Wars of Attrition with Budget Constraints

Wars of Attrition with Budget Constraints Gagan Ghosh Bingchao Huangfu Heng Liu October 19, 2017 (PRELIMINARY AND INCOMPLETE: COMMENTS WELCOME) Abstract We study wars of attrition between two bidders who have private budget constraints. In the baseline model, the bidders compete for one prize and their values of the prize are commonly known. We examine the impact of budgets on the bidding outcomes. In the case where a bidder s private budget is either high or low, we characterize the unique weak perfect Bayesian equilibrium that survives a refinement. In equilibrium, the high budget bidder always bids, which is reminiscent of the commitment type who is assumed to play a predetermined action in the reputation literature. Consequently, the low budget bidder, regardless of her value of the prize, never concedes with certainty as long as she has not depleted her budget. The results differ from the previous literature, which mostly assumes budgets are common knowledge. We then extend the model to a setting where the bidders compete for multiple prizes, either sequentially or simultaneously. We establish an equivalence result: the bidders equilibrium payoffs are the same regardless of the format of the wars of attrition, as long as the prizes are homogeneous. Finally, we consider the case where each bidder s budget is drawn from a continuum. We characterize the unique monotone equilibrium and show that the main features of the equilibrium in the binary type model continue to hold. Keywords: War of attrition, budgets, reputation JEL classification: C78, D82 Department of Economics, California State University Fullerton; gagan.ghosh@gmail.com Department of Economics, Zhongnan University of Economics and Law,; hfbcalan@gmail.com Department of Economics, University of Michigan; hengliu29@gmail.com 1

1 Introduction Many economic interactions where agents compete for a prize(s), such as patent races and lobbying activities in politics, are conveniently modeled as a war of attrition. In it s simplest form, a war of attrition is essentially a race to see who quits first. The bidder who does not quit, wins. Bidders may take turns in deciding to stay in the game or not. Each time a bidder decides to stay in the game, he commits some resources that are then sunk. In standard analysis the committed resources increase at a given rate per unit of time. Bidders competing in a war of attrition have value for the prize. The value may differ across bidders. We augment the standard model by also assuming that the bidders may have budget constraints that binds their ability to remain in the game in the style of Leininger (1991). 1 Valuations and budgets are common knowledge. We also introduce incomplete information in the game by assuming that bidders may be normal types, who quit when when the costs are higher than benefits, or they may be commitment types who never concede. They only stop competing once they run out budgets. These commitment or irrational types were introduced by Abreu and Gul (2000). We want to investigate the effect incomplete information has on a war of attrition game with budgets. Without incomplete information, Leininger (1991) showed that who wins the contest crucially depends on three things. Which bidder has the higher valuation, the higher budget and who moves first. Each one of these can lead to an advantage. For example a bidder with a higher valuation and who moves first, always wins. In all cases, he showed that there is no delay and agreement is reached immediately, which is usual with complete information bargaining games. By introducing incomplete information, we show that many of these results are reversed. There is often delay in the game, in the sense that both bidders remain in the game. Bidders with lower valuations and who move second may win, based on parameters. To be specific, we consider a baseline model that one of the bidders, say bidder 1 has private information of budgets: she has either a high budget or a low budget. Assume that bidder 2 s budget is common knowledge and is between the two possible budgets of bidder 1. Assume that two bidders bids alternatively and bidder 1 bids first. We derive a unique weak perfect Bayesian equilibrium as follows: 1. High budget bidder 1 always bids and guarantee a win, according to an equilibrium refinement 1 Also see Dekel, Jackson, and Wolinsky (2007). 2

which is a generalization of D1 criterion in signalling games. Therefore, high budget bidder 1 can be treated as a commitment type that always bids in the reputation literature. We call the probability of high budget bidder 1 as the reputation of high budget bidder 1. 2. If the reputation is high enough, the low budget bidder 1 bids for sure in period 1, and bidder 2 concedes immediately after she observes that bidder 1 bids. The idea is that since reputaton of high budget is already high enough, low budget bidder 1 can exploit the reputation by bidding for sure. 3. If the reputation is low, the low budgets bidder 1 mixes between bidding and conceding in period 1,and bidder 2 also mixes after observing that bidder 1 bids. The intuition is that when reputation is not high enough, lower budget bidder 1 builds up the reputation by randomizing the time of conceding, and the process of reputation building continues until the low budget bidder 1 runs out of budget. At the same time, bidder 2 also randomizes to guarantee that bidder 1 wants to build up reputation. In all, there is a war of attrition between low budget bidder 1 and bidder 2. The result of one-sided private budget can be generalized to two-sided private budgets. Either bidders has private information of her own budget: either a high budget or a low budget (two budget levels are the same for both bidders). Under the most interesting case that the probability of high budget for both bidders is low enough, then two results hold: 1. Both high budgets bidders behave as a commitment type by always bidding, according to the same equilibrium refinement; 2. Both low budget bidders engage in a war of attrition by randomizing between bidding and conceding at each period until the high budget is run out. We also study the war of attrition of N prizes if each bidder has a budget for all N prizes. Since N prizes are linked by the budget, we aims to explore the strategic influences among different wars of attritions. By assuming that N prizes are identical and the bidder s preference toward N prizes are separably additive. We show that the unique equilibrium has the following property: 1. If each bidder concedes one prize, then she will concede all the remaining prizes immediately. Therefore, two bidders bids as if they treated N prizes as one big prize; 3

2. In the continuous-time limit, players equilibrium payoffs are the same whether the prizes are allocated sequentially or simultaneously. All the above analysis can be extended to a continuum of budgets instead of two budget level. All the results hold qualitatively. 2 A baseline Model of One-Sided Private Budget We consider the following war of attrition game in the baseline model. Two bidders, 1 and 2, compete for an indivisible prize. Time proceeds in discrete periods t {1,2, }. Bidders alternate in their moves: bidder 1 bids on odd dates and bidder 2 bids on even dates. Bids are restricted to ε in each period. A bidder who is called upon to move can stay by bidding ε or concede. The game ends at the first time where either bidder 1 is called upon to bid ε and bidder 1 concedes by not bidding, in which case bidder 2 wins; or bidder 2 is called upon to bid ε and bidder 2 concedes, in which case bidder 1 wins. If the game never ends then both bidders payoffs are negative infinity. For i = 1,2, let V i denote bidder i s valuation of the prize. Let B i denote bidder i s budget. For simplicity, B i is a multiplication of ε. If bidder i wins the prize with a total bid of b i then her payoff is V i b i ; if she loses with a total bid of b i then her payoff is b i. We assume that bidders cannot bid more than their budgets. Assume that bidder 2 s budget B 2 is known to both bidders and bidder 1 s budget B 1 is her private information, which could take one of two values, B l or B h, with B l B 2 < B h. The prior probability that B 1 = B h is ρ 1 [0,1]. For t 2, let ρ t denote bidder 2 s updated belief in the beginning of period t. Let T = B l /ε. Finally, to avoid trivial cases we assume that V i > 2ε. The solution concepts are weak perfect Bayesian equilibrium (wpbe) and its refinement. 2.1 Preliminary results We first show that when it is common knowledge that bidder 1 has high budget, bidder 1 should always bid and bidder 2 should concede immediately. Lemma 1. In any wpbe, in any period t 2T + 1, the high budget bidder 1 should always bid and bidder 2 should concede immediately. 4

Proof. Note that the low budget bidder 1 can stay in the game for at most 2T periods, so from period 2T + 1 onward, it is a complete information game between the high budget bidder 1, whose remaining budget is B h T ε, and bidder 2, whose remaining budget is B 2 T ε. Since B h T ε > B 2 T ε, the result follows from backward induction (see Proposition 3 in Dekel et al. (2007)). Next we consider the behavior of the high budget bidder 1 in the first 2T periods. Without further refinements, there may exist an equilibrium where both high and low budget bidder 1 concedes immediately, which is supported by bidder 2 s off-equilibrium belief that assigns probability zero to the high type when bidder 1 bids. The next example illustrates such a possibility. Example 2. Assume V 1 = 2ε + δ with δ (0,ε), B l = 2ε, B 2 = 3ε and B h 4ε. Consider the strategy profile in which both types of bidder 1 concede in period 1, bidder 2 believes that bidder 1 has low budget with probability one after bidder 1 bids in period 1 and hence bidder 2 bids in periods 2 and 4. We now show that this strategy profile is a wpbe. For the low budget bidder 1, it is optimal to concede in period 1 since given bidder 2 s strategy, bidding yields a payoff no more than ε. For the high budget bidder 1, bidding in period 1 yields either ε or 2ε if she concedes in period 3 or 5 respectively; if she continues to bid in period 5 then bidder 2 will concede in period 6 since the low budget bidder 1 cannot bid in period 5, which leads to a payoff 3ε +V < 0. Therefore, it is also optimal for the high budget bidder 1 to concede in period 1. The equilibrium in the previous example relies on optimistic off-equilibrium beliefs of bidder 2 after bidder 1 bids. However such beliefs may not be reasonable given that it is more likely for the high budget bidder 1 to gain from bidding than for the low budget bidder 1. The next result shows that none of the equilibria in which both types of bidder 1 concede survives a generalization of the D1 refinement proposed by Cho and Kreps (1987). 2 (GD1) refinement, the high budget bidder 1 always bid in any period. Moreover, in any equilibrium that satisfies the generalized D1 Recall that in the canonical sender-receiver games where a sender first sends a signal, and a receiver chooses an action after learning the signal, D1 requires that the receiver puts zero weight a type t sender when a off-equilibrium signal m is sent if there is another type t sender such that t always strictly benefits from deviating to m whenever t benefits from this deviation. Since in this game bidder 1 can bid 2 However, these pooling equilibria survive the intuitive criterion proposed by Cho and Kreps (1987), since both types of bidder 1 can obtain positive payoffs if bidder 2 concedes in the next period. 5

or concede in every other period provided that she has not conceded yet, we consider a generalization of D1, which can be viewed as iterative applications of D1 refinement of wpbe. Let T = B 2 /ε. Note that the last relevant period of the game is 2T, in which bidder 2 chooses to bid her remaining budget ε or to concede. Notice that at period 2T bidder 2 assigns probability one to the event that bidder 1 has high budget and hence the continuation game has a unique wpbe that also satisfies D1 in which bidder 2 concedes. Denote this equilibrium by DE 2T (ρ 2T ). By backward induction, for any t 2T + 2, we have ρ t = 1 and the continuation game from period t has a unique wpbe that satisfies D1, which is denoted by DE t (ρ t ). Now for any t 2T + 1 and any ρ t [0,1], let DE t (ρ t ) be the set of wpbe that satisfies D1 in the continuation game from period t when the play from period t +1 is given by any element in DE t+1 (ρ t+1 ) for any feasible ρ t+1. 3 Definition 3. We say that a wpbe σ satisfies GD1 if σ DE 1 (ρ 1 ). Proposition 4. In any wpbe that satisfies GD1, for any t = 1,2,...,T, the high budget bidder 1 always bids in any period 2t 1. Proof. We prove this statement backwards. In period 2T 1, suppose the high budget bidder 1 bids, then given the previous lemma, the least payoff that she can obtain (which occurs when bidder 2 bids in period 2T ) is ε +V 1 ε = V 1 2ε > 0. Therefore, the high budget bidder 1 should bid in period 2T 1 regardless of bidder 2 s belief ρ 2T 1. This implies that in any equilibrium the low budget bidder will not concede for sure in period 2T 1. Consequently, in period 2T the probability that bidder 2 chooses to concede, β 2T, is no less than V ε 1. Note that the continuation payoff of the high budget bidder 1 from bidding in period 2T 1 is strictly positive and is weakly larger than that of the low budget bidder 1. In period 2T 3, suppose the high budget bidder 1 concedes with probability π 2T 3 > 0. If π 2T 3 < 1, then the high budget bidder 1 is indifferent between conceding and bidding. That is, 0 = ε + β 2T 2 V 1 + (1 β 2T 2 )( ε + β 2T V 1 + (1 β 2T )(V 1 ε)), where the right-hand side of the equality is her payoff from bidding in period 2T 3. Note that since 3 Note that for any t 2T + 2, ρ t 1, and for any t 2T + 1, ρ t maybe any real number in [0,1]. 6

V 1 > 2ε, we have β 2T 2 < 1. Then the low budget bidder 1 s payoff from bidding, ε + β 2T 2 V 1 + (1 β 2T 2 )( ε + β 2T V 1 + (1 β 2T ) 0), is strictly smaller than zero, which implies that in equilibrium she should concede in period 2T 3. However, if the low budget bidder 1 deviates to bidding, then the posterior belief of bidder 2 is ρ 2T 2 = 1; consequently, bidder 2 concedes in period 2T 2. Therefore, this is a profitable deviation for the low budget bidder 1. Hence, π 2T 3 (0,1) cannot happen in any equilibrium. If π 2T 3 = 1, that is, the high budget bidder 1 concedes in period 2T 3, then the low budget bidder 1 must also concede in period 2T 3. As a result, bidding in period 2T 3 is off the equilibrium path. Note that for any mixed action of bidder 2 in period 2T 2 that can result a strictly positive payoff for the low budget bidder 1, the high budget bidder 1 can also obtain a strictly positive payoff, but not vice versa. Therefore, according to GD1, the posterior belief of bidder 2 in period 2T 2 must satisfy ρ 2T 2 = 1. Consequently, bidder 2 concedes for sure in period 2T 2. But then it is profitable for both types of bidder 1 to deviate to bidding in period 2T 3. Hence, π 2T 3 = 1 cannot happen in any equilibrium that satisfies GD1. It follows that π 2T 3 = 0, that is, the high budget bidder 1 bids for sure in period 2T 3 in any equilibrium that satisfies GD1. Therefore, as in period 2T 1, the low budget bidder 1 will not concede for sure in period 2T 3 and hence bidder 2 concedes with probability at least V ε 1 in period 2T 2. Moreover, the continuation payoff of the high budget bidder 1 from bidding in period 2T 3 is strictly positive and is weakly larger than that of the low budget bidder 1. Now we proceed by induction backwards: in period 2T τ for any τ = 5,7,...,2T 1, given that the continuation play must be an equilibrium that satisfies GD1, by the same argument as in the previous paragraph, if the high budget bidder 1 does not bid for sure then it must be that both types of bidder 1 concedes for sure; consequently, the continuation payoff of the high budget bidder is strictly positive and is weakly above that of the low budget bidder 1, and that bidder 2 s off-equilibrium path belief after bidder 1 bids in period 2T τ must satisfy GD1, bidder 2 s off-equilibrium belief must assign probability one on bidder 1 having a high budget, in which case both types of bidder 1 would deviate to bidding in period 2T τ. Therefore, the high budget bidder 1 bids for sure in period 2T τ in any equilibrium that satisfies GD1. This completes the proof. 7

From now on, unless otherwise stated, we use equilibrium to denote wpbe that satisfies GD1. Because bidder 1 with a high budget (B h ) always bids in any equilibrium, in contrast to the war of attrition with complete information, bidder 1 with a low budget (B l ) may have an incentive to mimic the behavior of B h. In other words, the high budget bidder play the same role as a behavioral type bidder who never concedes, as in the reputation literature. 4 2.2 Equilibrium In this section, we provide a complete characterization of the equilibrium. We again proceed backwards. Notice that for each t, ρ 2t = ρ 2t+1, as only bidder 2 moves in period 2t. Since the high budget bidder 1 always bids, we focus on the strategies of the low budget bidder 1 and bidder 2. Let β t denote the probability that a bidder concedes in period t. Recall that T = B l /ε. In period 2T, bidder 2 s payoff from bidding is ε +ρ 2T 0+(1 ρ 2T ) V 2. Thus, bidder 2 concedes if ρ 2T > 1 ε/v 2 and bids if ρ 2T < 1 ε/v 2. She is indifferent if ρ 2T = 1 ε/v 2 ρ2t. In period 2T 1, suppose that bidder 1 concedes for sure. If she deviates to bidding ε (which is also her remaining budget), then in period 2T bidder 2 will concede with probability one since her updated belief will be ρ 2T = 1. Since V 1 > 2ε, this is a profitable deviation for bidder 1. Therefore, in any equilibrium β 2T 1 < 1. In addition, bidder 1 s equilibrium strategy depends on ρ 2T 1 : (1) if ρ 2T 1 1 ε/v 2, then bidder 1 bids for sure in period 2T 1, since otherwise the updated belief in period 2T, given bidder 1 bids in period 2T 1, is ρ 2T > ρ 2T 1 1 ε/v 2. From the previous paragraph bidder 2 concedes in period 2T, so it is optimal for bidder 1 to bid in period 2T 1, a contradiction. In all, β 2T 1 = 0 and hence ρ 2T = ρ 2T 1. In this case, bidder 1 s payoff is ε +V 1. (2) If ρ 2T 1 < 1 ε/v 2, then β 2T 1 > 0, otherwise bidder 2 will bid in period 2T in which case bidder 1 loses. Since β 2T 1 (0,1), bidder 1 s equilibrium payoff must be zero, which implies that bidder 2 must also randomize in period 2T. Thus, we have the following two equations: ρ 2T 1 ρ 2T 1 + (1 ρ 2T 1 )(1 β 2T 1 ) = ρ 2T = 1 ε V 2 and 0 = ε + β 2T V 1 + (1 β 2T ) 0. 4 See for example Abreu and Gul (2000). 8

The first equation follows from Bayesian updating and the second is bidder 1 s indifference condition. By induction, in period 2t for t = 1,...,T 1, when ρ 2t (= ρ 2t+1 ) ρ2t+2, bidder 2 loses for sure even if she bids by the argument in the previous paragraph, so bidder 2 concedes: β 2t = 1. When ρ 2t < ρ2t+2, bidder 1 will randomize in the next period if bidder 2 bids now. Thus, bidder 2 s payoff from bidding is ε + (1 ρ 2t )β 2t+1 V 2 = ε + (ρ 2t+2 ρ 2t)V 2 /ρ 2t+2. Therefore, bidder 2 concedes if ρ 2t > (1 ε/v 2 ) (T t+1) and bids if ρ 2t < (1 ε/v 2 ) (T t+1). She is indifferent if ρ 2t = (1 ε/v 2 ) (T t+1) ρ 2t. Finally, in period 2t 1 for t = 1,...,T 1, define ρ 2t ( 1 ε V 2 ) T t+1. Following the above argument, if ρ 2t 1 ρ 2t then bidder 1 bids for sure (β 2t 1 = 0) and obtains a payoff ε +V 1. If ρ 2t 1 < ρ 2t then bidder 1 concedes with probability β 2t 1 (0,1), which is given by ρ 2t 1 ρ 2t 1 + (1 ρ 2t 1 )(1 β 2t 1 ) = ρ 2t; moreover, for bidder 1 to be indifferent in period t 1, bidder 2 randomizes in period 2t: conceding with with probability β 2t given by 0 = ε + β 2t V 1 + (1 β 2t ) 0. Note that since ρ2t < ρ 2t+2, bidder 1 s continuation payoff from period 2t + 1 onward is zero by induction. The next proposition summarizes the above analysis. Define ρ 2t ( 1 ε V 2 ) T t+1. Theorem 5. There exists a unique wpbe that satisfies GD1 as follows: 1. Bidder 1 with budget B h always bids. 2. For 1 t T, in period 2t 1, where t T, bidder 1 with budget B l bids with probability one if ρ 2t 1 ρ 2t and concedes with probability β 2t 1 if ρ 2t 1 < ρ 2t, where β 2t 1 satisfies ρ 2t 1 ρ 2t 1 + (1 ρ 2t 1 )(1 β 2t 1 ) = ρ 2t, 9

In period 2t, bidder 2 concedes if ρ 2t > ρ2t, bids if ρ 2t < ρ2t, and concedes with probability β 2t = ε/v 2 if ρ 2t = ρ 2t. 3. On the equilibrium path, if ρ 1 > ρ 2, then bidder 1 with budget B l bids for sure and bidder 2 concedes; if ρ 1 < ρ2, then as long as the game has not ended, bidder 1 concedes with probability β 2t 1, and bidder 2 concedes with probability β 2t = ε/v 2, and the updated beliefs satisfy ρ 2t = ρ 2t+1 = ρ2t for each t = 1,...,T. We study the continuous time limit. Now suppose that the period length is and within each period bidders alternate in their bids. Each bidder incurs a per-period cost ε = from bidding. When we take the period length to zero, the game converges to a continuous-time war of attrition. To be specific, assume that each time period is composed by 2t and 2t + 1. Define τ at the real time, which is equal to the budgets that both bidders have depleted. We can show that ) ρ (τ) = lim ρ2t = lim (1 B εv2 l τ ε 0 0 = e B l τ V 2. On the equilibrium path, Bayes updating β where λ 1 (τ) lim 2t 1 0 ε of ρ (τ), we can solve dρ (τ) dτ ρ2t 2 ρ2t 2 +(1 ρ 2t 2 )(1 β 2t 1) = ρ 2t implies that = ρ (τ)(1 ρ (τ))λ 1 (τ), is low budget bidder 1 s Poisson rate of conceding at time τ. By the formula λ 1 (τ) = 1 V 2 (1 ρ (τ)). In other words, λ 1 (τ)(1 ρ (τ))v 2 = 1. LHS is bidder 2 s expected benefit from waiting at time τ, which is bidder 1 s expected probability of conceding at τ times the value from winning V 2 ; RHS is the cost of waiting. Moreover, bidder 2 s Poisson rate of conceding at τ satisfies λ 2 (τ)v 1 = 1. Therefore, λ 2 (τ) = 1 V 1. If the prior belief ρ > ρ (0) = e B l V 2 then bidder 2 concedes immediately. If ρ < ρ (0), bidder 1 with 10

budget B l concedes with a discrete probability α (0,1) initially to satisfy ρ ρ + (1 ρ)(1 α) = ρ (0). Therefore, α = ρ (0) ρ (1 ρ)ρ (0). We summary the result of continuous time limit as follows: Corollary 6. In the continuous-time limit, There exists a unique wpbe that satisfies GD1 as follows: 1. Bidder 1 with budget B h always bids. 2. If ρ ρ (0), then bidder 1 with budget B l bids for sure and bidder 2 concedes. 3. If ρ < ρ (0), then at the first instance, bidder 1 with budget B l concedes with a discrete probability ρ (0) ρ (1 ρ)ρ (0). As long as the game has not ended, bidder 1 concedes with a Poisson rate λ 1(τ), and bidder 2 concedes with a Poisson rate 1 V 1, and the updated beliefs satisfy ρ(τ) = ρ (τ) for τ B l. 3 Two-sided Private Budgets In this section we extend the analysis to case where each bidder s budget is private information. Suppose bidder i s budget B i {B l,b h }. The prior probability that B 1 = B h is ρ 1 [0,1] and the prior probability that B 2 = B h is η 1 [0,1]. For t 2, let ρ t and η t denote bidder 1 and 2 s updated belief in the beginning of period t respectively. Let T = B l /ε. We will refer to bidder i with the the high budget as ih and bidder i with the the low budget as il. To avoid trivial cases we again assume that V i > 2ε. 3.1 Preliminary results First, we establish that eventually, only 2h bids. Lemma 7. In any wpbe, in any period t 2T + 2, bidder 2h should always bid and bidder 1h should concede immediately. Furthermore, at t = 2T bidder 2h bids for sure. Proof. First, note that at t 2T + 2, only high budget bidders remain in the game. If at the start of t = 2T + 2, the budget remaining for bidder 2h is less or equal to her valuation V 2, then she will keep 11

bidding as eventually, at time period t = 2 B h ε she would defeat bidder 1h. Suppose to the contrary, B h T ε > V 2. Note, that using the same argument as above bidder 2h will eventually keep bidding at and after some 2 ˆT. Given this behavior of 2h, it is clear that bidder 1h would prefer to concede in t = 2 ˆT 1. Using backward induction, bidder 2h would want to bid at 2 ˆT 2 and so on, till we reach t = 2T + 2. To prove the final part of the lemma let us calculate the continuation payoff of bidder 2h in time period 2T, U2T 2h. Let β 1h 2T +1 be the probability with which bidder 1h concedes in 2T + 1. Then, ( ( ) ) U2T 2h = ε + (1 ρ 2T )V 2 + ρ 2T β2t 1h +1 V 2 + 1 β2t 1h +1 (V 2 ε) > 0 where we use the fact that bidder 1h will concede for sure after 2T + 2. The inequality follows from V 2 > 2ε. As was the case in the model with one-sided private information, without any further refinements, there may exist equilibria where both and 2h concede at the first opportunity. This equilibrium is supported by bidder 1 s off-equilibrium beliefs that assigns zero probability to the bidder 2 being 2h conditional on observing a bid from bidder 2 and parametric condition on the probability of bidder 1 s high type. We illustrate such an equilibrium using a similar example as 2. Example 8. Assume V 1 = V 2 = 2ε +δ with δ (αε,ε) where α +α 2 = 1, B l = 2ε,B h 4ε and ρ 1 > δ ε. Consider the strategy profile in which both types of bidder 2 concede in period 1, both types of bidder 1 believe that bidder 2 has low budget with probability one after bidder 2 bids in period 2 and both types of bidder 1 always bid till either they exhaust their budget or the game reaches round 7. We show that this strategy profile constitute a wpbe. First, consider bidder. This bidder can defeat 1l if the game moves to period 4. However, she will lose to bidder 2h. Therefore, this bidder s payoff is either ε if she bids in round 2 and concedes in 4, or it is 2ε + (1 ρ 1 )V 2 if she bids till she exhausts her budget. Note, that given bidder 1 s strategy there is no updating of bidder 1 s type. Finally, note that the latter payoff is also negative if ρ 1 > δ ε. Next, consider bidder 2h. This bidder can defeat 1l if she bids in round 4, and, she can defeat 1h if she bids in round 6, since if she bids in round 6, bidder 1h would know that bidder 2 is of the high type and hence would concede in round 7. If bidder 2h bids in round 2 and 4 but concedes in 6, then her payoff will be the same as who follows the same strategy, and the one we calculated in the previous 12

paragraph. Therefore, we only need to consider the case where 2h bids in rounds 2,4 and 6. In this case, her payoff will be 2ε + (1 ρ 1 )V 2 + ρ 1 (V 2 ε) < 0 where the inequality follows from ρ 1 > δ ε, this payoff is also negative. Hence, 2h also prefers to concede in round 2. Finally, we check whether the strategy followed by bidder 1 is sequentially rational. If the game reaches round 7, then 1h knows that she is competing against 2h and hence concedes. In round 5, given that 1h believes she is playing against, she will bid for sure. In round 3, bidder 1h will always bid as no matter what does in round 4, bidder 1h gets a positive payoff. What remains to check is whether bidder 1l will bid for sure in round 3. Given her belief that she is facing with probability one, bidder 1l will bid for sure if and only if she believes that will concede next round. Otherwise, will beat her and she gets a payoff of ε. Bidder will concede for sure in round 4 if and only if her payoff from bidding ε + (1 ρ 1 )V 2 < 0. That is ρ 1 > 1 ε V 2 = ε+δ 2ε+δ. We already know ρ 1 > δ ε. Note δ ε > ε+δ 2ε+δ if and only if εδ + δ 2 > ε 2 which is true if δ (αε,ε). Hence, given 1l s beliefs and the parametric condition on ρ 1, 1l will bid for sure in round 3. And finally, we already established that given bidder 1 s beliefs, both and 2h will concede in round 2, hence, both 1l and 1h will bid for sure in round 1. Again, the construction of the above example relied on the optimistic off-equilibrium beliefs of bidder 1. As we argued previously, such beliefs may not be reasonable and can be ruled out by appealing to the GD1 refinement. Using a similar argument as in the one-sided case, we have the following proposition. Proposition 9. In any wpbe that satisfies GD1, for any t = 1,2,...,T 1, the high budget bidder 2 (2h) always bids in any period 2t. A final point we want to make before proceeding with the analysis of the equilibrium is that there may be parameter values under which bidder 1h may bid in each period. Indeed we will show that such cases can occur. However, if bidder 2 has optimistic beliefs about the bidder 1 s type, then in these cases, other degenerate equilibria may also exist. However, an iterative application of D1, as done previously, ensures that such beliefs do not exist. As was the case before we use equilibrium to denote wpbe that satisfies GD1. Under two-sided incomplete information, bidder 2h always bids in equilibrium. Therefore, may have an incentive to mimic this behavior. Like the one-sided case, 2h behaves like a commitment type. Furthermore, we show that under some parametric conditions, 1h also always bids in equilibrium. This leads 1l to try and mimic this behavior. 13

3.2 Equillibrium In order to solve for the equilibrium we proceed backwards. Note that ρ 2t = ρ 2t+1, since only bidder 2 behaves in 2t, and, η 2t 1 = η 2t, since only bidder 1 behaves in 2t 1. Let βt i j be the probability bidder i type j concedes in period t. Also, since the payoff from conceding in any period is always zero, let Ut i j represent the payoff to bidder i type j in period t from bidding. 3.2.1 Period 2T + 1 Since bidder 2h always bids, the final round we need to consider is 2T + 1. In this round, only 1h can bid as 1l has exhausted her budget. In this round 1h s payoff from bidding is ε + (1 η 2T +1 )V 1. Therefore, if η 2T +1 > 1 V ε 1 then 1h concedes for sure and if η 2T +1 < 1 V ε 1 she bids fore sure. She is indifferent if η 2T +1 = 1 V ε 1 = η2t +1 which satisfies the initial inequality. 3.2.2 Period 2T In period 2T, 2h bids for sure. Hence, using the same argument as in the one-sided case, we can show that bidder will not concede for sure, no matter what her belief about bidder 1 s type. That is β2t < 1. Note that bidder s payoff form bidding is given by U 2T = ε + ( ρ 2T β2t 1h +1 + (1 ρ 2T ) ) V 2. Therefore, s payoff depends on both ρ 2T and η 2T 1. The latter due to the fact that the payoff depends on β 1h 2T +1 which depends on η 2T +1 which in turn depends on η 2T 1. Consider the following four cases. (1) η 2T 1 η 2T +1. First, suppose η 2T 1 > η 2T +1. Then, since η 2T +1 η 2T 1 bidder 1h will concede for sure in the next round. That is β2t 1h +1 = 1. In this case it is optimal for 1l to bid for sure. That is β2t = 0 and hence η 2T +1 = η 2T 1. Suppose η 2T 1 = η2t +1. If β 2T (0,1), then η 2T +1 > η2t +1 previous paragraph. leading to the conclusion of the (2) η 2T 1 < 1 V ε 1 and ρ 2T < 1 V ε 2 = ρ2t. For any β 1h 2T +1 [0,1],U 2T > 0. Therefore, β 2T = 0. (3) η 2T 1 < 1 V ε 1 and ρ 2T = 1 V ε 2. In this case, note that for any β2t 1h +1 [0,1),U 2T > 0 which would imply β2t = 0. However, this would imply η 2T +1 = η 2T 1 < 1 V ε 1, which would imply that bidder 1h bids for sure and β2t 1h +1 = 0. If β 1h 2T +1 = 0, then U 2T bidding and conceding. 14 = 0 and bidder is indifferent between

(4) η 2T 1 < 1 ε V 1 and ρ 2T > 1 ε V 2. In this case β 2T > 0. Otherwise η 2T +1 = η 2T 1 and bidder 1h will bid for sure in 2T +1 which would imply U2T < 0. Hence β 2T (0,1) such that 1h does not bid for sure in 2T + 1. This would occur if η 2T +1 = η 2T 1 η 2T 1 + (1 η 2T 1 ) ( 1 β2t ) = η2t +1 (1) In addition, if is randomizing in the current round it must be the case that U2T = 0. That is ( ) ρ 2T 1 β2t 1h +1 = 1 ε V 2 3.2.3 Period 2T 1 In period 2T 1 both 1l and 1h can bid. Note that this is the last period 1l can bid as after this she will not longer have any budget. Hence, we have the following payoffs from bidding for the two bidder types. U 1h 2T 1 = ε + ( η 2T 1 + (1 η 2T 1 )(1 β 2T ) ) U 1h 2T +1 + (1 η 2T 1)β 2T V 1 U 1l 2T 1 = ε + (1 η 2T 1)β 2T V 1 Clearly, the payoffs depend on η 2T 1 as well as on ρ 2T 2, the latter again due to the dependence of β 2T on ρ 2T. Consider the following four cases. ( ) 2 (1) η 2T 1 > 1 V ε 1 = η 2T 1. First, note that if η 2T 1 > η2t then we are in case (1) of 2T and bidder will bid for sure. Therefore, clearly, U2T 1l 1 < 0 and therefore, β 1l 2T 1 = 1. Furthermore, if β 2T = 0 then η 2T +1 = η 2T 1 > η 2T implying β 1h 2T +1 = 1 since U1h 2T +1 < 0. Therefore, U1h 2T 1 < 0. Hence β2t 1h 1 = 1. Therefore, we only need to consider the case η 2T η 2T 1 > η2t 1. We show that no matter what beliefs bidder has about bidder 1 s type in round 2T, bidders 1l and 1h always concede in 2T 1. There are three cases to consider. (a) Suppose ρ 2T < 1 V ε 2. Then we are in case (2) of round 2T and hence β2t = 0. This makes U2T 1l 1 < 0 and hence β 1l 2T 1 = 0. Therefore, ρ1 2T = 1 a contradiction to this case. (b) Next, suppose ρ 2T > 1 ε V 2. Then we are in case (4) of round 2T and will randomize such that equation (1) is satisfied. This has two implications. One, U2T 1h +1 = 0, and, two, given 15

(1 η 2T 1 )β2t < V ε 1 since η 2T 1 > η2t 1. Note, this implies U1h 2T 1 = U1l 2T 1 < 0. Therefore, β2t 2h 1 = β 2T 1 = 1. (c) Finally, suppose ρ 2T = 1 ε V 2. Then we are in case (3) of round 2T. Recall, in this case bidder is indifferent. If randomizes such that equation (1) is satisfied then we would reach the same conclusion as part (b) above. Suppose instead, randomizes in 2T such that (1 η 2T 1 )β2t = ε V 1, then, given η 2T 1 > η2t 1 the updated η 2T +1 > η2t +1 which implies β 1h 2T +1 = 1. In this case, 1l would bid for sure in 2T by backward induction. If β 2T β 1l 2T 1 = 1. Finally, this implies, ρ 2T = 1 a contradiction to this case. Therefore, in this case β 1l 2T 1 = β 1h 2T 1 = 1. = 0, then U1l 2T 1 < 0, therefore, (2) η 2T 1 = η2t 1. In this case the first thing to note is that it must be the case that bidders 1l and 1h play strategies such that ρ 2T ρ2t. Suppose to the contrary ρ 2T < ρ2t. Then we will be in case (2) of 2T in which case bidder bids for sure. This makes U2T 1 < 0 and hence β 2T 1 = 1. Also, if η 2T 1 = η2t 1 and bids for sure in round 2T, then U1h 2T +1 > 0. Therefore, ρ 2T = 1 contradicting the supposed inequality. Therefore, ρ2t 1 ρ 2T. Then, we will be in case (3) or (4) of round 2T. In either of these cases randomizes. Note that if β2t satisfies equation (1), then (1 η 2T 1)β2T = V ε 1. This equality further implies U2T 1l 1 = 0 and U1h 2T 1 = 0 the latter following from the above equality and η 2T +1 = η2t +1. Therefore, in this case both 1h and 1l are indifferent between bidding and conceding and may randomize on the equilibrium path. Before stating the next two cases which pertain to η 2T 1 < η2t 1, we make the following observation. Lemma 10. If η 2T 1 < η 2T 1 concede for sure in 2T 1. then on the equilibrium path it is never the case that both 1l and 1h Proof. Suppose to the contrary both 1h and 1l concede and bidding in 2T 1 is off the equilibrium path. Note that for any β2t [0,1) such that U1l 2T 1 > 0 implies U1h 2T 1 > 0 but not vice versa. Hence, by GD1, s posterior belief in 2T is ρ 2T = 1 and this leads to case (4) in 2T. However, if β2t satisfies equation (1) as is the case in (4) of 2T, then (1 η 2T 1 )β2t 1 > V ε 1 which implies U2T 1l 1 > 0 and U1h 2T 1 > 0. Therefore bidding in 2T 1 is a profitable deviation for 1l and 1h. 16

With this result, we can reduce the number of sub-cases we have to study in the remaining two cases. (3) η 2T 1 < η 2T 1 and ρ 2T 2 ρ 2T. First, suppose ρ 2T 2 > ρ 2T. Then, since ρ 2T ρ 2T 2, we will be in case (4) of round 2T. Since η 2T 1 < η2t 1, this implies (1 η 2T 1)β2T > V ε 1 which implies U2T 1l 1 = U1h 2T 1 > 0 and therefore, β2t 1l 1 = β 2T 1h 1 = 0. Next, suppose ρ 2T 2 = ρ 2T. Again, since ρ 2T ρ 2T 2, we will be in case (3) or (4) of 2T. If we move to case (4) then the reasoning in the previous paragraph applies and β 1l 2T 1 = β 1h 2T 1 = 0. The only way in which case (3) is on the equilibrium path, given ρ 2T 2 = ρ 2T To see this note that ( ) ρ 2T 2 1 β 1h 2T 1 ρ 2T = ( ) ρ 2T 2 1 β 1h 2T 1 + (1 ρ2t 2 ) ( 1 β 1l 2T 1 is if β 1l 2T 1 = β 1h 2T 1. ) (2) Recall, in case (3) is indifferent between bidding and conceding and on the equilibrium path β2t 1h +1 = 0. The latter equality implies that β 2T η 2T +1 = must satisfy η 2T 1 η 2T 1 + (1 η 2T 1 ) ( 1 β2t ) η2t +1 If η 2T +1 = η2t +1, then given η 2T 1 < η2t 1, the argument in the first paragraph of this case applies and therefore β 1l 2T 1 = β 1h 2T 1 = 0. If on the other hand η 2T +1 < η2t +1, but (1 η 2T 1)β2T > V ε 1, then again β2t 1l 1 = β 2T 1h 1 = 0. If (1 η 2T 1 )β2t = V ε 1 then U2T 1l 1 = 0 but U1h 2T 1 > 0 which implies β 1h 2T 1 = 0 and therefore on the equilibrium path β2t 1l 1 = 0. Finally, suppose β2t was such that (1 η 2T 1)β2T < V ε 1 and U2T 1l < U1h 2T < 0 In this case both 1h and 1l concede and bidding in 2T 1 is off the equilibrium path. However this violates lemma 10 and therefore can not be possible. Therefore, in this case 1l and 1h bid for sure, that is β 1l 2T 1 = β 1h 2T 1 = 0. (4) η 2T 1 < η2t 1 and ρ 2T 2 < ρ2t. As we proved in other cases, ρ 2T < ρ2t can not be on the equilibrium path in this case as this would imply we are in case (2) of 2T and therefore β2t = 0 leading to U2T 1l 1 < 0. Therefore, ρ 2T ρ2t 17

Suppose ρ 2T > ρ 2T then we move to case (4) of 2T and will randomize such that η 2T +1 = η 2T +1. However, since η 2T 1 < η2t 1, as before, this implies (1 η 2T 1)β2T > V ε 1 and U2T 1l > 0 and U1h 2T > 0. Therefore, β2t 1h 1 = β 2T 1h 1 = 0 which implies ρ 2T = ρ 2T 2 < ρ2t, a contradiction due to the previous paragraph. Therefore, it must be the case that ρ 2T = ρ2t. That is, we are in case (3) in 2T and is indifferent. Now, must randomize such that U 2T 1 = 0. (1 η 2T 1 )β 2T = ε V 1 (3) If not, then either 1l will concede for sure and ρ 2T = 1 which would be a contradiction due to the previous paragraph, or 1l will bid for sure and therefore 1h will bid for sure which would imply ρ 2T = ρ 2T 1 < ρ2t, a contradiction due the first paragraph. Note that if β 2T satisfies (3), then η 2T +1 < η2t +1 and therefore U1h 2T +1 > 0. This further implies U1h 2T 1 > 0. Therefore, β 1h 2T 1 = 0. Finally, since β2t 1h 1 = 0 and ρ 2T 2 < ρ2t and ρ 2T = ρ2t, 1l will randomize in 2T 1 such that β2t 1l 1 satisfies ρ 2T = ρ 2T 2 ρ 2T 2 + (1 ρ 2T 2 ) ( 1 β2t 1l ) = 1 ε = ρ2t (4) V 1 2 Therefore in this case β2t 1h 1 = 0 and β 1l 2T 1 satisfies (4) 3.2.4 Round 2T 2 Again, we know that 2h bids for sure in this round. Threfore, 2h will not concede for sure. This round for is slightly different than 2T as both 1l and 1h can bid in the next round. This feature shows up in the following payoff from bidding in this round ( ) ( ( )) U2T 2 = ε + ρ 2T 2 β2t 1h 1 + (1 ρ 2T 2)β2T 1l 1 V 2 + 1 ρ 2T 2 β2t 1h 1 + (1 ρ 2T 2)β2T 1l 1 U2T 1l However, the cases we need to study are similar to the cases in 2T. (1) η 2T 3 η2t 1. As before, if the inequality is strict for any β 2T 2 [0,1), η 2T 1 > η2t 1 and we will be in case (1) of 2T 1 where both 1l and 1h concede for sure. Therefore β2t 2 = 0. If 18

η 2T 3 = η2t 1 then any β 2T 2 (0,1) leads to the same conclusion, therefore, β 2T 2 = 0. ( ) 2 (2) η 2T 3 < η2t 1 and ρ 2T 2 < 1 V ε 2 = ρ 2T 2. Note, that if on the equilibrium path β2t 2 (0,1), then η 2T 1 η2t 1. If η 2T 1 > η2t 1 then we will move to case (1) of 2T 1 where both 1l and 1h concede for sure, propelling to bid for sure in 2T 2. If β2t 2 (0,1) on the equilibrium path then it must be the case that U 2T 2 = 0. Compare this to β2t 2 = 0. In this case, η 2T 1 = η 2T 3 < η2t 1. Since ρ 2T 2 < ρ2t 2, we will be in case (4) of 2T 1. In this case 1h bids for sure and β2t 1l 1 satisfies (4) which would imply U 2T = 0. However, (4) and ρ 2T 2 < ρ2t 2 imply (1 ρ 2T 2)β2T 1l 1 > V ε 2. This implies U2T 2 > 0. Therefore in this case β 2T 2 = 0. (3) η 2T 3 < η2t 1 and ρ 2T 2 = ρ2t 2. First, note that in this case β 2T 2 (0,1). If β 2T 2 = 0, then η 2T 1 = η 2T 3 < η2t 1 and we would move to case (3) in round 2T 1 in which case both 1l and 1h bid for sure, that is, β2t 1h 1 = β 2T 1l 1 = 0. Furthermore, after 2T 1 the game will move to case (3) or (4) in round 2T, where U2T = 0 and therefore U 2T 2 < 0. Therefore, β2t 2 (0,1) on the equilibrium path. In addition, it must be the case that β 2T 2 satisfies η 2T 1 η2t 1. To see this, note that if η 2T 1 > η2t 1 then 1l and 1h want to concede for sure in the next round (case (1) of 2T 1) and hence β2t 2 = 0 which can not occur in equilibrium. Since η 2T 1 η2t 1 and ρ 2T 2 = ρ2t 2 < ρ 2T we will move to either case (2) or case (4) of 2T 1. In both cases 1l and 1h choose strategies such that U2T = 0. In addition, if is randomizing in the current round it must be the case that U2T 2 = 0. Therefore, β 1l 2T 1 and β 1h 2T 1 must satisfy ρ 2T 2 β 1h 2T 1 + (1 ρ 2T 2)β 1l 2T 1 = ε V 2 (4) η 2T 3 < η2t 1 and ρ 2T 2 > ρ2t 2. First, note that β 2T 1 (0,1). If β 2T 1 = 0, then η 2T 1 = η 2T 3 < η2t 1 and we will move to case (2) in 2T 1 where both 1h and 1l bid for sure, making U2T 1 = ε. Therefore, in this case β 2T 2 (0,1) such that η 2T 1 = ( η 2T 3 η 2T 3 + (1 η 2T 1 )β2t = 1 ε ) 2 = η2t 1 V 2 1 19

To see this, if η 2T 1 > η2t 2, then we move to case (2) in 2T 1 and both 1h and 1l will concede. Therefore, β2t 2 = 0 a contradiction. If η 2T 1 < η2t 2 then we move to case (4) in 2T 1. In this case U2T 2 = ε + (1 ρ 2T 2)β2T 1l 1 V 2. However, since β2t 1l 1 satisfies (4) and ρ 2T 2 > ρ2t 2, U2T 2 < 0 and concedes for sure which can not occur in equilibrium. 3.2.5 Round 2t 1 Using identical arguments as above we can roll the equilibrium back. To describe equilibrium strategies in any round let us define two threshold posterior beliefs that depend on the round. For t = 2,...,T ) T t+2 η2t 1 = (1 εv1 and η 1 = η 3.5 For t = 1,...,T ) T t+1 ρ2t = (1 εv2 By induction, in period 2t 1 for t = 1,...,T the following describe optimal strategies for bidder 1. If η 2t 1 > η 2t 1 then both types of bidders 1 concede, that is β 1l 2t 1 = β 1h 2t 1 = 1. If η 2t 1 = η2t 1 then both types of bidders 1 may randomize such that ρ 2t ρ2t. Furthermore, for t > 1, to keep bidder indifferent in round 2t 2, β2t 1 1l and β 1l 2t 1 will satisfy ρ 2t 2 β 1h 2t 1 + (1 ρ 2t 2)β 1l 2t 1 = ε V 2 If η 2t 1 < η 2t 1 and ρ 2t 2 ρ 2t then both types of bidders 1 bid, that is β 1l 2t 1 = β 1h 2t 1 = 0 If η 2t 1 < η2t 1 and ρ 2t 2 < ρ2t then 1h bids for sure, that is β 1h 2t 1 = 0 and β 1l 2t 1 (0,1) such that ρ 2t = ρ2t ρ 2t 2 = ( ) ρ 2t 2 + ρ 2t 2 1 β 2t 1 5 The reason η1 = η 3 is that in round 1, both type of bidder 1 will concede for sure if and only if they believe that will bid for sure in the next round and every round after. This only happens if η 1 > η3. We illustrate this point in the next sub-section. 20

3.2.6 Round 2t Again, we can roll the equilibrium back to find bidder s equilibrium strategies in any round. Using the same definitions for the threshold posterior beliefs, we can describe optimal strategies in round 2t as follows. If η 2t 1 η 2t+1 or ρ 2t < ρ 2t then bidder bids for sure that is β 2t = 0. If η 2t 1 < η 2t+1 and ρ 2t > ρ 2t then bidder randomizes such that β 2t (0, 1) satisfies η 2t+1 = η 2t 1 η 2t 1 + (1 η 2t 1 ) ( 1 β 2t ) = η 2t+1 If η 2t 1 < η 2t+1 and ρ 2t = ρ 2t then bidder randomizes such that β 2t (0, 1) satisfies (1 η 2t 1 )β 2t = ε V 1 The next proposition summarizes the above analysis. Proposition 11. In equilibrium, bidder 2h always bids. As a consequence, for each t = 1,...,T (= B l /ε), in period 2t bidder never concedes with probability one. Furthermore, in period 2t 1, bidder 1 with any budget level concedes with probability one if η 2t 1 > η 2t 1 and in period 2t bidder bids with probability one if η 2t 1 > η2t+1. In all other cases, some or all the bidder types randomize, which we describe below. The equilibrium path takes one of three forms depending on the initial beliefs. (1) η 1 > η 1 = η 3 = (1 ε V 1 ) T. Then 1h and 1l concede for sure and bids for sure in all sub-games (2) η 1 < η 1 and ρ 1 < ρ 1 = (1 ε V 2 ) T. Then 1h bids for sure. Additionally, as long as the game has not ended, 1l and randomize such that β2t 1 1l and β 1l 2t satisfy ρ 2t 2 ( ) = ρ ρ 2t 2 + ρ 2t 2 1 β 2t; (1 η 2t 1 )β2t = ε V 2t 1 1 (3) η 1 < η1 and ρ 1 > ρ1. Then 1h and 1l bid for sure in round 1. Bidder will randomize in round 2t such that β2t satisfies η 2t+1 = η2t+1. For t 2, as long as the game continues bidder 1h and 1l 21

randomize such that β2t 1 1l and β 1h 2t 1 satisfy ρ 2t ρ2t and ρ 2T 2 β 1h 2T 1 + (1 ρ 2T 2)β 1l 2T 1 = ε V 2 There are two observations worth pointing out about the equilibrium. First, if the intial beleifs satisfy cases (1) or (2) then the equilibrium is unique. With case (3) the equilibrium can take different paths since the best responses for player 1 after round 1 are not unique. Second, note that under case (2) we get a game in which the high budget bidders of both bidders never concede. Hence this case is akin to game in which each player has a behavioral type who never quits. As before if take = ε to zero the game converges to a continuous time war of attrition. As before, note that B l lim 0 ρ 1 = e V 2 ; lim η1 = e 0 B l V 1. 4 One-Sided Private Budget with N Prizes In this section, we generalize the baseline model to study the case where two bidders compete for N prizes, either sequentially or simultaneously. In the sequential case, bidders compete for one prize at a time and the prizes are allocated one after another. In the simultaneous case, bidders bid for all prizes at the same time. We assume that bidder i s total budget is denoted by B i as before. The main result in this section is that the equilibrium outcomes are the same in both sequential and simultaneous wars of attrition. Assume that there are N 1 prizes and only bidder 1 s budget is private: B 1 {B l,b h } where B l B 2 < B h and the prior probability that B 1 = B h is ρ [0,1]. First, by essentially the same argument, we can show that bidder 1 with budget B h never concedes any prize in any wpbe that satisfies GD1. 4.1 Sequential wars of attrition Let T = B l /ε. First, note that in a k-prizes problem, once the first prize is allocated, the problem for the second good is identical to the k 1-prizes problem, where 1-prize problem is solved. The sequence of cutoff beliefs for the ith prize in a N-prizes problem case is defined as follows: 22

1. For T + i N t T, ρ i 2t = 1 ε V 2, 2. For 1 t T + i N 1, ρ i 2t = (1 ε (N + 1 i)v 2 ) T +i N t (1 ε V 2 ). Step 1: In period 2t, where T + 1 N t T, bidder 2 bids for sure if ρ 2t < ρ 1 2t 1 ε V 2 ; bidder 2 concedes for sure if ρ 2t > ρ2t 1 ε 2 ; bidder 2 concedes with probability (T t+1)v if ρ 1 2t = ρ2t 1. In period 2t 1, bidder 1 bids for sure if ρ 2t 1 ρ2t 1 ; bidder 1 concedes with probability β 2t 1 1 if ρ 2t 1 < ρ2t 1 so that the updated belief in period 2t is ρ2t 1. In period 2t, if ρ 2t 1 ε/v 2, then bidder 2 s payoff from conceding is (1 ρ 2t )(N (T t + 1))V 2, since low budget bidder 1 bids for the second prize in period 2t + 1 and bidder 2 concedes all the remaining prizes until low budget bidder 1 runs out of budget and gets N (T t + 1) prizes; bidder 2 s payoff from bidding is ε + (1 ρ 2t ) (N (T t))v 2, since low budget bidder 1 bids for sure in period 2t + 1 by Step 1, and consequently ρ 2t+1 = ρ 2t 1 ε/v 2, and thus bidder 2 concedes all the remaining prizes until low budget bidder 1 runs out of budget, and gets N (T t) prizes. Therefore, if ρ 2t 1 ε/v 2, then ε + (1 ρ 2t ) (N (T t))v 2 (1 ρ 2t )(N (T t + 1))V 2, bidder 2 concedes in period 2t. In period 2t, if ρ 2t < 1 ε/v 2, bidder 2 s payoff from conceding is 0 + (1 ρ 2t ) (β 2 2t+1(N 1)V 2 + (1 β 2 2t+1)(N (T t + 1))V 2 ) + ρ2t 0, where β2t+1 2 is the probability that the low budget bidder 1 concedes in period 2t +1 for the second prize. By induction, β2t+1 2 is given by ρ 2t ρ 2t + (1 β 2 2t+1 )(1 ρ 2t) = ρ2 2t+2 1 ε V 2. 23

As a result, we have β 1 2t+1 = β 2 2t+1. Bidder 2 s payoff from bidding is ε + (1 ρ 2t ) (β 1 2t+1 NV 2 + (1 β 1 2T 1)(N (T t))v 2 ) + ρ 2T 2 0, since in period 2t + 1, if bidder 1 concedes for the first prize, then bidder 1 concedes for all remaining N prizes. We can show that the cutoff belief to make bidder 2 indifferent is ρ 1 2t 1 ε V 2. In period 2t 1, suppose that bidder 1 concedes for sure. If she deviates to bidding, then in period 2t, bidder 2 will concede with probability one since her updated belief will be ρ 2t = 1. This is a profitable deviation for bidder 1. Therefore, in any equilibrium β2t 1 1 < 1. In addition, bidder 1 s equilibrium strategy depends on ρ 2t 1 : (1) if ρ 2t 1 ρ2t 1, then bidder 1 bids for sure in period 2t + 1, since otherwise the updated belief in period 2t, given bidder 1 bids in period 2t 1, is ρ 2t > ρ 2t 1 ρ2t 1. By backward induction, bidder 2 concedes in period 2t, so it is optimal for bidder 1 to bid in period 2t 1, a contradiction. In all, β2t 1 1 = 0 and hence ρ 2t = ρ 2t 1. (2) If ρ 2t 1 < ρ2t 1, then β 2t 1 1 > 0, otherwise bidder 2 will bid in period 2t in which case bidder 1 gets ε in period 2t 1, a contradiction. Since β 1 2t 1 (0,1), the fact that bidder 1 is indifferent between bidding and conceding in period 2t 1 implies that bidder 2 must also randomize in period 2t. Thus, Bayesian updating implies that ρ 2t 1 ρ 2t 1 + (1 ρ 2t 1 )(1 β 1 2t 1 ) = ρ1 2t. If bidder 1 bids in period 2t 1, bidder 2 concedes with probability β 1 2t in period 2t. Since ρ1 2t ρ 2 2t, then whenever bidder 2 concedes the first prize, she will immediately concedes the remaining prize too until bidder 1 runs out of budget. In all, by bidding in period 2t 1, bidder 1 gets ε + β2t 1 2 (T t + 1)V 1 + (1 β2t 1 ) 0. Moreover, bidder 1 s indifference condition in period 2t 1 implies that 0 = ε + β 1 2t(T t + 1)V 1 + (1 β 1 2t) 0, since in period 2t, low budget bidder 1 only has (T t + 1)ε budgets remains and gets (T t + 1)V 1 if bidder 2 concedes in period 2T 2. Therefore, we have β 1 2t = ε (T t+1)v 1. Step 2: In period 2t, where 1 t T N, bidder 2 bids for sure if ρ 2t < ρ2t 1 (1 NV ε 2 ) T +1 N t (1 V ε 2 ); bidder 2 concedes for sure if ρ 2t > ρ2t 1 ; bidder 2 concedes with probability NV ε 1 if ρ 2t = ρ2t 1. In period 24