Generalizations of the General Lotto and Colonel Blotto games

Abstract

In this paper, we generalize the General Lotto game (budget constraints satisfied in expectation) and the Colonel Blotto game (budget constraints hold with probability one) to allow for battlefield valuations that are heterogeneous across battlefields and asymmetric across players and for the players to have asymmetric resource constraints. We completely characterize Nash equilibrium in the generalized version of the General Lotto game and find that there exist sets of nonpathological parameter configurations of positive Lebesgue measure with multiple payoff nonequivalent equilibria. Across equilibria each player achieves a higher payoff when he more aggressively attacks battlefields in which he has lower relative valuations. Hence, the best defense is a good offense. We, then, show how this characterization can be applied to identify equilibria in the Colonel Blotto version of the game.

Appendices

Appendix 1

Appendix 1 contains the remaining two parts of the proof of Theorem 1: (i) for each equilibrium of the GGL game there exists a corresponding solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$) and (ii) for each solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$) each player in the GGL game has a unique set of Nash equilibrium univariate marginal distributions. We begin with the proof of part (i) and then conclude with the proof of part (ii).

The proof of the converse claim in Theorem 1, that for each equilibrium of the GGL game there exists a corresponding solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$), extends the arguments in Hart (2008) on the continuous General Lotto game and Hart (2016) on the relationship between the all-pay auction and the continuous General Lotto game. We begin by noting that the standard constant-sum continuous General Lotto game, denoted $L\{X_A, X_B\}$, is a special case of the GGL game in which $n=1$, $v_A=v_B= 1$, and a strategy is a univariate distribution function denoted $F_i$, for $i=A,B$, with $E_{F_i}(x)\le X_i$. Let $\widetilde{x}_i$ denote the realization of a random variable distributed according to the distribution function $F_i$. Player A’s expected payoff in the General Lotto game is given by

$$\begin{aligned} \pi _A(F_A, F_{B})={\mathrm{Pr}}(\widetilde{x}_A>\widetilde{x}_B) +\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_A=\widetilde{x}_B) \end{aligned}$$

and player B’s expected payoff is given by

$$\begin{aligned} \pi _B(F_B, F_{A})={\mathrm{Pr}}(\widetilde{x}_B>\widetilde{x}_A)+\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_B=\widetilde{x}_A)=1 -{\mathrm{Pr}}(\widetilde{x}_A>\widetilde{x}_B)-\frac{1}{2} {\mathrm{Pr}}(\widetilde{x}_A=\widetilde{x}_B). \end{aligned}$$

In this constant-sum game, player A chooses $F_A$ to maximize ${\mathrm{Pr}}(\widetilde{x}_A>\widetilde{x}_B)+\frac{1}{2} {\mathrm{Pr}}(\widetilde{x}_A=\widetilde{x}_B)$ and player B chooses $F_B$ to minimize ${\mathrm{Pr}}(\widetilde{x}_A>\widetilde{x}_B) +\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_A=\widetilde{x}_B)$.

Equilibrium in the (continuous) General Lotto game with strictly positive budgets is characterized by Sahuguet and Persico (2006) and Hart (2008). The following theorem extends that characterization to allow for one or both of the players to have a budget of 0. Unlike the case of $X_B\ge X_A>0$, if either $X_A=0$ and $X_B> 0$ or $X_A>0$ and $X_B=0$, then there are multiple equilibria.¹⁹ However, because the game is constant sum, the equilibrium expected payoffs are unique for all possible resource endowments $(X_A, X_B)$.

Theorem 2

For the General Lotto game $L\{X_A, X_B\}$ with $X_B\ge X_A>0$, the unique equilibrium strategies are

$$\begin{aligned} F_A(x)= & \left( 1-\frac{X_A}{X_B} \right) +\frac{x\cdot X_A}{2X_B^2} \quad {\mathrm{for}}\ x\in [0,2X_B]\\ F_B(x)= & \frac{x}{2X_B}\quad {\mathrm{for}}\ x\in [0,2X_B] \end{aligned}$$

and the equilibrium expected payoffs are $\frac{X_A}{2X_B}$ for player A and $1-\frac{X_A}{2X_B}$ for player B.

For the General Lotto game $L\{X_A, X_B\}$ with $X_B=0$ and/or $X_A=0$ :

• 1. If $X_A=0$ and $X_B> 0$, then the unique equilibrium expected payoffs are 0 for player A and 1 for player B.

• 2. If $X_A>0$ and $X_B=0$, then the unique equilibrium expected payoffs are 1 for player A and 0 for player B.

• 3. If $X_A=0$ and $X_B=0$, then the unique equilibrium strategies are $F_{A,j}^*(0)=F_{B,j}^*(0)=1$ and the equilibrium expected payoffs are $\frac{1}{2}$ for player A and $\frac{1}{2}$ for player B.

In moving from the General Lotto game $L\{X_A, X_B\}$ to the GGL game $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, recall that in the GGL game a strategy is an n-variate distribution function, $P_i$ for $i=A,B$, that satisfies the constraint that $\sum _{j=1}^nE_{F_{i,j}}(x)\le X_i$, where $F_{i,j}$ is the univariate marginal distribution of $P_i$ for battlefield j. Let $\widetilde{x}_{i,j}$ denote the realization of a random variable distributed according to the univariate marginal distribution $F_{i,j}$. Then, given the strategy profile $(P_A, P_B)$, player A’s expected payoff is given by

$$\begin{aligned} \pi _A(P_A, P_{B})=\sum _{j=1}^n v_{A,j}\left( {\mathrm{Pr}}(\widetilde{x}_{A,j} >\widetilde{x}_{B,j})+\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_{A,j} =\widetilde{x}_{B,j}) \right) \end{aligned}$$

and player B’s expected payoff is given by

$$\begin{aligned} \pi _B(P_B, P_{A})=\sum _{j=1}^n v_{B,j}\left( 1- {\mathrm{Pr}}(\widetilde{x}_{A,j} >\widetilde{x}_{B,j})-\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_{A,j} =\widetilde{x}_{B,j}) \right) . \end{aligned}$$

Given an equilibrium $(P_A^*, P_B^*)$, let $X_{i,j}^*\equiv E_{F_{i,j}^*}(x)$ for $i=A,B$ denote player i’s expected allocation of the resource to battlefield j under the strategy $P_i^*$.

Lemma 1

If $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then within each battlefield j, $(F_{A,j}^*, F_{B,j}^*)$ is an equilibrium of $L(X_{A,j}^*, X_{B,j}^*)$.

Proof

If $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then there are no payoff-increasing deviations for either player. But one feasible type of deviation for player i is to hold constant $X_{i,j}^*$ on each battlefield j and choose a feasible deviation $\widehat{P}_i$ with the set of univariate marginals $\{\widehat{F}_{i,j}\}_{j=1}^n$ with $E_{\widehat{F}_{i,j}}(x)=X_{i,j}^*$ for all j. Let $\widehat{x}_{i,j}$ denote the realization of a random variable distributed according to the univariate marginal distribution function $\widehat{F}_{i,j}$. Because in battlefield j each player i does not have a payoff increasing deviation $\widehat{F}_{i,j}$ with $E_{\widehat{F}_{i,j}}(x)=X_{i,j}^*$, it follows that

$$\begin{aligned} v_{i,j}\left( {\mathrm{Pr}}(\widetilde{x}_{i,j}>\widetilde{x}_{-i,j}) +\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_{i,j}=\widetilde{x}_{-i,j})\right) \ge v_{i,j}\left( {\mathrm{Pr}}(\widehat{x}_{i,j}>\widetilde{x}_{-i,j}) +\frac{1}{2}{\mathrm{Pr}}(\widehat{x}_{i,j}=\widetilde{x}_{-i,j})\right) \end{aligned}$$

(17)

for all possible univariate marginal distributions $\widehat{F}_{i,j}$ with $E_{\widehat{F}_{i,j}}(x)=X_{i,j}^*$. But it follows directly from (17) that

$$\begin{aligned} \left( {\mathrm{Pr}}(\widetilde{x}_{i,j}>\widetilde{x}_{-i,j}) +\frac{1}{2}{\mathrm{Pr}}(\widetilde{x}_{i,j}=\widetilde{x}_{-i,j})\right) \ge \left( {\mathrm{Pr}}(\widehat{x}_{i,j}>\widetilde{x}_{-i,j}) +\frac{1}{2}{\mathrm{Pr}}(\widehat{x}_{i,j}=\widetilde{x}_{-i,j}) \right) \end{aligned}$$

(18)

for all possible deviations $\widehat{F}_{i,j}$ with $E_{\widehat{F}_{i,j}}(x)=X_{i,j}^*$, and, thus, $(F_{A,j}^*, F_{B,j}^*)$ is an equilibrium of $L(X_{A,j}^*, X_{B,j}^*)$. $\square$

To complete the proof of the claim that if $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then there exists a corresponding solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$), Lemmas 2-4 collectively establish that in any equilibrium $(P_A^*, P_B^*)$ of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$ it must be the case that $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j. Because $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j, it follows from Lemma 1 and Theorem 2 that the equilibrium univariate marginal distributions are uniquely determined. Using the unique equilibrium univariate marginal distributions, Lemma 5 completes the proof that there exists a corresponding solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$).

Lemma 2

If $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then

$$\begin{aligned} \max \{X^*_{A,j},X^*_{B,j}\}>0\quad {\mathrm{for}}\ {\mathrm{all}}\ j. \end{aligned}$$

Proof

By way of contradiction, suppose that there exists an equilibrium $(P_A^*, P_B^*)$ in which for some battlefield $k\ \max \{X^*_{A,k},X^*_{B,k}\}=0$, which implies that $F_{A,k}^*(0)=F_{B,k}^*(0)=1$. We begin with the case in which $\sum _{j=1}^nE_{F^*_{A,j}}(x)<X_{A}$ and then examine the case in which $\sum _{j=1}^nE_{F^*_{A,j}}(x)=X_{A}$. If $\sum _{j=1}^nE_{F^*_{A,j}}(x)<X_{A}$, then player A can increase his payoff by $\frac{v_{A,k}}{2}$ by allocating a strictly positive level of the resource $X_{A,k}\le X_{A}-\sum _{j=1}^nE_{F^*_{A,j}}(x)$ to battlefield k and setting $F_{A,k}(0)=0$, a contradiction.

For $\sum _{j=1}^nE_{F^*_{A,j}}(x)=X_{A}>0$, there exists at least one battlefield $j'$ in which $X_{A,j'}^*>0$ and there are two cases to consider: (i) $X_{B,j'}^*=0$ and (ii) $X_{B,j'}^*>0$. In case (i), because $\sum _{j=1}^nE_{F^*_{A,j}}(x)=X_{A}>0$ and in battlefield $j'\ X_{A,j'}^*>0$ and $X_{B,j'}^*=0$, player A can increase his payoff by $\frac{v_{A,k}}{2}$ by shifting $X_{A,k}<X_{A,j'}^*$ of the resource from battlefield $j'$ to battlefield k and setting $F_{A,k}(0)=F_{A,j'}(0)=0$, a contradiction.

In case (ii), $X_{A,j'}^*>0$ and $X_{B,j'}^*>0$, and it follows from Lemma 1 and Theorem 2 that $F_{B,j'}^*(x)$ is the unique equilibrium strategy in the General Lotto game $L(X_{A,j'}^*, X_{B,j'}^*)$, where the support of $F_{B,j'}^*(x)$, denoted ${\mathrm{supp}}(F_{B,j'}^*(x))$, is $[0, 2\max \{X_{A,j'}^*, X_{B,j'}^*\}]$. Thus, player A can increase his total expected payoff by an amount arbitrarily close to $\frac{v_{A,k}}{2}$ by shifting, for a sufficiently small $\epsilon >0$, $\epsilon$ of the resource from battlefield $j'$ to battlefield k, in battlefield k choosing a distribution function $F_{A,k}(x)$ with $F_{A,k}(0)=0$ and $E_{F_{A,k}}(x)=\epsilon$, and in battlefield $j'$ choosing a distribution function $F_{A,j'}(x)$ with $F_{A,j'}(0)=0$, $E_{F_{A,j'}}(x)=X^*_{A,j'}-\epsilon$, and ${\mathrm{supp}}(F_{A,j'})\subseteq {\mathrm{supp}}(F_{B,j'}^*(x))$. In battlefield $j'$ player A’s expected payoff from the distribution function $F_{A,j'}(x)$ when player B’s distribution function is $F_{B,j'}^*(x)$ is given by

$$\begin{aligned} v_{A,j'}\int _{0}^\infty F_{B,j'}^*(x) \hbox {d}F_{A,j'}(x) ={\left\{ \begin{array}{ll} v_{A,j'}\left( \left( 1 -\frac{X^*_{B,j'}}{X^*_{A,j'}}\right) +\frac{\left( X_{A,j'}^*-\epsilon \right) X^*_{B,j'}}{2\left( X^*_{A,j'}\right) ^2}\right) & \ {\mathrm{if}}\ X^*_{A,j'}> X^*_{B,j'} \\ v_{A,j'}\left( \frac{\left( X_{A,j'}^*-\epsilon \right) }{2X^*_{B,j'}}\right) & \ {\mathrm{if}}\ X^*_{A,j'}\le X^*_{B,j'} \end{array}\right. } \end{aligned}$$

Thus, the loss in player A’s payoff in battlefield $j'$ approaches 0 as $\epsilon$ approaches 0, but the gain on battlefield k is $\frac{v_{A,k}}{2}$ for all $\epsilon >0$. This is a contradiction to the assumption that $(P_A^*, P_B^*)$ is an equilibrium and completes the proof that if $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$ then $\max \{X^*_{A,j},X^*_{B,j}\}>0$ for all j. $\square$

Lemma 3

If $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then $\sum _{j=1}^nE_{F^*_{i,j}}(x)>0$ for each player $i=A,B$.

Proof

By way of contradiction, suppose that there exists an equilibrium $(P_A^*, P_B^*)$ in which $\sum _{j=1}^nE_{F^*_{i,j}}(x)=0$ for some player i. From Lemma 2, it follows that for player $-i$, $X^*_{-i,j}>0$ for all j, which from Lemma 1 and Theorem 2 implies that player i earns an equilibrium expected payoff of 0. If $i=B$, then because $X_B\ge X_A>0$, it is clear that player B has a payoff increasing deviation that involves mimicking player A’s strategy, which yields B an expected payoff of $\frac{1}{2}\sum _{j=1}^n v_{B,j}$. Hence a contradiction. If $i=A$, then player A can mimic player B’s strategy with probability $\frac{X_{A}}{X_{B}}$ and bid 0 in every battlefield with probability $\left(1-\frac{X_{A}}{X_{B}}\right)$, which similarly yields A an expected payoff of $\frac{X_A}{2X_B}\sum _{j=1}^n v_{A,j}$. This yields a contradiction and completes the proof. $\square$

Lemma 4

If $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then

$$\begin{aligned} \min \{X^*_{A,j},X^*_{B,j}\}>0\quad {\mathrm{for}}\ {\mathrm{all}}\ j. \end{aligned}$$

Proof

By way of contradiction, suppose that there exists an equilibrium $(P_A^*, P_B^*)$ in which there is at least one battlefield k with $\min \{X^*_{A,k},X^*_{B,k}\}=0$. There are two cases to consider: (i) $\min \{X^*_{A,j},X^*_{B,j}\}=0$ for all j or (ii) $\min \{X^*_{A,j},X^*_{B,j}\}=0$ for at least one, but not all j. Beginning with case (i), because $\min \{X^*_{A,j},X^*_{B,j}\}=0$ for all j, from Lemma 3$\sum _{j=1}^nE_{F^*_{i,j}}(x)>0$ for each player i, and from Lemma 2$\max \{X^*_{A,j},X^*_{B,j}\}>0$ for all j, there exists at least one battlefield $j'$ with $X_{A,j'}^*>0$ and $X_{B,j'}^*=0$ and at least one battlefield $j''$ with $X_{A,j''}^*=0$ and $X_{B,j''}^*>0$. But, player B can strictly increase his total expected payoff by decreasing $X_{B,j''}^*$ by an $\epsilon \in (0, \min \{X_{B,j''}^*, X_{A,j'}^*\})$, allocating $\epsilon$ to battlefield $j'$, and utilizing a univariate marginal distribution on battlefield $j'$ that places mass $\left( 1-\frac{\epsilon }{X_{A,j'}^*} \right)$ on 0 and randomizes according to $F_{A,j'}^*$ with probability $\frac{\epsilon }{X_{A,j'}^*}$. Such a deviation would increase player B’s expected payoff on battlefield $j'$ by $\frac{\epsilon v_{B,j'}}{2 X_{A,j'}^*}$ with no decrease in the expected payoff on battlefield $j''$, a contradiction.

For case (ii), if $\min \{X^*_{A,j},X^*_{B,j}\}=0$ for at least one, but not all j, then there exists at least one battlefield $j'$ with $\min \{X^*_{A,j'},X^*_{B,j'}\}>0$ and at least one battlefield k with $\min \{X^*_{A,k},X^*_{B,k}\}=0$. Because from Lemma 2$\max \{X^*_{A,k},X^*_{B,k}\}>0$, there exists a player i with $X_{i,k}>0$ and a player $-i$ with $X_{-i,k}=0$. Then, because from Lemma 1, each player’s unique equilibrium univariate marginal distribution in battlefield $j'$ is given by Theorem 2, player i has a payoff increasing deviation that involves shifting $\epsilon \in (0,X_{i,k}^*)$ of the resource from battlefield k to battlefield $j'$, in battlefield k choosing a distribution function $F_{i,k}(x)$ with $F_{i,k}(0)=0$ and $E_{F_{i,k}}(x)=X_{i,k}^*-\epsilon$, in battlefield $j'$ choosing a distribution function $F_{i,j'}(x)$ with $F_{i,j'}(0)=0$, $E_{F_{i,j'}}(x)=X_{i,j'}^*+\epsilon$, and ${\mathrm{supp}}(F_{i,j'})\subseteq {\mathrm{supp}}(F_{-i,j'}^*(x))$. Such a deviation results in no loss to player i’s expected payoff in battlefield k. In battlefield $j'$, player i’s expected payoff from the distribution function $F_{i,j'}(x)$ when player $-i$’s distribution function is $F_{-i,j'}^*(x)$ is given by

$$\begin{aligned} v_{i,j'}\int _{0}^\infty F_{-i,j'}^*(x) \hbox {d}F_{i,j'}(x) ={\left\{ \begin{array}{ll} v_{i,j'}\left( \left( 1 - \frac{X^*_{-i,j'}}{X^*_{i,j'}}\right) +\frac{\left( X_{i,j'}^*+\epsilon \right) X^*_{-i,j'}}{2\left( X^*_{i,j'}\right) ^2}\right) & \ {\mathrm{if}}\ X^*_{i,j'}> X^*_{-i,j'} \\ v_{i,j'}\left( \frac{\left( X_{i,j'}^*+\epsilon \right) }{2X^*_{-i,j'}}\right) &\ {\mathrm{if}}\ X^*_{i,j'}\le X^*_{-i,j'} \end{array}\right. }. \end{aligned}$$

Thus, for all $\epsilon \in (0,X_{i,k}^*)$ player i’s expected payoff in battlefield $j'$ is strictly higher under the deviation, and there is no loss to player i’s expected payoff in battlefield k. This is a contradiction and completes the proof that if $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$ then $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j. $\square$

Lemma 5

If $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, then there exists a corresponding solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$).

Proof

From Lemma 4, $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j. Then, because $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$ and $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j, it follows from Lemma 1 that in each battlefield j the players’ unique equilibrium univariate marginal distributions are given by Theorem 2. Because, the unique equilibrium univariate marginal distributions given by Theorem 2 are linear, it follows that for player A any deviation $P_A$ that satisfies the following two conditions is payoff maximizing and feasible: (i) in each battlefield j the associated univariate marginal distribution function $F_{A,j}(x)$ satisfies $F_{A,j}(0)=0$ if $X_{A,j}^*>X_{B,j}^*$ and ${\mathrm{supp}}(F_{A,j})\subseteq {\mathrm{supp}}(F_{B,j}^*(x))$, and (ii) across battlefields $\sum _{j=1}^nE_{F_{A,j}}(x)=X_A$. Letting $X_{A,j}=E_{F_{A,j}}(x)$, player A’s total expected payoff from such a joint distribution function $P_A$, given that player B is using the joint distribution function $P_B^*$, is given by

$$\begin{aligned} \pi _A(P_A, P_B^*)&=\sum _{j=1}^n v_{A,j} \int _{0}^\infty F_{B,j}^*(x) \hbox {d}F_{A,j}(x) \\&=\sum _{j|X_{A,j}^*>X_{B,j}^*}v_{A,j}\left[ \left( 1-\frac{X^*_{B,j}}{X^*_{A,j}}\right) +\frac{X_{A,j}X^*_{B,j}}{2(X^*_{A,j})^2}\right] +\sum _{j|X_{A,j}^*\le X_{B,j}^*} v_{A,j}\left( \frac{X_{A,j}}{2X_{B,j}^*}\right) . \end{aligned}$$

(19)

Similarly, for player B the maximum achievable total expected payoff from a feasible deviation $P_B$ with $\{X_{B,j}\}_{j=1}^n$ is given by

$$\begin{aligned} \pi _B(P_B, P_A^*)&=\sum _{j=1}^n v_{B,j}\int _{0}^\infty F_{A,j}^*(x) \hbox {d}F_{B,j}(x) \\&=\sum _{j|X_{A,j}^*\ge X_{B,j}^*}v_{B,j}\left( \frac{X_{B,j}}{2X_{A,j}^*}\right) +\sum _{j|X_{A,j}^*< X_{B,j}^*}v_{B,j}\left[ \left( 1-\frac{X^*_{A,j}}{X^*_{B,j}}\right) +\frac{X_{B,j}X^*_{A,j}}{2(X^*_{B,j})^2}\right] \end{aligned}$$

(20)

Because $(P_A^*, P_B^*)$ is an equilibrium of $\hbox {GGL}(X_A, X_B, n, \{v_{A,j}, v_{B,j}\}_{j=1}^n)$, it must be the case that player A is maximizing Eq. (19) with respect to $\{X_{A,j}\}_{j=1}^n$ and player B is maximizing Eq. (20) with respect to $\{X_{B,j}\}_{j=1}^n$. Then, because Eqs. (19) and (20) are concave and continuously differentiable with respect to $\{X_{i,j}\}_{j=1}^n\in {\mathbb {R}}^n_{+}$, $i=A,B$, respectively, it follows that there exists a Lagrange multiplier $\lambda _i^*\ge 0$ such that the Kuhn–Tucker first-order conditions hold:²⁰

$$\begin{aligned} {\left\{ \begin{array}{ll} v_{i,j}\left( \frac{1}{2X_{-i,j}^*}\right) -\lambda _i^* =0 &\ {\mathrm{in}} \ {\mathrm{each}}\ {\mathrm{battlefield}}\ j\ {\mathrm{with}}\ X^*_{i,j}\le X_{-i,j}^* \\ v_{i,j}\left( \frac{X_{-i,j}^*}{2(X_{i,j}^*)^2}\right) -\lambda _i^*=0 &\ {\mathrm{in}}\ {\mathrm{each}}\ {\mathrm{battlefield}}\ j\ {\mathrm{with}}\ X^*_{i,j}> X_{-i,j}^* \end{array}\right. }, \end{aligned}$$

(21)

with complementary slackness condition $\lambda _i^*\ge 0$, $\sum _{j=1}^n X_{i,j}^*\le X_i$, and $\lambda _i^*\left( \sum _{j=1}^n X_{i,j}^*- X_i\right) =0$. Complementary slackness is clearly satisfied because from (19) and (20) it is clearly suboptimal to set $\sum _{j=1}^nX_{i,j}<X_i$.

From the first-order conditions in (21) we see that in each battlefield j with $X_{i,j}^*>X_{-i,j}^*$: (i) $X_{i,j}^*=\frac{v_{-i,j}}{2\lambda _{-i}^*}$ and (ii) $X_{i,j}^*=\left( \frac{v_{i,j}X_{-i,j}^*}{2\lambda _i^*}\right) ^{1/2}$ or equivalently $X_{-i,j}^*=\frac{\left( v_{-i,j}/2\lambda _{-i}^* \right) ^2}{\left( v_{i,j}/2\lambda _{i}^* \right) }$. Combining (i) and (ii), it follows from budget balance that $\lambda _A^*$ and $\lambda _B^*$ solve

$$\begin{aligned} \sum _{j|X_{A,j}^*>X_{B,j}^*} \frac{v_{B,j}}{2\lambda _B^*} +\sum _{j|X_{A,j}^*\le X_{B,j}^*}\frac{\left( \frac{v_{A,j}}{\lambda _A^*}\right) ^2}{2\left( \frac{v_{B,j}}{\lambda _B^*}\right) }=X_A \end{aligned}$$

(22)

and

$$\begin{aligned} \sum _{j|X_{A,j}^*\ge X_{B,j}^*} \frac{\left( \frac{v_{B,j}}{\lambda _B^*}\right) ^2}{2\left( \frac{v_{A,j}}{\lambda _A^*}\right) }+\sum _{j|X_{A,j}^*< X_{B,j}^*} \frac{v_{A,j}}{2\lambda _A^*}=X_B \end{aligned}$$

(23)

Because $X_{A,j}^*=\frac{v_{B,j}}{2\lambda _{B}^* }$ and $X_{B,j}^*=\frac{\left( v_{B,j}/2\lambda _{B}^* \right) ^2}{\left( v_{A,j}/2\lambda _{A}^* \right) }$ when $X_{A,j}^* >X_{B,j}^*$, and $X_{A,j}^*=\frac{\left( v_{A,j}/2\lambda _{A}^* \right) ^2}{\left( v_{B,j}/2\lambda _{B}^* \right) }$ and $X_{B,j}^*=\frac{v_{A,j}}{2\lambda _{A}^*}$ when $X_{A,j}^* \le X_{B,j}^*$, it follows that $\frac{v_{A,j}}{\lambda _{A}^*} >\frac{v_{B,j}}{\lambda _{B}^*}$ if and only if $X_{A,j}^* >X_{B,j}^*$. Thus, the system (22) and (23) is equivalent to system ($\star$).

This completes the proof of part (i), for each equilibrium of the GGL game, there exists a corresponding solution $(\lambda _A^*, \lambda _B^*)$ to system ($\star$).

We now conclude with the proof of part (ii), for each solution $(\lambda _A^*, \lambda _B^*)$ to system $(\star )$ each player in the GGL game has a unique set of Nash equilibrium univariate marginal distributions. From the argument utilized in the proof of Lemma 5, it follows that for each solution $(\lambda _A^*, \lambda _B^*)$ each player i’s n-tuple of the expected allocation of the resource to each of the n battlefields, $\{X_{i,j}^* \}_{j=1}^n$, is uniquely determined. Namely, $X_{A,j}^*=\frac{v_{B,j}}{2\lambda _{B}^* }$ and $X_{B,j}^*=\frac{\left( v_{B,j}/2\lambda _{B}^* \right) ^2}{\left( v_{A,j}/2\lambda _{A}^* \right) }$ when $X_{A,j}^* >X_{B,j}^*$, and $X_{A,j}^*=\frac{\left( v_{A,j}/2\lambda _{A}^* \right) ^2}{\left( v_{B,j}/2\lambda _{B}^* \right) }$ and $X_{B,j}^*=\frac{v_{A,j}}{2\lambda _{A}^*}$ when $X_{A,j}^* \le X_{B,j}^*$. From Lemma 4, each player’s expected allocation of the resource to each battlefield is strictly positive, $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j. Then, because $\{X_{A,j}^*, X_{B,j}^* \}_{j=1}^n$ is uniquely determined by $(\lambda _A^*, \lambda _B^*)$ and $\min \{X^*_{A,j},X^*_{B,j}\}>0$ for all j, it follows from Lemma 1 that in each battlefield j the players’ unique equilibrium univariate marginal distributions are given by Theorem 2. That is, for each solution $(\lambda _A^*, \lambda _B^*)$ each player in the GGL game has a unique set of Nash equilibrium univariate marginals, and this completes the proof of Theorem 1.

Appendix 2

Appendix 2 contains the proof of Proposition 3. As demonstrated by the Quasi-Symmetric Example, there may exist multiple equilibria with distinct payoffs, but there always exists an equilibrium with $\gamma ^*=1$. Note that if $\gamma ^*=1$, then $\lambda _A^*=\lambda _B^*$ which it will be convenient to denote by $\lambda ^{1}$.

We begin with part (1) of Proposition 3, where we provide the proof for case (iii), i.e., $n_{{\mathcal {A}}}\ge 2$ and $n_{{\mathcal {D}}}\ge \frac{4}{\epsilon }-4$. The proofs for the remaining two cases follow directly. Recall that Proposition 2 provides the following sufficient condition for an equilibrium set of univariate marginal distributions in a GGL game to be able to map, via a copula, into an equilibrium joint distribution in the corresponding GCB game: if for each distinct pair of battlefield valuations $(v_{A,{\mathfrak {j}}}, v_{B,{\mathfrak {j}}})$ with $\frac{v_{-i,{\mathfrak {j}}} \lambda _i^*}{v_{i,{\mathfrak {j}}} \lambda _{-i}^*}\le 1$, for some $i\in \{A,B\}$, it is the case that $\frac{2}{n_{{\mathfrak {j}}}}\le \frac{v_{-i,{\mathfrak {j}}} \lambda _i^*}{v_{i,{\mathfrak {j}}} \lambda _{-i}^*}$, then there exists a Nash equilibrium of the GCB game with the same set of univariate marginal distributions and expected payoffs as in the corresponding equilibrium in Theorem 1.

In a Quasi-Symmetric CB game, there are three distinct pairs of battlefield valuations to consider. We begin with the agreement set ${\mathcal {A}}$ and then examine the two halves of the disagreement set ${\mathcal {D}}$. At each battlefield $\mathfrak {j}\in {\mathcal {A}}$, $\left( v_{A,{\mathfrak {j}}}, v_{B,{\mathfrak {j}}}\right) =\left( \frac{1}{n}, \frac{1}{n}\right)$. For $\gamma ^*=1$, the Proposition 2 condition for the agreement set ${\mathcal {A}}$ may be stated as $n_{{\mathcal {A}}}\ge 2$. In the case that $\gamma ^*\ne 1$, the Proposition 2 condition for the agreement set ${\mathcal {A}}$ is $n_{{\mathcal {A}}}\ge \frac{2}{\gamma ^*}$ if $\gamma ^*\le 1$ and $n_{{\mathcal {A}}}\ge 2\gamma ^*$ if $\gamma ^*\ge 1$. Clearly the Proposition 2 condition for ${\mathcal {A}}$ becomes more difficult to satisfy as $\gamma ^*$ moves away from the value of 1. That is, if the Proposition 2 condition for ${\mathcal {A}}$ is not satisfied at $\gamma ^*=1$, then the Proposition 2 condition for ${\mathcal {A}}$ will not be satisfied for any $\gamma ^*\ne 1$.

Moving on to the disagreement set ${\mathcal {D}}$ where either $\left( v_{A,{\mathfrak {j}}}, v_{B,{\mathfrak {j}}}\right) =\left( \frac{2(1-\epsilon )}{n}, \frac{2\epsilon }{n}\right)$ or $\left( v_{A,{\mathfrak {j}}}, v_{B,{\mathfrak {j}}}\right) =\left( \frac{2\epsilon }{n}, \frac{2(1-\epsilon )}{n}\right)$, for $\gamma ^*=1$ the Proposition 2 condition for both halves of the disagreement set ${\mathcal {D}}$ may be stated as $\frac{4}{n_{{\mathcal {D}}}}\le \frac{\epsilon }{1-\epsilon }$, or equivalently as $n_{{\mathcal {D}}}\ge \frac{4}{\epsilon }-4$. In the case that $\gamma ^*\ne 1$, there are two cases to consider, which we label ($\lozenge$) for $1>\gamma ^*\ge \frac{\epsilon }{1-\epsilon }$ and ($\blacklozenge$) for $\frac{1-\epsilon }{\epsilon }>\gamma ^*\ge 1$. Note that in case ($\lozenge$) $\left( \frac{\epsilon }{1-\epsilon } \right) \gamma ^*< 1$ and in case ($\blacklozenge$) $\left( \frac{\epsilon }{1-\epsilon }\right) \frac{1}{\gamma ^*}< 1$. Thus for both cases ($\lozenge$) and ($\blacklozenge$), the Proposition 2 condition for half of the disagreement set ${\mathcal {D}}$ is $n_{{\mathcal {D}}}\ge \left( \frac{4}{\epsilon }-4\right) \gamma ^*$ and for the other half of the disagreement set ${\mathcal {D}}$ is $n_{{\mathcal {D}}}\ge \left( \frac{4}{\epsilon }-4\right) \frac{1}{\gamma ^*}$. Because both of these Proposition 2 conditions for the disagreement set ${\mathcal {D}}$ must be satisfied, it follows that regardless of the direction of the movement, one of these two conditions becomes more difficult to satisfy as $\gamma ^*$ moves away from the value of 1. This completes the proof of part (1) of Proposition 3.

For part (2) of Proposition 3, recall that Corollary 2 provides the following sufficient condition for an equilibrium set of univariate marginal distributions in the GGL game to not be able to map into an equilibrium joint distribution in the corresponding GCB game: $X_i <\max _{\mathfrak {j}}\left\{ \min \left\{ \frac{v_{-i,{\mathfrak {j}}}}{ \lambda _{-i}^*}, \frac{v_{i,{\mathfrak {j}}}}{\lambda _{i}^*} \right\} \right\}$. For $\gamma ^*=1$, we can use Eq. ($\star$) to solve for $\lambda ^1$ in the Quasi-Symmetric GL game, where

$$\begin{aligned} \frac{1}{\lambda ^1}\left( \frac{n_{{\mathcal {A}}}}{2n} +\frac{n_{{\mathcal {D}}}}{2n}\left( \frac{\epsilon }{1-\epsilon }\right) \right) =1. \end{aligned}$$

(24)

Next, note that for the Quasi-Symmetric GL game with $\gamma ^*=1$, it follows that

$$\begin{aligned} X_i =1< \frac{1}{n\lambda ^1}=\max _{\mathfrak {j}} \left\{ \min \left\{ \frac{v_{-i,{\mathfrak {j}}}}{ \lambda ^1}, \frac{v_{i,{\mathfrak {j}}}}{\lambda ^1} \right\} \right\} \end{aligned}$$

(25)

Combining $\lambda ^1$ from Eq. (24) with the Corollary 2 condition that $1< \frac{1}{n\lambda ^1}$ from Eq. (25), it follows that if $n_{{\mathcal {A}}} +\frac{\epsilon n_{{\mathcal {D}}}}{(1-\epsilon )}< 2$, then for the $\gamma ^*=1$ solution to system ($\star$) the Quasi-Symmetric CB game with $(\epsilon , n_{{\mathcal {A}}}, n_{{\mathcal {D}}})$ satisfies the condition in Corollary 2. Conversely, if $n_{{\mathcal {A}}}+\frac{\epsilon n_{{\mathcal {D}}}}{(1-\epsilon )}\ge 2$ then for the $\gamma ^*=1$ solution to system ($\star$) the Corollary 2 condition is not satisfied. We now show that if $n_{{\mathcal {A}}}+\frac{\epsilon n_{{\mathcal {D}}}}{(1-\epsilon )}\ge 2$, then there exists no $\gamma ^*\ne 1$ solution to system ($\star$) that satisfies the Corollary 2 condition.

For $\gamma ^*\ne 1$ we have the two cases ($\lozenge$) for $1>\gamma ^*\ge \frac{\epsilon }{1-\epsilon }$ and ($\blacklozenge$) for $\frac{1-\epsilon }{\epsilon }>\gamma ^*\ge 1$. Beginning with case ($\lozenge$), let $\lambda _i^{\lozenge }$ denote the value of the multiplier on player i’s expected resource expenditure constraint in a case ($\lozenge$) equilibrium and let $\gamma ^{\lozenge }=\frac{\lambda _A^{\lozenge }}{\lambda _B^{\lozenge }}$. Because $1>\gamma ^{\lozenge }\ge \frac{\epsilon }{1-\epsilon }$ in case ($\lozenge$), we have the following

$$\begin{aligned} \min \left\{ \frac{1}{n \lambda _A^{\lozenge }}, \frac{1}{n \lambda _B^{\lozenge }}\right\}&=\frac{1}{n\lambda _B^{\lozenge }}, \\ \min \left\{ \frac{2\epsilon }{n\lambda _A^{\lozenge }}, \frac{2(1-\epsilon )}{n\lambda _B^{\lozenge }}\right\}&=\frac{2\epsilon }{n\lambda _A^{\lozenge }}, \\ \min \left\{ \frac{2(1-\epsilon )}{n\lambda _A^{\lozenge }}, \frac{2\epsilon }{n\lambda _B^{\lozenge }}\right\}&=\frac{2\epsilon }{n\lambda _B^{\lozenge }}. \end{aligned}$$

(26)

It then follows from Eq. (26) that

$$\begin{aligned} \max _{\mathfrak {j}}\left\{ \min \left\{ \frac{v_{-i,{\mathfrak {j}}}}{\lambda _{-i}^{\lozenge }}, \frac{v_{i,{\mathfrak {j}}}}{\lambda _{i}^{\lozenge }} \right\} \right\} ={\left\{ \begin{array}{ll} \frac{1}{n \lambda _B^\lozenge } &\quad {\mathrm{if}}\quad \gamma ^\lozenge \in \left( 2\epsilon ,1\right] \\ \frac{2\epsilon }{n \lambda _A^\lozenge } &\quad {\mathrm{if}}\quad \gamma ^\lozenge \in \left[ \frac{\epsilon }{1-\epsilon }, 2\epsilon \right] \end{array}\right. } \end{aligned}$$

(27)

Recall that for the Quasi-Symmetric GL game Eq. (15) may be used to solve for all possible solutions to system ($\star$) with $\gamma ^\lozenge \in \left[ \frac{\epsilon }{1-\epsilon }, 1\right)$. For each solution $\gamma ^\lozenge$ to equation (15), we may use Eq. ($\star$) to solve for $(\lambda _A^{\lozenge }|_{\gamma ^\lozenge }, \lambda _B^{\lozenge }|_{\gamma ^\lozenge })$,

$$\begin{aligned} \lambda _A^{\lozenge }|_{\gamma ^\lozenge }&= (\gamma ^\lozenge )^2\left( \frac{n_{{\mathcal {A}}}}{2n}\right) +\frac{n_{{\mathcal {D}}}}{2n}\left( \epsilon +(\gamma ^\lozenge )^2 \left( \frac{\epsilon ^2}{(1-\epsilon )}\right) \right) , \\ \lambda _B^{\lozenge }|_{\gamma ^\lozenge }&= \frac{n_{{\mathcal {A}}}}{2n} +\frac{n_{{\mathcal {D}}}}{2n}\left( \epsilon +\left( \frac{1}{\gamma ^\lozenge } \right) ^2\left( \frac{\epsilon ^2}{(1-\epsilon )}\right) \right) . \end{aligned}$$

(28)

For each $\gamma ^\lozenge \in \left( 2\epsilon ,1\right)$ that is a solution to Eq. (15), it follows from Eqs. (24), (27) and (28) that

$$\begin{aligned} \frac{1}{n\lambda ^1}\ge \frac{1}{n\lambda _B^{\lozenge }|_{\gamma ^\lozenge }}. \end{aligned}$$

(29)

From the inequality in Eq. (29), it follows that the Corollary 2 condition for case ($\lozenge$) with $\gamma ^\lozenge \in \left( 2\epsilon ,1\right)$ (i.e., $\frac{1}{n\lambda _B^{\lozenge }}>1$) cannot be satisfied unless the Corollary 2 condition for $\gamma ^*=1$ is also satisfied, i.e., for $\gamma ^\lozenge \in \left( 2\epsilon ,1\right)$ it becomes more difficult to satisfy the Corollary 2 condition as $\gamma ^*$ moves away from the value of 1.

Next, for case ($\lozenge$) and any $\gamma ^\lozenge \in \left[ \frac{\epsilon }{1-\epsilon }, 2\epsilon \right]$ that is a solution to Eq. (15), the Corollary 2 condition, which follows from Eqs. (27) and (28), is given by,

$$\begin{aligned} 1<\frac{2\epsilon }{n \lambda _A^\lozenge |_{\gamma ^\lozenge }} =\frac{2\epsilon }{(\gamma ^\lozenge )^2\left( \frac{n_{{\mathcal {A}}}}{2n}\right) +\frac{n_{{\mathcal {D}}}}{2n}\left( \epsilon +(\gamma ^\lozenge )^2 \left( \frac{\epsilon ^2}{(1-\epsilon )}\right) \right) } \end{aligned}$$

(30)

Because the right-hand side of Eq. (30) is decreasing in $\gamma ^\lozenge$ it follows that,

$$\begin{aligned} \frac{2\epsilon }{n \lambda _A^\lozenge |_{\gamma ^\lozenge }} \le \frac{1}{\frac{n_{{\mathcal {A}}}}{2n}+\frac{n_{{\mathcal {D}}}}{2n} \left( \epsilon +\left( \frac{1}{2\epsilon } \right) ^2 \left( \frac{\epsilon ^2}{(1-\epsilon )}\right) \right) } \le \frac{1}{n\lambda ^1}. \end{aligned}$$

Thus, for any $\gamma ^\lozenge \in \left[ \frac{\epsilon }{1-\epsilon }, 2\epsilon \right]$ that is a solution to Eq. (15), the Corollary 2 condition for case ($\lozenge$) cannot be satisfied unless the Corollary 2 condition for $\gamma ^*=1$ is also satisfied. This completes the proof of case ($\lozenge$) of part (2) of Proposition 3. To complete the proof of Proposition 3, note that the proof for case ($\blacklozenge$) of part (2) of Proposition 3 follows from a symmetric argument.