11institutetext: Institute of Informatics, University of Warsaw, Banacha 2,
02-097 Warsaw, Poland 11email<EMAIL_ADDRESS>22institutetext:
Faculty of Mathematics and Information Science, Warsaw University of
Technology
22email<EMAIL_ADDRESS>
# Discrete Two Player All-Pay Auction with Complete Information††thanks: This
work was supported by Polish National Science Centre through grant
2018/29/B/ST6/00174.

Marcin Dziubiński 11 0000-0003-1756-2424 Krzysztof Jahn 22

###### Abstract

We study discrete two player all-pay auction with complete information. We
provide full characterization of mixed strategy Nash equilibria and show that
they constitute a subset of Nash equilibria of discrete General Lotto game. We
show that equilibria are not unique in general but they are interchangeable.
We also show that equilibrium payoffs are unique, unless valuation of at least
one of the players is an even integer number. If equilibrium payoffs are not
unique, continuum of equilibrium payoffs are possible. We compare equilibrium
characterization and equilibrium payoffs with the continuous variant of all-
pay auctions.

###### Keywords:

All-pay auctions Discrete bids Nash equilibria.

##  1 Introduction

All-pay auction constitutes a fundamental game theoretic model of contests
where players exert effort in order to win a prize and only the player
exerting the most effort wins the prize while other players effort is lost
without a reward. This form of strategic interaction underlies economic
activities such as political campaigns [15], R&D races [5], rent seeking and
lobbying activities [12, 11, 10, 1], competition for a monopoly position [7],
as well as sport competition [16].

Full characterisation of equilibria in a variant of all-pay auction where
effort of players is continuous was obtained by [2, 10]. In many situations,
however, it is natural to assume that effort is discrete: monetary
expenditure, the time spend on projects, or man-power are usually measured in
discrete units. This rises a number of questions How does the discrete
character of effort affect equilibrium behaviour of auction participants? Does
it benefit the stronger or the weaker side of an auction? How well does
equilibrium characterisation based on the continuous model approximates
equilibria in the discrete model? In this paper we address these and similar
questions by providing a complete characterisation of mixed strategy Nash
equilibria of discrete all-pay auctions. We show that certain qualitative
features of these equilibria are similar to features of equilibria in
continuous all-pay auctions. Let $v_{1}$ and $v_{2}$ be valuations of the
prize by the two auction participants. Suppose that $v_{1}\geq v_{2}$, so that
the second participant is the weaker one. In equilibrium of the continuous
model the weaker player chooses zero effort with probability $1-v_{2}/v_{1}$
and, with probability $v_{2}/v_{1}$ chooses her effort level by mixing
uniformly on the interval $[0,v_{2}]$. The stronger player mixes uniformly on
the interval $[0,v_{2}]$. We show that in the discrete model the weaker player
chooses zero effort with probability close to $1-v_{1}/v_{2}$ and chooses her
effort level by mixing on the interval $[0,v_{2}]$ with distributions which
are convex combinations of distributions which are uniform on even and odd
number in an interval close to $[0,v_{2}]$ or are distorted variants of such
distributions. Equilibrium payoffs are generically unique. Discreteness of
effort levels benefits the weaker player allowing her to obtain a positive
payoff when her prize valuation is close to the prize valuation of the
stronger player. In the case of the stronger player, discreteness of effort
levels may be beneficial or not, depending on the prize valuation of the
weaker player.

All-pay auctions with discrete effort levels were studied by [4]. The focus of
their paper is the effect of different tie-breaking policies. In the case of
the tie breaking policy where each of the two players wins the prize with
probability half in case of a tie, as considered in our paper, they provide
partial characterisation of equilibria in the cases where players valuations
are integer numbers. All-pay auctions are closely related to General Lotto
games [8]. Continuous variant of these games was considered by [3, 13, 14]. In
particular, [14] show that equilibria in these games are related to equilibria
in continous all pay auctions and exploit this connection to obtain
equilibrium characterisation. In this paper we take a reverse approach. We
show that equilibria in discrete all-pay auctions are a subset of equilibria
in the corresponding discrete General Lotto games. We then use the full
characterisation of equilibria in discrete General Lotto games obtained by [8]
and [6] to obtain full characterisation of equilibria in discrete all-pay
auctions.

The rest of the paper is organized as follows. In Section 2 we define the
model of discrete all-pay auctions. Section 3 contains the characterisation
results as well as discussion of relation to the continuous variant of all-pay
auctions. We conclude in Section 4. Most part of the proofs is given in the
Appendix.

##  2 The model

There are two players, $1$ and $2$, competing for a prize that is worth
$v_{2}>0$ for player $2$ and $v_{1}\geq v_{2}$ for player $1$. Player $2$, who
values the prize not more than player $1$ is called the _weaker player_ and
player $1$ is called the _stronger player_. Each player $i\in\\{1,2\\}$, not
observing the choice of other player, chooses an effort level
$x_{i}\in\mathbb{Z}_{\geq 0}$ in competition for the prize. The player
choosing the higher effort level wins the prize and, in the cases of a tie in
effort levels, one of the players receives the prize with probability $1/2$.
Both player pay the price equal to their chosen effort levels. Payoff to
player $i$ from effort the pair of effort choices
$(x_{1},x_{2})\in\mathbb{Z}^{2}_{\geq 0}$ is equal to

$p^{i}(x_{1},x_{2})=\begin{cases}v_{i}-x_{i},&\textrm{if $x_{i}>x_{-i}$,}\\\
\frac{v_{i}}{2}-x_{i},&\textrm{if $x_{i}=x_{-i}$,}\\\ -x_{i},&\textrm{if
$x_{i}<x_{-i}$,}\\\ \end{cases}$ (1)

where $x_{-i}$ denotes the effort level chosen by the other player. We allow
the players to make randomize choices, so that each player $i$ chooses a
probability distribution on non-negative integer numbers
$\xi_{i}\in\Delta(\mathbb{Z}_{\geq 0})$.111 Given a set $S$, $\Delta(S)$
denotes the set of all probability distributions on $S$. For simplicity and
notational convenience, with a probability distribution on $\mathbb{Z}_{\geq
0}$, $\xi$, we will identify a non-negative integer valued random variable
$X_{i}$ distributed according to $\xi_{i}$, so that for each
$k\in\mathbb{Z}_{\geq 0}$, $\mathbf{P}(X_{i}=k)=\xi^{i}_{k}$. We will also use
the random variables to refer to the associated probability distributions.
Expected payoff to player $i$ from randomized effort choices
$(X_{1},X_{2})\in\Delta(\mathbb{Z}_{\geq 0})^{2}$ is equal to

$P^{i}(X_{1},X_{2})=v_{i}\mathbf{P}(X_{i}>X_{-i})+\frac{v_{i}}{2}\mathbf{P}(X_{1}=X_{2})-\mathbf{E}(X_{i})$
(2)

We assume that the players are risk neutral and each of them aims to maximise
her expected payoff. We are interested in mixed strategy Nash equilibria of
this game, called Nash equilibria or equilibria, for short, throughout the
paper.

##  3 The analysis

Payoff to player $i$ from strategy profile $(X_{1},X_{2})$ can be written as

$\displaystyle P^{i}(X_{1},X_{2})$
$\displaystyle=v_{i}\mathbf{P}(X_{i}>X_{-i})+\frac{v_{i}}{2}\mathbf{P}(X_{i}=X_{-i})-\mathbf{E}(X_{i})$
(3) $\displaystyle=\frac{v_{i}}{2}\mathbf{P}(X_{i}\geq
X_{-i})+\frac{v_{i}}{2}\mathbf{P}(X_{i}>X_{-i})-\mathbf{E}(X_{i})$
$\displaystyle=\frac{v_{i}}{2}\left(\mathbf{P}(X_{i}>X_{-i})+1-\mathbf{P}(X_{i}<X_{-i})\right)-\mathbf{E}(X_{i})$
$\displaystyle=\frac{v_{i}}{2}\left(\mathbf{P}(X_{i}>X_{-i})-\mathbf{P}(X_{i}<X_{-i})\right)+\frac{v_{i}}{2}-\mathbf{E}(X_{i})$
$\displaystyle=\frac{v_{i}}{2}\left(H(X_{i},X_{-i})-\left(\frac{2\mathbf{E}(X_{i})}{v_{i}}-1\right)\right),$

where

$H(X_{i},X_{-i})=\mathbf{P}(X_{i}>X_{-i})-\mathbf{P}(X_{i}<X_{-i}).$

Given probability distributions $(X_{1},X_{2})$, the quantity
$H(X_{i},X_{-i})$ is payoff to the player choosing $X_{i}$ against the choice
$X_{-i}$ of the other player is the _discrete General Lotto game_ defined in
[8]. The game is played by two players, $1$ and $2$, who simultaneously and
independently chooses probability distributions on non-negative integers. Each
player $i$ is characterized by a number $b_{i}$ where $b_{1}\geq b_{2}>0$.
Choices of player $i$ are constrained so that the player chooses probability
distributions $X_{i}$ with $\mathbf{E}(X_{i})=b_{i}$.

The connection between continuous all-pay auctions and continuous General
Lotto games is well known in the literature and complete characterisation of
equilibria in continuous all-pay auctions, obtained by [2] was used by [13]
and [14] to obtain characterisation of equilibria in continuous General Lotto
games. In the case of discrete all-pay auctions we proceed in the reverse
direction and use the complete characterisation of equilibria obtained by [8]
and [6] to obtain complete characterisation of equilibria in discrete all-pay
auctions.

First, we establish that the set of equilibria in discrete all-pay auctions is
a subset of equilibria in discrete General Lotto games with properly chosen
constraints $(b_{1},b_{2})$.

###### Proposition 1

If a strategy profile $(X,Y)$ is a Nash equilibrium of all pay auction then it
is also a Nash equilibrium of the General Lotto game
$\mathit{\Gamma}(\mathbf{E}(X),\mathbf{E}(Y))$.

###### Proof

Suppose that $(X,Y)$ is a Nash equilibrium of all pay auction and let $v_{1}$
be the valuation of player $1$ and $v_{2}$ be the valuation of player $2$.
Since $(X,Y)$ is a Nash equilibrium so, for any strategy $X^{\prime}$ of
player $1$ with $\mathbf{E}(X^{\prime})=\mathbf{E}(X)$,

$P^{1}(X,Y)\geq P^{1}(X^{\prime},Y).$

By (3) and $\mathbf{E}(X^{\prime})=\mathbf{E}(X)$ this is equivalent to

$H(X,Y)\geq H(X^{\prime},Y).$

Similarly, for any strategy $Y^{\prime}$ of player $2$ with
$\mathbf{E}(Y^{\prime})=\mathbf{E}(Y)$,

$H(Y,X)\geq H(Y^{\prime},X).$

Since $H$ is the payoff function in the General Lotto game and strategies in
$\mathit{\Gamma}(\mathbf{E}(X),\mathbf{E}(Y))$ of each player are restricted
to distributions with the same expected value, $(X,Y)$ is a Nash equilibrium
of $\mathit{\Gamma}(\mathbf{E}(X),\mathbf{E}(Y))$.

The set of equilibria in a discrete all-pay auction is (usually, i.e. for most
values of $v_{1}$ and $v_{2}$) a proper subset of equilibria in the
corresponding General Lotto games. Before we provide the characterisation of
this set, we introduce the probability distributions that are the building
blocks of equilibria in General Lotto games.

[8] defines the following probability distributions. Given $m\geq 1$, let

$U_{\mathrm{O}}^{m}:=U(\\{1,3,\ldots,2m-1\\})=\sum_{i=1}^{m}\left(\frac{1}{m}\right)\mathbf{1}_{2i-1},$

and, given $m\geq 0$, let

$U_{\mathrm{E}}^{m}:=U(\\{0,2,\ldots,2m\\})=\sum_{i=1}^{m+1}\left(\frac{1}{m+1}\right)\mathbf{1}_{2i},$

where, given an integer $j$, $\mathbf{1}_{j}$ is the Dirac’s measure putting
probability $1$ on $j$. Distributions $U_{\mathrm{O}}^{m}$ and
$U_{\mathrm{E}}^{m}$ are “uniform on odd numbers” and “uniform on even
numbers”, respectively. We will use

$\mathcal{U}^{m}=\\{U_{\mathrm{E}}^{m},U_{\mathrm{O}}^{m}\\}$

to denote the set of these distributions. [6] defines the following
distributions. First, given $m\geq 1$, let

$U_{\mathrm{O}\uparrow
1}^{m}:=U(\\{2,4,\ldots,2m-2\\})=\sum_{i=1}^{m-1}\left(\frac{1}{m-1}\right)\mathbf{1}_{2i},$

which is a uniform distribution on even numbers from $2$ to $2m-2$. Given
$m\geq 2$ and $1\leq j\leq m-1$, let

$W_{j}^{m}:=\left(\frac{1}{2m}\right)\mathbf{1}_{0}+\sum_{i=1}^{j-1}\left(\frac{1}{m}\right)\mathbf{1}_{2i}+\left(\frac{1}{2m}\right)\mathbf{1}_{2j}+\sum_{i=j+1}^{m}\left(\frac{1}{m}\right)\mathbf{1}_{2i-1}.$

Each distribution $W_{j}^{m}$ is the distribution $U_{\mathrm{O}}^{m}$
distorted at the first $2j+1$ positions with a $2$-moving average, so that
$\mathbf{P}(W_{j}^{m}=i)=(\mathbf{P}(U_{\mathrm{O}}^{j}=i-1)+\mathbf{P}(U_{\mathrm{O}}^{j}=i+1))/2$,
for $0\leq i\leq 2j$ (where $\mathbf{P}(U_{\mathrm{O}}^{j}=-1)=0$). We will
use

$\mathcal{W}^{m}=\\{W_{1}^{m},\ldots,W_{m-1}^{m}\\}$

to denote the set of distributions $W_{j}^{m}$. Given $m\geq 1$ and $1\leq
j\leq m$, let

$V_{j}^{m}:=\sum_{i=1}^{j-1}\left(\frac{2}{2m+1}\right)\mathbf{1}_{2i-1}+\left(\frac{1}{2m+1}\right)\mathbf{1}_{2j-1}+\sum_{i=j}^{m}\left(\frac{2}{2m+1}\right)\mathbf{1}_{2i}.$

Each distribution $V_{j}^{m}$ is the distribution $U_{\mathrm{E}}^{m}$
distorted at the first $2j$ positions with a $2$-moving average, so that
$\mathbf{P}(V_{j}^{m}=i)=(\mathbf{P}(U_{\mathrm{E}}^{j-1}=i-1)+\mathbf{P}(U_{\mathrm{E}}^{j-1}=i+1))/2$,
for $0\leq i\leq 2j-1$ (where $\mathbf{P}(U_{\mathrm{E}}^{j-1}=-1)=0$). We
will use

$\mathcal{V}^{m}=\\{V_{1}^{m},\ldots,V_{m}^{m}\\}$

to denote the set of distributions $V_{j}^{m}$.

With these distributions in hand, we are ready to state our main results. We
divide the characterisation of equilibria in discrete all-pay auctions into
two cases, covered by Theorems 3.1 and 3.2 below. The first is the case where
half of the valuation of the prize by the second (weaker) player is an integer
number and the second is the case where it is not an integer number and it is
greater than $1$.222 For completeness, in the Appendix we provide an
additional Theorem 0.A.1 which covers the case of $v_{2}/2\in(0,1)$.

###### Theorem 3.1

Strategy profile $(X,Y)$ is a Nash equilibrium of all-pay auction with players
valuations $v_{1}\geq v_{2}$ and $v_{2}/2\in\mathbb{Z}_{\geq 1}$ if and only
if

  1. (i)

if $v_{1}=v_{2}$ then

$X=\alpha U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}\textrm{, }\quad
Y=\beta U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m},$

with $m=v_{2}/2-1$, $\alpha\in[0,1]$, and $\beta\in[0,1]$. Equilibrium payoffs
are

$P^{1}(X,Y)=1-\beta\ \textrm{ and }\ P^{2}(Y,X)=1-\alpha.$

  2. (ii)

if $v_{1}>v_{2}=2$ then

$X=U_{\mathrm{O}}^{1},\quad Y=(1-b)\mathbf{1}_{0}+b\left(\lambda
U_{\mathrm{O}}^{1}+(1-\lambda)U_{\mathrm{E}}^{1}\right),$

where $b\in(0,1]$ and

$\frac{4}{bv_{1}}-\frac{2}{b}+1\leq\lambda\leq\frac{4}{bv_{1}}-1.$

Equilibrium payoffs are

$P^{1}(X,Y)=v_{1}-\frac{bv_{1}}{2}-1\ \textrm{ and }\ P^{2}(Y,X)=0.$

  3. (iii)

if $v_{1}>v_{2}\geq 3$ then

$X=U_{\mathrm{O}}^{m},\quad
Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)Z,$

where $m=v_{2}/2$,
$b\in[v_{2}(v_{2}-2)/(2v_{1}),\min(m,v_{2}(v_{2}+2)/(2v_{1}))]$, and

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}$

with

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m-1}\geq 0\textrm{ and
}\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1,$ $\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{v_{2}^{2}}{2v_{1}b}-1,$

and

$\lambda_{\mathrm{O}}\geq\left(\frac{v_{2}}{2b}\right)\left(\frac{v_{2}(v_{2}+2)}{2v_{1}}+b-v_{2}\right)$

Equilibrium payoffs are

$P^{1}(X,Y)=v_{1}-\frac{bv_{1}}{v_{2}}-\frac{v_{2}}{2}\ \textrm{ and }\
P^{2}(Y,X)=0.$

The first point of the theorem covers the symmetric case where both players
value the prize equally. In this case each of the players uses a convex
combination of the uniform probability distribution on even numbers from $0$
to $2m$ and the uniform probability distribution on odd numbers from $0$ to
$2m+1$, where $m=v_{2}/2-1$. There are continuum of possible equilibrium
payoffs and payoff of each player is equal to $1$ minus the probability with
which the opponent uses the uniform on odd numbers probability distribution.
In particular, payoff of $0$ as well as payoff of $1$ is possible for each
player, depending on the strategy used by the opponent.

The second and the third points of the theorem cover the asymmetric case where
the valuation of player $2$ is strictly smaller than the valuation of player
$1$. The second point covers the subcase where $m=v_{2}/2$ takes value $1$ and
the third point covers the remaining subcases. In each case the stronger
player mixes uniformly on odd numbers between $1$ and $2m-1$ while the weaker
player chooses effort $0$ with probability $1-b/m$ and, with probability
$b/m$, uses a strategy which picks positive effort levels with probability
greater than $0$. Like in the case of the first point of the theorem, there is
continuum of equilibria and continuum of equilibrium payoffs. The equilibrium
expected payoff to the stronger player depends on the expected value, $b$, of
the strategy chosen by the weaker player. The expected equilibrium payoff of
the weaker player is equal to $0$ in all the cases.

Although there is continuum of equilibria and continuum of equilibrium payoffs
when valuation of the prize by the weaker player is an even number, the
equilibria exhibit the interchangeability property: if $(X,Y)$ and
$(X^{\prime},Y^{\prime})$ are both equilibria of all-pay auction,
$(X,Y^{\prime})$ and $(X^{\prime},Y)$ are equilibria as well. Thus so long as
each player chooses an equilibrium strategy, none of them has an incentive to
deviate to a different strategy.

The next theorem provides complete characterisation of equilibria in the case
where the valuation of the prize by the weaker player is not an even (integer)
number.

###### Theorem 3.2

Strategy profile $(X,Y)$ is a Nash equilibrium of all-pay auction with players
valuations $v_{1}\geq v_{2}>2$ and $v_{2}/2\notin\mathbb{Z}$ if and only if

  1. (i)

if $\lfloor v_{1}/2\rfloor=\lfloor v_{2}/2\rfloor$ then $X=\lambda
U_{\mathrm{O}}^{m}+(1-\lambda)U_{\mathrm{E}}^{m}$ and $Y=\kappa
U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m}$, with $m=\lfloor
v_{2}/2\rfloor$,

$\kappa=\frac{\lfloor
v_{1}/2\rfloor}{v_{1}/2}\left(\left\lceil\frac{v_{1}}{2}\right\rceil-\frac{v_{1}}{2}\right)\textrm{
and }\lambda=\frac{\lfloor
v_{2}/2\rfloor}{v_{2}/2}\left(\left\lceil\frac{v_{2}}{2}\right\rceil-\frac{v_{2}}{2}\right).$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=\frac{v_{1}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\ \textrm{
and }\ P^{2}(Y,X)=\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor.$

  2. (ii)

if $v_{1}/2=\lfloor v_{2}/2\rfloor+1$ then $Y=U_{\mathrm{E}}^{m}$, with
$m=\lfloor v_{2}/2\rfloor$, and

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right),$

with

$\delta=\frac{2\left\lfloor\frac{v_{2}}{2}\right\rfloor+1}{\left\lfloor\frac{v_{2}}{2}\right\rfloor+1},$

$\alpha\in[0,1/\delta]$,
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$,
and

$\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

or

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}\right),$

with

$\sigma=\frac{2\left\lfloor\frac{v_{2}}{2}\right\rfloor+1}{\left\lfloor\frac{v_{2}}{2}\right\rfloor},$

$\alpha\in(1/\delta,\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)]$,
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m}\geq
0$, $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}=1$,
and

$\lambda_{\mathrm{E}}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=1\ \textrm{ and }\
P^{2}(Y,X)=1-\frac{v_{2}}{v_{1}}\alpha-\frac{v_{1}-v_{2}}{2}.$

  3. (iii)

if $v_{1}/2>\lfloor v_{2}/2\rfloor+1$ then

$X\in\mathrm{conv}(\\{U^{m,\alpha}\\}\cup\mathcal{X}^{m,\alpha}),\quad
Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)U_{\mathrm{E}}^{m},$

where

$m=\left\lfloor\frac{v_{2}}{2}\right\rfloor,\quad
b=\frac{\left\lfloor\frac{v_{2}}{2}\right\rfloor\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{1}}{2}},\quad\alpha=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right),$

     * •

$U^{m,\alpha}=(1-\alpha)U_{\mathrm{O}}^{m}+\alpha U_{\mathrm{O}}^{m+1}$,

and

     * •

$\mathcal{X}^{m,\alpha}=\alpha\delta\mathcal{V}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}$,
if $v_{2}/2\leq\lceil v_{2}/2\rceil-1/2$,

     * •

$\mathcal{X}^{m,\alpha}=(1-\alpha)\sigma\mathcal{V}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}$,
if $v_{2}/2>\lceil v_{2}/2\rceil-1/2$,

where

$\delta=\frac{2\left\lfloor\frac{v_{2}}{2}\right\rfloor+1}{\left\lfloor\frac{v_{2}}{2}\right\rfloor+1},\quad\sigma=\frac{2\left\lfloor\frac{v_{2}}{2}\right\rfloor+1}{\left\lfloor\frac{v_{2}}{2}\right\rfloor}.$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=v_{1}+1-2\left\lceil\frac{v_{2}}{2}\right\rceil\ \textrm{ and }\
P^{2}(Y,X)=0.$

The first point of the theorem covers the case where valuations of the prize
for the two players are close to each other: the difference between them is
less than $1$ and the closest integer value not greater than half of each
valuation is the same for both of them. Similarly to the first point of
Theorem 3.1, each of the players uses a convex combination of the uniform
probability distribution on even numbers from $0$ to $2m$ and the uniform
probability distribution on odd numbers from $0$ to $2m+1$, where $m=\lfloor
v_{2}/2\rfloor$. This time, however, equilibrium is unique.

The second and the third point of the theorem cover the case where floors of
the valuations of the prize for the two players are not equal. The second
point covers the case where half of the valuation of the stronger player is
the smallest integer number greater than half of the valuation of the weaker
player. The weaker player has a unique equilibrium strategy: mixing uniformly
on even numbers between $0$ and $2\lfloor v_{2}/2\rfloor$. The stronger player
has continuum of equilibrium strategies depending on the fraction $\alpha$ by
which the expected value of the strategy of the stronger player exceeds the
expected value of the equilibrium strategy of the weaker player, $\lfloor
v_{2}/2\rfloor$. Payoff of the stronger player is equal to $1$ for all
equilibria. Payoff of the weaker player is positive unless $\alpha$ attains
its highest value. The third point covers the case where half of the valuation
of the stronger player exceed the ceiling of the half the valuation of the
weaker player. In this case there is a unique equilibrium strategy of the
weaker player: the player chooses effort $0$ with probability $1-b/m$ and,
with probability $b/m$, mixes uniformly on even numbers from $0$ to $2\lfloor
v_{2}/2\rfloor$. The stronger player has a continuum of equilibrium
strategies. All the equilibria are payoff equivalent and so equilibrium
payoffs are unique. Equilibrium payoff of the weaker player is $0$ and the
stronger player obtains a positive equilibrium payoff.

Notice that if the space of possible prize valuations, $v_{1}$ and $v_{2}$, is
a subset of real numbers then equilibrium payoffs are generically unique: the
cases where equilibrium payoffs are not unique require one of the players to
have a prize valuation which is an even number.

###  3.1 Comparison with continuous all-pay auction

In this section we compare equilibrium characterisation in discrete case with
equilibrium characterisation in the continuous case. The following result,
stated in [9] and [10] and rigorously proven in [2], provides full
characterisation of equilibria for continuous all-pay auction.

######  Theorem 3.3 (Hillman and Riley)

A strategy profile $(X,Y)$ is a Nash equilibrium of all-pay auction with
continuous strategies and players valuations $v_{1}\geq v_{2}>0$ if and only
if $X$ is distributed uniformly on the interval $[0,v_{2}]$ while $Y$ is
distributed on $[0,v_{2}]$ with a distribution with CDF
$F_{2}(x)=(x-v_{2})/v_{1}+1$. Equilibrium payoffs of the players are
$P_{\mathrm{cont.}}^{1}(X,Y)=v_{1}-v_{2}$ and $P_{\mathrm{cont.}}^{2}(X,Y)=0$.

One feature of equilibrium strategies that is present in both the discrete and
the continuous case is that the weaker player exerts zero effort with
probability close to $(v_{1}-v_{2})/2$ and then mixes with the remaining
probability on the interval close to $[0,v_{2}]$. In the case of the discrete
model the interval is $[0,v_{2}]$ when $v_{2}$ is even and $[0,2\lfloor
v_{2}/2\rfloor+1]$ when $v_{2}$ is not an even number. The probability
distribution on the interval, in the discrete case, is not uniform on discrete
values, in general. However it is a convex combination of probability
distributions which are uniform on odd number, uniform on even number, or
distorted such uniform distributions. In particular in the case of the
valuation of the weaker player, $v_{2}$, that is not even and is less than the
valuation of the stronger player by more than $2$ (point (iii) of Theorem 3.2)
the set of equilibrium strategies of the stronger player contains probability
distribution that is uniform on integer values in the interval $[0,2\lfloor
v_{2}/2\rfloor+1]$.

####  3.1.1 Comparative statics

Fixing the valuation of player $1$, we analyse the effect of increasing the
valuation of player $2$. Consider first the case when the valuation of player
$1$, $v_{1}$, is not an even number, illustrated in Figure 1. In this case
equilibrium payoffs of player $2$ are unique for all values of $v_{2}$. Notice
that the difference in payoffs under the discrete and continuous case is equal
to $0$ when $v_{2}\leq 2\lfloor v_{1}/2\rfloor$, equal to $v_{2}/2-\lfloor
v_{2}/2\rfloor$ when $2\lfloor v_{1}/2\rfloor<v_{2}\leq v_{1}$, equal to
$v_{1}-\lfloor v_{1}/2\rfloor-v_{2}/2$ when $v_{1}<v_{2}\leq 2\lceil
v_{1}/2\rceil$, and equal to $2(v_{1}/2-\lfloor v_{1}/2\rfloor-1/2)$ when
$v_{2}>2\lceil v_{1}/2\rceil$. In particular, discreteness of strategy space
benefits player $2$ when she is weaker but her valuation is close to the
valuation of player $1$: $v_{2}\in(2\lfloor v_{1}/2\rfloor,v_{1}]$. Depending
on whether the valuation of player $1$ is smaller or greater than the closest
odd integer number, the discreteness of strategy space disbenefits or benefits
player $2$, respectively, when she is stronger than player $1$, $v_{2}>v_{1}$.

Figure 1: Change in payoff of player $2$ when $v_{2}$ increases: the case of
$v_{1}$ not an even number. The case when $v_{1}/2-\lfloor v_{1}/2<1/2$ (left)
and the case of $v_{1}/2-\lfloor v_{1}/2\rfloor>1/2$ (right). Thick line
represents equilibrium payoffs in the discrete case and thin line represents
equilibrium payoffs in continuous case.

Second consider the case when valuation of player $1$, $v_{1}$, is an even
number, illustrated in Figure 2. In this case payoffs to player $2$ under the
discrete and the continuous case are equal when $v_{2}\leq 2\lfloor
v_{1}/2\rfloor$. When $v_{2}>2\lfloor v_{1}/2\rfloor$ there is a continuum of
possible equilibrium payoffs and, depending on the strategy chosen by player
$1$, player $2$ obtains lower or higher payoff under discrete strategy space
as compared to the payoff under the continous case.

Figure 2: Change in payoff of player $2$ when $v_{2}$ increases: the case of
$v_{1}$ an even number. Thin line represents payoffs in continuous case and
the are around it represents the set of equilibrium payoffs in the discrete
case.

##  4 Conclusions

In this paper we studied two player all-pay auctions with discrete strategies
of the players. We provided full characterisation of equilibria as well as of
equilibrium payoffs. We discussed how they are related to equilibria and
equilibrium payoffs in the continuous variant of all-pay auctions. We show
that equilibria in the discrete variant are not unique, in general, but the
are interchangeable. Equilibrium payoffs are unique, as long as none of the
players has valuation of the prize that is an even number. In case it is for
at least one of the players, there is continuum of possible equilibrium
payoffs. Equilibrium strategies involve convex combinations of uniform
distributions on even or odd numbers as well as distorted versions of such
distributions.

## References

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##  Appendix 0.A Appendix: Proofs

Before giving the proofs of the main results, we provide some useful
properties of function $H$. In particular we provide bounds on its values when
one of its arguments is a probability distribution constituting a building
block for equilibrium strategies.

For any strategies $X$ and $Y$,

$H(X,Y)=-H(Y,X)$ (4)

and for any strategies $X_{1}$, $X_{2}$, $Y$ and any $\lambda\in[0,1]$,

$H(\lambda X_{1}+(1-\lambda)X_{2},Y)=\lambda
H(X_{1},Y)+(1-\lambda)H(X_{2},Y).$ (5)

In addition, by (5),

$P(\lambda X_{1}+(1-\lambda)X_{2},Y)\\\ \quad\begin{aligned}
&{}=\frac{v}{2}\left(H(\lambda
X_{1}+(1-\lambda)X_{2},Y)-\left(\frac{2\mathbf{E}(\lambda
X_{1}+(1-\lambda)X_{2})}{v}-1\right)\right)\\\ &{}=\frac{v}{2}\left(\lambda
H(X_{1},Y)+(1-\lambda)H(X_{2},Y)-\left(\frac{2(\lambda\mathbf{E}(X_{1})+(1-\lambda)\mathbf{E}(X_{2}))}{v}-(\lambda+1-\lambda)\right)\right)\\\
&{}=\lambda P(X_{1},Y)+(1-\lambda)P(X_{2},Y)\end{aligned}$ (6)

As was shown in [8],

$H(X,Y)=1-\sum_{i=0}^{+\infty}\mathbf{P}(X=i)\left(\mathbf{P}(Y\geq
i)+\mathbf{P}(Y\geq i+1)\right).$ (7)

By (7),

$H(\mathbf{1}_{0},Y)=\mathbf{P}(Y=0)-1.$ (8)

We will use the following inequalities, shown in [8]: for any $Y$

$H(U_{\mathrm{O}}^{m},Y)=1-\left(\frac{1}{m}\right)\sum_{i=1}^{2m}\mathbf{P}(Y\geq
i)\geq 1-\frac{\mathbf{E}(Y)}{m},$ (9)

with equality if and only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y\geq j)=0$,
and

$H(U_{\mathrm{E}}^{m},Y)=1-\left(\frac{1}{m+1}\right)\left(1+\sum_{i=1}^{2m+1}\mathbf{P}(Y\geq
i)\right)\geq 1-\frac{\mathbf{E}(Y)+1}{m+1},$ (10)

with equality if and only if $\sum_{j=2m+2}^{+\infty}\mathbf{P}(Y\geq j)=0$.
In addition, we will use the following inequalities, shown in [6]: for every
$Y$, with $p_{i}=\mathbf{P}(Y=i)$:

$H(W_{j}^{m},Y)\geq 1-\frac{\mathbf{E}(Y)}{m}+\frac{p_{2j}-p_{0}}{2m}$ (11)

with equality if and only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y\geq j)=0$,

$H(U_{\mathrm{O}\uparrow 1}^{m},Y)\geq
1-\frac{\mathbf{E}(Y)-1}{m-1}-\frac{p_{0}}{m-1}$ (12)

with equality if and only if $\sum_{j=2m}^{+\infty}\mathbf{P}(Y\geq j)=0$, and

$H(V_{j}^{m},Y)\geq 1-\frac{2\mathbf{E}(Y)}{2m+1}+\frac{p_{2j-1}}{2m+1}$ (13)

with equality if and only if $\sum_{j=2m+2}^{+\infty}\mathbf{P}(Y\geq j)=0$.

Using (9) and (5), for any $\alpha\in[0,1]$,

$H((1-\alpha)U_{\mathrm{O}}^{m}+\alpha U_{\mathrm{O}}^{m+1},Y)\geq
1-\mathbf{E}(Y)\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)$ (14)

with equality if and only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y\geq j)=0$.

Using (9), (10), and (5), for any $\alpha\in[0,1]$,

$H((1-\alpha)U_{\mathrm{E}}^{m}+\alpha U_{\mathrm{O}}^{m+1},Y)\geq
1-\frac{\mathbf{E}(Y)+1}{m+1}+\frac{\alpha}{m+1}={}\\\
1-\mathbf{E}(Y)\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+(1-\alpha)\frac{\mathbf{E}(Y)-m}{m(m+1)}$
(15)

with equality if and only if $\sum_{j=2m+2}^{+\infty}\mathbf{P}(Y\geq j)=0$.

Using (9), (13), and (5), for any $\alpha\in[0,1]$ and $\delta=(2m+1)/(m+1)$,

$H(\alpha\delta V_{j}^{m}+(1-\alpha\delta)U_{\mathrm{O}}^{m},Y)\geq
1-\mathbf{E}(Y)\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{\alpha}{m+1}p_{2j-1}$
(16)

with equality if and only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y\geq j)=0$.

Using (10), (13), and (5), for any $\alpha\in[0,1]$ and $\delta=(2m+1)/(m+1)$,

$H(\alpha\delta V_{j}^{m}+(1-\alpha\delta)U_{\mathrm{E}}^{m},Y)\geq
1-\frac{\mathbf{E}(Y)+1}{m+1}\left(1+\frac{\alpha}{m+1}\right)+\frac{\alpha}{m+1}\left(2+p_{2j-1}\right)={}\\\
1-\mathbf{E}(Y)\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{\alpha}{m+1}p_{2j-1}+\frac{\mathbf{E}(Y)-m}{m+1}\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)$
(17)

with equality if and only if $\sum_{j=2m+2}^{+\infty}\mathbf{P}(Y\geq j)=0$.

Using (9), (13), and (5), for any $\alpha\in[0,1]$ and $\sigma=(2m+1)/m$,

$H((1-\alpha)\sigma V_{j}^{m}+(1-(1-\alpha)\sigma)U_{\mathrm{O}}^{m+1},Y)\geq
1-\mathbf{E}(Y)\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{1-\alpha}{m}p_{2j-1}$
(18)

with equality if and only if $\sum_{j=2m+2}^{+\infty}\mathbf{P}(Y\geq j)=0$.

The rest of the proof section is organized as follows. First we obtain
necessary and sufficient properties of equilibria of discrete all-pay auctions
for the following cases of expected values of the equilibrium strategies $X$
and $Y$:

  * •

$\mathbf{E}(X)=\mathbf{E}(Y)=m$ with $m\in\mathbb{Z}_{\geq 1}$ (Proposition
2),

  * •

$\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=m+\beta$ with
$m\in\mathbb{Z}_{\geq 0}$ and $\alpha,\beta\in(0,1)$ (Proposition 3),

  * •

$\mathbf{E}(X)=m$ and $\mathbf{E}(Y)=b$ with $m\in\mathbb{Z}_{\geq 1}$ and
$m>b>0$ (Propositions 4 and 5),

  * •

$\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=b$ with $m\in\mathbb{Z}_{\geq 1}$,
$m>b>0$, and $\alpha\in(0,1)$ (Propositions 6 and 7).

These cases cover all possible configurations of expected values of the
strategies in equilibrium strategy profiles. In the last section of the
appendix, we use these characterisations to provide proofs of the main
theorems 3.1 and 3.2.

### The case of $\mathbf{E}(X)=\mathbf{E}(Y)=m$ with $m\in\mathbb{Z}_{\geq 1}$

###### Proposition 2

A strategy profile $(X,Y)$ such that $\mathbf{E}(X)=\mathbf{E}(Y)=m$,
$m\in\mathbb{Z}_{\geq 1}$, is a Nash equilibrium of all pay auction with both
players valuations $v_{1}\geq v_{2}>0$ if and only if,

  * •

Either $m=\lfloor v_{2}/2\rfloor=\lfloor v_{1}/2\rfloor$ or $m=\lceil
v_{1}/2\rceil-1=\lceil v_{1}/2\rceil-1$ or $m=\lfloor v_{2}/2\rfloor=\lceil
v_{1}/2\rceil-1$,

  * •

$X=\lambda U_{\mathrm{O}}^{m}+(1-\lambda)U_{\mathrm{E}}^{m},\qquad Y=\kappa
U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m}.$

  * •

$\kappa=\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right)\textrm{ and
}\lambda=\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right).$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=\frac{v_{1}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\ \textrm{
and }\ P^{2}(Y,X)=\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor.$

###### Proof

For the left to right implication, suppose that $(X,Y)$ satisfying the
condition stated in the theorem is a Nash equilibrium of all pay auction with
both players valuations $v_{1},v_{2}>0$. Then, by Proposition 1, it is a Nash
equilibrium of general Lotto game $\Gamma(m,m)$. Hence, by [8, Theorem 2],

$X=\lambda U_{\mathrm{O}}^{m}+(1-\lambda)U_{\mathrm{E}}^{m},\qquad Y=\kappa
U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m},$ (19)

with $\lambda,\kappa\in[0,1]$, and

$H(X,Y)=0.$ (20)

Given $c\in(-m,m)$ let

$X^{\prime}=\gamma\mathbf{1}_{\lfloor m+c\rfloor}+(1-\gamma)\mathbf{1}_{\lceil
m+c\rceil},$

where

$\gamma=\begin{cases}1,&\textrm{if $m+c\in\mathbb{Z}$}\\\ \frac{\lceil
m+c\rceil-(m+c)}{\lceil m+c\rceil-\lfloor
m+c\rfloor},&\textrm{otherwise}\end{cases}$

so that $\mathbf{E}(X^{\prime})=m+c$. Since $c\in(-m,m)$ and
$m\in\mathbb{Z}_{\geq 1}$, so $\lfloor m+c\rfloor\geq 0$ and $\lceil
m+c\rceil\leq 2m$. Hence $\mathbf{P}(X^{\prime}\geq 2m+1)=0$. Thus, by (19),
(4), and (5),

$\displaystyle H(X^{\prime},Y)$ $\displaystyle=-\kappa
H(U_{\mathrm{O}}^{m},X^{\prime})-(1-\kappa)H(U_{\mathrm{E}}^{m},X^{\prime})$
(21)
$\displaystyle=\kappa\left(\frac{\mathbf{E}(X^{\prime})}{m}-1\right)+(1-\kappa)\left(\frac{\mathbf{E}(X^{\prime})+1}{m+1}-1\right)$
$\displaystyle=\mathbf{E}(X^{\prime})\frac{m+\kappa}{m(m+1)}+\frac{1-\kappa}{m+1}-1$
$\displaystyle=(m+c)\frac{m+\kappa}{m(m+1)}+\frac{1-\kappa}{m+1}-1.$

By (3), (20), and (21)

$\displaystyle P^{1}(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left((m+c)\frac{m+\kappa}{m(m+1)}+\frac{1-\kappa}{m+1}-\frac{2(m+c)}{v_{1}}\right),\textrm{
and}$ $\displaystyle P^{1}(X,Y)$
$\displaystyle=\frac{v_{1}}{2}\left(1-\frac{2m}{v_{1}}\right).$

Since $(X,Y)$ is a Nash equilibrium so $P(X,Y)\geq P(X^{\prime},Y)$. Hence

$\frac{v_{1}}{2}\left(1-\frac{2m}{v_{1}}\right)\geq\frac{v_{1}}{2}\left((m+c)\frac{m+\kappa}{m(m+1)}+\frac{1-\kappa}{m+1}-\frac{2(m+c)}{v_{1}}\right)$

and, since $v_{1}>0$, so

$\frac{2c}{v_{1}}\geq\frac{c(m+\kappa)}{m(m+1)}.$ (22)

Since $v_{1}>0$, $m\geq 0$, $\kappa\in(0,1)$, and (22) holds for any
$c\in(-m,m)$ so $v_{1}/2=m(m+1)/(m+\kappa)$. By analogous derivation we also
get $v_{2}/2=m(m+1)/(m+\lambda)$. From this we also get

$\kappa=\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right)\textrm{ and
}\lambda=\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right).$

Since $\kappa\in[0,1]$ so $m(m+1)/(m+\kappa)\in[m,m+1]$. Hence either
$m=\lfloor v_{1}/2\rfloor$ or $m=\lceil v_{1}/2\rceil-1$. Similarly, either
$m=\lfloor v_{2}/2\rfloor$ or $m=\lceil v_{2}/2\rceil-1$.

For the left to right implication, consider a strategy $X^{\prime}$ with
$\mathbf{E}(X^{\prime})\geq 0$ of player $1$. By (21) and (3),

$\displaystyle P^{1}(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(\mathbf{E}(X^{\prime})\frac{m+\kappa}{m(m+1)}+\frac{1-\kappa}{m+1}-1-\left(\frac{2\mathbf{E}(X^{\prime})}{v_{1}}-1\right)\right)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{1-\kappa}{m+1}+\mathbf{E}(X^{\prime})\left(\frac{m+\kappa}{m(m+1)}-\frac{2}{v_{1}}\right)\right).$

Similarly, by (20) and (3),

$P^{1}(X,Y)=\frac{v_{1}}{2}\left(1-\frac{2m}{v_{1}}\right)=\frac{v_{1}}{2}\left(\frac{1-\kappa}{m+1}+m\left(\frac{m+\kappa}{m(m+1)}-\frac{2}{v_{1}}\right)\right).$

Since $v_{1}/2=m(m+1)/(m+\kappa)$ so $P^{1}(X^{\prime},Y)=0=P^{1}(X,Y)$ and
there is no profitable deviation for player $1$ from $(X,Y)$. By analogous
derivation, using $v_{2}/2=m(m+1)/(m+\lambda)$, we conclude that there is no
profitable deviation for player $2$ from $(X,Y)$ either. Hence $(X,Y)$ is a
Nash equilibrium.

### The case of $\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=m+\beta$ with
$m\in\mathbb{Z}_{\geq 0}$ and $\alpha,\beta\in(0,1)$

###### Proposition 3

A strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=m+\beta$, $m\in\mathbb{Z}_{\geq 0}$, $0<\beta\leq\alpha<1$, is
a Nash equilibrium of all pay auction with both players valuations
$v_{1},v_{2}>0$ if and only if, $v_{1}/2=v_{2}/2=m+1$ and

$X=(1-\alpha)U_{\mathrm{E}}^{m}+\alpha U_{\mathrm{O}}^{m+1},\qquad
Y=(1-\beta)U_{\mathrm{E}}^{m}+\beta U_{\mathrm{O}}^{m+1}.$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=1-\beta\ \textrm{ and }\ P^{2}(Y,X)=1-\alpha.$

###### Proof

For the left to right implication, suppose that $(X,Y)$ satisfying the
condition stated in the theorem is a Nash equilibrium of all pay auction with
both players valuations $v_{1},v_{2}>0$. Then, by Proposition 1, it is a Nash
equilibrium of general Lotto game $\Gamma(m+\alpha,m+\beta)$. Hence, by [8,
Theorem 3],

$X=(1-\alpha)U_{\mathrm{E}}^{m}+\alpha U_{\mathrm{O}}^{m+1},\qquad
Y=(1-\beta)U_{\mathrm{E}}^{m}+\beta U_{\mathrm{O}}^{m+1},$ (23)

and

$H(X,Y)=\frac{\alpha-\beta}{m+1}.$ (24)

Given $c\in(-m-\alpha,m+1-\alpha)$ let

$X^{\prime}=\gamma\mathbf{1}_{\lfloor
m+\alpha+c\rfloor}+(1-\gamma)\mathbf{1}_{\lceil m+\alpha+c\rceil},$

where

$\gamma=\begin{cases}1,&\textrm{if $m+c+\alpha\in\mathbb{Z}$}\\\ \frac{\lceil
m+c+\alpha\rceil-(m+c+\alpha)}{\lceil m+c+\alpha\rceil-\lfloor
m+c+\alpha\rfloor},&\textrm{otherwise}\end{cases}$

so that $\mathbf{E}(X^{\prime})=m+\alpha+c$. Since
$c\in(-m-\alpha,m+1-\alpha)$, $m\in\mathbb{Z}_{\geq 0}$, and $\alpha\in(0,1)$
so $\lfloor m+c+\alpha\rfloor\geq 0$ and $\lceil m+c+\alpha\rceil\leq 2m+1$.
Hence $\mathbf{P}(X^{\prime}\geq 2m+2)=0$. Thus, by (23), (4), and (5),

$\displaystyle H(X^{\prime},Y)$
$\displaystyle=-\left(1-\beta\right)H(U_{\mathrm{E}}^{m},X^{\prime})-\beta
H(U_{\mathrm{O}}^{m+1},X^{\prime})$ (25)
$\displaystyle=\left(1-\beta\right)\left(\frac{\mathbf{E}(X^{\prime})+1}{m+1}-1\right)+\beta\left(\frac{\mathbf{E}(X^{\prime})}{m+1}-1\right)$
$\displaystyle=\frac{m+\alpha+c}{m+1}+\frac{1-\beta}{m+1}-1=\frac{\alpha-\beta+c}{m+1}.$

By (3), (24), and (25)

$\displaystyle P^{1}(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{\alpha-\beta+c}{m+1}-\left(\frac{2(m+c+\alpha)}{v_{1}}-1\right)\right),\textrm{
and}$ $\displaystyle P^{1}(X,Y)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{\alpha-\beta}{m+1}-\left(\frac{2(m+\alpha)}{v_{1}}-1\right)\right).$

Since $(X,Y)$ is a Nash equilibrium so $P(X,Y)\geq P(X^{\prime},Y)$. Hence

$\frac{v_{1}}{2}\left(\frac{\alpha-\beta}{m+1}-\left(\frac{2(m+\alpha)}{v_{1}}-1\right)\right)\geq\frac{v_{1}}{2}\left(\frac{\alpha-\beta+c}{m+1}-\left(\frac{2(m+c+\alpha)}{v_{1}}-1\right)\right)$

and, since $v_{1}>0$, so

$\frac{2c}{v_{1}}\geq\frac{c}{m+1}.$ (26)

Since $v_{1}>0$, $m\geq 0$, $\alpha\in(0,1)$, and (26) holds for any
$c\in(-m-\alpha,m+1-\alpha)$ so $v_{1}/2=m+1$. By analogous derivation we also
get $v_{2}/2=m+1$.

For the left to right implication, consider a strategy $X^{\prime}$ with
$\mathbf{E}(X^{\prime})\geq 0$ of player $1$. By (25) and (3),

$\displaystyle P^{1}(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{\mathbf{E}(X^{\prime})}{m+1}+\frac{1-\beta}{m+1}-1-\left(\frac{2\mathbf{E}(X^{\prime})}{v_{1}}-1\right)\right)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{1-\beta}{m+1}+\mathbf{E}(X^{\prime})\left(\frac{1}{m+1}-\frac{2}{v_{1}}\right)\right).$

Similarly, by (24) and (3),

$\displaystyle P^{1}(X,Y)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{\alpha-\beta}{m+1}-\left(\frac{2(m+\alpha)}{v_{1}}-1\right)\right)$
$\displaystyle=\frac{v_{1}}{2}\left(\frac{1-\beta}{m+1}+(m+\alpha)\left(\frac{1}{m+1}-\frac{2}{v_{1}}\right)\right)$

Since $v_{1}/2=m+1$ so $P^{1}(X^{\prime},Y)=0=P^{1}(X,Y)$ and there is no
profitable deviation for player $1$ from $(X,Y)$. By analogous derivation,
using $v_{2}/2=m+1$, we conclude that there is no profitable deviation for
player $2$ from $(X,Y)$ either. Hence $(X,Y)$ is a Nash equilibrium.

### The case of $\mathbf{E}(X)=m$ and $\mathbf{E}(Y)=b$ with
$m\in\mathbb{Z}_{\geq 1}$ and $m>b>0$

###### Proposition 4

A strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m=1$,
$\mathbf{E}(Y)=b\in(0,1)$, is a Nash equilibrium of all pay auction with both
players valuations $v_{1},v_{2}>0$ if and only if,

$v_{1}\geq v_{2}=2,\quad b\in\left(0,\frac{4}{v_{1}}\right]$
$X=U_{\mathrm{O}}^{1},\quad Y=(1-b)\mathbf{1}_{0}+b\left(\lambda
U_{\mathrm{O}}^{1}+(1-\lambda)U_{\mathrm{E}}^{1}\right),$

where

$\frac{4}{bv_{1}}-\frac{2}{b}+1\leq\lambda\leq\frac{4}{bv_{1}}-1.$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=\frac{v_{1}}{2}(2-b)-1\ \textrm{ and }\ P^{2}(Y,X)=0.$

###### Proposition 5

A strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m$, $\mathbf{E}(Y)=b$,
$m\in\mathbb{Z}_{\geq 2}$, and $m>b>0$, is a Nash equilibrium of all pay
auction with both players valuations $v_{1},v_{2}>0$ if and only if,

$v_{1}\geq v_{2},\quad m=\frac{v_{2}}{2},\quad
b\in\left[\frac{v_{2}(v_{2}-2)}{2v_{1}},\frac{v_{2}(v_{2}+2)}{2v_{1}}\right],$
$X=U_{\mathrm{O}}^{m},\quad
Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)Z,$

where

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}$

with

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m-1}\geq 0\textrm{ and
}\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1,$ $\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{v_{2}^{2}}{2v_{1}b}-1,$

and

$\lambda_{\mathrm{O}}\geq\left(\frac{v_{2}}{2b}\right)\left(\frac{v_{2}(v_{2}+2)}{2v_{1}}+b-v_{2}\right)$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=\frac{v_{1}-v_{2}}{2}+\frac{v_{1}}{v_{2}}\left(\frac{v_{2}}{2}-b\right)\
\textrm{ and }\ P^{2}(Y,X)=0.$

###### Lemma 1

If a strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m$, $\mathbf{E}(Y)=b$,
$m\in\mathbb{Z}_{\geq 1}$, and $m>b>0$, is a Nash equilibrium of all pay
auction with both players valuations $v_{1},v_{2}>0$ then $m=v_{2}/2$, $(X,Y)$
is a Nash equilibrium of $\Gamma(m,b)$, and $X=U_{\mathrm{O}}^{m}$.

###### Proof

Since $(X,Y)$ is a Nash equilibrium of all pay auction so, by Proposition 1,
it is a Nash equilibrium of general Lotto game $\Gamma(m,b)$. Hence, by [8,
Theorem 2],

$X=U_{\mathrm{O}}^{m}$ (27)

and

$H(X,Y)=1-\frac{b}{m}.$ (28)

Given $c\in(-b,2m-b)$ let

$Y^{\prime}=\gamma\mathbf{1}_{\lfloor b+c\rfloor}+(1-\gamma)\mathbf{1}_{\lceil
b+c\rceil},$

where

$\gamma=\begin{cases}1,&\textrm{if $b+c\in\mathbb{Z}$}\\\ \frac{\lceil
b+c\rceil-(b+c)}{\lceil b+c\rceil-\lfloor
b+c\rfloor},&\textrm{otherwise}\end{cases}$

so that $\mathbf{E}(X^{\prime})=b+c$. Since $c\in(-b,2m-b)$ and
$m\in\mathbb{Z}_{\geq 1}$ so $\lceil b+c\rceil\leq 2m$ and
$\mathbf{P}(X^{\prime}\geq 2m+1)=0$. Thus, by (27), (9), and (4),

$H(Y^{\prime},X)=\frac{b+c}{m}-1.$ (29)

By (28), (29), and (3),

$\displaystyle P(Y^{\prime},X)$
$\displaystyle=\frac{v_{2}}{2}(b+c)\left(\frac{1}{m}-\frac{2}{v_{2}}\right),\textrm{
and}$ $\displaystyle P(Y,X)$
$\displaystyle=\frac{v_{2}}{2}b\left(\frac{1}{m}-\frac{2}{v_{2}}\right)$

Since $(X,Y)$ is a Nash equilibrium so $P(Y,X)\geq P(Y^{\prime},X)$. Hence

$\frac{v_{2}}{2}b\left(\frac{1}{m}-\frac{2}{v_{2}}\right)\geq\frac{v_{2}}{2}(b+c)\left(\frac{1}{m}-\frac{2}{v_{2}}\right)$

and, consequently,

$\frac{2c}{v_{2}}\geq\frac{c}{m}.$ (30)

Since $v_{2}>0$, $b>0$, $m>b$, and (30) holds for all $c\in(-b,2m-b)$ so
$m=v_{2}/2$.

###### Lemma 2

If a strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m$, $\mathbf{E}(Y)=b$,
$m\in\mathbb{Z}_{\geq 1}$, and $m>b>0$, is a Nash equilibrium of all pay
auction with both players valuations $v_{1},v_{2}>0$ then
$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)Z$, where

  * •

if $m=1$ then

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}$

with

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}}\geq 0\textrm{,
}\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}=1,$
$\frac{\lambda_{\mathrm{E}}}{2}\geq 1-\frac{2}{bv_{1}}.$ (31)

and

$\frac{\lambda_{\mathrm{E}}}{2}\leq\frac{1}{b}-\frac{2}{bv_{1}}.$ (32)

  * •

if $m\geq 2$ then

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}$

with

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m-1}\geq 0\textrm{,
}\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1,$ (33)
$\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\leq\frac{m}{b}\left(1-\frac{v_{2}}{v_{1}}\right),$
(34)
$\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}=\frac{2m}{bv_{1}}.$ (35)

and, in the case of $b>m-1$,

$\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\leq\frac{m-b}{b}.$ (36)

###### Proof

Take any strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m$,
$\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq 1}$, and $m\geq b>0$, and suppose
that it is a Nash equilibrium of all pay auction with both players valuations
$v_{1},v_{2}>0$.

Since $(X,Y)$ is a Nash equilibrium of all pay auction so, by Proposition 1,
it is a Nash equilibrium of general Lotto game $\Gamma(m,b)$. Thus, by [8,
Theorem 2] and [6, Theorems 3 and 4],

$H(X,Y)=1-\frac{b}{m},$ (37) $X=U_{\mathrm{O}}^{m},$ (38)
$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)Z,$ (39)

with
$Z\in\mathrm{conv}\left(\mathcal{U}^{m}\cup\mathcal{W}^{m}\cup\\{U_{\mathrm{O}\uparrow
1}^{m}\\}\right)$, in the case of $m\geq 2$, and $Z\in\mathrm{conv}\
\mathcal{U}^{m}$, in the case of $m=1$. In addition, in the case of $m\geq 2$
and $b>m-1$, by [6, Theorems 4] (in particular (26) in proof of the theorem),

$\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\leq\frac{m-b}{b}.$

By (39),

$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)\left(\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}\right),$

where $\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m}\geq 0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1$, and $\lambda_{\mathrm{O}\uparrow 1}=0$, in
the case of $m=1$. In the case of $m\geq 2$, by (5), (9), (10), (11), and
(12), for any strategy of the first player, $X^{\prime}$

$\displaystyle H(Y,X^{\prime})$
$\displaystyle=\left(1-\frac{b}{m}\right)H(\mathbf{1}_{0},X^{\prime})+\frac{b}{m}\Bigg{(}\lambda_{\mathrm{O}}H(U_{\mathrm{O}}^{m},X^{\prime})+\lambda_{\mathrm{E}}H(U_{\mathrm{E}}^{m},X^{\prime})+\lambda_{\mathrm{O}\uparrow
1}H(U_{\mathrm{O}\uparrow 1}^{m},X^{\prime})+{}$ (40)
$\displaystyle\qquad\sum_{j=1}^{m-1}\lambda_{j}H(W_{j}^{m},X^{\prime})\Bigg{)}$
$\displaystyle\geq\left(1-\frac{b}{m}\right)(\mathbf{P}(X^{\prime}=0)-1)+\frac{b}{m}\Bigg{(}\lambda_{\mathrm{O}}\left(1-\frac{\mathbf{E}(X^{\prime})}{m}\right)$
$\displaystyle\quad{}+\lambda_{\mathrm{E}}\left(1-\frac{\mathbf{E}(X^{\prime})+1}{m+1}\right)+\lambda_{\mathrm{O}\uparrow
1}\left(1-\frac{\mathbf{E}(X^{\prime})-1}{m-1}-\frac{\mathbf{P}(X^{\prime}=0)}{m-1}\right)$
$\displaystyle\quad{}+\sum_{j=1}^{m-1}\lambda_{j}\left(1-\frac{\mathbf{E}(X^{\prime})}{m}+\frac{\mathbf{P}(X^{\prime}=2j)-\mathbf{P}(X^{\prime}=0)}{2m}\right)\Bigg{)}$
$\displaystyle=-1+\frac{b}{m}\Bigg{(}2-\mathbf{E}(X^{\prime})\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)-\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\Bigg{)}$
$\displaystyle\qquad{}+\left(1-\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\right)\mathbf{P}(X^{\prime}=0)+\frac{\sum_{j=1}^{m-1}\lambda_{j}\mathbf{P}(X^{\prime}=2j)}{2m}$

with equality only if $\sum_{j=2m}^{+\infty}\mathbf{P}(X^{\prime}\geq j)=0$.
In the case of $m=1$, by (5), (9), and (10), for any strategy of the first
player, $X^{\prime}$

$\displaystyle H(Y,X^{\prime})$
$\displaystyle=\left(1-\frac{b}{m}\right)H(\mathbf{1}_{0},X^{\prime})+\frac{b}{m}\left(\lambda_{\mathrm{O}}H(U_{\mathrm{O}}^{m},X^{\prime})+\lambda_{\mathrm{E}}H(U_{\mathrm{E}}^{m},X^{\prime})\right)$
(41)
$\displaystyle\geq\left(1-\frac{b}{m}\right)(\mathbf{P}(X^{\prime}=0)-1)+\frac{b}{m}\left(\lambda_{\mathrm{O}}\left(1-\frac{\mathbf{E}(X^{\prime})}{m}\right)+\lambda_{\mathrm{E}}\left(1-\frac{\mathbf{E}(X^{\prime})+1}{m+1}\right)\right)$
$\displaystyle=-1+\frac{b}{m}\left(2-\mathbf{E}(X^{\prime})\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}\right)-\frac{\lambda_{\mathrm{E}}}{m+1}\right)+\left(1-\frac{b}{m}\right)\mathbf{P}(X^{\prime}=0)$

with equality only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(X^{\prime}\geq j)=0$.

Using (38), (40), (41), (3), and (4), in the case of $m\geq 2$,

$\displaystyle P(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(H(X^{\prime},Y)-\left(\frac{2\mathbf{E}(X^{\prime})}{v_{1}}-1\right)\right)=\frac{v_{1}}{2}\left(1-H(Y,X^{\prime})-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\right)$
(42)
$\displaystyle\leq\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}\Bigg{(}2-\mathbf{E}(X^{\prime})\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)-\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\Bigg{)}$
$\displaystyle\qquad{}-\left(1-\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\right)\mathbf{P}(X^{\prime}=0)$
$\displaystyle\qquad{}-\frac{\sum_{j=1}^{m-1}\lambda_{j}\mathbf{P}(X^{\prime}=2j)}{2m}-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\Bigg{)},$

with equality only if $\sum_{j=2m}^{+\infty}\mathbf{P}(X^{\prime}\geq j)=0$,
and in the case of $m=1$,

$\displaystyle P(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(H(X^{\prime},Y)-\left(\frac{2\mathbf{E}(X^{\prime})}{v_{1}}-1\right)\right)=\frac{v_{1}}{2}\left(1-H(Y,X^{\prime})-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\right)$
(43)
$\displaystyle\leq\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}\Bigg{(}2-\mathbf{E}(X^{\prime})\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}\right)-\frac{\lambda_{\mathrm{E}}}{m+1}\Bigg{)}$
$\displaystyle\qquad{}-\left(1-\frac{b}{m}\right)\mathbf{P}(X^{\prime}=0)-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\Bigg{)},$

with equality only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(X^{\prime}\geq j)=0$.

On the other hand, by (3) and (37),

$\displaystyle P(X,Y)$
$\displaystyle=\frac{v_{1}}{2}\left(2-\frac{b}{m}-\frac{2m}{v_{1}}\right).$

In the case of $m=1$, let $X^{\prime}=\mathbf{1}_{2}$. Since
$\mathbf{P}(X^{\prime}=0)=0$ and
$\sum_{j=2m}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$ so, by (41),

$\displaystyle P(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(2-\frac{b}{m}\left(2-2\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}\right)-\frac{\lambda_{\mathrm{E}}}{m+1}\right)-\frac{4}{v_{1}}\right),$

Since $m=1$ and $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}=1$ so this can be
rewritten as

$\displaystyle P(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(2-b\left(2-2\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{4}{v_{1}}\right)$
$\displaystyle=\frac{v_{1}}{2}\left(2-b-\frac{2}{v_{1}}+b\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{2}{v_{1}}\right)$

Since $(X,Y)$ is a Nash equilibrium so $P(X,Y)\geq P(X^{\prime},Y)$ and so

$b\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{2}{v_{1}}\leq 0$

from which it follows that

$\frac{\lambda_{\mathrm{E}}}{2}\geq 1-\frac{2}{bv_{1}}.$

In the case of $m\geq 2$, given $c\in[-(m-1),m-1]$ let

$X^{\prime}=\gamma\mathbf{1}_{2\lfloor\frac{m+c+1}{2}\rfloor-1}+(1-\gamma)\mathbf{1}_{2\lceil\frac{m+c+1}{2}\rceil-1}$

where

$\gamma=\begin{cases}1,&\textrm{if $m+c+1$ is an even number},\\\
\frac{\lceil\frac{m+c+1}{2}\rceil-\frac{m+c+1}{2}}{\lceil\frac{m+c+1}{2}\rceil-\lfloor\frac{m+c+1}{2}\rfloor},&\textrm{otherwise}.\end{cases}$

Notice that $\mathbf{E}(X^{\prime})=m+c$, $\mathbf{P}(X^{\prime}=2j)=0$ for
all $j\geq 0$, and, since $c\in[-(m-1),m-1]$,
$\sum_{j=2m}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$. Hence, by (41),

$\displaystyle P(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}\Bigg{(}2-(m+c)\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)$
$\displaystyle\qquad{}-\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\Bigg{)}-\frac{2(m+c)}{v_{1}}\Bigg{)},$

Since $(X,Y)$ is a Nash equilibrium so $P(X,Y)\geq P(X^{\prime},Y)$ and so

$2-\frac{b}{m}-\frac{2m}{v_{1}}\geq
2-\frac{b}{m}\Bigg{(}2-(m+c)\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)\\\
{}-\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\Bigg{)}-\frac{2(m+c)}{v_{1}}$

which can be rewritten as

$\frac{2c}{v_{1}}+\frac{b}{m}\geq\frac{b(m+c)}{m}\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)+\frac{b}{m}\left(\frac{\lambda_{\mathrm{E}}}{m+1}-\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\right)$

and further as

$\frac{2c}{v_{1}}+\frac{b}{m}\geq\frac{bc}{m}\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)+\frac{b}{m}\left(\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}\right).$

Since $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1$ so this can be further rewritten as

$\frac{2c}{v_{1}}\geq\frac{bc}{m}\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right).$

Since $v_{1}>0$, $b>0$, $m>b$, and (30) holds for all $c\in(-b,2m-b)$ so

$\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}=\frac{2m}{bv_{1}}.$

Consider a strategy $X^{\prime}=\mathbf{1}_{0}$ of the first player. Since
$\mathbf{E}(X^{\prime}=0)$, $\mathbf{P}(X^{\prime}=0)$,
$\mathbf{P}(X^{\prime}=2j)=0$ for all $j>0$, and
$\sum_{j=2m}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$ so, by (42), in the case
of $m\geq 2$

$\displaystyle P(\mathbf{1}_{0},Y)$
$\displaystyle=\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}\Bigg{(}2-\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\Bigg{)}-1+\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\Bigg{)}$
$\displaystyle=\frac{v_{1}}{2}\left(1-\frac{b}{m}+\frac{b}{m}\left(\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\right)$

and, by (43), in the case of $m=1$,

$\displaystyle P(\mathbf{1}_{0},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(2-\frac{b}{m}\left(2-\frac{\lambda_{\mathrm{E}}}{m+1}\right)-1+\frac{b}{m}\right)$
$\displaystyle=\frac{v_{1}}{2}\left(1-\frac{b}{m}+\frac{b}{m}\left(\frac{\lambda_{\mathrm{E}}}{m+1}\right)\right)$

Since $(X,Y)$ is a Nash equilibrium so $P(X,Y)\geq P(X^{\prime},Y)$ and so, in
the case of $m\geq 2$,

$2-\frac{b}{m}-\frac{2m}{v_{1}}\geq
1-\frac{b}{m}+\frac{b}{m}\left(\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)$

from which (using $m=v_{2}/2$ from Lemma 1) it follows that

$\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\leq\frac{m}{b}\left(1-\frac{v_{2}}{v_{1}}\right).$

Similarly, in the case of $m=1$ we get

$\frac{\lambda_{\mathrm{E}}}{m+1}\leq\frac{m}{b}\left(1-\frac{v_{2}}{v_{1}}\right).$

Since $m=1$ and, by Lemma 1, $v_{2}=2m=2$ so this can be rewritten as

$\frac{\lambda_{\mathrm{E}}}{2}\leq\frac{1}{b}-\frac{2}{bv_{1}}.$

###### Lemma 3

If conditions in Lemma 1 and Lemma 2 are satisfied for a strategy profile
$(X,Y)$ such that $\mathbf{E}(X)=m$, $\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq
1}$, and $m\geq b>0$, then $(X,Y)$ is a Nash equilibrium of all pay auction
with both players valuations $v_{1},v_{2}>0$.

###### Proof

Take any strategy profile $(X,Y)$, as described in the lemma. By Lemma 1,
$(X,Y)$ is a Nash equilibrium of General Lotto game $\Gamma(m,b)$. Hence, by
[8, Theorem 2],

$H(X,Y)=1-\frac{b}{m}.$ (44)

We show first that there is no profitable deviation for player $1$. By (3) and
(44),

$P(X,Y)=\frac{v_{1}}{2}\left(2-\frac{b}{m}-\frac{2m}{v_{1}}\right).$ (45)

Take any strategy $X^{\prime}$ of player $1$ with $\mathbf{E}(X^{\prime})\geq
0$. We will consider the cases of $m\geq 2$ and $m=1$ separately.

Suppose first that $m\geq 2$. By (42),

$\displaystyle P(X^{\prime},Y)$
$\displaystyle\leq\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}\Bigg{(}2-\mathbf{E}(X^{\prime})\left(\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}\right)-\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}\Bigg{)}$
$\displaystyle\qquad{}-\left(1-\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\right)\mathbf{P}(X^{\prime}=0)-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\Bigg{)}.$

By Lemma 2, (33) and (35),

$\frac{1-\lambda_{\mathrm{E}}-\lambda_{\mathrm{O}\uparrow
1}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}=\frac{2m}{bv_{1}}$

from which it follows that

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{2m^{2}}{bv_{1}}-1.$ (46)

By (27) and Lemma 2, (35),

$\displaystyle P(X^{\prime},Y)$
$\displaystyle\leq\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}\left(2-\mathbf{E}(X^{\prime})\left(\frac{2m}{bv_{1}}\right)+\frac{2m^{2}}{bv_{1}}-1\right)$
$\displaystyle\qquad{}-\left(1-\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\right)\mathbf{P}(X^{\prime}=0)-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\Bigg{)}$
$\displaystyle=\frac{v_{1}}{2}\Bigg{(}2-\frac{b}{m}-\frac{2m}{v_{1}}-\left(1-\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\right)\right)\mathbf{P}(X^{\prime}=0)\Bigg{)}$

Lemma 2, (34), together with (46) yields

$\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\right)\leq
1-\frac{v_{2}}{v_{1}}+\frac{2m}{v_{1}}$

which, together with $m=v_{2}/2$ (from Lemma 1), yields

$1-\frac{b}{m}\left(1+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\right)\geq 0.$

Hence

$\displaystyle P(X^{\prime},Y)$
$\displaystyle\leq\frac{v_{1}}{2}\left(2-\frac{b}{m}-\frac{2m}{v_{1}}\right)=P(X,Y)$

and so player $1$ has no profitable deviation from $(X,Y)$.

Next, suppose that $m=1$. Since $\mathbf{E}(X^{\prime})=\sum_{j\geq
1}\mathbf{P}(X^{\prime}\geq j)\geq\mathbf{P}(X^{\prime}\geq
1)=1-\mathbf{P}(X^{\prime}=0)$ so
$\mathbf{P}(X^{\prime}=0)\geq\max(0,1-\mathbf{E}(X^{\prime}))$. This, together
with (43) and $m=1$ yields,

$\displaystyle P(X^{\prime},Y)$
$\displaystyle\leq\frac{v_{1}}{2}\Bigg{(}2-b\Bigg{(}2-\mathbf{E}(X^{\prime})\left(\lambda_{\mathrm{O}}+\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{\lambda_{\mathrm{E}}}{2}\Bigg{)}$
$\displaystyle\qquad{}-\left(1-b\right)\max(0,1-\mathbf{E}(X^{\prime}))-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\Bigg{)},$

By Lemma 2, $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}=1$. Hence

$P(X^{\prime},Y)\leq\frac{v_{1}}{2}\left(2-b-\frac{2}{v_{1}}+(\mathbf{E}(X^{\prime})-1)\left(b\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{2}{v_{1}}\right)-\left(1-b\right)\max(0,1-\mathbf{E}(X^{\prime}))\right),$

Suppose that $\mathbf{E}(X^{\prime})\geq 1$. Then, by Lemma 2, (31),

$b\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{2}{v_{1}}\leq 0$

and so

$\displaystyle
P(X^{\prime},Y)\leq\frac{v_{1}}{2}\left(2-b-\frac{2}{v_{1}}\right)=P(X,Y).$

Suppose that $\mathbf{E}(X^{\prime})<1$. Then,

$P(X^{\prime},Y)\leq\frac{v_{1}}{2}\left(2-b-\frac{2}{v_{1}}-(1-\mathbf{E}(X^{\prime}))\left(b\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{2}{v_{1}}+1-b\right)\right)$

and, by Lemma 2, (32)

$b\left(1-\frac{\lambda_{\mathrm{E}}}{2}\right)-\frac{2}{v_{1}}+1-b\geq 0.$

Hence

$P(X^{\prime},Y)\leq\frac{v_{1}}{2}\left(2-b-\frac{2}{v_{1}}\right)=P(X,Y).$

Thus player $1$ has no profitable deviation from $(X,Y)$.

Second, we show that there is no profitable deviation for player $2$. By Lemma
1, $m=v_{2}/2$. Thus, by (44), (4), (3),

$\displaystyle
P(Y,X)=\frac{v_{2}}{2}b\left(\frac{1}{m}-\frac{2}{v_{2}}\right)=0.$

Take any strategy $Y^{\prime}$ of player $2$ with $\mathbf{E}(Y^{\prime})\geq
0$. By Lemma 1, $X=U_{\mathrm{O}}^{m}$ and, by (9) and (4),

$H(Y^{\prime},X)\leq\frac{\mathbf{E}(Y^{\prime})}{m}-1.$

Thus, by (3) and $m=v_{2}/2$,

$\displaystyle P(Y^{\prime},X)$
$\displaystyle=\frac{v_{2}}{2}\left(H(Y^{\prime},X)-\left(\frac{2\mathbf{E}(Y^{\prime})}{v_{2}}-1\right)\right)$
$\displaystyle\leq\frac{v_{2}}{2}\left(\frac{\mathbf{E}(Y^{\prime})}{m}-\frac{2\mathbf{E}(Y^{\prime})}{v_{2}}\right)=0=P(Y,X).$

Hence player $2$ has no profitable deviation from $(X,Y)$.

######  Proof (Proof of Proposition 4)

Lemmas 1, 2, and 3 establish the sufficient and necessary properties of
equilibrium strategies. The condition on $v_{1}$, $v_{2}$, and $b$, follow
from the lemmas. By Lemma 1, $v_{2}=2m=2$. By Lemma 2, $\lambda\in[0,1]$ and

$\frac{1-\lambda}{2}\leq\frac{1}{b}-\frac{2}{bv_{1}}.$

Hence

$\frac{1}{b}-\frac{2}{bv_{1}}=\frac{v_{2}-2}{bv_{1}}\geq 0$

and so $v_{1}\geq 2=v_{2}$. Similarly, by Lemma 2,

$\frac{1-\lambda}{2}\geq 1-\frac{2}{bv_{1}}.$

Hence

$\frac{bv_{1}-2}{bv_{1}}\leq\frac{1}{2}$

from which it follows that $b\leq 4/v_{1}$.

######  Proof (Proof of Proposition 5)

Lemmas 1, 2, and 3 establish the sufficient and necessary properties of
equilibrium strategies.

By Lemma 2,

$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1$ (47)

and

$\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}=\frac{2m}{bv_{1}}.$ (48)

This, together with $m=v_{2}/2$, by Lemma 1, yields

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{v_{2}^{2}}{2bv_{1}}-1.$ (49)

Note that (49) together with

$\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\leq\frac{v_{2}}{2b}\left(1-\frac{v_{2}}{v_{1}}\right),$

holds if and only if (49) together with

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\leq\frac{v_{2}}{2b}-1$ (50)

holds.

Using (47), (49), and $m=v_{2}/2$, the left hand side of (50) can be rewritten
as

$\displaystyle\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}$
$\displaystyle=\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1-\lambda_{\mathrm{O}}-\lambda_{\mathrm{E}}-\lambda_{\mathrm{O}\uparrow
1}}{2m}=\frac{m+1}{2m}\left(\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}\right)-\frac{\lambda_{\mathrm{O}}}{2m}+\frac{1}{2m}$
$\displaystyle=\frac{m+1}{2m}\left(\frac{v_{2}^{2}}{2bv_{1}}-\frac{m}{m+1}\right)-\frac{\lambda_{\mathrm{O}}}{2m}=\frac{v_{2}+2}{2v_{2}}\left(\frac{v_{2}^{2}}{2bv_{1}}-\frac{v_{2}}{v_{2}+2}\right)-\frac{\lambda_{\mathrm{O}}}{v_{2}}.$

Thus (49) together with (50) holds if and only if (49) and

$\lambda_{\mathrm{O}}\geq\left(\frac{v_{2}}{2b}\right)\left(\frac{v_{2}(v_{2}+2)}{2v_{1}}+b-v_{2}\right)$

holds.

Now we show how the conditions on $v_{1}$, $v_{2}$, and $b$, follow from the
Lemmas 1, 2, and 3. By Lemma 2,

$\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{2m}\leq\frac{m}{b}\left(1-\frac{v_{2}}{v_{1}}\right).$
(51)

Since, by Lemma 2, $\lambda_{\mathrm{E}}\geq 0$, and $\lambda_{1}\geq 0$, …,
$\lambda_{m-1}\geq 0$, so

$\frac{m}{b}\left(1-\frac{v_{2}}{v_{1}}\right)=\frac{m}{b}\left(\frac{v_{1}-v_{2}}{v_{1}}\right)\geq
0$

Since $m>b>0$ so it follows that $v_{1}\geq v_{2}$.

For the conditions on $b$, by (48),

$b=\left(\frac{2m}{v_{1}}\right)\left(\frac{1}{\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}}\right).$

Since $\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m-1}\geq 0$ and
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+sum_{j=1}^{m-1}\lambda_{j}=1$ so

$\frac{\lambda_{\mathrm{O}}}{m}+\frac{\lambda_{\mathrm{E}}}{m+1}+\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{\sum_{j=1}^{m-1}\lambda_{j}}{m}$

is maximised when $\lambda_{\mathrm{O}\uparrow 1}=1$ and it is minimised when
$\lambda_{\mathrm{E}}=1$. Hence

$b\in\left[\frac{2m(m-1)}{v_{1}},\frac{2m(m+1)}{v_{1}}\right]=\left[\frac{v_{2}(v_{2}-2)}{2v_{1}},\frac{v_{2}(v_{2}+2)}{2v_{1}}\right].$

### The case of $\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=b$ with
$m\in\mathbb{Z}_{\geq 1}$, $m>b>0$, and $\alpha\in(0,1)$

###### Proposition 6

A strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq 1}$, $m>b>0$, and $\alpha\in(0,1)$,
is a Nash equilibrium of all pay auction with both players valuations
$v_{1},v_{2}>0$ if and only if,

$\frac{v_{1}}{2}>\left\lceil\frac{v_{2}}{2}\right\rceil>\frac{v_{2}}{2},\quad
m=\left\lceil\frac{v_{2}}{2}\right\rceil-1,\quad
b=\frac{\left(\left\lceil\frac{v_{2}}{2}\right\rceil-1\right)\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{1}}{2}},\quad\alpha=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)$
$X\in\mathrm{conv}(\mathcal{U}^{m,\alpha}\cup\mathcal{X}^{m,\alpha}),\quad
Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)U_{\mathrm{E}}^{m},$

  * •

$\mathcal{U}^{m,\alpha}=\left\\{(1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1}\right\\}$,

and

  * •

$\mathcal{X}^{m,\alpha}=\alpha\delta\mathcal{V}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}$,
if $v_{2}/2\leq\lceil v_{2}/2\rceil-1/2$,

  * •

$\mathcal{X}^{m,\alpha}=(1-\alpha)\sigma\mathcal{V}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}$,
if $v_{2}/2>\lceil v_{2}/2\rceil-1/2$,

where

$\delta=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil},\quad\sigma=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil-1}.$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=v_{1}+1-2\left\lceil\frac{v_{2}}{2}\right\rceil\ \textrm{ and }\
P^{2}(Y,X)=0.$

###### Proposition 7

A strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=m$, $m\in\mathbb{Z}_{\geq 1}$, and $\alpha\in(0,1)$, is a Nash
equilibrium of all pay auction with both players valuations $v_{1},v_{2}>0$ if
and only if,

$\frac{v_{1}}{2}=\left\lceil\frac{v_{2}}{2}\right\rceil,\quad
m=\left\lceil\frac{v_{2}}{2}\right\rceil-1,\quad\alpha\in\left(0,\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)\right],\quad
Y=U_{\mathrm{E}}^{m},$

  * •

if
$0<\alpha\leq\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}$
then

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right),$

where

$\delta=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil},$

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$,
and

$\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

  * •

if
$\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}<\alpha\leq\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)$
then

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}\right),$

where

$\sigma=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil-1},$

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m}\geq
0$, $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}=1$,
and

$\lambda_{\mathrm{E}}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

Equilibrium payoffs of the players are

$P^{1}(X,Y)=1\ \textrm{ and }\
P^{2}(Y,X)=1-\frac{v_{2}}{v_{1}}\alpha-\frac{v_{1}-v_{2}}{2}.$

###### Lemma 4

If a strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq 1}$, and $m\geq b>0$, is a Nash
equilibrium of all pay auction with both players valuations $v_{1},v_{2}>0$
then $m(m+1)/b=v_{1}/2$, $(X,Y)$ is a Nash equilibrium of
$\Gamma(m+\alpha,b)$, and

$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)U_{\mathrm{E}}^{m}.$

###### Proof

Since $(X,Y)$ is a Nash equilibrium of all pay auction so, by Proposition 1,
it is a Nash equilibrium of general Lotto game $\Gamma(m+\alpha,b)$. Hence, by
[8, Theorem 4],

$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)U_{\mathrm{E}}^{m}$
(52)

and

$H(X,Y)=1-\frac{(1-\alpha)b}{m}-\frac{\alpha
b}{m+1}=1-\frac{b}{m}+\frac{\alpha b}{m(m+1)}.$ (53)

Given $c\in(1-m-\alpha,m+1-\alpha)$ let

$X^{\prime}=\gamma\mathbf{1}_{\lfloor
m+\alpha+c\rfloor}+(1-\gamma)\mathbf{1}_{\lceil m+\alpha+c\rceil},$

where

$\gamma=\begin{cases}1,&\textrm{if $m+c+\alpha\in\mathbb{Z}$}\\\ \frac{\lceil
m+c+\alpha\rceil-(m+c+\alpha)}{\lceil m+c+\alpha\rceil-\lfloor
m+c+\alpha\rfloor},&\textrm{otherwise}\end{cases}$

so that $\mathbf{E}(X^{\prime})=m+\alpha+c$. Since
$c\in(1-m-\alpha,m+1-\alpha)$, $m\in\mathbb{Z}_{\geq 1}$, and $\alpha\in(0,1)$
so $\lfloor m+c+\alpha\rfloor\geq 1$ and $\lceil m+c+\alpha\rceil\leq 2m+1$.
Hence $\mathbf{P}(X^{\prime}=0)=0$ and $\mathbf{P}(X^{\prime}\geq 2m+2)=0$.
Thus, by (52), (4), and (5),

$\displaystyle H(X^{\prime},Y)$
$\displaystyle=-\left(1-\frac{b}{m}\right)H(\mathbf{1}_{0},X^{\prime})-\left(\frac{b}{m}\right)H(U_{\mathrm{E}}^{m},X^{\prime})$
(54)
$\displaystyle=-\left(1-\frac{b}{m}\right)(\mathbf{P}(X^{\prime}=0)-1)-\left(\frac{b}{m}\right)\left(1-\frac{\mathbf{E}(X^{\prime})+1}{m+1}\right)$
$\displaystyle=1-\frac{b}{m}+\frac{b(\alpha+c)}{m(m+1)}.$

By (3), (53), and (54)

$\displaystyle P(X^{\prime},Y)$
$\displaystyle=\frac{v_{1}}{2}\left(1-\frac{b}{m}+\frac{b(\alpha+c)}{m(m+1)}-\left(\frac{2(m+c+\alpha)}{v_{1}}-1\right)\right),\textrm{
and}$ $\displaystyle P(X,Y)$
$\displaystyle=\frac{v_{1}}{2}\left(1-\frac{b}{m}+\frac{\alpha
b}{m(m+1)}-\left(\frac{2(m+\alpha)}{v_{1}}-1\right)\right).$

Since $(X,Y)$ is a Nash equilibrium so $P(X,Y)\geq P(X^{\prime},Y)$. Hence

$\frac{v_{1}}{2}\left(1-\frac{b}{m}+\frac{\alpha
b}{m(m+1)}-\left(\frac{2(m+\alpha)}{v_{1}}-1\right)\right)\geq\\\
\frac{v_{1}}{2}\left(1-\frac{b}{m}+\frac{b(\alpha+c)}{m(m+1)}-\left(\frac{2(m+c+\alpha)}{v_{1}}-1\right)\right)$

and, consequently

$\frac{2c}{v_{1}}\geq\frac{bc}{m(m+1)}.$ (55)

Since $v_{1}>0$, $b>0$, $m>0$, $\alpha\in(0,1)$, and (55) holds for any
$c\in(1-m-\alpha,m+1-\alpha)$ so $v_{1}/2=m(m+1)/b$.

###### Lemma 5

If a strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq 1}$, and $m\geq b>0$, is a Nash
equilibrium of all pay auction with both players valuations $v_{1},v_{2}>0$
then

$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)U_{\mathrm{E}}^{m},$

  * •

if $0<\alpha\leq(m+1)/(2m+1)$ then

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right),$

where $\delta=(2m+1)/(m+1)$,

  * •

if $(m+1)/(2m+1)<\alpha<1$ then

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}\right),$

where $\sigma=(2m+1)/m$.

  * •

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$,
$\lambda_{\mathrm{E}}=0$ if $m>b$, and $\sum_{j=1}^{m}\kappa_{j}=0$ if $m>b$
or $(m+1)/(2m+1)<\alpha<1$.

  * •

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right),$

###### Proof

Take any strategy profile $(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq 1}$, and $m\geq b>0$, and suppose
that it is a Nash equilibrium of all pay auction with both players valuations
$v_{1},v_{2}>0$.

Since $(X,Y)$ is a Nash equilibrium of all pay auction so, by Proposition 1,
it is a Nash equilibrium of general Lotto game $\Gamma(m+\alpha,b)$. Thus, by
[8, Theorem 4] and [6, Theorem 6],

$H(X,Y)=1-\frac{(1-\alpha)b}{m}-\frac{\alpha
b}{m+1}=1-\frac{b}{m}+\frac{\alpha b}{m(m+1)},$ (56)
$X\in\mathrm{conv}(\mathcal{U}^{m,\alpha}\cup\mathcal{X}^{m,\alpha})$ (57)

and

$Y=\left(1-\frac{b}{m}\right)\mathbf{1}_{0}+\left(\frac{b}{m}\right)U_{\mathrm{E}}^{m},$
(58)

where

  * •

$\mathcal{U}^{m,\alpha}=(1-\alpha)\mathcal{U}^{m}+\alpha
U_{\mathrm{O}}^{m+1}$, if $b=m$,

  * •

$\mathcal{U}^{m,\alpha}=\left\\{(1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1}\right\\}$, if $b<m$

and

  * •

$\mathcal{X}^{m,\alpha}=\alpha\delta\mathcal{V}^{m}+\left(1-\alpha\delta\right)\mathcal{U}^{m}$,
if $0<\alpha\leq\frac{m+1}{2m+1}$ and $b=m$,

  * •

$\mathcal{X}^{m,\alpha}=\alpha\delta\mathcal{V}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}$,
if $0<\alpha\leq\frac{m+1}{2m+1}$ and $b<m$,

  * •

$\mathcal{X}^{m,\alpha}=(1-\alpha)\sigma\mathcal{V}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}$,
if $\frac{m+1}{2m+1}<\alpha<1$, where

$\delta=\frac{2m+1}{m+1},\quad\sigma=\frac{2m+1}{m}.$

If $0<\alpha\leq(m+1)/(2m+1)$ then, by (57),

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right)$

where
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$,
and $\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\kappa_{j}=0$ if $m>b$. By (5), (14),
and (18), for any strategy $Y^{\prime}$,

$\displaystyle H(X,Y^{\prime})$
$\displaystyle=\lambda_{\mathrm{O}}H((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1},Y^{\prime})+\lambda_{\mathrm{E}}H((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1},Y^{\prime})+{}$ (59)
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}H(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m},Y^{\prime})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}H(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m},Y^{\prime})+{}$
$\displaystyle\geq\lambda_{\mathrm{O}}\left(1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)\right)$
$\displaystyle\quad{}+\lambda_{\mathrm{E}}\left(1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+(1-\alpha)\frac{\mathbf{E}(Y^{\prime})-m}{m(m+1)}\right)$
$\displaystyle\quad{}+\sum_{j=1}^{m}\lambda_{j}\left(1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{\alpha}{m+1}\mathbf{P}(Y^{\prime}=2j-1)\right)$
$\displaystyle\quad{}+\sum_{j=1}^{m}\kappa_{j}\Bigg{(}1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{\alpha}{m+1}\mathbf{P}(Y^{\prime}=2j-1)$
$\displaystyle\hskip
170.71652pt{}+\frac{\mathbf{E}(Y^{\prime})-m}{m+1}\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)\Bigg{)}$
$\displaystyle=1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{\alpha}{m+1}\sum_{j=1}^{m}\left(\lambda_{j}+\kappa_{j}\right)\mathbf{P}(Y^{\prime}=2j-1)$
$\displaystyle\qquad{}+\lambda_{\mathrm{E}}(1-\alpha)\frac{\mathbf{E}(Y^{\prime})-m}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{\mathbf{E}(Y^{\prime})-m}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)$

with equality only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$.

Similarly, if $(m+1)/(2m+1)<\alpha<1$ then, by (57),

$X=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\sum_{j=1}^{m}\lambda_{j}\left((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}\right),$

where
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m}\geq
0$, $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}=1$,
and $\lambda_{\mathrm{E}}=0$ if $m>b$. By (5), (14), and (18),

$\displaystyle H(X,Y^{\prime})$
$\displaystyle=\lambda_{\mathrm{O}}H((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1},Y^{\prime})+\lambda_{\mathrm{E}}H((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1},Y^{\prime})+{}$ (60)
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}H((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1},Y^{\prime})$
$\displaystyle\geq\lambda_{\mathrm{O}}\left(1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)\right)$
$\displaystyle\qquad{}+\lambda_{\mathrm{E}}\left(1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+(1-\alpha)\frac{\mathbf{E}(Y^{\prime})-m}{m(m+1)}\right)$
$\displaystyle\qquad{}+\sum_{j=1}^{m}\lambda_{j}\left(1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{1-\alpha}{m}\mathbf{P}(Y^{\prime}=2j-1)\right)$
$\displaystyle=1-\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)+\frac{1-\alpha}{m}\sum_{j=1}^{m}\lambda_{j}\mathbf{P}(Y^{\prime}=2j-1)$
$\displaystyle\qquad{}+\lambda_{\mathrm{E}}(1-\alpha)\frac{\mathbf{E}(Y^{\prime})-m}{m(m+1)}$

with equality only if $\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$.

Using (59), (60), (3), and (4),

$\displaystyle P(Y^{\prime},X)$
$\displaystyle=\frac{v_{2}}{2}\left(H(Y^{\prime},X)-\left(\frac{2\mathbf{E}(Y^{\prime})}{v_{2}}-1\right)\right)=\frac{v_{2}}{2}\left(1-H(X,Y^{\prime})-\frac{2\mathbf{E}(Y^{\prime})}{v_{2}}\right)$
(61)
$\displaystyle\leq\frac{v_{2}}{2}\Bigg{(}\mathbf{E}(Y^{\prime})\left(\frac{1}{m+1}+\frac{1-\alpha}{m(m+1)}\right)-\frac{\alpha}{m+1}\sum_{j=1}^{m}\left(\lambda_{j}+\kappa_{j}\right)\mathbf{P}(Y^{\prime}=2j-1)$
$\displaystyle\qquad{}-\lambda_{\mathrm{E}}(1-\alpha)\frac{\mathbf{E}(Y^{\prime})-m}{m(m+1)}-\sum_{j=1}^{m}\kappa_{j}\left(\frac{\mathbf{E}(Y^{\prime})-m}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)-\frac{2\mathbf{E}(Y^{\prime})}{v_{2}}\Bigg{)}$
$\displaystyle=\frac{v_{2}}{2}\mathbf{E}(Y^{\prime})\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right)-\frac{v_{2}}{2}\Bigg{(}\frac{\alpha}{m+1}\sum_{j=1}^{m}\left(\lambda_{j}+\kappa_{j}\right)\mathbf{P}(Y^{\prime}=2j-1)$
$\displaystyle\qquad\qquad\qquad{}+\lambda_{\mathrm{E}}(1-\alpha)\frac{\mathbf{E}(Y^{\prime})-m}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{\mathbf{E}(Y^{\prime})-m}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)\Bigg{)},$

where $\lambda_{\mathrm{E}}=0$ if $m>b$ and $\sum_{j=1}^{m}\kappa_{j}=0$ if
either $m>0$ or $(m+1)/(2m+1)<\alpha<1$. (61) holds with equality only if
$\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$.

On the other hand,

$P(Y,X)=\frac{v_{2}}{2}\left(\frac{\alpha
b}{m+1}+\frac{(1-\alpha)b}{m}-1-\left(\frac{2b}{v_{2}}-1\right)\right)=\frac{v_{2}}{2}b\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right).$
(62)

For the remaining part of the proof we consider the cases of $m=b$ and $m>b$
separately.

Assume that $m=b$. Consider strategy
$Y^{\prime}=((j-\alpha)/(m+j))\mathbf{1}_{0}+((m+\alpha)/(m+j))\mathbf{1}_{m+j}$
with $j=1$, if $m\bmod 2=1$, and $j=2$, if $m\bmod 2=0$. Notice that
$\mathbf{E}(Y^{\prime})=m+\alpha$ and
$\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$. The latter follows
because if $m\geq 1$ then $m+j\leq 2m$. By (61),

$\displaystyle P(Y^{\prime},X)$
$\displaystyle=\frac{v_{2}}{2}(m+\alpha)\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right)$
$\displaystyle\qquad{}-\frac{v_{2}}{2}\Bigg{(}\lambda_{\mathrm{E}}(1-\alpha)\frac{\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{\alpha}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)\Bigg{)}.$

On the other hand, since $m=b$ so, by (62),

$P(Y,X)=\frac{v_{2}}{2}m\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right).$
(63)

Since $(X,Y)$ is a Nash equilibrium so $P(Y,X)\geq P(Y^{\prime},X)$. Hence it
follows that

$\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)\geq\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}.$
(64)

Consider strategy $Y^{\prime}=\mathbf{1}_{0}$. If $0<\alpha\leq(m+1)/(2m+1)$
then, by (61),

$\displaystyle P(Y^{\prime},X)$
$\displaystyle=\frac{v_{2}}{2}\Bigg{(}\lambda_{\mathrm{E}}\frac{1-\alpha}{m+1}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{m}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)\Bigg{)}.$

Since $(X,Y)$ is a Nash equilibrium so $P(Y,X)\geq P(Y^{\prime},X)$. Hence it
follows that

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\geq\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right).$
(65)

By (64) and (65),

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right).$
(66)

Assume that $m>b$. Consider any strategy
$Y^{\prime}=((j-m)/j)\mathbf{1}_{0}+(m/j)\mathbf{1}_{j}$ with
$j\in\\{m,\ldots,2m\\}$ and $j\bmod 2=0$ ($2m$ is an example of such a $j$).
Notice that $\mathbf{E}(Y^{\prime})=m$ and
$\sum_{j=2m+1}^{+\infty}\mathbf{P}(Y^{\prime}\geq j)=0$. By (61),

$P(Y^{\prime},X)=\frac{v_{2}}{2}m\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right).$
(67)

Since $(X,Y)$ is a Nash equilibrium so $P(Y,X)\geq P(Y^{\prime},X)$ and, by
(62) and (67), it follows that

$\frac{v_{2}}{2}b\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right)\geq\frac{v_{2}}{2}m\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right).$
(68)

Since $m>b$ so (68) is satisfied only if

$\frac{m+1-\alpha}{m(m+1)}\leq\frac{2}{v_{2}}.$ (69)

Consider any strategy $Y^{\prime}=\mathbf{1}_{0}$. By (61),
$P(Y^{\prime},X)=0$. Since $(X,Y)$ is a Nash equilibrium so $P(Y,X)\geq
P(Y^{\prime},X)$ and, by (62) and $b>0$, it follows that

$\frac{m+1-\alpha}{m(m+1)}\geq\frac{2}{v_{2}}.$

This, together with (69) yields

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=0$

which, given that $\lambda_{\mathrm{E}}=0$ and $\sum_{j=1}^{m}\kappa_{j}=0$ if
$m>b$ implies (66).

###### Lemma 6

If conditions in Lemma 4 and Lemma 5 are satisfied for a strategy profile
$(X,Y)$ such that $\mathbf{E}(X)=m+\alpha$, $\mathbf{E}(Y)=b$,
$m\in\mathbb{Z}_{\geq 1}$, and $m\geq b>0$, then $(X,Y)$ is a Nash equilibrium
of all pay auction with both players valuations $v_{1},v_{2}>0$

###### Proof

Take any strategy profile $(X,Y)$, as described in the lemma. By Lemma 4,
$(X,Y)$ is a Nash equilibrium of General Lotto game $\Gamma(m+\alpha,b)$.
Hence, by [8, Theorem 4],

$H(X,Y)=1-\frac{(1-\alpha)b}{m}-\frac{\alpha
b}{m+1}=1-\frac{b}{m}+\frac{\alpha b}{m(m+1)}.$ (70)

We first show that there is no profitable deviation for player $1$. By (3) and
(70),

$P(X,Y)=\frac{v_{1}}{2}\left(2-\frac{(1-\alpha)b}{m}-\frac{\alpha
b}{m+1}-\frac{2(m+\alpha)}{v_{1}}\right).$ (71)

Take any strategy $X^{\prime}$ with $\mathbf{E}(X^{\prime})\geq 0$. By 54,
(3), and $b\leq m$,

$P(X^{\prime},Y)\leq\frac{v_{1}}{2}\left(2-\left(\frac{b}{m}\right)\left(\frac{2m+1-\mathbf{E}(X^{\prime})}{m+1}\right)-\frac{2\mathbf{E}(X^{\prime})}{v_{1}}\right).$

By Lemma (4), $v_{1}/2=m(m+1)/b$ so

$\displaystyle P(X^{\prime},Y)$
$\displaystyle\leq\frac{v_{1}}{2}\left(2-\left(\frac{b}{m}\right)\left(\frac{2m+1-\mathbf{E}(X^{\prime})}{m+1}\right)-\frac{b\mathbf{E}(X^{\prime})}{m(m+1)}\right)$
$\displaystyle=\frac{v_{1}}{2}\left(2-b\left(\frac{2m+1}{m(m+1)}\right)\right)$

and, by (71),

$\displaystyle P(X,Y)$
$\displaystyle=\frac{v_{1}}{2}\left(2-\frac{(1-\alpha)b}{m}-\frac{\alpha
b}{m+1}-\frac{b(m+\alpha)}{m(m+1)}\right)$
$\displaystyle=\frac{v_{1}}{2}\left(2-b\left(\frac{2m+1}{m(m+1)}\right)\right).$

Hence $P(X,Y)\geq P(X^{\prime},Y)$. Thus player $1$ has no profitable
deviation from $(X,Y)$.

Second, we show that there is no profitable deviation for player $2$. Take any
strategy $Y^{\prime}$ with $\mathbf{E}(Y^{\prime})\geq 0$. By (61),

$\displaystyle P(Y^{\prime},X)$
$\displaystyle\leq\frac{v_{2}}{2}\mathbf{E}(Y^{\prime})\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right)$
$\displaystyle\qquad{}-\frac{v_{2}}{2}(\mathbf{E}(Y^{\prime})-m)\Bigg{(}\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)\Bigg{)}.$

By Lemma 5,

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right).$
(72)

Hence

$\displaystyle P(Y^{\prime},X)$
$\displaystyle\leq\frac{v_{2}}{2}m\left(\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\right).$

On the other hand, by (62),

$P(Y,X)=\frac{v_{2}}{2}b\left(\frac{m+1-\alpha}{m+1}-\frac{2}{v_{2}}\right).$
(73)

If $m=b$ then $P(Y,X)=P(Y^{\prime},X)$. If $m>b$ then, by Lemma 5,
$\lambda_{\mathrm{E}}=\sum_{j=1}^{m}\kappa_{j}=0$ and, by (72),

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=0.$

Hence in this case $P(Y,X)=P(Y^{\prime},X)$ as well. Thus there is no
profitable deviation for player $2$.

######  Proof (Proof of Proposition 6)

By Lemmas 4, 5, and 6, a strategy profile $(X,Y)$ with
$\mathbf{E}(X)=m+\alpha$, $\mathbf{E}(Y)=b$, $m\in\mathbb{Z}_{\geq 1}$, and
$m>b>0$, is a Nash equilibrium of all pay auction with both players valuations
$v_{1},v_{2}>0$ if and only if the conditions stated in the lemmas are
satisfied.

Thus

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)$
(74)

with $\lambda_{\mathrm{E}}=0$ and $\sum_{j=1}^{m}\kappa_{j}=0$, as $m>b$. Thus

$\frac{m+1-\alpha}{m(m+1)}=\frac{2}{v_{2}}$

and, since $\alpha\in(0,1)$ and $m>b$ so
$v_{2}=2m(m+1)/(m+1-\alpha)<2m(m+1)/b=v_{1}$ and

$m<\frac{v_{2}}{2}<m+1.$

Since $m\in\mathbb{Z}_{\geq 1}$ so it follows that $m=\lceil v_{2}/2\rceil-1$
and $\lceil v_{2}/2\rceil>v_{2}/2$. By conditions of Lemma 4,
$m(m+1)/b=v_{1}/2$. Hence

$b=\frac{\left(\left\lceil\frac{v_{2}}{2}\right\rceil-1\right)\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{1}}{2}}.$

Since $m>b$ so $v_{1}/2=m(m+1)/b>m+1>\lceil v_{2}/2\rceil$. Moreover

$\alpha=\frac{(m+1)\left(\frac{v_{2}}{2}-m\right)}{\frac{v_{2}}{2}}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)}{\frac{v_{2}}{2}}$

By Lemma 5 the characterization of $X$ and $Y$ is as stated in the theorem
(the cases in the definition of $\mathcal{X}^{\alpha,m}$ follow immediately by
substituting the formulas for $m$ and $\alpha$).

######  Proof (Proof of Proposition 7)

By Lemmas 4, 5, and 6, a strategy profile $(X,Y)$ with
$\mathbf{E}(X)=m+\alpha$, $\mathbf{E}(Y)=m$, $m\in\mathbb{Z}_{\geq 1}$, is a
Nash equilibrium of all pay auction with both players valuations
$v_{1},v_{2}>0$ if and only if the conditions stated in the lemmas are
satisfied.

Thus

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right).$

Since $\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\kappa_{j}\leq 1$ and

$\frac{1-\alpha}{m(m+1)}>\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right)$

so the RHS of (74) is maximised when $\lambda_{\mathrm{E}}=1$ and
$\sum_{j=1}^{m}\kappa_{j}=0$ and it is minimised when
$\lambda_{\mathrm{E}}=\sum_{j=1}^{m}\kappa_{j}=0$. Therefore

$0\leq\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}\leq\frac{1-\alpha}{m(m+1)}$

and it follows that

$\frac{m(m+1)}{m+1-\alpha}\leq\frac{v_{2}}{2}\leq m+1.$

Since $\alpha\in(0,1)$ so if further follows that

$m<\frac{v_{2}}{2}\leq m+1.$

Thus $m=\lceil v_{2}/2\rceil-1$. By Lemma 4, $v_{1}/2=m(m+1)/m=m+1=\lceil
v_{2}/2\rceil$.

By Lemma 5 the characterization of $X$ and $Y$ is as stated in the theorem
(the cases in the definition of $X$ follow immediately by substituting the
formulas for $m$ and $\alpha$).

By Lemma 5,

$\frac{m+1-\alpha}{m(m+1)}-\frac{2}{v_{2}}=\lambda_{\mathrm{E}}\frac{1-\alpha}{m(m+1)}+\sum_{j=1}^{m}\kappa_{j}\left(\frac{1}{m+1}\right)\left(\frac{1-\alpha}{m}-\frac{\alpha}{m+1}\right),$
(75)

with
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$,
and $\sum_{j=1}^{m}\kappa_{j}=0$ if $(m+1)/(2m+1)<\alpha<1$. Substituting
$\lceil v_{2}/2\rceil-1$ for $m$ and reorganizing we obtain

$\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

and the conditions for $\lambda_{\mathrm{E}}$ and $\sum_{i=1}^{m}\kappa_{i}$
stated in the theorem follow.

For the bounds on $\alpha$, by (75),

$\alpha=\frac{\frac{(v_{2}-2m)(m+1)}{v_{2}}-\lambda_{\mathrm{E}}-\sum_{j=1}^{m}\kappa_{j}}{1-\lambda_{\mathrm{E}}-\delta\sum_{j=1}^{m}\kappa_{j}}=1-\frac{(1-\delta)\sum_{j=1}^{m}\kappa_{j}+\frac{m(2(m+1)-v_{2})}{v_{2}}}{1-\lambda_{\mathrm{E}}-\delta\sum_{j=1}^{m}\kappa_{j}}.$

Since $\delta<1$, $v_{2}/2\leq(m+1)$, and
$\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\kappa_{j}\leq 1$ so it follows that the
RHS of the equality is decreasing when $\lambda_{\mathrm{E}}$ increases.
Similarly,

$\alpha=\frac{\frac{(v_{2}-2m)(m+1)}{v_{2}}-\lambda_{\mathrm{E}}-\sum_{j=1}^{m}\kappa_{j}}{1-\lambda_{\mathrm{E}}-\delta\sum_{j=1}^{m}\kappa_{j}}=\frac{1}{\delta}\left(1-\frac{(1-\delta)(1-\lambda_{\mathrm{E}})+\delta\left(\frac{m(2(m+1)-v_{2})}{v_{2}}\right)}{1-\lambda_{\mathrm{E}}-\delta\sum_{j=1}^{m}\kappa_{j}}\right).$

and the RHS of the equality is decreasing when $\lambda_{\mathrm{E}}<0$ and
$\sum_{j=1}^{m}\kappa_{j}$ increases. Thus the RHS attains its maximal value
when $\lambda_{\mathrm{E}}=\sum_{j=1}^{m}\kappa_{j}=0$. The RHS obtains it
minimal value, $0$, when $\sum_{j=1}^{m}\kappa_{j}=0$ and

$\lambda_{\mathrm{E}}=1-\frac{m(2(m+1)-v_{2})}{v_{2}}\geq
1-\left(m+1-\frac{v_{2}}{2}\right)=1-\left(\left\lceil\frac{v_{2}}{2}\right\rceil-\frac{v_{2}}{2}\right).$

Thus it follows that

$0<\alpha\leq\frac{(m+1)(v_{2}-2m)}{v_{2}}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right).$

### Proofs of Theorems 3.1 and 3.2

In this section we use the characterisations of equilibrium strategy profiles
for all the possible configurations of expected values of equilibrium
strategies, obtained in Propositions 2, 3, 4, 5, 6, and 7, to give proofs of
the two Theorems 3.1 and 3.2.

######  Proof (Proof of Theorem 3.1)

Point (i).
For point (i) assume that $v_{2}/2\in\mathbb{Z}$, $v_{1}=v_{2}>0$ and let
$m=v_{2}/2-1$.

For the right to left implication take any strategy profile $(X,Y)$ with
$X=\alpha U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$ and $Y=\beta
U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m}$. If $\alpha=\beta=0$ then
$\mathbf{E}(X)=\mathbf{E}(Y)=m$ and $(X,Y)$ is a Nash equilibrium by
Proposition 2. If $\alpha=\beta=1$ then $\mathbf{E}(X)=\mathbf{E}(Y)=v_{2}/2$
and again $(X,Y)$ is a Nash equilibrium by Proposition 2.

If $\alpha\in(0,1)$ and $\beta\in(0,1)$ then $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=m+\beta$ and $(X,Y)$ is a Nash equilibrium by Proposition 3.

In the case of $\alpha=1$ and $\beta\in(0,1)$ let $m^{\prime}=v_{2}/2$. Then
$\mathbf{E}(X)=m^{\prime}$ and $\mathbf{E}(Y)=m^{\prime}-(1-\beta)$,
$X=U_{\mathrm{O}}^{m^{\prime}}$, and $Y=\beta
U_{\mathrm{O}}^{m^{\prime}}+(1-\beta)U_{\mathrm{E}}^{m^{\prime}-1}$.

Suppose that $v_{2}/2=1$ and let $b=\beta$. Then $m^{\prime}=1$,
$\mathbf{E}(X)=1$, $\mathbf{E}(Y)=b$, and
$U_{\mathrm{E}}^{m^{\prime}-1}=\mathbf{1}_{0}$. Thus
$Y=bU_{\mathrm{O}}^{m^{\prime}}+(1-b)\mathbf{1}_{0}=(1-b)\mathbf{1}_{0}+b(\lambda
U_{\mathrm{O}}^{m^{\prime}}+(1-\lambda)U_{\mathrm{E}}^{m^{\prime}})$ with
$\lambda=1$. Since (by $v_{1}=v_{2}=2$)

$\frac{4}{bv_{1}}-\frac{2}{b}+1=1\leq\lambda\leq\frac{2-\beta}{\beta}=\frac{4}{bv_{1}}-1$

so, by Proposition 4 $(X,Y)$ is a Nash equilibrium.

Suppose that $v_{2}/2\geq 2$ and let $b=m^{\prime}-(1-\beta)$. Since

$U_{\mathrm{E}}^{m^{\prime}-1}=\left(1-\frac{m^{\prime}-1}{m^{\prime}}\right)\mathbf{1}_{0}+\left(\frac{m^{\prime}-1}{m^{\prime}}\right)U_{\mathrm{O}\uparrow
1}^{m^{\prime}}$ (76)

so

$\displaystyle Y$
$\displaystyle=(1-\beta)\left(1-\frac{m^{\prime}-1}{m^{\prime}}\right)\mathbf{1}_{0}+\beta
U_{\mathrm{O}}^{m^{\prime}}+(1-\beta)\left(\frac{m^{\prime}-1}{m^{\prime}}\right)U_{\mathrm{O}\uparrow
1}^{m^{\prime}}$
$\displaystyle=\left(1-\frac{b}{m^{\prime}}\right)\mathbf{1}_{0}+\left(\frac{b}{m^{\prime}}\right)\left(\lambda
U_{\mathrm{O}}^{m^{\prime}}+(1-\lambda)U_{\mathrm{O}\uparrow
1}^{m^{\prime}}\right)$

where $\lambda=\beta m^{\prime}/b$. Thus

$\displaystyle
Y=\left(1-\frac{b}{m^{\prime}}\right)\mathbf{1}_{0}+\left(\frac{b}{m^{\prime}}\right)Z$

where

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}$

and $\lambda_{\mathrm{O}}=\beta m^{\prime}/b$, $\lambda_{\mathrm{O}\uparrow
1}=1-\beta m^{\prime}/b$, and
$\lambda_{\mathrm{E}}=\lambda_{1}=\ldots=\lambda_{m-1}=0$. Since

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m^{\prime}-1}-\frac{\lambda_{\mathrm{E}}}{m^{\prime}+1}=\frac{b-\beta
m^{\prime}}{b(m^{\prime}-1)}=\frac{m^{\prime}(b-\beta)}{b(m^{\prime}-1)}-1=\frac{m^{\prime}}{b}-1=\frac{v_{2}^{2}}{2v_{1}b}-1,$
$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}=\frac{m^{\prime}}{b}-1=\frac{v_{2}}{2b}-1,$

and

$\frac{v_{2}(v_{2}-2)}{2v_{1}}=m^{\prime}-1<m^{\prime}-1+\beta=b<m^{\prime}<\frac{v_{2}(v_{2}+2)}{2v_{1}}$

so, by Proposition 5, $(X,Y)$ is a Nash equilibrium. Analogously, if
$\alpha\in(0,1)$ and $\beta=1$ then $(X,Y)$ is a Nash equilibrium by
Propositions 4 and 5.

If $\alpha\in(0,1)$ and $\beta=0$ then $\mathbf{E}(X)=m+\alpha$,
$\mathbf{E}(Y)=m$, and $Y=U_{\mathrm{E}}^{m}$. Suppose that
$0<\alpha\leq(m+1)/(2m+1)$. Then

$\displaystyle X$ $\displaystyle=\alpha
U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right),$

with $\lambda_{\mathrm{E}}=1$ and
$\lambda_{\mathrm{O}}=\lambda_{1}=\ldots=\lambda_{m}=\kappa_{1}=\ldots=\kappa_{m}=0$.
Since

$\frac{\lceil\frac{v_{2}}{2}\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)=1$
(77)

so

$\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=1=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

and so, by Proposition 7 $(X,Y)$ is a Nash equilibrium. Similarly, if
$(m+1)/(2m+1)\leq\alpha<1$ then

$\alpha<\frac{\lceil\frac{v_{2}}{2}\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)$

and

$\displaystyle X$ $\displaystyle=\alpha
U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}\right),$

with $\lambda_{\mathrm{E}}=1$, and
$\lambda_{\mathrm{O}}=\lambda_{1}=\ldots=\lambda_{m}=0$. Since

$\lambda_{\mathrm{E}}=1=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

so, by Proposition 7 $(X,Y)$ is a Nash equilibrium. Analogously, if $\alpha=0$
and $\beta\in(0,1)$ then $(X,Y)$ is a Nash equilibrium by Proposition 7. This
completes proof of the right to left implication.

For the left to right implication, suppose that $(X,Y)$ is a Nash equilibrium
of all-pay auction with $v_{1}=v_{2}>0$ and $v_{2}/2\in\mathbb{Z}$. Then, by
Propositions 2, 3, 4, 5, 6, and 7, one of the following cases holds:

  1. 1.

$\mathbf{E}(X)=\mathbf{E}(Y)=v_{2}/2-1$,

  2. 2.

$\mathbf{E}(X)=\mathbf{E}(Y)=v_{2}/2$,

  3. 3.

$\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=m+\beta$, where $m=v_{2}/2-1$ and
$\alpha,\beta\in(0,1)$,

  4. 4.

$\mathbf{E}(X)=1$ and $\mathbf{E}(Y)=b$, $v_{2}=2$, and $b\in(0,1)$.

  5. 5.

$\mathbf{E}(Y)=1$ and $\mathbf{E}(X)=b$, $v_{2}=2$, and $b\in(0,1)$.

  6. 6.

$\mathbf{E}(X)=m^{\prime}$ and $\mathbf{E}(Y)=b$, where
$m^{\prime}=v_{2}/2\geq 2$ and $m^{\prime}>b>0$.

  7. 7.

$\mathbf{E}(Y)=m^{\prime}$ and $\mathbf{E}(X)=b$, where
$m^{\prime}=v_{2}/2\geq 2$ and $m^{\prime}>b>0$.

  8. 8.

$\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=m$, where $m=v_{2}/2-1>0$ and
$\alpha\in(0,1)$.

  9. 9.

$\mathbf{E}(Y)=m+\beta$ and $\mathbf{E}(X)=m$, where $m=v_{2}/2-1>0$ and
$\beta\in(0,1)$.

In case 1, by Proposition 2, $X=Y=U_{\mathrm{E}}^{m}=\alpha
U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$ with $\alpha=0$. In case 2,
by Proposition 2, $X=Y=U_{\mathrm{O}}^{m+1}=\alpha
U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$ with $\alpha=1$. In case 3,
by Proposition 3 $X=\alpha U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$
and $Y=\beta U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m}$.

In case 4, $m=v_{2}/2-1=0$ and, by Proposition 4, $X=U_{\mathrm{O}}^{1}=\alpha
U_{\mathrm{O}}^{1}+(1-\alpha)U_{\mathrm{E}}^{0}$ with $\alpha=1$, and
$Y=(1-b)\mathbf{1}_{0}+b(\lambda
U_{\mathrm{O}}^{1}+(1-\lambda)U_{\mathrm{E}}^{1})$ with $\lambda\in[0,1]$ and

$\lambda\geq\frac{4}{bv_{1}}-\frac{2}{b}+1=1.$

Hence $\lambda=1$ and
$Y=(1-b)\mathbf{1}_{0}+bU_{\mathrm{O}}^{1}=bU_{\mathrm{O}}^{1}+(1-b)U_{\mathrm{E}}^{0}$
with $b\in(0,1)$. Analogously, in case 5, $Y=U_{\mathrm{O}}^{1}=\beta
U_{\mathrm{O}}^{1}+(1-\beta)U_{\mathrm{E}}^{0}$ with $\beta=1$, and $X=\alpha
U_{\mathrm{O}}^{1}+(1-\alpha)U_{\mathrm{E}}^{0}$ with $\alpha=b\in(0,1)$.

In case 6, $m=v_{2}/2-1=m^{\prime}-1$ and, by Proposition 5,
$X=U_{\mathrm{O}}^{m^{\prime}}=\alpha
U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$ with $\alpha=1$, and

$Y=\left(1-\frac{b}{m^{\prime}}\right)\mathbf{1}_{0}+\left(\frac{b}{m^{\prime}}\right)Z$

where

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m^{\prime}}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m^{\prime}}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow
1}^{m^{\prime}}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m^{\prime}},$

$0<b<m^{\prime}$,

$\frac{v_{2}(v_{2}-2)}{2v_{1}}\leq b\leq\frac{v_{2}(v_{2}+2)}{2v_{1}},$
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m^{\prime}-1}\geq 0\textrm{ and
}\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m^{\prime}-1}\lambda_{j}=1,$ $\frac{\lambda_{\mathrm{O}\uparrow
1}}{m^{\prime}-1}-\frac{\lambda_{\mathrm{E}}}{m^{\prime}+1}=\frac{v_{2}^{2}}{2v_{1}b}-1,$

and

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m^{\prime}-1}+\frac{1}{2m^{\prime}}\sum_{j=1}^{m^{\prime}-1}\lambda_{j}\leq\frac{v_{2}}{2b}-1.$

Hence

$\lambda_{\mathrm{O}\uparrow
1}=\frac{m^{\prime}(m^{\prime}-1)}{b}-(m^{\prime}-1)+\lambda_{\mathrm{E}}\frac{m^{\prime}-1}{m^{\prime}+1},$
$\frac{\lambda_{\mathrm{E}}}{m^{\prime}+1}+\frac{1}{2m^{\prime}}\sum_{j=1}^{m^{\prime}-1}\lambda_{j}\leq\frac{v_{2}v_{1}}{2v_{1}b}-\frac{v_{2}^{2}}{2v_{1}b}=0,$

and so $\lambda_{\mathrm{E}}=\lambda_{1}=\ldots=\lambda_{m^{\prime}-1}=0$,
$\lambda_{\mathrm{O}\uparrow 1}=1-\lambda_{\mathrm{O}}$, and

$\lambda_{\mathrm{O}}=m^{\prime}\left(1-\frac{m^{\prime}-1}{b}\right).$

This, together with (76) yields

$Y=\left(1-\frac{b}{m^{\prime}}\right)\mathbf{1}_{0}+\left(\frac{b}{m^{\prime}}\right)\left(\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m^{\prime}}+(1-\lambda_{\mathrm{O}})U_{\mathrm{O}\uparrow
1}^{m^{\prime}}\right)=(1-(m^{\prime}-b))U_{\mathrm{O}}^{m^{\prime}}+(m^{\prime}-b)U_{\mathrm{E}}^{m^{\prime}-1}.$

Since

$\frac{v_{2}(v_{2}-2)}{2v_{1}}=\frac{v_{2}}{2}-1=m^{\prime}-1\leq
b<m^{\prime}$

so $m^{\prime}-b\in(0,1]$ and so $Y=\beta
U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m}$ with
$\beta=m^{\prime}-b\in(0,1]$. Analogously, in case 7,
$Y=U_{\mathrm{O}}^{m+1}=\beta
U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m}$ with $\beta=1$, and
$X=\alpha U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$ with
$\alpha=m^{\prime}-b\in(0,1]$.

In case 8, by Proposition 7, $Y=U_{\mathrm{E}}^{m}=\beta
U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m}$ with $\beta=0$, and

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right),$

with

$\delta=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil},$

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$,
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$,
and

$\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha},$

in the case of
$0<\alpha\leq\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}$,
and

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left((1-\alpha)\sigma
V_{j}^{m}+\left(1-(1-\alpha)\sigma\right)U_{\mathrm{O}}^{m+1}\right),$

with

$\sigma=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil-1},$

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m}\geq
0$, $\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}=1$,
and

$\lambda_{\mathrm{E}}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha},$

in the case of
$\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}<\alpha\leq\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)$.
If
$0<\alpha\leq\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}$
then, by (77),

$\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=1.$

Hence $\lambda_{\mathrm{O}}=\lambda_{1}=\ldots\ldots\lambda_{m}=0$. Since
$v_{2}/2-1>0$ so $\delta>1$ and $(1-\alpha\delta)/(1-\alpha)<1$. Hence
$\lambda_{\mathrm{E}}=1$, $\kappa_{1}=\ldots=\kappa_{m}=0$, and
$X=(1-\alpha)U_{\mathrm{E}}^{m}+\alpha U_{\mathrm{O}}^{m+1}$ with
$\alpha\in(0,1)$. If
$\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}<\alpha\leq\frac{\left\lceil\frac{v_{2}}{2}\right\rceil}{\frac{v_{2}}{2}}\left(\frac{v_{2}}{2}-\left\lceil\frac{v_{2}}{2}\right\rceil+1\right)$
then, by (77), $\lambda_{\mathrm{E}}=1$ and, consequently,
$\lambda_{\mathrm{O}}=\lambda_{1}=\ldots\ldots\lambda_{m}=0$. Thus
$X=(1-\alpha)U_{\mathrm{E}}^{m}+\alpha U_{\mathrm{O}}^{m+1}$ with
$\alpha\in(0,1)$. Analogously, in case 9, $X=U_{\mathrm{E}}^{m}=\alpha
U_{\mathrm{O}}^{m+1}+(1-\alpha)U_{\mathrm{E}}^{m}$ with $\alpha=1$, and
$Y=\beta U_{\mathrm{O}}^{m+1}+(1-\beta)U_{\mathrm{E}}^{m}$ with
$\beta\in(0,1)$.

Thus we have shown that in all the cases 1–9, $(X,Y)$ is as stated in point
(i) which completes the proof of the left to right implication.

Point (ii).
For point (ii) assume that $v_{2}/2\in\mathbb{Z}$ and $v_{1}>v_{2}=2$.

For the right to left implication take any strategy profile $(X,Y)$ with
$X=U_{\mathrm{O}}^{1}$ and $Y=(1-b)\mathbf{1}_{0}+b(\lambda
U_{\mathrm{O}}^{1}+(1-\lambda)U_{\mathrm{E}}^{1})$, where $b\in(0,1]$ and

$\frac{4}{bv_{1}}-\frac{2}{b}+1\leq\lambda\leq\frac{4}{bv_{1}}-1.$

If $b=1$ then $X=U_{\mathrm{O}}^{1}=\alpha
U_{\mathrm{O}}^{1}+(1-\alpha)U_{\mathrm{E}}^{1}$ with

$\alpha=1=\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right)$

and $Y=\lambda U_{\mathrm{O}}^{1}+(1-\lambda)U_{\mathrm{E}}^{1}$ with

$\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right)=\frac{4}{v_{1}}-1=\frac{4}{bv_{1}}-\frac{2}{b}+1\leq\lambda\leq\frac{4}{bv_{1}}-1=\frac{4}{v_{1}}-1=\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right),$

where $m=v_{2}/2=1$. Thus $(X,Y)$ is a Nash equilibrium by Proposition 2. If
$b\in(0,1)$ then $(X,Y)$ is a Nash equilibrium by Proposition 4. This
completes proof of the right to left implication.

For the left to right implication, suppose that $(X,Y)$ is a Nash equilibrium
of all-pay auction with $v_{1}>v_{2}=2$. Then, by Propositions 2, 3, 4, 5, 6,
and 7, one of the following cases holds:

  1. 1.

$\lfloor v_{1}/2\rfloor=v_{2}/2$ or $\lceil v_{1}/2\rceil=v_{2}/2+1$,
$\mathbf{E}(X)=\mathbf{E}(Y)=v_{2}/2=1$,

  2. 2.

$\mathbf{E}(X)=1$ and $\mathbf{E}(Y)=b$ and $b\in(0,1)$.

In case 1, by Proposition 2, $X=\alpha
U_{\mathrm{O}}^{1}+(1-\alpha)U_{\mathrm{E}}^{1}$ with $\alpha=1$, so
$X=U_{\mathrm{O}}^{1}$, and $Y=\kappa
U_{\mathrm{O}}^{1}+(1-\kappa)U_{\mathrm{E}}^{1}=(1-b)\mathbf{1}_{0}+b(\kappa
U_{\mathrm{O}}^{1}+(1-\kappa)U_{\mathrm{E}}^{1})$ with $b=1$ and

$\kappa=\frac{2}{v_{1}}\left(2-\frac{v_{1}}{2}\right)=\frac{4}{bv_{1}}-1\leq\frac{4}{bv_{1}}-\frac{2}{b}+1$

with $b=1$. Hence $(X,Y)$ satisfies all the conditions stated in point (ii) of
the theorem with $b=1$. In case 2, by Proposition 4, $(X,Y)$ satisfies all the
conditions stated in point (ii) of the theorem. This completes proof of the
left to right implication.

Point (iii).
For point (iii) assume that $v_{2}/2\in\mathbb{Z}$ and $v_{1}>v_{2}=3$.

For the right to left implication take any strategy profile $(X,Y)$ with
$X=U_{\mathrm{O}}^{m}$ and $Y=(1-b/m)\mathbf{1}_{0}+(b/m)Z$, where
$m=v_{2}/2$, $b\in[v_{2}(v_{2}-2)/(2v_{1}),\min(m,v_{2}(v_{2}+2)/(2v_{1}))]$,
and

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}$

with

$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{\mathrm{O}\uparrow
1},\lambda_{1},\ldots,\lambda_{m-1}\geq 0\textrm{ and
}\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\lambda_{\mathrm{O}\uparrow
1}+\sum_{j=1}^{m-1}\lambda_{j}=1,$ $\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{v_{2}^{2}}{2v_{1}b}-1,$

and

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\leq\frac{v_{2}}{2b}-1.$

If $b=m$ then $X=U_{\mathrm{O}}^{m}=\alpha
U_{\mathrm{O}}^{m}+(1-\alpha)U_{\mathrm{E}}^{m}$ with

$\alpha=1=\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right)$

and $Y=Z$. Since

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}\leq\frac{v_{2}}{2b}-1=0$

so $\lambda_{\mathrm{O}\uparrow 1},\lambda_{1},\ldots,\lambda_{m-1}=0$. In
addition, since

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{v_{2}^{2}}{2v_{1}b}-1=\frac{2m}{v_{1}}-1,$

and $\lambda_{\mathrm{O}\uparrow 1}=0$ so

$\lambda_{\mathrm{E}}=(m+1)\left(1-\frac{2m}{v_{1}}\right)\textrm{ and
}\lambda_{\mathrm{O}}=1-\lambda_{\mathrm{E}}=\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right).$

Thus $(X,Y)$ is a Nash equilibrium by Proposition 2. If
$b\in[v_{2}(v_{2}-2)/(2v_{1}),v_{2}(v_{2}+2)/(2v_{1})]$ then $(X,Y)$ is a Nash
equilibrium by Proposition 5. This completes proof of the right to left
implication.

For the left to right implication, suppose that $(X,Y)$ is a Nash equilibrium
of all-pay auction with $v_{1}>v_{2}=3$. Then, by Propositions 2, 3, 4, 5, 6,
and 7, one of the following cases holds:

  1. 1.

$\lfloor v_{1}/2\rfloor=v_{2}/2$ or $\lceil v_{1}/2\rceil=v_{2}/2+1$,
$\mathbf{E}(X)=\mathbf{E}(Y)=v_{2}/2=m$,

  2. 2.

$\mathbf{E}(X)=m=v_{2}/2$ and $\mathbf{E}(Y)=b$ with $m>b>0$.

In case 1, by Proposition 2, $X=\alpha
U_{\mathrm{O}}^{m}+(1-\alpha)U_{\mathrm{E}}^{m}$ with $\alpha=1$, so
$X=U_{\mathrm{O}}^{m}$, and $Y=\kappa
U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m}=(1-b/m)\mathbf{1}_{0}+(b/m)(\kappa
U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m})$ with $b=m$ and

$\kappa=\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right).$

Thus $Y=(1-b/m)\mathbf{1}_{0}+(b/m)Z$ where

$Z=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\lambda_{\mathrm{O}\uparrow
1}U_{\mathrm{O}\uparrow 1}^{m}+\sum_{j=1}^{m-1}\lambda_{j}W_{j}^{m}$

with $\lambda_{\mathrm{O}\uparrow 1},\lambda_{1},\ldots,\lambda_{m-1}=0$,
$\lambda_{\mathrm{O}}=\kappa$, and $\lambda_{\mathrm{E}}=1-\kappa$. Thus

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}-\frac{\lambda_{\mathrm{E}}}{m+1}=\frac{\kappa-1}{m+1}=\frac{2m}{v_{1}}-1=\frac{v_{2}^{2}}{2v_{1}b}-1,$

and

$\frac{\lambda_{\mathrm{O}\uparrow
1}}{m-1}+\frac{1}{2m}\sum_{j=1}^{m-1}\lambda_{j}=0=\frac{v_{2}}{2b}-1.$

Hence $(X,Y)$ satisfies all the conditions stated in point (iii) of the
theorem with $b=m$. In case 2, by Proposition 5, $(X,Y)$ satisfies all the
conditions stated in point (iii) of the theorem. This completes proof of the
left to right implication.

######  Proof (Proof of Theorem 3.2)

For point (i) assume that $v_{2}/2\notin\mathbb{Z}$, $\lfloor
v_{1}/2\rfloor=\lfloor v_{2}/2\rfloor$, and $v_{2}>2$. The right to left
implication follows immediately from Proposition 2. For the left to right
implication, suppose that $(X,Y)$ is a Nash equilibrium of all-pay auction
with $v_{1}$ and $v_{2}$ satisfying the conditions listed above. Then, by
Propositions 2, 3, 4, 5, 6, and 7, $\mathbf{E}(X)=\mathbf{E}(Y)=\lfloor
v_{2}/2\rfloor=m$ and, by Proposition 2, $(X,Y)$ satisfies all the conditions
stated in point (i) of the theorem.

For point (iii) assume that $v_{2}/2\notin\mathbb{Z}$, $v_{1}/2>\lfloor
v_{1}/2\rfloor+1$, and $v_{2}>2$. The right to left implication follows
immediately from Proposition 6. For the left to right implication, suppose
that $(X,Y)$ is a Nash equilibrium of all-pay auction with $v_{1}$ and $v_{2}$
satisfying the conditions listed above. Then, by Propositions 2, 3, 4, 5, 6,
and 7, $\mathbf{E}(X)=m+\alpha$ with $\alpha\in(0,1)$ and $m=\lfloor
v_{2}/2\rfloor$, and $\mathbf{E}(Y)=b$ with $0<b<m$ . Thus, by Proposition 6,
$(X,Y)$ satisfies all the conditions stated in point (iii) of the theorem.

For point (ii) assume that $v_{2}/2\notin\mathbb{Z}$, $v_{1}/2=\lfloor
v_{1}/2\rfloor+1$, and $v_{2}>2$.

For the right to left implication take any strategy profile $(X,Y)$ satisfying
the conditions of point (ii) of the theorem. If $(X,Y)$ satisfies the
conditions with $\alpha>0$ then $(X,Y)$ is a Nash equilibrium by Proposition
7. If $(X,Y)$ satisfies the conditions with $\alpha=0$ then
$Y=U_{\mathrm{E}}^{m}=\kappa U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m}$
with

$\kappa=\frac{2m}{v_{1}}\left(m+1-\frac{v_{1}}{2}\right)=0$

and

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}U_{\mathrm{O}}^{m}+\lambda_{\mathrm{E}}U_{\mathrm{E}}^{m}+\sum_{j=1}^{m}\lambda_{j}U_{\mathrm{O}}^{m}+\sum_{j=1}^{m}\kappa_{j}U_{\mathrm{E}}^{m},$

with
$\lambda_{\mathrm{O}},\lambda_{\mathrm{E}},\lambda_{1},\ldots,\lambda_{m},\kappa_{1},\ldots,\kappa_{m}\geq
0$ and
$\lambda_{\mathrm{O}}+\lambda_{\mathrm{E}}+\sum_{j=1}^{m}\lambda_{j}+\sum_{j=1}^{m}\kappa_{j}=1$.
Let $\beta=\lambda_{\mathrm{O}}+\sum_{j=1}^{m}\lambda_{j}$. Then

$\displaystyle 1-\beta$
$\displaystyle=\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}=\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$
$\displaystyle=(m+1)\left(1-\frac{2m(m+1)}{v_{2}}\right)=1-\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right)$

Thus $(X,Y)$ is a Nash equilibrium by Proposition 2. This completes proof of
the right to left implication.

For the left to right implication, suppose that $(X,Y)$ is a Nash equilibrium
of all-pay auction with $v_{1}$ and $v_{2}$ satisfying the conditions listed
above. Then, by Propositions 2, 3, 4, 5, 6, and 7, one of the following cases
holds:

  1. 1.

$\lfloor v_{2}/2\rfloor=\lceil v_{1}/2\rceil-1$,
$\mathbf{E}(X)=\mathbf{E}(Y)=\lfloor v_{2}/2\rfloor=m$,

  2. 2.

$\mathbf{E}(X)=m+\alpha$ and $\mathbf{E}(Y)=m$ with $m=\lfloor
v_{2}/2\rfloor$.

In case 1, by Proposition 2, $Y=\kappa
U_{\mathrm{O}}^{m}+(1-\kappa)U_{\mathrm{E}}^{m}$ with $\kappa=0$, so
$Y=U_{\mathrm{E}}^{m}$, and $X=\beta
U_{\mathrm{O}}^{m}+(1-\beta)U_{\mathrm{E}}^{m}$ with

$\beta=\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right).$

Thus

$\displaystyle X$
$\displaystyle=\lambda_{\mathrm{O}}((1-\alpha)U_{\mathrm{O}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+\lambda_{\mathrm{E}}((1-\alpha)U_{\mathrm{E}}^{m}+\alpha
U_{\mathrm{O}}^{m+1})+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\lambda_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{O}}^{m}\right)+{}$
$\displaystyle\qquad\sum_{j=1}^{m}\kappa_{j}\left(\alpha\delta
V_{j}^{m}+\left(1-\alpha\delta\right)U_{\mathrm{E}}^{m}\right),$

where

$\delta=\frac{2\left\lceil\frac{v_{2}}{2}\right\rceil-1}{\left\lceil\frac{v_{2}}{2}\right\rceil},$

with $\alpha=0$,
$\lambda_{1}=\ldots=\lambda_{m}=\kappa_{1}=\ldots=\kappa_{m}=0$,
$\lambda_{\mathrm{O}}=\beta$ and $\lambda_{\mathrm{E}}=1-\beta$. Hence

$\displaystyle\lambda_{\mathrm{E}}+\sum_{i=1}^{m}\kappa_{i}\frac{1-\alpha\delta}{1-\alpha}$
$\displaystyle=\lambda_{\mathrm{E}}=1-\frac{2m}{v_{2}}\left(m+1-\frac{v_{2}}{2}\right)=(m+1)\left(1-\frac{2m(m+1)}{v_{2}}\right)$
$\displaystyle=\frac{\left\lceil\frac{v_{2}}{2}\right\rceil\left(\frac{v_{2}}{2}-\left\lfloor\frac{v_{2}}{2}\right\rfloor\right)}{\frac{v_{2}}{2}(1-\alpha)}-\frac{\alpha}{1-\alpha}$

with $\alpha=0$ so $(X,Y)$ satisfies all the conditions stated in point (iii)
of the theorem with $\alpha=0$. In case 2, by Proposition 7, $(X,Y)$ satisfies
all the conditions stated in point (ii) of the theorem. This completes proof
of the left to right implication.

### Proof of Theorem 0.A.1

###### Theorem 0.A.1

Strategy profile $(X,Y)$ is a Nash equilibrium of all-pay auction with players
valuations $v_{1}\geq v_{2}>0$ and $v_{2}<2$ if and only if $Y=\mathbf{1}_{0}$
and

  * •

$\frac{v_{1}}{2}>1$ and $X=\mathbf{1}_{1}$,

  * •

$\frac{v_{1}}{2}=1$ and $X=(1-\alpha)\mathbf{1}_{0}+\alpha\mathbf{1}_{1}$
where $\alpha\in[0,1]$,

  * •

$\frac{v_{1}}{2}<1$ and $X=\mathbf{1}_{0}$.

Equilibrium payoffs of the players are

  * •

if $\frac{v_{1}}{2}>1$ then $P^{1}(X,Y)=v_{1}-1$,

  * •

if $\frac{v_{1}}{2}=1$ then $P^{1}(X,Y)=(1+\alpha)\frac{v_{1}}{2}-\alpha$,

  * •

if $\frac{v_{1}}{2}<1$ then $P^{1}(X,Y)=0$,

and $P^{2}(X,Y)=0$.

###### Proof

For the left to right implication, let $(X,Y)$ be a Nash equilibrium of all-
pay auction with players valuations $v_{1}\geq v_{2}>0$ and $v_{2}<2$. By
Proposition 1, $(X,Y)$ is a Nash equilibrium of General Lotto game
$\mathit{\Gamma}(\mathbf{E}(X),\mathbf{E}(Y))$. Hence, by Propositions 2, 3,
4, 5, 6, and 7, either $\mathbf{E}(Y)=0$ or $\mathbf{E}(X)=0$. Suppose that
$\mathbf{E}(Y)=0$. Then $Y=\mathbf{1}_{0}$ and, by (8) and (4),

$H(X,\mathbf{1}_{0})=1-\mathbf{P}(X=0).$

Two cases of values of $\mathbf{E}(X)$ are possible: $\mathbf{E}(X)\in[0,1)$
and $\mathbf{E}(X)\geq 1$. Suppose that $\mathbf{E}(X)=m\geq 1$. By (3) payoff
to player $1$ from strategy profile $(X,Y)$ is

$P^{1}(X,Y)=\frac{v_{1}}{2}\left(H(X,Y)-\frac{2\mathbf{E}(X)}{v_{1}}+1\right).$

Since $m\geq 1$ so there exists $X$ with $\mathbf{P}(X=0)$ and
$\mathbf{E}(X)=m$ (e.g. a convex combination of $\mathbf{1}_{\lfloor
m\rfloor}$ and $\mathbf{1}_{\lceil m\rceil}$) so

$P^{1}(X,Y)\geq\frac{v_{1}}{2}\left(1-\frac{2\mathbf{E}(X)}{v_{1}}+1\right)=v_{1}-\mathbf{E}(X).$

Thus, in the case of $\mathbf{E}(X)=m\geq 1$, $P(X,Y)$ is maximised when
$\mathbf{E}(X)=1$ and $\mathbf{P}(X=0)=0$, in which case $X=\mathbf{1}_{1}$.

Suppose that $\mathbf{E}(X)\in[0,1)$. Then

$H(X,\mathbf{1}_{0})=1-\mathbf{P}(X=0)=\sum_{j\geq 1}\mathbf{P}(X\geq
j)-\sum_{j\geq 2}\mathbf{P}(X\geq j)=\mathbf{E}(X)-\sum_{j\geq
2}\mathbf{P}(X\geq j)\leq\mathbf{E}(X)$

with equality when $\mathbf{P}(X\geq 2)=0$. Thus the equality is attained by
strategy $(1-\alpha)\mathbf{1}_{0}+\alpha\mathbf{1}_{1}$, where
$\alpha=\mathbf{E}(X)$. By (3) payoff to player $1$ from strategy profile
$(X,Y)$ is

$P^{1}(X,Y)=\frac{v_{1}}{2}\left(H(X,Y)-\frac{2\mathbf{E}(X)}{v_{1}}+1\right)\geq\frac{v_{1}}{2}\left(\mathbf{E}(X)-\frac{2\mathbf{E}(X)}{v_{1}}+1\right)=\mathbf{E}(X)\left(\frac{v_{1}}{2}-1\right)+\frac{v_{1}}{2}.$

Since $(X,Y)$ is a Nash equilibrium so $\mathbf{E}(X)=0$ and
$X=\mathbf{1}_{0}$ if $v_{1}/2<1$, $\mathbf{E}(X)=\alpha\in[0,1]$ and
$X=(1-\alpha)\mathbf{1}_{0}+\alpha\mathbf{1}_{1}$, if $v_{1}/2=1$, and
$\mathbf{E}(X)=1$ and $X=\mathbf{1}_{1}$ if $v_{1}/2>1$.

It is elementary to verify that the strategy profiles stated in the theorem
are Nash equilibria for the corresponding values of $v_{1}$ and $v_{2}$.