# Reinforcement Strategies in General Lotto Games

Keith Paarporn  Rahul Chandan  Mahnoosh Alizadeh  Jason R. Marden K. Paarporn
is with the Department of Computer Science at the University of Colorado,
Colorado Springs. R. Chandan is with Amazon Robotics. M. Alizadeh and J. R.
Marden are with the Department of Electrical and Computer Engineering at the
University of California, Santa Barbara, CA. Contact<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>This work is supported by UCOP
Grant LFR-18-548175, ONR grant #N00014-20-1-2359, and AFOSR grants
#FA9550-20-1-0054 and #FA9550-21-1-0203. This manuscript extends a previous
conference version [3].

###### Abstract

Strategic decisions are often made over multiple periods of time, wherein
decisions made earlier impact a competitor’s success in later stages. In this
paper, we study these dynamics in General Lotto games, a class of models
describing the competitive allocation of resources between two opposing
players. We propose a two-stage formulation where one of the players has
reserved resources that can be strategically _pre-allocated_ across the
battlefields in the first stage of the game as reinforcements. The players
then simultaneously allocate their remaining _real-time_ resources, which can
be randomized, in a decisive final stage. Our main contributions provide
complete characterizations of the optimal reinforcement strategies and
resulting equilibrium payoffs in these multi-stage General Lotto games.
Interestingly, we determine that real-time resources are at least twice as
effective as reinforcement resources when considering equilibrium payoffs.

##  1 Introduction

System planners must make investment decisions to mitigate the risks posed by
disturbances or adversarial interference. In many practical settings, these
investments are made and built over time, leading up to a decisive point of
conflict. Security measures in cyber-physical systems and public safety are
deployed and accumulated over long periods of time. Attackers can consequently
use knowledge of the pre-deployed elements to identify vulnerabilities and
exploits in the defender’s strategy [7, 2, 31]. Many types of contests
involves deciding how much effort to exert over multiple rounds of competition
[15, 1, 25, 30, 16, 6].

Indeed, investment decisions are dynamic, where early investments affect how
successful a competitor is at later points in time. Many of these scenarios
involve the strategic allocation of resources, exhibiting trade-offs between
the costs of investing resources in earlier periods and reserving resources
for later stages. In particular, an adversary is often able to learn how the
resources were allocated in the earlier periods and can exploit this knowledge
in later periods.

In this manuscript, we seek to characterize the interplay between early and
late resource investments. We study these elements in General Lotto games, a
game-theoretic framework that describes the competitive allocation of
resources between opponents. The General Lotto game is a popular variant of
the classic Colonel Blotto game, wherein two budget-constrained players, $A$
and $B$, compete over a set of valuable battlefields. The player that deploys
more resources to a battlefield wins its associated value, and the objective
for each player is to win as much value as possible. Outcomes in the standard
formulations are determined by a single simultaneous allocation of resources,
i.e. they are typically studied as one-shot games.

The formulations considered in this paper focus on a multi-stage version of
the General Lotto game where one of the players can reinforce various
battlefields before the competition begins by pre-allocating resources to
battlefields; hence, we refer to these reinforcement strategies as pre-
allocation strategies. More formally, our analysis is centered on the
following multi-stage scenario: Player $A$ is endowed with $P\geq 0$ resources
to be pre-allocated, and both players possess real-time resources
$R_{A},R_{B}\geq 0$ to be allocated at the time of competition. In the first
stage, player $A$ decides how to deploy the pre-allocated resources $P$ over
the battlefields. The pre-allocation decision is binding and known to player
$B$. In the final stage, both players engage in a General Lotto game where
they simultaneously decide how to deploy their real-time resources, and
payoffs are subsequently derived.

The pre-allocated resources may represent, for example, the installation of
anti-virus tools on system servers. The capabilities of anti-virus software
are typically static and well-known, and thus a potential attacker would have
knowledge about the system’s base level of defensive capability. However, the
attacker would not generally have knowledge about the system’s placement of
intrusion-detection systems, which are often dynamic and part of a “moving
target defense” strategy [5, 32, 33]. Moreover, attackers’ strategies must be
unpredictable in an attempt to exploit defenses. Thus, the use of real-time
resources in our model represents such dynamic and unpredictable strategies. A
full summary of our contributions is provided below.

Our Contributions: Our main contribution in this paper is a full
characterization of equilibrium strategies and payoffs to both players in the
aforementioned two-stage General Lotto game (Theorem 3.1). By characterizing
these optimal reinforcement strategies, we are able to provide Pareto
frontiers for player $A$ as one balances a combination of real-time and pre-
allocated resources (Lemma 4.3). Interestingly, Theorem 4.2 demonstrates that
real-time resources are at least twice as effective as pre-allocated resources
when considering the equilibrium payoff of player $A$.

Our second set of results in this manuscript focus on the optimal investment
levels of pre-allocated and real-time resources. Rather than player $A$ being
equipped with a fixed budget of resources $(P,R_{A})$, we rather consider a
setting where player $A$ has a monetary budget $M_{A}$ and each type of
resource is associated with a given per-unit cost. Building upon the above
characterization of the optimal reinforcement strategies in Theorem 3.1, in
Theorem 4.5 we characterize the optimal investment strategies for this per-
unit cost variant of the two-stage General Lotto game. This provides an
understanding of the precise combination of pre-allocated and real-time
resources that optimize player $A$’s equilibrium payoff.

Our last contribution focuses on a variant of this General Lotto game where
both players can employ pre-allocated resources. In particular, we consider a
scenario where player $B$ is able to respond to player $A$’s pre-allocation
with its own pre-allocated resources, before engaging in the final-stage
General Lotto game. This is formulated as a Stackelberg game, where both
players have monetary budgets $M_{A},M_{B}$ and per-unit costs for investing
in the two types of resources. We fully characterize the Stackelberg
equilibrium (Proposition 5.3), which highlights that having the opportunity to
respond to an opponent’s early investments can significantly improve one’s
eventual performance.

Related works: This manuscript takes steps towards understanding the
competitive allocation of resources in multi-stage scenarios. There is
widespread interest in this research objective that involves the analysis of
zero-sum games [21, 14, 20], differential or repeated games [13, 27], and
Colonel Blotto games [29, 1, 19, 26]. The goal of many of these works is to
develop tools to compute decision-making policies for agents in adversarial
and uncertain environments. In comparison, our work provides explicit,
analytical characterizations of equilibrium strategies, which draws sharper
insights that relate the players’ performance with the various elements of
adversarial interaction. As such, our work is related to a recent research
thread in which allocation decisions are made over multiple stages [17, 10, 9,
19, 29, 23, 26, 4].

Our work also draws significantly from the primary literature on Colonel
Blotto and General Lotto games [8, 24, 18, 28]. In particular, the
simultaneous-move subgame played in the final stage of our formulations was
first proposed by Vu and Loiseau [28], and is known as the _General Lotto game
with favoritism_ (GL-F). Favoritism refers to the fact that pre-allocated
resources provide an incumbency advantage to one player’s competitive chances.
Their work establishes existence of equilibria and develops computational
methods to calculate them to arbitrary precision. However, this prior work
considers pre-allocated resources as exogenous parameters of the game. In
contrast, we model the deployment of pre-allocated resources as a strategic
element of the competitive interaction. Furthermore, we provide the first
analytical characterizations of equilibria and the corresponding payoffs in
GL-F games.

##  2 Problem formulation

Figure 1: (Left) The two-stage General Lotto game with Pre-allocations (GL-P).
Players $A$ and $B$ compete over $n$ battlefields, whose valuations are given
by $\\{w_{b}\\}_{b=1}^{n}$. In Stage 1, player $A$ decides how to deploy $P$
pre-allocated resources to the battlefields. Player $B$ observes the
deployment. In Stage 2, the players simultaneously decide how to deploy their
real-time resources $R_{A}$ and $R_{B}$ and final payoffs are determined.
(Center) This plot shows the SPE payoff to player $A$ under varying resource
endowments (Theorem 3.1). Obtaining more pre-allocated resources improves the
payoff with decreasing marginal returns. Here, we have fixed $R_{B}=1$.
(Right) The characterization of the SPE payoff is broken down into three
separate cases in the game’s parameters. These are shown as the three regions
in this plot, here parameterized by $P$ and $R_{A}$, which correspond to the
items in Theorem 3.1.

The _General Lotto game with pre-allocations_ (GL-P) is a two-stage game with
players $A$ and $B$, who compete over a set of $n$ battlefields, denoted as
$\mathcal{B}=\\{1,\ldots,n\\}$. Each battlefield $b\in\mathcal{B}$ is
associated with a known valuation $w_{b}>0$, which is common to both players.
Player $A$ is endowed with a pre-allocated resource budget $P>0$ and a real-
time resource budget $R_{A}>0$. Player $B$ is endowed with a budget $R_{B}>0$
of real-time resource, but no pre-allocated resources. The two stages are
played as follows:

– _Stage 1 (pre-allocation):_ Player $A$ decides how to allocate her $P$ pre-
allocated resources to the battlefields, i.e., it selects a vector
$\boldsymbol{p}=(p_{1},\ldots,p_{n})\in\Delta_{n}(P):=\\{\boldsymbol{p}^{\prime}\in\mathbb{R}_{+}^{n}:\|\boldsymbol{p}^{\prime}\|_{1}=P\\}$.
We term the vector $\boldsymbol{p}$ as player $A$’s _pre-allocation profile_.
No payoffs are derived in Stage 1, and $A$’s choice $\boldsymbol{p}$ becomes
binding and common knowledge.

– _Stage 2 (decisive point of conflict):_ Players $A$ and $B$ then compete in
a simultaneous-move sub-game with their real-time resource budgets $R_{A}$,
$R_{B}$. Here, both players can randomly allocate these resources as long as
their expenditure does not exceed their budgets in expectation. Specifically,
a strategy for player $i\in\\{A,B\\}$ is an $n$-variate (cumulative)
distribution $F_{i}$ over allocations
$\boldsymbol{x}_{i}\in\mathbb{R}_{+}^{n}$ that satisfies

$\mathbb{E}_{\boldsymbol{x}_{i}\sim
F_{i}}\left[\sum_{b\in\mathcal{B}}x_{i,b}\right]\leq R_{i}.$ (1)

We use $\mathcal{L}(R_{i})$ to denote the set of all strategies $F_{i}$ that
satisfy (1). Given that player $A$ chose $\boldsymbol{p}$ in Stage 1, the
expected payoff to player $A$ is given by

$U_{A}(\boldsymbol{p},F_{A},F_{B}):=\mathbb{E}_{\begin{subarray}{c}\boldsymbol{x}_{A}\sim
F_{A}\\\ \boldsymbol{x}_{B}\sim
F_{B}\end{subarray}}\left[\sum_{b\in\mathcal{B}}w_{b}\\!\cdot\\!\mathds{1}\\{x_{A,b}+p_{b}\geq
qx_{B,b}\\}\right]$ (2)

where $\mathds{1}\\{\cdot\\}$ is the usual indicator function taking a value
of 1 or 0.111The tie-breaking rule (i.e., deciding who wins if
$x_{A,b}+p_{b}=x_{B,b}$) can be assumed to be arbitrary, without affecting any
of our results. This property is common in the General Lotto literature, see,
e.g., [18, 28]. In words, player $B$ must overcome player $A$’s pre-allocated
resources $p_{b}$ as well as player $A$’s allocation of real-time resources
$x_{A,b}$ in order to win battlefield $b$. The parameter $q>0$ is the relative
quality of player $B$’s real-time resources against player $A$’s resources.
For simpler exposition, we will simply set $q=1$, noting that all of our
results are easily attained for any other value of $q$. The payoff to player
$B$ is
$U_{B}(\boldsymbol{p},F_{A},F_{B})=1-U_{A}(\boldsymbol{p},F_{A},F_{B})$, where
we assume without loss of generality that $\sum_{b\in\mathcal{B}}w_{b}=1$.

Stages 1 and 2 of GL-P are illustrated in Figure 1. We specify an
instantiation of the game as $\text{GL-P}(P,R_{A},R_{B},\boldsymbol{w})$. We
focus on the subgame-perfect equilibrium solution concept.

###### Definition 2.1

A profile
$(\boldsymbol{p}^{*},F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p}))$
where $\boldsymbol{p}^{*}\in\Delta_{n}(P)$ and
$F_{i}^{*}(\boldsymbol{p}):\Delta_{n}(P)\rightarrow\mathcal{L}(R_{i})$, for
$i=A,B$, is a _subgame-perfect equilibrium (SPE)_ if the following conditions
hold.

  * •

For any $\boldsymbol{p}\in\Delta_{n}(P)$,
$(F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p}))$ constitutes a Nash
equilibrium of the Stage 2 subgame:

$\displaystyle
U_{A}(\boldsymbol{p},F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p}))$
$\displaystyle\geq U_{A}(\boldsymbol{p},F_{A},F_{B}^{*}(\boldsymbol{p}))$ (3)
$\displaystyle\text{and
}U_{B}(\boldsymbol{p},F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p}))$
$\displaystyle\geq U_{B}(\boldsymbol{p},F_{A}^{*}(\boldsymbol{p}),F_{B})$

for any $F_{A}\in\mathcal{L}(R_{A})$ and $F_{B}\in\mathcal{L}(R_{B})$.

  * •

The pre-allocation $\boldsymbol{p}^{*}$ satisfies

$U_{A}(\boldsymbol{p}^{*},F_{A}^{*}(\boldsymbol{p}^{*}),F_{B}^{*}(\boldsymbol{p}^{*}))\geq
U_{A}(\boldsymbol{p},F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p}))$ (4)

for any $\boldsymbol{p}\in\Delta_{n}(P)$.

In an SPE, the players select their Stage 2 strategies conditioned on player
$A$’s choice of pre-allocation $\boldsymbol{p}$ in Stage 1, such that
$F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p})$ forms a Nash equilibrium
of the one-shot subgame of Stage 2. We stress the importance of the common
knowledge assumption for the pre-allocation choice $\boldsymbol{p}$ before
Stage 2 – over time, an opponent is likely to learn the placement of past
resources and would be able to exploit this knowledge at a later point in
time. The second condition in the above definition asserts that player $A$’s
SPE pre-allocation $\boldsymbol{p}^{*}$ in Stage 1 optimizes its equilibrium
payoff in the subsequent Stage 2 subgame.

We remark that the Stage 2 subgame has been studied in the recent literature,
where it is termed a _General Lotto game with Favoritism_ [28]. We denote it
as $\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$. There, a pre-allocation vector
$\boldsymbol{p}$ is viewed as an exogenous fixed parameter, whereas in our
GL-P formulation, it is an endogenous strategic choice. It is established in
[28] that a Nash equilibrium exists and its payoffs are unique for any
instance of $\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$. Consequently, the
players’ SPE payoffs in our GL-P game are necessarily unique. We will denote
$\pi^{*}_{i}(P,R_{A},R_{B}):=U_{i}(\boldsymbol{p}^{*},F_{A}^{*}(\boldsymbol{p}^{*}),F_{B}^{*}(\boldsymbol{p}^{*}))$,
$i\in\\{A,B\\}$, as the players’ payoffs in an SPE when the dependence on the
vector $\boldsymbol{w}$ is clear.

While [28] provided numerical techniques to compute an equilibrium of
$\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$ to arbitrary precision, analytical
characterizations of them (e.g. closed-form expressions) were not provided. In
the next section, we develop techniques to derive such characterizations, as
they are required to precisely express the SPE of the GL-P game.

##  3 Equilibrium characterizations

In this section, we present our main results regarding the characterization of
players’ SPE payoffs in the GL-P game. These results highlight the relative
effectiveness of pre-allocated vs real-time resources.

###  3.1 Main results

The result below provides an explicit characterization of the players’ payoffs
in an SPE of the two-stage GL-P game.

###### Theorem 3.1

Consider the game $\text{GL-P}(P,R_{A},R_{B},\boldsymbol{w})$. Player $A$’s
payoff $\pi^{*}_{A}(P,R_{A},R_{B})$ in a SPE is given as follows:

  1. 1.

If $R_{B}\leq P$, or $R_{B}>P$ and
$R_{A}\geq\frac{2(R_{B}-P)^{2}}{P+2(R_{B}-P)}$, then
$\pi^{*}_{A}(P,R_{A},R_{B})$ is

$1-\frac{R_{B}}{2R_{A}}\left(\frac{R_{A}+\sqrt{R_{A}(R_{A}+2P)}}{P+R_{A}+\sqrt{R_{A}(R_{A}+2P)}}\right)^{2}.$
(5)

  2. 2.

If $R_{B}>P$ and $0<R_{A}<\frac{2(R_{B}-P)^{2}}{P+2(R_{B}-P)}$, then
$\pi^{*}_{A}(P,R_{A},R_{B})$ is

$\frac{R_{A}}{2(qR_{B}-P)}.$ (6)

  3. 3.

If $R_{A}=0$, then $\pi^{*}_{A}(P,R_{A},R_{B})$ is

$\left(1-\min\left\\{\frac{R_{B}}{P},1\right\\}\right)$ (7)

Player $B$’s SPE payoff is given by
$\pi^{*}_{B}(P,R_{A},R_{B})=1-\pi^{*}_{A}(P,R_{A},R_{B})$. In all instances,
player $A$’s SPE pre-allocation is $\boldsymbol{p}^{*}=\boldsymbol{w}\cdot P$.

A visualization of the parameter regimes of the three cases above is shown in
the right Figure 1. Note that the standard General Lotto game (without pre-
allocations, [11]) is included as the vertical axis at $P=0$. An illustration
of the SPE payoffs to player $A$ is shown in the center plot of Figure 1. We
notice that given sufficiently high amount of pre-allocated resources (i.e.
$P>R_{B}$), player $A$ can attain a positive payoff even without any real-time
resources ($R_{A}=0$). For $P<R_{B}$, player $A$ receives zero payoff since
player $B$ can simply exceed the pre-allocation on every battlefield. Observe
that the SPE payoff $\pi_{A}^{*}(P,R_{A},R_{B})$ exhibits diminishing marginal
returns in $R_{A}$ and in $P$ for larger values of $P$, but is not in general
a concave function in $P$ – see the $R_{A}=0.5$ curve, which has an
inflection.

Proof approach and outline: The derivation of the SPE payoffs in Theorem 3.1
follows a backwards induction approach. First, for any fixed pre-allocation
vector $\boldsymbol{p}$, one characterizes the equilibrium payoff of the Stage
2 sub-game $\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$. We denote this payoff as

$\pi_{A}(\boldsymbol{p},R_{A},R_{B}):=U_{A}(\boldsymbol{p},F_{A}^{*}(\boldsymbol{p}),F_{B}^{*}(\boldsymbol{p}))$
(8)

where $F_{A}^{*}(\cdot),F_{B}^{*}(\cdot)$ satisfies the first condition of
Definition 2.1. Then, the SPE payoff is calculated by solving the following
optimization problem,

$\pi_{A}^{*}(P,R_{A},R_{B})=\max_{\boldsymbol{p}\in\Delta_{n}(P)}\pi_{A}(\boldsymbol{p},R_{A},R_{B}).$
(9)

The following proof outline is taken to derive the SPE strategies and payoffs.

Part 1: We first detail analytical methods to derive the equilibrium payoff
$\pi_{A}(\boldsymbol{p},R_{A},R_{B})$ to the second stage subgame
$\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$.

Part 2: We show that $\boldsymbol{p}^{*}=\boldsymbol{w}\cdot P$ is an SPE pre-
allocation, i.e. it solves the optimization problem (9).

Part 3: We derive the analytical expressions for $\pi_{A}^{*}$ reported in
Theorem 3.1.

Each one of the three parts has a corresponding Lemma that we present in the
following subsection.

###  3.2 Proof of Theorem 3.1

Part 1: The recent work of Vu and Loiseau [28] provides a method to derive an
equilibrium of the General Lotto game with Favoritism
$\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$. This method involves solving the
following system222The problem settings considered in [28] are more general,
which considers two-sided favoritism (i.e. $p_{b}<0$ for some $b$). However,
exact closed-form solutions were not provided. The paper [28] provided
computational approaches to calculate an equilibrium to arbitrary precision.
of two equations for two unknown variables
$(\kappa_{A},\kappa_{B})\in\mathbb{R}_{++}^{2}$:

$\displaystyle
R_{A}\\!=\\!\sum_{b=1}^{n}\frac{[h_{b}(\kappa_{A},\kappa_{B})-p_{b}]^{2}}{2w_{b}\kappa_{B}},\
R_{B}\\!=\sum_{b=1}^{n}\frac{h_{b}^{2}(\kappa_{A},\kappa_{B})-p_{b}^{2}}{2w_{b}\kappa_{A}}$
(10)

where
$h_{b}(\kappa_{A},\kappa_{B}):=\min\\{w_{b}\kappa_{B},w_{b}\kappa_{A}+p_{b}\\}$
for $b\in\mathcal{B}$. The two equations above correspond to the expected
budget constraint (1) for both players. There always exists a solution
$(\kappa_{A}^{*},\kappa_{B}^{*})\in\mathbb{R}_{++}^{2}$ to this system [28],
which allows one to calculate the following equilibrium payoffs.

######  Lemma 3.1 (Adapted from [28])

Suppose $(\kappa_{A}^{*},\kappa_{B}^{*})\in\mathbb{R}_{++}^{2}$ solves (10).
Let
$\mathcal{B}_{1}:=\\{b\in\mathcal{B}:h_{b}(\kappa_{A}^{*},\kappa_{B}^{*})=w_{b}\kappa_{B}\\}$
and $\mathcal{B}_{2}=\mathcal{B}\backslash\mathcal{B}_{1}$. Then there is a
Nash equilibrium $(F_{A}^{*},F_{B}^{*})$ of
$\text{GL-F}(\boldsymbol{p},R_{A},R_{B})$ where player $A$’s equilibrium
payoff is given by

$\displaystyle\pi_{A}(\boldsymbol{p},R_{A},R_{B})$
$\displaystyle=\sum_{b\in\mathcal{B}_{1}}\\!w_{b}\\!\left[1-\frac{\kappa_{B}^{*}}{2\kappa_{A}^{*}}\left(1-\frac{p_{i}^{2}}{(w_{b}\kappa_{B})^{2}}\right)\right]$
(11)
$\displaystyle\quad+\sum_{b\in\mathcal{B}_{2}}w_{b}\frac{\kappa_{A}^{*}}{2\kappa_{B}^{*}}$

and the equilibrium payoff to player $B$ is
$\pi_{B}(\boldsymbol{p},R_{A},R_{B})=1-\pi_{A}(\boldsymbol{p},R_{A},R_{B})$.

Lemma 3.1 provides an expression for $\pi_{A}(\boldsymbol{p},R_{A},R_{B})$ in
terms of a solution $(\kappa_{A}^{*},\kappa_{B}^{*})$ to the system of
equations (10). However, in order to study the optimization (9), we need to be
able to either find closed-form expressions for the solution
$(\kappa_{A}^{*},\kappa_{B}^{*})$ in terms of the defining game parameters
$\boldsymbol{p},R_{A},R_{B},\boldsymbol{w}$, or establish certain properties
about the payoff function (11), such as concavity in $\boldsymbol{p}$.
Unfortunately, we find that this function is not generally concave for
$\boldsymbol{p}\in\Delta_{n}(P)$. Our approach in Part 2 is to show that it is
always increasing in the direction pointing to $\boldsymbol{p}^{*}$.

Part 2: This part of the proof is devoted to showing that
$\boldsymbol{p}^{*}=\boldsymbol{w}\cdot P$ is an SPE pre-allocation for player
$A$. This divides the total pre-allocated resources $P$ among the battlefields
proportionally to their values $w_{b}$, $b\in\mathcal{B}$.

###### Lemma 3.2

The vector $\boldsymbol{p}^{*}=\boldsymbol{w}\cdot P$ is an SPE pre-
allocation.

Equivalently, $\boldsymbol{p}^{*}$ solves the optimization problem (9).

######  Proof 3.2.

The proof will follow two sub-parts, 2-a and 2-b. In part 2-a, we first
establish that $\boldsymbol{p}^{*}$ is a local maximizer of
$\pi_{A}(\boldsymbol{p},R_{A},R_{B})$, which necessarily occurs when either
$\mathcal{B}_{1}=\mathcal{B}$ or $\mathcal{B}_{2}=\mathcal{B}$. In part 2-b,
we show that no choice of $\boldsymbol{p}\in\Delta_{n}(P)$ that results in
both sets $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ being non-empty achieves a
higher payoff than $\pi_{A}(\boldsymbol{p}^{*},R_{A},R_{B})$, thus
establishing Lemma 3.2.

Part 2-a: $\boldsymbol{p}^{*}$ _is a local maximizer of_
$\pi_{A}(\boldsymbol{p},R_{A},R_{B})$.

From Lemma 3.1 and the definition of $h_{b}(\kappa_{A},\kappa_{B})$, we find
that the solution to (10) under the pre-allocation $\boldsymbol{p}^{*}$ is
always in one of two completely symmetric cases: 1)
$\mathcal{B}_{1}=\mathcal{B}$; or 2) $\mathcal{B}_{2}=\mathcal{B}$. Thus, we
need to show $\boldsymbol{p}^{*}$ is a local maximizer in both cases.

Case 1 ($\mathcal{B}_{1}=\mathcal{B}$): For $\boldsymbol{p}\in\Delta_{n}(P)$,
the system (10) is written

$\displaystyle
R_{A}=\sum_{b=1}^{n}\frac{(w_{b}\kappa_{B}-p_{b})^{2}}{2w_{b}\kappa_{B}}\text{
and
}R_{B}=\sum_{b=1}^{n}\frac{(w_{b}\kappa_{B})^{2}-p_{b}^{2}}{2w_{b}\kappa_{A}}$
(12) $\displaystyle\text{where }0<w_{b}\kappa_{B}-p_{b}\leq\kappa_{A}\text{
holds }\forall b\in\mathcal{B}.$

It yields the algebraic solution

$\displaystyle\kappa_{B}^{*}$
$\displaystyle=P+R_{A}+\sqrt{(P+R_{A})^{2}-\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}}$
(13) $\displaystyle\kappa_{A}^{*}$
$\displaystyle=\frac{(P+R_{A})\kappa_{B}^{*}-\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}}{R_{B}}.$

where
$\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}:=\sum_{b=1}^{n}\frac{p_{b}^{2}}{w_{b}}$.
This solution needs to satisfy the set of conditions
$0<w_{b}\kappa_{B}-p_{b}\leq\kappa_{A}\ \forall b\in\mathcal{B}$, but the
explicit characterization of these conditions is not needed to show that
$\boldsymbol{p}^{*}$ is a local maximum. Indeed, first observe that the
expression for $\kappa_{B}^{*}$ is required to be real-valued, which we can
write as the condition

$\boldsymbol{p}\in
R^{(1n)}:=\left\\{\boldsymbol{p}\in\Delta_{n}(P):\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}<(P+R_{A})^{2}\right\\}.$
(14)

We thus have a region $R^{(1n)}$ for which player $A$’s equilibrium payoff
(Lemma 3.1) is given by the expression

$\pi_{A}^{(1n)}(\boldsymbol{p}):=1-\frac{R_{B}}{f(\|\boldsymbol{p}\|_{\boldsymbol{w}})}\\!\left(1-\frac{\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}}{(P\\!+\\!R_{A}\\!+\\!f(\|\boldsymbol{p}\|_{\boldsymbol{w}}))^{2}}\right)$
(15)

where
$f(\|\boldsymbol{p}\|_{\boldsymbol{w}}):=\sqrt{(P+R_{A})^{2}-\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}}$.
The partial derivatives are calculated to be

$\frac{\partial\pi_{A}^{(1n)}}{\partial
p_{b}}(\boldsymbol{p})=\frac{p_{b}}{w_{b}}\cdot\frac{2R_{B}}{f(\|\boldsymbol{p}\|_{\boldsymbol{w}})(P+R_{A}+f(\|\boldsymbol{p}\|_{\boldsymbol{w}}))^{2}}$
(16)

A critical point of $\pi_{A}^{(1n)}$ must satisfy
$\boldsymbol{z}^{\top}\nabla\pi_{A}^{(1n)}(\boldsymbol{p})=0$ for any
$\boldsymbol{z}\in\mathbb{T}_{n}$, where we define
$\mathbb{T}_{n}:=\\{\boldsymbol{z}\in\mathbb{R}^{n}:\sum_{b=1}^{n}z_{b}=0\\}$
as the tangent space of $\Delta_{n}(P)$. Indeed for any $\boldsymbol{p}\in
R^{(1n)}$, we calculate

$\displaystyle(\boldsymbol{p}-\boldsymbol{w}\cdot
P)^{\top}\nabla\pi_{A}^{(1n)}(\boldsymbol{p})$
$\displaystyle=g(\|\boldsymbol{p}\|_{\boldsymbol{w}})\cdot\left(\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}-P^{2}\right)$
(17) $\displaystyle\geq 0$

where
$g(\|\boldsymbol{p}\|_{\boldsymbol{w}}):=\frac{2R_{B}}{f(\|\boldsymbol{p}\|_{\boldsymbol{w}})(P+R_{A}+f(\|\boldsymbol{p}\|_{\boldsymbol{w}}))^{2}}>0$
for any $\boldsymbol{p}\in R^{(1n)}$. The inequality above is met with
equality if and only if $\boldsymbol{p}=\boldsymbol{p}^{*}$. This is due to
the fact that
$\min_{\boldsymbol{p}\in\Delta_{n}(P)}\|\boldsymbol{p}\|_{\boldsymbol{w}}^{2}=\|\boldsymbol{p}^{*}\|_{\boldsymbol{w}}^{2}=P^{2}$.
Thus, $\boldsymbol{p}^{*}$ is the unique maximizer of
$\pi_{A}^{(1n)}(\boldsymbol{p})$ on $R^{(1n)}$.

Case 2 ($\mathcal{B}_{2}=\mathcal{B}$): For $\boldsymbol{p}\in\Delta_{n}(P)$,
the system is written as

$R_{A}=\sum^{n}_{b=1}\frac{(w_{b}\kappa_{A})^{2}}{2w_{b}\kappa_{B}}\text{ and
}R_{B}=\sum^{n}_{b=1}\frac{(w_{b}\kappa_{A}-p_{b})^{2}-(p_{b})^{2}}{2w_{b}\kappa_{A}},$

where $w_{b}\kappa_{B}-p_{b}>w_{b}\kappa_{A}$ holds for all $b\in\mathcal{B}$.
This readily yields the algebraic solution:

$\kappa_{B}^{*}=2\frac{(R_{B}-P)^{2}}{R_{A}}\text{ and
}\kappa_{A}^{*}=2(R_{B}-P).$ (18)

For this solution to be valid, the following conditions are required:

$\bullet$ $\kappa_{A}^{*},\kappa_{B}^{*}\in\mathbb{R}_{++}$: This requires
that $R_{B}-P>0$.

$\bullet$ $w_{b}\kappa_{B}^{*}-p_{b}>w_{b}\kappa_{A}^{*}$ for all
$b\in\mathcal{B}$: This requires that

$2\frac{(R_{B}-P)^{2}}{R_{A}}-2(R_{B}-P)-\max_{b}\\{\frac{p_{b}}{w_{b}}\\}>0.$

The left-hand side is quadratic in $R_{B}-P$, and thus requires that either

$\displaystyle
R_{B}-P<\frac{R_{A}}{2}\left(1-\sqrt{1+\frac{2}{R_{A}}\max_{b}\\{\frac{p_{b}}{w_{b}}\\}}\right)$

or

$R_{B}-P>\frac{R_{A}}{2}\left(1+\sqrt{1+\frac{2}{R_{A}}\max_{b}\\{\frac{p_{b}}{w_{b}}\\}}\right).$
(19)

The former cannot hold since the numerator on the right-hand side is strictly
negative, but $\kappa_{A}^{*},\kappa_{B}^{*}\in\mathbb{R}_{++}$ requires
$R_{B}-P>0$. Thus, (19) must hold, Clearly, this is more restrictive than
$R_{B}-P>0$. This dictates the boundary of Case 2.

For any $\boldsymbol{p}\in\Delta_{n}(P)$ such that all battlefields are in
Case 2, the expression for player $A$’s payoff in (11) simplifies to

$\pi_{A}(\boldsymbol{p},R_{A},R_{B})=\sum^{n}_{b=1}w_{b}\frac{\kappa_{A}^{*}}{2\kappa_{B}^{*}}=\frac{R_{A}}{2(R_{B}-P)},$

where we use the expression for $\kappa_{B}^{*}$ and $\kappa_{A}^{*}$ in (18).
Observe that player $A$’s payoff is constant in the quantity $\boldsymbol{p}$.
Thus, for any $\boldsymbol{p}$ that satisfies (19), it holds that all
battlefields are in Case 2, and that player $A$’s payoff is the above. We
conclude sub-part 2-a noting that, for given quantities $R_{A}$ and $P$, if
there exists any $\boldsymbol{p}\in\Delta_{n}(P)$ such that (19) is satisfied,
then $\boldsymbol{p}^{*}=\boldsymbol{w}\cdot P$ must also satisfy (19), since
$||\boldsymbol{p}||_{\infty}\geq||\boldsymbol{p}^{*}||_{\infty}$ and the
right-hand side in (19) is increasing in $||\boldsymbol{p}||_{\infty}$.

Part 2-b: _Any pre-allocation $\boldsymbol{p}$ that corresponds to a solution
of (10) with $\mathcal{B}_{1},\mathcal{B}_{2}\neq\varnothing$ satisfies
$\pi_{A}(\boldsymbol{p},R_{A},R_{B})\leq\pi_{A}(\boldsymbol{p}^{*},R_{A},R_{B})$._

For easier exposition, the proof of Part 2-b is presented in the Appendix.
Together, Parts 2-a and 2-b imply that $\boldsymbol{p}^{*}$ is a global
maximizer of the function $\pi_{A}(\boldsymbol{p},R_{A},R_{B})$, completing
the proof of Lemma 3.2.

Part 3: In the third and final part, we obtain the formulas for SPE payoffs
reported in Theorem 3.1.

######  Proof 3.3 (Proof of Theorem 3.1).

We proceed to derive closed-form solutions for the SPE payoff
$\pi_{A}^{*}(P,R_{A},R_{B})$. From Lemmas 3.1 and 3.2, the SPE payoff is
attained by evaluating $\pi_{A}(\boldsymbol{p}^{*},R_{A},R_{B})$, i.e. from
equation (11). From the discussion of Part 2-a, this amounts to analyzing the
two completely symmetric cases $\mathcal{B}_{1}=\mathcal{B}$ and
$\mathcal{B}_{2}=\mathcal{B}$.

Case 1 ($\mathcal{B}_{1}=\mathcal{B}$): Substituting
$\boldsymbol{p}^{*}=\boldsymbol{w}\cdot P$ into (13) and simplifying, we
obtain

$\displaystyle\kappa_{B}^{*}$ $\displaystyle=P+R_{A}+\sqrt{R_{A}(R_{A}+2P)}$
(20) $\displaystyle\kappa_{A}^{*}$
$\displaystyle=\frac{(P+R_{A})\kappa_{B}^{*}-P^{2}}{R_{B}}.$

Next, we verify that this solution satisfies the conditions
$0<\kappa_{B}^{*}-P\leq\kappa_{A}^{*}$ imposed by the case
$\mathcal{B}_{1}=\mathcal{B}$.

$\bullet$ $\kappa_{B}^{*}-P>0$: This holds by inspection.

$\bullet$ $\kappa_{B}^{*}-P\leq\kappa_{A}$: We can write this condition as

$R_{B}-P\leq R_{A}+\frac{PR_{A}}{R_{A}+\sqrt{R_{A}(R_{A}+2P)}}$ (21)

We note that whenever $R_{B}\leq P$, this condition is always satisfied. When
$R_{B}>P$, this condition does not automatically hold, and an equivalent
expression of (21) is given by

$R_{A}\geq\frac{2(R_{B}-P)^{2}}{P+2(R_{B}-P)}.$ (22)

Observe that $R_{A}=\frac{2(R_{B}-P)^{2}}{P+(R_{B}-P)}$ satisfies (21) with
equality, and is in fact the only real solution (one can reduce it to a cubic
polynomial in $R_{A}$).

When these conditions hold, the equilibrium payoff
$\pi_{A}^{*}(P,R_{A},R_{B})=\pi_{A}(\boldsymbol{p}^{*},R_{A},R_{B})$ can be
directly computed from Lemma 3.1, i.e. (11). It is given by the expression
(5).

Case 2 ($\mathcal{B}_{2}=\mathcal{B}$): Substituting
$\boldsymbol{p}=\boldsymbol{w}\cdot P$ into (18) and simplifying, we obtain

$\kappa_{A}^{*}=\frac{2(R_{B}-P)}{W}\text{ and
}\kappa_{B}^{*}=\frac{2(R_{B}-P)^{2}}{R_{A}}.$ (23)

The solution satisfies the conditions $0<\kappa_{A}^{*}<\kappa_{B}^{*}-P$
imposed by the case $\mathcal{B}_{2}=\mathcal{B}$ if and only if $R_{B}>P$ and
$R_{A}>\frac{2(R_{B}-P)^{2}}{P+(R_{B}-P)}$. When this holds, the SPE payoff is
calculated from (11) to be
$\pi_{A}^{*}(P,R_{A},R_{B})=W\cdot\frac{R_{A}}{2(R_{B}-P)}$.

##  4 Interplay between resource types

Figure 2: (Left) A plot of the effectiveness ratio $E(R_{A},R_{B})$ (Theorem
4.2), which quantifies the multiplicative factor of pre-allocated resources
needed to achieve the same performance as an amount of real-time resources
$R_{A}$. Notably, real-time resources are at least twice as effective as an
equivalent amount of pre-allocated resources. (Center) This plot shows a
collection of level curves for player $A$’s SPE payoff. A level curve
corresponds to a fixed performance level $\Pi$, and any point $(P,R_{A})$ on
the level curve satisfies $\pi_{A}^{*}(P,R_{A},R_{B})=\Pi$ (Lemma 4.3).
(Right) This plot shows player $A$’s optimal investment in pre-allocated
resources $P^{*}$ when it has a per-unit cost of $c_{A}$ and a fixed monetary
budget of $M_{A}$ to invest in both types of resources (Theorem 4.5). Player
$A$ invests the remaining $M_{A}-c_{A}P^{*}$ in real-time resources. In these
plots, we set $R_{B}=1$, and $W=1$.

In this section, we present some implications from Theorem 3.1 regarding the
interplay between pre-allocated and real-time resources. Specifically, we seek
to compare the relative effectiveness of the two types of resources in the
GL-P game by first quantifying an _effectiveness ratio_. We then leverage this
analysis to address how player $A$ should invest in both types of resources
when they are costly to acquire.

###  4.1 The effectiveness ratio

We define the effectiveness ratio as follows.

######  Definition 4.1.

For a given $R_{A},R_{B}>0$, let $P^{\text{eq}}(R_{A},R_{B})>0$ be the unique
value such that
$\pi_{A}^{*}(P^{\text{eq}},0,R_{B})=\pi_{A}^{*}(0,R_{A},R_{B})$. The
_effectiveness ratio_ is defined as

$E(R_{A},R_{B}):=\frac{P^{\text{eq}}}{R_{A}}$ (24)

In words, $P^{\text{eq}}$ is the amount of pre-allocated resources required to
achieve the same level of performance as the amount $R_{A}$ of real-time
resources in the absence of pre-allocated resources. The effectiveness ratio
$E$ thus quantifies the multiplicative factor of pre-allocated resources
needed compared to real-time resources $R_{A}$.

The following result establishes the effectiveness ratio for any given
parameters.

######  Theorem 4.2.

For a given $R_{A},R_{B}>0$, the effectiveness ratio is

$E(R_{A},R_{B})=\begin{cases}2&\quad\text{if }R_{A}\geq R_{B},\\\
\frac{2(R_{B})^{2}}{R_{A}(2R_{B}-R_{A})}&\quad\text{if
}R_{A}<R_{B}.\end{cases}$ (25)

Here, it is interesting to note that the ratio $E$ is lower-bounded by $2$ –
real-time resources are at least twice as effective as pre-allocated
resources. Additionally, as $R_{A}\to 0^{+}$, the ratio grows unboundedly
$E\to\infty$. This is due to the fact that without any real-time resources,
player $A$ needs $P\geq R_{B}$ pre-allocated resources to obtain a positive
payoff (see third case of Theorem 3.1). A plot of the ratio $E$ is shown in
the left Figure 2.

The proof of Theorem 4.2 relies on the following technical lemma, which
provides the level curves of the SPE payoff $\pi_{A}^{*}(P,R_{A},R_{B})$. A
level curve with fixed performance level $\Pi\in[0,1]$ is defined as the set
of points

$L_{\Pi}:=\\{(P,R_{A})\in\mathbb{R}_{+}^{2}:\pi_{A}^{*}(P,R_{A},R_{B})=\Pi\\}.$
(26)

######  Lemma 4.3.

Given any $R_{B}>0$ and $\boldsymbol{w}\in\mathbb{R}^{n}_{++}$, fix a desired
performance level $\Pi\in[0,1]$. The level curve $L_{\Pi}$ is given by

$L_{\Pi}=\bigcup_{P\in\left[0,\frac{R_{B}}{1-\Pi}\right]}\left(P,R_{\Pi}(P)\right)$
(27)

where if $0\leq\Pi<\frac{1}{2}$,

$R_{\Pi}(P)=\\!\\!\begin{cases}2\Pi(R_{B}-P)&\,\text{for
}P\in\Big{[}0,\frac{(1-2\Pi)R_{B}}{1-\Pi}\Big{)}\\\
\\!\\!\frac{(R_{B}-(1-\Pi)P)^{2}}{2R_{B}(1-\Pi)}&\,\text{for
}P\in\Big{[}\frac{(1-2\Pi)R_{B}}{1-\Pi},\frac{WR_{B}}{1-\Pi}\Big{]}\end{cases}$
(28)

and if $\frac{1}{2}\leq\Pi\leq 1$,

$R_{\Pi}(P)=\frac{(R_{B}-(1-\Pi)P)^{2}}{2R_{B}(1-\Pi)}$ (29)

If $P>\frac{R_{B}}{1-\Pi}$, then $\pi^{*}_{A}(P,R_{A},R_{B})>\Pi$ for any
$R_{A}\geq 0$.

The proof of this Lemma directly follows from the expressions in Theorem 3.1
and is thus omitted. In the center Figure 2, we illustrate level curves
associated with varying performance levels $\Pi$. We can now leverage the
above Lemma to complete the proof of Theorem 4.2.

######  Proof 4.4 (Proof of Theorem 4.2).

First, suppose $R_{A}<R_{B}$. Then
$\pi_{A}^{*}(0,R_{A},R_{B})=\frac{R_{A}}{2R_{B}}<1/2$. Focusing on the level
curve associated with the value $\Pi=\frac{R_{A}}{2R_{B}}$, the quantity
$P^{\text{eq}}$ is determined as the endpoint of this curve where there are
zero real-time resources. From (28) , this occurs when
$P=\frac{R_{B}}{1-\Pi}=\frac{2(R_{B})^{2}}{2R_{B}-R_{A}}$.

Now, suppose $R_{A}\geq R_{B}$. Then
$\pi_{A}^{*}(0,R_{A},R_{B})=(1-\frac{R_{B}}{2R_{A}}\geq 1/2$. Similarly, the
quantity $P^{\text{eq}}$ is determined as the endpoint of the level curve
associated with $\Pi=(1-\frac{R_{B}}{2R_{A}})$. From (29) , this occurs when
$P=\frac{R_{B}}{1-\Pi}=2R_{A}$.

###  4.2 Optimal investment in resources

In addition to the effectiveness ratio, the interplay between the two types of
resources is also highlighted by the following scenario: player $A$ has an
opportunity to make an investment decision regarding its resource endowments.
That is, the pair $(P,R_{A})\in\mathbb{R}^{2}_{+}$ is a strategic choice made
by player $A$ before the game $\text{GL-P}(P,R_{A},R_{B},\boldsymbol{w})$ is
played. Given a monetary budget $M_{A}>0$ for player $A$, any pair $(P,R_{A})$
must belong to the following set of feasible investments:

$\mathcal{I}(M_{A}):=\\{(P,R_{A}):R_{A}+c_{A}P\leq M_{A}\\}$ (30)

where $c_{A}\geq 0$ is the per-unit cost for purchasing pre-allocated
resources, and we assume the per-unit cost for purchasing real-time resources
is 1 without loss of generality. We are interested in characterizing player
$A$’s optimal investment subject to the above cost constraint, and given
player $B$’s resource endowment $R_{B}>0$. This is formulated as the following
optimization problem:

$\pi_{A}^{\mathrm{opt}}:=\max_{(P,R_{A})\in\mathcal{I}(M_{A})}\pi_{A}^{*}(P,R_{A},R_{B}).$
(31)

In the result below, we derive the complete solution to the optimal investment
problem (31).

######  Theorem 4.5.

Fix a monetary budget $M_{A}>0$, relative per-unit cost $c_{A}>0$, and
$R_{B}>0$ real-time resources for player $B$. Then, player $A$’s optimal
investment in pre-allocated resources in (31) is

$P^{*}=\begin{cases}\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}},&\text{if
}c_{A}<t\\\ \in[0,\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}],&\text{if
}c_{A}=t\\\ 0,&\text{if }c_{A}>t\end{cases}.$ (32)

where $t:=\min\\{1,\frac{M_{A}}{R_{B}}\\}$. The optimal investment in real-
time resources is $R_{A}^{*}=M_{A}-c_{A}P^{*}$. The resulting payoff
$\pi_{A}^{\mathrm{opt}}$ to player $A$ is given by

$\begin{cases}1-\frac{R_{B}}{2M_{A}}c_{A}(2-c_{A}),&\text{if }c_{A}<t\\\
1-\frac{R_{B}}{2M_{A}},&\text{if }c_{A}\geq t\text{ and
}\frac{M_{A}}{R_{B}}\geq 1\\\ \frac{M_{A}}{2R_{B}},&\text{if }c_{A}\geq
t\text{ and }\frac{M_{A}}{R_{B}}<1\end{cases}.$ (33)

A plot of the optimal investment $P^{*}$ (32) is shown in the right Figure 2.
If the cost $c_{A}$ exceeds 1, then there is no investment in pre-allocated
resources since they are less effective than real-time resources. Thus,
$c_{A}$ must necessarily be cheaper than real-time resources in order to
invest in any positive amount. We note that while an optimal investment can
purely consist of real-time resources, no optimal investment from Theorem 4.5
can purely consist of pre-allocated resources. Interestingly, when the
monetary budget is small ($M_{A}<1$), there is a discontinuity in the
investment level $P^{*}$ at $c_{A}=R_{B}$. A visual illustration of how the
optimal investments are determined is shown in Figure 3, which is detailed in
the proof below.

Figure 3: This plot illsutrates how to determine the optimal investment
$(P^{*},R^{*}_{A})\in\mathbb{R}^{2}_{+}$ subject to the cost constraint in
(30). The set of feasible investments $\mathcal{I}(M_{A})$ is the line segment
connecting $(0,M_{A})$ and $(M_{A}/c_{A},0)$. The optimal investment lies on
the level curve tangent to this line segment. For example, when $c=0.423$, the
optimal investment is $(2.309,0.357)$ (unfilled circle), which gives a
performance level of $\Pi=0.75$. For sufficiently high cost $c_{A}$,
$\mathcal{I}(M_{A})$ will not be tangent to any level curve, and the optimal
investment is $(0,M_{A})$. For example, when $c_{A}=1.333$, the highest level
curve that intersects $\mathcal{I}(M_{A})$ is $\Pi=0.625$, and the optimal
investment is $(0,4/3)$ (filled square).

######  Proof 4.6.

We first observe that for any $\Pi\in(0,1)$, the level curve $R_{\Pi}(P)$
(from Lemma 4.3) is strictly decreasing and convex in
$P\in[0,\frac{R_{B}}{1-\Pi}]$. Hence, the function $\pi_{A}(P,R_{A},R_{B})$ is
quasi-concave in $(P,R_{A})$. Observe that the set of points
$(P,R_{A})\in\mathbb{R}^{2}_{+}$ that satisfy $R_{A}+c_{A}P=M_{A}$ consists of
the line segment $R_{A}=M_{A}-c_{A}P$, $P\in[0,M_{A}/c_{A}]$, with slope
$-c_{A}$, and end-points $(M_{A},0)$ and $(0,M_{A}/c_{A})$. Thus, the
optimization amounts to finding the highest level curve that intersects with
$R_{A}=M_{A}-c_{A}P$, $P\in[0,M_{A}/c_{A}]$.

The slope of a level curve $R_{\Pi}(P)$ at $P=0$ is

$\frac{\partial R_{\Pi}}{\partial P}(0)=\begin{cases}-2\Pi,&\text{if
}\Pi<\frac{1}{2}\\\ -1,&\text{if }\Pi\geq\frac{1}{2}\end{cases}.$ (34)

Let $M_{A}\geq 0$ such that $\pi^{*}_{A}(0,M_{A},R_{B})=\Pi\geq 1/2$. Then,
note that, if $-c_{A}<-1$ (or, equivalently $c_{A}>1$), then
$R_{A}=M_{A}-c_{A}P$ shrinks faster in $P$ than the level curve $R_{\Pi}(P)$
by monotonicity and convexity of the level curve in $P$. Thus, the allocation
$(P,R_{A})=(0,M_{A})$ maximizes $A$’s payoff, as all other points on the line
segment $R_{A}=M_{A}-c_{A}P$, $P\in[0,M_{A}/c_{A}]$, intersect with strictly
lower level curves. Similarly, for $M_{A}\geq 0$ such that
$\pi^{*}_{A}(0,M_{A},R_{B})=\Pi<1/2$, the allocation $(P,R_{A})=(0,M_{A})$
maximizes $A$’s payoff when $c_{A}>2\Pi$. Since the condition
$\pi^{*}_{A}(0,M_{A},R_{B})=\Pi\geq 1/2$ is equivalent to $M_{A}\geq R_{B}$
and $\pi^{*}_{A}(0,M_{A},R_{B})=\frac{R_{A}}{2R_{B}}$ when $M_{A}<R_{B}$, it
follows that $P^{*}=0$ if $c_{A}>\min\\{1,\frac{M_{A}}{R_{B}}\\}$. For the
remainder of the proof, we use $t=1$ (resp. $t=\frac{M_{A}}{R_{B}}$) and
$M_{A}\geq R_{B}$ (resp. $M_{A}<R_{B}$) interchangeably.

Suppose that $c_{A}=t$ and $t=\frac{M_{A}}{R_{B}}$. Then the level curve
corresponding to $\Pi(M_{A})=\frac{W}{2}c_{A}$ has an interval of budget-
feasible points $(P,R_{A})$ parameterized by
$P\in[0,(1-\frac{c_{A}}{2-c_{A}})\frac{M_{A}}{c_{A}}]$ with $R_{A}=M_{A}-cP$.
If $t=1$, then there is a single budget-feasible point $(P,R_{A})=(0,M_{A})$
for the level curve corresponding to $\Pi(M_{A})=(1-\frac{R_{B}}{2M_{A}})$. In
both cases, there are no budget-feasible points for any level curve
corresponding to $\Pi>\Pi(M_{A})$.

Now, suppose $t=1$ and $c_{A}<t$. We wish to find the level curve for which
the line $(P,M_{A}-c_{A}P)$, $P\in[0,M_{A}/c_{A}]$, lies tangent. The point(s)
of tangency yields the optimal solution due to the quasi-concavity of
$\pi_{A}^{*}$. Furthermore, since $M_{A}\geq R_{B}$ and $c_{A}<1$, a solution
$(P,R_{A})$ must satisfy $\Pi\in[\frac{1}{2},1]$ and

$\displaystyle\frac{\partial R_{\Pi}}{\partial P}(P^{*})$
$\displaystyle=\frac{P^{*}(1-\Pi)}{R_{B}W}-1$ $\displaystyle=-c_{A}$ (35)
$\displaystyle R_{\Pi}(P^{*})$
$\displaystyle=\frac{(R_{B}-(1-\Pi)P^{*})^{2}}{2R_{B}(1-\Pi)}$
$\displaystyle=M_{A}-c_{A}P^{*}$

From the first equation, we obtain $P^{*}=\frac{R_{B}(1-c_{A})}{1-\Pi}$.
Plugging this expression into the second equation, we obtain
$\Pi=(1-\frac{R_{B}}{2M_{A}}c_{A}(2-c_{A}))\in[\frac{1}{2},1]$, which leads to
the unique solution
$P^{*}=(1-\frac{c_{A}}{2-c_{A}})\frac{M_{A}}{c_{A}}\leq\frac{M_{A}}{c_{A}}$.

Lastly, suppose $c_{A}<t$ and $t=\frac{M_{A}}{R_{B}}$ ($M_{A}<R_{B}$). Similar
to the preceding case, we seek the highest level curve for which the budget
constraint is tangent. Due to the assumption that $M_{A}<R_{B}$ and
$c<\frac{M_{A}}{R_{B}}$, we observe that tangent points cannot exist for
$P<\frac{1-2\Pi}{1-\Pi}R_{B}$ and $\Pi<\frac{1}{2}$, i.e. in the region where
the level curve is linear. Thus, it must be that either $\Pi<\frac{1}{2}$ and
$P\geq\frac{1-2\Pi}{1-\Pi}R_{B}$, or $\Pi\geq\frac{1}{2}$. In either case, a
solution must also satisfy the equations in (35), from which we obtain an
identical expression for $P^{*}$.

##  5 Two-sided pre-allocations

The scenarios studied thus far have considered one-sided pre-allocations,
where only player $A$ has the opportunity for early investments. The goal in
this section is to take preliminary steps in understanding how multiple rounds
of early investments, on the part by both competitors, impacts the players’
performance in the final stage. We will consider a scenario where player $B$
has an opportunity to respond to the pre-allocation decision of player $A$
with its own pre-allocated resources, which we formulate as a Stackelberg
game.

######  Remark 5.1.

Before formalizing this game, we remark that such a scenario admits positive
and negative pre-allocations, i.e. $p_{b}>0$ for some subset of battlefields
and $p_{b}<0$ on the others. Here, $p_{b}<0$ means that the amount $|p_{b}|$
of pre-allocated resources favors player $B$. While the work in [28]
establishes existence of equilibrium for any such pre-allocations as well as
numerical approaches to compute equilibria to arbitrary precision, it does not
provide analytical characterizations of them. Indeed, while our current
techniques (i.e. from Theorem 3.1) analytically derive the equilibria for any
positive pre-allocation vector, they are yet unable to account for such two-
sided favoritism. Developing appropriate methods is subject to future study.

In light of the aforementioned limitations, we may still investigate the
impact of player $B$’s response in the context of a single-battlefield
environment333In contrast to Colonel Blotto games, General Lotto games with a
single battlefield still provides rich insights that often generalize to
multi-battlefield scenarios [11, 12, 22].. The Stackelberg game is defined as
follows. Player $A$ has a monetary budget $M_{A}$ with per-unit cost
$c_{A}\in(0,1)$ for stationary resources. Similarly, player $B$ has a monetary
budget $M_{B}$ with per-unit cost $c_{B}\in(0,1)$. The players compete over a
single battlefield of unit value.

– _Stage 1:_ Player $A$ chooses its pre-allocation investment
$p_{A}\in[0,\frac{M_{A}}{c_{A}}]$. This becomes common knowledge.

– _Stage 2:_ Player $B$ chooses its pre-allocation investment
$p_{B}\in[0,\frac{M_{B}}{c_{B}}]$.

– _Stage 3:_ The players engage in the General Lotto game with favoritism
$\text{GL-F}(p_{A}-p_{B},M_{A}-c_{A}p_{A},M_{B}-c_{B}p_{B})$. Players derive
the final payoffs

$\displaystyle u_{A}(p_{A},p_{B})$
$\displaystyle:=\pi_{A}^{*}(p_{A}-p_{B},M_{A}-c_{A}p_{A},M_{B}-c_{B}p_{B})$
(36) $\displaystyle u_{B}(p_{A},p_{B})$ $\displaystyle:=1-u_{A}(p_{A},p_{B})$

Note that $p_{A}-p_{B}$ is the favoritism to player $A$. When it is non-
negative, $\pi_{A}^{*}$ is given precisely by Theorem 3.1. When it is
negative, $\pi^{*}_{A}=1-\pi^{*}_{B}$ where $\pi_{B}^{*}$ is given as in
Theorem 3.1 with the indices switched.

Let us denote this game as $\text{GL-S}(\\{M_{i},c_{i}\\}_{i=A,B})$. We seek
to characterize the following equilibrium concept.

######  Definition 5.2.

The investment profile $(p_{A}^{*},p_{B}^{*})$ is a _Stackelberg equilibrium_
if

$p_{A}^{*}\in\arg\max_{p_{A}\in[0,M_{A}/c_{A}]}\left(\min_{p_{B}\in[0,M_{B}/c_{B}]}u_{A}(p_{A},p_{B})\right)$
(37)

and

$p_{B}^{*}\in\arg\min_{p_{B}\in[0,M_{B}/c_{B}]}u_{A}(p_{A}^{*},p_{B}).$ (38)

Note that the definition in (37) is in a max-min form, since the final payoffs
in GL-S are constant-sum. The characterization of the Stackelberg equilibrium
is given in the result below.

######  Proposition 5.3.

The Stackelberg equilibrium of $\text{GL-S}(\\{M_{i},c_{i}\\}_{i=A,B})$ is
given as follows.

  1. 1.

Suppose $\frac{M_{B}}{c_{B}}\leq M_{A}$. Then $p_{A}^{*}$ is given according
to Theorem 4.5 and $p_{B}^{*}=0$.

  2. 2.

Suppose $M_{A}<\frac{M_{B}}{c_{B}}\leq\frac{M_{A}}{c_{A}}$. If
$p_{A}^{\dagger}<\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}$, then
$p_{A}^{*}$ is given according to Theorem 4.5 and $p_{B}^{*}=0$, where
$p_{A}^{\dagger}\in(0,\frac{M_{A}}{c_{A}}]$ is the unique value that satisfies
$u_{B}(p_{A},0)=u_{B}(p_{A},\hat{p}_{B})$, with

$\hat{p}_{B}:=\frac{M_{B}}{c_{B}}-\frac{M_{B}-c_{B}p_{A}}{2-c_{B}}.$ (39)

If $p_{A}^{\dagger}\geq\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}$, then
$p_{A}^{*}=p_{A}^{\dagger}$, and $p_{B}^{*}=0$ or $\hat{p}_{B}$.

  3. 3.

Suppose $\frac{M_{A}}{c_{A}}<\frac{M_{B}}{c_{B}}$. Then $p_{A}^{*}=0$ and
$p_{B}^{*}=\hat{p}_{B}$.

Several comments are in order. In the first interval $\frac{M_{B}}{c_{B}}\leq
M_{A}$, player $B$ is sufficiently weak such that it does not respond with any
of its own stationary resources against any player $A$ investment
$p_{A}\in[0,\frac{M_{A}}{c_{A}}]$. Thus, the Stackelberg solution recovers the
result from Theorem 4.5. For a large part of the middle interval
$M_{A}<\frac{M_{B}}{c_{B}}\leq\frac{M_{A}}{c_{A}}$, $p_{A}^{*}$ coincides with
the investment from Theorem 4.5, which forces player $B$ to respond with zero
stationary resources (i.e. when
$p_{A}^{\dagger}<\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}$). However,
when $p_{A}^{\dagger}\geq\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}$,
player $A$’s optimal investment makes player $B$ indifferent between
responding with zero or with an amount $\hat{p}_{B}>p_{A}^{*}$ of stationary
resources that exceeds the pre-allocation of player $A$. In the last interval
$\frac{M_{A}}{c_{A}}<\frac{M_{B}}{c_{B}}$, player $B$ is sufficiently strong
such that it is optimal for player $A$ to invest zero stationary resources.

###  5.1 The impact of responding

We illustrate the implications of Proposition 5.3 regarding the responding
player’s ($B$) performance in the numerical example shown in Figure 4. Here,
we compare player $B$’s Stackelberg equilibrium payoff to the payoff it would
have obtained if it did not have an opportunity to respond. We recall that
this payoff was characterized in Theorem 4.5. By definition, the Stackelberg
payoff is necessarily at least as high as the non-responding payoff – one
performs better being able to respond to a pre-allocation. However, what is
notable in Figure 4 is a significant, discontinuous increase in payoff for
player $B$ once its budget $\frac{M_{B}}{c_{B}}$ is sufficiently high (i.e.
exceeds $\frac{M_{A}}{c_{A}}$).

The presence of the discontinuity strongly suggests that being in a resource-
advantaged position with regards to the monetary budget $\frac{M_{B}}{c_{B}}$
is a crucial factor for performance in multi-stage resource allocation. At the
same time, Proposition 5.3 asserts that no response is optimal if player $B$
is resource-disadvantaged (first item). This conclusion is contrasted with the
classic simultaneous-move General Lotto games, in which there is no such
discontinuity in the equilibrium payoffs – they vary continuously in the
players’ budgets [11, 18].

Figure 4: This plot illustrates the Stackelberg equilibrium payoff (red line,
Proposition 5.3) to player $B$ contrasted with its payoff if it did not have
the opportunity to respond with pre-allocated resources, i.e. setting
$p_{B}=0$ (green dashed line, Theorem 4.5). Notably, there is a dramatic
improvement in performance when player $B$ is sufficiently budget-rich,
$\frac{M_{B}}{c_{B}}=\frac{M_{A}}{c_{A}}$. In this example, we set
$M_{A}=0.5$, $c_{A}=0.2$, and $c_{B}=0.5$. We vary $M_{B}$ from $0$ to $3$.

The remainder of this section provides the proof of Proposition 5.3, which
utilizes two supporting Lemmas.

###  5.2 Follower’s best-response

We begin by analyzing player $B$’s best-response to any player $A$ pre-
allocation $p_{A}$. We need to find $p_{B}^{*}$ that solves

$\max_{p_{B}\in[0,M_{B}/c_{B}]}u_{B}(p_{A},p_{B}).$ (40)

Let us denote $\textbf{p}=(p_{A},p_{B})$, $R_{A}=M_{A}-c_{A}p_{A}$, and
$R_{B}=M_{B}-c_{B}p_{B}$. Define
$f_{A}(\textbf{p}):=R_{A}+\sqrt{R_{A}(R_{A}+2(p_{A}-p_{B}))}$,
$f_{B}(\textbf{p}):=\sqrt{R_{B}}$, and
$g_{B}(\textbf{p}):=\sqrt{R_{B}+2(p_{B}-p_{A})}$. Define

$\displaystyle u_{B}^{1A}(\textbf{p})$
$\displaystyle:=\frac{M_{B}-c_{B}p_{B}}{2R_{A}}\left(\frac{f_{A}(\textbf{p})}{p_{A}-p_{B}+f_{A}(\textbf{p})}\right)^{2}$
(41) $\displaystyle u_{B}^{2A}(\textbf{p})$
$\displaystyle:=1-\frac{R_{A}}{2(M_{B}-c_{B}p_{B}-(p_{A}-p_{B}))}$
$\displaystyle u_{B}^{1B}(\textbf{p})$
$\displaystyle:=1-\frac{R_{A}}{2}\left(\frac{f_{B}(\textbf{p})+g_{B}(\textbf{p})}{p_{B}-p_{A}+f_{B}(\textbf{p})(f_{B}(\textbf{p})+g_{B}(\textbf{p}))}\right)^{2}$
$\displaystyle u_{B}^{2B}(\textbf{p})$
$\displaystyle:=\frac{M_{B}-c_{B}p_{B}}{2(R_{A}-(p_{B}-p_{A}))}$

From Theorem 3.1, one can write player $B$’s payoff in GL-S as

$u_{B}(\textbf{p})=\begin{cases}u_{B}^{1A}(\textbf{p})&\text{if
}\textbf{p}\in\mathcal{R}^{1A}\\\ u_{B}^{2A}(\textbf{p})&\text{if
}\textbf{p}\in\mathcal{R}^{2A}\\\ u_{B}^{1B}(\textbf{p})&\text{if
}\textbf{p}\in\mathcal{R}^{1B}\\\ u_{B}^{2B}(\textbf{p})&\text{if
}\textbf{p}\in\mathcal{R}^{2B}\end{cases}$ (42)

where

$\displaystyle\mathcal{R}^{1A}$ $\displaystyle=\\{p_{A}\geq
p_{B}:R_{B}<p_{A}-p_{B},\text{ or }$ (43) $\displaystyle R_{B}\geq
p_{A}-p_{B}\text{ and
}R_{A}\geq\frac{2(R_{B}-(p_{A}-p_{B}))^{2}}{2R_{B}-(p_{A}-p_{B})}\\}$
$\displaystyle\mathcal{R}^{2A}$ $\displaystyle=\\{p_{A}\geq
p_{B}\\}\backslash\mathcal{R}^{1A}$ $\displaystyle\mathcal{R}^{1B}$
$\displaystyle=\\{p_{B}>p_{A}:R_{A}<p_{B}-p_{A},\text{ or }$ $\displaystyle
R_{A}\geq p_{B}-p_{A}\text{ and
}R_{B}\geq\frac{2(R_{A}-(p_{B}-p_{A}))^{2}}{2R_{A}-(p_{B}-p_{A})}\\}$
$\displaystyle\mathcal{R}^{2B}$
$\displaystyle=\\{p_{B}>p_{A}\\}\backslash\mathcal{R}^{1B}$

The following Lemma details player $B$’s payoff for any response
$p_{B}\in[0,\frac{M_{B}}{c_{B}}]$.

######  Lemma 5.4.

Consider any fixed strategy $(p_{A},R_{A})$ for player $A$. Player $B$’s
payoff is given as follows.

  1. a)

If $\frac{M_{B}}{c_{B}}\leq R_{A}+p_{A}$, then $u_{B}(\textbf{p})$ is
decreasing for all $p_{B}\in[0,\frac{M_{B}}{c_{B}}]$.

  2. b)

If $R_{A}+p_{A}<\frac{M_{B}}{c_{B}}\leq\frac{R_{A}}{c_{B}}+p_{A}$, then

$u_{B}(\textbf{p})=\begin{cases}u_{B}^{1A}(\textbf{p}),&\text{if
}p_{B}\in[0,p_{A}]\\\ u_{B}^{2B}(\textbf{p}),&\text{if
}p_{B}\in(p_{A},p_{B}^{1B}]\\\ u_{B}^{1B}(\textbf{p}),&\text{if
}p_{B}\in(p_{B}^{1B},\frac{M_{B}}{c_{B}}]\end{cases}$ (44)

where $p_{B}^{1B}$ is the unique solution to

$F(p_{B}):=\frac{(R_{A}+p_{A}-p_{B})^{2}}{2R_{A}+p_{A}-p_{B}}=M_{B}-c_{B}p_{B}.$
(45)

  3. c)

If
$\frac{R_{A}}{c_{B}}+p_{A}<\frac{M_{B}}{c_{B}}\leq\frac{1}{c_{B}}\left(p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2}\right)$,
then

$u_{B}(\textbf{p})=\begin{cases}u_{B}^{1A}(\textbf{p}),&p_{B}\in[0,p_{B}^{1A}]\\\
u_{B}^{2A}(\textbf{p}),&p_{B}\in(p_{B}^{1A},p_{A}]\\\
u_{B}^{1B}(\textbf{p}),&p_{B}\in(p_{A},\frac{M_{B}}{c_{B}}]\end{cases}$ (46)

where $p_{B}^{1A}$ is the unique solution to

$G(p_{B}):=\frac{2(M_{B}+(1-c_{B})p_{B}-p_{A})^{2}}{2(M_{B}-c_{B}p_{B})-(p_{A}-p_{B})}=R_{A}.$
(47)

on $p_{B}\in(0,p_{A})$

  4. d)

If
$\frac{M_{B}}{c_{B}}>\frac{1}{c_{B}}\left(p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2}\right)$,
then

$u_{B}(\textbf{p})=\begin{cases}u_{B}^{2A}(\textbf{p}),&\text{if
}p_{B}\in[0,p_{A}]\\\ u_{B}^{1B}(\textbf{p}),&\text{if
}p_{B}\in(p_{A},\frac{M_{B}}{c_{B}}].\end{cases}$ (48)

The proof is deferred to Appendix .2. The final Lemma characterizes player
$B$’s best response to any fixed player $A$ strategy.

######  Lemma 5.5.

Consider any fixed strategy $(p_{A},R_{A})$ for player $A$. Player $B$’s best-
response

$p_{B}^{*}:=\underset{p_{B}\in[0,\frac{M_{B}}{c_{B}}]}{\operatorname{arg}\,\operatorname{max}}\;u_{B}(p_{A},p_{B})$
(49)

is determined according to

$p_{B}^{*}=\begin{cases}0,&\text{if }\frac{M_{B}}{c_{B}}<h(R_{A},p_{A})\\\
\hat{p}_{B},&\text{if }\frac{M_{B}}{c_{B}}>h(R_{A},p_{A})\\\ 0\text{ or
}\hat{p}_{B},&\text{if }\frac{M_{B}}{c_{B}}=h(R_{A},p_{A})\end{cases}$ (50)

where $\hat{p}_{B}$ was defined in (39), and
$h(R_{A},p_{A})\in(R_{A}+p_{A},p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2})$
is the unique value of $\frac{M_{B}}{c_{B}}$ at which
$u_{B}^{1A}(p_{A},0)=u_{B}^{1B}(p_{A},\hat{p}_{B})$.

The proof is deferred to Appendix .3, where we thoroughly analyze the
properties of the functions $u_{B}^{1A}$, $u_{B}^{2A}$, $u_{B}^{1B}$, and
$u_{B}^{2B}$ as characterized by Lemma 5.4. This allows us to identify the
maximizer $p_{B}^{*}\in[0,\frac{M_{B}}{c_{B}}]$ of $u_{B}$ for all possible
cases stated in Lemma 5.4.

###  5.3 Proof of Proposition 5.3

We are now ready to establish Proposition 5.3.

######  Proof 5.6.

Let us define

$\displaystyle T_{1}(p_{A})$ $\displaystyle:=p_{A}+R_{A}(p_{A})$ (51)
$\displaystyle T_{2}(p_{A})$
$\displaystyle:=p_{A}+\frac{R_{A}(p_{A})+\sqrt{R_{A}(p_{A})(R_{A}(p_{A})+2p_{A})}}{2}.$

where $R_{A}(p_{A})=M_{A}-c_{A}p_{A}$. It holds that
$T_{1}(0)=T_{2}(0)=M_{A}$,
$T_{1}(\frac{M_{A}}{c_{A}})=T_{2}(\frac{M_{A}}{c_{A}})=\frac{M_{A}}{c_{A}}$,
and $T_{1}(p_{A})<T_{2}(p_{A})$ on the interval
$p_{A}\in(0,\frac{M_{A}}{c_{A}})$. We prove the result item by item.

1) Suppose $\frac{M_{B}}{c_{B}}\leq M_{A}$. In this case, we have
$\frac{M_{B}}{c_{B}}\leq T_{1}(p_{A})$ for all
$p_{A}\in[0,\frac{M_{A}}{c_{A}}]$, with equality at $p_{A}=0$ if and only if
$\frac{M_{B}}{c_{B}}=M_{A}$. Thus, player $B$’s best-response against any
$p_{A}$ is $p_{B}^{*}=0$ (Lemma 5.5). The scenario reduces to the optimization
problem from Corollary 4.1.

2) Suppose $M_{A}<\frac{M_{B}}{c_{B}}\leq\frac{M_{A}}{c_{A}}$.

By Lemma 5.5, the threshold at which player $B$’s best-response switches is
given by the value of $p_{A}$ that satisfies
$h(M_{A}-c_{A}p_{A},p_{A})=\frac{M_{B}}{c_{B}}$. Equivalently, this is the
value of $p_{A}$ that satisfies
$u_{B}^{1A}(p_{A},0)=u_{B}^{1B}(p_{A},\hat{p}_{B})$ with
$\frac{M_{B}}{c_{B}}\leq\frac{1}{c_{B}}T_{2}(p_{A})$. The latter condition
ensures that either $u_{B}^{1A}(p_{A},0)$ or $u_{B}^{1B}(p_{A},\hat{p}_{B})$
is player $B$’s best-response payoff (Lemma 5.4). One can write

$u_{B}^{1B}(p_{A},\hat{p}_{B})=1-\frac{c_{A}(2-c_{B})}{2}\frac{\frac{M_{A}}{c_{A}}-p_{A}}{\frac{M_{B}}{c_{B}}-p_{A}}$
(52)

This is a decreasing and concave function on
$p_{A}\in[0,\frac{M_{B}}{c_{B}})$, and it decreases to $-\infty$ as
$p_{A}\rightarrow\frac{M_{B}}{c_{B}}$. The payoff $u_{B}^{1A}(p_{A},0)$ is
given by

$u_{B}^{1A}(p_{A},0)=\frac{M_{B}}{2\sqrt{R_{A}(p_{A})}}\left(\frac{2(T_{2}(p_{A})-p_{A})}{T_{1}(p_{A})+2(T_{2}(p_{A})-p_{A})}\right)^{2}$
(53)

This function has a single critical point in the interval
$[0,\frac{M_{A}}{c_{A}}]$ at
$\bar{p}_{A}=\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}$, which is a local
minimum. It is decreasing on $[0,\bar{p}_{A})$ and increasing on
$(\bar{p}_{A},\frac{M_{A}}{c_{A}}]$.

There is a unique value $p_{A}^{\dagger}\in(0,\frac{M_{B}}{c_{B}})$ such that
$\frac{M_{B}}{c_{B}}\leq\frac{1}{c_{B}}T_{2}(p_{A}^{\dagger})$ where these two
functions intersect. Note that from Lemma 5.5, $p_{A}^{\dagger}$ is the unique
value that satisfies

$h(M_{A}-c_{A}p_{A}^{\dagger},p_{A}^{\dagger})=\frac{M_{B}}{c_{B}}.$ (54)

Player $B$’s best-response is $p_{B}^{*}=\hat{p}_{B}$ for
$p_{A}<p_{A}^{\dagger}$, and $p_{B}^{*}=0$ for $p_{A}>p_{A}^{\dagger}$ (Lemma
5.5). Consequently, player $A$’s payoff (under player $B$’s best-response) is
given by

$u_{A}(p_{A},p_{B}^{*})=\begin{cases}1-u_{B}^{1B}(p_{A},\hat{p}_{B}),&\text{if
}p_{A}\in[0,p_{A}^{\dagger})\\\ 1-u_{B}^{1A}(p_{A},0),&\text{if
}p_{A}\in(p_{A}^{\dagger},\frac{M_{A}}{c_{A}}]\end{cases}$ (55)

We observe that $u_{A}(p_{A},p_{B}^{*})$ is increasing on
$p_{A}\in[0,p_{A}^{\dagger})$. If $p_{A}^{\dagger}\geq\bar{p}_{A}$, then
$u_{A}(p_{A},p_{B}^{*})$ is decreasing on
$(p_{A}^{\dagger},\frac{M_{A}}{c_{A}}]$, and hence player $A$’s security
strategy is $p_{A}^{*}=p_{A}^{\dagger}$. The resulting payoff to player $A$ is

$1-u_{B}^{1B}(p_{A}^{\dagger},\hat{p}_{B})=\frac{c_{A}(2-c_{B})}{2}\frac{\frac{M_{A}}{c_{A}}-p_{A}^{\dagger}}{\frac{M_{B}}{c_{B}}-p_{A}^{\dagger}}.$
(56)

If $p_{A}^{\dagger}<\bar{p}_{A}$, then $u_{A}(p_{A},p_{B}^{*})$ is increasing
on $(p_{A}^{\dagger},\bar{p}_{A}]$ and decreasing on
$(\bar{p}_{A},\frac{M_{A}}{c_{A}}]$. Hence, player $A$’s security strategy is
$p_{A}^{*}=\bar{p}_{A}=\frac{2(1-c_{A})}{2-c_{A}}\frac{M_{A}}{c_{A}}$. The
resulting payoff to player $A$ is given by Corollary 4.1.

3) Suppose $\frac{M_{B}}{c_{B}}>\frac{M_{A}}{c_{A}}$.

In this case, the function $u_{B}^{1B}(p_{A},\hat{p}_{B})$ is strictly
increasing on $p_{A}\in[0,\frac{M_{A}}{c_{A}}]$. Any intersection (if any)
with $u_{B}^{1A}(p_{A},0)$ must occur on $p_{A}>\bar{p}_{A}$, where
$u_{B}^{1A}(p_{A},0)$ is increasing. Therefore, $u_{A}(p_{A},p_{B}^{*})$ must
be strictly decreasing on $[0,\frac{M_{A}}{c_{a}}]$ and hence $p_{A}^{*}=0$.
The resulting payoff to player $A$ is

$1-u_{B}^{1B}(0,\hat{p}_{B})=\frac{c_{B}(2-c_{B})}{2}\frac{M_{A}}{M_{B}}.$
(57)

##  6 Conclusion

In this manuscript, we studied the strategic role of pre-allocations in
competitive interactions under a two-stage General Lotto game model, where one
of the players can place resources before a decisive point of conflict. Our
main contribution fully provided subgame-perfect equilibrium characterizations
to this formulation. This result revealed a rich interplay between the
effectiveness of pre-allocated resources and real-time resources, which
allowed us to quantify optimal investments in both types of resources. We then
analyzed a Stackelberg game scenario where the other player is able to respond
to the pre-allocation before the final decisive round. This highlights the
significance that more dynamic and sequential interactions can have on a
player’s eventual performance. Future work will involve studying these dynamic
interactions in richer environmental contexts, e.g. with multiple fronts of
battlefields.

## References

  * [1] T. Aidt, K. A. Konrad, and D. Kovenock.  Dynamics of conflict.  European journal of political economy, (60):101838, 2019.
  * [2] G. Brown, M. Carlyle, J. Salmerón, and K. Wood.  Defending critical infrastructure.  Interfaces, 36(6):530–544, 2006.
  * [3] R. Chandan, K. Paarporn, M. Alizadeh, and J. R. Marden.  Strategic investments in multi-stage general lotto games.  In 2022 IEEE 61st Conference on Decision and Control (CDC), pages 4444–4448. IEEE, 2022.
  * [4] R. Chandan, K. Paarporn, D. Kovenock, M. Alizadeh, and J. R. Marden.  The art of concession in general lotto games.  In Game Theory for Networks: 11th International EAI Conference, GameNets 2022, Virtual Event, July 7–8, 2022, Proceedings, pages 310–327. Springer, 2023.
  * [5] A. Chowdhary, S. Sengupta, A. Alshamrani, D. Huang, and A. Sabur.  Adaptive mtd security using markov game modeling.  In 2019 International Conference on Computing, Networking and Communications (ICNC), pages 577–581. IEEE, 2019.
  * [6] E. Dechenaux, D. Kovenock, and R. M. Sheremeta.  A survey of experimental research on contests, all-pay auctions and tournaments.  Experimental Economics, 18:609–669, 2015.
  * [7] S. R. Etesami and T. Başar.  Dynamic games in cyber-physical security: An overview.  Dynamic Games and Applications, 9(4):884–913, 2019.
  * [8] O. Gross and R. Wagner.  A continuous Colonel Blotto game.  Technical report, RAND Project, Air Force, Santa Monica, 1950.
  * [9] A. Gupta, T. Başar, and G. Schwartz.  A three-stage Colonel Blotto game: when to provide more information to an adversary.  In International Conference on Decision and Game Theory for Security, pages 216–233. Springer, 2014.
  * [10] A. Gupta, G. Schwartz, C. Langbort, S. S. Sastry, and T. Başar.  A three-stage colonel blotto game with applications to cyberphysical security.  In 2014 American Control Conference, pages 3820–3825, 2014.
  * [11] S. Hart.  Discrete Colonel Blotto and General Lotto games.  International Journal of Game Theory, 36(3-4):441–460, 2008.
  * [12] S. Hart.  Allocation games with caps: from captain lotto to all-pay auctions.  International Journal of Game Theory, 45(1):37–61, 2016.
  * [13] R. Isaacs.  Differential Games: A Mathematical Theory with Applications to Warfare and Pursuit, Control and Optimization.  Wiley, 1965.
  * [14] D. Kartik and A. Nayyar.  Upper and lower values in zero-sum stochastic games with asymmetric information.  Dynamic Games and Applications, 11(2):363–388, 2021.
  * [15] K. A. Konrad et al.  Strategy and dynamics in contests.  OUP Catalogue, 2009.
  * [16] D. Kovenock and B. Roberson.  Electoral poaching and party identification.  Journal of Theoretical Politics, 20(3):275–302, 2008.
  * [17] D. Kovenock and B. Roberson.  Coalitional colonel blotto games with application to the economics of alliances.  Journal of Public Economic Theory, 14(4):653–676, 2012.
  * [18] D. Kovenock and B. Roberson.  Generalizations of the General Lotto and Colonel Blotto games.  Economic Theory, pages 1–36, 2020.
  * [19] V. Leon and S. R. Etesami.  Bandit learning for dynamic colonel blotto game with a budget constraint.  In 2021 60th IEEE Conference on Decision and Control (CDC), pages 3818–3823, 2021.
  * [20] L. Li and J. S. Shamma.  Efficient strategy computation in zero-sum asymmetric information repeated games.  IEEE Transactions on Automatic Control, 65(7):2785–2800, 2020.
  * [21] A. Nayyar, A. Gupta, C. Langbort, and T. Başar.  Common information based markov perfect equilibria for stochastic games with asymmetric information: Finite games.  IEEE Transactions on Automatic Control, 59(3):555–570, 2013.
  * [22] K. Paarporn, R. Chandan, M. Alizadeh, and J. R. Marden.  A General Lotto game with asymmetric budget uncertainty.  arXiv preprint arXiv:2106.12133, 2021.
  * [23] K. Paarporn, R. Chandan, D. Kovenock, M. Alizadeh, and J. R. Marden.  Strategically revealing intentions in General Lotto games.  arXiv preprint arXiv:2110.12099, 2021.
  * [24] B. Roberson.  The Colonel Blotto game.  Economic Theory, 29(1):1–24, 2006.
  * [25] A. Sela.  Resource allocations in the best-of-k (k= 2, 3) contests.  Journal of Economics, pages 1–26, 2023.
  * [26] D. Shishika, Y. Guan, M. Dorothy, and V. Kumar.  Dynamic defender-attacker blotto game.  In 2022 American Control Conference (ACC), pages 4422–4428. IEEE, 2022.
  * [27] A. Von Moll, M. Pachter, D. Shishika, and Z. Fuchs.  Guarding a circular target by patrolling its perimeter.  In 2020 59th IEEE Conference on Decision and Control (CDC), pages 1658–1665, 2020.
  * [28] D. Q. Vu and P. Loiseau.  Colonel blotto games with favoritism: Competitions with pre-allocations and asymmetric effectiveness.  In Proceedings of the 22nd ACM Conference on Economics and Computation, page 862–863, 2021.
  * [29] D. Q. Vu, P. Loiseau, and A. Silva.  Combinatorial bandits for sequential learning in colonel blotto games.  In 2019 IEEE 58th Conference on Decision and Control (CDC), pages 867–872. IEEE, 2019.
  * [30] H. Yildirim.  Contests with multiple rounds.  Games and Economic Behavior, 51(1):213–227, 2005.
  * [31] C. Zhang and J. E. Ramirez-Marquez.  Protecting critical infrastructures against intentional attacks: A two-stage game with incomplete information.  IIE Transactions, 45(3):244–258, 2013.
  * [32] Q. Zhu and T. Başar.  Game-theoretic approach to feedback-driven multi-stage moving target defense.  In International conference on decision and game theory for security, pages 246–263. Springer, 2013.
  * [33] R. Zhuang, S. A. DeLoach, and X. Ou.  Towards a theory of moving target defense.  In Proceedings of the first ACM workshop on moving target defense, pages 31–40, 2014.

###  .1 Proof of Part 2-b

Here, we present the proof of Part 2-b from the proof outline of Theorem 3.1.
It states that: _Any pre-allocation $\boldsymbol{p}$ that corresponds to a
solution of (10) with $\mathcal{B}_{1},\mathcal{B}_{2}\neq\varnothing$
satisfies
$\pi_{A}(\boldsymbol{p},R_{A},R_{B})\leq\pi_{A}(\boldsymbol{p}^{*},R_{A},R_{B})$._

Throughout the proof, we will use the short-hand notation
$W_{j}=\sum_{b\in\mathcal{B}_{j}}w_{b}$,
$P_{j}=\sum_{b\in\mathcal{B}_{j}}p_{b}$ and
$\boldsymbol{p}_{j}=(p_{b})_{b\in\mathcal{B}_{j}}$, for $j=1,2$. For
$\boldsymbol{p}\in\Delta_{n}(P)$, we obtain the system of equations

$\displaystyle R_{A}$
$\displaystyle=\sum_{b\in\mathcal{B}_{1}}\frac{(w_{b}\kappa_{B}-p_{b})^{2}}{2w_{b}\kappa_{B}}+\sum_{b\in\mathcal{B}_{2}}\frac{(w_{b}\kappa_{A})^{2}}{2w_{b}\kappa_{B}},$
$\displaystyle R_{B}$
$\displaystyle=\sum_{b\in\mathcal{B}_{1}}\frac{(w_{b}\kappa_{B})^{2}-(p_{b})^{2}}{2w_{b}\kappa_{A}}+\sum_{b\in\mathcal{B}_{2}}\frac{(w_{b}\kappa_{A}+p_{b})^{2}-(p_{b})^{2}}{2w_{b}\kappa_{A}},$

where $0<w_{b}\kappa_{B}-p_{b}\leq w_{b}\kappa_{A}$ holds for all
$b\in\mathcal{B}_{1}$, and $w_{b}\kappa_{B}-p_{b}>w_{b}\kappa_{A}$ holds for
all $b\in\mathcal{B}_{2}$. The system of equations readily gives the
expression:

$\displaystyle W_{1}\kappa_{B}^{2}+W_{2}\kappa_{A}^{2}$
$\displaystyle=2\kappa_{B}(X_{A}+P_{2})-||\boldsymbol{p}_{1}||^{2}_{w}$ (58)
$\displaystyle=2\kappa_{A}(X_{B}-P_{2})+||\boldsymbol{p}_{1}||^{2}_{w},$

where recall that
$||\boldsymbol{p}_{1}||^{2}_{w}=\sum_{b\in\mathcal{B}_{2}}[(p_{b})^{2}/w_{b}]$.
The solution to the above system of equations is

$\displaystyle\kappa^{*}_{B}$
$\displaystyle=\frac{C_{1}H_{2}\pm\sqrt{C_{2}^{2}H_{1}H_{2}}}{W_{1}C_{2}^{2}+W_{2}C_{1}^{2}},$
(59) $\displaystyle\kappa^{*}_{A}$
$\displaystyle=\frac{C_{2}H_{1}\pm\sqrt{C_{1}^{2}H_{1}H_{2}}}{W_{1}C_{2}^{2}+W_{2}C_{1}^{2}},$

where we denote $C_{1}:=R_{A}+P_{1}$, $C_{2}:=R_{B}-P_{2}$,
$H_{1}:=C_{1}^{2}-W_{1}||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}$ and
$H_{2}:=C_{2}^{2}+W_{2}||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}$. We
consider only the scenario where $\pm=+$ in (59), since the expression for
$\kappa^{*}_{A}$ is strictly negative when $\pm=-$. Simply observe that
$C_{1}>0$, $C_{1}^{2}>H_{1}$, $0<C_{2}^{2}<H_{2}$ and, thus, that either (i)
$H_{1}>0$, $C_{2}>0$ and $0<C_{2}H_{1}<C_{1}\sqrt{H_{1}H_{2}}$, (ii)
$H_{1}<0$, $C_{2}<0$ and
$0<C_{2}H_{1}=|C_{2}||H_{1}|<C_{1}\sqrt{|H_{1}||H_{2}|}$, or (iii) only one of
$H_{1}$ or $C_{2}$ is negative, in which case $C_{2}H_{1}<0$.

Substituting (59) into (11) and simplifying, we obtain

$\displaystyle\pi_{A}(\boldsymbol{p},R_{A},R_{B})$
$\displaystyle=W_{1}+\frac{\sqrt{H_{1}H_{2}}-C_{1}C_{2}}{||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}},$
(60)

and the partial derivatives of $\pi_{A}(\boldsymbol{p},R_{A},R_{B})$ with
respect to $p_{b}$ are as follows. For $b\in\mathcal{B}_{1}$,

$\displaystyle\frac{\partial\pi_{A}}{\partial p_{b}}$
$\displaystyle=\frac{-p_{b}/w_{b}}{(||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}})^{2}\sqrt{H_{1}H_{2}}}(C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}})^{2}$
(61)
$\displaystyle\quad+\frac{1}{||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}\sqrt{H_{1}}}(C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}})$

and for $b\in\mathcal{B}_{2}$,

$\frac{\partial\pi_{A}}{\partial
p_{b}}=\frac{1}{||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}\sqrt{H_{2}}}(C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}}).$
(62)

We first consider critical points $\boldsymbol{p}$ strictly in the interior of
$\Delta_{n}(P)$, and resolve the points on the boundary later. One necessary
condition for a critical point is that $(\frac{\partial}{\partial
p_{b}}-\frac{\partial}{\partial p_{c}})\pi_{A}=0$ for all
$b\in\mathcal{B}_{1}$ and $c\in\mathcal{B}_{2}$. Firstly, observe that
$C_{1}>\sqrt{H_{1}}$ and $\sqrt{H_{2}}>C_{2}$, and, thus, it must be that
$C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}}>0$. We can thus divide the expression
$\frac{\partial\pi_{A}}{\partial p_{b}}=\frac{\partial\pi_{A}}{\partial
p_{c}}$ on both sides by $C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}}$ and rearrange
to obtain

$(p_{b}/w_{b})(C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}})=||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}(\sqrt{H_{2}}-\sqrt{H_{1}})>0.$

Observe that the left-hand side is strictly greater than zero, and, thus, the
right-hand side must be as well. This immediately requires
$\sqrt{H_{2}}-\sqrt{H_{1}}>0$, since
$||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}>0$. Re-arranging the above
expression, note that we also require

$\sqrt{H_{1}}[C_{2}(p_{b}/w_{b})-||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}]=\sqrt{H_{2}}[C_{1}(p_{b}/w_{b})-||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}].$

Since we have just shown that $\sqrt{H_{2}}>\sqrt{H_{1}}$ must hold, it
follows that each $b\in\mathcal{B}_{1}$ satisfies either (i)
$C_{2}(p_{b}/w_{b})-||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}<C_{1}(p_{b}/w_{b})-||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}<0$;
or (ii)
$C_{2}(p_{b}/w_{b})-||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}>C_{1}(p_{b}/w_{b})-||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}>0$.
Observe that $C_{1}(p_{b}/w_{b})>||\boldsymbol{p}_{1}||^{2}_{1}$ must hold for
$b^{\prime}\in\arg\,\max_{b\in\mathcal{B}_{1}}p_{b}/w_{b}$, and thus
$b^{\prime}$ must satisfy scenario (ii) and $C_{2}>C_{1}$ (or, equivalently,
$R_{B}-P>R_{A}$). This last inequality then implies that scenario (ii) must be
satisfied for all $b\in\mathcal{B}_{1}$.

We have shown that, in order for $(\frac{\partial}{\partial
p_{b}}-\frac{\partial}{\partial p_{c}})\pi_{A}=0$ to hold for all
$b\in\mathcal{B}_{1}$ and $c\in\mathcal{B}_{2}$, a critical point
$\boldsymbol{p}$ must satisfy

$\frac{p_{b}}{w_{b}}=\bar{p}:=\frac{\sqrt{H_{2}}-\sqrt{H_{1}}}{C_{1}\sqrt{H_{2}}-C_{2}\sqrt{H_{1}}}||\boldsymbol{p}_{1}||^{2}_{w},$

for each $b\in\mathcal{B}_{1}$. Expanding this expression, and solving for
$\bar{p}$ explicitly, we obtain the following two possible (real) solutions
for $\bar{p}$:

$\bar{p}=0\text{ or }\bar{p}=\frac{2(R_{B}-P)(qR_{B}-R_{A}-P)}{R_{A}},$

where we use $P_{1}=W_{1}\bar{p}$, $P_{2}=P-P_{1}$, and
$||\boldsymbol{p}_{1}||^{2}_{\boldsymbol{w}}=W_{1}\bar{p}^{2}$. As $\bar{p}=0$
is inadmissible, we consider the latter expression for $\bar{p}$. After
inserting this expression for $\bar{p}$ into the right-hand side of (19),
where $\max_{b}\\{p_{b}/w_{b}\\}=\bar{p}$, we obtain

$\frac{R_{A}}{2}\left[1+\sqrt{1+\frac{2W}{R_{A}}\bar{p}}\right]=\frac{R_{A}}{2}+R_{B}-P-\frac{R_{A}}{2}=R_{B}-P,$

which follows since we showed above that $R_{B}-P>R_{A}$ must hold. Thus, the
only critical point sits at the boundary of the region where all battlefields
are in Case 2, since decreasing $\bar{p}$ even slightly will satisfy the
condition in (19). We can further verify that the payoff at this critical
point is equal to the constant payoff in the region where all battlefields are
in Case 2, but omit this for conciseness.

We conclude the proof by resolving the scenario where $\boldsymbol{p}$ lies on
the boundaries of $\Delta_{n}(P)$. Observe that the conditions on
$q\kappa^{*}_{B}$ and $\kappa^{*}_{A}$ immediately imply that
$p_{b}/w_{b}>p_{c}/w_{c}$ for any $b\in\mathcal{B}_{1}$ and
$c\in\mathcal{B}_{2}$. Thus, on the boundaries of $\Delta_{n}(P)$, it must
either be that all battlefields with $p_{b}=0$ (and possibly more) are in Case
2, or that all battlefields in $\mathcal{B}$ are in Case 1 (which is covered
by Lemma 3.2, part 2-a).

In the scenario where all battlefields with $p_{b}=0$ are in Case 2, note that
the necessary condition $(\frac{\partial}{\partial
p_{i}}-\frac{\partial}{\partial p_{j}})\pi_{A}\geq 0$ for
$i\in\arg\,\min_{b\in\mathcal{B}_{1}}\\{p_{b}/w_{b}\\}$ and
$j\in\arg\,\max_{b\in\mathcal{B}_{1}}\\{p_{b}/w_{b}\\}$ only holds with
equality if $p_{b}/w_{b}=P_{1}/W_{1}$ for all $b\in\mathcal{B}_{1}$. If
$P_{1}/W_{1}<\bar{p}$, then the inequality in (19) is satisfied implying that
all battlefields are in Case 2, and Lemma 3.2 shows that $\boldsymbol{p}^{*}$
must correspond with the same payoff to player $A$. Otherwise, if
$P_{1}/W_{1}=\bar{p}$, then we showed above that the global maximum sits at
the boundary where all battlefields are in Case 2 and $\boldsymbol{p}^{*}$
achieves the same payoff.

Finally, if $P_{1}/W_{1}>\bar{p}$, then, from (61) and (62), we know that
$(\frac{\partial}{\partial p_{b}}-\frac{\partial}{\partial p_{c}})\pi_{A}<0$
must hold for all $b\in\mathcal{B}_{1}$ and $c\in\mathcal{B}_{2}$, since the
choice $p_{b}/w_{b}=\bar{p}$ satisfies $(\frac{\partial}{\partial
p_{b}}-\frac{\partial}{\partial p_{c}})\pi_{A}=0$, and
$\frac{\partial\pi_{A}}{\partial p_{b}}$ is decreasing with respect to
$p_{b}/w_{b}$ while $\frac{\partial\pi_{A}}{\partial p_{c}}$ is constant. This
violates a necessary condition for a critical point, and implies that $A$’s
payoff is increasing in the direction of decreasing $p_{b}$ and increasing
$p_{c}$, as expected.

###  .2 Proof of Lemma 5.4

We prove the result in two parts, considering the two separate intervals
$p_{B}\in[0,p_{A}]$ and $p_{B}\in(p_{A},\frac{M_{B}}{c_{B}}]$.

Part 1:  $p_{B}\in[0,p_{A}]$. Define the function

$G(p_{B})=\frac{2(M_{B}-p_{A}+p_{B}(1-c_{B}))^{2}}{2(M_{B}-p_{A}+p_{B}(1-c_{B}))+p_{A}-p_{B}}$
(63)

The definition of $\mathcal{R}^{1A}$ may be restated as

$\mathcal{R}^{1A}=\left\\{p_{B}<p_{0},\text{ or }p_{B}\geq p_{0}\text{ and
}G(p_{B})<R_{A}\right\\}.$ (64)

where $p_{0}=\frac{p_{A}-M_{B}}{1-c_{B}}$.

$\bullet$ Suppose $\frac{M_{B}}{c_{B}}\leq R_{A}+p_{A}$. Since $M_{B}\leq
c_{B}p_{A}$, we have $p_{0}>p_{A}$, and consequently
$u_{B}(\textbf{p})=u_{B}^{1A}(\textbf{p})$ $\forall p_{B}\in[0,p_{A}]$. The
sign of $\frac{\partial u_{B}^{1A}}{\partial p_{B}}$ is equivalent to the sign
of

$\displaystyle-f_{A}(p_{B})$
$\displaystyle\left[c_{B}(2p_{A}^{2}+R_{A}(2R_{A}-3p_{B}))\right.$ (65)
$\displaystyle+p_{A}(c_{B}(5R_{A}-2p_{B})-2M_{B})$
$\displaystyle+(f_{A}(p_{B})-R_{A})(c_{B}(3p_{A}+2R_{A}-p_{B})-2M_{B})$
$\displaystyle\left.-2M_{B}(R_{A}-p_{B})\right]$

where $f_{A}(p_{B})=R_{A}+\sqrt{R_{A}(R_{A}+2(p_{A}-p_{B}))}$. It then holds
that $\frac{\partial u_{B}^{1A}}{\partial p_{B}}(p_{A},p_{B})\leq 0$ for
$p_{B}\in[0,p_{A}]$ if and only if $\frac{M_{B}}{c_{B}}\leq R_{A}+p_{A}$. It
follows that $u_{B}$ is decreasing on the interval $p_{B}\in[0,p_{A}]$.

$\bullet$ Suppose
$R_{A}+p_{A}<\frac{M_{B}}{c_{B}}\leq\frac{R_{A}}{c_{B}}+p_{A}$. First we
consider the case that $\frac{R_{A}}{c_{B}}+p_{A}<\frac{p_{A}}{c_{B}}$. Here,
$p_{0}\in[0,p_{A})$ and thus we immediately have
$u_{B}(\textbf{p})=u_{B}^{1A}(\textbf{p})$ for $p_{B}\in[0,p_{0})$. We also
have $G(p_{0})=0$, $G(p_{B})$ is strictly increasing on
$p_{B}\in(p_{0},p_{A})$, and $G(p_{A})<R_{A}$ (from $M_{B}\leq
R_{A}+c_{B}pA$). Therefore, $G(p_{B})<R_{A}$ for all $p_{B}\in(p_{0},p_{A}]$,
and hence $u_{B}(\textbf{p})=u_{B}^{1A}(\textbf{p})$.

Now, we consider the case that
$\frac{R_{A}}{c_{B}}+p_{A}\geq\frac{p_{A}}{c_{B}}$. The range
$p_{A}<\frac{M_{B}}{c_{B}}\leq\frac{p_{A}}{c_{B}}$ follows the above argument
identically. Now, when
$\frac{p_{A}}{c_{B}}<\frac{M_{B}}{c_{B}}\leq\frac{R_{A}}{c_{B}}+p_{A}$, we
have $p_{0}<0$, $G(p_{B})$ is strictly increasing on $[0,p_{A}]$, and
$G(p_{A})<R_{A}$. Therefore, $G(p_{B})<R_{A}$ for all $p_{B}\in(p_{0},p_{A}]$,
and hence $u_{B}(\textbf{p})=u_{B}^{1A}(\textbf{p})$.

$\bullet$ Suppose
$\frac{R_{A}}{c_{B}}+p_{A}<\frac{M_{B}}{c_{B}}\leq\frac{1}{c_{B}}\left(p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2}\right)$.
First we consider the case $\frac{R_{A}}{c_{B}}+p_{A}<\frac{p_{A}}{c_{B}}$. In
the range
$\frac{R_{A}}{c_{B}}+p_{A}<\frac{M_{B}}{c_{B}}\leq\frac{p_{A}}{c_{B}}$, we
have $p_{0}\in[0,p_{A})$ and thus we immediately have
$u_{B}(\textbf{p})=u_{B}^{1A}(\textbf{p})$ for $p_{B}\in[0,p_{0})$. We also
have $G(p_{0})=0$ and $G(p_{B})$ is strictly increasing on
$p_{B}\in(p_{0},p_{A})$. Now, the condition
$\frac{R_{A}}{c_{B}}+p_{A}<\frac{M_{B}}{c_{B}}$ asserts that $G(p_{A})>R_{A}$.
Thus, there is a unique value, $p_{B}^{1A}\in(p_{0},p_{A})$ that solves
$G(p_{B}^{1A})=R_{A}$.

In the range
$\frac{p_{A}}{c_{B}}<\frac{M_{B}}{c_{B}}\leq\frac{1}{c_{B}}\left(p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2}\right)$,
$G(p_{B})$ is strictly increasing, $G(0)<R_{A}$, and $G(p_{A})>R_{A}$ which
yields the result. To see this, we first note that $p_{0}<0$. To show $G$ is
strictly increasing, the sign of $G^{\prime}(p_{B})$ is equal to the sign of
$(2c_{B}^{2}-3c_{B}+1)p_{B}+(3-2c_{B})M_{B}-p_{A}$ (from $M_{B}>p_{A}$).
Observe that this value is linear in $p_{B}$. At $p_{B}=0$, it is
$(3-2c_{B})M_{B}-p_{A}>0$, and at $p_{B}=p_{A}$, it is
$(3-2c_{B})(M_{B}-c_{B}p_{A})>0$. Thus, $G^{\prime}(p_{B})>0$ for
$p_{B}\in(0,p_{A})$. The condition
$G(0)=\frac{2(M_{B}-p_{A})^{2}}{2(M_{B}-p_{A})+p_{A}}<R_{A}$ is equivalent to
$M_{B}\leq\frac{1}{c_{B}}\left(p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2}\right)$.

The case $R_{A}+c_{B}p_{A}\geq p_{A}$ follows identical arguments to the
above.

$\bullet$ Suppose
$\frac{M_{B}}{c_{B}}>\frac{1}{c_{B}}\left(p_{A}+\frac{R_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}{2}\right)$.
Here, $p_{0}<0$, $G(p_{B})$ is strictly increasing on $(0,p_{A})$, and
$G(p_{A})<R_{A}$. This yields the result.

Part 2:  $p_{B}\in(p_{A},\frac{M_{B}}{c_{B}}]$. Define the functions

$\displaystyle F(p_{B})$
$\displaystyle:=\frac{2(t_{A}-p_{B})^{2}}{2R_{A}+p_{A}-p_{B}}$ (66)
$\displaystyle L(p_{B})$ $\displaystyle:=M_{B}-c_{B}p_{B}$

where $t_{A}:=R_{A}+p_{A}$. The definition of $\mathcal{R}^{1B}$ may be
restated as

$\mathcal{R}^{1B}=\\{p_{B}>t_{A},\text{ or }p_{B}\leq t_{A}\text{ and
}F(p_{B})\leq L(p_{B})\\}.$ (67)

$\bullet$ Suppose $\frac{M_{B}}{c_{B}}\leq t_{A}$. Here, $F(p_{B})$ is
strictly decreasing on $(p_{A},\frac{M_{B}}{c_{B}})$,
$F(p_{A})=R_{A}>L(p_{A})$, and
$F(\frac{M_{B}}{c_{B}})>L(\frac{M_{B}}{c_{B}}=0$. The solutions to the
equation $F(p_{B})=L(p_{B})$ are given by

$\displaystyle
r_{\pm}(M_{B}):=\frac{1}{2(2-c_{B})}\left[2t_{A}(2-c_{B})\\!-\\!L(p_{A})\right.$
(68)
$\displaystyle\left.\pm\sqrt{(2t_{A}(2-c_{B})-L(p_{A}))^{2}\\!-\\!4(2-c_{B})(2t_{A}(t_{A}\\!-\\!M_{B})+M_{B}p_{A})}\right]$

The roots are complex if and only if
$M_{B}<c_{B}p_{A}-2(2-c_{B})R_{A}+2R_{A}\sqrt{2(2-c_{B})}$, in which case
$u_{B}=u_{B}^{2B}$ for all $p_{B}\in[p_{A},\frac{M_{B}}{c_{B}}]$ (since
$F(p_{B})>L(p_{B})$).

So, we consider $c_{B}p_{A}-2(2-c_{B})R_{A}+2R_{A}\sqrt{2(2-c_{B})}\leq
M_{B}\leq c_{B}t_{A}$, in which the roots are real-valued, and only $r_{+}$
can be in the interval $(p_{A},\frac{M_{B}}{c_{B}})$. We thus have

$u_{B}(\textbf{p})=\begin{cases}u_{B}^{2B}(\textbf{p}),&p_{B}\in[p_{A},r_{-}]\\\
u_{B}^{1B}(\textbf{p}),&p_{B}\in(r_{-},r_{+}]\\\
u_{B}^{2B}(\textbf{p}),&p_{B}\in(r_{+},\frac{M_{B}}{c_{B}}]\end{cases}$ (69)

Both payoff functions $u_{B}^{2B}$ and $u_{B}^{1B}$ are decreasing in $p_{B}$
in their respective intervals. Indeed,

$\frac{\partial u_{B}^{2B}}{\partial
p_{B}}=\frac{L(p_{B})-c_{B}(t_{A}-p_{B})}{2(t_{A}-p_{B})^{2}}\leq 0$ (70)

where the inequality is due to $M_{B}\leq c_{B}t_{A}$.

To show $u_{B}^{1B}$ is decreasing, we observe that the payoff $u_{B}^{1B}$
admits a single critical point in the interval $(p_{A},\frac{M_{B}}{c_{B}})$.
Indeed, denoting $\ell(p_{B})=\sqrt{L(p_{B})}\geq 0$ and
$g(p_{B})=\sqrt{L(p_{B})+2(p_{B}-p_{A})}>0$, we can calculate

$\frac{\partial u_{B}^{1B}}{\partial
p_{B}}=-\frac{R_{A}}{2}\frac{(\ell+g)(c_{B}(\ell+g)-2\ell)}{\ell g(\ell
g+L(p_{B})+p_{B}-p_{A})^{2}}.$ (71)

It holds that $\frac{\partial u_{B}^{1B}}{\partial p_{B}}(p_{A})>0$,
$\frac{\partial u_{B}^{1B}}{\partial p_{B}}(\frac{M_{B}}{c_{B}})<0$, and
$\frac{\partial u_{B}^{1B}}{\partial p_{B}}(p_{B})=0$ if and only if

$p_{B}=\hat{p}_{B}:=\frac{M_{B}}{c_{B}}-\frac{M_{B}-c_{B}p_{A}}{2-c_{B}}\in(p_{A},\frac{M_{B}}{c_{B}}).$
(72)

Thus, $\hat{p}_{B}$ is a local maximizer, with $u_{B}^{1B}$ decreasing for
$p_{B}>\hat{p}_{B}$. Now, observe that $r_{-}(c_{B}t_{A})=\hat{p}_{B}$,
$r_{+}(c_{B}t_{A})=\frac{M_{B}}{c_{B}}$, $r_{-}(M_{B})$ is decreasing in
$M_{B}$, and $\hat{p}_{B}$ is increasing in $M_{B}$. We have $\hat{p}_{B}\leq
r_{-}$, and therefore $u_{B}(\textbf{p})$ is decreasing for all
$p_{B}\in[p_{A},\frac{M_{B}}{c_{B}}]$.

$\bullet$ Suppose $t_{A}<\frac{M_{B}}{c_{B}}\leq\frac{R_{A}}{c_{B}}+p_{A}$.
Here, $u_{B}=u_{B}^{1B}$ for $p_{B}\in(t_{A},\frac{M_{B}}{c_{B}})$. The
function $F(p_{B})$ is strictly decreasing on $(p_{A},t_{A})$ with
$F(p_{A})=R_{A}\geq L(p_{A})$ and $F(t_{A})=0$. Thus, we have
$F(p_{B})>L(p_{B})$ for $p_{B}\in[p_{A},p_{B}^{1B}]$ and $F(p_{B})\leq
L(p_{B})$ for $p_{B}\in[p_{B}^{1B},t_{A}]$, where $p_{B}^{1B}$ is the unique
value in the interval $[p_{A},t_{A})$ that satisfies $F(p_{B})=L(p_{B})$. We
thus get the stated characterization for $u_{B}(\textbf{p})$.

$\bullet$ Suppose $\frac{R_{A}}{c_{B}}+p_{A}<M_{B}$. Here, $u_{B}=u_{B}^{1B}$
for $p_{B}\in(t_{A},\frac{M_{B}}{c_{B}})$. The function $F(p_{B})$ is strictly
decreasing on $(p_{A},t_{A})$ with $F(p_{A})=R_{A}<M_{B}-c_{B}p_{A}$ and
$F(t_{A})=0$. Thus, we have $F(p_{B})<M_{B}-c_{B}p_{B}$ for
$p_{B}\in[p_{A},t_{A}]$. Therefore, $u_{B}=u_{B}^{1B}$ for all
$p_{B}\in[p_{A},\frac{M_{B}}{c_{B}}]$.

From identical arguments from the first bullet point ($\frac{M_{B}}{c_{B}}\leq
t_{A}$), we know that $u_{B}^{1B}$ admits the local maximizer
$\hat{p}_{B}\in(p_{A},\frac{M_{B}}{c_{B}})$. Combining the characterizations
from Part 1 and Part 2 yields the result.

###  .3 Proof of Lemma 5.5

We first address the two extreme lower and upper intervals. We will denote
$f_{A}(p_{B})=R_{A}+\sqrt{R_{A}(R_{A}+2(p_{A}-p_{B}))}$.

  * •

Suppose $\frac{M_{B}}{c_{B}}\leq R_{A}+p_{A}$. By Lemma 5.5
$u_{B}(p_{A},p_{B})$ is decreasing for all $p_{B}\in[0,\frac{M_{B}}{c_{B}}]$,
and hence $\max_{p_{B}\in[0,\frac{M_{B}}{c_{B}}]}u_{B}(p_{A},p_{B})=u_{B}(0)$.

  * •

Suppose $p_{A}+\frac{f_{A}(0)}{2}<\frac{M_{B}}{c_{B}}$. Here, $u_{B}$ is
strictly increasing on $p_{B}\in[0,p_{A})$, and the maximum value in the
interval $p_{B}\in(p_{A},\frac{M_{B}}{c_{B}}]$ is given by
$u_{B}^{1B}(\hat{p}_{B})$. Therefore,

$\max_{p_{B}\in[0,\frac{M_{B}}{c_{B}}]}u_{B}(p_{B})=u_{B}^{1B}(\hat{p}_{B}).$
(73)

  * •

Now, consider the middle interval $R_{A}+p_{A}<\frac{M_{B}}{c_{B}}\leq
p_{A}+\frac{f_{A}(0)}{2}$. By Lemma 5.4, $u_{B}(0)=u_{B}^{1A}(0)$. Here,
$u_{B}$ is decreasing at $p_{B}=0$ but is increasing at $p_{B}=p_{A}$ (whether
in region $\mathcal{R}^{1A}$ or $\mathcal{R}^{2A}$). By Lemma 5.4, the maximum
value in the interval $p_{B}\in(p_{A},\frac{M_{B}}{c_{B}}]$ is given by
$u_{B}^{1B}(\hat{p}_{B})$. Therefore,

$\max_{p_{B}\in[0,\frac{M_{B}}{c_{B}}]}u_{B}(p_{B})=\max\\{u_{B}^{1A}(0),u_{B}^{1B}(\hat{p}_{B})\\}.$
(74)

We can identify the existence of a threshold value
$h(R_{A},p_{A})\in(R_{A}+p_{A},p_{A}+\frac{f_{A}(0)}{2})$ for which
$p_{B}^{*}=0$ for $\frac{M_{B}}{c_{B}}<h(p_{A})$, $p_{B}^{*}=\hat{p}_{B}$ for
$\frac{M_{B}}{c_{B}}>h(p_{A})$, and $p_{B}^{*}=\hat{p}_{B}$ or $0$ for
$\frac{M_{B}}{c_{B}}=h(p_{A})$. Indeed the payoff $u_{B}^{1B}(\hat{p}_{B})$
may be written as

$u_{B}^{1B}(\hat{p}_{B})=1-\frac{c_{B}(2-c_{B})}{2}\frac{R_{A}}{M_{B}-c_{B}p_{A}}$
(75)

The payoff $u_{B}^{1A}(0)$ is given by

$u_{B}^{1A}(0)=\frac{M_{B}}{2}\left(\frac{\sqrt{R_{A}}+\sqrt{R_{A}+2p_{A}}}{R_{A}+p_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}\right)^{2}$
(76)

Denoting $K_{1}=\frac{c_{B}(2-c_{B})}{2}$ and
$K_{2}=\left(\frac{\sqrt{R_{A}}+\sqrt{R_{A}+2p_{A}}}{R_{A}+p_{A}+\sqrt{R_{A}(R_{A}+2p_{A})}}\right)^{2}$,
we have $u_{B}^{1B}(\hat{p}_{B})=u_{B}^{1A}(0)$ precisely at

$\displaystyle\frac{M_{B}}{c_{B}}=\frac{1}{c_{B}K_{2}}\left[\frac{K_{2}}{2}c_{B}p_{A}+1\right.$
(77)
$\displaystyle\quad\quad\left.-\sqrt{(\frac{K_{2}}{2}c_{B}p_{A}+1)^{2}-2K_{2}(c_{B}p_{A}+K_{1}R_{A})}\right]$

Denoting $h(R_{A},p_{A})$ as the right-hand side above, we obtain the result.

[] Keith Paarporn is an Assistant Professor in the Department of Computer
Science at the University of Colorado, Colorado Springs. He received a B.S. in
Electrical Engineering from the University of Maryland, College Park in 2013,
an M.S. in Electrical and Computer Engineering from the Georgia Institute of
Technology in 2016, and a Ph.D. in Electrical and Computer Engineering from
the Georgia Institute of Technology in 2018. From 2018 to 2022, he was a
postdoctoral scholar in the Electrical and Computer Engineering Department at
the University of California, Santa Barbara. His research interests include
game theory, control theory, and their applications to multi-agent systems and
security.

[] Rahul Chandan is a Research Scientist at Amazon Robotics. He received a
B.A.Sc. in Engineering Science from the University of Toronto in 2017, and an
M.S. and PhD in Electrical and Computer Engineering from the University of
California, Santa Barbara in 2019 and 2022, respectively. His research
interests include game theory, multi-agent systems, optimization, and
economics and computation.

[] Mahnoosh Alizadeh is an Associate Professor of Electrical and Computer
Engineering at the University of California Santa Barbara. She received the
B.Sc. degree (’09) in Electrical Engineering from Sharif University of
Technology and the M.Sc. (’13) and Ph.D. (’14) degrees in Electrical and
Computer Engineering from the University of California Davis, where she was
the recipient of the Richard C. Dorf award for outstanding research
accomplishment. From 2014 to 2016, she was a postdoctoral scholar at Stanford
University. Her research interests are focused on designing scalable control,
learning, and market mechanisms to promote efficiency and resiliency in
societal-scale cyber-physical systems. Dr. Alizadeh is a recipient of the NSF
CAREER award and the best paper award from HICSS-53 power systems track.

[] Jason Marden is a Professor in the Department of Electrical and Computer
Engineering at the University of California, Santa Barbara. Jason received a
BS in Mechanical Engineering in 2001 from UCLA, and a PhD in Mechanical
Engineering in 2007, also from UCLA, under the supervision of Jeff S. Shamma,
where he was awarded the Outstanding Graduating PhD Student in Mechanical
Engineering. After graduating from UCLA, he served as a junior fellow in the
Social and Information Sciences Laboratory at the California Institute of
Technology until 2010 when he joined the University of Colorado. Jason is a
recipient of the NSF Career Award (2014), the ONR Young Investigator Award
(2015), the AFOSR Young Investigator Award (2012), the American Automatic
Control Council Donald P. Eckman Award (2012), and the SIAG/CST Best SICON
Paper Prize (2015). Jason’s research interests focus on game theoretic methods
for the control of distributed multiagent systems.