Extremism drives out moderation

Abstract

This article examines the impact of the distribution of preferences on equilibrium behavior in conflicts modeled as all-pay auctions with identity-dependent externalities. Centrists and radicals are defined using a willingness-to-pay criterion that admits preferences more general than a simple ordering on the line. Extremism, characterized by a higher per capita expenditure by radicals than centrists, may persist and generate higher aggregate expenditure by radicals, even when they are relatively small in number. Our results demonstrate the importance of the institutions of conflict in determining the role of extremism and moderation in economic, political, and social environments.

Appendices

Appendix 1: Proof of Proposition 1

Proof

In a first step we show existence by constructing an equilibrium, we then show uniqueness of the equilibrium described before in a second step involving multiple lemmas.

The strategy profile in which $2$ stays out completely (puts mass 1 on zero) and players $1$ and $3$ randomize uniformly over $[0,r_{jk}]$ ($j,k\in \{1,3\}$, $j\ne k$) is a Nash equilibrium. Assume that $2$ uses the strategy $F_2(x)=\left\{ \!\!\begin{array}{cll}0&{}\quad \hbox { for }&{}x<0\\ 1&{}\quad \hbox { for }&{}x\ge 0\end{array}\right.$. Then (by Baye et al. 1996) it is optimal for players $1$ and $3$ to randomize over $[0,r_{jk}]$ according to

$$\begin{aligned}F(x)=\left\{ \!\! \begin{array}{cll} 0&{}\quad \hbox { for }&{} x<0\\ \frac{x}{r_{jk}}&{}\quad \hbox { for }&{} 0\le x\le r_{jk}\\ 1&{}\quad \hbox { for }&{} x>r_{jk}\\ \end{array}.\right. \end{aligned}$$

Given that $1$ and $3$ apply this strategy player $2$’s payoff if he submits a strictly positive bid $x\in (0,r_{2j}]$ is:

$$\begin{aligned} u_2^*(x)&=[F(x)]^2v_{22}+\left( 1-[F(x)]^2\right) v_{2j}-x\\&=v_{2j}+F(x)^2r_{2j}-x\\&=v_{2j}-x\left( 1-\frac{x}{r_{jk}}\frac{r_{2j}}{r_{jk}}\right) \\&<v_{2j}. \end{aligned}$$

It is therefore a best response for player 2 to stay out of the conflict.

Next we prove the uniqueness of the equilibrium described in Proposition 1, by first showing that any equilibrium of $\varGamma _{21}$ is symmetric in the sense that both radicals (players 1 and 3) choose identical strategies. In a second step we then show that the set of symmetric equilibria of $\varGamma _{21}$ is a singleton, given by the cut-throat competition equilibrium described above.

Let $\underline{s}_i$ and $\bar{s}_i$ be the lower and upper bound, respectively, of the support of an equilibrium strategy for player $i$, $i\in I$, and define $\bar{s}=\max _{i\in I}\{\bar{s}_i\}$. In the following, the indices $j$ and $k$ refer to two different radical players, i.e. $j\!\in \!\{1,3\}, k\!\in \!\{1,3\}\backslash \{j\}$.

$\square$

Lemma 1

In any equilibrium of $\varGamma _{21}$, both radicals actively participate in the conflict.

Proof

By way of contradiction, assume that one of the radical players stays out of the conflict; without loss of generality let that player be player $1$, i.e. $F_1(0)=1$. Given player $1$’s strategy players $2$ and $3$ would randomize up to $r_{23}=r_{32}<r_{31}=r_{13}$. Player $1$’s payoff if he bids zero will be in the interval $(v_{13},v_{12})$ and he could strictly improve upon this by bidding $r_{23}$ which would guarantee him a payoff of $v_{11}-r_{23}=v_{11}-r_{12}=v_{12}$.$\square$

Lemma 2

$\underline{s}_i=0$ for all $i\in I$, and for at least one player $l\in I$, $F_l(0)=0$.

Proof

Assume $\underline{s}_i>\underline{s}_l\ge \underline{s}_m\ge 0$ for some $i,l,m\in I$. Any bid $x\in [0,\underline{s}_i)$ results in a loss with certainty. Therefore, players $l$ and $m$ do not put mass anywhere over $(0,\underline{s}_i)$. Moreover, no player $l$ or $m$ can place a mass point at $\underline{s}_i$, because if two or more players had a mass point at $\underline{s}_i$, then one could improve by moving mass up, and if only one player had a mass point at $\underline{s}_i$, then he would improve by moving the mass down. Altogether players $l$ and $m$ do not put mass anywhere over $(0,\underline{s}_i]$, but then player $i$ would improve by moving mass down. This contradiction implies that there exist mutually different $i,l,m\in I$ such that $\underline{s}_i=\underline{s}_l\ge \underline{s}_m\ge 0$. Assume that $\underline{s}_i=\underline{s}_l> \underline{s}_m\ge 0$ for some $i,l,m\in I$. It cannot be the case that both players, $i$ and $l$, have a mass point at $\underline{s}_i=\underline{s}_l$ (otherwise one could improve by moving mass up slightly), but then at least one of them would win with probability arbitrarily close to zero in some neighborhood above $\underline{s}_i$ and would be better off by moving mass down to zero. It follows that in equilibrium $\underline{s}_1=\underline{s}_2=\underline{s}_3=\underline{s}$. It cannot be the case that all three players have a mass point at $\underline{s}$ otherwise a player could improve by moving this mass up slightly. Therefore, at least one player loses with certainty at $\underline{s}$. Altogether this shows that $\underline{s}_1=\underline{s}_2=\underline{s}_3=0$.$\square$

Lemma 3

There are no mass points at $x$ in any player’s equilibrium distribution $\forall x\in (0,\bar{s}]$.

Proof

Suppose player $i\in I$ has a mass point at $x\in (0,\bar{s}]$. Since, from Lemma 2, $F_l(x)>0$ for every $l\in I$, for sufficiently small $\epsilon >0$ no player $j\ne i$ would place mass in $(x-\epsilon ,x]$ since that player could improve his payoff by moving mass from that interval to infinitesimally above $x$. But then it is not optimal for $i$ to put mass at $x$.$\square$

Lemma 4

$\bar{s}_1=\bar{s}_3>\bar{s}_2$.

Proof

Obviously, it cannot be the case that $\bar{s}_i>\bar{s}_l\ge \bar{s}_m$ for some $i,l,m\in I$, because player $i$ would strictly improve his payoff by moving mass from $\left( \frac{1}{2}(\bar{s}_l+\bar{s}_i),\bar{s}_i\right]$ down to $\frac{1}{2}(\bar{s}_l+\bar{s}_i)$. Suppose, $\bar{s}_1=\bar{s}_2=\bar{s}_3=\bar{s}>0$. Since any bid $b_2>r_{2j}$ of player 2 is strictly dominated by $b_2=0$ it follows that $\bar{s}\le r_{2j}$. By Lemma 2, $\underline{s}_i=0$ for all $i\in I$ and at most two players may have a mass point at zero. Therefore, there exists a radical player $j$, who is outbid with certainty when bidding zero and whose payoff from bidding zero is $u_j^*(0)=\alpha v_{j2}+(1-\alpha )v_{jk}$ for some $\alpha \in (0,1)$. By assumption $r_{jk}>r_{j2}$ which implies by definition that $v_{jk}<v_{j2}$. Then, (by Lemma 2) player $j$’s expected equilibrium payoff would be $u_j^*<v_{j2}$. On the other hand, by submitting a bid $\bar{s}+\epsilon$ greater than $\bar{s}$ player $j$ would receive $u_j^*(\bar{s}+\epsilon )=v_{jj}-\bar{s}-\epsilon \ge v_{jj}- r_{j2}-\epsilon =v_{j2}-\epsilon$. Therefore, by choosing $\epsilon >0$ small enough, he would improve his payoff. Thus, $\bar{s}_1=\bar{s}_2=\bar{s}_3$ cannot hold true. By the same argument it cannot be the case that $\bar{s}_j<\bar{s}_2=\bar{s}_k=\bar{s}$. Hence, $\bar{s}_j=\bar{s}_k>\bar{s}_2$.$\square$

Lemma 5

$\bar{s}_2<r_{2j}, j\in \{1,3\}$.

Proof

By Lemma 4 player $2$ loses with strictly positive probability at $\bar{s}_2$. Suppose $\bar{s}_2\ge r_{2j}$, then player $2$’s equilibrium payoff at $\bar{s}_2$ is

$$\begin{aligned}u_2^*(\bar{s}_2, F_j,F_k)&=[F_j(\bar{s}_2)\cdot F_k(\bar{s}_2)]v_{22}+\left( 1-[F_j(\bar{s}_2)\cdot F_k(\bar{s}_2)]\right) v_{2j}-\bar{s}_2\\&\le v_{2j}-\underbrace{\left( 1-[F_j(\bar{s}_2)\cdot F_k(\bar{s}_2)]\right) }_{>0 \text { by Lemma 4}}r_{2j}<v_{2j}. \end{aligned}$$

This is a contradiction, because player 2 could guarantee himself a payoff of at least $v_{2j}$ by bidding zero.$\square$

Lemma 6

Players $j\in \{1,3\}$ earn expected equilibrium payoffs $v_{jj}-\bar{s}$.

Proof

From Lemmas 3 and 4 players $1$ and $3$ must earn their expected equilibrium payoff at the upper bound of the support of their mixed strategies, $\bar{s}$, and neither has a mass point at $\bar{s}$. Therefore, their expected equilibrium payoff is $u_j^*=v_{jj}-\bar{s}$.$\square$

Lemma 7

$F_1(x)=F_3(x)$ for all $x\in [\bar{s}_2,\bar{s}]$.

Proof

Notice that $F_2(x)=1$ for all $x\in [\bar{s}_2,\bar{s}]$, and $F_1(\bar{s})=F_3(\bar{s})=1$. From Lemma 4, for $x\in (\bar{s}_2,\bar{s}]$

$$\begin{aligned}u_j(x, F_2, F_k)=F_k(x)v_{jj}+(1-F_k(x))v_{jk}-x =v_{jk}+F_k(x)r_{jk}-x.\end{aligned}$$

By Lemma 6 it follows that

$$\begin{aligned}&v_{jk}+F_k(x)r_{jk}-x= v_{jj}-\bar{s}\\&\quad \Leftrightarrow F_k(x)=1-\frac{\bar{s}-x}{r_{jk}} \end{aligned}$$

and by Assumption 2 (symmetric inter-agent antagonism) follows that players $j$ and $k$ use identical strategies $F_j(x)=F_k(x)=1-\frac{\bar{s}-x}{r_{jk}}$ over the interval $(\bar{s}_2,\bar{s}]$. If $\bar{s}_2>0$, then by Lemma 3 this holds over $[\bar{s}_2,\bar{s}]$. If $\bar{s}_2=0$ right-continuity of $F_i, i\in I,$ implies $F_1(0)=F_3(0)$.$\square$

Lemma 8

For any nondegenerate interval $[\underline{t},\bar{t}]\in [0,\bar{s}]$ ($\underline{t}<\bar{t}$) there are at least two players, $i,j\in I$, such that $F_l(\bar{t})-F_l(\underline{t})>0$ for $l=i,j$.

Proof

Suppose there is a $t>\underline{t}$ such that $F_i(t)-F_i(\underline{t})=0$ for all $i\in I$, and let $\bar{t}$ be the supremum over all $t$ with this property, i.e. define $\bar{t}=\sup \{t>\underline{t} : F_i(t)-F_i(\underline{t})=0 \hbox { for all } i\in l\}.$ Notice that by Lemma 2 $\underline{t}>0$. Since $\bar{t}>\underline{t}\ge 0$ no player has a mass point at $\bar{t}$ by Lemma 3. Let player $i\in I$ and $m,l\in I\backslash \{i\}$, then player $i$’s payoff from a bid $\bar{t}+\epsilon$ is

$$\begin{aligned} \begin{aligned} u_i(\bar{t}+\epsilon ,F_l,F_m)=\,&v_{ii}\cdot F_l(\bar{t}+\epsilon )F_m(\bar{t}+\epsilon )+ v_{il}\int _{\bar{t}+\epsilon }^{\bar{s}}F_m(y)f_l(y)dy\\&+ v_{im}\int _{\bar{t}+\epsilon }^{\bar{s}}F_l(y)f_m(y)dy-\bar{t}-\epsilon . \end{aligned} \end{aligned}$$

On the other hand player $i$’s payoff from bidding $\underline{t}$ is

$$\begin{aligned} u_i(\underline{t},F_l,F_m)&\!=\!v_{ii}\cdot F_l(\underline{t})F_m(\underline{t})\!+\!v_{il}\int _{\underline{t}}^{\bar{s}}F_m(y)f_l(y)dy\!+\!v_{im}\int _{\underline{t}}^{\bar{s}}F_l(y)f_m(y)dy-\underline{t}\\&\!=\!v_{ii}\cdot F_l(\bar{t})F_m(\bar{t})\!+\!v_{il}\int _{\bar{t}}^{\bar{s}}F_m(y)f_l(y)dy\!+\!v_{im}\int _{\bar{t}}^{\bar{s}}F_l(y)f_m(y)dy-\underline{t}, \end{aligned}$$

which is strictly greater than $u_i(\bar{t}+\epsilon ,F_l,F_m)$ for $\epsilon >0$ sufficiently small. Thus, for small enough $\epsilon >0$ a player would improve his payoff by moving mass from $[\bar{t},\bar{t}+\epsilon ]$ to $\underline{t}$. Therefore, no $t>\underline{t}$ such that $F_i(t)-F_i(\underline{t})=0$ for all $i\in I$ exists.

Suppose that there is only one player $i\in I$ with $F_i(\bar{t})-F_i(\underline{t})>0$, and denote the other two players by $l,m\in I\backslash \{i\}$. Note that for players $p\in \{l,m\}$, $f_p(t)=0$ for all $t\in (\underline{t},\bar{t})$ and $F_p(\underline{t})=F_p(t)=F_p(\bar{t})$ for all $t\in (\underline{t},\bar{t})$. Player $i$’s expected payoff from a bid $t\in (\underline{t},\bar{t})$ is

$$\begin{aligned} u_i(t,F_l,F_m)&=v_{ii}\cdot F_l(t)F_m(t)\!+\! v_{il}\int _t^{\bar{s}}F_m(y)f_l(y)dy\!+\! v_{im}\int _t^{\bar{s}}F_l(y)f_m(y)dy-t\\&=v_{ii}\cdot F_l(\underline{t})F_m(\underline{t})\!+\! v_{il}\int _{\underline{t}}^{\bar{s}}F_m(y)f_l(y)dy\!+\! v_{im}\int _{\underline{t}}^{\bar{s}}F_l(y)f_m(y)dy-t\\&<v_{ii}\cdot F_l(\underline{t})F_m(\underline{t})\!+\! v_{il}\int _{\underline{t}}^{\bar{s}}F_m(y)f_l(y)dy\!+\! v_{im}\int _{\underline{t}}^{\bar{s}}F_l(y)f_m(y)dy-\underline{t}\\&=u_i(\underline{t},F_l,F_m). \end{aligned}$$

Therefore, player $i$ could improve his payoff by moving mass from the interval $(\underline{t},\bar{t}]$ to its lower bound $\underline{t}$.$\square$

Lemma 9

$F_1(x)=F_3(x)$ for all $x\in [0,\bar{s}]$.

Proof

If $\bar{s}_2=0$ then $F_1(x)=F_3(x)$ for all $x\in [0,\bar{s}]$ by Lemma 7, thus we assume in the following that $\bar{s}_2>0$.

For any bid $b_j>0$ in the support of player $j$’s equilibrium strategy his expected payoff must be equal to $v_{jj}-\bar{s}$ (by Lemma 6). That is:¹³

$$\begin{aligned} \begin{aligned} v_{jj}-\bar{s}=\,&v_{jj}\cdot \left( 1- p\{\hbox {2 wins }|\, b_j\}- p\{\hbox {k wins }|\, b_j\}\right) +v_{j2}\cdot p\{\hbox {2 wins }|\, b_j\}\\&+v_{jk}\cdot p\{\hbox {k wins }|\, b_j\}-b_j\\ =\,&v_{jj}-r_{j2}\cdot p\{\hbox {2 wins }|\, b_j\}-r_{jk}\cdot p\{\hbox {k wins }|\, b_j\}-b_j\\ =\,&v_{jj} -r_{j2}\cdot \int _{b_j}^{\bar{s }}F_k(y) f_2(y) dy -r_{jk}\cdot \int _{b_j}^{\bar{s}} F_2(y) f_k(y) dy-b_j\\ =\,&v_{jj} -r_{j2}\cdot \int _{b_j}^{\bar{s }}F_k(y) f_2(y) dy\\&-r_{jk}\cdot \left( \left[ F_2(y)F_k(y)\right] _{b_j}^{\bar{s}}- \int _{b_j}^{\bar{s}} F_k(y) f_2(y) dy\right) -b_j\\ =\,&v_{jj} -(r_{j2}-r_{jk})\cdot \int _{b_j}^{\bar{s }}F_k(y) f_2(y) dy -r_{jk}\cdot \left( 1-F_2(b_j)F_k(b_j)\right) -b_j\\ \Leftrightarrow \bar{s}-b_j =\,&(r_{j2}-r_{jk})\cdot \int _{b_j}^{\bar{s }}F_k(y) f_2(y) dy +r_{jk}\cdot \left( 1-F_2(b_j)F_k(b_j)\right) \end{aligned} \end{aligned}$$

Define $\alpha ,\beta , \gamma$ such that $\alpha \equiv r_{12}=r_{21}=r_{32}=r_{23}$, $\beta \equiv r_{13}=r_{31}$, and $\gamma =\alpha -\beta$. Note that $\alpha , \beta >0$ and $\gamma <0$. Then for any $b_j,b_k\in (0,\bar{s}]$:

$$\begin{aligned} \bar{s}-b_j\le \gamma \cdot \int _{b_j}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(b_j)F_k(b_j)\right) , \end{aligned}$$

(A.1.1)

and

$$\begin{aligned} \bar{s}-b_k\le \gamma \cdot \int _{b_k}^{\bar{s }}F_j(s) f_2(s) ds +\beta \cdot \left( 1-F_2(b_k)F_j(b_k)\right) , \end{aligned}$$

(A.1.2)

where equality must hold in A.1.1 for bids $b_j$ in the support of player $j$’s equilibrium strategy and in A.1.2 for $b_k$ in the support of player $k$’s equilibrium strategy.

By way of contradiction, assume that there exists some $b_0>0$ such that $F_1(b_0)\ne F_3(b_0)$. By Lemma 3 $F_1$ and $F_3$ are continuous everywhere on $(0,\bar{s}]$ and by Lemma 7 $F_1(\bar{s}_2)=F_3(\bar{s}_2)$. This implies that either there exists an interval $[x,y]\subset (0,\bar{s}_2]$ such that $F_1(x)=F_3(x)$, $F_1(y)=F_3(y)$, and $F_1(b)\ne F_3(b) \forall b\in (x,y)$, or there exists $\bar{x}>b_0$ such that $F_1(b)=F_3(b)\; \forall b\ge \bar{x}$ and $F_1(b)\ne F_3(b) \forall b\in [0,\bar{x})$.

Suppose that $[x,y]$ is an interval such that $F_1(x)=F_3(x)$, $F_1(y)=F_3(y)$, and $F_1(b) \ne F_3(b) \forall b\in (x,y)$. We treat the following four cases separately:

• 1. $x,y\in supp_j\cap supp_k$, where $supp_i$ denotes the support of player $i$’s equilibrium strategy. Without loss of generality let $F_j(b)>F_k(b)$ for all $b\in (x,y)$. In this case by (A.1.1) and (A.1.2) at $b=y$

  $$\begin{aligned} \gamma \cdot \int _{y}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(y)F_k(y)\right)&= \gamma \cdot \int _{y}^{\bar{s}}F_j(s) f_2(s) ds\\&\quad +\,\beta \cdot \left( 1-F_2(y)F_j(y)\right) \end{aligned}$$

  if and only if

  $$\begin{aligned} \gamma \cdot \int _{y}^{\bar{s }}(F_j(s)-F_k(s)) f_2(s) ds =\beta F_2(y)\cdot \underbrace{\left( F_j(y)-F_k(y)\right) }_{=0}=0 \end{aligned}$$

  By definition $\gamma <0$, hence

  $$\begin{aligned}\int _{y}^{\bar{s }}(F_j(s)-F_k(s)) f_2(s) ds =0. \end{aligned}$$

  Similarly, at $b=x$

  $$\begin{aligned} \int _{x}^{\bar{s }}(F_j(s)-F_k(s)) f_2(s) ds =0. \end{aligned}$$

  Then, by

  $$\begin{aligned} \int _{x}^{\bar{s }}(F_j(s)\!-\!F_k(s)) f_2(s) ds\!=\!\int _{x}^{y}(F_j(s)\!-\!F_k(s)) f_2(s) ds\!+\!\int _{y}^{\bar{s }}(F_j(s)\!-\!F_k(s)) f_2(s) ds \end{aligned}$$

  follows that

  $$\begin{aligned} \int _{x}^{y}(F_j(s)-F_k(s)) f_2(s) ds =0. \end{aligned}$$

  If $f_2(s)>0$ for any $s\in (x,y)$ this contradicts $F_j(s)>F_k(s) \forall s\in (x,y)$. If $f_2(s)=0$ for all $s\in (x,y)$, then by Lemma 8 $[x,y]\subseteq supp_j\cap supp_k$ and $F_2(x)=F_2(y)$. In this case (A.1.1) and (A.1.2) simplify to

  $$\begin{aligned} \bar{s}-b_j&=\gamma \cdot \int _{y}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(y)F_k(b_j)\right) , \hbox { and}\\ \bar{s}-b_k&=\gamma \cdot \int _{y}^{\bar{s }}F_j(s) f_2(s) ds +\beta \cdot \left( 1-F_2(y)F_j(b_k)\right) \end{aligned}$$

  respectively for all $b_j,b_k\in (x,y)$. Notice that in both expressions the integral is constant in the player’s own bid. Since $F_j$ and $F_k$ coincide at $x$ and $y$, $\int _{y}^{\bar{s }}F_k(s) f_2(s) ds= \int _{y}^{\bar{s }}F_j(s) f_2(s) ds$. This shows that $F_j(b)=F_k(b)\; \forall b\in (x,y)$, which contradicts our assumption.

• 2. $y\in supp_j\cap supp_k, x\in supp_j\backslash supp_k .$ Then by (A.1.1) and (A.1.2) at $b=x$

  $$\begin{aligned}&\gamma \cdot \int _{x}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(x)F_k(x)\right) \le \gamma \cdot \int _{x}^{\bar{s}}F_j(s) f_2(s) ds\\&\quad +\beta \cdot \left( 1-F_2(x)F_j(x)\right) \\&\Leftrightarrow \gamma \cdot \int _{x}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \le \beta F_2(x)\cdot \underbrace{\left( F_k(x)-F_j(x)\right) }_{=0}=0 \end{aligned}$$

  By definition $\gamma <0$, hence

  $$\begin{aligned}\int _{x}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \ge 0. \end{aligned}$$

  If $f_2(s)>0$ for any $s\in (x,y)$ this implies $F_j(s)< F_k(s)$, because by assumption $F_j(b)\ne F_k(b)\, \forall b\in (x,y)$ and by Lemma 3 (no mass points) $F_j$ and $F_k$ are continuous. By assumption $x\not \in supp_k$. So there exists an $\epsilon >0$ such that $F_k(x+\delta )=F_k(x)$ for all $\delta$ such that $0<\delta <\epsilon .$ But then

  $$\begin{aligned} F_j(x+\delta )<F_k(x+\delta )=F_k(x)=F_j(x), \end{aligned}$$

  which is a contradiction, because $F_j$ is a cumulative distribution function and as such is non-decreasing. If $f_2(s)=0\, \forall s\in (x,y)$, then from Lemma 8 $[x,y]\subseteq supp_j\cap supp_k$, a contradiction to the assumption $x\not \in supp_k$.

• 3. $x\in supp_j\cap supp_k, y\in supp_j\backslash supp_k .$ By (A.1.1) and (A.1.2) at $b=y$

  $$\begin{aligned}&\gamma \cdot \int _{y}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(y)F_k(y)\right) \le \gamma \cdot \int _{y}^{\bar{s}}F_j(s) f_2(s) ds\\&\quad +\beta \cdot \left( 1-F_2(y)F_j(y)\right) \\&\Leftrightarrow \gamma \cdot \int _{y}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \le \beta F_2(y)\cdot \underbrace{\left( F_k(y)-F_j(y)\right) }_{=0}=0 \end{aligned}$$

  By definition $\gamma <0$, hence

  $$\begin{aligned}\int _{y}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \ge 0. \end{aligned}$$

  By (A.1.1) and (A.1.2) at $b=x$

  $$\begin{aligned} \begin{array}{c} \displaystyle \gamma \cdot \int _{x}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(x)F_k(x)\right) =\gamma \cdot \int _{x}^{\bar{s}}F_j(s) f_2(s) ds\\ \displaystyle +\beta \cdot \left( 1-F_2(x)F_j(x)\right) \\ \displaystyle \Leftrightarrow \gamma \cdot \int _{x}^{\bar{s }}(F_j(s)-F_k(s)) f_2(s) ds =\beta F_2(x)\cdot \underbrace{\left( F_j(x)-F_k(x)\right) }_{=0}=0.\\ \end{array} \end{aligned}$$

  $\gamma <0$, hence

  $$\begin{aligned}&\int _{x}^{y}(F_j(s)-F_k(s)) f_2(s) ds+\underbrace{\int _{y}^{\bar{s }}(F_j(s)-F_k(s)) f_2(s) ds }_{\le 0} =0\\&\quad \Rightarrow \int _{x}^{y}(F_j(s)-F_k(s)) f_2(s) ds \ge 0 \end{aligned}$$

  If $f_2(s)>0$ for any $s\in (x,y)$, then this implies $F_j(s)\ge F_k(s)$. By assumption, $F_j(b)\ne F_k(b)\, \forall b\in (x,y)$, thus $F_j(b)> F_k(b)\, \forall b\in (x,y)$. By assumption $y\not \in supp_k$. Hence, there exists an $\epsilon >0$ such that $F_k(y-\delta )=F_k(y)\, \forall 0<\delta <\epsilon .$ But then

  $$\begin{aligned}F_j(y-\delta )>F_k(y-\delta )=F_k(y)=F_j(y),\end{aligned}$$

  a contradiction to the fact that $F_j$ is a cumulative distribution function and as such is non-decreasing. If $f_2(s)=0\, \forall s\in (x,y)$, then from Lemma 8 $[x,y]\subseteq supp_j\cap supp_k$, a contradiction to the assumption $y\not \in supp_k$.

• 4. $x\in supp_j\backslash supp_k, y\in supp_k\backslash supp_j .$ By (A.1.1) and (A.1.2) at $b=x$

  $$\begin{aligned}&\gamma \cdot \int _{x}^{\bar{s }}F_k(s) f_2(s) ds +\beta \cdot \left( 1-F_2(x)F_k(x)\right) \le \gamma \cdot \int _{x}^{\bar{s}}F_j(s) f_2(s) ds\\&\quad +\beta \cdot \left( 1-F_2(x)F_j(x)\right) \\&\Leftrightarrow \gamma \cdot \int _{x}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \le \beta F_2(x)\cdot \underbrace{\left( F_k(x)-F_j(x)\right) }_{=0}=0 \end{aligned}$$

  By definition $\gamma <0$, hence

  $$\begin{aligned}\int _{x}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \ge 0. \end{aligned}$$

  A similar argument shows that at $b=y$

  $$\begin{aligned}\int _{y}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \le 0. \end{aligned}$$

  Consequently,

  $$\begin{aligned} \begin{aligned} 0\le&\int _{x}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds \\=&\int _{x}^{y}(F_k(s)-F_j(s)) f_2(s) ds +\underbrace{\int _{y}^{\bar{s }}(F_k(s)-F_j(s)) f_2(s) ds}_{\le 0}\\ \Rightarrow&\int _{x}^{y}(F_k(s)-F_j(s)) f_2(s) ds \ge 0. \end{aligned} \end{aligned}$$

  If $f_2(s)>0$ for any $s\in (x,y)$, then this implies $F_k(s)\ge F_j(s)$. By assumption $F_j(b)\ne F_k(b)\, \forall b\in (x,y)$, thus $F_k(b)> F_j(b)\, \forall b\in (x,y)$. By assumption $x\not \in supp_k$. Hence, there exists an $\epsilon >0$ such that $F_k(x+\delta )=F_k(x)\, \forall 0<\delta <\epsilon .$ But then

  $$\begin{aligned}F_j(x+\delta )<F_k(x+\delta )=F_k(x)=F_j(x),\end{aligned}$$

  a contradiction to the fact that $F_j$ is a cumulative distribution function and as such is non-decreasing. If $f_2(s)=0 \,\forall s\in (x,y)$, then from Lemma 8 $[x,y]\subseteq supp_j\cap supp_k$, a contradiction to the assumption $x\not \in supp_k, y\not \in supp_j$.

Taking these four possible cases together, there cannot exist any interval $[x,y]$ with $F_1(x)=F_3(x)$, $F_1(y)=F_3(y)$, and $F_1(b)\ne F_3(b)\, \forall b\in (x,y)$.

Assume now that there exists an $\bar{x}>b_0$ such that $F_j(b)=F_k(b)\, \forall b\ge \bar{x}$ and $F_j(b)> F_k(b), \forall b\in [0,\bar{x})$. Players $1$ and $3$ must earn their equilibrium payoff at (or arbitrarily close to) zero, so by (A.1.1) and (A.1.2)

If $f_2(s)=0$ for all $s\in (0,\bar{x})$, then (A.1.3) simplifies to

$$\begin{aligned} -\beta F_2(0)\underbrace{\left[ F_j(0)-F_k(0)\right] }_{>0}=0.\end{aligned}$$

This implies $F_2(0)=0$, which is a contradiction, because Lemma 2 and $f_2(s)=0$ for all $s\in (0,\bar{x})$ imply $F_2(0)>0$.

If $f_2(s)>0$ for some $s\in (0,\bar{x})$, then $\beta >0$ and $\gamma <0$ imply that

$$\begin{aligned} \underbrace{\gamma \cdot \int _{0}^{\bar{x }}(F_j(s)-F_k(s)) f_2(s) ds}_{<0} -\underbrace{\beta F_2(0)\left[ F_j(0)-F_k(0)\right] }_{\ge 0}<0, \end{aligned}$$

a contradiction to (A.1.3). Consequently, there can exist no $b_0>0$ such that $F_1(b_0)\ne F_3(b_0)$.$\square$

Lemma 10

$F\equiv F_1=F_3$ first order stochastically dominates $F_2$.

Proof

If $\bar{s}_2=0$, then $F_2(x)=1\,\forall x\ge 0$. Hence, $F$ first order stochastically dominates $F_2$.

Therefore, assume in the following that $\bar{s}_2>0$. By way of contradiction assume that there exists some $b_0\in [0,\bar{s}_2)$ such that $F_2(b_0)<F(b_0)$. Note that by Lemmas 8 and 9 $supp_j=[0,\bar{s}], j\in \{1,3\}$. Furthermore, by Lemma 4 $F_2(\bar{s}_2)>F(\bar{s}_2)$ and by Lemma 3 no player’s equilibrium strategy has a mass point at any strictly positive bid. Then, there must exist an interval $[\underline{t},\bar{t}\,]\subseteq (0,\bar{s}_2]$ such that $[\underline{t},\bar{t}\,]\subseteq \bigcap _{i\in I} supp_i$, $F_2(\underline{t})<F(\underline{t})$, and $F_2(\bar{t})>F(\bar{t}).$ $[\underline{t},\bar{t}\,]\subseteq supp_2.$ Therefore, player 2 must earn his equilibrium expected payoff at any bid $x\in [\underline{t},\bar{t}\,];$ that is, for every $x\in [\underline{t},\bar{t}\;]$

$$\begin{aligned} u_2^*(x, F, F)&=v_{22}[F(x)]^2+v_{2j}\left( 1-[F(x)]^2\right) -x\\&=v_{2j}+\alpha [F(x)]^2-x\\&=v_{2j}+\alpha [F(0)]^2, \end{aligned}$$

where the last equality follows from Lemma 2. Hence,

Similarly, $[\underline{t},\bar{t}\,]\subseteq supp_j, j\in \{1,3\},$ implies that player $j$, $j\in \{1,3\}$, must earn his equilibrium expected payoff at any bid $x\in [\underline{t},\bar{t}\,]$. Player $j$’s expected payoff from a bid, $x\in [\underline{t},\bar{t}\,]$, is

$$\begin{aligned} u_j^*(x, F_2, F)=v_{jk}-\gamma \int _x^{\bar{s}_2}f_2(s)F(s)ds+\beta F(x)F_2(x)-x. \end{aligned}$$

Player $j$’s payoff must be constant on $[\underline{t},\bar{t}]$, that is,

$$\begin{aligned} \frac{du_j^*(x)}{dx}\!=\!\gamma F_2'(x)F(x)\!+\!\beta \left( F_2'(x)F(x)\!+\!F_2(x)F'(x)\right) -1=0 \quad \hbox { for all }\; x\in [\underline{t},\bar{t}\,]. \end{aligned}$$

This yields the following linear first order differential equation, which must hold for all $x\in [\underline{t},\bar{t}\,]$

Since $F$ takes the form described in (A.1.4), the solution to (A.1.5) is

$$\begin{aligned}F_2(x)=\frac{2\alpha }{\alpha +\beta }F(x)+c\cdot [F(x)]^{-\frac{\beta }{\alpha }},\end{aligned}$$

where $c\in \mathbb {R}$ is a constant of integration.

By assumption $\beta >\alpha$, thus there exists a $\delta >0$ such that $\beta =(1+\delta )\alpha$ and we can write

By differentiating (A.1.6) we obtain

Suppose $F(\bar{t})<F_2(\bar{t})$, then by continuity of the equilibrium strategies (Lemma 3) $F(\bar{t}-\epsilon )<F_2(\bar{t}-\epsilon )$ for sufficiently small $\epsilon >0$. Considering $x=\bar{t}-\epsilon$ in (A.1.6) yields the necessary condition $c>0.$ Using this in (A.1.7) shows that $F_2'(x)<F'(x)$ for $x\in [\underline{t},\bar{t}]$. Hence, $F(\underline{t})<F_2(\underline{t}),$ a contradiction to the assumption that $F(\underline{t})>F_2(\underline{t}).$ Therefore, there exists no point $b_0\in [0,\bar{s}_2]$ such that $F_2(b_0)<F(b_0)$, and $F$ first order stochastically dominates $F_2$. $\square$

Lemma 11

$F_2(x)=1 \quad {\text{ for } \text{ all }} \; x\ge 0$

Proof

Lemmas 2, 9, and 10 together imply $F(0)=0$, hence by Lemma 2 player 2’s expected payoff in equilibrium is $v_{2j}$. By way of contradiction assume that $\bar{s}_2>0$. Then, by the same argument as in the proof of Lemma 10 equation (A.1.6) must hold at every $x\in supp_2$ with $F(x)=\left( \frac{x}{\alpha }\right) ^\frac{1}{2}$. Player 2 may not randomize over strictly positive bids arbitrarily close to zero. Indeed, if such randomization did occur, because all players’ equilibrium strategies are continuous over $(0,\bar{s}]$ by Lemma 3, $F(0)=0$ and therefore

$$\begin{aligned} \lim _{\epsilon \rightarrow 0} F(\epsilon )^{-(1+\delta )}=\infty , \end{aligned}$$

and $F_2(0)<1$ (under the assumption that $\bar{s}_2>0$), then (A.1.6) would imply that $c=0$, which is a contradiction to Lemma 10. Given that player 2 does not randomize over strictly positive bids arbitrarily close to zero, there exists a $\underline{t}>0$ such that $\underline{t}=\inf \{t>0\,| t\in supp_2\}.$ Then, $F_2(\underline{t})=F_2(0)$. By (A.1.1) player $j$’s expected payoff from a bid $x\in (0,\underline{t}]$ is

$$\begin{aligned} u_j^*(x,F_2,F)=v_{jj}-\gamma \int _{x}^{\bar{s}}F(y)f_2(y)dy-\beta (1-F_2(0)F(x))-x.\end{aligned}$$

$x\in (0,\underline{t}]$ is a best response for player $j$ therefore $u_j^*(x,F_2,F)$ must be constant over $(0,\underline{t}]$. It follows that $F'(x)=\frac{1}{F_2(0)\beta }$ for $x\in (0,\underline{t}]$. From $F(0)=0$ follows that players $1$ and $3$ randomize uniformly over $[0,\underline{t}]$ according to

$$\begin{aligned}F(x)=\frac{x}{F_2(0)\beta },\, x\in [0,\underline{t}].\end{aligned}$$

Continuity of $F$ at $\underline{t}$ yields

$$\begin{aligned}\frac{\underline{t}}{F_2(0)\beta }=\left( \frac{\underline{t}}{\alpha }\right) ^\frac{1}{2}\Leftrightarrow \underline{t}=\frac{\beta ^2}{\alpha }\cdot [F_2(0)]^2.\end{aligned}$$

Consequently,

$$\begin{aligned}F(\underline{t})=(1+\delta )F_2(0).\end{aligned}$$

Using this and $F_2(\underline{t})=F_2(0)$ in (A.1.6) yields

$$\begin{aligned} F_2(0)&=F_2(\underline{t})\\&=\frac{2}{2+\delta }F(\underline{t})+c\cdot [F(\underline{t})]^{-(1+\delta )}\\&=\frac{2}{2+\delta }(1+\delta )F_2(0)+c\cdot [(1+\delta ) F_2(0)]^{-(1+\delta )}, \end{aligned}$$

which implies

$$\begin{aligned}c=\left( -\frac{\delta }{2+\delta }F_2(0)\right) \left[ (1+\delta )F_2(0)\right] ^{1+\delta }\le 0.\end{aligned}$$

This contradicts Lemma 10; therefore $\bar{s}_2=0$.$\square$

Altogether, this shows that player 2 stays out of the conflict in equilibrium. Hence, the equilibrium described in Proposition 1 is the unique equilibrium of $\varGamma _{21}$.

Appendix 2: Proof of Proposition 5

Proof

Under the assumption that all three players make positive bids with strictly positive probability and players $1$ and $3$ use identical strategies, i.e. $F_1=F_3=:F$, we know that $\underline{s}_1=\underline{s}_2=\underline{s}_3=0$ and $\bar{s}_2=\bar{s}_1=\bar{s}_3=: \bar{s}$. Moreover, $\bar{s}\in (r_{jk},r_{2j})$, $j,k\in \{1,3\}, j\ne k,$ and player $2$ cannot have a masspoint at zero. Assume that all players randomize continuously over $[0,\bar{s}]$. All players must earn their equilibrium payoff at $\bar{s}$, therefore player $2$’s expected payoff from a bid $b\in (0,\bar{s}]$, $u_2(b,F)=v_{22}[F(b)]^2+v_{2i}(1-[F(b)]^2)$, must be $v_{22}-\bar{s}$. This yields

$$\begin{aligned}F(x)=\left\{ \!\! \begin{array}[]{lll} 0&{}\quad x<0\\ \left[ \left( 1-\frac{\bar{s}}{r_{2j}}\right) +\frac{x}{r_{2j}}\right] ^{\frac{1}{2}}&{}\quad 0\le x\le \bar{s}\\ 1&{}\quad x>\bar{s} \end{array}\right. .\end{aligned}$$

Player $j$’s payoff must be $v_{jj}-\bar{s}$. Moreover, player $j$ chooses his equilibrium strategy such that his expected payoff, $u_j(b,F_2,F)=-b+v_{j2}+[v_{jj}-v_{j2}]F(b)F_2(b)+[v_{jk}-v_{j2}]\int _b^{\bar{s}}F_2(s)F'(s)ds$, is maximized. The first order condition yields the first order differential equation

$$\begin{aligned}0=F(x)F_2'(x)r_{j2}+F'(x)F_2(x)r_{jk}-1.\end{aligned}$$

Using the boundary conditions $F_2(0)=0$ and $F_2(\bar{s})=1$ this yields

$$\begin{aligned}F_2(x)=\kappa F(x)-(\kappa -1)F(x)^{-\frac{r_{jk}}{r_{j2}}}\end{aligned}$$

with $\kappa =\frac{2r_{j2}}{r_{j2}+r_{jk}}>1$ and $\bar{s}=r_{j2}\left[ 1-\left( 1-\frac{1}{\kappa }\right) ^\kappa \right]$. Note that $\bar{s}\in (r_{jk},r_{j2})$ and $F_2$ is strictly increasing.

In order to show that this equilibrium exhibits extremism, we need to show that $F_2(x)\le F(x)\, \forall x$. All players’ cdfs coincide for $x<0$ and $x\ge \bar{s}$. The centrist players put strictly positive mass on zero, thus $F_2(0)<F(0)$. For $x\in (0,\bar{s})$,

$$\begin{aligned}F_2(x)=\kappa F(x)-(\kappa -1)F(x)^{-\frac{r_{13}}{r_{12}}}=F(x)\underbrace{\Bigl [\kappa -(\kappa -1)\overbrace{F(x)^{-\left( 1+\frac{r_{13}}{r_{12}}\right) }}^{>1}\Bigr ]}_{<\kappa -(\kappa -1)=1}<F(x).\end{aligned}$$

Therefore, $F_2$ first order stochastically dominates $F$.$\square$