution. We proceed by induction. Table 1 shows the unique solution to the system when \(k \in \{1,\,2,\,3\}\), and it is also relevant that the unique solution is symmetric in each case. Now suppose that for \(k \ge 3\) that the \(r_{k}\) rows of \(\varvec{A_{_{k}}}\) are linearly independent. In the matrix \(\varvec{A_{_{k+1}}}\), rows 1 through \(r_{k+1}-4\) are still linearly independent. This follows because the last two columns of rows 3 through \(r_{k+1}-4\) only contain zeros, and rows 1 and 2 are always linearly independent of each other because of their second and third elements. Since rows 1, 2, \(r_{k+1}-3\), and \(r_{k+1}\) are the only rows with nonzero elements in column \(c_{k+1}-1\), and since rows 1, 2, \(r_{k+1}-2\), and \(r_{k+1}-1\) are the only rows with nonzero elements in column \(c_{k}\), then if each of these groups of four rows are linearly independent, then all \(r_{k+1}\) rows are linearly independent. For the first group (rows 1, 2, \(r_{k+1}-3\), and \(r_{k+1}\)), linear independence can be seen by looking at columns 2, 3, \(c_{k+1}-2\), and \(c_{k+1}-1\):

$$\begin{aligned} \left( \begin{array}{@{}c@{}c@{}cc} 1 &{} 0 &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v}\\ 0 &{} 1 &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v}\\ 0 &{} 0 &{} 1 &{} \frac{\beta }{1-\beta }\\ 0 &{} 0 &{} 1 &{} 1\\ \end{array} \right) \end{aligned}$$
Table 1 Unique solution to the system \(\varvec{A_{_{k}}x\!=\!0}\) for \(k \in \{1,\,2,\,3\}\)

Linear independence for the second group (rows 1, 2, \(r_{k+1}-2\), and \(r_{k+1}-1\)) can be seen from columns 2, 3, \(c_{k+1}-3\), and \(c_{k+1}\):

$$\begin{aligned} \left( \begin{array}{@{}c@{}c@{}cc} 1 &{} 0 &{} \frac{1}{(1-\beta )v} &{} \frac{1}{\beta v} \\ 0 &{} 1 &{} \frac{1}{\beta v} &{} \frac{1}{(1-\beta )v} \\ 0 &{} 0 &{} 1 &{} \frac{\beta }{1-\beta }\\ 0 &{} 0 &{} 1 &{} 1\\ \end{array} \right) \end{aligned}$$

Thus, the \(r_{k+1}\) rows of \(\varvec{A_{_{k+1}}}\) are linearly independent, so \(\varvec{A_{_{k+1}}x\!=\!0}\) has a unique solution. We next demonstrate that the unique solution is symmetric. That is, \(\alpha _{y}(0)=\alpha _{w}(0)\) and \(y_{1}=w_{1},\, \ldots ,\, y_{k}=w_{k}\). As we stated earlier, for \(k \in \{1,\,2,\,3\}\), the unique solution to \(\varvec{A_{_{k}}x\!=\!0}\) is indeed symmetric (see Table 1). With symmetry, the system \(\varvec{A_{_{k}}x\!=\!0}\) collapses to the following system, which we call \(\varvec{B_{_{k}}x\!=\!0}\):

$$\begin{aligned} {\left( \begin{array}{ccccccccccc} -1\; &{} 1 &{} \frac{1}{(1-\beta )\beta v} &{} \frac{1}{(1-\beta )\beta v} &{} \frac{1}{(1-\beta )\beta v} &{} \frac{1}{(1-\beta )\beta v} &{} \cdots &{} \frac{1}{(1-\beta )\beta v}\\ -\delta \; &{} (1\!-\!\beta )v &{} 1 &{} \frac{\beta }{1-\beta } &{} 0 &{} 0 &{} \cdots &{} 0 \\ -\delta \; &{} 0 &{} \frac{1-\beta }{\beta } &{} 1 &{} \frac{\beta }{1-\beta } &{} 0 &{} \cdots &{} 0\\ -\delta \; &{} 0 &{} 0 &{} \frac{1-\beta }{\beta } &{} 1 &{} \frac{\beta }{1-\beta } &{} &{} 0\\ \,\vdots &{} \vdots &{} \vdots &{} &{} \ddots &{} &{} \ddots &{} \vdots \\ -\delta \; &{} 0 &{} 0 &{} 0 &{} 0 &{} \frac{1-\beta }{\beta } &{} 1 &{} \frac{\beta }{1-\beta } \\ -\delta \; &{} 0 &{} 0 &{} 0 &{} 0 &{} 0 &{} \frac{1-\beta }{\beta } &{} 1 \\ \end{array} \right) \left( \begin{array}{c} 1\\ \alpha _{y}(0)\\ y_{1}\\ y_{2}\\ y_{3}\\ \vdots \\ y_{k}\\ \end{array} \right) = \varvec{0} } \end{aligned}$$

By construction, if \(\varvec{B_{_{k}}x\!=\!0}\) has a unique solution, then it must coincide with the unique solution of \(\varvec{A_{_{k}}x\!=\!0}\). The matrix \(\varvec{B_{_{k}}}\) has \(k+1\) rows and \(k+2\) columns, and \(\varvec{B_{_{k-1}}}\) makes up the first k rows and \(k+1\) columns of \(\varvec{B_{_{k}}}\). The last column only has three nonzero elements: \(1/[(1-\beta )\beta v]\) in the first row, \(\beta /(1-\beta )\) in the second to last row, and 1 in the last row. The only other nonzero elements of the last row are the first and second to last elements: \(-\delta \) and \((1-\beta )/\beta \). Since the linear independence of \(\varvec{B_{_{k-1}}}\) guarantees that the first k rows of \(\varvec{B_{_{k}}}\) are still linearly independent, we just need to check that the last row is also linearly independent. Since the last row is only one of three rows with a nonzero element in the last column (the other two rows being the first and second to last), it suffices to check that these three rows are linearly independent. This can be seen from the above matrix. Therefore, \(\varvec{B_{_{k}}}\) has a unique solution. Sufficient conditions for equilibria when \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta < 1/2\) are easily checked and roughly sketched in Sects. 3 and 4.1. \(\square \)

Proofs specific to \(\beta > 1/2\)

Lemma 6

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta > 1/2\). In any equilibrium, for any \(z \ge 0\), any subset of the support of \(G_{i}\) over \([z,\;z+2\delta ]\) with a density rate of \(1/[(1-\beta )v]\) is connected.

Proof

Suppose to the contrary that for some \(z \ge 0\), there is a subset of the support of \(G_{i}\) with a density rate of \(1/[(1-\beta )v]\) over \([z,\;z+2\delta ]\) which is disconnected. That is, \(G_{i}\) contains at least two intervals, \([a_{1},\, b_{1}]\), \([a_{2},\,b_{2}] \subset [z,\;z+2\delta ]\) where \(b_{1} < a_{2}\), with density rates of \(1/[(1-\beta )v]\); furthermore, the density rates in a neighborhood immediately above \(b_{1}\) and immediately below \(a_{2}\) differ from \(1/[(1-\beta )v]\). By Corollary 1, the intervals \([a_{1}+\delta ,\, b_{1}+\delta ]\) and \([a_{2}+\delta ,\,b_{2}+\delta ]\) have a density rate of \(1/(\beta v)\) in \(G_{-i}\). So in equilibrium, \(u_{-i}(b_{1}+\delta ,\; G_{i}) = u_{-i}(a_{2}+\delta ,\; G_{i})\):

$$\begin{aligned} \left[ G_{i}(a_{2}+2\delta ) - G_{i}(b_{1}+2\delta )\right] \beta v + \left[ G_{i}(a_{2}) - G_{i}(b_{1})\right] (1-\beta )v&= a_{2} - b_{1} \end{aligned}$$

This can be rewritten using Lemma 3 as:

$$\begin{aligned}&\left[ \frac{r}{(1-\beta )v} + \frac{s}{\beta v} \right] \beta v + \left[ \frac{m}{(1-\beta )v} + \frac{n}{\beta v}\right] (1-\beta )v \nonumber \\&\quad = r\left( \frac{\beta }{1-\beta }\right) + s + m + n\left( \frac{1-\beta }{\beta }\right) = a_{2} - b_{1} \end{aligned}$$
(18)

Here, r and s denote the respective lengths of the support of \(G_{i}\) in \([b_{1}+2\delta ,\, a_{2}+2\delta ]\) with density rates of \(1/[(1-\beta )v]\) and \(1/(\beta v)\); m and n are defined similarly for \([b_{1},\; a_{2}]\). We next show that \(r > 0\). This will ultimately lead to demonstrating that if \(b_{1} < a_{2}\), \(G_{i}\) must have mass in higher and higher intervals, eventually rising above v. Suppose by way of contradiction that \(r = 0\). Corollary 1 and Eq. 18 imply that \([b_{1}+\delta ,\, a_{2}+\delta ]\) in \(G_{-i}\) must contain a portion of length \(s \ge 0\) with density rate \(1/[(1-\beta )v]\), a portion of length \(m \ge 0\) with density rate of \(1/(\beta v)\), and a portion of length \(n \ge 0\) with no mass. Thus,

$$\begin{aligned} s + m + n \le a_{2} - b_{1} \end{aligned}$$
(19)

With \(r=0\) and \(\beta >1/2\), Eqs. 18 and 19 can only be jointly satisfied if \(n=0\) and if \(s+m = a_{2} - b_{1}\). We obtain a contradiction by looking at the distribution of \(G_{i}\) over \([b_{1},\; a_{2}]\). From Eq. 18, this interval has a portion of length m with a density rate \(1/[(1-\beta )v]\), and since \(G_{-i}\) has a portion of length s over \([b_{1}+\delta ,\, a_{2}+\delta ]\) with a density rate of \(1/[(1-\beta )v]\), by Corollary 1, \(G_{i}\) also has a portion of length s with no mass over \([b_{1},\; a_{2}]\). With \(s+m = a_{2} - b_{1}\), \(G_{i}\) has no room for anything else over \([b_{1},\; a_{2}]\). Let \([{\underline{c}},\, {\overline{c}}]\) be a subset of the portion of length s in \(G_{-i}\) with a density rate of \(1/[(1-\beta )v]\) over \([b_{1}+\delta ,\, a_{2}+\delta ]\). It must be that \(G_{i}\) has no mass over \((\,{\overline{c}} - ({\overline{c}} - {\underline{c}})(\beta /(1-\beta )) - \delta ,\; {\overline{c}} - \delta \,] \subseteq [b_{1},\; a_{2}]\). This follows from observing that \(u_{-i}({\overline{c}}-\delta ) > u_{-i}(x-\delta )\) for \(x \in (\,{\overline{c}} - ({\overline{c}} - {\underline{c}})(\beta /(1-\beta )),\; {\overline{c}}\,]\) for \(\beta > 1/2\). Specifically,

$$\begin{aligned}&\left[ G_{i}({\overline{c}}) - G_{i}(x)\right] \beta v + \left[ G_{i}({\overline{c}}-2\delta ) - G_{i}(x-2\delta )\right] (1-\beta )v - {\overline{c}} + x\\&\quad \ge ({\overline{c}} - {\underline{c}})\left( \frac{\beta }{1-\beta }\right) - {\overline{c}} + x > 0 \end{aligned}$$

Moreover, for \(\beta > 1/2\), the length of the interval \((\,{\overline{c}} - ({\overline{c}} - {\underline{c}})(\beta /(1-\beta )) - \delta ,\; {\overline{c}} - \delta \,]\) is greater than \([{\underline{c}},\, {\overline{c}}]\), implying that \(G_{i}\) has a portion greater than length s in \([b_{1},\; a_{2}]\) with no mass. This is a contradiction, and so it must be that \(r>0\). Hence, there exists at least one interval \([a_{3},\,b_{3}] \subset [b_{1}+2\delta ,\, a_{2}+2\delta ]\) in \(G_{i}\) with a density rate of \(1/[(1-\beta )v]\) (with different density rates immediately above \(b_{3}\) and below \(a_{3}\)). Since \([a_{2},\,b_{2}]\) and \([a_{3},\,b_{3}]\) are within a \(2\delta \) interval (i.e., \(b_{3} - a_{2} \le 2\delta \)), then the same argument we just used for \([a_{1},\,b_{1}]\) and \([a_{2},\,b_{2}]\) implies that there is an interval \([a_{4},\,b_{4}] \subset [b_{2}+2\delta ,\, a_{3}+2\delta ]\) in \(G_{i}\) with a density rate of \(1/[(1-\beta )v]\). In general, \([a_{k},\,b_{k}] \subset [b_{k-2}+2\delta ,\, a_{k-1}+2\delta ]\) is in the support of \(G_{i}\) and has a density rate of \(1/[(1-\beta )v]\), where \(k \in \{3, 4, \ldots \}\). The sequence of \(a_{k}\) is then unbounded (since \(a_{k+2} - a_{k} > 2\delta \)), rising to bids that are strictly dominated. This contradicts our original supposition that the subset of \(G_{i}\) with a density rate of \(1/[(1-\beta )v]\) in \([z,\;z+2\delta ]\) is disconnected. \(\square \)

For Lemmata 7 through 9, we define \(\phi _{1}\) (if it exists) as the lowest value for which either player has a gap in their support of at least \(\delta \) in length, followed by a density rate of \(1/[(1-\beta )v]\). The gap may not include a mass point (that is, if \(\phi _{1} \in [0,\ \delta )\), then \(\alpha _{\ell }(0)=0\)). Formally:

$$\begin{aligned}&\phi _{1} \equiv \,\min \{\,x \ge 0\ |\ \text {for some player }\ell , \ G_{\ell }(x)=G_{\ell }(x-\delta )\ \text {and}\ [x,\ x+z)\\&\quad \quad \text {has a density rate of}\ 1/[(1-\beta )v]\ \text {in }G_{\ell } \text { for some}\ z > 0\,\} \end{aligned}$$

If this definition for \(\phi _{1}\) is satisfied by only one player’s distribution, we refer to that player as player i (and to the opponent as player \(-i\)). If both players’ distributions satisfy \(\phi _{1}\), we designate player i as the player whose distribution has the longest connected segment of density rate \(1/[(1-\beta )v]\) immediately following \(\phi _{1}\). The length of this segment is denoted c. That is:

$$\begin{aligned} c \equiv \, \sup \{\,z \in (0,\, 2\delta ]\ |\ [\phi _{1},\ \phi _{1}+z)\ \text {has a density rate of}\ 1/[(1-\beta )v]\ \text {in}\ G_{\ell }\,\} \end{aligned}$$

If both players’ distributions satisfy \(\phi _{1}\) and c (or if \(\phi _{1}\) does not exist), then player i is chosen arbitrarily. When \(\phi _{1}\) exists, player i has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1},\ \phi _{1}+c)\), so by Corollary 1, player \(-i\) has no mass over \([\phi _{1}-\delta ,\ \phi _{1}+c-\delta )\). We next define \(\varphi _{1}\) as the difference between the lowest value in the support of player \(-i\)’s distribution following \(\phi _{1}+c-\delta \) and \(\phi _{1}\):

$$\begin{aligned} \varphi _{1} \equiv \, \min \{\, x \in \text {supp}(G_{-i})\ |\ x \ge \phi _{1} + c - \delta \, \} - \phi _{1} \end{aligned}$$

Since player \(-i\) has no mass over \([\phi _{1}-\delta ,\ \phi _{1}+c-\delta )\) and a density rate of \(1/(\beta v)\) over \([\phi _{1}+\delta ,\ \phi _{1}+c+\delta )\) (both by Corollary 1), then \(\varphi _{1} \in [c-\delta ,\ \delta ]\) (that is, \(\phi _{1}+c-\delta \le \phi _{1} + \varphi _{1} \le \phi _{1}+\delta \)). Finally, for \(j \in \{1,\ldots ,k-1\}\), for some \(k \in {\mathbb {N}}\), we define \(\phi _{j+1}\) and \(\varphi _{j+1}\) recursively:

$$\begin{aligned}&\phi _{j+1} \equiv \, \min \{\, x \in \, \text {supp}(G_{i})\ |\ x \ge \phi _{j} + \varphi _{j} + 3\delta \, \} \nonumber \\&\varphi _{j+1} \equiv \, \min \{\, x \in \text {supp}(G_{-i})\ |\ x \ge \phi _{j} + 3\delta \, \} - \phi _{j+1} \end{aligned}$$
(20)

If \(\phi _{1}\) exists, Lemma 7 specifies a structure that each player’s distribution must follow for all values greater than \(\phi _{1}\) in \(G_{i}\) and for all values greater than \(\phi _{1}+\varphi _{1}\) in \(G_{-i}\). Specifically, all mass in each player’s distribution above these points is made up of k connected intervals of length-\(2\delta \). The lower \(\delta + \varphi _{j}\) of player i’s jth such interval has a density rate of \(1/[(1-\beta )v]\), while the upper \(\delta - \varphi _{j}\) has a density rate of \(1/(\beta v)\). Following Corollary 1, the lower \(\delta - \varphi _{j}\) of player \(-i\)’s jth such interval has a density rate of \(1/[(1-\beta )v]\), while the upper \(\delta + \varphi _{j}\) has a density rate of \(1/(\beta v)\).

Lemma 8 affirms that \(\phi _{1}\) necessarily exists whenever either player has any mass in their support above \(2\delta \). It further states that any mass below \(\phi _{1}\) in \(G_{i}\) and below \(\phi _{1}+\varphi _{1}\) in \(G_{-i}\) must take one of three forms:

  1. A.

    Both players have a mass point at zero, followed by an optional pair of intervals that, if present, have lower bounds of zero and \(\delta \).

  2. B.

    Player i has a mass point at zero and a single interval with a lower bound of zero; player \(-i\) may or may not have a mass point at zero, but has a single interval with a lower bound of \(\delta \).

  3. C.

    Player i has a mass point at zero, followed by an interval that has no gap between its \(1/[(1-\beta )v]\) and \(1/(\beta v)\) portions; player \(-i\) does not have a mass point, but has an interval in accordance with Corollary 1.

Lemma 8 additionally states that form (A) above is the only form that can hold in equilibrium when \(\phi _{1}\) does not exist. (If \(\phi _{1}\) does not exist, neither player has mass above \(2\delta \).)

Lemma 9 establishes symmetry. Below \(\phi _{1}\), each player must have a mass point of the same size, and if the mass points are immediately followed by an interval pair, the intervals must be of the same length and begin at zero and \(\delta \). In other words, form (A) from Lemma 8 must hold and be symmetric. Above \(\phi _{1}\), symmetry entails that \(\varphi _{j} = 0\) for all \(j \in \{1,\ldots ,k\}\). With \(\varphi _{j} = 0\), each length-\(2\delta \) interval above \(\phi _{1}\) has a total mass of \(\delta /[(1-\beta )\beta v]\) (that is, \(\delta /[(1-\beta )v]\) over the lower half plus \(\delta /(\beta v)\) over the upper half). Uniqueness is then pinned down by determining the number of these intervals that can fit within the distribution, as described in Sect. 4.3.

We now proceed with Lemmata 7 through 9 for \(\beta > 1/2\), using the definitions in this overview for \(\phi _{j}\), \(\varphi _{j}\), c, and the naming conventions for players i and \(-i\).

Lemma 7

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta > 1/2\). In equilibrium, if \(\phi _{1}\) exists, then any portion of the support weakly greater than \(\phi _{1}\) in \(G_{i}\) and weakly greater than \(\phi _{1}+\varphi _{1}\) in \(G_{-i}\) must have a distribution of the following form. For \(j \in \{1,\ldots ,k\}\), for some \(k \in {\mathbb {N}}\), \(G_{i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{j},\; \phi _{j} + \varphi _{j} + \delta ]\) and \(1/(\beta v)\) over \([\phi _{j} + \varphi _{j} + \delta ,\; \phi _{j} + 2\delta ]\); and \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{j} + \varphi _{j},\; \phi _{j} + \delta ]\) and \(1/(\beta v)\) over \([\phi _{j} + \delta ,\; \phi _{j} + \varphi _{j} + 2\delta ]\). All other bids weakly greater than \(\phi _{1}\) in \(G_{i}\) and \(\phi _{1}+\varphi _{1}\) in \(G_{-i}\) are not in the supports of \(G_{i}\) or \(G_{-i}\).

Proof

Since \([\phi _{1},\; \phi _{1} + c)\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\), then by Corollary 1, \(G_{-i}(\phi _{1} - \delta ) = G_{-i}(\phi _{1} + c - \delta )\) and \([\phi _{1} + \delta ,\; \phi _{1} + c + \delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{-i}\). With \(\phi _{1} + \delta \in \text {supp}(G_{-i})\), equilibrium requires that \(u_{-i}(\phi _{1}+\delta ,\; G_{i}) \ge u_{-i}(\phi _{1},\; G_{i})\). Using the density rates permitted by Lemma 3, and recalling from the definition of \(\phi _{1}\) that \(G_{i}(\phi _{1}-\delta ) = G_{i}(\phi _{1})\), we have:

$$\begin{aligned} u_{-i}(\phi _{1}+\delta ,\; G_{i}) - u_{-i}(\phi _{1},\; G_{i})&= \left[ G_{i}(\phi _{1}+2\delta ) - G_{i}(\phi _{1}+\delta )\right] \beta v - \delta \nonumber \\&= \left[ \frac{\tau }{\beta v} + \frac{\mu }{(1-\beta )v} \right] \beta v - \delta \ge 0 \end{aligned}$$
(21)

Here, \(\tau \) and \(\mu \) are the lengths of the support over \([\phi _{1}+\delta ,\; \phi _{1}+2\delta ]\) in \(G_{i}\) that have density rates of \(1/(\beta v)\) and \(1/[(1-\beta )v]\), respectively. So \(\tau + \mu \le \delta \). Suppose first that \(\mu = 0\). To satisfy Equation 21, it must be that \(\tau = \delta \) (or rather \([\phi _{1}+\delta ,\; \phi _{1}+2\delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{i}\), and by the definition of c, we have \(\phi _{1}+c \le \phi _{1}+\delta \)). So Corollary 1 implies that \(G_{-i}(\phi _{1}+2\delta ) = G_{-i}(\phi _{1}+3\delta )\) and \([\phi _{1},\; \phi _{1}+\delta ]\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{-i}\). Note that \(\phi _{1} + c\) and \(\phi _{1}+\delta \in \text {supp}(G_{i})\): the former by the definitions of \(\phi _{1}\) and c, since supports are closed sets; the latter to satisfy Equation 21 when \(\mu = 0\). So \(u_{i}(\phi _{1}+c,\; G_{-i}) = u_{i}(\phi _{1}+\delta ,\; G_{-i})\), or rather:

$$\begin{aligned}&[G_{-i}(\phi _{1}+2\delta ) - G_{-i}(\phi _{1}+c+\delta )]\beta v \nonumber \\&\quad +\, [G_{-i}(\phi _{1}) - G_{-i}(\phi _{1}+c-\delta )](1-\beta )v = \delta - c \end{aligned}$$
(22)

By Lemma 6, portions of a support with a density rate of \(1/[(1-\beta )v]\) that are within \(2\delta \) of each other must be connected. As previously stated, \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1},\; \phi _{1}+\delta ]\) (for \(\mu =0\)) and a density rate of \(1/(\beta v)\) over \([\phi _{1} + \delta ,\; \phi _{1} + c + \delta ]\) (for any \(\mu \)). These rates, together with Lemmata 3 and 6, require that any mass in \(G_{-i}\) over \([\phi _{1}+c+\delta ,\; \phi _{1}+2\delta ]\) must have a density rate of \(1/(\beta v)\). Since \([\phi _{1},\; \phi _{1} + c)\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\) (from the definition of \(\phi _{1}\)), the same argument implies that any mass in \(G_{i}\) over \([\phi _{1} + c - 2\delta ,\; \phi _{1} - \delta ]\) must likewise have a density rate of \(1/(\beta v)\). Corollary 1 then implies that any mass in \(G_{-i}\) over \([\phi _{1}+c-\delta ,\; \phi _{1}]\) necessarily has a density rate of \(1/[(1-\beta )v]\). Equation 22 can then be written as follows, where \(q, r \in [0,\; \delta - c]\):

$$\begin{aligned} \left[ \frac{q}{\beta v}\right] \beta v + \left[ \frac{r}{(1-\beta )v}\right] (1-\beta )v = \delta - c \end{aligned}$$

We first focus on r, which represents the portion of the support of \(G_{-i}\) over \([\phi _{1} + c - \delta ,\ \phi _{1}]\). As we have shown, \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1},\; \phi _{1}+\delta ]\) (for \(\mu =0\)), so to be connected (in line with Lemma 6), the length-r mass with a density rate of \(1/[(1-\beta )v]\) must be placed over \([\phi _{1} - r,\; \phi _{1}]\). A similar argument applies to q. The portions in \(G_{i}\) with a density rate of \(1/[(1-\beta )v]\) that are within \(2\delta \) of each other must be connected, so by Corollary 1, the portions of \(G_{-i}\) with a density rate of \(1/(\beta v)\) that are within \(2\delta \) of each other must likewise be connected. Thus, with a density rate of \(1/(\beta v)\) in \(G_{-i}\) over \([\phi _{1} + \delta ,\ \phi _{1} + c + \delta )\), the length-q mass with a density of \(1/(\beta v)\) must be placed over \([\phi _{1} + c + \delta ,\ \phi _{1} + c + \delta + q]\). Summarizing these results, \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1} - r,\; \phi _{1} + \delta ]\) and a density rate of \(1/(\beta v)\) over \([\phi _{1} + \delta ,\; \phi _{1} + 2\delta - r]\) (where we substitute \(q = \delta - c - r\)), and from Corollary 1, \(G_{i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1},\; \phi _{1} + \delta - r]\) and \(1/(\beta v)\) over \([\phi _{1} + \delta - r,\; \phi _{1} + 2\delta ]\).

Now suppose instead that \(\mu > 0\). Lemma 6 implies that the interval \([\phi _{1},\; \phi _{1} + \delta + \mu ]\) has a density rate of \(1/[(1-\beta )v]\) in \(G_{i}\). Then, by Corollary 1, \([\phi _{1} + \delta ,\; \phi _{1} + 2\delta + \mu ]\) has a density rate of \(1/(\beta v)\) in \(G_{-i}\). Since \(\phi _{1} + \delta \in \text {supp}(G_{-i})\), \(u_{-i}(\phi _{1}+\delta ,\; G_{i}) \ge u_{-i}(\phi _{1}+\mu ,\; G_{i})\), or rather:

$$\begin{aligned} \left[ G_{i}(\phi _{1}+2\delta ) - G_{i}(\phi _{1} + \delta + \mu )\right] \beta v = \left[ \frac{\tau }{\beta v}\right] \beta v \ge \delta - \mu \end{aligned}$$

To jointly satisfy the above inequality and the definition specifying that \(\tau + \mu \le \delta \), \(G_{i}\) must have a density rate of \(1/(\beta v)\) over \([\phi _{1} + \delta + \mu ,\; \phi _{1}+2\delta ]\). And by Corollary 1, \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1} + \mu ,\; \phi _{1}+\delta ]\).

Combining the results for the cases where \(\mu = 0\) and \(\mu > 0\), \(G_{i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1},\; \phi _{1} + \varphi _{1} + \delta ]\) and \(1/(\beta v)\) over \([\phi _{1} + \varphi _{1} + \delta ,\; \phi _{1} + 2\delta ]\), and \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{1} + \varphi _{1},\; \phi _{1} + \delta ]\) and \(1/(\beta v)\) over \([\phi _{1} + \delta ,\; \phi _{1} + \varphi _{1} + 2\delta ]\).

We must next show that this pattern holds moving up the remainder of the distribution. If we are already at the top of the distribution, then we are done. Otherwise, the arguments used thus far can be applied whenever at least one player has a gap in their support of at least \(\delta \), followed by an interval with a density rate of \(1/[(1-\beta )v]\). Since \(G_{-i}\) has a density rate of \(1/(\beta v)\) over \([\phi _{1} + \delta ,\; \phi _{1} + \varphi _{1} + 2\delta ]\) and \(G_{i}\) has a density rate of \(1/(\beta v)\) over \([\phi _{1} + \varphi _{1} + \delta ,\; \phi _{1} + 2\delta ]\), then by Corollary 1, \(G_{i}(\phi _{1}+2\delta )=G_{i}(\phi _{1}+\varphi _{1}+3\delta )\) and \(G_{-i}(\phi _{1} + \varphi _{1} + 2\delta ) = G_{-i}(\phi _{1}+3\delta )\). At least one of these two gaps must be at least \(\delta \) in length. (Depending on the value of c, it is possible for \(\varphi _{1}\) to be weakly positive or negative; \(\varphi _{1} \in [c-\delta ,\ \delta ]\) and \(c \in (0,\ 2\delta ]\).)

As previously defined in Eq. 20, let \(\phi _{2} \equiv \min \{ x \in \text {supp}(G_{i})\ |\ x \ge \phi _{1} + \varphi _{1} + 3\delta \}\) and \(\varphi _{2} \equiv \min \{ x \in \text {supp}(G_{-i})\ |\ x \ge \phi _{1} + 3\delta \} - \phi _{2}\). Also define \({\widehat{x}} \equiv \min \{\phi _{2},\; \phi _{2} + \varphi _{2}\}\). Given that neither player’s distribution has had a density rate of \(1/[(1-\beta )v]\) for more than a space of \(\delta \), any mass in \([{\widehat{x}},\ {\widehat{x}} + \delta ]\) in either player’s distribution must have a density rate of \(1/[(1-\beta )v]\) (by Lemma 3 and Corollary 1). At least one player must have mass beginning at \({\widehat{x}}\) at a density rate of \(1/[(1-\beta )v]\). To see that the other player’s distribution also has some mass in \([{\widehat{x}},\; {\widehat{x}}+\delta ]\), suppose that \({\widehat{x}} \in \text {supp}(G_{-i})\). Since \(u_{-i}({\widehat{x}},\; G_{i})=u_{-i}(\phi _{1} + \varphi _{1} + 2\delta ,\; G_{i})\), then:

$$\begin{aligned}&\left[ G_{i}({\widehat{x}}+\delta ) - G_{i}(\phi _{1}+\varphi _{1}+3\delta ) \right] \beta v + \left[ \frac{\delta - \varphi _{1}}{\beta v}\right] (1-\beta )v \\&\quad = {\widehat{x}} - (\phi _{1} + \varphi _{1} + 2\delta )\\&\quad \ge (\phi _{1}+3\delta ) - (\phi _{1} + \varphi _{1} + 2\delta ) =\delta - \varphi _{1} \end{aligned}$$

From the \((\delta - \varphi _{1})\) term on each side, since \(\beta > 1/2\), the equation can only be satisfied if \(G_{i}({\widehat{x}}+\delta ) - G_{i}(\phi _{1}+\varphi _{1}+3\delta ) > 0\). An analogous argument holds for \({\widehat{x}} \in \text {supp}(G_{i})\), where the relevant equation is \(u_{i}({\widehat{x}},\; G_{-i}) = u_{i}(\phi _{1}+2\delta ,\; G_{-i})\). Thus, both \(G_{i}\) and \(G_{-i}\) have some mass in \([{\widehat{x}},\; {\widehat{x}}+\delta ]\) with a density rate of \(1/[(1-\beta )v]\). Applying the earlier arguments of this proof, \(G_{i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{2},\; \phi _{2} + \varphi _{2} + \delta ]\) and \(1/(\beta v)\) over \([\phi _{2} + \varphi _{2} + \delta ,\; \phi _{2} + 2\delta ]\). Likewise, \(G_{-i}\) has a density rate of \(1/[(1-\beta )v]\) over \([\phi _{2} + \varphi _{2},\; \phi _{2} + \delta ]\) and \(1/(\beta v)\) over \([\phi _{2} + \delta ,\; \phi _{2} + \varphi _{2} + 2\delta ]\). More generally, with \(\phi _{j+1}\) and \(\varphi _{j+1}\) defined as in Eq. 20, the pattern continues as long as there is any remaining mass in \(G_{i}\) or \(G_{-i}\). \(\square \)

Lemma 8

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta > 1/2\). In equilibrium, if \(\min \{G_{i}(2\delta ), G_{-i}(2\delta )\} < 1\), then \(\phi _{1}\) exists. Moreover, any mass in either player’s distribution below \(\min \{\phi _{1},\ \phi _{1} + \varphi _{1}\}\) is limited to a mass point at zero, a lower interval at a density rate of \(1/[(1-\beta )v]\), and an upper interval at a density rate of \(1/\beta v\). The mass points and lower intervals may be distributed according to one of three forms, with the upper intervals following Corollary 1:

  1. A.

    \(\alpha _{i}(0) > 0\) and \(\alpha _{-i}(0) > 0\). Lower intervals begin at zero and have length \(\psi _{k} \in [0,\; \delta )\) for \(k \in \{i,\; -i\}\). This is also the only possible equilibrium form when \(\phi _{1}\) does not exist.

  2. B.

    \(\alpha _{i}(0) > 0\) and \(\alpha _{-i}(0) \ge 0\). Player i’s lower interval begins at zero and has length \(\psi _{i} \in [0,\; 2\delta ]\). Player \(-i\) has no lower interval.

  3. C.

    \(\alpha _{i}(0) > 0\) and \(\alpha _{-i}(0) = 0\). Player i’s lower interval begins at zero and has length \(\psi _{i} \in [\delta ,\; 2\delta )\). Player \(-i\)’s lower interval begins at \(\psi _{i} - \delta \) and has length \(\psi _{-i} \in (0,\ 2\delta - \psi _{i})\).

Proof

Suppose that \(\min \{G_{i}(2\delta ),\ G_{-i}(2\delta )\} < 1\); that is, at least one player’s distribution has mass above \(2\delta \). For player \(\ell \in \{i,\, -i\}\) define:

$$\begin{aligned} \mu _{\ell }&\! \equiv \!\inf \{x\ge 0\ \, |\ \, (x,\ x+z)\ \text {has a density rate of}\ 1/[(1\!-\!\beta )v]\ \text {in}\ G_{\ell }\ \text {for some}\ z>0\} \\ M_{\ell }&\! \equiv \!\sup \{z > \mu _{\ell }\ \, |\ \, (\mu _{\ell },\ z)\ \text {has a density rate of}\ 1/[(1-\beta )v]\ \text {in}\ G_{\ell }\} \end{aligned}$$

Without loss of generality with respect to the naming of players, there are six cases to consider.

Case 1: \(\alpha _{i}(0) = \alpha _{-i}(0) = 0\). If neither player has a mass point at zero, then by Lemma 3 and Corollary 1, at least one player’s distribution must begin with a segment that has a density rate \(1/[(1-\beta )v]\). Thus, \(\phi _{1}\) exists, and neither player’s distribution has any mass below it.

Case 2: \(\mu _{i} \ge \delta \), \(\mu _{-i} \ge \delta \). By Lemma 3 and Corollary 1, any mass in \((0,\; \delta ]\) must necessarily have a density rate of \(1/[(1-\beta )v]\). So with \(\mu _{i} \ge \delta \) and \(\mu _{-i} \ge \delta \), it must be that \(G_{i}(\delta )=G_{i}(0)\) and \(G_{-i}(\delta )=G_{-i}(0)\). Following the gap between 0 and \(\delta \), the first nonzero density rate in at least one player’s distribution must be \(1/[(1-\beta )v]\) (again by Lemma 3 and Corollary 1). Hence, \(\phi _{1}\) exists, and the only possible mass in either player’s distribution below it is a mass point at zero.

Case 3: \(\alpha _{i}(0)>0\), \(\alpha _{-i}(0) > 0\), \(\mu _{i} < \delta \), \(\mu _{-i} < \delta \). With a mass point at zero in each player’s distribution, players would prefer to shift any mass immediately below \(\delta \) to immediately above it to beat the opponent’s mass point. This implies that with \(\mu _{i},\, \mu _{-i} < \delta \), the upper bounds of these connected segments must be strictly below \(\delta \). That is, \(M_{i}\), \(M_{-i} < \delta \). Suppose by way of contradiction that a player (say player i) has \(\mu _{i} > 0\). Then for \(u_{i}(0,\; G_{-i}) = u_{i}(\mu _{i},\; G_{-i})\) to hold, \(G_{-i}\) must have some mass in \([\delta ,\ \mu _{i}+\delta ]\) that a bid of \(\mu _{i}\) can tie. Since player i has no mass in \((0,\ \mu _{i})\), then by Lemma 3 and Corollary 1, any mass in \(G_{-i}\) in \([\delta ,\ \mu _{i}+\delta ]\) must have a density rate of \(1/[(1-\beta )v]\). Since \(M_{-i} < \delta \), that would imply that \(G_{-i}\) has multiple, disconnected segments with a density rate of \(1/[(1-\beta )v]\) that are within \(2\delta \) of each other, contradicting Lemma 6. Hence, \(\mu _{i} = \mu _{-i} = 0\). By Lemma 6, the next segment after \([\mu _{i},\ M_{i}]\) in \(G_{i}\) with a density rate of \(1/[(1-\beta )v]\) must have a lower bound weakly greater than \(M_{i} + 2\delta \). (Likewise, the lower bound in \(G_{-i}\) must be weakly greater than \(M_{-i} + 2\delta \).) After the segments with a density rate of \(1/(\beta v)\) that follow from Corollary 1 (specifically, \([\delta ,\ M_{-i}+\delta ]\) in \(G_{i}\) and \([\delta ,\ M_{i}+\delta ]\) in \(G_{-i}\)), player i has a gap with no mass over \([M_{-i}+\delta ,\ M_{i}+2\delta ]\), and player \(-i\) has a similar gap with no mass over \([M_{i} + \delta ,\ M_{-i} + 2\delta ]\). For players to bid above these gaps, there must be mass that they can tie; beating rather than tying the segment below the gap with a density rate of \(1/(\beta v)\) does not of itself compensate for the added bidding cost. (The marginal benefit is \((1-\beta )/\beta < 1\) when \(\beta > 1/2\).) Given Lemma 3 and Corollary 1, above these gaps, each player’s distribution must then begin with a segment of density rate \(1/[(1-\beta )v]\) that the other player can tie. Since one of the gaps is at least \(\delta \) in length, \(\phi _{1}\) exists (provided that \(G_{i}(M_{-i} + \delta )<1\) or \(G_{-i}(M_{i} + \delta )<1\)). Any mass below \(\min \{\phi _{1},\ \phi _{1} + \varphi _{1}\}\) is distributed according to Part A of Lemma 8.

Case 4: \(\alpha _{i}(0)>0\), \(\alpha _{-i}(0) \ge 0\), \(\mu _{i} < \delta \), \(\mu _{-i} \ge \delta \). By Corollary 1, \([\mu _{i} + \delta ,\ M_{i} + \delta ]\) has a density rate of \(1/(\beta v)\) in \(G_{-i}\), so \(\mu _{-i} \notin [\mu _{i} + \delta ,\ M_{i} + \delta ]\). There are two subcases to consider for \(\mu _{-i} \ge \delta \): either \([\mu _{-i},\; M_{-i}] \subseteq [\delta ,\ \mu _{i} + \delta ]\) or \(\mu _{-i} \ge M_{i} + \delta \). Suppose first that \([\mu _{-i},\; M_{-i}] \subseteq [\delta ,\ \mu _{i} + \delta ]\), which requires \(\mu _{i} > 0\). For player \(-i\), \(\mu _{-i} > \delta \) is strictly dominated by \(\mu _{-i} = \delta \) since \(G_{i}(0) = G_{i}(\mu _{-i} - \delta )\) and \(G_{i}(2\delta ) = G_{i}(\mu _{-i} + \delta )\) (see Corollary 1; Lemma 6 further ensures that \(G_{i}\) has no mass between \(M_{i}\) and \(\mu _{-i} + \delta \)). In other words, \(\mu _{-i} > \delta \) has no additional mass to tie or beat than \(\mu _{-i} = \delta \), but \(\mu _{-i} > \delta \) comes at a higher cost. In equilibrium, \(u_{i}(0,\; G_{-i}) = u_{i}(\mu _{i},\; G_{-i})\), so:

$$\begin{aligned} \left[ G_{-i}(\mu _{i} + \delta ) - G_{-i}(\delta ) \right] \beta v = \mu _{i} \quad \Rightarrow \quad \left( \frac{M_{-i} - \mu _{-i}}{(1-\beta )v} \right) \beta v = \mu _{i} \end{aligned}$$

Since \(\beta > 1/2\), the last equality implies that \(M_{-i} - \mu _{-i} < \mu _{i}\). This places a further restriction on the upper bound of \([\mu _{-i},\; M_{-i}] \subseteq [\delta ,\ \mu _{i} + \delta ]\). We have already established that \(\mu _{-i} = \delta \), so \(M_{-i} - \mu _{-i} < \mu _{i}\) is equivalent to \(M_{-i} < \mu _{i} + \delta \). Player i could profitably deviate by shifting \(\mu _{i}\) down to \(M_{-i} - \delta \). The first subcase is therefore not an equilibrium. Moving to the second subcase, suppose now that \(\mu _{-i} \ge M_{i} + \delta \). By Lemma 6 and Corollary 1, \(G_{i}(M_{i} + 2\delta ) = G_{i}(M_{i})\). If \(\mu _{-i} = M_{i} + \delta \), player i could deviate by moving mass from a neighborhood below \(M_{i}\) to a neighborhood above it, since tying a segment with a density rate of \(1/[(1-\beta )v]\) would more than cover the increased bidding cost when \(\beta >1/2\). If \(\mu _{-i} > M_{i} + \delta \), player \(-i\) will only be indifferent between \(\mu _{-i}\) and \(M_{i} + \delta \), both of which are in the support of \(G_{-i}\), if there is some mass in \(G_{i}\) that a bid of \(\mu _{-i}\) could tie (i.e., \(G_{i}(\mu _{-i}+\delta ) - G_{i}(M_{i} + 2\delta ) > 0\); \(\mu _{-i}\) and \(M_{i} + \delta \) beat the same mass). From Lemma 3 and Corollary 1, this mass in \(G_{i}\) would need to have a density rate of \(1/[(1-\beta )v]\). In which case, so long as \(G_{i}(M_{i})<1\), then \(\phi _{1}\) exists. Furthermore, \(\mu _{-i} > M_{i} + \delta \) implies that \(G_{-i}(\delta ) = G_{-i}(\mu _{i}+\delta )\), so we can only have \(u_{i}(0,\; G_{-i}) = u_{i}(\mu _{i},\; G_{-i})\) if \(\mu _{i}=0\). If \(\alpha _{-i}(0) > 0\), then \(M_{i} < \delta \); otherwise, Corollary 1 sets the upper bound of \(M_{i}\) at \(2\delta \). Thus, any mass below \(\min \{\phi _{1},\ \phi _{1} + \varphi _{1}\}\) is distributed according to Part B of Lemma 8.

Case 5: \(\alpha _{i}(0)>0\), \(\alpha _{-i}(0) = 0\), \(\mu _{i} < \delta \), \(\mu _{-i} < \delta \). By Corollary 1 and Lemma 6, player i has no mass in \([M_{-i} + \delta ,\ M_{i} + 2\delta ]\), and player \(-i\) has no mass in \([M_{i} + \delta ,\ M_{-i} + 2\delta ]\). At least one of these intervals has a length weakly greater than \(\delta \), so if \(G_{i}(M_{-i} + \delta ) < 1\) or \(G_{-i}(M_{i} + \delta ) < 1\), then \(\phi _{1}\) exists. (This is similar to Case 3.) We next show that any mass below \(\min \{\phi _{1},\ \phi _{1} + \varphi _{1}\}\) must be distributed according to Part C of Lemma 8. This is seen by demonstrating that \(M_{-i} <\delta \), \(\mu _{i} = 0\), and \(\mu _{-i}= M_{i} - \delta \), and \(M_{i} \ge \delta \). As in Case 3, since \(\alpha _{i}(0)>0\) and \(\mu _{-i} < \delta \), then \(M_{-i} < \delta \). The result that \(\mu _{i} = 0\) likewise follows directly from the argument that \(\mu _{i} = 0\) in Case 3. \(G_{i}\) thus has a density rate of \(1/[(1-\beta )v]\) over \([0,\ M_{i}]\), and by Corollary 1 and Lemma 6, \(G_{i}\) no mass in \((M_{i},\ \mu _{-i}+\delta )\). Given this high density rate, it is optimal for Player \(-i\) to tie all of the \([0,\ M_{i}]\) segment at the lowest cost, and so \(\mu _{-i}=\max \{0,\ M_{i} - \delta \}\). Finally, unless \(M_{i} \ge \delta \), there would be a gap in \(G_{i}\) between \([0,\ M_{i}]\) and \([\mu _{-i}+\delta ,\ M_{-i}+\delta ]\), and since \(\alpha _{-i}(0) = 0\), player i’s payoff above the gap would be strictly lower than below the gap. Thus, \(M_{i} \ge \delta \).

Case 6: \(\alpha _{i}(0)>0\), \(\alpha _{-i}(0) = 0\), \(\mu _{i} \ge \delta \), \(\mu _{-i} < \delta \). Let \(\zeta \equiv \min \{x \in \text {supp}(G_{i})\ |\ x \ge \delta \}\). We will first show that there is no mass in \(G_{-i}\) over \((\delta ,\ \delta + \zeta )\). By Corollary 1, since \(\mu _{i} \ge \delta \), player i has no mass in \((0,\ \delta )\). Combined with the definition of \(\zeta \), player i has no mass in \((0,\ \zeta )\). Corollary 1 further implies that player \(-i\) has no mass at a density rate of \(1/(\beta v)\) in \((\delta ,\ \delta + \zeta )\). Since \(\alpha _{i}(0)>0\), player \(-i\) would strictly prefer to move any mass in a neighborhood below \(\delta \) to a neighborhood above it. Thus, with \(\mu _{-i} < \delta \) and no mass in \(G_{-i}\) in a neighborhood below \(\delta \), then \(M_{-i} < \delta \). So by Lemma 6, player \(-i\) has no mass at a density rate of \(1/[(1-\beta )v]\) in \((\delta ,\ 2\delta + M_{-i})\), where \(2\delta + M_{-i} \ge \delta + \zeta \). With no mass at the density rates of \(1/[(1-\beta )v]\) or \(1/(\beta v)\) in \((\delta ,\ \delta + \zeta )\), then by Lemma 3, \(G_{-i}\) has no mass in \((\delta ,\ \delta + \zeta )\). In equilibrium, \(u_{i}(0,\; G_{-i}) = u_{i}(\zeta ,\; G_{-i})\), but this cannot be. A bid of \(\zeta \) beats no mass (\(\zeta \le \mu _{-i} + \delta \) by Corollary 1), and it ties the same mass as a bid of 0, but at a higher cost.

When \(\phi _{1}\) does not exist: We will show that Parts B and C of Lemma 8 cannot hold if \(\phi _{1}\) does not exist. Beginning with Part B, in order for player i’s total mass of \(\alpha _{i}(0) + [\psi _{i}/(1-\beta )v]\) and player \(-i\)’s total mass of \(\alpha _{-i}(0) + [\psi _{i}/(\beta v)]\) to each equal one, we must have \(\alpha _{-i}(0) > 0\). With this positive mass point, preventing a jump in player i’s expected payoff near \(\delta \) requires that \(\psi _{i} < \delta \). Player \(-i\)’s indifference condition between 0 and \(\delta \) (i.e., \(u_{-i}(0,\; G_{i}) = u_{-i}({\overline{\delta }},\; G_{i})\)) equates to \(\alpha _{i}(0)=\delta /[(1-\beta )v]\). This, combined with the two mass constraints, implies that \(\alpha _{-i}(0)=[\delta -(1-2\beta )v]/\beta v\) and \(\psi _{i} = (1-\beta )v - \delta \). However, since \(\delta < (1-\beta )v\), player i could profitably deviate with a bid of \(\delta \). For Part C, the two mass constraints and player \(-i\)’s indifference condition between \(\psi _{i}-\delta \) and \(\delta \) imply that \(\alpha _{i}(0) = 2[\delta - (1-\beta )\beta v]/[(1-\beta )^{2}v]\); \(\psi _{i} = \delta + [(3\beta -1)((1-\beta )\beta v - \delta ) / (1-\beta )(2\beta -1)]\); and \(\psi _{-i}=[2\delta - (1-\beta )v]\beta /(2\beta - 1)\). Note, however, that the conditions in Part C for \(\alpha _{i}(0) > 0\) and \(\psi _{i} \in [\delta ,\, 2\delta )\) cannot be jointly satisfied. (The closest case is \(\delta =(1-\beta )\beta v\) so that \(\alpha _{i}(0)=0\) and \(\psi _{i}=\psi _{-i}=\delta \), but then \(\phi _{1}\) would exist.) \(\square \)

Lemma 9

Let \(\delta \in (0,\; (1-\beta )v\,)\) and \(\beta > 1/2\). Equilibrium must be symmetric. Specifically, \(\alpha _{i}(0)=\alpha _{-i}(0)\), \(\psi _{i}=\psi _{-i}\), and \(\varphi _{j}=0\) for \(j \in \{1,\ldots ,k\}\) for some \(k \in {\mathbb {N}}\).

Proof

From Lemmata 7 and 8, mass in \(G_{i}\) and \(G_{-i}\) have the following forms:

$$\begin{aligned} 1&=\ \alpha _{i}(0) + \frac{\psi _{i}}{(1-\beta )v} + \frac{\psi _{-i}}{\beta v} + \frac{\delta + \varphi _{1}}{(1-\beta )v} + \frac{\delta - \varphi _{1}}{\beta v} + \cdots + \frac{\delta + \varphi _{k}}{(1-\beta )v} + \frac{\delta - \varphi _{k}}{\beta v} \\ 1&=\ \alpha _{-i}(0) + \frac{\psi _{-i}}{(1-\beta )v} + \frac{\psi _{i}}{\beta v} + \frac{\delta - \varphi _{1}}{(1-\beta )v} + \frac{\delta + \varphi _{1}}{\beta v} + \cdots + \frac{\delta - \varphi _{k}}{(1-\beta )v} + \frac{\delta + \varphi _{k}}{\beta v} \end{aligned}$$

Combining these two equations yields:

$$\begin{aligned} \alpha _{-i}(0) - \alpha _{i}(0) = \left[ \frac{4\beta - 2}{(1\!-\!\beta )\beta v}\right] \Big (\varphi _{1} + \varphi _{2} + \cdots + \varphi _{k} \Big ) - \left[ \frac{2\beta - 1}{(1\!-\!\beta )\beta v}\right] (\psi _{-i} - \psi _{i})\nonumber \\ \end{aligned}$$
(23)

For \(j \in \{1,\ldots ,k-1\}\) where \(k \ge 2\), \(u_{i}(\phi _{j+1}) = u_{i}(\phi _{j})\) and \(u_{-i}(\phi _{j+1} + \varphi _{j+1}) = u_{-i}(\phi _{j} + \varphi _{j})\), respectively, imply the following:

$$\begin{aligned} \phi _{j+1} - \phi _{j}&= \frac{\delta }{(1-\beta )\beta } + \left( \frac{1-\beta }{\beta }\right) \varphi _{j} - \left( \frac{\beta }{1-\beta } \right) \varphi _{j+1} \\ \phi _{j+1} - \phi _{j}&= \frac{\delta }{(1-\beta )\beta } + \left( \frac{2\beta -1}{\beta }\right) \varphi _{j} + \left( \frac{2\beta -1}{1-\beta } \right) \varphi _{j+1} \end{aligned}$$

We then obtain the following from the above two equations:

$$\begin{aligned} \varphi _{j+1} = \left[ \frac{2 - 3\beta }{3\beta - 1}\right] \left( \frac{1-\beta }{\beta }\right) \varphi _{j} \end{aligned}$$
(24)

Since Eq. 24 takes the form \(\varphi _{j+1} = H \varphi _{j}\), we can write all \(\varphi _{j}\) in terms of \(\varphi _{1}\). In particular:

$$\begin{aligned} \left( \varphi _{1} + \varphi _{2} + \cdots + \varphi _{k}\right) \ =\ \varphi _{1}\left[ 1 + H + H^{2} + \cdots + H^{k-1} \right] \ =\ \varphi _{1}\left[ \frac{1- H^{k}}{1 - H}\right] \end{aligned}$$

So Eq. 23 becomes:

$$\begin{aligned} \left[ \alpha _{-i}(0) - \alpha _{i}(0)\right] (1-\beta )v = \left[ \frac{1- H^{k}}{1 - H}\right] \left[ \frac{4\beta \!-\! 2}{\beta }\right] \varphi _{1} - \left[ \frac{2\beta \! -\! 1}{\beta }\right] (\psi _{-i} - \psi _{i})\nonumber \\ \end{aligned}$$
(25)

If \(\alpha _{i}(0) = \alpha _{-i}(0)\) and \(\psi _{i} = \psi _{-i}\), then Eq. 25 simplifies to:

$$\begin{aligned} 0 = \left[ \frac{1- H^{k}}{1 - H}\right] \left[ \frac{4\beta - 2}{\beta }\right] \varphi _{1} \end{aligned}$$

This can only be satisfied by \(\varphi _{1} = 0\) since \(1 = H^{k}\) and \(4\beta = 2\) both require \(\beta = 1/2\). So by Eq. 24, \(\varphi _{j}=0\) for \(j \in \{1,\ldots ,k\}\).

Now suppose that either \(\alpha _{i}(0) \ne \alpha _{-i}(0)\) or \(\psi _{i} \ne \psi _{-i}\). Based on Lemma 8, if there is any mass below \(\phi _{1}\), then at least one player has a strictly positive mass point at zero. Without loss of generality, assume this is player i. Then \(u_{i}(0,\; G_{-i}) = u_{i}(\phi _{1},\; G_{i})\). For player \(-i\), we have \(u_{-i}(0,\; G_{-i}) \le u_{-i}(\phi _{1}+\varphi _{1},\; G_{i})\), which holds with strict equality in equilibrium whenever \(\alpha _{-i}(0)>0\). These imply the following:

$$\begin{aligned} \phi _{1}&\ =\ \alpha _{-i}(0)(1-\beta )v + \psi _{-i} + \left( \frac{\psi _{i}}{\beta }\right) + (\delta - \varphi _{1})\left( \frac{\beta }{1 - \beta } \right) \\ \phi _{1}&\ \le \ \alpha _{i}(0)(1-\beta )v + \psi _{i} + \left( \frac{\psi _{-i}}{\beta }\right) + (\delta + \varphi _{1})\left( \frac{\beta }{1 - \beta } \right) - \varphi _{1} \end{aligned}$$

Combined, we have:

$$\begin{aligned} \left[ \alpha _{-i}(0) - \alpha _{i}(0)\right] (1-\beta )v \ \le \ \left( \psi _{-i} - \psi _{i}\right) \left( \frac{1-\beta }{\beta }\right) + \left( \frac{3\beta - 1}{1 - \beta }\right) \varphi _{1} \end{aligned}$$
(26)

Substituting \(1-H = (4\beta -2)/[(3\beta -1)\beta ]\), Eqs. 25 and 26 imply the following, which holds with strict equality whenever \(\alpha _{-i}(0)>0\):

$$\begin{aligned} \left( \psi _{-i} - \psi _{i}\right) \ge (3\beta - 1) \left[ (1-H^{k}) - \left( \frac{1}{1-\beta }\right) \right] \varphi _{1} \end{aligned}$$
(27)

Consider the subcase where \(\alpha _{i}(0) > 0\), \(\alpha _{-i}(0) \ge 0\), \(\psi _{i} > 0\), and \(\psi _{-i} \ge 0\) (and either \(\alpha _{i}(0) \ne \alpha _{-i}(0)\) or \(\psi _{i} \ne \psi _{-i}\)). Here, \(u_{-i}({\overline{\delta }},\; G_{i}) = u_{-i}(\phi _{1} + \varphi _{1},\ G_{i})\) and \(u_{i}({\overline{\delta }},\; G_{-i}) \le u_{i}(\phi _{1},\ G_{-i})\); the latter holds with strict equality if \(\psi _{-i} > 0\). These can be rewritten as follows:

$$\begin{aligned} \phi _{1}&\ =\ \psi _{i} + \psi _{-i}\left( \frac{1-\beta }{\beta }\right) + \delta \left( \frac{1}{1-\beta }\right) + \varphi _{1}\left( \frac{2\beta - 1}{1 - \beta }\right) \\ \phi _{1}&\ \le \ \psi _{-i} + \psi _{i}\left( \frac{1-\beta }{\beta }\right) + \delta \left( \frac{1}{1-\beta }\right) - \varphi _{1}\left( \frac{\beta }{1-\beta }\right) \end{aligned}$$

Combining them yields:

$$\begin{aligned} \left( \psi _{-i} - \psi _{i}\right) \ge \left( \frac{3\beta - 1}{2\beta - 1}\right) \left( \frac{\beta }{1-\beta }\right) \varphi _{1} \end{aligned}$$
(28)

We will show that Eqs. 27 and 28 can only jointly hold if \(\psi _{-i} = \psi _{i}\) and \(\varphi _{1} = 0\). Suppose instead that \(\psi _{-i} \ne \psi _{i}\), and without loss of generality, suppose that \(\psi _{-i} < \psi _{i}\). With \(\beta \in (1/2,\; 1)\), the coefficient on \(\varphi _{1}\) in Eq. 28 is strictly positive. So since the left-hand side of Eq. 28 is negative, it must be that \(\varphi _{1} < 0\). Next, note that the coefficient on \(\varphi _{1}\) in Eq. 27 is strictly negative for \(\beta \in (1/2,\; 1)\). Showing that this coefficient is negative is equivalent to showing that \(H^{k}(1-\beta )+\beta > 0\) for \(\beta \in (1/2,\; 1)\). This holds since the minimal value of \(H^{k}\) over this range is \(4\sqrt{3} - 7 \approx -0.0718\). (The minimum is obtained at \(\beta = (3 + \sqrt{3})/6 \approx 0.7887\) and \(k=1\).) With a left-hand side that is negative, and a coefficient on \(\varphi _{1}\) that is also negative, Eq. 28 can only hold if \(\varphi _{1}>0\). Hence, we have a contradiction. Since the coefficients on \(\varphi _{1}\) in Eqs. 27 and 28 are nonzero for \(\beta \in (1/2,\; 1)\), these equations can only hold simultaneously if \(\psi _{-i} = \psi _{i}\) and \(\varphi _{1} = 0\). If this is the case, then by Eq. 24, \(\varphi _{j}=0\) for \(j \in \{1,\ldots ,k\}\), and by Eq. 25, \(\alpha _{i}(0) = \alpha _{-i}(0)\). The equilibrium must therefore be symmetric.

For the subcase where \(\alpha _{i}(0),\ \alpha _{-i}(0) > 0\) for \(\alpha _{i}(0) \ne \alpha _{-i}(0)\), and \(\psi _{i} = \psi _{-i} = 0\), Eq. 27 holds with strict equality. The left-hand side is zero, and we already demonstrated that the coefficient on the right-hand side is strictly negative, so \(\varphi _{1}=0\). Consequently, all mass above zero is then symmetric, so \(\alpha _{i}(0) \ne \alpha _{-i}(0)\) cannot hold.

The final subcase of \(\alpha _{i}(0)>0\), \(\alpha _{-i}(0) =0\), and \(\psi _{i} = \psi _{-i} = 0\) likewise cannot hold. The indifference conditions of \(u_{i}(0,\; G_{-i}) = u_{i}(\phi _{1},\; G_{i})\) and \(u_{-i}({\overline{\delta }},\; G_{i}) = u_{-i}(\phi _{1} + \varphi _{1},\ G_{i})\), which appear with strict equality before Eqs. 26 and 28, together require \(\varphi = -\delta \). However, this is below the possible range for \(\varphi \), since \(\varphi \in [c-\delta ,\ \delta ]\) where \(c>0\). \(\square \)

Proofs specific to \(\beta = 1/2\)

Lemma 10

Let \(\beta = 1/2\) and \(\delta \in (0,\ v/4]\). If \((G_{i},\ G_{-i})\) is a pair of equilibrium strategies for players i and \(-i\), then \(u_{i}(a,\ G_{-i}) = u_{i}(a-2\delta ,\ G_{-i})\) for all \(a \in \text {{supp}}(G_{i})\) such that \(a > 3\delta \). Equivalently, \([G_{-i}(a+\delta ) - G_{-i}(a-3\delta )] = 4\delta /v\).

Proof

When \(\beta = 1/2\), Eq. 14 stipulates that for any \(x, y \in \text {supp}(G_{i})\) such that \(x > y\):

$$\begin{aligned} G_{-i}(x+\delta ) - G_{-i}(y+\delta ) + G_{-i}(x-\delta ) - G_{-i}(y-\delta ) = (2/v)(x-y) \ \end{aligned}$$

The condition of having mass in \(G_{-i}\) over \([y-\delta ,\ x-\delta ]\) and \([y+\delta ,\ x+\delta ]\) sum to \((2/v)(x-y)\) continues to hold for arbitrarily close x and y. Hence, \(g_{-i}(x+\delta ) + g_{-i}(x-\delta ) = 2/v\), where \(g_{-i}(x)\) is the density rate of \(G_{-i}\) at x. An implication is that, for any given \(\kappa >0\), the interval \([\kappa ,\ \kappa +4\delta ]\) in \(G_{-i}\) cannot have more mass than \((2/v)\times 2\delta \). That is, \([G_{-i}(\kappa +4\delta ) - G_{-i}(\kappa )] \le 4\delta /v\). This follows because \(g_{-i}(q) + g_{-i}(q-2\delta ) = 2/v\) for any \(q \in \text {supp}(G_{-i})\cap [\kappa + 2\delta ,\; \kappa + 4\delta ]\) and \(g_{-i}(r) + g_{-i}(r+2\delta ) = 2/v\) for any \(r \in \text {supp}(G_{-i})\cap [\kappa ,\; \kappa + 2\delta ]\). With this upper bound on the mass in \([\kappa ,\ \kappa +4\delta ]\), the main result follows quickly.

If \((G_{i},\ G_{-i})\) is a pair of equilibrium strategies, then for any \(a \in \text {supp}(G_{i})\) such that \(a > 3\delta \), equilibrium implies that \(u_{i}(a,\ G_{-i}) \ge u_{i}(a-2\delta ,\ G_{-i})\). To show that this must hold with equality, suppose by way of contradiction that \(u_{i}(a,\ G_{-i}) > u_{i}(a-2\delta ,\ G_{-i})\) for some a. Since \(u_{i}(a,\ G_{-i}) = [G_{-i}(a+\delta ) + G_{-i}(a-\delta )](v/2) - a\) and \(u_{i}(a-2\delta ,\ G_{-i}) = [G_{-i}(a-\delta ) + G_{-i}(a-3\delta )](v/2) - (a-2\delta )\), then \(u_{i}(a,\ G_{-i}) > u_{i}(a-2\delta ,\ G_{-i})\) implies that \([G_{-i}(a+\delta ) - G_{-i}(a-3\delta )] > 4\delta /v\). This, however, contradicts the upper bound of \(4\delta /v\), and so \(u_{i}(a,\ G_{-i}) = u_{i}(a-2\delta ,\ G_{-i})\). \(\square \)

Lemma 11

Let \(\beta = 1/2\) and \(\delta \in (0,\ v/4]\). Property \({\mathscr {P}}\) must hold in any equilibrium.

Proof

For added clarity, we will refer to \(G_{i}\) and \(G_{-i}\) as \(G_{w}\) and \(G_{y}\). We will also denote \({\overline{w}}_{1} = \max \{x \in \text {supp}(G_{w})\}\) and \({\overline{y}}_{1} = \max \{x \in \text {supp}(G_{y})\}\). Without loss of generality, assume that \({\overline{w}}_{1} \ge {\overline{y}}_{1}\), so \({\overline{y}}_{1} \in [{\overline{w}}_{1} - \delta , {\overline{w}}_{1}]\). Our first step is to show that \(G_{w}({\overline{w}}_{1}-3\delta ) = G_{w}({\overline{w}}_{1}-4\delta )\) and that \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\overline{w}}_{1}-4\delta )\). By Lemma 10, since \({\overline{w}}_{1} \in \text {supp}(G_{w})\) and \(G_{y}({\overline{w}}_{1}+\delta )=G_{y}({\overline{y}}_{1})\), then:

$$\begin{aligned}{}[G_{y}({\overline{y}}_{1}) - G_{y}({\overline{w}}_{1}-3\delta )]=4\delta /v \end{aligned}$$
(29)

For \({\overline{y}}_{1} \in ({\overline{w}}_{1} - \delta ,\ {\overline{w}}_{1}]\), Lemma 2.B implies that \({\overline{y}}_{1} - \delta \in \text {supp}(G_{w})\).Footnote 25 Then by Lemma 10:

$$\begin{aligned}{}[G_{y}({\overline{y}}_{1}) - G_{y}({\overline{y}}_{1}-4\delta )]=4\delta /v \end{aligned}$$

Hence, \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\overline{y}}_{1}-4\delta )\), and by a similar argument, \(G_{w}({\overline{y}}_{1}-3\delta ) = G_{w}({\overline{w}}_{1}-4\delta )\). If \({\overline{w}}_{1} = {\overline{y}}_{1}\), then we are done. Otherwise, for \({\overline{w}}_{1} > {\overline{y}}_{1}\), we must still show that \(G_{w}({\overline{w}}_{1}-3\delta ) = G_{w}({\overline{y}}_{1}-3\delta )\). Suppose instead that \(G_{w}({\overline{w}}_{1}-3\delta ) > G_{w}({\overline{y}}_{1}-3\delta )\). Denote \({\underline{w}} = \min \{x \in \text {supp}(G_{w})\ |\ x \ge {\overline{y}}_{1}-3\delta \}\). Since \({\overline{w}}_{1}-\delta \in \text {supp}(G_{y})\), then by Lemma 10, \(u_{y}({\overline{w}}_{1}-3\delta ,\ G_{w}) = u_{y}({\overline{w}}_{1}-\delta ,\ G_{w})\). Hence, in equilibrium, \(u_{y}({\overline{w}}_{1}-3\delta ) \ge u_{y}({\underline{w}})\), or rather from Eq. 2:

$$\begin{aligned}&[G_{w}({\overline{w}}_{1}\!-\!2\delta ) - G_{w}({\underline{w}}\!+\!\delta )]v/2 \ + \ [G_{w}({\overline{w}}_{1}\!-\!4\delta ) \\&\quad - G_{w}({\underline{w}}\!-\!\delta )]v/2 \ \ge \ {\overline{w}}_{1}\!-\!3\delta \!-\!{\underline{w}} \end{aligned}$$

Since density rates are bounded between 0 and 2 / v (inclusive), the above inequality requires \(G_{w}\) to have mass in \([{\underline{w}} - \delta , {\overline{w}}_{1}-4\delta ]\) unless the entire interval of \([{\underline{w}} + \delta ,\ {\overline{w}}_{1}-2\delta ]\) has a density rate of 2 / v in \(G_{w}\). But that cannot be the case: There must be some mass in a neighborhood below \({\overline{w}}_{1}\) in \(G_{w}\), which by Lemma 2.B implies that there is mass in a neighborhood below \({\overline{w}}_{1}-\delta \) in \(G_{y}\). So to satisfy Equation 14, there must be a neighborhood below \({\overline{w}}_{1}-2\delta \) in \(G_{w}\) that has density rates less than 2 / v. Hence, it must be that \(G_{w}({\overline{w}}_{1}-4\delta ) - G_{w}({\underline{w}}-\delta ) > 0\).

Let \(\ell _{1},\; \ell _{2} \in \text {supp}(G_{w}) \cap [{\underline{w}}-\delta , {\overline{w}}_{1}-4\delta ]\) be given such that \(\ell _{1} > \ell _{2}\) and \(G_{w}(\ell _{1}) - G_{w}(\ell _{2}) > 0\). (Such pairs of \(\ell _{1}\) and \(\ell _{2}\) necessarily exist since \(G_{w}({\overline{w}}_{1}-4\delta ) - G_{w}({\underline{w}}-\delta ) > 0\).) Recall that \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\overline{y}}_{1}-4\delta )\) and \({\underline{w}} \in [{\overline{y}}_{1}-3\delta ,\ {\overline{w}}_{1}-3\delta )\), so \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\underline{w}})\). Lemma 2.B implies that \(\ell _{1}-\delta ,\; \ell _{2}-\delta \in \text {supp}(G_{y})\) and that any mass in \(G_{y}\) over \([{\underline{w}}-2\delta , {\overline{w}}_{1}-5\delta ]\) has a density rate of 2 / v. By Lemma 10, \(u_{y}(\ell _{1}-\delta ,\ G_{w}) = u_{y}(\ell _{1}-3\delta ,\ G_{w})\) and \(u_{y}(\ell _{2}-\delta ,\ G_{w}) = u_{y}(\ell _{2}-3\delta ,\ G_{w})\). Equivalently, \([G_{w}(\ell _{1}) - G_{w}(\ell _{1}-4\delta )]=4\delta /v\) and \([G_{w}(\ell _{2}) - G_{w}(\ell _{2}-4\delta )]=4\delta /v\), or rather, \(G_{w}(\ell _{1}) - G_{w}(\ell _{2}) = G_{w}(\ell _{1}-4\delta ) - G_{w}(\ell _{2}-4\delta )\). Since \(G_{w}(\ell _{1}) - G_{w}(\ell _{2}) > 0\), it must be that \(G_{w}(\ell _{1}-4\delta ) - G_{w}(\ell _{2}-4\delta ) > 0\). Therefore, \(G_{w}({\overline{w}}_{1}-8\delta ) - G_{w}({\underline{w}}-5\delta ) > 0\) since \([\ell _{2}-4\delta ,\ \ell _{1}-4\delta ] \subseteq [{\underline{w}}-5\delta , {\overline{w}}_{1}-8\delta ]\).

By Lemma 10, \(u_{w}(\ell _{1},\ G_{y}) = u_{w}(\ell _{1}-2\delta ,\ G_{y})\) and \(u_{w}(\ell _{2},\ G_{y}) = u_{w}(\ell _{2}-2\delta ,\ G_{y})\). Since these are equilibrium expected payoffs, they must equal each other: \(u_{w}(\ell _{1}-2\delta ,\ G_{y})=u_{w}(\ell _{2}-2\delta ,\ G_{y})\). Recall that any mass in \(G_{y}\) over \([{\underline{w}}-2\delta ,\ {\overline{w}}_{1}-5\delta ]\) has a density rate of 2 / v. Applying Eq. 14 to \(u_{w}(\ell _{1}-2\delta ,\ G_{y})=u_{w}(\ell _{2}-2\delta ,\ G_{y})\) implies that \(G_{y}\) has no mass over \([\ell _{2}-3\delta ,\ \ell _{1}-3\delta ]\subseteq [{\underline{w}}-4\delta , {\overline{w}}_{1}-7\delta ]\). Moreover, since \(\ell _{1}\) and \(\ell _{2}\) can be arbitrarily chosen within the stated constraints, \(G_{y}\) must have no mass in \([{\underline{w}}-4\delta , {\overline{w}}_{1}-7\delta ]\). Or rather, \(G_{y}({\overline{w}}_{1}-7\delta ) = G_{y}({\underline{w}}-4\delta )\).

We now have a situation analogous to the one we had previously: Just as \(G_{w}({\overline{w}}_{1}-4\delta ) - G_{w}({\underline{w}}-\delta ) > 0\), we now have \(G_{w}({\overline{w}}_{1}-8\delta ) - G_{w}({\underline{w}}-5\delta ) > 0\), and just as \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\underline{w}})\), we now have \(G_{y}({\overline{w}}_{1}-7\delta ) = G_{y}({\underline{w}}-4\delta )\). The argument then repeats until the bottom of the distribution. At that point, payoffs can no longer be sustained by additional mass from below, which contradicts the supposition that \(G_{w}({\overline{w}}_{1}-3\delta ) > G_{w}({\overline{y}}_{1}-3\delta )\). Since \(G_{w}\) is weakly monotonic, it must be that \(G_{w}({\overline{w}}_{1}-3\delta ) = G_{w}({\overline{y}}_{1}-3\delta )\). Combined with the earlier finding that \(G_{w}({\overline{y}}_{1}-3\delta ) = G_{w}({\overline{w}}_{1}-4\delta )\), we have \(G_{w}({\overline{w}}_{1}-3\delta ) = G_{w}({\overline{w}}_{1}-4\delta )\). We also previously showed that \(G_{y}({\overline{w}}_{1}-3\delta ) = G_{y}({\overline{y}}_{1}-4\delta )\) (with \({\overline{w}}_{1} \ge {\overline{y}}_{1}\)).

Just as there is no mass within \(\delta \) above \({\overline{w}}_{1}\) and \({\overline{y}}_{1}\), since \(G_{w}\) and \(G_{y}\) have no mass for at least \(\delta \) below \({\overline{w}}_{1}-3\delta \), then all of the above arguments apply to mass below \({\overline{w}}_{1}-4\delta \). Let \({\overline{w}}_{2} = \max \{x \in \text {supp}(G_{w})\ |\ x \le {\overline{w}}_{1}-4\delta \}\), \({\overline{y}}_{2} = \max \{x \in \text {supp}(G_{y})\ |\ x \le {\overline{y}}_{1}-4\delta \}\), and \(m_{2} = \max \{{\overline{w}}_{2}, {\overline{y}}_{2}\}\). By the same arguments, \(G_{w}(m_{2}-3\delta ) = G_{w}(m_{2}-4\delta )\) and \(G_{y}(m_{2}-3\delta ) = G_{y}(m_{2}-4\delta )\). Or more generally, for \({\overline{w}}_{z} = \max \{x \in \text {supp}(G_{w})\ |\ x \le {\overline{w}}_{z-1}-4\delta \}\), \({\overline{y}}_{z} = \max \{x \in \text {supp}(G_{y})\ |\ x \le {\overline{y}}_{z-1}-4\delta \}\), and \(m_{z} = \max \{{\overline{w}}_{z}, {\overline{y}}_{z}\}\), where \(z \in \{2,3,\ldots \}\), then \(G_{w}(m_{z}-3\delta ) = G_{w}(m_{z}-4\delta )\) and \(G_{y}(m_{z}-3\delta ) = G_{y}(m_{z}-4\delta )\). Moreover,

$$\begin{aligned}{}[G_{w}(m_{z}) - G_{w}(m_{z}-3\delta )] = [G_{y}(m_{z}) - G_{y}(m_{z}-3\delta )] = 4\delta /v \end{aligned}$$

To satisfy the constraint that all mass must sum to one, \(G_{w}\) and \(G_{y}\) each have \(p \equiv \left\lfloor v/4\delta \right\rfloor \) such intervals. That is, p intervals of length \(4\delta \), each having a total mass of \(4\delta /v\), with no mass in the bottom \(\delta \) of each length-\(4\delta \) interval. (The remaining \(1 - [4\delta p/v]\) is then at the bottom of the distribution, as is explained below.) Placing a mass of \(4\delta /v\) within \(3\delta \), with no mass \(\delta \) above or below, requires that \(G_{w}\) and \(G_{y}\) each have \(2\delta /v\) over \([m_{z}-2\delta ,\ m_{z}-\delta ]\) and \(2\delta /v\) over \([m_{z}-3\delta ,\ m_{z}-2\delta ]\, \bigcup \, [m_{z}-\delta ,\ m_{z}]\) for \(z \in \{1,\ldots ,p\}\). Mass over \([m_{z}-2\delta ,\ m_{z}-\delta ]\) must be at a density rate of 2 / v, while the density rates at x and \(x-2\delta \) for \(x \in [m_{z}-\delta ,\ m_{z}]\) must sum to 2 / v (see Lemma 2 and Eq. 14).

We can also state how successive length-\(4\delta \) intervals fit together. Since \([m_{z}-2\delta ,\ m_{z}-\delta ]\) is in the support of \(G_{w}\) and \(G_{y}\), then by Lemma 10, players are indifferent between any bid in \([m_{z}-2\delta ,\ m_{z}-\delta ]\) and any bid in \([m_{z}-3\delta ,\ m_{z}-4\delta ]\). In particular, \(u_{y}(m_{z}-3\delta ,\ G_{w}) = u_{y}(m_{z}-4\delta ,\ G_{w})\) and \(u_{w}(m_{z}-3\delta ,\ G_{y}) = u_{w}(m_{z}-4\delta ,\ G_{y})\), which, respectively, imply:

$$\begin{aligned}{}[G_{w}(m_{z}-2\delta )-G_{w}(m_{z}-5\delta )]&=2\delta /v \nonumber \\ [G_{y}(m_{z}-2\delta )-G_{y}(m_{z}-5\delta )]&=2\delta /v \end{aligned}$$
(30)

So \(G_{w}\) and \(G_{y}\) each have \(2\delta /v\) over \([m_{z}-5\delta ,\ m_{z}-4\delta ]\, \bigcup \, [m_{z}-3\delta ,\ m_{z}-2\delta ]\), and the density rates at \(x \in [m_{z}-3\delta ,\ m_{z}-2\delta ]\) and \(x-2\delta \) must sum to 2 / v to support the expected payoffs in \([m_{z}-3\delta ,\ m_{z}-4\delta ]\). Consequently, players must be indifferent between bids in \([m_{z}-3\delta ,\ m_{z}-4\delta ]\) and bids in \([m_{z}-4\delta ,\ m_{z}-5\delta ]\).Footnote 26 We therefore have \(u_{y}(m_{z}-3\delta ,\ G_{w}) = u_{y}(m_{z}-5\delta ,\ G_{w})\) and \(u_{w}(m_{z}-3\delta ,\ G_{y}) = u_{w}(m_{z}-5\delta ,\ G_{y})\), and so \([G_{w}(m_{z}-2\delta )-G_{w}(m_{z}-6\delta )]=4\delta /v\) and \([G_{y}(m_{z}-2\delta )-G_{y}(m_{z}-6\delta )]=4\delta /v\). Combined with Eq. 30, \([G_{w}(m_{z}-5\delta )-G_{w}(m_{z}-6\delta )]=2\delta /v\) and \([G_{y}(m_{z}-5\delta )-G_{y}(m_{z}-6\delta )]=2\delta /v\), which can only hold if the mass is distributed at a rate of 2 / v. Finally, \(u_{y}(m_{z}-4\delta ,\ G_{w}) = u_{y}(m_{z}-6\delta ,\ G_{w})\) and \(u_{y}(m_{z}-5\delta ,\ G_{w}) = u_{y}(m_{z}-7\delta ,\ G_{w})\) give us \(G_{w}(m_{z}-7\delta ) = G_{w}(m_{z}-8\delta )\); the corresponding equations for \(u_{w}\) yield \(G_{y}(m_{z}-7\delta ) = G_{y}(m_{z}-8\delta )\).

Lemma 12

Let \(\beta = 1/2\) and \(\delta \in (0,\ v/4]\). Also, let \(p = \left\lfloor v/4\delta \right\rfloor \) be the number of length-\(4\delta \) intervals with the properties specified by \({\mathscr {P}}\). In any equilibrium, the top of the pth such interval (i.e., the top of the bottommost length-\(4\delta \) interval) is in \([2\delta ,\ 3\delta ]\) if \(\delta = v/4p\); in \((3\delta ,\ 4\delta ]\) if \(\delta \in [v/(4p+2),\ v/4p)\); and at \((v/2) - \delta (2p-4)\) if \(\delta \in (v/(4p+4),\ v/(4p+2))\). Below the pth interval, a total mass of \(1-[4\delta p/v]\) is distributed as follows:

A.:

If \(\delta \in [v/(4p+2),\ v/4p)\), the remaining \(1-[4\delta p/v]\) is at zero, neither player has mass in \((0,\ \delta )\), and all equilibria have an expected payoff of \((v/2) - 2\delta p\). For \(\delta = v/(4p+2)\), the equilibrium is unique: There is no mass in \((0,\ 2\delta )\) and the top of the pth interval is at \(4\delta \).

B.:

If \(\delta \in (v/(4p+4),\ v/(4p+2))\), each player has a mass point at zero of \([4\delta (p+1)/v] - 1\), a uniform density rate of 2 / v over the intervals \([0,\ v/2 - \delta (2p+1)]\) and \([\delta ,\ v/2 - 2\delta p]\), and an expected payoff of \(\delta \). This also holds for \(\delta \in (v/4,\ v/2)\) (i.e., \(p=0\)).

Proof

Following the notation from the proof of Lemma 11, let \(m_{p} = m_{1} - 4\delta (p-1)\), where \(m_{1} = \max \{{\overline{w}}_{1},\ {\overline{y}}_{1}\}\). That is, \(m_{p}\) is the top of the pth length-\(4\delta \) interval (specifically for the player whose support contains the highest element; alternatively, \(m_{p} = \max \{{\overline{w}}_{p}, {\overline{y}}_{p}\}\)). By Lemma 11, these p intervals satisfy \({\mathscr {P}}\), and so the remaining mass of \(1-[4\delta p/v]\) must be distributed below them at the bottom of the distribution. There are two bounds that we can quickly place on \(m_{p}\). First, \(\min \{{\overline{w}}_{p}, {\overline{y}}_{p}\} \ge 2\delta \); otherwise, with a maximal density rate of 2 / v it is not possible to have \(4\delta /v\) of continuously distributed mass. Second, \(m_{p} < 6\delta \), or it would be possible to have \(4\delta /v\) of continuously distributed mass below \(m_{p}-4\delta \). We will show that \(m_{p} \in [2\delta ,\ 3\delta ]\) when \(\delta = v/4p\); \(m_{p} \in (3\delta ,\ 4\delta ]\) when \(\delta \in [v/(4p+2),\ v/4p)\); and \(m_{p} \in (5\delta ,\ 6\delta )\) when \(\delta \in (v/(4p+4),\ v/(4p+2))\). This covers the complete range of \(\delta \) for any given p. Furthermore, we will show how the remaining \(1-[4\delta p/v]\) is distributed, as well as a uniqueness result for \(\delta = v/(4p+2)\). Without loss of generality, we will suppose throughout that \(m_{p} ={\overline{w}}_{p}\).

If \({\overline{w}}_{p} \in [2\delta ,\ 3\delta )\), \(4\delta /v\) must be distributed according to property \({\mathscr {P}}\) over \((0, {\overline{w}}_{p})\) in \(G_{w}\) and over \((0, {\overline{y}}_{p})\) in \(G_{y}\); the remaining mass of \(1-[4\delta p/v]\) must then be at zero. We will show that neither player has a mass point at zero (that is, \(1-[4\delta p/v]=0\), or equivalently, \(\delta = v/4p\)). Suppose by way of contradiction that \(1-[4\delta p/v]>0\). According to property \({\mathscr {P}}\), \([{\overline{w}}_{p}-2\delta ,\ {\overline{w}}_{p}-\delta ]\) has a density rate of 2 / v in \(G_{w}\), and \([{\overline{y}}_{p}-2\delta , {\overline{y}}_{p}-\delta ]\) has a density rate of 2 / v in \(G_{y}\). However, with \({\overline{w}}_{p},\ {\overline{y}}_{p} \in [2\delta ,\ 3\delta )\), then \(\delta \in ({\overline{w}}_{p}-2\delta , {\overline{w}}_{p}-\delta ]\) and \(\delta \in ({\overline{y}}_{p}-2\delta , {y}_{p}-\delta ]\). If the opponent has a strictly positive mass point at zero, placing mass immediately below \(\delta \) cannot hold in equilibrium for either player. Thus, we can only have \({\overline{w}}_{p} \in [2\delta ,\ 3\delta )\) or \({\overline{y}}_{p} \in [2\delta ,\ 3\delta )\) when \(\delta = v/4p\).

If \({\overline{w}}_{p} \in (3\delta ,\ 4\delta ]\), \(4\delta /v\) must likewise be distributed according to property \({\mathscr {P}}\) over \((0, {\overline{w}}_{p})\) in \(G_{w}\) and over \((0, {\overline{y}}_{p})\) in \(G_{y}\), and the remaining mass of \(1-[4\delta p/v]\) must be at zero. We will show that both players must have a mass point at zero (that is, \(1-[4\delta p/v]>0\), or equivalently, \(\delta < v/4p\)). Suppose by way of contradiction that \(1-[4\delta p/v]=0\). We will first demonstrate that \(2\delta \in \text {supp}(G_{w})\) and \(2\delta \in \text {supp}(G_{y})\). From property \({\mathscr {P}}\), \([{\overline{w}}_{p}-2\delta ,\ {\overline{w}}_{p}-\delta ]\) has a density rate of 2 / v in \(G_{w}\) and \([{\overline{y}}_{p}-2\delta , {\overline{y}}_{p}-\delta ]\) has a density rate of 2 / v in \(G_{y}\). In particular, \(2\delta \in [{\overline{w}}_{p}-2\delta , {\overline{w}}_{p}-\delta ]\), so \(2\delta \in \text {supp}(G_{w})\). Likewise, if \({\overline{y}}_{p} \ge 3\delta \), then \(2\delta \in [{\overline{y}}_{p}-2\delta ,\ {\overline{y}}_{p}-\delta ]\), so \(2\delta \in \text {supp}(G_{y})\). If instead \({\overline{y}}_{p} \in [{\overline{w}}_{p} - \delta ,\ 3\delta )\), then since \({\overline{y}}_{p}-2\delta \in \text {supp}(G_{y})\), we must have \(u_{y}({\overline{y}}_{p}-2\delta ,\ G_{w}) \ge u_{y}(0,\ G_{w})\). By property \({\mathscr {P}}\), \(G_{w}({\overline{y}}_{p}-3\delta )=0\) when \(1-[4\delta p/v]=0\), so \(u_{y}({\overline{y}}_{p}-2\delta ,\ G_{w}) \ge u_{y}(0,\ G_{w})\) is equivalent to \([G_{w}({\overline{y}}_{p}-\delta ) - G_{w}(\delta )](v/2) \ge ({\overline{y}}_{p}-2\delta )\) (by Eq. 2). This inequality can only be satisfied if \([\delta ,\ {\overline{y}}_{p}-\delta ]\) has a density rate of 2 / v in \(G_{w}\). With \(\delta \in \text {supp}(G_{w})\) and no mass point at zero in \(G_{y}\), we have \(u_{w}(\delta ,\ G_{y}) \ge u_{w}(0,\ G_{y})\). By Eq. 2, this is equivalent to \([G_{y}(2\delta ) - G_{y}(\delta )](v/2) \ge \delta \), which can only be satisfied if \([\delta ,\ 2\delta ]\) has a density rate of 2 / v in \(G_{y}\). Thus, \(2\delta \in \text {supp}(G_{y})\) (for all possible ranges of \({\overline{y}}_{p}\)). We must then have \(u_{y}(2\delta ,\ G_{w}) \ge u_{y}(0,\ G_{w})\), which is equivalent to \(G_{w}(3\delta ) \ge 4\delta /v\) (by Eq. 2). This leads to a contradiction: Without a mass point at zero, \(G_{w}({\overline{w}}_{p}) = 4\delta /v\) where \({\overline{w}}_{p} > 3\delta \). But there must be some mass in a neighborhood immediately below \({\overline{w}}_{p}\) in \(G_{w}\) (i.e., \(G_{w}({\overline{w}}_{p}) - G_{w}(3\delta ) > 0\), but \(G_{w}(3\delta ) \ge G_{w}({\overline{w}}_{p})\)). Hence, we can only have \({\overline{w}}_{p} \in (3\delta ,\ 4\delta ]\) when \(\delta < v/4p\), and since both players must have a mass point at zero, it must also be the case that \({\overline{y}}_{p} \in [3\delta ,\ 4\delta ]\).

We next show that we can only have \({\overline{w}}_{p} \in (3\delta ,\ 4\delta ]\) when \(\delta \ge v/(4p+2)\). With zero and \(2\delta \) in each player’s support, \(u_{w}(2\delta ,\ G_{y}) = u_{w}(0,\ G_{y})\) and \(u_{y}(2\delta ,\ G_{w}) = u_{y}(0,\ G_{w})\). Equating these yields:

$$\begin{aligned} G_{w}(3\delta ) = G_{y}(3\delta ) = 4\delta /v \end{aligned}$$
(31)

By property \({\mathscr {P}}\), at least \(2\delta /v\) of this \(4\delta /v\) is distributed at a density rate of 2 / v over \([{\overline{w}}_{p}-2\delta , {\overline{w}}_{p}-\delta ]\) in \(G_{w}\) and over \([{\overline{y}}_{p}-2\delta ,\ {\overline{y}}_{p}-\delta ]\) in \(G_{y}\). So the mass point of \(1-[4\delta p/v]\) must be weakly less than \(2\delta /v\). Equivalently, \(\delta \ge v/(4p+2)\), and so \({\overline{w}}_{p} \in (3\delta ,\ 4\delta ]\) implies that \(\delta \in [v/(4p+2),\ v/4p)\). Equation 31 also rules out the possibility of a strictly positive mass point when \({\overline{w}}_{p}=3\delta \) or \({\overline{y}}_{p}=3\delta \). (By property \({\mathscr {P}}\) and Eq. 31, there is already \(4\delta /v\) in \((0,\ 3\delta ]\), so there is no room for a mass point.)

Next, suppose that \({\overline{w}}_{p} \in (4\delta ,\ 5\delta ]\). We will show that this cannot hold in equilibrium. By property \({\mathscr {P}}\), there is no mass in \(G_{w}\) over \([{\overline{w}}_{p} - 4\delta , {\overline{w}}_{p} - 3\delta ]\) and no mass in \(G_{y}\) over \((\max \{{\overline{y}}_{p} - 4\delta ,\ 0\}, {\overline{y}}_{p} - 3\delta ]\). Moreover, since \({\overline{w}}_{p} - 4\delta \le \delta \) and \(\max \{{\overline{y}}_{p} - 4\delta ,\ 0\} \le \delta \), Lemma 2 prohibits any continuously distributed mass over \((0,\ {\overline{w}}_{p} - 4\delta ]\) in \(G_{w}\) and over \((0,\ {\overline{y}}_{p} - 4\delta ]\) in \(G_{y}\) (if \({\overline{y}}_{p} - 4\delta > 0\)). So the remaining mass of \(1-[4\delta p/v]\) is at zero in \(G_{w}\) and \(G_{y}\). Since \(G_{w}\) has no mass in \((0,\ {\overline{w}}_{p} - 3\delta )\), where \({\overline{w}}_{p} - 3\delta > \delta \), and \(G_{y}\) has no mass in \((0, {\overline{y}}_{p} - 3\delta )\), it must be that the mass point at zero is strictly positive (i.e., \(\alpha (0) = 1-[4\delta p/v] > 0)\); otherwise, both players could profitably deviate with a bid of zero. For property \({\mathscr {P}}\) to hold, due to this mass point at zero, any mass in \(G_{y}\) over \([{\overline{y}}_{p} - 3\delta , {\overline{y}}_{p}\)) must be above \(\delta \). With no mass over \((0,\ \delta )\) in \(G_{w}\) or \(G_{y}\), the players have the same equilibrium expected payoffs: \(u_{w}(0,\ G_{y}) = u_{y}(0,\ G_{w}) = \alpha (0)(v/2)\). By property \({\mathscr {P}}\), all bids in \((\delta , {\overline{w}}_{p}]\) in \(G_{w}\) and all bids in \((\delta , {\overline{y}}_{p}]\) in \(G_{y}\) yield this same equilibrium expected payoff. This follows because the density rates \(g_{w}(x+\delta ) + g_{w}(x-\delta ) = 2/v\) for all \(x \in ({\overline{w}}_{p}-3\delta , {\overline{w}}_{p}-\delta ]\), and \(g_{y}(z+\delta ) + g_{y}(z-\delta ) = 2/v\) for all \(z \in (\max \{{\overline{y}}_{p} - 3\delta ,\ \delta \},\ {\overline{y}}_{p}-\delta ]\). In particular, since the players have the same expected payoff, \(u_{w}(x,\ G_{y}) = u_{y}(x,\ G_{w})\) for all \(x \in (\delta ,\ 2\delta ]\). By Eq. 2, and with no mass in \((0,\ \delta )\), this implies that \(G_{y}(x +\delta ) - G_{y}(\delta ) = G_{w}(x +\delta ) - G_{w}(\delta )\) for all \(x \in (\delta ,\ 2\delta ]\). Thus, \(G_{y}\) and \(G_{w}\) have the same distribution over \((2\delta ,\ 3\delta ]\). Let \(\zeta \equiv \min \{x \in \text {supp}(G_{w})\ |\ x \ge {\overline{w}}_{p}-3\delta \}\). Since \(\zeta > \delta \) and \(G_{w}\) has no mass in \((0,\ \zeta )\), \(u_{w}(\zeta ,\ G_{y}) = u_{w}({\overline{\delta }},\ G_{y})\) can only hold if \(G_{y}\) has the maximal density rate of 2 / v over \([2\delta ,\ \zeta +\delta ]\). \(G_{w}\) has this same distribution. Then by property \({\mathscr {P}}\), \(G_{y}\) and \(G_{w}\) have the minimal density rate of zero over \([4\delta ,\ \zeta +3\delta ]\). However, with \(\zeta +3\delta \ge {\overline{w}}_{p}\), this implies that there is no mass in \(G_{w}\) over \([4\delta , {\overline{w}}_{p}]\), which contradicts \({\overline{w}}_{p} \in (4\delta ,\ 5\delta ]\).

Finally, if \({\overline{w}}_{p} \in (5\delta ,\ 6\delta )\), then \({\overline{w}}_{p}-4\delta \in (\delta ,\ 2\delta )\). Following \({\mathscr {P}}\), there is no mass in \(G_{w}\) over \([{\overline{w}}_{p}-4\delta , {\overline{w}}_{p}-3\delta ]\), and consequently, by Lemma 2, no mass in \(G_{y}\) over \([{\overline{w}}_{p}-5\delta ,\ \delta )\). Likewise, there is no mass in \(G_{y}\) over \([{\overline{y}}_{p}-4\delta ,\ {\overline{y}}_{p}-3\delta ]\), and no mass in \(G_{w}\) over \([{\overline{y}}_{p}-5\delta ,\ \delta )\).Footnote 27 Applying Parts A and B of Lemma 2, any continuously distributed mass over \([0,\ {\overline{y}}_{p}-5\delta ]\bigcup \,[\delta ,\ {\overline{w}}_{p}-4\delta ]\) in \(G_{w}\) and over \([0,\ {\overline{w}}_{p}-5\delta ]\bigcup \,[\delta ,\ {\overline{y}}_{p}-4\delta ]\) in \(G_{y}\) must have a density rate of 2 / v. It follows then that \(G_{w}\) and \(G_{y}\) must have the same amount of continuously distributed mass, and so \(\alpha _{w}(0) = \alpha _{y}(0) \equiv \alpha (0)\). Since \(G_{w}\) has no mass over \([{\overline{y}}_{p}-5\delta ,\ \delta )\), and \(G_{y}\) has no mass over \([{\overline{w}}_{p}-5\delta ,\ \delta )\), for players to be indifferent between bids above and below these gaps, we need \(\alpha (0) > 0\). Since a bid of \({\overline{w}}_{p}-4\delta \) for player w and of \({\overline{y}}_{p}-4\delta \) for player y have the same expected payoffs as a bid \(2\delta \) above that or any other bid in their support (see Lemmata 10 and 11), it must be that \(u_{w}({\overline{w}}_{p}-4\delta ,\ G_{y}) \ge u_{w}({\overline{\delta }},\ G_{y})\) and \(u_{y}({\overline{y}}_{p}-4\delta ,\ G_{w}) \ge u_{y}({\overline{\delta }},\ G_{w})\). These can only be satisfied if \(G_{y}\) has a density rate of 2 / v over \([0,\ {\overline{w}}_{p}-5\delta ]\) and if \(G_{w}\) has a density rate of 2 / v over \([0, {\overline{y}}_{p}-5\delta ]\). By Lemma 2, these in turn imply a density rate of 2 / v over \([\delta , {\overline{w}}_{p}-4\delta ]\) in \(G_{w}\) and \([\delta , {\overline{y}}_{p}-4\delta ]\) in \(G_{y}\). With a common mass point, each player will only be indifferent between these two intervals if \({\overline{w}}_{p} = {\overline{y}}_{p}\). This constrains the mass below \({\overline{w}}_{p}-4\delta \) in \(G_{w}\) and \(G_{y}\) to \(1-[4\delta p/v] = \alpha (0) + 2\,[{\overline{w}}_{p}-5\delta ](2/v)\). From this mass constraint and \(u_{w}({\overline{w}}_{p}-4\delta ,\ G_{y})=u_{w}(0,\ G_{y})\), we can identify \({\overline{w}}_{p} = (v/2) - 2\delta (p-2)\) and \(\alpha (0)=[4\delta (p+1)/v]-1\). We can establish bounds on \(\delta \): from \(\alpha (0)>0\), \(\delta > v/(4p+4)\), and from \({\overline{w}}_{p}-5\delta > 0\), \(\delta < v/(4p+2)\). That is, \(\delta \in (v/(4p+4),\ v/(4p+2))\). This argument also applies to the case of \(\delta \in (v/4,\ v/2)\).

Having established the various bounds for \(\delta \) for each potential value of \({\overline{w}}_{p}\), we conclude with a uniqueness result at \(\delta = v/(4p+2)\), which falls under the case of \({\overline{w}}_{p} \in (3\delta ,\ 4\delta ]\). We have already shown that there is a mass point at zero of \(1-[4\delta p/v]\) and no mass in \((0,\ \delta )\). In equilibrium, we must have \(u_{i}(0,\ G_{-i}) \ge u_{i}({\overline{\delta }},\ G_{-i})\), which equates to \([\delta (4p+2)/v] - 1 \ge G_{-i}(2\delta ) - G_{-i}(\delta )\) (by Eq. 2). The left-hand side is zero when \(\delta = v/(4p+2)\). Hence, there is no mass in \((0,\ 2\delta )\). The \(4\delta /v\) in the pth interval must then be distributed over \([2\delta ,\ 4\delta ]\) at a rate of 2 / v. The rest of the distribution follows from \({\mathscr {P}}\). \(\square \)

Asymmetric equilibria

For \(\delta \in (0,\, (1-\beta )v)\) and \(\beta \in (0,\, 1/2)\), Algorithm 1 identifies the complete set of asymmetric equilibria for the game \(APT\{\delta , \beta , v\}\). (The labels are the same as in Fig. 8.) With players arbitrarily assigned as player w or player y, the algorithm systematically varies the first \(1/[(1-\beta )v]\) segment in player w’s distribution to omit, as well as the uppermost interval pair in the two distributions. Then for each combination of omitted interval pairs, there are at most four conditions that must be checked to verify the existence of an asymmetric equilibrium. First, the system of equations formed from the indifference conditions between the intervals in each player’s distribution needs to produce strictly positive lengths for each of the non-omitted interval pairs and strictly positive mass for the non-excluded mass points. (The system of equations also includes two equations which specify that the mass in each player’s distribution must sum to one.) Second, if player w’s mass point was excluded, player w cannot profitably deviate by bidding zero.

figure a

All other profitable deviations are captured by the third and fourth conditions. Third, bidding \(\delta \) above the first omitted \(1/[(1-\beta )v]\) segment in player w’s distribution cannot be profitable for player y. Below this point, bids within gaps in either player’s distribution can be ruled out by arguments similar to those for the symmetric case (see the paragraphs leading up to Theorem 1). Above this point, the gaps are so large that bidding within a gap does not adequately increase the amount of mass a player is tying or beating. Precisely at this point, however, player y beats all of the \(1/(\beta v)\) segment that is \(\delta \) below it, and so the expected payoff rises to a peak—the only peak in this gap. It therefore suffices to check that this peak is not too high. The fourth condition similarly pertains to a peak. As in the symmetric case, it merely specifies that outbidding the opponent’s distribution by \(\delta \) cannot be profitable. Here, however, one player’s upper bound is a \(1/[(1-\beta )v]\) segment, while the other’s is a \(1/(\beta v)\) segment, which is already \(\delta \) above the first. So we simply need to verify that the player with the smaller upper bound cannot profit by outbidding their opponent’s distribution and winning with certainty.

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Gelder, A., Kovenock, D. & Roberson, B. All-pay auctions with ties. Econ Theory 74, 1183–1231 (2022). https://doi.org/10.1007/s00199-019-01195-7

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