# Battlefield transfers in coalitional Blotto games

Vade Shah and Jason R. Marden V. Shah<EMAIL_ADDRESS>and J.R. Marden
<EMAIL_ADDRESS>are with the Department of Electrical and Computer
Engineering at the University of California, Santa Barbara, CA. This work is
supported by AFOSR grant FA9550-20-1-0054.

###### Abstract

In this work, we consider the framework of coalitional Blotto games in which
two players compete against a common adversary by allocating their budgeted
resources across disjoint sets of valued battlefields; the agent that
allocates a higher amount wins the corresponding battlefield value. At the
beginning of the game, the budgets of the agents and the values of the
battlefields are specified. In the first stage, the players are allowed to
perform a battlefield transfer in which one player offloads a number of its
battlefields onto the other player. In the second stage, the adversary
observes this transfer and determines how to allocate their budget
accordingly. Finally, in the third stage, the players and the adversary
allocate their budgets to their battlefields, the game is played, and their
payoffs are realized. We provide necessary and sufficient conditions for the
existence of a battlefield transfer that strictly increases the payoff of each
player. We then augment the model, allowing players to not only transfer
subsets of battlefields, but also portions of their budget, in the first
stage. We also provide sufficient conditions for the existence of a joint
transfer of battlefields and budgets. The results demonstrate that in almost
all game instances, both players would benefit from such a joint transfer.

##  1 Introduction

In adversarial resource allocation settings with multiple competitors,
individual agents often seek to form partnerships to obtain some degree of
mutual benefit. Sometimes, these alliances are formed through an agreement
where players offer resources to one another; in military conflicts, for
example, countries provide aid to alliance members to defend against a common
enemy [1]. However, in other situations, alliances may decide upon some kind
of non-compete clause instead; fisheries, for example, establish agreements to
decide when and where they can fish, and they ally to defend these zones from
poachers [2], [3]. Depending on the scenario, one method of alliance formation
may be more suitable than another. When considering their odds of success
against other opponents, the agents’ chosen method of alliance formation is
pivotal.

In this work, we study the potential for collaborative relationships in the
context of a well-studied model of strategic resource allocation in
adversarial environments termed Colonel Blotto (or General Lotto) games. In
these games, two competitors allocate their scarce budgets across a number of
valued battlefields, as shown on the right in Figure 1. At each battlefield,
the player having allocated a greater amount of budget wins the associated
battlefield value; accordingly, players distribute their budgets with the aim
of maximizing their accrued battlefield values. Indeed, the framework of the
game is general and has been studied in contexts ranging from political
campaigns [4], [5] to cybersecurity [6], [7].

To analyze methods of partnership formation, we utilize the framework of the
coalitional Colonel Blotto game, through which we seek to identify when
opportunities emerge for different types of collaborations between players. A
foundational model of the coalitional Blotto game is given in [8], where two
players compete in disjoint Blotto games against a common adversary. Their
work demonstrates a seemingly counterintuitive result: in some scenarios, if
one of the two players (say, Player 1) unilaterally transfers some of their
budget to the other player (Player 2), both players experience greater payoffs
in their individual games against the adversary. That is, Player 1 becomes
weaker, Player 2 becomes stronger, but in their separate competitions against
the adversary, each of them accrue more battlefield value than they would have
if no such budgetary transfer had occurred.

This foundational result has inspired much of the existing literature in
coalitonal Blotto games. Many studies have explored other strategic
opportunities to gain a competitive advantage in the same two-versus-one
setting, each with different variations on the actions available to the
players and the information structure of the game. In the existing literature,
scholars have analyzed budgetary transfers [8], [9], battlefield additions
[10], informational asymmetries [11], [12], and various combinations of these
strategies and structures [7], [13]. In this work, we consider a different
strategic maneuver: battlefield transfers. In a battlefield transfer, one
player abstains from competing on a subset of their battlefields, allowing the
other player to compete against the adversary for these battlefields instead.
Battlefield transfers, to the best of our knowledge, have not received
attention in the literature.

To study these transfers, we adopt the coalitional Blotto framework presented
in [8]. The model, as shown in Figure 1, describes a three-stage coalitional
Colonel Blotto game between two players, who are not necessarily allies, and a
common adversary. The players, referred to as Players 1 and 2, are equipped
with budgets $X_{1}$ and $X_{2}$, respectively, and the adversary is equipped
with a budget $X_{A}$. In the first stage of the game, Player 1 transfers a
portion of their budget $\tau_{1\to 2}^{b}$ and/or a number of their
battlefields $\tau_{1\to 2}^{v}$ to Player 2 if doing so would be beneficial.
In the second stage, the common adversary, having observed this transfer,
accordingly determines how to allocate their budget to each of its games
against Players 1 and 2. In the third and final stage, the disjoint Blotto
games are played and the payoffs are realized.

Figure 1: Coalitional Colonel Blotto game depiction. The items enclosed by the
black dashed box represent a standard Colonel Blotto game between Player 2 and
the adversary.

Within this setting, our work seeks to address the following questions:

  * •

Are there ever opportunities for mutually beneficial battlefield transfers,
i.e., transfers that strictly benefit both players individually?

  * •

If so, how does the space where such transfers exist compare to the space
where mutually beneficial budgetary transfers exist?

  * •

Are there increased opportunities for mutual improvement if players can
transfer both their budgets and battlefields?

We answer each of these three questions in Theorems 1, 2, and 3, respectively.
In Theorem 1, we show that there exist mutually beneficial battlefield
transfers, and we fully characterize the subset of the parameter space in
which there are opportunities for these transfers. In Theorem 2, we show that
the region where mutually beneficial battlefield transfers exist is different
from the region where mutually beneficial budgetary transfers exist; that is,
in certain parameter regimes, there are opportunities for mutually beneficial
battlefield transfers, but there are not opportunities for mutually beneficial
budgetary transfers, and in other parameter regimes, the opposite is true.
Finally, in Theorem 3, we show that if players are allowed to transfer budgets
and battlefields simultaneously, then for almost all game instances, there
exists an opportunity for a transfer of both budgets and battlefields that is
mutually beneficial.

These results lend insight into the nature of transfers in coalitional Blotto
games. They suggest that in settings where budgetary transfers are infeasible
or impractical, mutually beneficial battlefield transfers can serve as a
favorable alternative. Moreover, we demonstrate that for a subset of game
instances, mutually beneficial battlefield transfers exist whereas mutually
beneficial budgetary transfers do not exist. However, our final results show
that neither budgetary nor battlefield transfers are enough individually to
guarantee the existence of a mutually beneficial transfer; however, if players
can perform both types of transfers simultaneously, then there is almost
always an opportunity for both players to improve.

##  2 Model and results

First, in Section II.A, we present the model of the three-stage coalitional
Blotto game. In Section II.B, we characterize the set of parameters for which
there exist mutually beneficial battlefield transfers. Lastly, we present
simulations that align with these characterizations in Section II.C.

###  2.1 Model

In this section, we formulate the three-player, three-stage full information
Colonel Blotto game. Two players, referred to as Players 1 and 2, compete
against a common adversary. Each Player $i\in\\{1,2\\}$ competes against the
adversary on their own set of battlefields; that is, Players 1 and 2 only
compete against the adversary, not one another, as depicted graphically in
Figure 1.

Player $i$ competes across $n_{i}$ battlefields, where we assume $n_{i}\geq
3$. Each of the $n_{i}$ battlefields is of positive value111We assume that all
battlefields in one front are of equal value for consistency with previous
models, but this assumption can be relaxed when players follow their
equilibrium strategies, so these results still hold. $v_{i}>0$, so the total
value of the battlefields on each front is $\phi_{i}\triangleq n_{i}v_{i}$.
Player $i$ is equipped with a normalized budget $X_{i}\in\mathbb{R}_{>0}$, and
the adversary is equipped with a normalized budget $X_{A}=1$. A game instance
$G$ is thus parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}=\mathbb{R}_{>0}^{4}$.

The adversary competes against both Players 1 and 2 across all of the
$n_{1}+n_{2}$ battlefields. On each battlefield, the player that allocates a
higher level of resources will win that battlefield. In the event of a tie,
the two players on that front share the payoff equally222When players follow
their equilibrium strategies, a tie is a measure zero event, so the tie-
breaking rule does not affect the equilibrium payoffs..

The game proceeds in three distinct stages, described as follows:

####  2.1.1 Stage 1

In the first stage, Players 1 and 2 decide whether to engage in a transfer. We
consider three forms of transfers: budgetary, valuation, and joint.

##### Budgetary

We define a budgetary transfer as a number $\tau_{1\to
2}^{b}\in(-X_{2},X_{1})$ that is the net amount of budget transferred from
Player 1 to 2 (a negative $\tau_{1\to 2}^{b}$ indicates that the net transfer
goes in the opposite direction); the post-transfer budgets of Players 1 and 2
are thus given by $\overline{X}_{1}\triangleq X_{1}-\tau_{1\to 2}^{b}$ and
$\overline{X}_{2}\triangleq X_{2}+\tau_{1\to 2}^{b}$, respectively.

##### Valuation

We define a battlefield transfer as a number $\tau_{1\to
2}^{v}\in(-\phi_{2},\phi_{1})$ that is the net amount of total battlefield
valuation333It is assumed that the number of battlefields is large enough such
that any desired amount of battlefield valuation can be feasibly transferred.
Alternatively, a transfer is performed by decreasing the value of each
battlefield on one front and increasing the value of each battlefield on the
other front correspondingly. transferred from Player 1 to 2 (a negative
$\tau_{1\to 2}^{v}$ indicates that the net transfer goes in the opposite
direction); the post-transfer battlefield valuations of Players 1 and 2 are
thus given by $\overline{\phi}_{1}\triangleq\phi_{1}-\tau_{1\to 2}^{v}$ and
$\overline{\phi}_{2}\triangleq\phi_{2}+\tau_{1\to 2}^{v}$. Henceforth, the
terms ’battlefield transfer’ and ’valuation transfer’ will be used
interchangeably.

##### Joint

We define a joint transfer as a pair $\tau\triangleq(\tau_{1\to
2}^{v},\tau_{1\to 2}^{b})\in(-\phi_{2},\phi_{1})\times(-X_{1},X_{2})$ of
simultaneous budgetary and battlefield transfers that go from Player 1 to
Player 2. Note that if the constituent battlefield transfer is zero, then it
is simply a budgetary transfer, and vice versa.

Players only perform a transfer if doing so would improve both of their
outcomes in their final stage games against the adversary. Any transfer that
occurs in this stage is binding and the post-transfer battlefield valuations
and budgets become publicly known.

####  2.1.2 Stage 2

In the second stage, after having observed any transfers that occurred between
Players 1 and 2, the adversary must decide how to allocate their budget
$X_{A}=1$ to each of its competitions against Players 1 and 2. The adversary’s
allocation to each front is given by $X_{A,i}(G,\tau)$, which we express as
$X_{A,i}$ for brevity when the dependence is clear. The explicit form of the
adversary’s best response allocation is provided in Lemma 1.

####  2.1.3 Stage 3

In the third and final stage, the adversary and Players 1 and 2 deploy their
resources to the corresponding fronts, the two separate Colonel Blotto games
are played, and the payoffs are realized. Each player’s equilibrium payoff, as
derived in [8], is given by

$\begin{split}u_{1}(\tau)&=\begin{cases}(\phi_{1}-\tau_{1\to
2}^{v})\left(\frac{X_{1}-\tau_{1\to 2}^{b}}{2X_{A,1}(G,\tau)}\right)&X_{1}\leq
X_{A,1}\\\ (\phi_{1}-\tau_{1\to
2}^{v})\left(1-\frac{X_{A,1}(G,\tau)}{2(X_{1}-\tau_{1\to
2}^{b})}\right)&\text{otherwise}\end{cases}\\\
u_{2}(\tau)&=\begin{cases}(\phi_{2}+\tau_{1\to
2}^{v})\left(\frac{X_{2}+\tau_{1\to 2}^{b}}{2X_{A,2}(G,\tau)}\right)&X_{2}\leq
X_{A,2}\\\ (\phi_{2}+\tau_{1\to
2}^{v})\left(1-\frac{X_{A,2}(G,\tau)}{2(X_{2}+\tau_{1\to
2}^{b})}\right)&\text{otherwise}\end{cases}\end{split}$ (1)

The focus of this paper is on understanding where there exists any mutually
beneficial transfer. Specifically, we call a transfer $\tau=(\tau_{1\to
2}^{v},\tau_{1\to 2}^{b})$ mutually beneficial if

$u_{i}(\tau)>u_{i}(0,0)\quad\forall\;i\in\\{1,2\\}$ (2)

where $u_{i}(0,0)$ is Player $i$’s payoff when no transfer occurs.

###  2.2 Results

We first characterize $\mathcal{G}^{v}$, the subset of $\mathcal{G}$ where
mutually beneficial battlefield transfers exist. Then, we characterize
$\mathcal{G}^{b,v}$, the subset of $\mathcal{G}$ where mutually beneficial
joint transfers exist.

####  2.2.1 Valuation Transfers

Consider the scenario where players are only allowed to perform valuation
transfers in the first stage of the game, i.e., $\tau_{1\to 2}^{b}=0$. Theorem
1 defines the subset of the parameter space where mutually beneficial
valuation transfers exist.

######  Theorem 1.

Consider any coalitional Blotto game parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}$. There exists a mutually
beneficial valuation transfer $\tau_{1\to 2}^{v}$ if and only if
$G\in\mathcal{G}^{v}\subseteq\mathcal{G}$, where the explicit form of
$\mathcal{G}^{v}$ is given in the Appendix. Furthermore, for any game
$G\in\mathcal{G}^{v}$, $\tau_{1\to 2}^{v}>0$ if and only if
$\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$.

Before delving into the proof of Theorem 1, we will present an important Lemma
from [8] that explicitly characterizes the adversary’s optimal allocation of
their normalized budget $X_{A}=1$ to each front.

######  Lemma 1.

For given post-transfer battlefield valuations $\overline{\phi}_{i}$,
$\overline{\phi}_{-i}$ and post-transfer budgets $\overline{X}_{i}$,
$\overline{X}_{-i}$, where $i\in\\{1,2\\}$ and $-i$ is the other player, the
adversary’s optimal allocation of resources is given by the pair $(X_{A,i}$,
$X_{A,-i}=1-X_{A,i})$ and is determined by the cases summarized in Table 1.

Table 1: The adversary’s optimal budget allocation Case | Condition | $X_{A,i}$
---|---|---
1 | $\frac{\overline{\phi}_{i}}{\overline{\phi}_{-i}}>\frac{\overline{X}_{i}}{\overline{X}_{-i}}\text{ and }\frac{\overline{\phi}_{i}}{\overline{\phi}_{-i}}\geq\frac{1}{\overline{X}_{i}\overline{X}_{-i}}$ | $1$
2 |  $\frac{\overline{\phi}_{i}}{\overline{\phi}_{-i}}>\frac{\overline{X}_{i}}{\overline{X}_{-i}}$ and $0<1-\sqrt{\frac{\overline{\phi}_{i}\overline{X}_{i}\overline{X}_{-i}}{\overline{\phi}_{-i}}}\leq\overline{X}_{-i}$ | $\sqrt{\frac{\overline{\phi}_{i}\overline{X}_{i}\overline{X}_{-i}}{\overline{\phi}_{-i}}}$
3 |  $\frac{\overline{\phi}_{i}}{\overline{\phi}_{-i}}\geq\frac{\overline{X}_{i}}{\overline{X}_{-i}}$ and $1-\sqrt{\frac{\overline{\phi}_{i}\overline{X}_{i}\overline{X}_{-i}}{\overline{\phi}_{-i}}}>\overline{X}_{-i}$ | $\frac{\sqrt{\overline{\phi}_{i}\overline{X}_{i}}}{\sqrt{\overline{\phi}_{i}\overline{X}_{i}}+\sqrt{\overline{\phi}_{-i}\overline{X}_{-i}}}$
4 |  $\frac{\overline{\phi}_{i}}{\overline{\phi}_{-i}}=\frac{\overline{X}_{i}}{\overline{X}_{-i}}$ and $1\leq\overline{X}_{i}+\overline{X}_{-i}$ | $X_{A,i}\leq X_{i}$

We briefly describe each of the cases below:

  * •

Case 1: The adversary allocates all of their resources against Player 1, who
is poorly endowed in comparison to Player 2.

  * •

Case 2: The adversary reaches a point of diminishing returns in its game
against Player 1 and thus diverts some of its resources towards its game
against Player 2.

  * •

Case 3: The adversary has a larger budget than Players 1 and 2 combined (i.e.,
$X_{1}+X_{2}<1$), so the adversary allocates their budget such that the
marginal payoffs in the regions of diminishing returns in each game are equal.

  * •

Case 4: The adversary’s marginal payoff in each game is equal so long as the
adversary’s allocation to each game is less than the corresponding player’s
budget.

Note that these cases describe the adversary’s optimal allocation to each
front when
$\frac{\overline{\phi}_{1}}{\overline{\phi}_{2}}\geq\frac{\overline{X}_{1}}{\overline{X}_{2}}$,
i.e., when Player 1 is relatively weaker than or as strong as Player 2. The
analysis is entirely analogous when Player 1 is relatively stronger.

In general, the full proof of Theorem 1 (provided in the Appendix) requires a
specific analysis for various different regimes in the parameter space
$\mathcal{G}$. For brevity, we purely analyze one regime in the proof
presented below; the proofs for the other regimes follow a similar structure.

###### Proof.

Consider a game parameterized by $G\in\mathcal{G}$ such that $X_{1}\geq 1$,
$X_{2}\geq 1$, and $\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$. First,
observe that by Lemma 1, the initial allocation $G$ satisfies the conditions
of Case 1, meaning that the adversary would concentrate all of their resources
against Player 1 (i.e., $X_{A,1}=1$, $X_{A,2}=0$). Thus, according to Equation
1, the payoff function of each player is given by

$\displaystyle u_{1}(\tau_{1\to 2}^{v},0)$ $\displaystyle=(\phi_{1}-\tau_{1\to
2}^{v})\left(1-\frac{1}{2X_{1}}\right)$ $\displaystyle u_{2}(\tau_{1\to
2}^{v},0)$ $\displaystyle=\phi_{2}+\tau_{1\to 2}^{v}$

for any set of post-transfer parameters that satisfies the conditions of Case
1. Similarly, notice that by Lemma 1, if $\tau_{1\to 2}^{v}$ is chosen such
that

$\displaystyle\frac{\phi_{1}-\tau_{1\to 2}^{v}}{\phi_{2}+\tau_{1\to
2}^{v}}=\frac{\overline{\phi}_{1}}{\overline{\phi}_{2}}<\frac{X_{1}}{X_{2}}$
(3) $\displaystyle\iff\tau_{1\to
2}^{v}>\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\triangleq\alpha,$
(4)

i.e., if the valuation transfer is large enough such that the adversary
concentrates all of their resources against Player 2 instead, then the payoffs
are given by

$\displaystyle u_{1}(\tau_{1\to 2}^{v},0)$ $\displaystyle=\phi_{1}-\tau_{1\to
2}^{v}$ $\displaystyle u_{2}(\tau_{1\to 2}^{v},0)$
$\displaystyle=(\phi_{2}+\tau_{1\to 2}^{v})\left(1-\frac{1}{2X_{2}}\right).$

That is, the payoff functions for any $\tau_{1\to
2}^{v}\in(-\phi_{2},\phi_{1})$ are given by

$\displaystyle u_{1}(\tau_{1\to 2}^{v},0)$
$\displaystyle=\begin{cases}\phi_{1}-\tau_{1\to 2}^{v}&\tau_{1\to
2}^{v}>\alpha\\\ (\phi_{1}-\tau_{1\to
2}^{v})\left(1-\frac{1}{2X_{1}}\right)&\text{otherwise}\end{cases}$
$\displaystyle u_{2}(\tau_{1\to 2}^{v},0)$
$\displaystyle=\begin{cases}(\phi_{2}+\tau_{1\to
2}^{v})\left(1-\frac{1}{2X_{2}}\right)&\tau_{1\to 2}^{v}>\alpha\\\
\phi_{2}+\tau_{1\to 2}^{v}&\text{otherwise}.\end{cases}$

Notice that for any $\tau_{1\to 2}^{v}<0$, the post-transfer payoff of Player
2 is strictly smaller than their payoff if no transfer had occurred; thus, any
mutually beneficial transfer must be strictly positive. Similarly, for any
positive $\tau_{1\to 2}^{v}<\alpha$, the post-transfer payoff of Player 1 is
strictly smaller than their payoff if no transfer had occurred; thus, any
mutually beneficial transfer must be strictly greater than $\alpha$. By simple
algebraic manipulation, it can be shown that there exists a transfer
$\tau_{1\to 2}^{v}$ satisfying Inequalities 2 and 4 if and only if

$\frac{2X_{1}X_{2}-X_{1}-X_{2}}{2X_{1}^{2}}<\frac{\phi_{2}}{\phi_{1}}<\frac{2X_{2}-1}{2X_{1}}$
(5)

∎

First, notice that Theorem 1 establishes the existence of mutually beneficial
battlefield transfers. That is, there exist strategic opportunities in which
one player elects not to compete with the adversary on a subset of its
battlefields, the other player competes for them instead, and both players
obtain strictly greater payoffs than if no transfer had occurred. The
existence of such transfers validates competitive agreements as a method of
alliance formation in this model.

Second, observe that Theorem 1 establishes that mutually beneficial valuation
transfers must always go from the weaker player to the stronger player. This
result is consistent with that of [8], in which they show that budgetary
transfers always go from the stronger player to the weaker player; in other
words, either form of mutually beneficial transfer must bring the ratio of the
player’s budgets and battlefields closer to one another. Moreover, this result
seems to align with intuition; a weaker player would seek to concentrate their
limited resources on a smaller set of battlefields, whereas a stronger player
would hope to compete for a larger valuation.

Finally, note that akin to Theorem 1, the work of [8] identifies
$\mathcal{G}^{b}$, the subset of $\mathcal{G}$ in which mutually beneficial
budgetary transfers exist. To better understand the differences between these
methods of alliance formation, the result provided in Theorem 2 states whether
each type of transfer offers unique opportunities for mutual improvement.

######  Theorem 2.

Consider $\mathcal{G}^{v}$, the set in which mutually beneficial valuation
transfers exist, and $\mathcal{G}^{b}$, the set in which mutually beneficial
budgetary transfers exist. It holds that

$\mathcal{G}^{v}\neq\mathcal{G}^{b}$

Moreover, $\mathcal{G}^{v}\not\subset\mathcal{G}^{b}$ and
$\mathcal{G}^{b}\not\subset\mathcal{G}^{v}$.

The full proof of Theorem 2 is presented in the Appendix. It should be noted
that, by means of a counterexample, one can deduce that
$\mathcal{G}^{v}\neq\mathcal{G}^{b}$; however, explicitly characterizing the
sets $\mathcal{G}^{v}\setminus\mathcal{G}^{b}$ and
$\mathcal{G}^{b}\setminus\mathcal{G}^{v}$ is lengthy and requires the analysis
of several cases, similar to Theorem 1. Nevertheless, the important point to
note is that there are parameters for which there exist opportunities for
battlefield transfers but not budgetary transfers and vice versa.

This phenomenon is depicted graphically in Figures 2 and 2 for different
initial battlefield values. The set $\mathcal{G}^{v}\setminus\mathcal{G}^{b}$,
shown in blue, represents the subset of the parameter space where there exist
opportunities for mutually beneficial battlefield transfers, but there are no
opportunities for mutually beneficial budgetary transfers. Similarly, the set
$\mathcal{G}^{b}\setminus\mathcal{G}^{v}$ is shown in red for mutually
beneficial budgetary transfers. Their intersection,
$\mathcal{G}^{v}\cap\mathcal{G}^{b}$, is checkered in red and blue. Note that
when both players are relatively well endowed, i.e., when $X_{1}\geq 1$ and
$X_{2}\geq 1$, players may have an incentive to transfer battlefields, but
they have no incentive to transfer budgets. Conversely, when one player is
much weaker than the other, players may have an incentive to transfer budgets,
but not battlefields. These results imply that in adversarial resource
allocation environments, the method of alliance formation is indeed critical.
As such, players should consider both budgetary and battlefield transfers to
determine the mutual advantages and individual disadvantages associated with
each.

Figure 2: Depiction of the subsets where mutually beneficial transfers exist.
The horizontal and vertical axes represent the budgets of Players 1 and 2,
respectively; the initial battlefield valuations $\phi_{1}$, $\phi_{2}$ are
fixed. Note that the subsets are depicted only for parameters satisfying
$\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$ to avoid redundancy. (a) The
subsets $\mathcal{G}^{v}\setminus\mathcal{G}^{b}$, i.e., the subset where only
mutually beneficial valuation transfers exist (shown in blue),
$\mathcal{G}^{b}\setminus\mathcal{G}^{v}$, i.e., the subset where only
mutually beneficial budgetary transfers exist (shown in red), and
$\mathcal{G}^{v}\cap\mathcal{G}^{b}$, i.e., the subset where both types of
mutually beneficial transfers exist (shown checkered), when
$1.2=\phi_{1}>\phi_{2}=1$. (b) The subsets
$\mathcal{G}^{v}\setminus\mathcal{G}^{b}$,
$\mathcal{G}^{b}\setminus\mathcal{G}^{v}$, and
$\mathcal{G}^{v}\cap\mathcal{G}^{b}$ when $1=\phi_{1}<\phi_{2}=1.2$ (c) The
subset $\mathcal{G}^{b,v}=\mathcal{G}\setminus\mathcal{G}^{b,v}_{0}$ where
mutually beneficial joint transfers exist, shown in green. The lines shown in
solid black depicts $\mathcal{G}^{b,v}_{0}$.

####  2.2.2 Joint Transfers

Given the existence of mutually beneficial battlefield and budgetary
transfers, we seek to study the existence of mutually beneficial joint
transfers. Certainly, since budgetary and battlefield transfers are specific
types of joint transfers, the space of mutually beneficial joint transfers
must be at least as large as the space of mutually beneficial budgetary
transfers and battlefield transfers, i.e.,
$\mathcal{G}^{b,v}\supseteq\mathcal{G}^{b}\cup\mathcal{G}^{v}$; however, the
space may well be larger. Theorem 3 describes the full subset of $\mathcal{G}$
in which there exists a mutually beneficial joint transfer.

######  Theorem 3.

Consider any coalitional Blotto game parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}$. Then, there exists a
mutually beneficial joint transfer almost everywhere in $\mathcal{G}$; in
particular, there exists a mutually beneficial joint transfer if
$G\in\mathcal{G}\setminus\mathcal{G}^{b,v}_{0}$, where $\mathcal{G}^{b,v}_{0}$
is a measure zero subset of $\mathcal{G}$.

In general, the full proof of Theorem 3 (provided in the Appendix, along with
a characterization of $\mathcal{G}^{b,v}_{0}$) requires a specific analysis
for various different parameter regimes of the space $\mathcal{G}$. For
brevity, we purely analyze one case in the proof below. It is worth noting
that the proofs for the other parameter regimes follow a similar structure.

###### Proof.

Consider a game parameterized by $G\in\mathcal{G}$ such that

$\frac{\phi_{1}}{\phi_{2}}>\max\left\\{\frac{X_{1}}{X_{2}},\frac{1}{X_{1}X_{2}}\right\\}$
(6)

First, notice that by Lemma 1, any $G$ satisfying Inequality 6 also satisfies
the conditions of Case 1. Thus, according to Equation 1, the payoff functions
of each player are given by

$\displaystyle u_{1}(\tau)=(\phi_{1}-\tau_{1\to
2}^{v})\left(1-\frac{1}{2(X_{1}+\tau_{1\to 2}^{b})}\right)$ $\displaystyle
u_{2}(\tau)=\phi_{2}+\tau_{1\to 2}^{v}$

These functions are smooth on the subset of $\mathcal{G}$ defined by
Inequality 6, so the gradient of $u_{i}$ with respect to $\tau$ is well-
defined. If the gradients of the payoff functions of each player evaluated at
$\tau=(0,0)$ are not diametrically opposed, i.e., if

$\nabla u_{1}(0,0)\neq-\beta\nabla u_{2}(0,0)$

for some $\beta\in\mathbb{R}_{>0}$, then there exists some direction
$\tau^{*}$ such that, by the definition of the gradient, the payoff functions
of each player increase along $\tau^{*}$. Thus, all that remains to be shown
is that the gradients of the payoff functions are not diametrically opposed.
We proceed by contradiction. Suppose that $\nabla u_{1}(0,0)=-\beta\nabla
u_{2}(0,0)$. Then, we have

$\displaystyle\nabla u_{1}(0,0)=\begin{bmatrix}\frac{\phi_{1}}{2X_{1}^{2}}\\\
\frac{1}{2X_{1}}-1\end{bmatrix}=-\beta\begin{bmatrix}0\\\
1\end{bmatrix}=-\beta\nabla u_{2}(0,0)$

This would imply $\phi_{1}=0$, which contradicts the assumption that
$\phi_{1}>0$. Thus, there exists a mutually beneficial joint transfer in any
game $G$ whose parameters satisfy Inequality 6. ∎

The statement in Theorem 3 is depicted graphically in Figure 2. As shown, a
mutually beneficial joint transfer exists almost everywhere444Along the
measure zero subset $\mathcal{G}^{b,v}_{0}$, the payoff functions of the
players are not smooth, so the gradients are undefined; thus, the proof
techniques used in Theorem 3 are not applicable. For a full characterization
of this subset, refer to the Appendix. in the parameter space
$\mathcal{G}^{b,v}$. It is important to note that here, we make no assumptions
regarding the direction of the transfer; that is, the constituent battlefield
transfer need not go from Player 1 to Player 2, and the constituent budgetary
transfer need not go from Player 2 to Player 1.

The implications of this result are significant. When players are limited to
either budgetary or battlefield transfers, there exist nontrivial subsets of
the parameter space for which there are no opportunities for mutual
improvement. However, if players are allowed to perform both forms of
transfers simultaneously, then there is an opportunity for mutual improvement
in almost all game instances. This suggests that in more general adversarial
resource allocation settings, if multiple methods of alliance formation are
available, players may derive greater rewards by utilizing each method instead
of adhering to one.

###  2.3 Simulation

Figure 3: Simulation results of battlefield transfers when initial battlefield
values are fixed at $\phi_{1}=1.2$ and $\phi_{2}=1$. (a) Existence of mutually
beneficial battlefield transfers for 1,000 pairs of randomly selected initial
budgets. Initializations for which there exist mutually beneficial battlefield
transfers are shown as filled black circles, and initializations for which
there do not are shown as unfilled black circles. The shaded region shown in
blue is given by Equation 5. (b) Change in payoff for each player $\Delta
u_{i}(\tau_{1\to 2}^{v})\triangleq u_{i}(\tau_{1,2}^{v},0)-u_{i}(0,0)$ as a
function of the battlefield transfer amount for initial budgets $X_{1}=1.5$
and $X_{2}=2$, shown as a green star in 3. The range of mutually beneficial
transfers is highlighted in blue.

In this section, we present numerical simulations that align with the
statements of Theorems 1 and 2. Figure 3 depicts whether or not there exists a
mutually beneficial battlefield transfer for random pairs of initial budgets.
First, notice that mutually beneficial battlefield transfers only exist inside
the blue region, which is defined by Inequality 5. Consistent with Theorem 1,
these results indicate when both players are significantly stronger than the
adversary, they may or may not have an incentive to perform a battlefield
transfer. However, for all of the random initializations, they do not have an
incentive to perform a budgetary transfer. It is worth noting that the
boundaries of these regions are determined by both the battlefield donor (the
weaker player) and the recipient (the stronger player); that is, the existence
of a mutually beneficial transfer is not constrained by any individual player.
This result is explained by the arguments in the proof of Theorem 1, but it is
also consistent with intuition; the existence of mutually beneficial transfers
is contingent upon not only the donor’s willingness to sacrifice valuable
battlefields, but also the recipient’s ability to compete on additional
fronts. This lies in contrast to budgetary transfers, which are always
constrained by the stronger player [8].

To better understand this phenomenon, Figure 3 depicts the change in each
player’s payoff as the transfer amount varies, where the initial budgets and
battlefield values (depicted as a green star in both panels) are such that
Player 1 is relatively weaker than Player 2. Notice that Player 1 experiences
a positive change in payoff for a limited range of transfers, whereas Player 2
experiences a positive change in a larger range. Thus, the existence of
mutually beneficial transfers is limited by the donor in part because
exorbitantly large transfer amounts excessively diminish the value that the
donor can compete for.

Furthermore, the results depicted align with the statement made in Theorem 2.
The entire range of transfers that are mutually beneficial, shown in blue, is
strictly positive. However, for any negative transfer amount (i.e., a transfer
from Player 2 to Player 1), Player 2 experiences a strictly lower payoff. This
is consistent with intuition; the stronger player would only suffer from
abstaining to compete for valuation.

##  3 Conclusion

In this paper, we considered a three-stage three-player coalitional Blotto
game where two players compete against an adversary. First, analogous to the
work of [8], we allowed the two players to transfer battlefields between one
another if doing so would strictly improve each of their payoffs. We provided
necessary and sufficient conditions for the existence of such mutually
beneficial battlefield transfers. Then, we allowed the players to transfer
both battlefields and budgets in a so-called joint transfer, and we provided
sufficient conditions for the existence of mutually beneficial joint
transfers. We showed that these transfers would occur almost everywhere in the
parameter space.

In the future, we would like to extend this work to explicitly characterize
the Nash equilibrium battlefield transfers. We can also study and compare
these methods of alliance formation in a different model, such as when players
budgets’ are parameterized by a per unit cost instead of a fixed amount.

##  4 Appendix

In the Appendix, we prove the necessary and sufficient conditions for the
existence of mutually beneficial valuation transfers and mutually beneficial
joint transfers; that is, we characterize the sets $\mathcal{G}^{v}$ and
$\mathcal{G}^{b,v}$. In Sections 1 and 2, the set $\mathcal{G}^{v}$ is
characterized in a series of propositions; these propositions allow us to
prove the first portion of Theorem 1 as stated in the paper, and they also
allow us to prove Theorem 2 as stated in the paper. In Section 3, we show that
any mutually beneficial valuation transfer must go from the relatively weaker
player to the relatively strongly player; these arguments allow us to prove
the second portion of Theorem 1 as stated in the paper. In Section 4, the set
$\mathcal{G}^{b,v}$ is characterized in a series of propositions; these
arguments allow us to prove Theorem 3 as stated in the paper.

For brevity, we define the following sets which will be used in the proofs:

$\displaystyle C_{1a}$
$\displaystyle\equiv\left\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}},\frac{\phi_{1}}{\phi_{2}}\geq\frac{1}{X_{1}X_{2}}\right\\}$
$\displaystyle C_{1b}$
$\displaystyle\equiv\left\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:\frac{\phi_{2}}{\phi_{1}}>\frac{X_{2}}{X_{1}},\frac{\phi_{2}}{\phi_{1}}\geq\frac{1}{X_{1}X_{2}}\right\\}$
$\displaystyle C_{2a}$
$\displaystyle\equiv\left\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}},0<1-\left(\frac{\phi_{1}X_{1}X_{2}}{\phi_{2}}\right)^{\frac{1}{2}}\leq
X_{2}\right\\}$ $\displaystyle C_{2b}$
$\displaystyle\equiv\left\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:\frac{\phi_{2}}{\phi_{1}}>\frac{X_{2}}{X_{1}},0<1-\left(\frac{\phi_{2}X_{1}X_{2}}{\phi_{1}}\right)^{\frac{1}{2}}\leq
X_{1}\right\\}$ $\displaystyle C_{3a}$
$\displaystyle\equiv\left\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:\frac{\phi_{1}}{\phi_{2}}\geq\frac{X_{1}}{X_{2}},1-\left(\frac{\phi_{1}X_{1}X_{2}}{\phi_{2}}\right)^{\frac{1}{2}}>X_{2}\right\\}$
$\displaystyle C_{3b}$
$\displaystyle\equiv\left\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:\frac{\phi_{2}}{\phi_{1}}\geq\frac{X_{2}}{X_{1}},1-\left(\frac{\phi_{2}X_{1}X_{2}}{\phi_{1}}\right)^{\frac{1}{2}}>X_{1}\right\\}$
$\displaystyle A_{1}$
$\displaystyle\equiv\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:X_{1}\geq
1,X_{2}\geq 1\\}$ $\displaystyle A_{2}$
$\displaystyle\equiv\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:X_{1}\geq
1,X_{2}<1\\}$ $\displaystyle A_{3}$
$\displaystyle\equiv\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:X_{1}<1,X_{2}\geq
1\\}$ $\displaystyle A_{4}$
$\displaystyle\equiv\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:X_{1}<1,X_{2}<1,X_{1}+X_{2}\geq
1\\}$ $\displaystyle A_{5}$
$\displaystyle\equiv\\{G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}:X_{1}+X_{2}<1\\}$

Unless otherwise specified, $C_{1}$, $C_{2}$, and $C_{3}$ refer to $C_{1a}$,
$C_{2a}$, and $C_{3a}$, respectively. Additionally, for notational simplicity,
in Sections 1-3, we simply use the variable $\tau$ to represent a mutually
beneficial battlefield transfer, instead of $\tau_{1\to 2}^{v}$. In Section 4,
we use the definition of $\tau\equiv(\tau_{1\to 2}^{b},\tau_{1\to 2}^{v})$
adopted in the paper.

###  4.1 Intra-case proofs

In this section, we characterize the regions in which arbitrarily small
mutually beneficial transfers can occur, i.e., transfers such that if the pre-
transfer parameters satisfy the conditions of Case $i$, then the post-transfer
parameters also satisfy the conditions of Case $i$. Additionally, we show that
these transfers can only occur unilaterally; that is, transfers only go from
the weaker player (Player 1) to the stronger player (Player 2). We begin our
analysis with Case 1.

######  Proposition 1.

For any $(\phi_{1},\phi_{2},X_{1},X_{2})\in{\text{int}}(C_{1})$, there does
not exist an arbitrarily small mutually beneficial valuation transfer that
remains within Case 1.

###### Proof.

In Case 1, the payoff functions of each player are given by

$\displaystyle\pi_{1}(\tau)=(\phi_{1}-\tau)\left(\frac{\underline{X}_{1}}{2\overline{X}_{1}}\right)$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(1-\frac{\underline{X}_{2}}{2\overline{X}_{2}}\right)$
where we use the notation defined in [8]. The derivatives of each function are
given by
$\displaystyle\pi_{1}^{\prime}(\tau)=-\frac{\underline{X}_{1}}{2\overline{X}_{1}}$
$\displaystyle\pi_{2}^{\prime}(\tau)=1-\frac{\underline{X}_{2}}{2\overline{X}_{2}}$

Since $X_{1}$, $X_{2}$, and $X_{A}=1$ are all strictly positive,
$\pi_{1}^{\prime}(\tau)<0$ and $\pi_{2}^{\prime}(\tau)>0$, so the payoff
functions of each player cannot simultaneously increase in either transfer
direction. ∎

######  Proposition 2.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in{\text{int}}(C_{2})$, then there exists
an arbitrarily small mutually beneficial valuation transfer from Player 1 to
Player 2 if and only if $\tau>0$, $\frac{\phi_{2}}{\phi_{1}}<1$, and
$\frac{2-4X_{2}}{\phi_{1}-\phi_{2}}<\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}$.

###### Proof.

In Case 2, the payoff functions of each player are given by

$\displaystyle\pi_{1}(\tau)=\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{\frac{1}{2}}$
The derivatives of each function are given by
$\displaystyle\pi_{1}^{\prime}(\tau)=\frac{1}{4}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{-\frac{1}{2}}(-2\tau+\phi_{1}-\phi_{2})$
$\displaystyle\pi_{2}^{\prime}(\tau)=1-\frac{1}{2X_{2}}+\frac{1}{4}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{-\frac{1}{2}}(-2\tau+\phi_{1}-\phi_{2})$
Evaluated at $\tau=0$, the derivatives are
$\displaystyle\pi_{1}^{\prime}(0)=\frac{1}{4}\left(\frac{X_{2}\phi_{1}\phi_{2}}{X_{1}}\right)^{-\frac{1}{2}}(\phi_{1}-\phi_{2})$
$\displaystyle\pi_{2}^{\prime}(0)=1-\frac{1}{2X_{2}}+\frac{1}{4}\left(\frac{X_{2}\phi_{1}\phi_{2}}{X_{1}}\right)^{-\frac{1}{2}}(\phi_{1}-\phi_{2})$

Observe that $\pi_{1}^{\prime}(0)>0\iff\frac{\phi_{2}}{\phi_{1}}<1$. Thus, we
seek to derive a condition such that guarantees $\pi_{2}^{\prime}(0)>0$ when
$\frac{\phi_{2}}{\phi_{1}}<1$. Clearly, if $X_{2}\geq\frac{1}{2}$ and
$\frac{\phi_{2}}{\phi_{1}}<1$, then $\pi_{2}^{\prime}(0)>0$. Thus, assuming
$\tau>0$, $X_{2}<\frac{1}{2}$ and $\frac{\phi_{2}}{\phi_{1}}<1$, we obtain the
following sufficient condition:

$\displaystyle
0<1-\frac{1}{2X_{2}}+\frac{1}{4}\left(\frac{X_{2}\phi_{1}\phi_{2}}{X_{1}}\right)^{-\frac{1}{2}}(\phi_{1}-\phi_{2})$
$\displaystyle\iff\frac{2-4X_{2}}{\phi_{1}-\phi_{2}}<\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}$
(7)

It remains to be shown that if
$(\phi_{1},\phi_{2},X_{1},X_{2})\in{\text{int}}(C_{2})$, then either
$\pi_{1}^{\prime}(0)>0$ or $\pi_{2}^{\prime}(0)>0$, i.e., any mutually
beneficial valuation transfer must go from Player 1 to Player 2. Notice that
$\pi_{1}^{\prime}(0)\leq 0\iff\frac{\phi_{2}}{\phi_{1}}\geq 1$. However, when
$\frac{\phi_{2}}{\phi_{1}}=1$, Player 1 has no incentive to perform any
transfer, so this case is trivial. Thus, we must show that
$\frac{\phi_{2}}{\phi_{1}}>1\implies\pi_{2}^{\prime}(0)>0$. First, notice that
if $\frac{\phi_{2}}{\phi_{1}}>1$ and
$(\phi_{1},\phi_{2},X_{1},X_{2})\in{\text{int}}(C_{2})$, then

$\displaystyle\frac{X_{1}}{X_{2}}<\frac{\phi_{1}}{\phi_{2}}\leq 1\implies
X_{1}<X_{2}\implies\frac{\phi_{1}X_{1}X_{2}}{\phi_{2}}<X_{2}^{2}$ However,
this implies $\displaystyle
1-X_{2}<1-\left(\frac{\phi_{1}X_{1}X_{2}}{\phi_{2}}\right)^{\frac{1}{2}}\leq
X_{2}\implies X_{2}>\frac{1}{2}$ Thus, we have $\displaystyle
0<\pi_{2}^{\prime}(0)\iff\frac{2-4X_{2}}{\phi_{1}-\phi_{2}}>\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}$

which is the opposite of the condition for $\frac{\phi_{2}}{\phi_{1}}<1$. We
will show that this inequality is satisfied for any set of parameters in Case
2 where $\frac{\phi_{2}}{\phi_{1}}>1$. First, notice that for fixed $\phi_{1}$
and $\phi_{2}$, the point that satisfies the conditions of Case 2 with the
smallest value of $X_{2}$ lies arbitrarily close to the intersection of the
curves $\frac{\phi_{1}}{\phi_{2}}=\frac{X_{1}}{X_{2}}$ and
$1-\left(\frac{\phi_{1}X_{1}X_{2}}{\phi_{2}}\right)^{\frac{1}{2}}=X_{2}$,
which follows from the fact that the former and latter equations are strictly
increasing and decreasing, respectively, in $X_{1}$ when written as a function
of $X_{2}$. The intersection of the curves is given by the pair
$\left(X_{1}=\frac{\phi_{1}}{\phi_{1}+\phi_{2}},X_{2}=\frac{\phi_{2}}{\phi_{1}+\phi_{2}}\right)$.
Next, notice that the previously-derived condition can be solved for $X_{2}$:

$\displaystyle\frac{2-4X_{2}}{\phi_{1}-\phi_{2}}>\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}\iff
X_{2}>\frac{1}{2}(aX_{1}-1)+\sqrt{(1-aX_{1})^{2}-1}$ where
$a=\frac{16\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}>0$. Notice that this
lower bound is concave and strictly increasing in $X_{1}$. The intersection of
this curve with the line $\frac{\phi_{1}}{\phi_{2}}=\frac{X_{1}}{X_{2}}$ is
given by the pair
$\left(X_{1}=\frac{2\phi_{1}}{\phi_{1}+3\phi_{2}},X_{2}=\frac{2\phi_{2}}{\phi_{1}+3\phi_{2}}\right)$.
Finally, we make the following observations:
$\displaystyle\frac{\phi_{2}}{\phi_{1}}>1\iff\phi_{2}>\phi_{1}\iff\phi_{1}+3\phi_{2}>2\phi_{1}+2\phi_{2}\iff\frac{\phi_{2}}{\phi_{1}+\phi_{2}}>\frac{2\phi_{2}}{\phi_{1}+3\phi_{2}}$

Thus, for a fixed $\phi_{1}$ and $\phi_{2}$, the curve of the lower bound
(which is strictly increasing and concave) intersects the line
$\frac{\phi_{1}}{\phi_{2}}=\frac{X_{1}}{X_{2}}$ strictly below the point in
Case 2 with the minimum value of $X_{2}$. Consequentially, any point in Case 2
satisfies
$\frac{2-4X_{2}}{\phi_{1}-\phi_{2}}>\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}$,
i.e., $\frac{\phi_{2}}{\phi_{1}}>1\implies\pi_{2}^{\prime}(0)>0$. Thus,
arbitrarily small mutually beneficial transfers exists in Case 2 if and only
if $\tau>0$, $\frac{\phi_{2}}{\phi_{1}}<1$, and (1) are satisfied. ∎

######  Proposition 3.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in{\text{int}}(C_{3})$, then there exists
an arbitrarily small mutually beneficial valuation transfer from Player 1 to
Player 2 if and only if $\tau>0$, $\frac{\phi_{2}}{\phi_{1}}<1$, and
$X_{2}>\frac{\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$.

###### Proof.

In Case 3, the payoff functions of each player are given by

$\displaystyle\pi_{1}(\tau)=\frac{1}{2}(\phi_{1}-\tau)\left(X_{1}+\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}\right)$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}(\phi_{2}+\tau)\left(X_{2}+\left(\frac{X_{1}X_{2}(\phi_{1}-\tau)}{\phi_{2}+\tau}\right)^{\frac{1}{2}}\right)$
The derivatives of each function are given by
$\displaystyle\pi_{1}^{\prime}(\tau)=-\frac{1}{2}X_{1}+\frac{1}{2}\left(\frac{X_{1}X_{2}}{(\phi_{1}-\tau)(\phi_{2}+\tau)}\right)^{\frac{1}{2}}(-2\tau+\phi_{1}-\phi_{2})$
$\displaystyle\pi_{2}^{\prime}(\tau)=\frac{1}{2}X_{2}+\frac{1}{2}\left(\frac{X_{1}X_{2}}{(\phi_{1}-\tau)(\phi_{2}+\tau)}\right)^{\frac{1}{2}}(-2\tau+\phi_{1}-\phi_{2})$
Evaluated at $\tau=0$, the derivatives are
$\displaystyle\pi_{1}^{\prime}(0)=-\frac{1}{2}X_{1}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})$
$\displaystyle\pi_{2}^{\prime}(0)=\frac{1}{2}X_{2}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})$

First, notice that $\frac{\phi_{2}}{\phi_{1}}>1\implies\pi_{1}^{\prime}(0)<0$.
Thus, assuming $\frac{\phi_{2}}{\phi_{1}}<1$ and $\tau>0$, we derive the
following sufficient condition for the existence of a positive mutually
beneficial transfer:

$\displaystyle
0<-\frac{1}{2}X_{1}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})$
$\displaystyle\iff
X_{2}>\frac{4\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$ (8)

It remains to be shown that negative mutually beneficial valuation transfers
do not exist. Now, notice that $\pi_{2}^{\prime}(0)<0$ only if
$\frac{\phi_{2}}{\phi_{1}}>1$. Assuming instead $\frac{\phi_{2}}{\phi_{1}}>1$,
we find that

$\displaystyle
0>\frac{1}{2}X_{2}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})$
$\displaystyle\iff
X_{2}<\frac{(\phi_{1}-\phi_{2})^{2}}{4\phi_{1}\phi_{2}}X_{1}$ However, observe
that
$\displaystyle\phi_{1}>0\iff\phi_{2}-\phi_{1}<\phi_{2}\iff(\phi_{1}-\phi_{2})^{2}<\phi_{2}^{2}\iff\frac{(\phi_{1}-\phi_{2})^{2}}{\phi_{1}\phi_{2}}<\frac{\phi_{2}}{\phi_{1}}$

which implies that there are no negative mutually beneficial valuation
transfers in Case 3. Thus, arbitrarily small mutually beneficial transfers
exist in Case 3 if and only if $\tau>0$, $\frac{\phi_{2}}{\phi_{1}}<1$, and
(2) are satisfied. ∎

Finally, note that in Case 2,

$\displaystyle\pi_{1}^{\prime\prime}(\tau)=\pi_{2}^{\prime\prime}(\tau)=-\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{-\frac{1}{2}}\left(\frac{1}{4(\phi_{1}-\tau)(\phi_{2}+\tau)}+1\right)$
which is strictly negative for all possible $\tau$. Similarly, in Case 3,
$\displaystyle\pi_{1}^{\prime\prime}(\tau)=\pi_{2}^{\prime\prime}(\tau)=-\frac{1}{4}(X_{1}X_{2})^{\frac{1}{2}}((\phi_{1}-\tau)(\phi_{2}+\tau))^{-\frac{3}{2}}(-2\tau+\phi_{1}-\phi_{2})^{2}$

which is also strictly negative for all possible $\tau$. Thus, the payoff
functions for both players are concave (with the same concavity) in each case.
This allows us to eliminate the possibility of intra-case mutually beneficial
transfers that can only be achieved by performing a large enough transfer;
that is, within a given case, we only need to study arbitrarily small
transfers.

###  4.2 Inter-case proofs

Figure 4: Cartoon depiction of different possible forms of transfer depending
on the players’ initial budgets. The five subplots indicate which cases exist
in the valuation space when the players’ budgets are fixed; for example,
$A_{3}$ corresponds to the situation in which only Case 1a, 2a, and 1b exist.
The dashed lines in the subplots indicate the line
$\phi_{2}=\frac{X_{2}}{X_{1}}\phi_{1}$.

In this section, we characterize the regions in which mutually beneficial
transfers occur such that the original and post-transfer parameters satisfy
the conditions of two different cases. Additionally, we show that these
transfers can only occur unilaterally. We begin with a brief discussion of the
symmetry of the parameter space.

Our discussion so far has centered has focused on the $(X_{1},X_{2})$ space.
Analogous versions of Cases 1 and exist on both sides of the line
$\phi_{2}=\frac{X_{2}}{X_{1}}\phi_{1}$ in the $(\phi_{1},\phi_{2})$ space. We
will refer to Cases 1a and 2a as the regions that satisfy
$\phi_{2}<\frac{X_{2}}{X_{1}}\phi_{1}$, and Cases 1b and 2b as the
corresponding regions that do not satisfy this inequality. The payoff function
for each player is smooth in Case 3 across both sides of this line, so we
refer to the region on both sides of the line as Case 3. We present this
change because the results in this section are better understood visually in
the valuation space, and will thus be represented pictorially in this space.

The following conditions determine the existence of the various cases in the
valuation space:

  * •

For any fixed $X_{1}$, $X_{2}$, there exist $\phi_{1}$, $\phi_{2}$ such that
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cup C_{1b}$

  * •

For any fixed $X_{1}$, $X_{2}$, there exist $\phi_{1}$, $\phi_{2}$ such that
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}$ if and only if $X_{1}<1$

  * •

For any fixed $X_{1}$, $X_{2}$, there exist $\phi_{1}$, $\phi_{2}$ such that
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2b}$ if and only if $X_{2}<1$

  * •

For any fixed $X_{1}$, $X_{2}$, there exist $\phi_{1}$, $\phi_{2}$ such that
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{3}$ if and only if $X_{1}+X_{2}<1$

These conditions separate the $X_{1}-X_{2}$ space as shown in Figure 4.

This figure illustrates the existence of several possible inter-case mutually
beneficial valuation transfers. Take, for example, the top left region
$A_{3}$, where $X_{1}<1$ and $X_{2}>1$. Here, there are 3 possible forms of
inter-case transfers: Case 1a to 2a, Case 1a to 1b, and Case 2a to 1b. Indeed,
mutually beneficial valuation transfers are possible in all of these forms. In
this section, we provide proofs for all possible forms of inter-case
transfers. Note that, similar to the proof given in [8], there are no mutually
beneficial valuation transfers for initial allocations that satisfy the
conditions of Case 4. Thus, we only focus on Cases 1, 2, and 3 in the
following proofs.

#### Region 1: $X_{1}\geq 1$, $X_{2}\geq 1$

######  Proposition 4.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{1}$ and
$\frac{2X_{1}X_{2}-X_{1}-X_{2}}{2X_{1}^{2}}<\frac{\phi_{2}}{\phi_{1}}<\frac{2X_{2}-1}{2X_{1}}$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(1-\frac{1}{2X_{1}}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\phi_{1}-\tau$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(1-\frac{1}{2X_{2}}\right)$
For the existence of a mutually beneficial valuation transfer from Case 1a to
1b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{X_{2}}{X_{1}}$ (3)
where (3) insists that the post-transfer parameters satisfy the conditions of
Case 1b. A necessary and sufficient condition for the existence of a mutually
beneficial transfer that satisfies (1) and (2) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{2X_{2}-1}{2X_{1}}$ and a
necessary and sufficient condition for the existence of a mutually beneficial
transfer that satisfies (1) and (3) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}>\frac{2X_{1}X_{2}-X_{1}-X_{2}}{2X_{1}^{2}}$

∎

#### Region 2: $X_{1}\geq 1$, $X_{2}<1$

######  Proposition 5.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{2}$ and
$\frac{1}{X_{1}X_{2}}\left(1-\frac{1+X_{1}X_{2}}{2X_{1}}\right)<\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}}{2X_{1}(2-X_{2})}$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(1-\frac{1}{2X_{1}}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\phi_{1}-\tau$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(\frac{X_{2}}{2}\right)$ For
the existence of a mutually beneficial valuation transfer from Case 1a to 1b,
the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{1}{X_{1}X_{2}}$ (3)
where (3) insists that the post-transfer parameters satisfy the conditions of
Case 1b. A necessary and sufficient condition for the existence of a mutually
beneficial transfer that satisfies (1) and (2) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}}{2X_{1}(2-X_{2})}$ and a
necessary and sufficient condition for the existence of a mutually beneficial
transfer that satisfies (1) and (3) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}>\frac{1}{X_{1}X_{2}}\left(1-\frac{1+X_{1}X_{2}}{2X_{1}}\right)$

∎

######  Proposition 6.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{2}$ and
$\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}+4(X_{1}X_{2})^{\frac{1}{2}}}{16X_{1}-X_{2}}$
and

$\max\left\\{z_{2}^{-},\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

, where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of the equation
$\tau^{2}+(\phi_{2}-\phi_{1})\tau+4\frac{X_{1}}{X_{2}}\phi_{2}^{2}-\phi_{1}\phi_{2}=0$,
and $z_{2}^{+}$ is the second zero of the equation
$\left(1+\frac{(2X_{1}-1)^{2}}{X_{1}X_{2}}\right)^{2}\tau^{2}+(\phi_{2}-\phi_{1})\tau-\phi_{1}\phi_{2}=0$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(1-\frac{1}{2X_{1}}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=(\phi_{1}-\tau)\left(1-\frac{1}{2X_{1}}\right)+\frac{1}{2}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{\frac{1}{2}}$
For the existence of a mutually beneficial valuation transfer from Case 1a to
2b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{X_{2}}{X_{1}}$ (3)
$\displaystyle
0<1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}$
(4) where (3) and (4) insist that the post-transfer parameters satisfy the
conditions of Case 2b. First, notice that (1) and (2) can be rewritten as
$\displaystyle\left(1+\frac{(2X_{1}-1)^{2}}{X_{1}X_{2}}\right)^{2}\tau^{2}+(\phi_{2}-\phi_{1})\tau-\phi_{1}\phi_{2}<0$
$\displaystyle\tau^{2}+(\phi_{2}-\phi_{1})\tau+4\frac{X_{1}}{X_{2}}\phi_{2}^{2}-\phi_{1}\phi_{2}<0$
These conditions can only be satisfied if these quadratic inequalities have
real solutions. Notice that the discriminant of the first inequality is always
positive, so this is trivially true. However, the second quadratic has real
solutions if and only if the condition
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}+4(X_{1}X_{2})^{\frac{1}{2}}}{16X_{1}-X_{2}}$
is satisfied. Notice that (3) and (4) can be written as
$\displaystyle\tau>\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})$
$\displaystyle\tau<\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})$ Thus,
if the previous two conditions hold and
$\displaystyle\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})<\frac{2X_{1}}{2X_{1}-1}\phi_{2}<\frac{1}{X_{1}+X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})$
then a mutually beneficial valuation transfer exists. A comprehensive
necessary and sufficient condition is thus given by
$\displaystyle\max\left\\{z_{2}^{-},\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{X_{1}+X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

where $z_{1}^{+}$ is the second zero of the quadratic form of (2) and
$z_{2}^{-}$ and $z_{2}^{+}$ are the first and second zeros, respectively, of
the quadratic form of (2). ∎

#### Region 3: $X_{1}<1$, $X_{2}\geq 1$

######  Proposition 7.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{3}$ and
$\frac{\phi_{2}}{\phi_{1}}>2(X_{1}X_{2})^{\frac{1}{2}}-1$ and

$\max\left\\{z^{-},\frac{1}{1+X_{1}X_{2}}(X_{1}X_{2}\phi_{1}-\phi_{2})\right\\}<\min\left\\{z^{+},\frac{1}{X_{1}+X_{2}}(X_{2}\phi_{1}-X_{1}\phi_{2})\right\\}$

, where $z^{-}$ and $z^{+}$ are the first and second zeros, respectively, of
the equation
$\tau^{2}+(\phi_{2}-\phi_{1})\tau+X_{1}X_{2}\phi_{1}^{2}-\phi_{1}\phi_{2}=0$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2a}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(\frac{X_{1}}{2}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{\frac{1}{2}}$
For the existence of a mutually beneficial valuation transfer from Case 1a to
2a, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{1}-\tau}{\phi_{2}+\tau}>\frac{X_{1}}{X_{2}}$ (3)
$\displaystyle
0<1-\left(\frac{X_{1}X_{2}(\phi_{1}-\tau)}{\phi_{2}+\tau}\right)^{\frac{1}{2}}$
(4) where (3) and (4) insist that the post-transfer parameters satisfy the
conditions of Case 2a. First, notice that (1) can also be rewritten as
$\displaystyle\tau^{2}+(\phi_{2}-\phi_{1})\tau+X_{1}X_{2}\phi_{1}^{2}-\phi_{1}\phi_{2}<0$
This condition can only be satisfied if this quadratic inequality has real
solutions, which is true if and only if the condition
$\displaystyle\frac{\phi_{2}}{\phi_{1}}>2(X_{1}X_{2})^{\frac{1}{2}}-1$ is
satisfied. Next, notice that (1) and (2) can be rewritten as
$\displaystyle\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{\frac{1}{2}}>\phi_{1}\left(\frac{X_{1}}{2}\right)$
$\displaystyle\frac{1}{2}\left(\frac{X_{1}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{2}}\right)^{\frac{1}{2}}>\frac{1}{2X_{2}}(\phi_{2}+\tau)-\tau$
Observe that if (1) is true, then (2) is true because
$\displaystyle\phi_{1}\left(\frac{X_{1}}{2}\right)>\frac{1}{2X_{2}}(\phi_{2}+\tau)-\tau\iff
X_{1}X_{2}\phi_{1}-\phi_{2}>(1-2X_{2})\tau$ where the right side of the
inequality is strictly positive because the initial allocation belongs to Case
1a, and the left side of the inequality is strictly negative because
$X_{2}\geq 1$ by assumption and $\tau>0$ by Theorem 1. Finally, (3) and (4)
can be written as
$\displaystyle\tau>\frac{1}{1+X_{1}X_{2}}(X_{1}X_{2}\phi_{1}-\phi_{2})$
$\displaystyle\tau<\frac{1}{X_{1}+X_{2}}(X_{2}\phi_{1}-X_{1}\phi_{2})$ Thus, a
comprehensive necessary and sufficient condition is given by
$\displaystyle\max\left\\{z^{-},\frac{1}{1+X_{1}X_{2}}(X_{1}X_{2}\phi_{1}-\phi_{2})\right\\}<\min\left\\{z^{+},\frac{1}{X_{1}+X_{2}}(X_{2}\phi_{1}-X_{1}\phi_{2})\right\\}$

where $z^{-}$ and $z^{+}$ are the first and second zeros, respectively, of the
quadratic form of (1). ∎

######  Proposition 8.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{3}$ and
$\frac{1}{2}(X_{1}+X_{2}-2)<\frac{\phi_{2}}{\phi_{1}}<\frac{1}{2}(2-X_{1})(2X_{2}-1)$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(\frac{X_{1}}{2}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\phi_{1}-\tau$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(1-\frac{1}{2X_{2}}\right)$
For the existence of a mutually beneficial valuation transfer from Case 1a to
1b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{X_{2}}{X_{1}}$ (3)
where (3) insists that the post-transfer parameters satisfy the conditions of
Case 1b. A necessary and sufficient condition for the existence of a mutually
beneficial transfer that satisfies (1) and (2) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{1}{2}(2-X_{1})(2X_{2}-1)$ and a
necessary and sufficient condition for the existence of a mutually beneficial
transfer that satisfies (1) and (3) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}>\frac{1}{2}(X_{1}+X_{2}-2)$

∎

######  Proposition 9.

Take any $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}\cap A_{3}$ and consider
the quadratic equation
$q(r)=r^{2}+\left(2-\frac{(X_{1}+X_{2})^{2}}{4X_{1}X_{2}}\right)r+1=0$. Let
$r^{-}$ and $r^{+}$ be the first and second zeros, respectively, of $q$.
Define $(z^{-},z^{+})=(r^{-},r^{+})$ if $(r^{-},r^{+})$ are real and
$(z^{-},z^{+})=(\infty,0)$ otherwise. If
$\frac{\phi_{2}}{\phi_{1}}\in\\{(0,z^{-})\cup(z^{+},\infty)\cap(0,2(X_{1}X_{2})^{\frac{1}{2}}-1)\\}$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(0)=\phi_{2}\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}\phi_{2}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\phi_{1}-\tau$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(1-\frac{1}{2X_{2}}\right)$
For the existence of a mutually beneficial valuation transfer from Case 2a to
1b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{X_{2}}{X_{1}}$ (3)
where (3) insists that the post-transfer parameters satisfy the conditions of
Case 1b. Conditions (1) and (2) can be rewritten as
$\displaystyle\tau<\phi_{1}-\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\tau>\frac{1}{2X_{1}-1}(X_{1}X_{2}\phi_{1}\phi_{2})^{\frac{1}{2}}$
Thus, a necessary and sufficient condition for the existence of a mutually
beneficial transfer that satisfies (1) and (2) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}}{X_{1}}\left(\frac{4X_{2}-2}{4X_{2}-1}\right)^{2}$
Similarly, (3) can be rewritten as
$\displaystyle\tau>\frac{1}{X_{1}+X_{2}}(X_{2}\phi_{1}-X_{1}\phi_{2})$ and a
corresponding necessary and sufficient condition for the existence of a
mutually beneficial transfer that satisfies (1) and (3) is given by
$\displaystyle\phi_{2}^{2}+\left(2-\frac{(X_{1}+X_{2})^{2}}{4X_{1}X_{2}}\right)\phi_{1}\phi_{2}+\phi_{1}^{2}>0$

Notice that if this quadratic has imaginary solutions, then the inequality is
trivially satisfied; otherwise, it is satisfied for values of
$\frac{\phi_{2}}{\phi_{1}}$ that lie outside of the interval $[z^{-},z^{+}]$,
where $z^{-}$ is the first zero and $z^{+}$ is the second zero. ∎

#### Region 4: $X_{1}<1$, $X_{2}<1$, $X_{1}+X_{2}\geq 1$

######  Proposition 10.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{4}$ and
$\frac{X_{1}X_{2}-2X_{2}+1}{2X_{2}}<\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}(2-X_{1})}{2(2-X_{2})}$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(\frac{X_{1}}{2}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\phi_{1}-\tau$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(\frac{X_{2}}{2}\right)$ For
the existence of a mutually beneficial valuation transfer from Case 1a to 1b,
the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{1}{X_{1}X_{2}}$ (3)
where (3) insists that the post-transfer parameters satisfy the conditions of
Case 1b. A necessary and sufficient condition for the existence of a mutually
beneficial transfer that satisfies (1) and (2) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}(2-X_{1})}{2(2-X_{2})}$ and
a necessary and sufficient condition for the existence of a mutually
beneficial transfer that satisfies (1) and (3) is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}>\frac{X_{1}X_{2}-2X_{2}+1}{2X_{2}}$

∎

######  Proposition 11.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{4}$ and
$\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}+4(X_{1}X_{2})^{\frac{1}{2}}}{16X_{1}-X_{2}}$
and

$\max\left\\{z_{1}^{-},\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

, where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of the equation
$\tau^{2}+(\phi_{2}-\phi_{1})\tau+4\frac{X_{1}}{X_{2}}\phi_{2}^{2}-\phi_{1}\phi_{2}=0$,
and $z_{2}^{+}$ is the second zero of the equation
$\left(1+\frac{(2X_{1}-1)^{2}}{X_{1}X_{2}}\right)^{2}\tau^{2}+\left(\phi_{2}+\phi_{1}\left(\frac{2(X_{1}-1)^{2}(2X_{1}-1)}{X_{1}X_{2}}-1\right)\right)\tau+\frac{(X_{1}-1)^{4}}{X_{1}X_{2}}\phi_{1}^{2}-\phi_{1}\phi_{2}=0$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(\frac{X_{1}}{2}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=(\phi_{1}-\tau)\left(1-\frac{1}{2X_{1}}\right)+\frac{1}{2}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{\frac{1}{2}}$
For the existence of a mutually beneficial valuation transfer from Case 1a to
2b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{X_{2}}{X_{1}}$ (3)
$\displaystyle
0<1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}$
(4) where (3) and (4) insist that the post-transfer parameters satisfy the
conditions of Case 2b. First, observe that (3) and (4) can be written as
$\displaystyle\tau>\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})$
$\displaystyle\tau<\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})$
Consider any large enough transfer
$\tau^{\prime}=\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})+\epsilon$
where $\epsilon>0$. We have $\displaystyle\lim_{\epsilon\to
0}\pi_{1}(\tau^{\prime})=\frac{\phi_{1}+\phi_{2}}{2(X_{1}+X_{2})}(2X_{1}+X_{2}-1)$
(5) which is strictly positive because $X_{1}+X_{2}\geq 1$. Then, notice that
(1) and (2) can be rewritten as
$\displaystyle\left(1+\frac{(2X_{1}-1)^{2}}{X_{1}X_{2}}\right)^{2}\tau^{2}+\left(\phi_{2}+\phi_{1}\left(\frac{2(X_{1}-1)^{2}(2X_{1}-1)}{X_{1}X_{2}}-1\right)\right)\tau+\frac{(X_{1}-1)^{4}}{X_{1}X_{2}}\phi_{1}^{2}-\phi_{1}\phi_{2}<0$
$\displaystyle\tau^{2}+(\phi_{2}-\phi_{1})\tau+4\frac{X_{1}}{X_{2}}\phi_{2}^{2}-\phi_{1}\phi_{2}<0$
These conditions can only be satisfied if these quadratic inequalities have
real solutions. However, since $\pi_{1}(\tau^{\prime})>0$ and $\pi_{1}$ is a
continuous, concave function, (1) must have real solutions; thus, we only need
$\pi_{1}(\tau)<z^{+}$ in order to ensure that (1) is satisfied. However, the
second quadratic has real solutions if and only if the condition
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}+4(X_{1}X_{2})^{\frac{1}{2}}}{16X_{1}-X_{2}}$
is satisfied. Thus, a necessary and sufficient condition for the existence of
a mutually beneficial transfer is given by
$\displaystyle\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}+4(X_{1}X_{2})^{\frac{1}{2}}}{16X_{1}-X_{2}}$
$\displaystyle\text{and}\max\left\\{z_{1}^{-},\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

∎

######  Proposition 12.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}\cap A_{4}$ and

$\max\left\\{\frac{1}{X_{2}}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-\frac{(X_{2}-1)^{2}}{X_{2}^{2}}\phi_{2},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}<\phi_{1}-\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$

then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(0)=\phi_{2}\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}\phi_{2}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\phi_{1}-\tau$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau)\left(\frac{X_{2}}{2}\right)$ For
the existence of a mutually beneficial valuation transfer from Case 2a to 1b,
the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{1}{X_{1}X_{2}}$ (3)
where (3) insists that the post-transfer parameters satisfy the conditions of
Case 1b. The conditions above can be rewritten as
$\displaystyle\tau<\phi_{1}-\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\tau>\frac{1}{X_{2}}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-\frac{(X_{2}-1)^{2}}{X_{2}^{2}}\phi_{2}$
$\displaystyle\tau>\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})$ Thus,
a necessary and sufficient condition for the existence of a mutually
beneficial transfer is given by
$\displaystyle\max\left\\{\frac{1}{X_{2}}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-\frac{(X_{2}-1)^{2}}{X_{2}^{2}}\phi_{2},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}<\phi_{1}-\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$

∎

######  Proposition 13.

Consider the quadratic equations

$\displaystyle q_{1}(\tau)=$
$\displaystyle\left(\left(\frac{1}{X_{1}}-2\right)^{2}+\frac{X_{2}}{X_{1}}\right)\tau^{2}$
$\displaystyle-$
$\displaystyle\left(2\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}+\left(\frac{1}{X_{1}}-2\right)\phi_{1}\right)\left(\frac{1}{X_{1}}-2\right)+\frac{X_{2}}{X_{1}}(\phi_{1}-\phi_{2})\right)\tau$
$\displaystyle+$
$\displaystyle\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}+\left(\frac{1}{X_{1}}-2\right)\phi_{1}\right)^{2}-\frac{X_{2}}{X_{1}}\phi_{1}\phi_{2}$
and $\displaystyle q_{2}(\tau)=$
$\displaystyle\tau^{2}+(\phi_{2}-\phi_{1})\tau+\frac{X_{1}}{X_{2}}\left(\left(2-\frac{1}{X_{2}}\right)\phi_{2}+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)^{2}-\phi_{1}\phi_{2}$

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}\cap A_{4}$, if the discriminants
of $q_{1}$ and $q_{2}$ are positive, and if

$\max\left\\{z_{1}^{-},z_{2}^{-},\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of $q_{1}$, and $z_{2}^{-}$ and $z_{2}^{+}$ are the first and
second zeros, respectively, of $q_{2}$, then there exists a mutually
beneficial valuation transfer from Player 1 to Player 2 such that
$(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2b}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(0)=\phi_{2}\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=(\phi_{1}-\tau)\left(1-\frac{1}{2X_{1}}\right)+\frac{1}{2}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}\left(\frac{X_{2}(\phi_{1}-\tau)(\phi_{2}+\tau)}{X_{1}}\right)^{\frac{1}{2}}$
For the existence of a mutually beneficial valuation transfer from Case 2a to
2b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2)
$\displaystyle\frac{\phi_{2}+\tau}{\phi_{1}-\tau}>\frac{X_{2}}{X_{1}}$ (3)
$\displaystyle
0<1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}$
(4) where (3) and (4) insist that the post-transfer parameters satisfy the
conditions of Case 2b. First, observe that (3) and (4) can be rewritten as
$\displaystyle\tau>\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})$
$\displaystyle\tau<\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})$ Next,
notice that (1) and (2) can be rewritten as
$\displaystyle\left(1+\left(\frac{1-2X_{1}}{X_{2}}\right)^{2}\right)\tau^{2}-\left(\phi_{1}-\phi_{2}+2\left(\frac{1-2X_{1}}{X_{2}}\right)^{2}\phi_{1}+\frac{2X_{1}(1-2X_{1})}{X_{2}^{2}}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)\tau$
$\displaystyle+\frac{X_{1}^{3}}{X_{2}^{3}}\phi_{1}\phi_{2}+\frac{2X_{1}(1-2X_{1}}{X_{2}^{2}}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\phi_{1}+\left(\frac{1-2X_{1}}{X_{2}}\right)^{2}\phi_{1}^{2}-\phi_{1}\phi_{2}<0$
$\displaystyle\tau^{2}+(\phi_{2}-\phi_{1})\tau+\left(\frac{X_{1}}{X_{2}}\left(\phi_{2}\left(2-\frac{1}{X_{2}}\right)+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)\right)^{2}<0$
These inequalities can only be satisfied if the discriminant of each quadratic
equation is positive. Thus, assuming this is true, a necessary and sufficient
condition is given by
$\displaystyle\max\left\\{z_{1}^{-},z_{2}^{-},\frac{1}{X_{1}+X_{2}}(\phi_{1}X_{2}-\phi_{2}X_{1})\right\\}<\min\left\\{z_{1}^{+},z_{2}+,\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of the quadratic form of (1), and $z_{2}^{-}$ and $z_{2}^{+}$
are the first and second zeros, respectively, of the quadratic form of (2). ∎

#### Region 5: $X_{1}+X_{2}<1$

######  Proposition 14.

Consider the quadratic equations

$\displaystyle
q_{1}(\tau)=\left(1+\frac{X_{1}}{X_{2}}\right)\tau^{2}+(\phi_{1}-\phi_{2})\tau-\phi_{1}\phi_{2}$
and $\displaystyle
q_{2}(\tau)=X_{2}(X_{1}+X_{2})\tau^{2}+2X_{2}((X_{1}+X_{2}-2)\phi_{2}-X_{1}\phi_{1})\tau+(2-X_{2})^{2}\phi_{2}^{2}-X_{1}X_{2}\phi_{1}\phi_{2}$

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{5}$, if the discriminant
of $q_{2}$ is positive, and if

$\max\left\\{z_{1}^{-},z_{2}^{-},\frac{X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}}{X_{1}X_{2}+(1-X_{2})^{2}}\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{(1-X_{1})^{2}\phi_{1}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of $q_{1}$, and $z_{2}^{-}$ and $z_{2}^{+}$ are the first and
second zeros, respectively, of $q_{2}$, then there exists a mutually
beneficial valuation transfer from Player 1 to Player 2 such that
$(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{3}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\phi_{1}\left(\frac{X_{1}}{2}\right)$
$\displaystyle\pi_{2}(0)=\phi_{2}$ The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\frac{1}{2}(\phi_{1}-\tau)\left(X_{1}+\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}\right)$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}(\phi_{2}+\tau)\left(X_{2}+\left(\frac{X_{1}X_{2}(\phi_{1}-\tau)}{\phi_{2}+\tau}\right)^{\frac{1}{2}}\right)$
For the existence of a mutually beneficial valuation transfer from Case 2a to
1b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2) $\displaystyle
1-\left(\frac{X_{1}X_{2}(\phi_{1}-\tau}{\phi_{2}+\tau}\right)^{\frac{1}{2}}>X_{2}$
(3) $\displaystyle
1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau}{\phi_{1}-\tau}\right)^{\frac{1}{2}}>X_{1}$
(4) where (3) and (4) insist that the post-transfer parameters satisfy the
conditions of Case 3. Notice that (1) and (2) can be rewritten as
$\displaystyle\left(1+\frac{X_{1}}{X_{2}}\right)\tau^{2}+(\phi_{1}-\phi_{2})\tau-\phi_{1}\phi_{2}<0$
$\displaystyle
X_{2}(X_{1}+X_{2})\tau^{2}+2X_{2}((X_{1}+X_{2}-2)\phi_{2}-X_{1}\phi_{1})\tau+(2-X_{2})^{2}\phi_{2}^{2}-X_{1}X_{2}\phi_{1}\phi_{2}<0$
The discriminant of the first quadratic is positive, so it will have real
solutions, but we must have that the discriminant of the second quadratic is
also positive so that (2) has real solutions. Next, (3) and (4) can be
rewritten as
$\displaystyle\tau>\frac{X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}}{X_{1}X_{2}+(1-X_{2})^{2}}$
$\displaystyle\tau<\frac{(1-X_{1})^{2}\phi_{1}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}$
A necessary and sufficient condition for the existence of a mutually
beneficial transfer is given by
$\displaystyle\max\left\\{z_{1}^{-},z_{2}^{-},\frac{X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}}{X_{1}X_{2}+(1-X_{2})^{2}}\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{(1-X_{1})^{2}\phi_{1}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of the quadratic form of (1), and $z_{2}^{-}$ and $z_{2}^{+}$
are the first and second zeros, respectively, of the quadratic form of (2). ∎

######  Proposition 15.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{1a}\cap A_{5}$ and
$\frac{\phi_{2}}{\phi_{1}}<\frac{X_{2}+4(X_{1}X_{2})^{\frac{1}{2}}}{16X_{1}-X_{2}}$
and

$\max\left\\{z_{2}^{-},\frac{(1-X_{1})^{2}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of the equation
$\tau^{2}+(\phi_{2}-\phi_{1})\tau+4\frac{X_{1}}{X_{2}}\phi_{2}^{2}-\phi_{1}\phi_{2}=0$,
and $z_{2}^{+}$ is the second zero of the equation
$\left(1+\frac{(2X_{1}-1)^{2}}{X_{1}X_{2}}\right)^{2}\tau^{2}+(\phi_{2}-\phi_{1})\tau-\phi_{1}\phi_{2}=0$,
then there exists a mutually beneficial valuation transfer from Player 1 to
Player 2 such that $(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2b}$.

###### Proof.

The proof is the same as the proof of Proposition 6, except now, the lower
bound on $\tau$ is determined by

$\displaystyle
1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}>X_{1}$
which can be rewritten as
$\displaystyle\tau>\frac{(1-X_{1})^{2}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}$

∎

######  Proposition 16.

Consider the quadratic equations

$\displaystyle q_{1}(\tau)=$
$\displaystyle(X_{1}^{2}+X_{1}X_{2})\tau^{2}+\left(2X_{1}\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-X_{1}\phi_{1}\right)-X_{1}X_{2}(\phi_{1}-\phi_{2})\right)\tau$
$\displaystyle+$
$\displaystyle\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-X_{1}\phi_{1}\right)^{2}-X_{1}X_{2}\phi_{1}\phi_{2}$
and $\displaystyle 0>q_{2}(\tau)=$
$\displaystyle(X_{2}^{2}+X_{1}X_{2})\tau^{2}$ $\displaystyle-$
$\displaystyle\left(2X_{2}\left(\phi_{2}\left(2-\frac{1}{X_{2}}-X_{2}\right)+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)+X_{1}X_{2}(\phi_{1}-\phi_{2})\right)\tau$
$\displaystyle+$
$\displaystyle\left(\phi_{2}\left(2-\frac{1}{X_{2}}-X_{2}\right)+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)^{2}-X_{1}X_{2}\phi_{1}\phi_{2}$

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}\cap A_{5}$, if the discriminants
of $q_{1}$ and $q_{2}$ are positive, and if

$\max\left\\{z_{1}^{-},z_{2}^{-},\frac{X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}}{X_{1}X_{2}+(1-X_{2})^{2}}\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{(1-X_{1})^{2}\phi_{1}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of $q_{1}$, and $z_{2}^{-}$ and $z_{2}^{+}$ are the first and
second zeros, respectively, of $q_{2}$, then there exists a mutually
beneficial valuation transfer from Player 1 to Player 2 such that
$(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{3}$.

###### Proof.

The initial payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(0)=\phi_{2}\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}\phi_{2}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau)=\frac{1}{2}(\phi_{1}-\tau)\left(X_{1}+\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}\right)$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}(\phi_{2}+\tau)\left(X_{2}+\left(\frac{X_{1}X_{2}(\phi_{1}-\tau)}{\phi_{2}+\tau}\right)^{\frac{1}{2}}\right)$
For the existence of a mutually beneficial valuation transfer from Case 2a to
1b, the following inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau)>\pi_{1}(0)$ (1)
$\displaystyle\pi_{2}(\tau)>\pi_{2}(0)$ (2) $\displaystyle
1-\left(\frac{X_{1}X_{2}(\phi_{1}-\tau}{\phi_{2}+\tau}\right)^{\frac{1}{2}}>X_{2}$
(3) $\displaystyle
1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau}{\phi_{1}-\tau}\right)^{\frac{1}{2}}>X_{1}$
(4)

where (3) and (4) insist that the post-transfer parameters satisfy the
conditions of Case 3. Notice that (1) and (2) can be rewritten as

$\displaystyle 0>$
$\displaystyle(X_{1}^{2}+X_{1}X_{2})\tau^{2}+\left(2X_{1}\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-X_{1}\phi_{1}\right)-X_{1}X_{2}(\phi_{1}-\phi_{2})\right)\tau$
$\displaystyle+$
$\displaystyle\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}-X_{1}\phi_{1}\right)^{2}-X_{1}X_{2}\phi_{1}\phi_{2}$
$\displaystyle 0>$
$\displaystyle(X_{2}^{2}+X_{1}X_{2})\tau^{2}-\left(2X_{2}\left(\phi_{2}\left(2-\frac{1}{X_{2}}-X_{2}\right)+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)+X_{1}X_{2}(\phi_{1}-\phi_{2})\right)\tau$
$\displaystyle+$
$\displaystyle\left(\phi_{2}\left(2-\frac{1}{X_{2}}-X_{2}\right)+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)^{2}-X_{1}X_{2}\phi_{1}\phi_{2}$

In order for these inequalities to be satisfied, the discriminant of each
quadratic must be positive. Next, (3) and (4) can be rewritten as

$\displaystyle\tau>\frac{X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}}{X_{1}X_{2}+(1-X_{2})^{2}}$
$\displaystyle\tau<\frac{(1-X_{1})^{2}\phi_{1}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}$
A necessary and sufficient condition for the existence of a mutually
beneficial transfer is given by
$\displaystyle\max\left\\{z_{1}^{-},z_{2}^{-},\frac{X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}}{X_{1}X_{2}+(1-X_{2})^{2}}\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{(1-X_{1})^{2}\phi_{1}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of the quadratic form of (1), and $z_{2}^{-}$ and $z_{2}^{+}$
are the first and second zeros, respectively, of the quadratic form of (2). ∎

######  Proposition 17.

Consider the quadratic equations

$\displaystyle q_{1}(\tau)=$
$\displaystyle\left(\left(\frac{1}{X_{1}}-2\right)^{2}+\frac{X_{2}}{X_{1}}\right)\tau^{2}$
$\displaystyle-$
$\displaystyle\left(2\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}+\left(\frac{1}{X_{1}}-2\right)\phi_{1}\right)\left(\frac{1}{X_{1}}-2\right)+\frac{X_{2}}{X_{1}}(\phi_{1}-\phi_{2})\right)\tau$
$\displaystyle+$
$\displaystyle\left(\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}+\left(\frac{1}{X_{1}}-2\right)\phi_{1}\right)^{2}-\frac{X_{2}}{X_{1}}\phi_{1}\phi_{2}$
and $\displaystyle q_{2}(\tau)=$
$\displaystyle\tau^{2}+(\phi_{2}-\phi_{1})\tau+\frac{X_{1}}{X_{2}}\left(\left(2-\frac{1}{X_{2}}\right)\phi_{2}+\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)^{2}-\phi_{1}\phi_{2}$

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}\cap A_{5}$, if the discriminants
of $q_{1}$ and $q_{2}$ are positive, and if

$\max\left\\{z_{1}^{-},z_{2}^{-},\frac{(1-X_{1})^{2}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}\right\\}<\min\left\\{z_{1}^{+},z_{2}^{+},\frac{1}{1+X_{1}X_{2}}(\phi_{1}-X_{1}X_{2}\phi_{2})\right\\}$

where $z_{1}^{-}$ and $z_{1}^{+}$ are the first and second zeros,
respectively, of $q_{1}$, and $z_{2}^{-}$ and $z_{2}^{+}$ are the first and
second zeros, respectively, of $q_{2}$, then there exists a mutually
beneficial valuation transfer from Player 1 to Player 2 such that
$(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2b}$.

###### Proof.

The proof is the same as the proof of Proposition 13, except now, the lower
bound on $\tau$ is determined by

$\displaystyle
1-\left(\frac{X_{1}X_{2}(\phi_{2}+\tau)}{\phi_{1}-\tau}\right)^{\frac{1}{2}}>X_{1}$
which can be rewritten as
$\displaystyle\tau>\frac{(1-X_{1})^{2}-X_{1}X_{2}\phi_{2}}{(1-X_{1})^{2}+X_{1}X_{2}}$

∎

Gathering the results of Propositions 2-17, we can now restate Theorem 1:

######  Theorem 1.

Consider any coalitional Blotto game parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}$ such that
$\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$. There exists a mutually
beneficial battlefield transfer if and only if $g$ satisfies at least one of
the conditions from Propositions 2 through Proposition 17.

The proof of Theorem 1 follows directly from the proofs of each of the
individual propositions. Moreover, the propositions cover the entire space
$\mathcal{G}$ where $\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$, so the
conditions are both necessary and sufficient.

###  4.3 Directional proofs

####  4.3.1 Intra-case directional proofs

######  Proposition 18.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{2a}$, then there does not exist a
mutually beneficial valuation transfer from Player 2 to Player 1 such that
$(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{1a}$.

###### Proof.

First, notice that a transfer of this form must be negative; thus, for this
proof, we define a positive variable $\tau^{\prime}=-\tau>0$. The initial
payoffs for each player are given by

$\displaystyle\pi_{1}(0)=\frac{1}{2}\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(0)=\phi_{2}\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}\phi_{2}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}$
The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau^{\prime})=\frac{1}{2}(\phi_{1}+\tau^{\prime})X_{1}$
$\displaystyle\pi_{2}(\tau^{\prime})=\phi_{2}-\tau^{\prime}$ For the existence
of a mutually beneficial valuation transfer from Case 2a to 1b, the following
inequalities must be satisfied:
$\displaystyle\pi_{1}(\tau^{\prime})>\pi_{1}(0)$
$\displaystyle\pi_{2}(\tau^{\prime})>\pi_{2}(0)$ which can be rewritten as
$\displaystyle\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}-\phi_{1}<\tau^{\prime}<\frac{1}{2}\left(\frac{\phi_{2}}{X_{2}}-\left(\frac{X_{1}\phi_{1}\phi_{2}}{X_{2}}\right)^{\frac{1}{2}}\right)$
A transfer satisfying these conditions can exist if and only if
$\displaystyle\frac{\phi_{2}}{\phi_{1}}X_{1}\left(X_{1}X_{2}-\frac{\phi_{2}}{\phi_{1}}\right)<4X_{2}\left(X_{1}X_{2}-\frac{\phi_{2}}{\phi_{1}}\right)$
Since the initial allocation is in Case 2, we know that
$\left(X_{1}X_{2}-\frac{\phi_{2}}{\phi_{1}}\right)<0$, so we can divide each
side and rewrite the condition as
$\displaystyle\frac{\phi_{2}}{\phi_{1}}X_{1}>4X_{2}$

which is not satisfied for any initial allocation in Case 2. ∎

######  Proposition 19.

If $(\phi_{1},\phi_{2},X_{1},X_{2})\in C_{3a}$, then there does not exist a
mutually beneficial valuation transfer from Player 2 to Player 1 such that
$(\phi_{1}^{\tau},\phi_{2}^{\tau},X_{1},X_{2})\in C_{2a}$.

###### Proof.

First, notice that a transfer of this form must be negative; thus, for this
proof, we define a positive variable $\tau^{\prime}=-\tau>0$. However, by
Propositions 2 and 3, within Cases 2 and 3 we know that the payoff functions
of each player are simultaneously increasing if and only if $\tau>0$. Thus, we
know that within each case, the payoffs of each player can never be
simultaneously increasing. It is therefore sufficient to establish that there
does not exist a transfer from Case 3a to 2a such that the payoffs of each
player are equal and positive, since if this intersection were negative, then
there cannot exist any such transfer. First, the initial payoffs of each
player are given by

$\displaystyle\pi_{1}(\tau)=\frac{1}{2}\phi_{1}\left(X_{1}+\left(\frac{X_{1}X_{2}\phi_{2}}{\phi_{1}}\right)^{\frac{1}{2}}\right)$
$\displaystyle\pi_{2}(\tau)=\frac{1}{2}\phi_{2}\left(X_{2}+\left(\frac{X_{1}X_{2}\phi_{1}}{\phi_{2}}\right)^{\frac{1}{2}}\right)$
The post-transfer payoffs are given by
$\displaystyle\pi_{1}(\tau^{\prime})=\frac{1}{2}\left(\frac{X_{1}(\phi_{1}+\tau^{\prime})(\phi_{2}-\tau^{\prime})}{X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau^{\prime})=(\phi_{2}-\tau^{\prime})\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}(\phi_{1}+\tau^{\prime})(\phi_{2}-\tau^{\prime})}{X_{2}}\right)^{\frac{1}{2}}$
The transfer that equates the two payoffs is given by
$\displaystyle\tau_{\text{eq}}^{\prime}=\frac{1}{2X_{2}-1}\left(X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}\right)$
Notice that in order for $\tau_{\text{eq}}^{\prime}$ to be positive, we need
that $X_{2}<\frac{1}{2}$ since the initial allocation must satisfy
$X_{1}X_{2}\phi_{1}-(1-X_{2})^{2}\phi_{2}<0$ because it lies in Case 3. Thus,
the conditions
$\displaystyle\frac{1}{2}\left(\frac{X_{1}(\phi_{1}+\tau_{\text{eq}}^{\prime})(\phi_{2}-\tau_{\text{eq}}^{\prime})}{X_{2}}\right)^{\frac{1}{2}}>\frac{1}{2}\phi_{1}\left(X_{1}+\left(\frac{X_{1}X_{2}\phi_{2}}{\phi_{1}}\right)^{\frac{1}{2}}\right)$
$\displaystyle(\phi_{2}-\tau_{\text{eq}}^{\prime})\left(1-\frac{1}{2X_{2}}\right)+\frac{1}{2}\left(\frac{X_{1}(\phi_{1}+\tau_{\text{eq}}^{\prime})(\phi_{2}-\tau_{\text{eq}}^{\prime})}{X_{2}}\right)^{\frac{1}{2}}>\frac{1}{2}\phi_{2}\left(X_{2}+\left(\frac{X_{1}X_{2}\phi_{1}}{\phi_{2}}\right)^{\frac{1}{2}}\right)$

must be satisfied. However, notice that the intersection of $C_{3a}$ and the
set defined by the first condition is lower bounded by $X_{2}=\frac{1}{2}$;
that is, the first inequality can only be satisfied if $X_{2}\geq\frac{1}{2}$.
However, this contradicts the assumption that $\tau<0$. ∎

Gathering Propositions 1, 2, 3, 18, and 19, we can now restate and prove
Theorem 2, which establishes the directions in which mutually beneficial
battlefield transfers occur.

######  Theorem 2.

Let $\tau_{1\to 2}^{v}$ be a mutually beneficial valuation transfer. Then,
$\tau_{1\to 2}^{v}>0$ if and only if
$\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$.

###### Proof.

First, observe that by Proposition 1, there cannot exist a negative transfer
that remains within Case 1a such that the payoffs of both players strictly
increase; that is, Player 2’s payoff will strictly decrease by a negative
transfer.
Next, observe that by Propositions 2 and 3, in Cases 2 and 3, the
instantaneous derivatives of the payoff functions are never simultaneously
negative. Since the payoff functions are concave, this implies that one player
will always experience a strictly decreasing payoff for transfers in the
negative direction.
The previous arguments establish that there exists a positive mutually
beneficial battlefield transfer from Player 1 to Player 2 that remains within
the same case if and only if Player 1 is weaker than Player 2. By Propositions
18 and 19, we know that there are no mutually beneficial negative transfers
that go from Case 3a to Case 2a, and similarly, there are no mutually
beneficial negative transfers that go from Case 2a to 1a. Since the payoff
functions are continuous along Cases 3a, 2a, and 1a, we can state that there
are also no mutually beneficial negative transfers that go from Case 3a to
Case 1a. Thus, there are no mutually beneficial battlefield transfers where
$\tau<0$. ∎

###  4.4 Joint transfer proofs

Here, we provide sufficient conditions for the existence of a mutually
beneficial joint transfer. Note that these conditions are only sufficient, not
necessary; we only prove the existence of arbitrarily small mutually
beneficial joint transfers, but there may be other forms of mutually
beneficial joint transfers (namely, inter-case transfers as described in
Section 2). The proof of Case 1 is provided in the paper, so we begin our
analysis below with Case 2. Note that, for the sake of consistency with the
work of [8], we assume that budgetary transfers go from Player 2 to Player 1
instead.

######  Proposition 20.

Consider any coalitional Blotto game parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}$ such that
$g\in\text{int}(C_{2a})$. If $\frac{\phi_{1}}{\phi_{2}}>\frac{X_{1}}{X_{2}}$
and $\frac{\phi_{1}}{\phi_{2}}\geq\frac{1}{X_{1}X_{2}}$, then there exists a
mutually beneficial joint transfer if
$\frac{\phi_{1}}{\phi_{1}-\phi_{2}}(1-2X_{2})\neq\frac{X_{1}}{X_{1}+X_{2}}$.

###### Proof.

In Case 2, the payoff functions of each player are given by

$\displaystyle\pi_{1}(\tau)=\frac{1}{2}\left(\frac{(X_{1}+\tau_{1\to
2}^{b})(\phi_{1}-\tau_{1\to 2}^{v})(\phi_{2}+\tau_{1\to
2}^{v})}{X_{2}-\tau_{1\to 2}^{b}}\right)^{\frac{1}{2}}$
$\displaystyle\pi_{2}(\tau)=(\phi_{2}+\tau_{1\to
2}^{v})\left(1-\frac{1}{2(X_{2}-\tau_{1\to
2}^{b})}\right)+\frac{1}{2}\left(\frac{(X_{1}+\tau_{1\to
2}^{b})(\phi_{1}-\tau_{1\to 2}^{v})(\phi_{2}+\tau_{1\to
2}^{v})}{X_{2}-\tau_{1\to 2}^{b}}\right)^{\frac{1}{2}}$ The gradients of each
function are given by
$\displaystyle\nabla\pi_{1}(\tau)=\begin{bmatrix}\frac{((\phi_{1}-\tau_{1\to
2}^{v})(\phi_{2}+\tau_{1\to
2}^{v}))^{\frac{1}{2}}(X_{1}+X_{2})}{4(X_{2}-\tau_{1\to
2}^{b})^{\frac{3}{2}}(X_{1}+\tau_{1\to 2}^{b})^{\frac{1}{2}}}\\\
\frac{1}{4}(\phi_{1}-\phi_{2}-\tau_{1\to 2}^{v})\left(\frac{X_{1}+\tau_{1\to
2}^{b}}{(X_{2}-\tau_{1\to 2}^{b})(\phi_{1}-\tau_{1\to
2}^{v})(\phi_{2}+\tau_{1\to 2}^{v})}\right)^{\frac{1}{2}}\end{bmatrix}$
$\displaystyle\nabla\pi_{2}(\tau)=\begin{bmatrix}-\frac{(\phi_{2}+\tau_{1\to
2}^{v})}{2(X_{2}-\tau_{1\to 2}^{b})^{2}}+\frac{((\phi_{1}-\tau_{1\to
2}^{v})(\phi_{2}+\tau_{1\to
2}^{v}))^{\frac{1}{2}}(X_{1}+X_{2})}{4(X_{2}-\tau_{1\to
2}^{b})^{\frac{3}{2}}(X_{1}+\tau_{1\to 2}^{b})^{\frac{1}{2}}}\\\
1-\frac{1}{2(X_{2}-\tau_{1\to
2}^{b})}+\frac{1}{4}(\phi_{1}-\phi_{2}-\tau_{1\to
2}^{v})\left(\frac{X_{1}+\tau_{1\to 2}^{b}}{(X_{2}-\tau_{1\to
2}^{b})(\phi_{1}-\tau_{1\to 2}^{v})(\phi_{2}+\tau_{1\to
2}^{v})}\right)^{\frac{1}{2}}\end{bmatrix}$ Evaluated at $\tau=0$, the
gradients are
$\displaystyle\nabla\pi_{1}(0)=\begin{bmatrix}\frac{(\phi_{1}\phi_{2})^{\frac{1}{2}}(X_{1}+X_{2})}{4X_{2}^{\frac{3}{2}}X_{1}^{\frac{1}{2}}}\\\
\frac{1}{4}(\phi_{1}-\phi_{2})\left(\frac{X_{1}}{X_{2}\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}\end{bmatrix}$
$\displaystyle\nabla\pi_{2}(0)=\begin{bmatrix}-\frac{\phi_{2}}{2X_{2}^{2}}+\frac{(\phi_{1}\phi_{2})^{\frac{1}{2}}(X_{1}+X_{2})}{4X_{2}^{\frac{3}{2}}X_{1}^{\frac{1}{2}}}\\\
1-\frac{1}{2X_{2}}+\frac{1}{4}(\phi_{1}-\phi_{2})\left(\frac{X_{1}}{X_{2}\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}\end{bmatrix}$
The gradients are diametrically opposed to one another if and only if
$\displaystyle\frac{(\phi_{1}\phi_{2})^{\frac{1}{2}}(X_{1}+X_{2})}{4X_{2}^{\frac{3}{2}}X_{1}^{\frac{1}{2}}}=-\alpha\left(-\frac{\phi_{2}}{2X_{2}^{2}}+\frac{(\phi_{1}\phi_{2})^{\frac{1}{2}}(X_{1}+X_{2})}{4X_{2}^{\frac{3}{2}}X_{1}^{\frac{1}{2}}}\right)$
$\displaystyle\frac{1}{4}(\phi_{1}-\phi_{2})\left(\frac{X_{1}}{X_{2}\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}=-\alpha\left(1-\frac{1}{2X_{2}}+\frac{1}{4}(\phi_{1}-\phi_{2})\left(\frac{X_{1}}{X_{2}\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}\right)$
First, observe that since the first entry of the gradient of the first player
is always nonzero, Player 1 will never be at a maximum. However, the second
player can have a gradient of zero, which occurs precisely at the point where
the two curves defined by $\displaystyle
X_{1}+X_{2}=2\left(\frac{P_{2}X_{1}}{P_{1}X_{2}}\right)^{\frac{1}{2}}$
$\displaystyle\frac{2-4X_{2}}{\phi_{1}-\phi_{2}}=\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}$
intersect, and only if this intersection satisfies the conditions of Case 2.
Note that the first equation above is derived in [8] and the second equation
is derived in Proposition 2. Next, observe that the gradients of Player 1 and
2 can be diametrically opposed if $\phi_{1}=\phi_{2}$, $X_{2}=\frac{1}{2}$,
and
$\displaystyle\frac{\phi_{2}}{2X_{2}^{2}}>\frac{(\phi_{1}\phi_{2})^{\frac{1}{2}}(X_{1}+X_{2})}{4X_{2}^{\frac{3}{2}}X_{1}^{\frac{1}{2}}}$
However, this is not of concern, since when $\phi_{1}=\phi_{2}$,
$X_{2}>\frac{1}{2}$ on the interior of Case 2. Henceforth, we can thus assume
that $\phi_{1}\neq\phi_{2}$. This allows us to combine the two conditions of
diametric opposition into one, given by
$\displaystyle\frac{4}{\phi_{1}-\phi_{2}}\left(1-\frac{1}{2X_{2}}\right)\left(\frac{\phi_{1}\phi_{2}X_{2}}{X_{1}}\right)^{\frac{1}{2}}=-\frac{2}{X_{1}+X_{2}}\left(\frac{\phi_{2}X_{1}}{\phi_{1}X_{2}}\right)^{\frac{1}{2}}$
which can be rewritten as
$\displaystyle\frac{\phi_{1}}{\phi_{1}-\phi_{2}}(1-2X_{2})=\frac{X_{1}}{X_{1}+X_{2}}$

∎

######  Proposition 21.

Consider any coalitional Blotto game parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}$ such that
$g\in\text{int}(C_{3a})$. If $\frac{\phi_{1}}{\phi_{2}}\neq 3$ or
$X_{2}\neq\frac{4\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$, then there
exists a mutually beneficial joint transfer.

###### Proof.

In Case 2, the payoff functions of each player are given by

$\displaystyle\pi_{1}(\tau)=\frac{\phi_{1}-\tau_{1\to
2}^{v}}{2}\left(X_{1}+\tau_{1\to 2}^{b}+\left(\frac{(X_{1}+\tau_{1\to
2}^{b})(X_{2}-\tau_{1\to 2}^{b})(\phi_{2}+\tau_{1\to
2}^{v})}{\phi_{1}-\tau_{1\to 2}^{v}}\right)^{\frac{1}{2}}\right)$
$\displaystyle\pi_{2}(\tau)=\frac{\phi_{2}+\tau_{1\to
2}^{v}}{2}\left(X_{2}-\tau_{1\to 2}^{b}+\left(\frac{(X_{1}+\tau_{1\to
2}^{b})(X_{2}-\tau_{1\to 2}^{b})(\phi_{1}-\tau_{1\to
2}^{v})}{\phi_{2}+\tau_{1\to 2}^{v}}\right)^{\frac{1}{2}}\right)$ The
gradients of each function are given by
$\displaystyle\nabla\pi_{1}(\tau)=\begin{bmatrix}\frac{1}{2}(\phi_{1}-\tau_{1\to
2}^{v})+\frac{1}{4}\left(\frac{(\phi_{1}-\tau_{1\to
2}^{v})(\phi_{2}+\tau_{1\to 2}^{v})}{(X_{1}+\tau_{1\to
2}^{b})(X_{2}-\tau_{1\to 2}^{b})}\right)^{\frac{1}{2}}(X_{2}-X_{1}-2\tau_{1\to
2}^{b})\\\ -\frac{1}{2}(X_{1}+\tau_{1\to
2}^{b})+\frac{1}{4}\left(\frac{(X_{1}+\tau_{1\to 2}^{b})(X_{2}-\tau_{1\to
2}^{b})}{(\phi_{1}-\tau_{1\to 2}^{v})(\phi_{2}+\tau_{1\to
2}^{v})}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2}-2\tau_{1\to
2}^{v})\end{bmatrix}$
$\displaystyle\nabla\pi_{2}(\tau)=\begin{bmatrix}-\frac{1}{2}(\phi_{2}+\tau_{1\to
2}^{v})+\frac{1}{4}\left(\frac{(\phi_{1}-\tau_{1\to
2}^{v})(\phi_{2}+\tau_{1\to 2}^{v})}{(X_{1}+\tau_{1\to
2}^{b})(X_{2}-\tau_{1\to 2}^{b})}\right)^{\frac{1}{2}}(X_{2}-X_{1}-2\tau_{1\to
2}^{b})\\\ \frac{1}{2}(X_{2}-\tau_{1\to
2}^{b})+\frac{1}{4}\left(\frac{(X_{1}+\tau_{1\to 2}^{b})(X_{2}-\tau_{1\to
2}^{b})}{(\phi_{1}-\tau_{1\to 2}^{v})(\phi_{2}+\tau_{1\to
2}^{v})}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2}-2\tau_{1\to
2}^{v})\end{bmatrix}$ Evaluated at $\tau=0$, the gradients are
$\displaystyle\nabla\pi_{1}(0)=\begin{bmatrix}\frac{1}{2}\phi_{1}+\frac{1}{4}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})\\\
-\frac{1}{2}X_{1}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})\end{bmatrix}$
$\displaystyle\nabla\pi_{2}(0)=\begin{bmatrix}-\frac{1}{2}\phi_{2}+\frac{1}{4}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})\\\
\frac{1}{2}X_{2}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})\end{bmatrix}$
The gradients are diametrically opposed to one another if and only if
$\displaystyle\frac{1}{2}\phi_{1}+\frac{1}{4}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})=-\alpha\left(-\frac{1}{2}\phi_{2}+\frac{1}{4}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})\right)$
$\displaystyle-\frac{1}{2}X_{1}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})=-\alpha\left(\frac{1}{2}X_{2}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})\right)$
Notice that the right hand side of the first equation is equal to zero if and
only if $X_{2}=\frac{4\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$, as
derived in Proposition 3, and similarly, the right hand side of the second
equation is equal to zero if and only if
$1-\frac{X_{1}}{X_{2}}=2\left(\frac{\phi_{2}X_{1}}{\phi_{1}X_{2}}\right)^{\frac{1}{2}}$,
as derived in [8]. However, note that
$\displaystyle-\frac{1}{2}\phi_{2}+\frac{1}{4}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})=0\implies\frac{1}{2}\phi_{1}+\frac{1}{4}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})\neq
0$
$\displaystyle\frac{1}{2}X_{2}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})=0\implies-\frac{1}{2}X_{1}+\frac{1}{4}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})\neq
0$ Thus, if either the first or second entry of $\nabla\pi_{2}(0)$ are equal
to zero, then the corresponding entry of $\nabla\pi_{1}(0)$ must be nonzero;
therefore, the gradients are not diametrically opposed. However, it is
possible for the gradients to be simultaneously equal to zero; this occurs
when $\frac{\phi_{1}}{\phi_{2}}=3$ and
$X_{2}=\frac{4\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$. In this
scenario, there does not exist any mutually beneficial valuation transfer,
since player 2 is already achieving their maximum payoff. After discussing
those scenarios, we can now consider the case where the entries of
$\nabla\pi_{2}(0)$ are nonzero. The conditions of diametric opposition can be
rewritten as
$\displaystyle\frac{X_{1}-\frac{1}{2}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})}{X_{2}+\frac{1}{2}\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{1}-\phi_{2})}=\frac{\phi_{1}+\frac{1}{2}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})}{\phi_{2}-\frac{1}{2}\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})}$
Applying the componendo dividendo rule, this can be rewritten as
$\displaystyle\frac{X_{1}}{\left(\frac{X_{1}X_{2}}{\phi_{1}\phi_{2}}\right)^{\frac{1}{2}}(\phi_{2}-\phi_{1})}=\frac{\phi_{1}}{\left(\frac{\phi_{1}\phi_{2}}{X_{1}X_{2}}\right)^{\frac{1}{2}}(X_{2}-X_{1})}$
which is equivalent to $\displaystyle X_{2}=\frac{\phi_{2}}{\phi_{1}}X_{1}$

which lies on the boundary of $C_{3a}$. ∎

These propositions allow us to restate and prove Theorem 3:

######  Theorem 3.

Consider any coalitional Blotto game parameterized by
$G=(\phi_{1},\phi_{2},X_{1},X_{2})\in\mathcal{G}$. Then, there exists a
mutually beneficial joint transfer almost everywhere in $\mathcal{G}$.

###### Proof.

The proof follows from the proofs of Propositions 20 and 21, as well as the
joint transfer proof provided in the paper. The proof provided in the paper
shows that mutually beneficical joint transfers exist on the interior of Case
1. Proposition 20 shows that mutually beneficial joint transfers exist on the
interior of Case 2 so long as
$\frac{\phi_{1}}{\phi_{1}-\phi_{2}}(1-2X_{2})\neq\frac{X_{1}}{X_{1}+X_{2}}$.
Proposition 21 shows that mutually beneficial joint transfers exist on the
interior of Case 3 so long as $\frac{\phi_{1}}{\phi_{2}}\neq 3$ or
$X_{2}\neq\frac{4\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$. The sets
where mutually beneficial joint transfers may not exist (namely, the
boundaries of Cases 1, 2, and 3, the subset $\\{G\in
C_{2a}:\frac{\phi_{1}}{\phi_{1}-\phi_{2}}(1-2X_{2})=\frac{X_{1}}{X_{1}+X_{2}}\\}$,
and the subset $\\{G\in
C_{3a}:\frac{\phi_{1}}{\phi_{2}}=3,X_{2}=\frac{4\phi_{1}\phi_{2}}{(\phi_{1}-\phi_{2})^{2}}X_{1}$
}) are all of measure zero, and their union is of measure zero. These
arguments hold analogously for any $G\in C_{1b}\cup C_{2b}\cup C_{3b}$; thus,
a mutually beneficial joint transfer exists almost everywhere in
$\mathcal{G}$. ∎

## References

  * [1] R. K. Fleck and C. Kilby, “Changing aid regimes? us foreign aid from the cold war to the war on terror,” Journal of Development Economics, vol. 91, no. 2, pp. 185–197, 2010.
  * [2] L. R. De Geest, J. K. Stranlund, and J. M. Spraggon, “Deterring poaching of a common pool resource,” Journal of Economic Behavior & Organization, vol. 141, pp. 254–276, 2017.
  * [3] C. A. Chávez, J. J. Murphy, F. J. Quezada, J. K. Stranlund, et al., “The endogenous formation of common pool resource coalitions,” tech. rep., 2021\.
  * [4] S. Behnezhad, A. Blum, M. Derakhshan, M. HajiAghayi, M. Mahdian, C. H. Papadimitriou, R. L. Rivest, S. Seddighin, and P. B. Stark, “From battlefields to elections: Winning strategies of blotto and auditing games,” in Proceedings of the Twenty-Ninth Annual ACM-SIAM Symposium on Discrete Algorithms, pp. 2291–2310, SIAM, 2018.
  * [5] C. Thomas, “N-dimensional blotto game with heterogeneous battlefield values,” Economic Theory, vol. 65, no. 3, pp. 509–544, 2018.
  * [6] A. Ferdowsi, W. Saad, B. Maham, and N. B. Mandayam, “A colonel blotto game for interdependence-aware cyber-physical systems security in smart cities,” in Proceedings of the 2nd international workshop on science of smart city operations and platforms engineering, pp. 7–12, 2017.
  * [7] A. Gupta, G. Schwartz, C. Langbort, S. S. Sastry, and T. Bařar, “A three-stage colonel blotto game with applications to cyberphysical security,” in 2014 American Control Conference, pp. 3820–3825, IEEE, 2014\.
  * [8] D. Kovenock and B. Roberson, “Coalitional colonel blotto games with application to the economics of alliances,” Journal of Public Economic Theory, vol. 14, no. 4, pp. 653–676, 2012.
  * [9] J. L. Heyman and A. Gupta, “Colonel blotto game with coalition formation for sharing resources,” in Decision and Game Theory for Security: 9th International Conference, GameSec 2018, Seattle, WA, USA, October 29–31, 2018, Proceedings 9, pp. 166–185, Springer, 2018.
  * [10] D. Kovenock, M. J. Mauboussin, B. Roberson, et al., “Asymmetric conflicts with endogenous dimensionality,” Korean Economic Review, vol. 26, pp. 287–305, 2010.
  * [11] R. Chandan, K. Paarporn, and J. R. Marden, “When showing your hand pays off: Announcing strategic intentions in colonel blotto games,” in 2020 American Control Conference (ACC), pp. 4632–4637, IEEE, 2020.
  * [12] K. Paarporn, R. Chandan, M. Alizadeh, and J. R. Marden, “Characterizing the interplay between information and strength in blotto games,” in 2019 IEEE 58th Conference on Decision and Control (CDC), pp. 5977–5982, IEEE, 2019\.
  * [13] A. Gupta, T. Başar, and G. A. Schwartz, “A three-stage colonel blotto game: when to provide more information to an adversary,” in Decision and Game Theory for Security: 5th International Conference, GameSec 2014, Los Angeles, CA, USA, November 6-7, 2014. Proceedings 5, pp. 216–233, Springer, 2014\.