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All-pay-all aspects of political decision making


Decision-making processes are studied using non-standard all-pay structures. Our interest is motivated by regulatory, political, legal, military, and economic applications in which individual actions determine the consequences for a larger group or the general public. The common features of these examples are a competitive environment, a winner-takes-all reward structure, and some form of all-pay-all payment rule.

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  1. See, for instance, the examples in Baye et al. (1998, 2012).

  2. In this game, the players’ bids are interpreted as proposals, rather than efforts. Only the winning proposal is ex post payoff relevant.

  3. The EU commission’s recent regulation of mobile phone roaming charges may serve as an example, documented in, for example, King (2011). The payoff from being perceived as consumer friendly may be the motivation behind mobile carrier Vodafone’s announcement of the ‘travel promise’ (the statement of 3-Aug-2005 begins with ‘Commitment to simplify and increase value to Vodafone customers’) and competitor T-Mobile’s ‘Un-carrier’ strategy, both of which can be seen as widely advertised announcements of ‘consumer-friendly’ price reductions in order to generate future payoffs.

  4. Three examples from politics, business, and the public sector, respectively, are Kasperowicz (2013), Chazan (2013), and Foster (2010).

  5. This idea is not new. The two-player complete information case of this setup has recently been analyzed by Baye et al. (2012). In particular, see their section ‘Territorial contests with injuries.’ Moreover, Baye et al. (1998) specify a ‘civil war’ parametrization of their general contest class, which is a special case of our auction model.

  6. Higgins et al. (1985) provide the first analysis of endogenous participation in contests.

  7. The beautiful theory of these games has been developed, among others, by Dasgupta (1986), who discusses R&D races, Hillman and Samet (1987), who introduce rent-seeking and lobbying contests, Hillman and Riley (1989) and Baye et al. (1993), who analyze competitions for monopoly, Clark and Riis (1998), who study the competition for promotions, and Moldovanu and Sela (2001), who discuss optimal prize structures. The availability of recent and comprehensive surveys of the contests literature such as Garfinkel and Skaperdas (2006), Congleton et al. (2007), Corchón (2007), and Konrad (2008) allows us to focus our discussion on only the most directly related parts of the literature.

  8. For further discussions of asymmetric all-pay auctions, see, for instance, Hillman and Riley (1989), Fibich et al. (2006), Parreiras and Rubinchik (2010), or Szech (2011). For a detailed discussion, we appeal, again, to the above-mentioned literature surveys.

  9. A later published variant Goeree et al. (2005), is less related to the present work because, there, the payment rules do not exhibit an ‘all-pay-all’ structure.

  10. The parameters (α,β,δ,θ,γ) appear in the winner’s and loser’s linear payoff functions in Baye et al. (1998, 2012). In Baye et al. (2005), the parameters are essentially the same, with γ=0.

  11. So called after former US Vice President Dan Quayle, who chaired a commission that recommended a reform of the US legal system. According to this proposal, the loser should reimburse the winner’s legal costs up to the amount actually spent by the loser. The Quayle system is one of the legal systems analyzed in Baye et al. (1998, 2005).

  12. Ties have zero probability in the equilibria we derive.

  13. In a simple symmetric oligopoly model, the price cap is proportional to the firm’s profits. In an asymmetric setting, the firms’ bids can be understood as proposals for direct lump-sum reductions of profits.

  14. Given full participation, this boundary condition is intuitive: if the valuation is zero, then winning has no upside. The zero type would focus on minimizing payments: by making the lowest feasible bid, the price reduction in the event of winning is minimized.

  15. The case of taxation is mentioned in Baye et al. (2005) as a way of reducing wasteful legal expenditures.

  16. Baye et al. (2005) show that these performance measures are a function of their β only, with β=1 and β=0 in the American and British systems, respectively, while our example uses β=1/2. Apart from this case, we cannot compare results because, as soon as we deviate from α=1/2 and linear payoffs, we violate assumption (A3) in Baye et al. (2005), which is incorporated in their solution from the beginning. Moreover, note that these results of Baye et al. (2005) are comparable with ours given only that there is litigation, i.e., participation.

  17. Similarly, replace f(x)x/F(x) in (9) by α/(n−1), simplify, and solve \(\bar{u}=\varPi(0,0)\) to obtain the required upper bound of the outside option.

  18. We cannot compare results with Baye et al. (2005), however, because we violate their assumption (A3) ‘Internalized legal cost,’ on which their relevant results rest.

  19. Recall that g(x)=G′(x)=(n−1)F n−2(x)f(x).

  20. Consider full participation and F(θ)=θ t, for some t>0, and for which the auction parameters are k=t(n−1)+1 and α=(k−1)/k, then the equilibrium exhibits truth-telling β(θ i )=θ i : F(x)=x t implies that g(x)x=t(n−1)x t(n−1). Inserting this into (12) and setting the term equal to θ i (truthful bidding), we get

    $$ {\theta_i = \biggl( \frac{1}{\alpha} \int _0^{\theta_i} \bigl(x^{t(n-1)} \bigr)'x dx \biggr)^{\frac{1}{k}}\quad \Rightarrow\quad \alpha \theta_i^k = \frac{t(n-1)}{t(n-1)+1}\theta_i^{t(n-1)+1}.} $$

    This equation is satisfied if we insert the proposed α=(k−1)/k and k=t(n−1)+1. This result is nongeneric, in the sense that it relies on a special distributional assumption.

  21. Recall that the equilibrium bid function of the FP-APA under our assumptions is \(\beta^{\text{FP-APA}}(\theta_{i}) = \int_{0}^{\theta_{i}} g(x) x dx\). The corresponding equilibrium payoff is \(G(\theta _{i})\theta_{i} - \beta^{\text{FP-APA}}(\theta_{i}) = \int_{0}^{\theta_{i}} G(x) dx\).

  22. Consider (11). Under full participation (\(\hat{\theta}=0\)), the first term is zero, the second term is the FP-APA-payoff, and the last term is negative. Therefore, the payoff in our auction is lower than under the FP-APA.

  23. For instance, consider the uniform distribution and full participation (\(\hat{\theta}=0\)). Then (9) is equal to \(-(n-2+1/n)+ \theta_{i}^{2}/n\), where the first term is negative and the second term is the FP-APA-payoff. The payoff to our auction thus is smaller.

  24. For a brief discussion of the state of the literature, see Chawla and Hartline (2013). Note that a large part of the all-pay auction literature deals with the complete-information case, whereas we are concerned with the standard private-values setting.

  25. Negative equilibrium payoffs are discussed in, e.g., Krishna and Morgan (1997).


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Thanks to Bipasa Datta, Alex Gershkov, Arye Hillman, Bettina Klose, Peter Simmons, Jacco Thijssen, Zaifu Yang, and two anonymous referees for helpful comments. We are grateful for financial support by the University of York Research and Impact Support Fund.

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Correspondence to Paul Schweinzer.



Proof of Lemma 1

We need to show that we can rule out symmetric, pure-strategy bid functions that prescribe a strictly positive bid for the lowest participating type \(\hat{\theta}\), say b 0>0 plus a strictly monotone, type-dependent part. In order to see this, consider first the case of full participation. Then, under a monotone bidding function, a bidder with a value between zero and b 0 bids more than b 0 and wins with positive probability. In case of winning, she would have to pay more than her value, while in case of losing her payoff is independent of her own bid. Thus, she can strictly improve her payoff by reducing her bid, contradicting the candidate equilibrium strategy.

Second, consider the case of partial participation and focus on the indifferent bidder with valuation \(\theta_{i}=\hat{\theta}\). With positive probability, the indifferent bidder is the only bidder in the auction. Then she wins with certainty and her payoff is decreasing in her bid. Thus, a smaller bid is better. On the other hand, if there are other active bidders, then, by the definition of being the indifferent bidder, these other bidders have higher values and bid more. Then the indifferent bidder never wins and her payoff is independent of her bid. An incentive to reduce the bid therefore exists. We conclude that every strictly monotone bid function must satisfy \(\beta(\hat{\theta})=0\). □

Proof of Proposition 1

If \(\bar{u} \ge1\), the outside option is superior to the auction. Thus, we restrict attention to \(\bar{u}\in(-\infty, 1)\). We moreover focus on symmetric, pure-strategy equilibria with continuous, strictly monotone bid functions. Therefore, participation is governed by a comparison between the common outside option and the expected equilibrium payoff from participation. This implies a common cutoff valuation, denoted by \(\hat{\theta}\in[0,1]\), below which a player prefers the outside option. This includes the cases of full and no participation. From Lemma 1, we know that \(\beta(\hat{\theta})=0\).

Suppose that an equilibrium exists wherein all bidders take their outside options if their valuations are below the common threshold value \(\hat{\theta}\). If their valuations exceed this value, then they bid according to the strictly monotone bidding function β. We first determine this bidding function, taking the threshold value \(\hat{\theta}\in [0,1]\) as given. We start by considering bidder i’s decision problem when her value exceeds the threshold value, \(\theta_{i}>\hat{\theta}\). W.l.o.g. we only consider i’s positive bids β(z) for \(z\in(\hat{\theta},1]\). Bidder i’s expected profit from bidding β(z), denoted by π(z,θ i ), is

$$\begin{aligned} \pi(z,\theta_i) =& G(z) \bigl( \theta_i-\alpha\beta(z)\bigr) - \bigl(1-G(z)\bigr) \mathbb {E}\bigl[\alpha \beta (\varTheta _{(1:n-1)})| \varTheta _{(1:n-1)}>z\bigr] \\ =& F^{n-1}(z) \bigl(\theta_i-\alpha\beta(z)\bigr) - \int _{z}^1 \alpha\beta(s) d F^{n-1}(s); \\ \frac{\partial\pi(z,\theta_i)}{\partial z} =& \bigl(F^{n-1}(z)\bigr)'\bigl( \theta_i-\alpha \beta(z)\bigr) - \bigl(F^{n-1}(z)\alpha \beta'(z)\bigr) + \alpha\beta(z) \bigl(F^{n-1}(z) \bigr)' = 0 \\ &{}\iff\quad \beta'(z) =\frac{1}{\alpha} \frac {(F^{n-1}(z))'}{F^{n-1}(z)} \theta_i = \frac{(n-1)}{\alpha}\frac {F(z)^{n-2}(z)f(z)}{F^{n-1}(z)}\theta_i \\ =& \frac{(n-1)}{\alpha}\frac {f(z)}{F(z)}\theta_i. \end{aligned}$$

As argued above, \(\beta(\hat{\theta})=0\). Using this as the boundary condition, our equilibrium bid function is found by integrating the above differential equation, with one exception: Because F(0)=0, the differential equation is not defined at \(\hat{\theta}=0\). In that case, we integrate over the open interval (0,θ i ] and define β(0)=0, corresponding to \(\beta(\hat{\theta})=0\). In symmetric equilibrium, z=θ i . Thus,

$$ \beta(\theta_i)= \frac{(n-1)}{\alpha}\int_{\hat{\theta}}^{\theta_i} \frac {f(x)x}{F(x)} dx. $$

This corresponds to (10). Note that, at x=0, the numerator as well as the denominator of the integrand vanish. Thus, the integrand is bounded by l’Hôspital’s rule,

$$ {\lim_{x\to0} \frac{f(x)x}{F(x)} = \lim_{x \to0} \frac {f'(x)x}{f(x)}+1 = 1,} $$

and, therefore, the integrand of (17) is bounded everywhere. Since every bounded function can be integrated, the resulting bidding function is well defined.

We now turn to i’s expected equilibrium profit, denoted by \(\varPi(\hat{\theta}, \theta_{i})\), where all players follow (17) and \(\theta_{i}>\hat{\theta}\).

$$\begin{aligned} \varPi(\hat{\theta},\theta_i) =& F^{n-1}(\theta_i) \bigl(\theta_i-\alpha\beta( \theta_i)\bigr) - \int _{\theta_i}^1 \alpha\beta(x) d F^{n-1}(x) \\ =& F^{n-1}(\theta_i) \biggl(\theta_i - (n-1)\int _{\hat{\theta}}^{\theta_i} \frac {f(y)y}{F(y)} dy \biggr) \\ &{} - \underbrace{\int _{\theta_i}^1 (n-1)\int _{\hat{\theta} }^{x} \frac{f(y)y}{F(y)} dy \bigl(F^{n-1}(x)\bigr)' dx}_{=:(A)}. \end{aligned}$$

Interchanging the order of integration, the last term can be written as

$$\begin{aligned} (A) =& \int _{\hat{\theta}}^{\theta_i}\!\!\! \int _{\theta_i}^1 \bigl(F^{n-1}(x)\bigr)' dx (n-1)\frac{f(y)y}{F(y)} dy + \int _{\theta_i}^{1}\!\!\! \int_{y}^1 \bigl(F^{n-1}(x)\bigr)' dx (n-1)\frac{f(y)y}{F(y)} dy \\ =& \int _{\hat{\theta}}^{\theta_i} \bigl(1-F^{n-1}( \theta_i)\bigr) (n-1)\frac{f(y)y}{F(y)} dy + \int _{\theta_i}^{1} \bigl(1-F^{n-1}(y)\bigr) (n-1)\frac{f(y)y}{F(y)} dy. \end{aligned}$$

After reinserting (A) into (19) and straightforward simplification, we obtain

$$ \varPi(\hat{\theta},\theta_i) = F^{n-1}( \theta_i)\theta_i - (n-1)\int_{\hat{\theta}}^1 \frac{f(x)x}{F(x)} dx + \int_{\theta_i}^1\underbrace{ F^{n-1}(x) (n-1)\frac{f(x)x}{F(x)} }_{=(F^{n-1}(x))'x} dx. $$

Integrating the last term by parts, we get

$$\begin{aligned} \varPi(\hat{\theta},\theta_i) =&F^{n-1}(\theta_i)\theta_i - (n-1)\int _{\hat{\theta}}^1 \frac {f(x)x}{F(x)} dt + 1-F^{n-1}(\theta_i)\theta_i -\int _{\theta_i}^1 F^{n-1}(x) dx \\ =& 1-(n-1)\int _{\hat{\theta}}^1 \frac{f(x)x}{F(x)} dx - \int _{\theta_i}^1 F^{n-1}(x) dx, \end{aligned}$$

which equals \(\varPi(\hat{\theta},\theta_{i})\) in (9). Given this participation payoff, we determine participation behavior by taking into account the commonly known outside options. Since \(\varPi(\hat{\theta},\theta _{i})\) is obviously strictly increasing in both, θ i and \(\hat{\theta}\), the minimum of (9) is Π(0,0) and the maximum is Π(1,1)=1. Therefore, we have full participation for \(\bar{u}\le\varPi (0,0)\), no participation for \(\bar{u} \ge\varPi(1,1)=1\), and participation according to the cutoff value \(\hat{\theta}\in(0,1)\) for \(\varPi(0,0)< \bar{u} < \varPi (1,1)=1\). In the latter case, the cutoff is defined by \(\bar{u}=\varPi(\hat{\theta} ,\hat{\theta})\), the expected payoff of the indifferent bidder.

It remains to be shown that (17) does not only satisfy the first-order condition of i’s best response problem but that it is indeed a maximizer for \(\theta_{i}>\hat{\theta}\).

Similar to the above, consider \({\theta},z\in(\hat{\theta},1]\). Start from (16) and insert (17), canceling out α:

$$ \pi(z,{\theta})= F^{n-1}(z) \biggl({\theta}-(n-1)\int_{\hat{\theta}}^z \frac {f(x)x}{F(x)}dx \biggr) - \underbrace{\int_{z}^1 (n-1)\int_{\hat{\theta}}^{y} \frac {f(x)x}{F(x)} dx \bigl(F^{n-1}(y)\bigr)' dy}_{=:(A)}. $$

Interchanging the order of integration, the term (A) can be written as

$$ (A)=(n-1) \biggl(\int_{\hat{\theta}}^z \frac{f(x)x}{F(x)} \underbrace{\int_z^1 \bigl(F^{n-1}(y) \bigr)' dy}_{=1-F^{n-1}(z)} dx + \int_z^1 \frac{f(x)x}{F(x)} \underbrace{\int_{x}^1 \bigl(F^{n-1}(y)\bigr)' dy}_{=1-F^{n-1}(x)} dx \biggr). $$

Now, reinsert into (23), cancel out the terms \(\pm F^{n-1}(z) (n-1)\int _{\hat{\theta}}^{z} \frac{f(x)x}{F(x)}dx\), and get

$$\begin{aligned} \pi(z,{\theta}) =& \underbrace{F^{n-1}(z)\theta}_{=:(B)}- (n-1) \biggl(\int_{\hat{\theta}}^z \frac{f(x)x}{F(x)} dx + \int _z^1 \frac{f(x)x}{F(x)} \bigl(1-F^{n-1}(x) \bigr) dx \biggr) \\ =& \underbrace{{\theta}-\int _{z}^{1} \bigl(F^{n-1}(x)\bigr)'{\theta} dx}_{=(B)} -(n-1) \int _{\hat{\theta}}^1 \frac{f(x)x}{F(x)} dx + \underbrace{(n-1)\int _z^1 \frac{f(x)x}{F(x)}\bigl(F^{n-1}(x)\bigr) dx}_{\int_z^1 (F^{n-1}(x))'x dx} \\ =& {\theta} -(n-1)\int _{\hat{\theta}}^1 \frac{f(x)x}{F(x)} dx + \underbrace{\int _z^1 \bigl(F^{n-1}(x) \bigr)'(x-{\theta}) dx}_{=:(C)}. \end{aligned}$$

Note, that only (C) depends on z. Thus,

$$ \pi({\theta},{\theta})-\pi(z,{\theta}) = \int_{\theta}^z \bigl(F^{n-1}(x)\bigr)'(x-{\theta}) dx. $$

Since this is strictly positive for all zθ, where z,θ∈[0,1], this completes the proof. □

Proof of Proposition 2

Obviously, if \(\bar{u}\ge1\), the outside option is superior to participation. Thus, we only need to consider \(\bar{u} \in(-\infty, 1)\). We restrict attention to symmetric, pure-strategy equilibria with continuous, monotone bid functions. Therefore, participation is governed by a comparison of (candidate) equilibrium payoffs from participation, and the outside option. This implies a common cutoff valuation, denoted by \(\hat{\theta}\in[0,1]\), above which a player participates in the auction. This includes the cases of full and no participation. Similar to the proof of Lemma 1, we can show that \(\beta(\hat{\theta})=0\). The only difference is that, here, losing bidders’ payoff is decreasing in (rather than independent of) the own bid.

Suppose that there is an equilibrium wherein player i takes the outside option if \(\theta_{i}<\hat{\theta}\) and otherwise participates and follows the common bid function β. Obviously, (12) is strictly increasing in θ i . Suppose that player i’s rivals j participate if \(\theta_{j} \ge\hat{\theta}\) and, if so, bid according to (12). W.l.o.g. we only consider bidder i’s deviating bids β(z) where \(z\in[\hat{\theta},1]\) (and discuss outside options below). Recalling that G=F n−1, and denoting a valuation by the random variable Θ, bidder i’s expected payoff is

$$ \begin{aligned} &\pi(z,{\theta}) = F^{n-1}(z){\theta}- \alpha\beta^k(z) - (n-1) \mathbb {E}\bigl[\alpha \beta({\varTheta })^k\bigr],\\ &\frac{\partial\pi(z,{\theta})}{\partial z} = \bigl(F^{n-1}(z)\bigr)'{\theta}- \bigl(\alpha\beta^k(z)\bigr)'. \end{aligned} $$

Applying the symmetric equilibrium condition, z=θ, and the boundary condition \(\beta(\hat{\theta})=0\) (see above), integration delivers (12). As a proof of sufficiency, we show that π i (θ,θ)>π i (z,θ) for all \(z\in[\hat{\theta},1]\), zθ. Note that the term \(- \alpha (n-1) \mathbb {E}[\beta({\varTheta })^{k}]\) cancels out in the first line of (27) when we compute the difference π i (θ,θ)−π i (z,θ). Inserting the candidate payoff, we get

$$\begin{aligned} \pi({\theta},{\theta})-\pi(z,{\theta}) =& {\theta}\bigl(F^{n-1}({\theta})-F^{n-1}(z)\bigr)- \int _z^{\theta} \bigl(F^{n-1}(x) \bigr)'x dx \\ =& {\theta} \int _z^{\theta} \bigl(F^{n-1}(x) \bigr)' dx - \int _z^{\theta} \bigl(F^{n-1}(x) \bigr)'x dx \\ =& \int _z^{\theta} \bigl(F^{n-1}(x) \bigr)'({\theta}-x) dx >0. \end{aligned}$$

We continue by computing a participating bidder i’s (candidate) expected equilibrium payoff, denoted by \(\varPi(\hat{\theta}, \theta_{i})\), i.e., we take the cutoff value \(\hat{\theta}\) as given, and assume \(\theta_{i} \ge\hat{\theta}\) and consider participating rivals’ bids for \(\theta_{j}\ge\hat{\theta}\), ji. Inserting the candidate payoff into the first line of (27), we get

$$ \varPi(\hat{\theta}, \theta_i) = \underbrace{F^{n-1}( \theta_i) \theta_i - \int_{\hat{\theta}}^{\theta _i} \bigl(F^{n-1}(x)\bigr)'x dx}_{=:(A)} - (n-1) \underbrace{\int_0^1 \int_{\hat{\theta}} ^{\theta_j\ge\hat{\theta}} \bigl(F^{n-1}(x)\bigr)'x dx f( \theta_j) d \theta_j }_{=:(B)}. $$

Using integration by parts, term (A) in the above can be written as

$$ (A) = F^{n-1}(\theta_i) \theta_i - \biggl( \bigl[F^{n-1}(x)x \bigr]_{\hat{\theta} }^{\theta_i} - \int _{\hat{\theta}}^{\theta_i} F^{n-1}(x) dx \biggr) = F^{n-1}(\hat{\theta} )\hat{\theta}+ \int_{\hat{\theta}}^{\theta_i} F^{n-1}(x) dx. $$

Interchanging the order of integration, term (B) can be written as

$$\begin{aligned} (B) = \int_{\hat{\theta}}^1\int_{x}^{1}f( \theta_j)d\theta_j \bigl(F^{n-1}(x) \bigr)'x dx = \int_{\hat{\theta}}^1\bigl(1-F(x) \bigr) \bigl(F^{n-1}(x)\bigr)'x dx. \end{aligned}$$

Reinserting these expressions for (A) and (B) gives (11).

Finally, given this participation payoff, we determine participation behavior by taking into account the value of the commonly known outside options. Note that (11) is strictly increasing in both θ i and \(\hat{\theta}\). The former is obvious and the latter can be seen in the first derivative,

$$\begin{aligned} \frac{\partial\varPi(\hat{\theta}, \theta_i)}{\partial\hat{\theta}} &= \bigl(F^{n-1}(\hat{\theta})\bigr)' \hat{\theta} + F^{n-1}(\hat{\theta}) - F^{n-1}(\hat{\theta}) + (n-1) \bigl(1-F(\hat{\theta})\bigr) \bigl(F^{n-1}(\hat{\theta}) \bigr)'\hat{\theta}, \end{aligned}$$

where the second and third term cancel out, while the remaining terms are positive. Thus, for \(\hat{\theta}, \theta_{i}\in[0,1]\) and \(\theta_{i} \ge\hat{\theta}\), the maximum of (11) is Π(1,1)=1 while the minimum is Π(0,0). Therefore, we have full participation for \(\bar{u} \le\varPi (0,0)\), and participation according to the cutoff value \(\hat{\theta}\in (0,1)\) whenever \(\varPi(0,0)<\bar{u} < \varPi(1,1) = 1\). In the latter case, the cutoff is defined by \(\bar{u} = \varPi(\hat{\theta}, \hat{\theta})\), the payoff of the indifferent bidder. □

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Giebe, T., Schweinzer, P. All-pay-all aspects of political decision making. Public Choice 161, 73–90 (2014).

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  • Auctions
  • Contests
  • Regulation
  • Conflict

JEL Classification

  • C7
  • D7
  • L5