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\begin{document}

\title{An Algorithmic Analysis of Parallel Contests}
\date{\today }
\author{Chen Cohen\thanks{%
Department of Public Policy and Administration, Guilford Glazer Faculty of
Business and Management, Ben-Gurion University of the Negev, 84105 Beer
Sheva, Israel. Chen chencohe@post.bgu.ac.il}, Ishay Rabi\thanks{%
Department of Economics, Ben-Gurion University of the Negev, 84105 Beer
Sheva, Israel. ishay.rabi@gmail.com}, Aner Sela\thanks{%
Department of Economics, Ben-Gurion University of the Negev, 84105 Beer
Sheva, Israel. anersela@bgu.ac.il}}
\maketitle

\begin{abstract}
We consider two-stage parallel contests with a finite set of agents, as well
as a finite set of heterogeneous contests with varying winning prizes. In
the first stage, each agent simultaneously chooses in which contest he wants
to compete, and in the second stage, the agents who chose the same contest
compete against each other in a contest for a single prize. We first assume
a general Tullock contest success function with nonlinear cost functions,
and then we present an algorithm that converges to equilibrium, namely, the
algorithm organizes the allocation of homogeneous or heterogeneous agents of
two types, each with a different cost of effort, among the contests until no
agent wishes to change his current contest.\ Later, we assume the Tullock
contest success function with linear cost functions, and we show that for
homogeneous agents, our model and a potential game are equivalent, allowing
us to explicitly characterize the subgame perfect equilibrium with pure
strategies.

\emph{JEL classification:} D44, D72, D82, J31
\end{abstract}

\newpage

\section{Introduction}

In our world, agents frequently encounter multiple parallel contests in
which they can only participate in one, and they must then decide which
contest to enter. In that case, each agent who needs to choose one contest
may confront the famous dilemma that was attributed to Julius Caesar: "I had
rather be the first in this village than the second in Rome" (see Damiano et
al. 2010). For example, researchers from the same country or even from
several countries compete to win one of several competitive grants with the
constraint that each researcher cannot submit his proposal to more than one
competitive grant. If a researcher submits his proposal to one of the highly
competitive grants, he will have to compete against other excellent
researchers, and therefore his chance to win the grant will not be very
high, but if he wins the value of the prize is high particularly because of
the reputation of winning that grant. On the other hand, if he chooses to
submit his proposal to an ordinary competitive grant, the reputation of
winning this grant will be lower, but the competition with his opponents
will be easier such that his chance to win will be higher. This is a classic
problem of parallel contests where the prizes (grants) are different among
the contests, and the heterogeneous agents (researchers) have to choose in
which one they want to compete.

There are many other real-life situations that can be modeled as parallel
contests. For example, consider a professional tennis player who has to
choose which tournament to compete in when several of them are taking place
simultaneously. A tennis player who is ranked in the top 50 can choose to
compete in almost every tournament. However, if he has to choose between an
ATP-1000 tournament in which the prizes are relatively large and an ATP-250
tournament in which the prizes are smaller, he knows that in the ATP-1000
tournament he is going to match stronger tennis players than in the ATP-250
tournament such that his chance to win the ATP-1000 tournament is smaller
than to win the ATP-250 tournament. Thus, in parallel contests an agent
should consider his expected payoff in each of them, and since it is not
possible to compete in more than one contest, he should choose the one in
which he has the highest expected payoff. Since an agent's expected payoff
is based on the reward in each contest as well as his chance to win it, the
choice may not be that simple. Thus, the relevant question is if there
exists an equilibrium in parallel contests at all, and if so, how to find
it? In this paper, we develop a model of parallel contests to provide
answers to these questions.

Specifically, we consider a two-stage model of parallel contests, each of
which has a single prize. In the first stage, the agents simultaneously
choose the contest in which they want to compete with the condition that
each agent can compete in at most one contest. In the second stage, after
the agents choose their contests, the agents in each contest compete against
each other, and the winner of each contest wins a prize where all the other
agents bear the costs of their efforts.

We first consider parallel contests each of which has a generalized Tullock
contest success function (see Yamazaki 2008). There are $n$ heterogeneous
agents who have two possible types, each with a different cost of effort.
The costs of effort are non-linear, increasing and weakly convex in effort.
We also assume some ranking including the agents' costs of effort that
clarify the relative strength of the agents of different types. In that
case, we find the subgame perfect equilibrium even if the effort costs are
not explicitly specified, making it impossible to calculate the agents'
expected payoffs from the parallel contests. To overcome this obstacle, we
use revealed preference techniques to demonstrate the existence of a subgame
perfect equilibrium with pure strategies and provide an algorithm that
enables us to calculate this equilibrium, namely, the allocation of agents
across the contests. The algorithm is based on finite sequential steps each
of which includes an instruction that enables some types of agents to move
across the contests. One might think that these steps can repeat in an
infinite loop. However, the algorithm reaches a point according to which no
agent wants to change his position, and then we say that the algorithm
converges to equilibrium and our parallel contests have a subgame perfect
equilibrium with pure strategies. Our algorithms not only find a pure
strategy equilibrium, which is impossible to calculate because not all of
the parameters (effort costs) are explicitly specified, but they also
present dynamic processes that simulate agent behavior, in which agents
change their strategies until they reach the equilibrium point. These
algorithms show how quickly agents learn to use the equilibrium strategies,
demonstrating the equilibrium's reliability as a predictor of agent behavior
in our parallel contests. We first show that every model of parallel
contests with two contests and two types of agents has a subgame perfect
equilibrium with pure strategies, and then we generalize this result to
parallel contests with any finite number of contests.

Later, we consider parallel contests, each with a Tullock contest success
function (Tullock 1980). When the agents are homogeneous and have linear
effort costs, we show that our parallel contest model is a potential game,
which is well known to have a pure-strategy equilibrium (Rosenthal 1973,
Monderer and Shapley 1996). A game is considered a potential game if the
incentive for all agents to change their strategy can be expressed using a
single global function known as the potential function. As a result, a
potential game is a simple game with a pure strategy equilibrium in which
all agents have the same interests, and by demonstrating the equivalence of
the first stage of our parallel contest and a potential game, we demonstrate
that the analysis of our parallel contest model with linear cost of efforts
is as simple as that of potential games. When the agents are heterogeneous,
it is unclear whether our model is comparable to a potential game, and more
specifically, whether our model has a subgame perfect equilibrium with pure
strategies. However, we demonstrate that if such a subgame perfect
equilibrium exists, it has the counterintuitive property of equilibrium,
namely that agents of the same type in different contests with different
prizes never have the same expected payoff.

The main findings of this paper are that in our two-stage parallel contests
with heterogeneous agents of two types (cost of effort), whether agents have
linear or nonlinear cost functions, and regardless of the number of contests
or the prize values in these contests, there is always subgame perfect
equilibrium in pure strategies. Furthermore, we provide algorithms for
calculating these equilibrium points, specifically the agent distribution
across contests, so that no agent wants to switch to another contest. Our
model also has a subgame-perfect equilibrium in mixed strategies, but we are
interested in the presence of a subgame-perfect equilibrium in pure
strategies. We believe that there is a distinction between pure strategy
equilibrium and mixed strategy equilibrium in our two-stage parallel
contests. The reason is that if agents use mixed strategies in the first
stage to decide which contest to compete in, and if at the start of the
second stage, after seeing how all agents are allocated among the contests,
some of them will have an incentive to change their decisions, while agents
who choose pure strategies in the first stage, when they observe the
allocation of agents among the contests, will not regret their decisions
and, in particular, do not want to change their location at the start of the
second stage. As a result, a subgame-perfect equilibrium with pure
strategies in our model's first stage is a more stable solution, making it
more realistic and valuable.

\subsection{Related literature}

We are not the first to deal with a model of parallel contests, although
thus far they have received quite limited attention despite their widespread
presence in our lives. The majority of research on parallel contests comes
from fields other than traditional economics, such as operation research and
computer science, and this is most likely due to the complexity of
equilibrium calculations in these models.\footnote{%
Segev (2020) provides a survey of crowdsourcing contests, including parallel
contests.} Korpeo\u{g}lu et al. (2022) study parallel innovation contests,
where agents can compete and improve their solutions for each. Each contest
is won by the agent with the highest output, which is a noisy function of
the agent's effort. They look into the circumstances in which the organizers
benefit from an agent competing in multiple contests. Deng et al. (2023)
study parallel contests and assume that agents using mixed strategies can
participate in any number of contests, while Chan et al. (2020) study
parallel contests and assume that agents in their model can choose their
contests using pure strategies; however, like the previous papers, in their
model, each agent can choose more than one contest to compete in rather than
being limited to one contest as in ours.

The assumption in these models that an agent can participate in multiple
contests simplifies the analysis, as opposed to our assumption that each
agent can only participate in one contest. The reason for this is that if an
agent can participate in multiple contests, his decision to participate in
any one of them is solely determined by the agents who chose to participate
in the same contest as him. On the other hand, when an agent can only
participate in one contest, his decision to do so is determined by the
distribution of agents across all contests, i.e., by the choices of the
other agents, including those who do not participate in the contest in which
he chooses to compete.

Deng et al. (2022) investigate the parallel contest model and, like us,
assume that each contestant participates in only one contest. However, in
their model, the agents only choose which contest to compete in and are not
involved in the subsequent stages; in particular, they make no costly
efforts. Moreover, in their model, the agents select their contests using
mixed strategies, whereas we focus on equilibrium using pure strategies. It
is also worth noting that Deng et al. (2022) and the other models mentioned
above assume either linear or no costs of effort, whereas in our model, the
costs of effort are not necessarily linear and can be any type of weakly
convex cost of effort.

There are also several works on parallel contests in the economics theory
literature; for example, Azmat and M\"{o}ller (2009) investigate two
parallel Tullock contests with homogeneous agents that have linear costs of
effort where each agent chooses a contest and in each contest, there are
several heterogenous prizes. Konrad and Kovenock (2012) study a model of
parallel all-pay contests with homogeneous agents who have linear costs of
effort where each agent first chooses a contest and then in each contest the
agents compete in an all-pay contest for several identical prizes. Juang et
al. (2020) also study the same model with two parallel all-pay contests with
homogenous agents who have linear costs of effort. However, in contrast to
Konrad and Kovenock they assume that the number of identical prizes in each
contest is not the same. They also assume that each agent chooses his
contest when he does not necessarily know the number of his opponents in
that contest. Azmat and M\"{o}ller (2018) study parallel equilibrium
contests with continuum heterogeneous agents who first choose the contest
and then compete in all-pay contests with several identical prizes. Morgan
et al. (2018) analyze two parallel contests with a continuum of heterogenous
agents where each agent's performance is deterministic but noisily measured.
In all the above models, the authors usually assume that agents are
homogenous with a linear cost of effort, and those who consider
heterogeneous agents also assume a continuum of agents. Our current work is
unique in that we are the first in the economics literature to deal with a
finite number of heterogeneous agents. When there is a continuum of agents,
the movement of a single agent between two contests has no effect on the
expected payoff of the other agents, but the opposite is true when there are
a finite number of agents. Moreover, while in the above papers it is assumed
that the agents have linear costs of effort, we also assume that the agents'
costs of effort are nonlinear and, moreover, are not even explicitly given
such that the agents' payoffs cannot be calculated.

In our model of two-stage parallel contests, the agents face several prizes,
each of which is in a different contest, and they have to choose the prize
they want to compete for. In most models of contests with several prizes
either under complete or incomplete information (see, among others, Clark
and Riis 1996, 1998, Moldovanu and Sela 2001, 2006, Archak and Sundararajan
2009, Moldovanu et al. 2012, Ghosh and Hummel 2012, Fu et al. 2014, Ghosh
and Kleinberg 2016, and Sela 2020) it is assumed that the prizes are
identical to all the agents, and if they are not identical, then the ratio
of the values for every pair of prizes is the same for all the agents who
differ from each other by their ability or, alternatively, their marginal
cost of effort. In contrast, in our model when we assume non-linear costs of
effort, the agents of the different types also have different ratios of
prizes over different contests. In other words, the heterogeneity of the
prizes between the two types of agents in our model with non-linear costs of
effort is unrestricted.

In the second stage of our model, in each contest, the agents compete in a
Tullock contest (see, among others, Baye and Hoppe 2003, Corchon and Dahm
2010, and Ewerhart 2015). The convergence of our algorithm to the
equilibrium in the case of non-linear costs of effort is based on the
uniqueness of the equilibrium in one-stage Tullock contests (for the
existence and uniqueness of equilibrium in contests with complete
information, see Szidarovszky and Okuguchi 1997 and Yamazaki 2008; for
contests with incomplete information, see Einy et al. 2015, 2020 and
Ewerhart and Quartieri 2020). The reason is that we assume that for any
change in the allocation of agents that is solved by the algorithm, there is
a unique equilibrium in every contest in the second stage. Nonetheless, in
the first stage, we demonstrate that the equilibrium allocation of agents is
not necessarily unique; consequently, the equilibrium in our model of
parallel contests with homogeneous agents and linear effort costs is not
necessarily unique.

When agents are homogeneous and have linear cost functions, we show that the
first stage of our two-stage parallel contests corresponds to a potential
game (see, for example, Monderer and Shapley 1996; Ui 2001, Gopalakrishnan
et al. 2014) and, more specifically, a congestion game (see, for example,
Milchtaich 1996; Holzman and Law-Yone 1997, Harks and Klimm 2012). In this
case, we explicitly specify the potential function in our model. When the
agents are heterogeneous and have linear cost functions, we cannot
demonstrate this equivalence with potential games.

Our paper focuses on determining a pure Nash equilibrium, which is an
important concept in economic theory for predicting agent behavior in a
variety of situations. Nash equilibrium strategies may be obtained after
agents engage in adaptive processes such as learning, imitation, or
evolution (Kandori et al. 1993; Roth and Erev, 1995; Fudenberg and Levine,
1998). Similarly, we not only show the existence of equilibrium in pure
strategies and develop algorithms to compute it, but we also describe
dynamic processes by which fully rational agents learn to play the pure
strategy Nash equilibrium.

The rest of the paper is organized as follows: In Section 2, we describe the
model of parallel contests. In Section 3, we analyze the subgame perfect
equilibrium when the agents have non-linear effort cost functions. In
subsection 3.1, we assume that there are two contests while in subsection
3.2 we generalize the analysis for any finite number of contests. In Section
4, we analyze the subgame perfect equilibrium in this model where agents
have linear costs of effort. In subsection 4.1, we assume that the agents
are homogeneous while in subsection 4.2 we assume that the agents are
heterogeneous. Section 5 concludes. The proofs appear in the Appendix.

\section{Two-stage parallel contests}

Consider a set $N=\{1,2,3,...,n\}$ of $n$ heterogeneous agents, and a set $%
M=\{1,2,3,...,m\}$ of $m$, $m<n$, heterogeneous contests. In the first stage
each agent chooses to compete in one of the $m$ contests where each contest $%
k\in M$ \ has a single prize $v_{k}$, and, without loss of generality, we
assume that $v_{k}\geq v_{k+1},$ $k=1,...,m-1$. The set of agents that
choose to compete in contest $k\in M$ will be defined as $N_{k}$, where $%
n_{k}$ is the number of agents in that contest. In the second stage, after
the agents were allocated across the contests, the agents observe the other
agents who chose the same contest, and then the agents in each contest
compete against each other to win the prize.

The agents in contest $k\in M$ simultaneously exert their efforts. Let $%
x_{i} $ denote the effort of agent $i\in N_{k}$. Then, each agent $i$'s
probability to win in contest $k$ is $p_{i}=\frac{f_{i}(x_{i})}{\sum_{j\in
N_{k}}f_{j}(x_{j})}$ where $f_{i}(x_{i})$, which be called the production
function, is assumed to be zero at $x_{i}=0$, continuous, strictly
increasing, and weakly concave in $x_{i}$ for all $i\in N$. Each agent $i\in
N$ bears the cost of his effort whether he wins or not, where his effort
cost function $c_{i}(x):\mathbb{R}_{+}\rightarrow \mathbb{R}_{+}$ is
strictly increasing and weakly convex in the effort $x$. The expected payoff
of agent $i\in N_{k}$ is: 
\begin{equation}
v_{k}\frac{f_{i}(x_{i})}{\sum_{j\in N_{k}}f_{j}(x_{j})}-c_{i}(x_{i}).
\label{eq:expected-payoff}
\end{equation}%
Let $y_{j}=f_{j}(x_{j})$ for all $j\in N_{k}$. Then by (\ref%
{eq:expected-payoff}), we obtain that 
\begin{equation}
v_{k}\frac{y_{i}}{\sum_{j\in N_{k}}y_{j}}-\hat{c}_{i}(y_{i}),
\label{eq:transformed-expected-payoff}
\end{equation}%
where $\widehat{c}_{i}=f_{i}^{-1}\circ c_{i}$.\footnote{%
This transformation was proposed by Szidarovszky and Okuguchi (1997), and
Yamazaki (2008).} The assumptions on $f_{i}$ and $c_{i}$ imply that $%
\widehat{c}_{i}$ is a well-defined function with $\widehat{c}_{i}(0)=0$ and
that $\widehat{c}$ is strictly increasing and weakly convex with respect to $%
y_{i}$. The initial maximization problem (\ref{eq:expected-payoff}) with
respect to $x_{i}$ is equivalent to the maximization of (\ref%
{eq:transformed-expected-payoff}) with respect to $y_{i}$.

We assume that the agents' costs of effort can be ranked such that for all $%
x\geq 0,$ $\widehat{c}_{i}(x)\leq \widehat{c}_{i+1}(x)$, $i=1,...,n-1$. This
competition will be referred to as the model of two-stage parallel contests.%
\footnote{%
We can assume here a much more general contest success function than the one
given by (\ref{eq:expected-payoff}). However, we also need the uniqueness of
the equilibrium in the second stage in each contest, and for this reason, we
consider the contest success function given by (\ref{eq:expected-payoff}).}

\section{An algorithmic analysis of parallel contests}

For this section, we assume for simplicity that there are two types of
agents, strong ($h$) and weak ($l$), such that the costs of effort satisfy $%
\widehat{c}_{h}(x)\leq \widehat{c}_{l}(x)$ for all $x\geq 0$. By this
assumption, we know that in any competition in the second stage with strong
and weak agents there is a unique equilibrium (see Szidarovszky and Okuguchi
1997 and Yamazaki 2008) in which the strong agents exert higher efforts and
have higher expected payoffs than the weak agents. We will use the property
that the equilibrium in every contest is unique because it implies the
intuitive result that if any agent enters another contest, the expected
payoffs of all the other agents in that contest decrease while the expected
payoffs of all the other agents in the contest that he just left increase.
Likewise, the uniqueness of the equilibrium in every contest implies that if
an agent leaves and then returns to the same contest and no other agent left
or joined that contest, his expected payoff will not change.\footnote{%
For example, (Nti 1997) and Cornes and Hartley (2005) discuss and establish
changes in agents' efforts and payoffs as a result of other agents entering
or exiting the contest.} In the subgame perfect equilibrium of the model of
two-stage parallel contests agents do not want to move to another contest.
However, because we allow a general form of non-linear costs of effort, we
cannot prove that the equilibrium exists by comparing agents' expected
payoffs. Instead, we use revealed preference techniques to demonstrate
unequivocally that our model has a subgame perfect equilibrium where in the
first stage the agents choose their contests by pure strategies. In the next
section, we will begin by analyzing the case of two parallel contests, and
then we will generalize our result to any finite number of parallel
contests. The results for the case of two parallel contests are included in
the general case of any finite number of contests. However, because the
equilibrium analysis in the case of two parallel contests is slightly
simpler than the equilibrium analysis in the general case, we find it easier
to explain the equilibrium analysis of the two parallel contests first,
followed by the equilibrium analysis of any number of parallel contests.

\subsection{Two parallel contests}

Consider a model of two-stage parallel contests with two contests $1$ and $2$%
, and two prizes $v_{1},v_{2}$ respectively, where $v_{1}\geq v_{2}$ (see
Figure \ref{fig:figure1}).\ For any allocation of agents between the two
contests, each agent's expected payoff is well defined by the uniqueness of
the equilibrium of the competition in each contest. Thus, we can assume that
if all of the agents remain in their current positions but one agent moves
from contest $i$ to contest $j$, his expected payoff in contest $j$ is
greater than his expected payoff in contest $i$. The first result describes
the response of $h$-type agents to a move of $l$-type agents between the two
contests.

\begin{lemma}
\label{le:movement-2} Consider a model of two-stage parallel contests with
two contests and with two types of agents $h$ and $l$ where $\widehat{c}%
_{h}(x)\leq \widehat{c}_{l}(x)$ for all $x\geq 0$. If for some allocation of
agents between the two contests there is no $h$-type agent who wants to move
between the two contests, but there is one $l$-type agent who moves from
contest $i$ to contest $j,i\neq j,i,j\in \{1,2\}$. Then, in response to the $%
l$-type move, no more than one $h$-type agent will want to move in the
opposite direction from contest $j$ to contest $i$.
\end{lemma}

Lemma \ref{le:movement-2} implies that the number of moves between the two
contests is not increasing in time such that in response to each move of one 
$l$-type agent there will be at most a move of one $h$-type agent.
Accordingly, we construct an algorithm based on finite sequential steps each
of which includes an instruction that enables a type of agents to move
across the contests. In this algorithm, we allow one $l$-type agent to move
and then we allow all the $h$-type agent to move. By Lemma \ref%
{le:movement-2}, at most one $h$-type agent will move in the opposite
direction of the $l$-type agents. One might think that these steps can
repeat in an infinite loop and never reach a stable state. However, the
algorithm does reach a point according to which the agents do not react,
namely, they do not want to change their positions and the algorithm stops.\
Then, we say that the algorithm converges to equilibrium and our model of
two-stage parallel contests has a subgame perfect equilibrium with pure
strategies in both stages. To find the equilibrium, consider the following
algorithm:

\begin{algorithm}
\label{alg:two-types-two-contests} The algorithm has the following four
steps:

\begin{tabular}{ll}
Step 1 & Allocate the $h$-type agents in contest 1 and the $l$-type agents
in contest 2. \\ 
Step 2 & $H$-type agents have the option to move between the contests.%
\QQfnmark{%
In this step, the $h$-type agents may move in any order, either
simultaneously or sequentially, and there are no limits on the number of
possible movements for each agent. This step is complete when no one desires
to move. {}} \\ 
Step 3 & A single $l$-type agent has the option to move between the contests.
\\ 
Step 4 & Repeat steps 2-3 until in step 3 no agent wants to move.%
\end{tabular}%
\QQfntext{0}{
In this step, the $h$-type agents may move in any order, either
simultaneously or sequentially, and there are no limits on the number of
possible movements for each agent. This step is complete when no one desires
to move. {}}
\end{algorithm}

The following result shows that the above algorithm always converges to
equilibrium.

\begin{proposition}
\label{prop:alg-1-equilibrium} Algorithm \ref{alg:two-types-two-contests}
always converges to equilibrium and therefore a two-stage parallel contests
with two contests and two types of agents $h$ and $l$ where $\widehat{c}%
_{h}(x)\leq \widehat{c}_{l}(x)$ for all $x\geq 0$ has a subgame perfect
equilibrium in which agents choose pure strategies in both stages.
\end{proposition}

The result of Proposition \ref{prop:alg-1-equilibrium} is not intuitive
because, as shown in the proof, strong agents (h-type agents) move from
contest 1 with the larger prize to contest 2 with the smaller prize and
never return to contest 1.

\begin{figure}[h]
\centering
\includegraphics[scale=0.3]{version 9/Figure1}
\caption[fe:alg-1-step-1]{An illustration of the two contests at the end of
Step 1 of Algorithm \protect\ref{alg:two-types-two-contests}}
\label{fig:figure1}
\end{figure}

Furthermore, the expected payoffs of the strong agents in contest 2 will be
higher than the strong agents' expected payoffs in contest 1. The opposite
holds for the weak agents ($l$-type agents) who move from contest 2 to
contest 1 and never go back to contest 2, namely, the weak agents in contest
1 will have a higher expected payoff than the weak agents in contest 2.

\begin{figure}[h]
\centering
\includegraphics[scale=0.3]{version 9/Figure2}
\caption[fe:alg-1-step-4]{An illustration of the two contests at the end of
Step 4 of Algorithm \protect\ref{alg:two-types-two-contests}}
\label{fe:alg-1-step-1}
\end{figure}

\subsection{More than two parallel contests}

Consider a set $M=\{1,2,3,...,m\}$ of $m$ contests. Each contest $j\in M$
has a prize $v_{j}$ where $v_{j}\geq v_{j+1},$ $j=1,...,m-1$ (see Figure \ref%
{fig:figure3}). \bigskip\ Similarly to the previous subsection, regardless
of the allocation of the agents among the $m$ contests, by the uniqueness of
the equilibrium in each of the $m$ contests, we can assume that if an agent
moves from contest $i$ to contest $j$, then his expected payoff in contest $%
j $ is larger than his expected payoff in contest $i$. We can generalize
Lemma \ref{le:movement-2} as follows:

\begin{lemma}
\label{le:movement-m} Consider a model of two-stage parallel contests with $%
m $ contests and two types of agents $h$ and $l$ where $\widehat{c}%
_{h}(x)\leq \widehat{c}_{l}(x)$ for all $x\geq 0$. If for some allocation of
agents among the $m$ contests, there is not a $h$-type agent who wants to
move across the contests but there is a $l$-type agent who wants to move
from contest $i$ to contest $j$, then no more than $(m-1)$ $h$-type agents
will want to move across the contests where at most one of them will join
contest $i$.
\end{lemma}

By Lemma \ref{le:movement-m}, a move of one $l$-type agent implies at most $%
m-1$ moves of $h$-type agents. But, in fact, the move of one $l$-type agent
implies at most one move of a $h$-type agent since except for one move of a $%
h$-type agent to the initial place of the $l$-type agent, the other moves
are of $h$-type agents to contests that other $h$-type agent just left. By
using this argument, we construct the following algorithm:

\begin{algorithm}
\label{alg:two-types-m contests} The algorithm has the following four steps:

\begin{tabular}{ll}
Step 1 & Allocate all the $h$-type agents in contest 1 and all the $l$-type
agents \\ 
& in contest 2 where all the other contests do not include agents. \\ 
Step 2 & $H$-type agents have the option to move across the contests.%
\QQfnmark{%
Similar to step 2 of the preceding algorithm, neither the order nor the
number of possible movements for each agent is constrained. When no one
wishes to proceed, this step concludes.} \\ 
Step 3 & A single $l$-type agent has the option to move across the contests.
\\ 
Step 4 & Repeat steps 2-3 until in step 3 no agent wants to move.%
\end{tabular}%
\QQfntext{0}{
Similar to step 2 of the preceding algorithm, neither the order nor the
number of possible movements for each agent is constrained. When no one
wishes to proceed, this step concludes.}
\end{algorithm}

The next result shows that this algorithm has to be stopped at some point
since agents do not want to move across the contests, and then we say that
the algorithm converges to equilibrium.

\begin{proposition}
\label{prop:alg-2-equilibrium} Algorithm \ref{alg:two-types-m contests}
always converges to equilibrium and therefore every model of two-stage
parallel contests with $m$ contests and two types of agents $h$ and $l$
where $\widehat{c}_{h}(x)\leq \widehat{c}_{l}(x)$ for all $x\geq 0$, has a
subgame perfect equilibrium in which agents choose pure strategies in both
stages.
\end{proposition}

\begin{figure}[h]
\centering
\includegraphics[scale=0.3]{version 9/Figure3}
\caption[fe:alg-2-step-1]{An illustration of the $m$ contests at the end of
Step 1 of Algorithm \protect\ref{alg:two-types-m contests}}
\label{fig:figure3}
\end{figure}

Proposition \ref{prop:alg-2-equilibrium} demonstrates that in a model of
two-stage parallel contests with non-linear contests, independent of the
number of contests, there is a subgame perfect equilibrium. In the initial
point of our algorithm, all the $h$-type agents are in contest 1 and all the 
$l$-type agents are in contest 2. The initial point might affect the agents'
equilibrium strategies in the first stage such that we might have
equilibrium points that depend on the initial point of algorithm \ref%
{alg:two-types-m contests}.

\begin{figure}[h]
\centering
\includegraphics[scale=0.3]{version 9/Figure4}
\caption[fe:alg-2-step-4]{An illustration of the $m$ contests at the end of
Step 4 of Algorithm \protect\ref{alg:two-types-m contests}}
\label{fe:alg-2-step-4}
\end{figure}

So far, we have assumed that the contests in the second stage have a
generalized Tullock contest success function (CSF) and non-linear costs of
effort. In the following section we study the model of two-stage parallel
contests with a Tullock CSF (Tullock 1980) and assume linear costs of effort
for the agents. We will demonstrate that, given these assumptions, we can
explicitly characterize the equilibrium and generalize our findings to any
number of agents and contests.

\section{Parallel contests with linear cost functions}

We assume here that agent $i$'s production function is $f_{i}(x_{i})=x_{i}$
and that the effort cost function is linear, that is, $c_{i}(x)=c_{i}x$
where $c_{i}$ $\in \mathbb{R}_{+}$ and $c_{i}\leq c_{i+1},i\in \{1,...,n-1\}$%
. We refer to all agents with a marginal effort cost $c_{i}$ as $i$-type
agents. Thus, the number of agent types could be any number less than or
equal to the number of agents. In order to analyze the subgame perfect
equilibrium in this case, we begin with the analysis of the second stage and
go backwards to the first one (the necessary conditions for existence of a
unique equilibrium in the second stage is satisfied according to
Szidarovszky and Okuguchi 1997 and Yamazaki 2008).

\subsection{The second stage}

In the second stage, the agents in contest $k,k=1,2,...,m$ compete against
each other. The maximization problem of agent $i\in N_{k}$ is%
\begin{equation}
\max_{x_{i}}v_{k}\frac{x_{i}}{\sum_{j\in N_{k}}x_{j}}-c_{i}x_{i},
\label{eq:max-second stage-tullock}
\end{equation}%
When agents are heterogeneous, some of the agents with high marginal effort
costs may stay out of the competition in equilibrium. Without loss of
generality, we assume that all the agents are active in contest $k$,
otherwise, only the active agents will be considered. Then, the well-known
solution of (\ref{eq:max-second stage-tullock}) yields the following
well-known result\footnote{%
For more information on this solution, see Konrad (2009), and Corch\'{o}n
and Serena (2018).}:

\begin{proposition}
\label{prop:equilibrium second stage} In the second stage of the model of
two-stage parallel contests, if $i\in N_{k}$, then agent $i$'s equilibrium
effort is%
\begin{equation}
x_{i}=p_{i}\sum_{j\in N_{k}}x_{j},  \label{eq:equilibrium-second stage}
\end{equation}%
where $p_{i}$ is agent $i$'s probability of winning and is given by 
\begin{equation}
p_{i}=1-\frac{(n_{k}-1)c_{i}}{\sum_{j\in N_{k}}c_{j}},
\label{eq:probability-second stage}
\end{equation}%
and agent $i$'s expected payoff is given by%
\begin{equation}
u_{i}=v_{k}(p_{i})^{2}.  \label{eq:payoff- second stage}
\end{equation}
\end{proposition}

So far, we have analyzed the agents' equilibrium efforts in the second
stage. Next, we analyze the first stage where each agent chooses the contest
in which he wants to compete

\subsection{The first stage}

First, we assume that the agents are homogeneous, which means that their
marginal costs of effort are the same, i.e., $c_{i}=c_{j},\forall i,j\in N$.
By the symmetry of the agents, since the equilibrium in each contest is
unique (see Szidarovszky and Okuguchi 1997 and Yamazaki 2008), in the second
stage, the agents in each contest choose the same effort, and then each of
them has the same probability to win the contest. When agents are
homogeneous, the first stage of our parallel contest model corresponds to a
potential game, specifically a congestion game. Potential games are well
known to have a pure-strategy equilibrium (Rosenthal 1973, Monderer and
Shapley 1996). Consider the set $M=\{1,2\}$\ of two contests such that the $%
n $\ homogeneous agents have to choose to compete either in contest $1$\ or
contest $2$. Then, the potential function\bigskip\ is%
\begin{equation*}
P(n_{1},n_{2})=\sum_{i=1}^{n_{1}}\frac{v_{1}}{(i)^{2}}+\sum_{i=1}^{n_{2}}%
\frac{v_{2}}{(i)^{2}},
\end{equation*}%
when $n_{j}$\ agents compete for a prize of $v_{j}$\ in a contest $j,j=1,2$.
If there are $n_{j}$\ agents in contest $j$\ and an agent moves from contest
1 to contest 2, there exists

\begin{equation*}
P(n_{1},n_{2})-P(n_{1}-1,n_{2}+1)=u_{i}(n_{1},n_{2})-u_{i}(n_{1}-1,n_{2}+1),
\end{equation*}%
where agent $i$\ is the one who switched between the contests, and $u_{i}$\
is his payoff function, which is given by (\ref{eq:payoff- second stage}).

Thus, in our model of two-stage parallel contests with two contests and $n$\
homogeneous agents, there is a subgame perfect equilibrium in which, in the
first stage, $n_{k}^{\ast }$\ agents choose to compete in contest $k=1,2$.
We will now characterize this pure-strategy equilibrium. Assume that $%
v_{1}\geq v_{2}$ and $\frac{v_{1}}{(n-1)^{2}}<\frac{v_{2}}{4}$ such that
both contests $1$ and $2$ are active. Otherwise, only contest 1 is active,
namely, in the first stage, all the agents choose to compete in contest 1.
If the agents' expected payoffs in both contests would be the same, then by
Proposition \ref{prop:equilibrium second stage} we have 
\begin{equation*}
u_{i}^{1}=v_{1}(p_{i}^{1})^{2}=\frac{v_{1}}{n_{1}^{2}}=\frac{v_{2}}{n_{2}^{2}%
}=v_{2}(p_{i}^{2})^{2}=u_{i}^{2}.
\end{equation*}%
In that case, the number of agents in the both contests will be%
\begin{eqnarray}
\tilde{n_{1}} &=&\frac{n\sqrt{v_{1}}}{\sqrt{v_{1}}+\sqrt{v_{2}}}
\label{eq:equilibrium numbers-2} \\
\tilde{n_{2}} &=&\frac{n\sqrt{v_{2}}}{\sqrt{v_{1}}+\sqrt{v_{2}}}.  \notag
\end{eqnarray}

Thus, $u_{i}^{1}=u_{i}^{2}$ when $\widetilde{n}_{k},k=1,2$ are positive
integers, and exactly $\widetilde{n}_{1}$ agents choose to compete in
contest 1 and $\widetilde{n}_{2}$choose to compete in contest 2, but $%
\widetilde{n}_{k},k=1,2$ are not necessarily positive integers.
Nevertheless, in that case there is a subgame perfect equilibrium in which
in the first stage, the agents use pure strategies. To see that, assume that
in equilibrium $\tilde{n_{1}}-2$ agents choose to compete in contest 1 and $%
\tilde{n_{2}}+2$ agents choose to compete in contest 2 . Then, there exists
that $u^{1}>u^{2}$, namely, the agents' expected payoff in contest 1 is
larger than in contest 2. But, then, an agent from contest 2 could move to
contest 1 and he will have a larger expected payoff than an agent in contest
2. This is because if the allocation of agents is $\tilde{n_{1}}-1$ in
contest 1 and $\tilde{n_{2}}+1$ agents in contest 2 , there exists that $%
u^{1}>u^{2}$. This argument reduces the number of possible equilibrium
allocations to only two combinations of $\left\lfloor \tilde{n_{k}}%
\right\rfloor $-the largest integer that smaller than $\tilde{n_{i}}$, and $%
\left\lceil \tilde{n_{k}}\right\rceil $ the smallest integer that larger
than $\tilde{n_{k}},$ $k=1,2$. Hence, in the model of two-stage parallel
contests with two contests and $n$ homogeneous agents, there is a subgame
perfect equilibrium in which in the first stage $n_{k}^{\ast }$ agents
choose to compete in contest $k=1,2$ where $(n_{1}^{\ast },n_{2}^{\ast })\in
\{(\left\lfloor \tilde{n_{1}}\right\rfloor ,\left\lceil \tilde{n_{2}}%
\right\rceil ),(\left\lceil \tilde{n_{1}}\right\rceil ,\left\lfloor \tilde{%
n_{2}}\right\rfloor )\}$ and $\tilde{n_{1}},\tilde{n_{2}}$ are given by (\ref%
{eq:equilibrium numbers-2}).

When there are more than two sets, by the same argument, in the model of
two-stage parallel contests with $m$\ contests and $n$\ homogeneous agents,
there is a subgame perfect equilibrium in which $n_{k}^{\ast }$\ agents
choose to compete in contest $k,k=1,2,...,m$\ where $n_{k}^{\ast
}=\left\lfloor \tilde{n_{k}}\right\rfloor $\ or $n_{k}^{\ast }=\left\lceil 
\tilde{n_{k}}\right\rceil $, $\sum_{k=1}^{m}n_{k}^{\ast }=n$\ and $\tilde{%
n_{k}}$\ are given by 
\begin{equation}
\tilde{n_{k}}=\frac{n\sqrt{v_{k}}}{\sum_{i=1}^{m}\sqrt{v_{i}}}.
\label{eq:equilibrium numbers-m}
\end{equation}%
When agents are heterogeneous, there is no clear equivalence between our
parallel contest model and potential games. We do not find the subgame
perfect equilibrium in that case, but we show that in any subgame perfect
equilibrium in which agents use pure strategies in the first stage, so that
an agent in any contest has no incentive to move to another contest, agents
of the same type in different contests with different prizes never have the
same expected payoff.

\begin{proposition}
\label{prop: non existence-mix} In a model of two-stage parallel contests
with $m\geq 2$ contests and $n$ heterogenous agents, no subgame perfect
equilibrium exists in pure strategies in which agents of every type in
different contests with different prizes have the same expected payoff.
\end{proposition}

\section{Conclusion}

Parallel contests where each agent can participate in only one contest might
be complicated especially when we assume that contests are asymmetric in
size as well as in the value of the prize(s) and the agents are also
asymmetric in their costs of effort. In this paper, we analyze the subgame
perfect equilibrium of two-stage parallel contests with a finite number of
agents under asymmetric assumptions on the contests as well on the agents.

We first assumed non-linear costs of effort and a generalized Tullock
contest success function. Furthermore, we assumed two types of effort costs,
but these costs are given in the most general form under the condition of
increasing weakly convex functions. In that case, we can find the
equilibrium even if we do not know the exact form of the effort costs;
however, we do know the agents' preferences for different allocations, so we
can design an algorithm that converges to equilibrium, namely, the algorithm
allocates the agents across the contests such that no agent wants to move
from his position. Thus, we do not just prove that there is a subgame
perfect equilibrium with pure strategies in both stages, but also provide an
efficient mechanism to find it.

Next, we consider two-stage parallel contests with linear effort costs and
the Tullock contest success function. When the agents are homogeneous, our
model is equivalent to a potential game with a pure-strategy equilibrium,
and we explicitly define the potential function. Thus, in that case, our
model of parallel contest has a subgame-perfect equilibrium in which agents
in the first stage choose their contests using pure strategies. In that
case, we explicitly describe the subgame perfect equilibrium. When the
agents are heterogeneous and come in a variety of types, with any two agents
having different costs of effort, it is unclear whether our model is
equivalent to a potential game, and thus whether there is a subgame perfect
equilibrium. We show that if our model has a subgame perfect equilibrium
with heterogeneous agents, it has the counterintuitive property that agents
of the same type in different contests with different prizes never have the
same expected payoff.

One of the consequences of demonstrating the presence of subgame perfect
equilibrium in pure strategies in our two-stage parallel contests is that
agents who use pure strategies in the first stage have no incentive to
change their strategies in the second stage. In contrast, if the agents
employ mixed strategies in the first stage, they may come to regret
implementing their first-stage strategies at the beginning of the second
stage. In other words, the subgame perfect equilibrium of multi-stage
parallel contests with pure strategies is more likely to be accepted as a
predictor of agent behavior than the same type of equilibrium with mixed
strategies.

To simplify the algorithms in our model of two-stage parallel contests with
non-linear costs of effort, we assume two types of costs of effort. We
believe that the subgame perfect equilibrium exists in our model for any
number of costs of effort, but the appropriate algorithms to demonstrate the
equilibrium's existence will be much more complex. Finally, by demonstrating
the equivalence of our model and potential games when effort costs are
linear and agents are homogeneous, we demonstrate that parallel contest
analyses are feasible and applicable.

\section{Appendix}

\subsection{Proof of Lemma \protect\ref{le:movement-2}}

Let $u_{t}(v_{k},h_{k},l_{k})$ be the expected payoff for an agent of type $%
t\in \{h,l\}$ in contest $k$, given an allocation of $h_{k}$ high-type
agents and $l_{k}$ low-type agents in that contest. Additionally, let the
initial allocation be $(h_{i},l_{i})$ in contest $i$ and $(h_{j},l_{j})$ in
contest $j$. If a $l$-type agent chooses to move from contest $i$ to contest 
$j$, it means that his expected payoff in contest $j$ is strictly greater
than his current expected payoff in contest $i$. Formally, 
\begin{equation}
u_{l}(v_{j},h_{j},l_{j}+1)>u_{l}(v_{i},h_{i},l_{i}).
\end{equation}%
Furthermore, because no $h$-type wants to move, for any $h$-type agent in
contest $j$, \ there exists, 
\begin{equation}
u_{h}(v_{j},h_{j},l_{j})\geq u_{h}(v_{i},h_{i}+1,l_{i}).
\end{equation}%
Following the move of the $l$-type agent, the new allocation is $%
(h_{i},l_{i}-1)$ in contest $i$ and $(h_{j},l_{j}+1)$ in contest $j$.

Consider the following scenario: in response to the $l$-type agent's move,
more than one $h$-type agent wants to move from contest $j$ to contest $i$.
This implies that a second $h$-type agent would also desire to move after
the first $h$-type agent has already moved. The incentive for this second $h$%
-type agent to move would require 
\begin{equation}
u_{h}(v_{i},h_{i}+2,l_{i}-1)>u_{h}(v_{j},h_{j}-1,l_{j}+1).
\end{equation}%
However, an agent's expected payoff decreases as he faces stronger
opponents, specifically, those with a lower cost of effort. Since a $h$-type
agent is a stronger competitor than a $l$-type agent, replacing a $l$-type
agent with a $h$-type agent in contest $i$ makes the competition more
intense for the $h$-type agents. Thus, we have%
\begin{equation}
u_{h}(v_{i},h_{i}+1,l_{i})>u_{h}(v_{i},h_{i}+2,l_{i}-1).  \label{eq:in-eq-1}
\end{equation}%
Furthermore, replacing a $h$-type agent with a $l$-type agent in contest $j$
reduces the intensity of competition for $h$-type agents. Thus, we have. 
\begin{equation}
u_{h}(v_{j},h_{j}-1,l_{j}+1)>u_{h}(v_{j},h_{j},l_{j}).  \label{eq:in-eq-2}
\end{equation}%
Combining \eqref{eq:in-eq-1} and \eqref{eq:in-eq-2} with the condition above
that the second $h$-type agent moves, we deduce that 
\begin{equation}
u_{h}(v_{i},h_{i}+1,l_{i})>u_{h}(v_{j},h_{j},l_{j}).
\end{equation}%
This inequality contradicts the initial assumption:$u_{h}(v_{j},h_{j},l_{j})%
\geq u_{h}(v_{i},h_{i}+1,l_{i})$. Therefore, two or more $h$-type agents
cannot choose to move. As a result, we conclude that any $l$-type agent move
will be followed by no more than one $h$-type agent move.

\subsection{Proof of Proposition \protect\ref{prop:alg-1-equilibrium}}

To demonstrate that Algorithm \ref{alg:two-types-two-contests} converges to
equilibrium, we show that a $l$-type agent who moves from contest 2 to
contest 1 will not move back in a subsequent step, ensuring the algorithm
will terminate. Let the allocation of agents prior to Step 3 be $%
(h_{1},l_{1})$ in contest 1 and $(h_{2},l_{2})$ in contest 2. At this point,
no $h$-type agent wishes to move.

In Step 3, a $l$-type agent moves from contest 2 to 1. This implies that his
expected payoff from this move is strictly greater than his current payoff.
Thus, we have 
\begin{equation}
u_{l}(v_{1},h_{1},l_{1}+1)>u_{l}(v_{2},h_{2},l_{2}).  \label{eq:l-move-cond}
\end{equation}%
The new allocation is $(h_{1},l_{1}+1)$ in contest 1 and $(h_{2},l_{2}-1)$
in contest 2.

In the subsequent Step 2, $h$-type agents may respond. According to Lemma %
\ref{le:movement-2}, only one $h$-type agent will move from Contest 1 to
Contest 2. We consider two cases:

\textbf{Case A: No $h$-type agent moves.} The allocation remains $%
(h_{1},l_{1}+1)$ and $(h_{2},l_{2}-1)$. To return to contest 2 in Step 3,
the $l$-type agent needs $%
u_{l}(v_{2},h_{2},(l_{2}-1)+1)>u_{l}(v_{1},h_{1},l_{1}+1)$, which simplifies
to $u_{l}(v_{2},h_{2},l_{2})>u_{l}(v_{1},h_{1},l_{1}+1)$. This contradicts
the condition in \eqref{eq:l-move-cond} that prompted the initial move.
Thus, the l-type agent does not move back.

\textbf{Case B: One $h$-type agent moves from contest 1 to 2.} The
allocation after this response is $(h_{1}-1,l_{1}+1)$ in contest 1 and $%
(h_{2}+1,l_{2}-1)$ in contest 2. For the $l$-type agent to move back to
contest 2 in the next Step 3, it would require 
\begin{equation}
u_{l}(v_{2},h_{2}+1,(l_{2}-1)+1)>u_{l}(v_{1},h_{1}-1,l_{1}+1).
\label{eq:l-return-cond}
\end{equation}%
We will show that this last condition cannot hold: When a $h$-type agent
enters, an agent's payoff decreases, but it increases when one leaves.
Therefore, we have the following inequalities. 
\begin{align}
u_{l}(v_{1},h_{1}-1,l_{1}+1)& >u_{l}(v_{1},h_{1},l_{1}+1)\quad (\text{due to
the }h\text{-type agent leaving contest 1}) \\
u_{l}(v_{2},h_{2},l_{2})& >u_{l}(v_{2},h_{2}+1,l_{2})\quad (\text{due to the 
}h\text{-type agent entering contest 2})
\end{align}%
Combining these with the initial move condition from \eqref{eq:l-move-cond}
gives the following: 
\begin{equation}
u_{l}(v_{1},h_{1}-1,l_{1}+1)>u_{l}(v_{1},h_{1},l_{1}+1)>u_{l}(v_{2},h_{2},l_{2})>u_{l}(v_{2},h_{2}+1,l_{2}).
\end{equation}%
This implies that $u_{l}(v_{1},h_{1}-1,l_{1}+1)>u_{l}(v_{2},h_{2}+1,l_{2})$,
which contradicts \eqref{eq:l-return-cond}. Thus, the $l$-type agent will
not move back.

In both cases, once a $l$-type agent switches from contest 2 to contest 1,
he will not return. The algorithm starts with all $l$-type agents in contest
2. Each run of Steps 2-3 cycle results in either no move or one $l$-type
agent moving from Contest 2 to 1. Therefore, the algorithm must terminate in
a finite number of steps when no $l$-type agent wishes to move in Step 3 and
the resulting allocation is a subgame perfect equilibrium.

In both cases, once a $l$-type agent switches from contest 2 to contest 1,
he will not return. The algorithm starts with all $l$-type agents in contest
2. Each run of Steps 2-3 cycle results in either no move or one $l$-type
agent moving from Contest 2 to 1. Therefore, the algorithm must terminate in
a finite number of steps when no $l$-type agent wishes to move in Step 3 and
the resulting allocation is a subgame perfect equilibrium.

\subsection{Proof of Lemma \protect\ref{le:movement-m}}

Consider the initial allocation be $(h_{k},l_{k})$ for each contest $k\in M$%
. A $l$-type agent that moves from contest $i$ to $j$ implies 
\begin{equation*}
u_{l}(v_{j},h_{j},l_{j}+1)>u_{l}(v_{i},h_{i},l_{i}).
\end{equation*}%
The fact that no $h$-type wants to move implies that for any $h$-type agent
in contest $k_{1}$, its payoff is greater than or equal to the payoff it
would get by moving to any other contest $k_{2}$: 
\begin{equation}
u_{h}(v_{k_{1}},h_{k_{1}},l_{k_{1}})\geq
u_{h}(v_{k_{2}},h_{k_{2}}+1,l_{k_{2}})\quad \forall k_{1},k_{2}\in
M,k_{1}\neq k_{2}.  \label{eq:condition-lemma-2}
\end{equation}%
After the $l$-type agent's move, the allocation in contest $i$ becomes $%
(h_{i},l_{i}-1)$. The departure of a low-type agent makes contest $i$ more
attractive. A $h$-type agent, $H_{1}$, from some contest $k_{1}$ will be
induced to move to contest $i$ if 
\begin{equation}
u_{h}(v_{i},h_{i}+1,l_{i}-1)>u_{h}(v_{k_{1}},h_{k_{1}},l_{k_{1}}).
\end{equation}%
This change in preferences of $h$-type agents may happen due to the change
in contest $i$. By Lemma \ref{le:movement-2}, at most one such $h$-type
agent will move to contest $i$.

The departure of $H_{1}$ from contest $k_{1}$ changes its allocation to $%
(h_{k_{1}}-1,l_{k_{1}})$. This can trigger a sequential chain of movements.
A second $h$-type agent, $H_{2}$, from contest $k_{2}$ may now be induced to
move to contest $k_{1}$, if its expected payoff there is now higher 
\begin{equation}
u_{h}(v_{k_{1}},(h_{k_{1}}-1)+1,l_{k_{1}})>u_{h}(v_{k_{2}},h_{k_{2}},l_{k_{2}}),
\end{equation}%
which simplifies to 
\begin{equation*}
u_{h}(v_{k_{1}},h_{k_{1}},l_{k_{1}})>u_{h}(v_{k_{2}},h_{k_{2}},l_{k_{2}}).
\end{equation*}
Following the movement of $H_{2}$ to contest $k_{1}$, any other $h$-type
agent from contest $r$ will have no incentive to move to contest $k_{1}$.
This is because his expected payoff would be $%
u_{h}(v_{k_{1}},h_{k_{1}}+1,l_{k_{1}})$, which by condition %
\eqref{eq:condition-lemma-2} is less than or equal to his current payoff $%
u_{h}(v_{r},h_{r},l_{r})$.

This chain continues, with an agent $H_p$ from contest $k_p$ moving to
contest $k_{p-1}$ (where $k_0 = i$), as each move may create an incentive
for at most one $h$-type agent to move to the last departing agent's
contest. Thus, the chain of movements, $k_p \to k_{p-1} \to \dots \to k_1
\to i$, must consist of distinct contests. Since there are $m$ contests, the
longest possible non-cyclical path of moves involves $m-1$ steps. Therefore,
the total number of $h$-type agents who move in response to the initial $l$%
-type agent's move is at most $m-1$.

\subsection{Proof of Proposition \protect\ref{prop:alg-2-equilibrium}}

To prove that Algorithm \ref{alg:two-types-m contests} converges to
equilibrium, we show that a $l$-type agent who leaves a contest will not
return to it in a subsequent step. Because the number of agents and contests
is finite, the algorithm must reach equilibrium in a finite number of steps.

In Step 3, a $l$-type agent moves from contest $i$ to $j$, which implies: 
\begin{equation}
u_{l}(v_{j},h_{j},l_{j}+1)>u_{l}(v_{i},h_{i},l_{i}).
\label{eq:l-move-m-cond}
\end{equation}%
In the subsequent Step 2, $h$-type agents respond. By Lemma \ref%
{le:movement-m}, this may induce a chain of at most $m-1$ moves, where each $%
h$-type agent moves into the contest that the previous agent just left. Let
this chain of moves be $k_{p}\rightarrow k_{p-1}\rightarrow \dots
\rightarrow k_{1}\rightarrow i$, where a $h$-type agent moves from contest $%
k_{1}$ to $i$, another from $k_{2}$ to $k_{1}$, and so on. The final
allocation in contest $i$ becomes $(h_{i}+1,l_{i}-1)$, and the allocation in
contest $j$ is $(h_{j},l_{j}+1)$ if $j$ was not part of the move chain, or $%
(h_{j}-1,l_{j}+1)$ if $j=k_{s}$ for some $s\in \{1,\dots ,p\}$.

For the original $l$-type agent to consider moving back from $j$ to $i$ in a
future Step 3, the following condition must be met after the $h$-type agents
have settled where $h_{j}^{\prime }$ denotes the number of $h$-type agents
in contest $j$ after their movements 
\begin{equation}
u_{l}(v_{i},h_{i}+1,(l_{i}-1)+1)>u_{l}(v_{j},h_{j}^{\prime },l_{j}+1).
\label{eq:l-return-m-cond}
\end{equation}%
If we analyze the expected payoffs, we conclude that the entry of a $h$-type
agent into contest $i$ makes it less attractive for a $l$-type agent than
when he left 
\begin{equation}
u_{l}(v_{i},h_{i},l_{i})>u_{l}(v_{i},h_{i}+1,l_{i}).
\end{equation}%
Now consider contest $j$. If $j$ was not in the move chain of $h$-type
agents, its composition for the returning $l$-type agent would be $%
(h_{j},l_{j}+1)$. If $j$ was in the move chain (i.e., $j=k_{s}$), a $h$-type
agent has left, making it less competitive, and the composition is $%
(h_{j}-1,l_{j}+1)$. In either scenario for contest $j$, an agent's payoff
increases when another agent leaves. Thus, 
\begin{equation*}
u_{l}(v_{j},h_{j}-1,l_{j}+1)>u_{l}(v_{j},h_{j},l_{j}+1).
\end{equation*}%
This gives us a general inequality for the payoff in contest $j$ after the
moves 
\begin{equation}
u_{l}(v_{j},h_{j}^{\prime },l_{j}+1)\geq u_{l}(v_{j},h_{j},l_{j}+1).
\end{equation}%
By combining these inequalities with the initial condition %
\eqref{eq:l-move-m-cond}, we establish a clear ordering 
\begin{equation}
u_{l}(v_{j},h_{j}^{\prime },l_{j}+1)\geq
u_{l}(v_{j},h_{j},l_{j}+1)>u_{l}(v_{i},h_{i},l_{i})>u_{l}(v_{i},h_{i}+1,l_{i}).
\end{equation}%
This demonstrates that 
\begin{equation*}
u_{l}(v_{j},h_{j}^{\prime },l_{j}+1)>u_{l}(v_{i},h_{i}+1,l_{i}),
\end{equation*}
which contradicts the condition required for the $l$-type agent to return to
contest $i$ \eqref{eq:l-return-m-cond}.

Because a $l$-type agent never returns to a contest he has left, and the
number of agents and contests is finite, the algorithm must terminate in a
finite number of steps when no $l$-type agent wishes to move in Step 3 and
the resulting allocation is a subgame perfect equilibrium.

\subsection{Proof of Proposition \protect\ref{prop:equilibrium second stage}}

The FOC of agent $i$'s maximization problem (\ref{eq:max-second
stage-tullock}) is 
\begin{equation}
v_{k}\frac{\sum_{j=1}^{n_{k}}x_{j}-x_{i}}{\left(
\sum_{j=1}^{n_{k}}x_{j}\right) ^{2}}\leq c_{i}.
\label{eq:foc-equilibrium-linear}
\end{equation}%
Without loss of generality, assume that for all agents in $N_{k}$ the FOC is
non-negative, and then we obtain: 
\begin{align*}
\sum_{i\in N_{k}}\frac{v_{k}(\sum_{j\in N_{k}}x_{j}-x_{i})}{(\sum_{j\in
N_{k}}x_{j})^{2}}& =v_{k}\left( \sum_{i\in N_{k}}\frac{\sum_{j\in N_{k}}x_{j}%
}{(\sum_{j\in N_{k}}x_{j})^{2}}-\frac{x_{i}}{(\sum_{j\in N_{k}}x_{j})^{2}}%
\right) = \\
v_{k}\left( \frac{n_{k}}{\sum_{j\in N_{k}}x_{j}}-\frac{\sum_{j\in N_{k}}x_{j}%
}{(\sum_{j\in N_{k}}x_{j})^{2}}\right) & =v_{k}\left( \frac{n_{k}}{%
\sum_{j\in N_{k}}x_{j}}-\frac{1}{\sum_{j\in N_{k}}x_{j}}\right) =\sum_{i\in
N_{k}}c_{i}.
\end{align*}%
Thus, we have

\begin{equation*}
v_{k}\left( \frac{n_{k}-1}{\sum_{j\in N_{k}}x_{j}}\right) =\sum_{i\in
N_{k}}c_{i}.
\end{equation*}%
which implies that the total effort in contest $k$ is 
\begin{equation}
\sum_{j\in N_{k}}x_{j}=\frac{v_{k}(n_{k}-1)}{\sum_{i\in N_{k}}c_{i}}.
\label{eq:total effort-linear}
\end{equation}%
Inserting (\ref{eq:total effort-linear}) into (\ref%
{eq:foc-equilibrium-linear}) yields 
\begin{equation*}
\frac{v_{k}\left( \frac{v_{k}(n_{k}-1)}{\sum_{j\in N_{k}}c_{j}}-x_{i}\right) 
}{\left( \frac{v_{k}(n_{k}-1)}{\sum_{j\in N_{k}}c_{j}}\right) ^{2}}=c_{i},
\end{equation*}%
or, alternatively,

\begin{equation*}
\frac{v_{k}(n_{k}-1)}{\sum_{j\in N_{k}}c_{j}}-x_{i}=v_{k}c_{i}{\left( \frac{%
n_{k}-1}{\sum_{j\in N_{k}}c_{j}}\right) ^{2}.}
\end{equation*}%
Thus, agent $i$'s equilibrium effort in contest $k$ is 
\begin{equation}
x_{i}=\frac{v_{k}(n_{k}-1)}{\sum_{j\in N_{k}}c_{j}}\left( 1-\frac{%
(n_{k}-1)c_{i}}{\sum_{j\in N_{k}}c_{j}}\right) .
\label{eq:effort-equilibrium -linear}
\end{equation}%
By (\ref{eq:total effort-linear}) and (\ref{eq:effort-equilibrium -linear})
agent $i$'s probability of winning in contest $k$ is 
\begin{equation}
p_{i}=\frac{x_{i}}{\sum_{j\in N_{k}}x_{j}}=\frac{\frac{v_{k}(n_{k}-1)}{%
\sum_{j\in N_{k}}c_{j}}\left( 1-\frac{(n_{k}-1)c_{i}}{\sum_{j\in N_{k}}c_{j}}%
\right) }{\frac{v_{k}(n_{k}-1)}{\sum_{j\in N_{k}}c_{j}}}=1-\frac{%
(n_{k}-1)c_{i}}{\sum_{j\in N_{k}}c_{j}}.  \label{eq:probability-linear}
\end{equation}%
Then, by (\ref{eq:effort-equilibrium -linear}) and (\ref%
{eq:probability-linear}), agent $i$'s expected payoff is 
\begin{eqnarray*}
u_{i} &=&v_{k}p_{i}-x_{i} \\
&=&v_{k}(1-\frac{(n_{k}-1)c_{i}}{\sum_{j\in N_{k}}c_{j}})-\frac{%
v_{k}(n_{k}-1)}{\sum_{j\in N_{k}}c_{j}}\left( 1-\frac{(n_{k}-1)c_{i}}{%
\sum_{j\in N_{k}}c_{j}}\right) \\
&=&v_{k}(1-\frac{(n_{k}-1)c_{i}}{\sum_{j\in N_{k}}c_{j}}%
)^{2}=v_{k}(p_{i})^{2}.
\end{eqnarray*}

\subsection{Proof of Proposition \protect\ref{prop: non existence-mix}}

In order to prove that agents with the same type from different contest do
not have the same expected payoff in equilibrium, without loss of
generality, we consider two contests and two types of agents. \ In the case
of more than two contests and two types of agents, the following argument
holds for any choice of two contests from the set of contests.

The agents have marginal costs of effort either $c_{h}$\ or $c_{l}$. Assume
that $h_{i}$ agents with a marginal effort cost of $c_{h}$\ chooses to
participate in contest $i,i=1,2$\ where $h_{1}+h_{2}=h$. Similarly, $l_{i}$
agents with a marginal effort cost of $c_{l}$\ chooses to participate in
contest $i,i=1,2$\ where $l_{1}+l_{2}=l$. We assume that the prizes in both
contests are different $v_{1}\neq v_{2}$\ and the we will show that agents
of the same type could not have the same expected payoff in both contests.
For this purpose, suppose that agents of the same type have the same
expected payoff in two different contests. Then, by Proposition \ref%
{prop:equilibrium second stage} the agents' expected payoffs in contest $1$, 
$u_{j}^{1},j=l,h$ and in contest $2$, $u_{j}^{1},j=l,h$ satisfy: 
\begin{equation}
u_{h}^{1}=v_{1}\left( 1-\frac{(h_{1}+l_{1}-1)c_{h}}{h_{1}c_{h}+l_{1}c_{l}}%
\right) ^{2}=u_{h}^{2}=v_{2}\left( 1-\frac{(h_{2}+l_{2}-1)c_{h}}{%
h_{2}c_{h}+l_{2}c_{l}}\right) ^{2},  \label{eq:payoff -h}
\end{equation}%
and%
\begin{equation}
u_{l}^{1}=v_{1}\left( 1-\frac{(h_{1}+l_{1}-1)c_{l}}{h_{1}c_{h}+l_{1}c_{l}}%
\right) ^{2}=u_{l}^{2}=v_{2}\left( 1-\frac{(h_{2}+l_{2}-1)c_{l}}{%
h_{2}c_{h}+l_{2}c_{l}}\right) ^{2}.  \label{eq:payoff-l}
\end{equation}%
We can divide both equations (\ref{eq:payoff -h}) and (\ref{eq:payoff-l}) by
each other and obtain that, 
\begin{equation*}
\frac{c_{h}-(c_{h}-c_{l})l_{1}}{c_{l}+(c_{h}-c_{l})h_{1}}=\frac{%
c_{h}-(c_{h}-c_{l})l_{2}}{c_{l}+(c_{h}-c_{l})h_{2}}.
\end{equation*}%
From the last equation we get 
\begin{equation*}
h_{2}=\frac{c_{h}h_{1}+c_{l}l_{1}-c_{l}l_{2}-c_{h}h_{1}l_{2}+c_{l}h_{1}l_{2}%
}{c_{h}-c_{h}l_{1}+c_{l}l_{1}}.
\end{equation*}%
Since $h_{1}+h_{2}=h$, we have

\begin{equation}
h_{2}=\frac{c_{h}h+c_{l}l_{1}-c_{l}l_{2}-c_{h}hl_{2}+c_{l}hl_{2}}{%
2c_{h}-c_{h}l_{1}-c_{h}l_{2}+c_{l}l_{1}+c_{l}l_{2}},  \label{eq:h2}
\end{equation}%
and 
\begin{equation}
h_{1}=\frac{c_{h}h-c_{l}l_{1}+c_{l}l_{2}-c_{h}hl_{1}+c_{l}hl_{1}}{%
2c_{h}-c_{h}l_{1}-c_{h}l_{2}+c_{l}l_{1}+c_{l}l_{2}}.  \label{eq:h1}
\end{equation}%
Inserting (\ref{eq:h2}) and (\ref{eq:h1}) into (\ref{eq:payoff -h}), yields 
\begin{equation*}
v_{1}\left( \frac{(l_{2}+l_{1})c_{l}-(l_{2}+l_{1}-2)c_{h}}{%
c_{h}h+(l_{2}+l_{1})c_{l}}\right) ^{2}=v_{2}\left( \frac{%
(l_{2}+l_{1})c_{l}-(l_{2}+l_{1}-2)c_{h}}{c_{h}h+(l_{2}+l_{1})c_{l}}\right)
^{2}.
\end{equation*}%
The last equation implies that $v_{1}=v_{2}$, and this contradicts our
assumption that $v_{1}\neq v_{2}$.

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\end{document}
