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\begin{document}
 
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\title{Splitting a Pie: Mixed Strategies in Bargaining under Complete Information}

%
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% \bigskip
%Christopher Connell and Eric Rasmusen
% \bigskip
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\author[Connell]{Chris Connell}
\address[Chris Connell]{Department of Mathematics, Rawles Hall, Indiana University, Bloomington, IN 47405.}
\email{connell@indiana.edu}

\author[Rasmusen]{Eric Rasmusen$^\dagger$}
\address[Eric Rasmusen]{Department of Business Economics and Public Policy, Kelley School of Business, Indiana University 
Bloomington, Indiana, 47405.}
\email{erasmuse@indiana.edu}

\begin{abstract}
 We study the mixed-strategy equilibria for the bargaining game in which two players simultaneously bid $p$ and $q$ for a share of a pie and receive shares proportional to their bids or zero if the bids sum to more than 100\%. Of particular interest is the equilibrium in which each player's support is a single interval. We show that this consists of convex increasing densities $f_1(p)$ on $[{a},b]$ and $f_2(q)$ on $[1-b,1-a]$ together with atoms of probability at $p=a$ and $q=1-b$. These are unique for any given $0<a<b<1$. The two single outcomes with highest probability are breakdown and an $\set{\frac{a}{a+1-b}, \frac{1-b}{a+1-b}}$ split. We provide formulas for the probability weights of the atoms and exact power series representations for the densities $f_1$ and $f_2$.
\end{abstract}

\thanks{${}^\dagger$We would like to thank Rick Harbaugh and participants in Indiana University's BEPP Brown Bag  for their comments.}
%\noindent
%Connell:  Department of
%Mathematics,  Indiana University,  Rawles Hall,
%RH 351,   Bloomington,
%Indiana, 47405.     (812) 855-1883. Fax:(812) 855-0046.
%\href{mailto:connell@indiana.edu}{connell@indiana.edu}.\\
% Rasmusen:  Department of Business Economics and Public Policy, Kelley
%School
%of Business, Indiana University.   1309 E. 10th Street,
%Bloomington,
%Indiana, 47405-1701. (812) 855-9219. 
%\href{mailto:erasmuse@indiana.edu}{ erasmuse@indiana.edu}.   


%\subjclass[JEL Codes]{C72,C78,D70}
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\maketitle

 
 
\section{Introduction}\label{sec:1}
 
  A  fundamental problem  in game theory is how to model bargaining between two players who must agree on shares  of surplus or obtain zero payoff. In the classic bargaining problem ``Splitting a Pie'' (also known as ``Dividing the Dollar"), two players bid simultaneously for shares of the pie. If their bids add up to more than one,   both get a breakdown payoff of zero. The game has a continuum of pure strategy equilibria,  for example,  (0.00, 1.00) and (.30,.70).  A multitude of other models of bargaining exist, of which the two best known are the Nash bargaining solution of Nash (1950) and  the alternating offer model of Rubinstein (1982). Nash takes the approach of cooperative game theory and finds axioms which guarantee a 50-50 split. Rubinstein takes the approach of a noncooperative alternating-offer model   and finds close to a 50-50 split, with a small advantage to whichever player makes the first offer.   In both models, the players always agree  in equilibrium. Other models incorporate  incomplete information, in which case failure to agree can occur in equilibrium when a player refuses to back down because he thinks, wrongly, that the other player's payoff function will cause him to agree to accept his proposal.

  Muthoo (1999) is the standard book-length treatment of  bargaining under complete information and the  Ausubel,    Cramton \&   Deneckere  (2002)  {\it Handbook of Game Theory} chapter treats bargaining under incomplete information. Since Nash  there has been much work on this game  and its relation to the Nash bargaining solution, notably  Binmore, Rubinstein  \&   Wolinsky  (1986),   
Binmore   (1987), and   
Carlsson   (1991).  More recently, Anbarci \& Boyd (2011) relate the game to cooperative game theory more broadly
 and  Bastianello \& LiCalzi (2019) use a different axiomatic approach generalizing Nash by the introduction of a hypothetical mediator and probabilistic acceptances. 
 Andersson,  Argenton \& Weibull  (2018) continue with the idea of using a different kind of uncertainty than incomplete information, and a number of authors have further explored Rubinstein's 1982 model of offers over time under complete information--- see  
Britz,   Herings   \& Predtetchinski    (2010), 
Kawamori  (2014), and Friedenburg (2019).  
 

 In this paper we return to the classic  ``Splitting a Pie''  and  investigate its mixed-strategy equilibria.  Although examples of mixed-strategy equilibria appear as early as Roth (1985),  they do not seem to have been studied systematically until Malueg (2010), who showed that a huge variety  exist. In particular, he showed that for any   ``balanced'' pair of  supports,  such that for each bid in a player's support there is a bid in the other player's support that sums to one,    there exists an equilibrium.   He  does this for general utility functions and general  sharing rules for what happens if the two players' bids add up to   less than 1. 
 
  The easiest sharing rule is the   ``take what you bid''  rule used in the ``Nash Demand Game'' of Nash (1953): if the bids are .20 and .30,  Player 1   receives  20\%, Player 2  receives 30\%, and the other 50\% is abandoned.  Nash himself used the ``split the difference'' sharing rule that if the bids were .20 and .30, the shares would be 45\% (=  .2 + .5/2) and   55\% (= .3+ .5/2).  
  
    We believe   the most  natural specification is  proportional sharing (also called ``the Tullock rule''):   if the bids  are .20 and .30, the shares are  40\% (= .2/(.2+.3)) and  60\% (= .3/(.2+.3)).     Proportional sharing is suitable if  the  model is to be  a metaphor  for  players deciding   how ``tough'' to be  in   negotiations. If both choose to be too tough, they fail to reach agreement, which is represented by complete  breakdown and payoffs of zero.  If they  are less tough,  it is  their relative toughness that matters, and they split the pie  in proportion to toughness. Proportional sharing has  desirable properties that ``take what you bid'' lacks:  (a) if agreement is reached,    one player's share  is  greater if  the other player is less tough,  and  (b) if the players  come to an agreement, they share the entire pie rather than leaving some of it to waste. 
    
     Other sharing rules such as ``split the difference''  also satisfy (a) and (b), and since ``toughness'' is an artificial construct without natural units, the differences between sharing rules that do allocate the entire pie upon agreement with interaction between the players' shares are less important.   Of course, all these sharing rules maintain the assumption that there is a discontinuity when bids sum to more than 1, in which case bargaining breaks down and both players receive zero. That assumption is what prevents the players from each bidding an infinite amount; in the bargaining game, unlike, for example, rentseeking, bidding higher is costless in terms of effort or expenditure, so there needs to be some reason why bidding higher is not always better.  
     
 Unfortunately,  under proportional sharing    the mathematics is  difficult   because a player's share depends not just on his own bid  but  on the other player's. Nonetheless, Malueg (2010) proved existence  for general sharing rules, though he restricted himself to rules like ``take what you bid''  for uniqueness and describing the equilibrium.
 
    We focus on the proportional sharing rule, and we completely characterize a particularly interesting class of equilibria from an economic point of view, the ``interval equilibria," in which each player mixes over a single interval. An example   is  the symmetric equilibrium in which both players mix over  [.30, .70].  We  show that  there  is a unique mixing distribution for this support, consisting  of a probability atom at .30 and a convexly increasing density over (.30, .70].  The resulting combination of  a modal 50-50 split, occasional breakdown, and a large range of other possible shares  is consistent with real-world behavior. As we will see, there are also asymmetric interval equilibria.
  
  Section \ref{sec:2} lays   out the model and discusses two simple approaches to mixed-strategy equilibria: the Hawk-Dove equilibria  in which each player mixes over two bids, and interval equilibria in the ``Nash Demand Game.'' Section \ref{sec:interval}    analyzes interval equilibria for the game with proportional sharing. We first characterize  interval   equilibria, both symmetric and asymmetric, using a relatively  nontechnical approach that provides intuition (Proposition \ref{prop:1}).  We  turn to  a more technical methodology  to derive uniqueness and the  exact atom size  and density function (Proposition \ref{prop:main}).
    
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% \newpage
 
 
 
\section{The Model and Two Easy Special Cases}\label{sec:2} %SECTION 2

Players 1 and 2 simultaneously choose bids ${p}$ and ${q}$ in the interval $[0,1]$.  If their bids add up to more than $100\%$, then we say {\em breakdown} occurs, and  both get zero. Otherwise, they split a pie of size $1$ according to a sharing rule $\set{v_1,v_2}$. Thus, Player $i$'s   payoff function is (for $i=1,2$)
\[
u_i(p,q)=\begin{cases} v_i(p,q) & \text{ if }p+q\leq 1 \\ 0 & \text{ otherwise }\end{cases}.
\]
Throughout this paper we assume that the players' strategies are Borel probability measures $(\mu_1,\mu_2)$ with supports $(A,B)$ with $A$ and $B$ subsets of $[0,1]$.\footnote{For a nonnegative Borel measure $\mu$, its {\em support} $\supp(\mu)$ is defined to be the smallest closed set whose complement has $0$ $\mu$-measure, i.e. $\ds{\supp(\mu)=\bigcap_{\set{K=\overline{K}:\mu(K^c)=0}} K}$. For a general signed measure $\mu$ its support is $\supp(\mu^+) \bigcup \supp(\mu_-)$ where $\mu=\mu_+ -\mu_-$ for the unique nonnegative measures $\mu_+$ and $\mu_-$ given by the Hahn decomposition theorem.}  Thus, the expected payoff for player $i=1,2$ is:
\[
\pi_i(\mu_1,\mu_2)=\int_{[0,1]\times[0,1]} u_i(p,q) d\mu_1({p})d\mu_2(q).
\]

At equilibrium measures $(\mu_1,\mu_2)$, the expected payoff of   player $i$ will be equal to his expected payoff when $\mu_i$ is replaced by a single atom in the support of $\mu_i$, so one can solve for  his equilibrium measure by requiring a constant payoff across the pure strategies he mixes among. When either $\mu_1$ or $\mu_2$ is an atom at a point in $[0,1]$ and the other measure is understood,   we simplify the notation so that, for instance,   Player 2's expectation of the payoff if he bids $q$  is:
\begin{equation}\label{eq:gen_payoff}
\pi_2(q)=\int_{0}^{1} u_2(p,q) d\mu_1({p})=\int_{a}^{1-q} v_2(p,q) d\mu_1({p}),
\end{equation}
where we have assumed the greatest lower bound of the support of $\mu_1$ is $a=\inf A$.

Starting with sets $A\subset [0,1]$ and $B\subset[0,1]$, we think of the expected payoff $\pi_2(q)$ as a function on the set $B$ and the problem is to find all signed measures $\mu_1$ supported on $A$ so that equality \eqref{eq:gen_payoff}  holds for some constant $\pi_2^*=\pi_2(q)$ for all $q\in B$. (And similarly for $\mu_2$ on $B$.) Since equation \eqref{eq:gen_payoff} is invariant under scaling the measure $\mu_1$ and payoff $\pi_2(q)$ by any constant $r\in\R$, without loss of generality we may assume $\norm{\mu}=1$. In other words, we wish to find signed measures $\mu$ such that when Player 1's strategy measure $\mu_1$ equals $\mu$, there is a constant $\pi_2^*$ such that: 
\begin{equation}\label{eq:fund1_gen}
\pi_2^* =\pi_2(q)=\int_a^{1-{q}} v_2(p,q) d\mu_1({p})\quad\text{and}\quad \norm{\mu_1}=1,
\end{equation}
and similarly for $\mu_2$.

%We observe that the zero measure $\mu_1=0$ with $\pi_2^*=0$ is always a solution, and more generally that if $\mu_1=\mu$ with $\pi_2^*=C$ is a solution then for any constant $r\in \R$ the measure $r\mu_1$ with $\pi_2^*=rC$ is also a solution. Hence we will always assume that our solutions 
    
    
    
%     Instead, while retaining that the pie explodes if the shares sum to more than 1, let us assume the following: 

For his general existence result, Malueg (2010) makes the following assumptions on the sharing rules. 
\begin{Assumptions}\label{assump}
For players $i=1$ and $2$ bidding $p$ and $q$ the sharing rule $\set{v_1,v_2}$ satisfies the following properties:
%The payoff function 
%$$u_i(p,q)=\begin{cases} v_i(p,q) & \text{ if }p+q\leq 1 \\ 0 & \text{ otherwise }\end{cases}$$
%    for players $i=1$ and $2$ bidding $p$ and $q$ satisfies the following properties:
\begin{enumerate}
\item $v_i:[0,1]\times [0,1]\to [0,\infty)$ are Borel measurable and continuous on $(0,1]\times (0,1]$.
\item $v_1(p,0)>v_1(0,0)$ and $v_2(0,q)>v_2(0,0)$ for all $p,q\in (0,1]$.
\item $v_1(p,q)$ is nondecreasing in $p$ and nonincreasing in $q$, and strictly increasing in $p$ if $q>0$. 
\item $v_2(p,q)$ is nonincreasing in $p$ and nondecreasing in $q$, and strictly increasing in $q$ if $p>0$. 
\end{enumerate}
\end{Assumptions}

Assumptions (1) are technical assumptions ensuring payoffs are well defined and that certain approximation arguments hold. Assumptions (2) guarantee that simultaneous bids of 0 are not pure strategy equilibria, and Assumptions (3)-(4) guarantee that while bids sum to less than one, increasing one player's bid helps that player while not helping the other. 
%Our second assumption on the sharing rule will be needed only for uniqueness results. Let $D(A)$ be the set of bounded functions $f$ on $A\subset [0,1]$ which at each point $x\in A$ have $\lim_{t\nearrow x}f(t)=f(x)$ and $\lim_{t\searrow x}f(t)$ exists. (Such functions go by the name {\em caglad}: ``continu \`a gauche, limites \`a droite.'') Under assumptions \ref{assump}  and for compact $A\subset (0,1)$, the functions $u_2(\cdot,q)$ belong to $D(A)$ for all $q\in B$.
%
%\begin{AssumptionB}\label{assump2}
%The spaces
%\[
%H_1=\Span(\set{u_1(p,\cdot)\,:\, p\in A}\cup\set{1})\quad\text{and}\quad H_2=\Span(\set{u_2(\cdot,q)\,:\, q\in B}\cup\set{1})
%\]
%have closures in $D(B)$ (respectively $D(A)$) containing $C^0(B)$ (respectively, $C^0(A)$, respectively.
%\end{AssumptionB}
%
%Assumption \ref{assump2} excludes exotic Borel signed measures with unit mass that integrate to 0 on all payoff functions (see Lemma \ref{lem:exotic}).

As discussed in   Section \ref{sec:1}, proportional sharing--- players receiving shares proportional to their bargaining intensity---  is our favored bargaining rule. 

     
%\noindent
\begin{Definition}
The {\bf proportional sharing rule}, also called the {Tullock Rule}, is the sharing rule  $v_1= {p}/  ({p}+{q})$   and   $v_2={q}/  ({p}+{q})$  unless  ${p}={q}=0$, in which case $v_1=v_2 =.5$. 
\end{Definition}

It is easy to verify that the proportional sharing rule satisfies assumptions \ref{assump}. Much of our analysis will also be looking at one intuitive category of equilibrium:   interval equilibria.

 

%\noindent
\begin{Definition}
An {\bf interval equilibrium} is a Nash equilibrium in which Player 1 bids by mixing over the support $A=[a, b]$ with $b>a$ and Player 2 bids by mixing over the support $B=[c,d]$ with $d>c$.
%
%An {\bf interval equilibrium} is a Nash equilibrium in which Player 1 bids by mixing  with density $f_1(p)>0$ over the interval $[a, b]$ and Player 2 bids with density $f_2(q)>0$ over  $[c,d]$, possibly with atoms somewhere in the interval or at the end points.\footnote{We actually consider arbitrary (Borel) equilibrium measures, provided they are supported on the entire closed interval. However, we will later show that any such equilibrium measure will consist of a measure in the Lebesgue class plus a single atom at the lower endpoint.}
\end{Definition}

%(In the above definition, recall that the support of a measure is the complement of the union of all open sets with measure zero.) 
In particular, every nonempty open subinterval of $[a,b]$ and $[c,d]$ has positive probability for their  corresponding measures.
A key feature of any    mixed-strategy equilibrium    will be  ``balanced supports'':

%\noindent
\begin{Definition}  
The player's bid supports are {\bf balanced} if and only if   element $s$ being in  Player 1's  support  $A$ implies that   $1-s$ is in Player 2's   support  $B$, and vice versa.
\end{Definition}

A balanced support for an interval equilibrium can be either symmetric, e.g.  both players mixing over $[.2, .8]$, or asymmetric, e.g. Player 1 mixing over $[.2, .4]$ and Player 2 over $[.6, .8]$.

    Splitting a Pie has a continuum of pure strategy Nash  equilibria: every pair of bids with ${p}+{q} = 1$.  There are also many mixed-strategy equilibria, some with bidding over a finite set of points, some with bidding over a continuum. Before we proceed to characterizing interval equilibria, it will be helpful to  understand two special cases:  Hawk-Dove equilibria for Splitting a Pie, and  interval equilibria for the Nash Demand Game. 

  
 
\subsection{Hawk-Dove Equilibria for Splitting a Pie}

The simplest mixed strategy for Splitting a Pie has the two players  each  mixing over  two bids.  This is a 
Hawk-Dove equilibrium,   mathematically the same as the well-known  biological model   of birds deciding whether to pursue aggressive or timid strategies.   
  
    In a symmetric Hawk-Dove equilibrium, each player chooses $a$ with probability $\mu(a)$ and $1-a$ with probability $\mu(1-a)=1-\mu(a)$ for $a < .5$.  The two bids must add up to 1, because otherwise
 it would be a profitable deviation  for one player to choose a bigger number for his lower bid,    increasing his share without  any  greater likelihood of breakdown.   This is an immediate example of the importance of ``balanced supports.'' Let us assume that the sharing rule is proportional sharing.   The mixing probabilities  must make  each action's expected payoff    the same in equilibrium, so  for $i = 1,2$, 
 \begin{equation}
 \pi_i(a) =   \mu(a) (.5) +(1-\mu(a))a = \pi_i(1-a) =   \mu(a)  (1-a) +(1-\mu(a)) (0), 
\end{equation}
 which solves to 
$\mu(a)  =   2a$ and    $ \pi_1=\pi_2  = 2a(1-a)$.

  The players share the pie equally in equilibrium with probability $ 4a^2$ and  bargaining breaks down with probability $(1- 2a)^2$.  Note that there is a continuum of equilibria, since any value of $a$ in (0,.5) can support an equilibrium, and they can be pareto-ranked, with higher payoffs if $ a$ is closer to .5.  In the  case  when $a=0$, both players choose  $a=0$ with probability 0 and $1-a=1$ with probability 1,  and the expected payoff is zero.  Observe that $\mu(a) <1/2$ for    $a <1/4$,  so in that case the players actually choose the higher of their two bids (hawk) with the highest probability,  in contrast to the result in  biology's Hawk-Dove game.
  
 
   There also  exist asymmetric Hawk-Dove equilibria, as Malueg (2010) points out. Player 1 chooses $a$ with probability $\mu_1(a)$ and $b$ with probability $(1-\mu_1(a))$,  for $a<b < .5$. Player 2 chooses $1-b$ with  probability $\mu_2(1-b)$ and $1-a$ with probability $\mu_2(1-a)=(1-\mu_2(1-b))$. Player 1's expected payoffs  from his two pure strategies must be equal for him to be willing to mix between them, so 

\[
\pi_1(a) = \mu_2(1-b) \left(  \frac{a}{a+1-b}\right)   + (1-\mu_2(1-b))   \left(  \frac{a}{a+1-a}\right)
 =  \pi_1(b) = \mu_2(1-b)  \left(  \frac{b}{b+1-b}\right)    + (1-\mu_2(1-b))\cdot 0,
\]
and  $  \mu_2(1-b)  = \frac{a(a+1-b)}{a^2-b^2+b}$ with payoff $\pi_1=\frac{a b (a+1-b)}{a^2-b^2+b}$. 
    
Similarly, Player 2's expected payoffs from his two pure strategies must   be equal, so 
\[
\pi_2(1-b) = \mu_1(a)  \left(  \frac{1-b}{a+1-b}\right)    + (1- \mu_1(a))  \left(  \frac{1-b}{b+1-b}\right) = \pi_2(1-a) =   \mu_1(a)  \left(  \frac{1-a}{a+1-a}\right)   + (1- \mu_1(a))\cdot 0,
\]
and  $  \mu_1(a) = \frac{(1-b)(a+1-b)}{(1-b)^2-a^2+a}$ with payoff $\pi_2=\frac{(1-a) (1-b) (a+1-b)}{(1-b)^2-a^2+a}$. 
 
 Thus we might have Player 1 choosing  .10 and .40 with probabilities  of    .28 and .72  for an expected payoff of   .11;  and Player 2 choosing .60 and .90 with probabilities of    .93  and .07  for an expected payoff of   .84 (rounding to two digits).  
 
The Hawk-Dove equilibrium is of interest since it is a stepping stone to an interval equilibrium where the probability mass must be concentrated at the two endpoints. While it cannot tell us about the densities, it does illustrate why supports must be balanced and how asymmetric equilibria can arise.

\subsection{The Nash Demand Game}

  In   the  
  ``Nash Demand Game,''     players  use  the ``take what you bid'' sharing rule.  The players bid $p$ and $q$, and they receive $p$ and $q$ as payoffs if $p +q \leq 1$ but zero if $p+q>1$ (Nash (1953)).   As discussed in Section 1, this means that if both players are   mild in their bargaining strategy, it will happen that $p+q <1$ and in their agreement they will specify that  some of the pie goes to neither of them, a strange outcome for successful bargaining. The sharing rule also has the feature that unless there is breakdown, each player's payoff depends only on his own bid, not the other player's. This feature will make it easy to solve for equilibrium. 
  
    In one   equilibrium of  the Nash Demand Game, each player bids $a$ with probability $K$ and then mixes using density $f=f_1=f_2$ over the interval $[a, 1-a]$. 
       Player  2 can guarantee a payoff of $a$ by bidding $a$, since   $p+a \leq 1$ for any bid $p$   Player 1 might play. Player 2 will have a payoff of $K(1-a)$ from bidding $q=(1-a)$, since Player 1 bids $p=a$ with probability $K$ and otherwise Player 1 bids $p>a$, whence $p + (1-a)>1$ and both players receive $0$. Since Player 2 is only willing to mix between bids if they have equal expected payoffs, this implies $\pi_2(a) = a = \pi_2(1-a) = K (1-a)$ and we can conclude that $K = \frac{a}{1-a}$. 
  
    For bids between $a$ and $1-a$, Player 2's expected payoff is 
    \begin{equation} \label{e4}
\pi_2 ({q})= Kq  +\int_a^{1-{q}}q f ({p})  d{p},    
\end{equation}
    which we can rewrite using $F$ as the cumulative distribution for $f$ and   our knowledge that $K = \frac{a}{1-a}$, and combine with the requirement that $\pi_2 ({q}) = \pi(a) = a$ to yield
        \begin{equation} \label{e5}
\pi_2 ({q})= \left( \frac{a}{1-a} \right) q  + qF(1-q) = a.    
\end{equation}
    Using the change-of-variables $p = 1-q$, this becomes
$
 \left( \frac{a}{1-a} \right) (1-p) + F(p) (1-p) = a$,    
which solves to 
  $F(p) =    \frac{a}{1-p}  -  \frac{a}{1-a}$, which can be differentiated to yield 
  the equilibrium mixing density,
$
  f(p) =    \frac{a}{(1-p)^2}  .  
$

 The Nash Demand Game is easy to solve because each player's payoff function depends on the other player's  bid only if breakdown occurs. If the bids add up to less than one, a player's payoff is entirely independent of what the other player does. This is what allows us to move smoothly from equation  \eqref{e4}'s $\int_a^{1-{q}}q f ({p})  d{p}$  to equation  \eqref{e5}'s $qF(1-q)$. If Player 2's share of the pie depended on Player 1's bid   instead of  being just $v_2(q,p)=q$,   equation  \eqref{e4} would have the expression $\int_a^{1-{q}} v_2(q,p) f ({p})  d{p}$ and it would no longer be straightforward to extract $f(p)$ from the integral. We will do that extraction  in Section 3 for the proportional sharing rule.  
 
    
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
 
 
\section{Interval Equilibria under Proportional Sharing}\label{sec:interval} %Section 3
 
 
   We will now consider interval equilibria for Splitting a Pie,   e.g.,  both players mix over [.2, .8],  or Player 1 mixes over [.2, .4]  and Player 2   over [.6, .8].  Under proportional sharing we characterize interval equilibria (Proposition 1) and find an explicit formula for their mixing distributions (Proposition 2).  


 
\begin{Proposition}\label{prop:1}
Any interval equilibrium consists of exactly one probability atom of size $K_1$ at $a$ and density $f_1$ on $[a,b]\subset (0,1)$ for Player 1, and probability atom $K_2$ at $1-b$ and density $f_2$ on $[1-b,1-a]$ for Player 2. The densities start strictly positive with $f_1(a) = \left(  \frac{a   }{1-a}\right) K_1 $ and $f_2(1-b) =  \left( \frac{1-b  }{b}\right) K_2 $ and are convex increasing with positive derivatives of every order.
\end{Proposition}
  
%\noindent 
\begin{proof}{(Included here rather than the appendix because it is  relatively simple  and helpful for  intuition.)}
 
\noindent
{\it (1) The lower  and upper bounds  are strictly between zero and one: $0<a<b<1$ for Player 1's strategy $p$ and $0 < c<d<1$ for Player 2's strategy $q$.   }\\

    \hspace*{16pt} In an interval equilibrium, Player 2 will place positive probability on bids in some proper subinterval $(c',d') \subset [c,d]$. If Player 1 plays $p=0$, his payoff will be 0 if Player 2 bids anything but $q=0$, and will be positive (at 1/2) if Player 2 bids $q=0$. If Player 1 bids $1-d'$, on the other hand, his payoff will be positive if Player  2 bids in $[c',d' ]$ and 1 if Player 2 bids $q=0$. Hence, Player 1 should never bid $p=0$ and the lower bound of his support is $a>0$ (with $c>0$ by similar reasoning).

%    \hspace*{16pt} In a mixed-strategy equilibrium, Player 1's expected payoff from each of the strategies in his support must be equal.   Player 1's  expected payoff from the  strategy of ${p}=0$  is zero unless Player  2 bids $q=0$ with an atom of probability $K_2>0$, in which case it is   $ \frac{K_2}{2} $ because they will split the pie evenly. However in that case, Player 1 could increase his payoff by $\frac{K_2}{2}$ if he switched to putting zero probability on $p=0$, so we can conclude Player 1 does not include $p=0$ in his support.   
    
     Since Player 2 puts zero probability on $[0,c)$, if Player 1 bids $p=1$ there is always breakdown and a payoff of $0$. On the other hand, if Player 1 bids $p=1-c$ he always receives a positive payoff, and thus he should never choose $p=1$, showing that $b<1$ (and $d<1$ by parallel reasoning).

%      however,  then since he is not putting probability  1 on $q=1$ (recall we have assumed this is a mixed-strategy equilibrium), 
%  Player 1's payoff is lower from $p=0$ than from some slightly higher bid.  Thus,  Player 1  will not include  ${p}=0$ in  his mixing  support.    If Player 1's support does not include $p=0$, however, we can deduce that Player 2's support will not include $q=1$, because that strategy would always result in breakdown and a zero payoff for Player 2.     The same argument shows that Player 2's support will not include ${q}=0$ and Player 1's will not include $p=1$. 
    
  \noindent
 {\it (2)    The supports are  balanced:   Player 2's support is    $[c,d]=[1-b, 1-a]$}. \\  
  \hspace*{16pt}   Suppose $c<1-b$. Player 2 would deviate because a bid of ${q}\in [c,1-b)$ would earn a lower share than ${q}=1-b$ and he could therefore move all the mass in $[c,1-b)$ to $1-b$ without risking breakdown. Suppose $c > 1-b$. In that case there is breakdown whenever Player 1 bids $p$ in the interval $(1-c, b]$. Player 1 can avoid that zero payoff by  reducing $b$ so that $c = 1-b$ and reassigning the probability in any way. Thus, $c =1-b$. 
 
 Suppose $d<1-a$. Player 1 would deviate because a bid of ${p}\in [a,1-d)$ would earn a lower share than ${p}=1-d$ and he could therefore move all the mass in $[a,1-d)$ to $1-d$ without risking breakdown. Suppose $d > 1-a$. In that case there is breakdown whenever Player 2 bids $q$ in the interval $(1-a, d]$. Player 2 can avoid that zero payoff by  reducing $d$ so that $d = 1-a$ and reassigning the probability in any way. Thus, $d =1-a$. \todo{recheck again}
 
% Suppose $d >1-a$. Player 2 would deviate because a bid of $q\in(1-a , d]$   yields zero payoff,  so he could do better by moving that mass to $q=1-a$ which has positive payoff. Suppose $d< 1-a$. 
% 
% 
%  because if Player 1 bids $a +\epsilon$ for small enough $\epsilon$,  it will happen that $a+\epsilon +d<1$. If this happens, Player 1's bid of $p=a+\epsilon$ will fail to cause breakdown for any value of $q$ within Player 2's support and would yield Player 1 a higher share than would the bid ${p}=a$. Thus, if $d<1-a$, Player 1 could increase his expected payoff by raising the lower bound of his support to $a+\epsilon$. 
%  
% 

    
    
    \noindent
 {\it (3)  The mixing distributions  have probability atoms $K_1>0$  and $K_2>0$ at $a$ and $1-b$, and only there.} \\  
   \hspace*{16pt}     Player 1's expected payoff is positive because  Player 2 will not bid more than ${q}=1-a$,  and    if  Player 1  bids  ${p}= a$ breakdown will not result and  his expected payoff  will be at least $a$.  Consider Player 1's pure strategy of bidding ${p}= b$, the upper end of his support. This will cause  breakdown unless Player 2 bids ${q}= 1-b$, the lower end of his support. But if there  is no  probability atom for Player 2 at $1-b$, Player 1's expected payoff from bidding $b$ is zero. That is impossible in equilibrium, since zero is  less than Player 1's positive payoff from bidding $a$. Hence, in equilibrium, Player 2 must have a probability atom at $1-b$, the low end of his support.  
   
   Suppose Player 2 has an atom of size $K$ at some other  point $x$  in $(1-b, 1-a]$. This would induce Player 1 to deviate.    Player 1's pure strategy of bidding $1-x$ must have positive expected payoff, since that share is positive and  breakdown will not occur unless ${q} > x$, which has probability less than 1. Player 1's pure strategy of bidding ${p}=1-x+ \epsilon$ for sufficiently small $\epsilon$, however, will have a lower expected payoff than from  $1-x$  because the expected loss from the increased probability of breakdown will be at least $(1-x)K$ but the expected gain from having an $\epsilon$ higher share tends to $0$ as  $\epsilon$ tends to $0$. Note that this argument fails  to apply if $x=1-b$, because then a pure-strategy bid  of $1-x+\epsilon$ is not in Player 1's support and  does not  require a  payoff equal to that of $p=1-x$.
   
    A parallel argument shows that Player 1 must have an atom at $p=a$ and only at $p=a$. 
  
  
  
     
     
      \noindent   
     {\it (4) The equilibrium strategies apart from the atoms consist of densities $f_1$ and $f_2$ that start at the positive levels $f_1(a) = \left( \frac{a  }{1-a} \right) K_1 $ and $f_2(1-b) =  \left( \frac{  1-b   }{b} \right) K_2 $. } \\     
       \hspace*{16pt}  
Player 2's expected payoff from the pure strategy of bidding $q$ is
 \begin{equation} \label{payoff-p2}
\pi_2({q})  =\left( \frac{{q}}{{q}+ {a}}  \right) K_1  +\int_{{a}}^{1-{q}} \left(   \frac{{q}}{{p}+{q}}\right) d\mu_1(p)    
\end{equation}
for $q\in [{1-b},1-a]$. Since  the payoffs from  Player 2's pure-strategy best responses     are constant for all $q$ in the support, the derivative of  payoff   \eqref{payoff-p2} identically equals zero, and by the generalized Leibniz rule we may differentiate under the integral sign:
\begin{equation} \label{payoffderivative2}
0\equiv\frac{d \pi_2({q} )}{d{q}} =  \left( \frac{{a}   }{   ({q}+{a})^2}  \right)K_1  +   \int_{{a}}^{1-{q}}  \left(   \frac{{p}}{({p}+{q})^2}  \right) d\mu_1(p)  -  \frac{d}{dx}_{x=1-q}\int_a^x \left(   \frac{{q}}{{p}+{q}}\right) d\mu_1(p).
\end{equation}
Since the left hand side is $0$ and $\mu_1$ is nonatomic in $(a,b]$, the first two terms on the right hand side are continuous in $q$. Hence the derivative $\frac{d}{dx}_{x=1-q}\int_a^x \left(   \frac{{q}}{{p}+{q}}\right) d\mu_1(p)$ must exist and be continuous for all $q\in [1-b,1-a]$. Since $\frac{p}{q+p}$ is smooth in $p$ and $q$, the Borel measure $\mu_1$ must be in the Lebesgue class with continuous density $f_1$, i.e. $d\mu_1(p)=f_1(p)dp$ and $\frac{d}{dx}_{x=1-q}\int_a^x \left(   \frac{{q}}{{p}+{q}}\right) d\mu_1(p)=q f_1(1-q)$.
Setting $x\equiv 1-{q}$ for $x \in [ a,b]$ and solving for $f_1(x)$, we see that for  it to be the equilibrium density, it must be true that: 
\begin{equation} \label{density-g2}
  f_1(x)   =    \left(  \frac{1}{1-x} \right)\left(  \left( \frac{ a    }{   ( 1-x+a)^2 } \right.      \right)K_1 
  \left.  + \int_{{a}}^{x}  \left(   \frac{{p}}{(1-x+{p})^2}  \right) f_1({p}) d{p}        \right).
\end{equation}
By setting $x=a$ we can see that the  starting density is $f_1({a} ) =  \left( \frac{a}{1-a } \right) K_1.$  We can derive Player 2's density similarly by starting with $\pi_1(p)$, differentiating with respect to $p$, solving for $f_2(x)$, and setting $x=1-b$ to yield  $f_2(1-b) =  \left( \frac{ 1-b  }{b} \right) K_2$.
 
 
      
      \noindent
     {\it (5) The densities $f_1({p})$ and $f_2({q})$ have strictly positive derivatives of all orders, e.g.,  $f_1'>0, f_1''>0, f_1'''>0$, ...  } \\ 
       \hspace*{16pt} Noting that the right hand side of (\ref{density-g2}) is differentiable, we obtain that $f_1$ is differentiable, and we can differentiate equation (\ref{density-g2}) to obtain%\marg{Chceck this deriviafve; wrong in last draft.}
 \begin{align}
 \begin{split} \label{payoffderivative11}
   f_1'(x)=&    \left(  \frac{  1  }{ (1-x)^2  ( 1-x+a)^2 }+\frac{  2  }{ (1-x)   ( 1-x+a)^3 } \right) aK_1   +  \frac{x f_1(x)}{(1-x)} \\
 & +  \int_{{a}}^{x}    \left( \frac{ {p}}{ (1-x)^2(1-x+{p})^2}+  \frac{2{p}}{ (1-x)(1-x+{p})^3} \right)   f_1({p}) d{p}
\end{split}
\end{align}

The right hand side of \eqref{payoffderivative11} is differentiable for the same reason as in \eqref{density-g2}, and so $f_1'$ is differentiable and we obtain,

 \begin{align}
 \begin{split}  \label{payoffderivative12}
    f_1''(x) =&\left(\frac{2 }{(1-x)^3 (1-x+a)^2}+\frac{4}{(1-x)^2(1-x+a)^3}+\frac{6}{(1-x)(1-x+a)^4}\right)aK_1\\ &+\left(\frac{1}{1-x}+\frac{2 x}{1-x}+\frac{2
       x}{(1-x)^2}\right)f_1(x)  +\frac{x f_1'(x)}{1-x}\\
     & +   \int_a^x \left( \frac{2p}{(1-x)^3(1-x+p)^2}+\frac{4p}{(1-x)^2(1-x+p)^3}+\frac{6 p}{(1-x)(1-x+p)^4}\right)f_1(p)\, dp.
    \end{split}
 \end{align}
  
 Note that $f_1'>0$ and $f_1''>0$, since $0<x <1$.    
Continuing inductively, we obtain that derivatives of every order for $f_1$  since $f_1^{(k)}$ consists of rational functions and integrals of rational functions in $x$ times lower order derivatives of $f_1$ as in expressions \eqref{payoffderivative11} and \eqref{payoffderivative12}. 

We claim that for each $k$, $f_1^{(k)}$ is positive. At the $k$-th stage of differentiation one obtains that the rational function coefficients for $aK_1$ or $f_1^{(\ell)}$ with $0\leq\ell<k$ appearing in the expression for $f_1^{(k)}$, when kept unsimplified, have summands whose denominators all have the form  $(1-x)^m(1-x+a)^n$ for nonnegative integers $m$ and   $n$,   with   the numerators  being positive integers or positive integer multiples of $x$. Similarly, for the coefficient $f_1$ in the integrand, the summands are rational functions whose denominators have the form  $(1-x)^m(1-x+p)^n$ for nonnegative integers $m$ and $n$ and whose numerators are positive integer multiples of $p$. By inductive hypothesis the lower order derivatives of $f_1$ are positive and so each of these terms is positive for $0<a\leq x<1$. The positive sign of the derivatives of $f_2({q})$ can be shown similarly.  \end{proof}

 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
 
 
 
 We have not yet  demonstrated existence or uniqueness or  found   solutions for  
  the atoms and the mixing densities, but it will be appropriate to discuss the intuition here and explain the difficulties that will be involved in finding the distributions. 
  Figure \ref{fig-four-examples} show what we will find when the densities are computed using the formula from Proposition 2 below. It shows  symmetric equilibria    for four different supports:       $[.1, .9]$,     $[.2, .8]$,   $[.3, .7]$,   and  $[.4, .6]$.  As ${a}$ gets smaller, the solution begins flatter and ends steeper and with a smaller atom.   



% \begin{minipage}{1.0\textwidth}
\begin{figure}[ht!]
%\begin{centering}
	\includegraphics[width=4in]{fig-four-examples}
% \end{centering}
\caption{\sc    The  Symmetric-Equilibrium Mixing Density    $f_1({p})$  and Atom $K_1$  for ${a}=.10,.20,.30,.40 $ }\label{fig-four-examples}   
 \end{figure} 
%\end{minipage}
  
 The derivative expression \eqref{payoffderivative2}, which we expand here,  is helpful for intuition about the tradeoffs the players are making.  It must hold for every ${q}$ in Player 2's support. 
\begin{align}\label{payoffderivative2m}
0=\frac{d \pi_2({q} )}{d{q}} =   \frac{{a} K_1 }{   ({q}+{a})^2}  +   \int_{{a}}^{1-{q}}  \left(   \frac{{p}}{({p}+{q})^2}  \right) f_1({p} ) d{p}     -  {q} f_1(1-{q}).
\end{align}
  
 
 

    The first two terms of (\ref{payoffderivative2m}) are the advantages of using a higher bid ${q}$. Term 1  represents the marginal payoff from  Player 2's share increasing when Player 1  chooses  the atom bid of $a$, which never  causes breakdown.  Term 2 represents  Player 2's marginal payoff  as the result of his share  increasing when he raises $q$ if  Player 1 bids between $a$ and $q$ so raising $q$ does not cause breakdown. 
Term 3 represents Player 1's disadvantage from bidding higher. It is the increase in the probability of breakdown, and the resulting loss of the  share $q$.     The crucial level of $p$ is $p= 1-q$, since that is at the edge of breakdown.  The probability density for Player 1 choosing that bid is  $f_1(1-q)$,  and since the two player's bids add up to 1 at that bid, Player 2's lost pie share is  $q$ if he edges past breakdown. 
 
 To see why the  equilibrium  density is increasing, think of how the marginal benefits and costs change for Player 2 as $q$ rises. Imagine if the $f_1$ density were uniform instead of increasing.  
 Both benefits, terms 1 and 2,  in equation  (\ref{payoffderivative2m})  would be larger for small $q$, for the reason that with small $q$, increasing the bid a given amount from that low level  (and thus starting from a small share) has a bigger effect on the share. For term 2, the integral, there is an additional reason why the benefit would be bigger unless the density increases: at small $q$, term 2   includes a larger probability range $[a, 1-q]$ over which the shares add up to less than one and breakdown is avoided.   If we start with a large $q$, near the top of the support, then increasing $q$ yields a   greater share, but obtained with small probability; most of the expected payoff from a high $q$ is coming from the atom of probability with which Player 1 bids $p=a$. This gives the second reason $f_1 (p)$ must be rising--- so that a small $q$'s wider interval of $p$'s that avoid breakdown is offset by  smaller values of the density $f_1$ over that wide interval.
 
   Term 3 provides another reason why $f_1 $ must increase:  the rise in the marginal cost of breakdown,     $q f_1(1-q) $. The density takes the argument $1-q$ because   $p=1-q$  is the threshold for breakdown, and hence $f_1(1-q)$ is the rate of increase of the probability of breakdown as $q$ increases.     If $f_1$ were uniform,  the marginal cost would increase with $q$, because a constant rate of increase   in the probability of breakdown causes  Player 2 more harm when his share is bigger. Thus, $f_1$ must be  rising,  to make the  marginal cost term 3 of   $q f_1(1-q) $ smaller for the critical $1-q$ threshold  when $q$ is large. 
   
   \todo{Eric revised these last two paragraphsa dn they need checking}
 
 
 
OLD VERSION Both of the advantage terms  in equation  (\ref{payoffderivative2m})  would be larger for small $q$, because the effect on the share   of increasing 
the bid is bigger from a lower base. Also, the second advantage term, the integral, is  bigger for small $q$ because it includes a larger probability range $[a, 1-q]$ over which the shares add up to less than one and breakdown is avoided.   If we start with a large $q$, near the top of the support, then increasing $q$ yields a   greater share, but obtained with small probability; most of the expected payoff from a high $q$ is coming from the atom of probability with which Player 1 bids $p=a$. This gives one reason why $f_1 (p)$ has to be rising--- so that a small $q$'s wider interval of $p$'s that avoid breakdown is offset by  smaller values of the density $f_1$ over that wide interval.

 OLD VERSION  The second reason  why $f_1 $ must increase  is  the  third-term marginal cost of increasing $q$ and increasing the probability of breakdown,  a marginal cost of   $q f_1(1-q) $. The density takes the argument $1-q$ because   $p=1-q$  is the threshold for breakdown, and hence $f_1(1-q)$ is the rate of increase of the probability of breakdown as $q$ increases.     If $f_1$ were uniform,  the marginal cost would increase with $q$, because a constant rate of increase   in the probability of breakdown causes  Player 2 more harm when his share is bigger. Thus, $f_1$ must be  rising, so it is smaller for the critical $1-q$ threshold  when $q$ is large. 
   
  \bigskip
  
    There also exist asymmetric interval equilibria, as shown in 
Figure \ref{fig-Asymmetric} (again, using an  explicit formula from Proposition 2 below).  Player 1 bids an atom  of approximately .94 at $a=.2$ and  mixes   over $[.2, .4]$, while  Player 2 bids an atom of approximately .33 at $a=.6$ and mixes  over $[.6, .8]$.  The supports are balanced because $.2+.8$ and $.4+.6$ equal 1.   Player 2 always bids more and  will  have a higher expected payoff, and the modal outcome will be a share of .25 for Player 1 and .75 for Player 2  as a result of bids of $p=.2$ and $q=.6$.   Each player's mixing density is rising convexly, as Proposition 1 requires. 

 
 
% \begin{minipage}{1.0\textwidth}
\begin{figure}[ht!]
\begin{centering}
 \includegraphics[width=4in,height=2.5in]{fig-Asymmetric}
\end{centering}
\caption{\sc    The Asymmetric-Equilibrium Mixing Densities    $f_1({p})$ over [.2,.4] and $f_2(q)$  over $[.6,.8]$ }\label{fig-Asymmetric}   
 \end{figure} 
  
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
\bigskip
 
%\section{Finding the Equilibrium Mixing Distribution for Interval Equilibria}

We will now proceed to find a solution for the mixing density for an interval equilibrium. Proposition 1 tells us that the mixing distribution for Player 1  is a measure $\mu_1$ on $[{a}, b ]$ consisting of a single atom  at ${a}$  together with a  positive  measure in the Lebesgue class,
\[
\mu_1({p})=K_1 \delta_a({p}) + f_1({p})d{p},
\] 
where $K_1=\frac{\pi_1}{1-a}$, with expected payoff $\pi_1$, and $f_1$ is some smooth function with all positive derivatives and subject to the normalization that $\mu_1$ is a probability measure. The fundamental   equation   that characterizes the equilibrium then says that Player 1's $f_1$ and  $K_1$ must be such that Player 2's expected   payoff  from any ${q}$ in his support    equals that from playing ${q}=1-a$:  
 \begin{equation} \label{crucial3}
\pi_2({q})= K_1 \left( \frac{{q}}{{q}+{a}} \right)+\int_{{a}}^{1-{q}} \left( \frac{{q}}{{q}+{p}} \right)f_1({p})  d{p}   = K_1 (1-a). 
\end{equation}

The purpose of the following proposition is to solve equation \eqref{crucial3} for $f_1$ and $K_1$.

\bigskip

\begin{Proposition}\label{prop:main}
Splitting a Pie with the proportional sharing rule has  unique  interval equilibrium mixing strategies among Borel measures for given supports. Given any $0<a<b<1$, Players 1 and 2 use real-analytic densities $f_1({p})$ on $[a,b]$ and $f_2({q})$ on   $[1-b,1-a]$ with atoms of probability $K_1$ at $a$  and $K_2$ at $1-b$ such that: 
\begin{equation}
f_1({p})=a K_1\sum_{i\geq 0} \frac{m^{(i)}(a)}{i!}({p}-a)^i,   \;\;\; K_1 = \frac{1}{1+ a \sum_{i\geq 0} \frac{m^{(i)}(a)}{(i+1)!}(b-a)^{i+1}},  
\end{equation}
and 
\begin{equation}
f_2({q})=(1-b)K_2  \sum_{i\geq 0} \frac{m^{(i)}(1-b)}{i!}({q}-(1-b))^i, \;\;\; K_2 =    \frac{1}{1+  (1-b) \sum_{i\geq 0} \frac{m^{(i)}(1-b) }{(i+1)!}(b-a)^{i+1}}  
\end{equation} 
with $m^{(i)}(\cdot)$ for both densities recursively expressed as: 
\begin{equation}\label{eq:coeffrecur}
m^{(i)}(x)=\frac{ (i+1)!}{ 1-x }+\sum_{j=0}^{i-1} \left(\frac{ (j+1)(i-1-j)! +(i-j)! x }{1-x}\right) m^{(j)}(x).
\end{equation}
\end{Proposition}

\noindent
For the proof, see Appendix \ref{app:PropositionInterval}. 

If we expand out the first few terms (see the end of Appendix \ref{app:PropositionInterval} for calculations),   $f_1$ evaluates to:
\begin{align}\label{eq:exampExpand}
\begin{split}
f_1({p})  = & a K_1  \left(  \frac{1}{(1-a) } + \left(   \frac{ 3-a}{(1-a)^{2}} \right)  (p-a) + \left(   \frac{ 13-10a+3a^2 }{2(1-a)^{3}}  \right) (p-a)^2  \right.  
+\\ 
& \left(  \frac{71 - 89 a + 55 a^2 - 13 a^3}{6(1-a)^{4}}   \right)  (p-a)^3 
    \left.   + \sum_{i\geq 4} \left(   \frac{m^{(i)}(a)}{i!}   \right)(p-a)^i \right).
\end{split}
\end{align}


The equations for $f_1 $ and $f_2 $ in Proposition \ref{prop:main} are   not approximations, but  exact: the infinite series converges. Showing this is the hardest part of the proof.  If only a finite number of terms are included,   it becomes   an approximation. In this, it is like the formula $\pi = \sqrt{\Sigma_{n=1}^\infty \frac{6}{n^2}}$, since the exact value of $\pi$ can  be represented by an infinite number of rational terms. 

The equilibrium densities have two surprising features. First, both $f_1({p})$ and $f_2({q})$ are composed of the same $m^{(i)}(\cdot)$ functions, even in an asymmetric equilibrium where the two players use different mixing supports. Second, the upper bounds of the support only enter into the expression for the atom weights $K_1$ and $K_2$ and thus the overall scaling of $f_1$ and $f_2$. 


%\begin{Remark}
%As $a\to 0$,   $f_1$ tends to the constant $0$ function on $[0,b]$. Hence, $K_1=1$, and $\mu_1=\delta_1$. Proposition \ref{prop:cases}\todo{FIX THIS} will show that $\mu_2=\delta_1$ in this case. In particular $K_2=0$ and $f_2$ vanishes on $[1-b,1)$ pointwise but limits to $\delta_1$ in the weak-* sense. Similarly, as $b\to 1$ $f_2$ tends to the constant $0$ function on $[0,1-a]$ and $K_2=1$ so $\mu_2=\delta_0$ and Proposition \ref{prop:cases} will show that $\mu_1=\delta_1$. In particular $K_1=0$ and $f_1$ vanishes on $[a,1)$ pointwise but weak-* limits to $\delta_1$. When both $a\to 0$ and $b\to 1$ then $f_1$ and $f_2$ both vanish on $[0,1)$ pointwise. Proposition \ref{prop:cases} will again show that depending on how $a$ and $b$ limit as a pair we obtain that either $K_1=0$ and $f_1$ limits to $\mu_1=\delta_1$ and $K_2\delta_a+f_2$ limits to $\mu_2=\alpha\delta_0+(1-\alpha)\delta_1$ for any $\alpha\in [0,1]$ or else $f_2$ limits to $\mu_2=\delta_1$ and $K_1\delta_a+f_1$ limits to $\mu_1=\alpha\delta_0+(1-\alpha)\delta_1$ for any $\alpha\in [0,1]$.
%\end{Remark}
%\todo{Do we need Remark 1? It essentially states Proposition 4, which is in the next section. Removing it would shorten and streamline. If we kept it, we would need to make clear that $\delta$ refers to the Dirac measure. }


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 

%\newpage



 


\section{Concluding Remarks}\label{sec:6}

     ``Splitting a  Pie'' is the simplest of bargaining games but it  generates surprisingly complex equilibria.  We have explored those  equilibria  when players bid over interval supports with proportional sharing.  We have seen that any ``balanced'' pair of  intervals for the two players supports a unique equilibrium, and have derived explicit formulae for the probabilities and densities of the equilibria. These equilibria will always consist of a probability atom at the lowest bid in the player's interval plus an increasing  probability density over the rest of the interval that begins strictly positive and is convex with positive derivatives of all orders.  Although the qualitative features of the equilibria are simple, the derivations were technical, so let us now return to the economic motivation for the model.  
     
       The situation being modeled is that of two players choosing how tough to be with each other in sharing something of value. In the pure-strategy equilibria, they begin with an expectation of how tough each other will be, and their behavior confirms that expectation. In the mixed-strategy equilibria, they begin with an expectation of a range of toughness  that the other player might choose, and their behavior, while different from realization to realization,  does not contradict that expectation. Indeed, time and repetition would confirm it.  To model mixed strategies, one  must also make some assumption as to how the pie is shared when both players bid conservatively so that the bids---the toughness levels--- add up to less than one. Our preferred assumption is   that the shares are proportional to the bids and that the shares add up to one. 
     
 We found, as Malueg and others have, that mixed-strategy equilibria do exist. Although we must start with the players having a common expectation of the interval supports of the bids, the players can then deduce the unique mixing distributions that will be chosen in equilibrium. If both players mix over the same interval, the most common choice for each player will be to bid the minimum level of toughness in the support, which means that a common outcome will be for both to choose equal toughness and   receive equal shares. Also common will be breakdown, when each player guesses wrong as to what the other player will choose and wishes, ex post, that he had not been so tough. The remaining case will be a continuous range of equilibrium shares, not from 0\% to 100\%, but over a range depending on their expectations of the minimum and maximum toughness of their rival. If players choose unequal support intervals, this asymmetric equilibrium models the case where one player tends to choose tougher bids and so will have a higher expected share.
 
     We think  these are good  features to find for   a model of bargaining,   but the solution is intricate because each player's share depends continuously on both how tough he is and how tough his rival is.  The Nash Demand Game, with its ``take what you bid'' sharing rule, is similar in its qualitative features and far simpler to solve. Its downside is that   players who agree  usually  agree only over part of the pie and discard the rest.    Still
      simpler, and without that downside, is the  Hawk-Dove  equilibrium, in which each player mixes over a single tough and a single mild bid. This equilibrium is easy to analyze even under a proportional sharing rule. The interval equilibria we have described turn out to have qualitative features in some ways like those of Hawk-Dove. Both types of equilibria have the two outcomes of (a) breakdown and (b) a split depending on the players' expectation of their relative toughnesses as being common outcomes. The difference is that with interval equilibria there is also a continuous range of other splits. Thus, the lesson for applied work may be that  the simpler Hawk-Dove  equilibria have  results not dissimilar from interval equilibria.
      
      
%      
%      In the end, for applied work the Hawk-Dove equilibrum is what may be most useful, but  we hope that this paper has shown that its qualitative features are not so very different from   a model that uses a continuum of toughness levels.  










 \newpage
%
%\noindent
\appendix



\section{Proof of Proposition 2.}\label{app:PropositionInterval}

\begin{repProposition}{prop:main} % 2. 
Splitting a Pie with the proportional sharing rule has  unique  interval equilibrium mixing strategies among Borel measures for given supports. Given any $0<a<b<1$, Players 1 and 2 use real-analytic densities $f_1({p})$ on $[a,b]$ and $f_2({q})$ on   $[1-b,1-a]$ with atoms of probability $K_1$ at $a$  and $K_2$ at $1-b$ such that: 
\begin{equation}
f_1({p})=a K_1\sum_{i\geq 0} \frac{m^{(i)}(a)}{i!}({p}-a)^i,   \;\;\; K_1 = \frac{1}{1+ a \sum_{i\geq 0} \frac{m^{(i)}(a)}{(i+1)!}(b-a)^{i+1}},  
\end{equation}
and 
\begin{equation}
f_2({q})=(1-b)K_2  \sum_{i\geq 0} \frac{m^{(i)}(1-b)}{i!}({q}-(1-b))^i, \;\;\; K_2 =    \frac{1}{1+  (1-b) \sum_{i\geq 0} \frac{m^{(i)}(1-b) }{(i+1)!}(b-a)^{i+1}}  
\end{equation} 
with $m^{(i)}(\cdot)$ for both densities recursively expressed as: 
\begin{equation}\label{eq:coeffrecur2}
m^{(i)}(x)=\frac{ (i+1)!}{ 1-x }+\sum_{j=0}^{i-1} \left(\frac{ (j+1)(i-1-j)! +(i-j)! x }{1-x}\right) m^{(j)}(x).
\end{equation}
\end{repProposition}
 
%\noindent
\begin{proof}
It will later reduce clutter to   define   $m({p})\equiv\frac{1}{aK_1}f_1(p)$ and rewrite  the crucial  equation (\ref{crucial3})  that describes equilibrium  as 
 \begin{equation} 
K_1 \left(\frac{  {q} }{q+a } \right) +  \int_a^{1-{q}} \left( \frac{{q}}{{q}+{p}}\right) a K_1  m({p})d{p} = K_1(1-a).
\end{equation}

 Our aim will be to find the derivatives of $m$  and use them to construct a power series representation of the function. 
Moving the  first term of the previous equation, the contribution from the point mass, to the right-hand side  and dividing by $aK_1$ yields 
\begin{equation}\label{eq:main_int}
  \int_a^{1-{q}}  \left(\frac{{q}}{q + p}  \right)   m({p}) d{p}=    \left(\frac{1-a}{a}  -\frac{{q}}{  a({q}+a) }\right)=\frac{1-q-a}{q+a}  
\end{equation}
for all ${q}\in [0, 1-a]$. 

By Proposition \ref{prop:1}, $m$ is infinitely differentiable and differentiating both sides of \eqref{eq:main_int} with respect to the variable ${q}$ yields, since the equation is true for all ${q}$ in the interval $[0,1-a]$, 
\begin{equation}
-{q} m(1-{q})+ \int_a^{1-{q}} \left( \frac{1}{{q}+{p}} - \frac{{q}}{({q}+{p})^2}\right) m({p})\, d{p}= -\frac{1-q-a}{(q+a)^2}-\frac{1}{q+a}
\end{equation}
 which simplifies to 
\begin{align}\label{eq:recurse}
{q} m( 1-{q})-\int_a^{1-{q}} \left( \frac{{p}}{({q}+{p})^2} \right) m({p})\, d{p}= \frac{1}{ (q+a)^2}
\end{align}

We obtain, for example by setting $1-{q}= a$, that $m(a)=\frac{1}{1-a}$, which gives the $f_1(a) = \frac{aK_1}{1-a}$ we found in Proposition 1. 
Again taking the first derivative with respect to ${q}$ of both sides of equation \eqref{eq:recurse}, and multiplying through by $-1$, we obtain
\[
{q} m'(1-{q}) -(2-{q}) m(1-{q})-2\int_a^{1-{q}} \left( \frac{{p}}{({q}+{p})^{3}} \right) m({p})\, d{p}=2 ({q}+a)^{-3}
\]

Recursively define $r_i({q})$ by 
\begin{align}\label{eq:p-rec}
r_0({q})={q} m(1-{q})\quad\text{and}\quad r_{i}({q})=-r_{i-1}'({q})-i!\, (1-{q})\, m(1-{q}) \quad\text{for}\quad i>0.
\end{align}

Our objective is to obtain expressions for the derivatives of $m^{(i)} (1-{q})$ that we can use to construct  the $f_1$ density.   The linear recursion relation \eqref{eq:p-rec} can be  solved  to get  $r_n({q})$  in terms of all the $m^{(i)}$ from  $i=0$ to $n$ instead of in terms of    $ r_{n-1}'({q})$, yielding 
\begin{align}\label{eq:p-exp1}
r_n({q})={q}m^{(n)}(1-{q})-\sum_{i=0}^{n-1} \left( (i+1)(n-1-i)! +(n-i)! (1-{q}) \right) m^{(i)}(1-{q}).
\end{align} 


Taking the $n$-th derivative of equation \eqref{eq:recurse} in $1-{q}$    and multiplying by (-1)  we obtain 
\begin{align}\label{eq:p-integ}
r_n({q}) +(n+1)!\int_a^{1-{q}}  \left(\frac{{p}}{({q}+{p})^{n+2}} \right)  m({p})\, d{p}= (n+1)!({q}+a)^{-(n+2)}.
\end{align}
 
 
Evaluating \eqref{eq:p-integ} at ${q}=1-a$ gives $r_n(1-a)= (n+1)!$ since the integral vanishes.  We can substitute $r_n(1-a)= (n+1)!$ into \eqref{eq:p-exp1} in evaluating $r_n({q})$  at ${q}=1-a$ to get 
\begin{equation}\label{eq:lin}
 (n+1)! = (1-a)m^{(n)}(a)-\sum_{i=0}^{n-1} ((i+1)(n-1-i)! +(n-i)! a)m^{(i)}(a), 
\end{equation}
which after rearranging terms becomes 
\begin{equation}\label{eq:lin2}
m^{(n)}(a)=\frac{ (n+1)!}{1-a}+ \sum_{i=0}^{n-1}  \left(\frac{ (i+1)(n-1-i)! +(n-i)! a }{1-a} \right) m^{(i)}(a).
\end{equation}

Equation   \eqref{eq:lin2} is a  recursive formula yielding the derivative $m^{(n)}(a)$ in terms of $m^{(i)}(a)$ for $i <n$, and hence ultimately in terms of   $a$. This immediately implies that all of the $m^{(i)}(a)$ are positive,   since $m^{(0)}(a)=m(a)$ is positive. Hence, the mixing density rises   and is  convex at $a$.   We will able to use this to find a power series solution for $m({p})$ if we can find an appropriate bound for the sum of its derivatives. 

Assuming that bound, we can also find the formula for the size of the atom, $K_1$. Since $\mu_1$ is a probability measure, integrating  $d\mu_1({p})=K_1\delta_a({p}) +f_1({p})d{p}=K_1\delta_a +aK_1 m({p})d{p}$ over the full support $[a,b]$ gives us $1$. Therefore $K_1=\frac{1}{ 1+ a\int_a^{b}   m({p})d{p}}$. We can calculate the integral in the denominator in terms of  the power series expansion for $m({p})$: 
\begin{align}
\int_a^{b}  m({p})d{p}&=\int_a^b \sum_{i\geq 0}  \left( \frac{m^{(i)}(a)}{i!}  \right) ({p}-a)^{i}d{p}
=\sum_{i\geq 0}\left(\frac{m^{(i)}(a)}{i!} \right)\int_a^b  ({p}-a)^{i}d{p} \nonumber\\
&=\sum_{i\geq 0}\left(\frac{m^{(i)}(a)}{i!}\right)\left( \frac{(b-a)^{i+1}}{i+1}\right)
=\sum_{i\geq 0}\left( \frac{m^{(i)}(a)}{(i+1)!}\right) (b-a)^{i+1}.
\end{align}

It remains to establish the key bounds, which we do in Lemma \ref{lem:bounds} below. Corollary \ref{cor:1.2} of Lemma \ref{lem:bounds} will then establish the main statement of Proposition \ref{prop:main}, namely the existence, uniqueness and real-analyticity of the power series solution whose coefficients are given by \eqref{eq:coeffrecur2}. The statements and formulas for $f_2$ and $K_2$ follow immediately from the symmetry forced by the balance requirement.

% \todo{put the lemmas after prop and before proof of prop.}
%
\end{proof}



%We will continue the proof of the proposition after this lemma and its corollaries below.



\begin{Lemma}\label{lem:bounds} 
The derivatives $m^{(n)}(a)$ arising in the formal power series for $m({p})$ at $v=a$ satisfy the double-sided bounds,
\begin{equation} \label{lemma1equation}
\frac{n!}{(1-a)^n}< m^{(n)}(a)< \frac{(n+3)! (n+1)}{(1-a)^{n+1}}.
\end{equation}
\end{Lemma}

\begin{proof}
 We will proceed inductively for both bounds, and we treat the upper bound first. First observe that the proposition is true for the base case, the original function's value $m^{(0)}(a)$:
\[
m^{(0)}(a)=m(a)=\frac{1}{ 1-a }< \frac{(0+3)!(0+1)}{(1-a)^{0+1}}
\]
  
   We will show that if the proposition holds for all derivatives lower than $n$, then it holds for  $m^{(n)}(a)$ too.  Thus, the  inductive step that we assume is that 
    \begin{equation}\label{eq:step}
m^{(i)}(a)<\frac{(i+3)!(i+1)}{(1-a)^{i+1}}  \; {\rm for\; all}\;  0\leq i \leq n-1
\end{equation}
 
  
 Equation  \eqref{eq:lin2} gave us an expression for $m^{(n)}(a)$ in terms of  a sum of the lower derivatives $m^{(i)}(a)$ multiplied by
 $ \frac{ (i+1)(n-1-i)! +(n-i)! a }{1-a}$. Let us multiply  both sides of \eqref{eq:step} by that coefficient and rearrange:
\begin{align*}
\frac{ (i+1)(n-1-i)! +(n-i)! a }{1-a}& m^{(i)}(a)< 
\left( \frac{ (i+1)(n-1-i)! +(n-i)! a }{(1-a)^{i+2}}  \right)(i+3)!(i+1) \\
&=  \frac{(n-i)!(\frac{i+1}{n-i} + a)(i+3)!(i+1)}{(1-a)^{i+2}}  \\
&= \frac{(i+1)(n-i)!(i+3)!(\frac{i+1}{n-i} + a)(1-a)^{n-1-i}}{(1-a)^{n+1}} \\
&= (n+3)!  \left(\frac{(i+1)(n+3-(i+3))!(i+3)! (\frac{i+1}{n-i}+a) (1-a)^{n-1-i}}{(n+3)!(1-a)^{n+1}} \right) \\
&= (n+3)!  \left(\frac{(i+1)(\frac{i+1}{n-i} + a) (1-a)^{n-1-i}}{\binom{n+3}{i+3}(1-a)^{n+1}}  \right) \\
&\leq (n+3)!  \left(\frac{(i+1)(i+2)}{\binom{n+3}{i+3}(1-a)^{n+1}} \right) 
\end{align*}
This last inequality holds for all positive $n$ and $i<n$, and establishes a bound for each of the terms in the sum of derivatives making up $m^{(n)}(a)$   in  \eqref{eq:lin2}. To check the bound for  $m^{(n)}(a)$,  we sum the right hand side  bounds for  $i=0,\dots,n-1$, and add $\frac{(n+1)!}{ 1-a }$ from  \eqref{eq:lin2}  to get 
\begin{align}
m^{(n)}(a)&=\frac{(n+1)!}{ 1-a }+\sum_{i=0}^{n-1} \left( \frac{((i+1)(n-1-i)! +(n-i)! a)}{1-a}  \right)m^{(i)}(a) \\
&< \frac{ (n+1)!}{ 1-a }+ \sum_{i=0}^{n-1} (n+3)!  \left( \frac{(i+1)(i+2)}{\binom{n+3}{i+3}(1-a)^{n+1}} \right) \label{e34}\\
&\leq\frac{(n+3)!(n+1)}{(1-a)^{n+1}} \left( \frac{(1-a)^n}{(n+3)(n+2)(n+1)} + \sum_{i=0}^{n-1} (i+1)\binom{n+3}{i+3}^{-1}\right)   \label{e35}\\
&= \frac{(n+3)!(n+1)}{(1-a)^{n+1}} \left( \frac{(1-a)^n}{(n+3)(n+2)(n+1)}+\frac{6}{(n+3)(n+2)(n+1)}+ \frac{n}{n+3}\right. \label{e36}\\
&\left.\phantom{ (1-a)(n+3)!} +\frac{2(n-1)}{(n+3)(n+2)}+\frac{6(n-2)}{(n+3)(n+2)(n+1)}+\sum_{i=1}^{n-4} (i+1)\binom{n+3}{i+3}^{-1}\right)  \nonumber\\
&\leq \frac{(n+3)!(n+1)}{(1-a)^{n+1}} \left( \frac{1}{(n+3)(n+2)(n+1)}+\frac{6}{(n+3)(n+2)(n+1)}+ \frac{n}{n+3}\right. \label{e38}\\
&\left.\phantom{ (1-a)(n+3)!} +\frac{2(n-1)}{(n+3)(n+2)}+\frac{6(n-2)}{(n+3)(n+2)(n+1)}+(n-3)\sum_{i=1}^{n-4} \binom{n+3}{4}^{-1}\right) \nonumber \\
&= \frac{(n+3)!(n+1)}{(1-a)^{n+1}} \left( \frac{7}{(n+3)(n+2)(n+1)} + \frac{n}{n+3}+\frac{2(n-1)}{(n+3)(n+2)}\right.  \label{e40}\\
&\left.\phantom{ (1-a)(n+3)!} +\frac{6(n-2)}{(n+3)(n+2)(n+1)}+\frac{24(n-3)(n-4)}{(n+3)(n+2)(n+1)n}\right)  \nonumber\\
&=  \frac{(n+3)!(n+1)}{(1-a)^{n+1}} \left(1+\frac{-n^3+21 n^2-181 n+288}{(n+3)(n+2)(n+1)n}\right)  \label{e42}\\
&\leq  \frac{(n+3)!(n+1)}{(1-a)^{n+1}}.\label{e43}
\end{align}

Inequality \eqref{e35} results from pulling out the $\frac{(n+3)!(n+1)}{(1-a)^{n+1}}$ factor and noting that $i+2\leq n+1 $ for $i\leq n-1$. Equality \eqref{e36}  results from expanding the $i=0,n-1,n-2,n-3$ terms of the sum. Inequality \eqref{e38}   results from noting that $(1-a)^n\leq 1$ and also in the range $1\leq i\leq n-4$ we have both $i+1\leq n-3$ and $\binom{n+3}{i+3}=\binom{n+3}{n-i}\geq \binom{n+3}{4}$. Equality \eqref{e40}   simply records the fact that $(n-3)(n-4)\binom{n+3}{4}^{-1}=\frac{24(n-3)(n-4)}{(n+3)(n+2)(n+1)n}$, and the ensuing algebra gives  equality \eqref{e42}. Inequality  \eqref{e43}  follows from the fact that the numerator polynomial $-n^3+21 n^2-181 n+288$ has a leading negative term and a single real root in the interval $(2,3)$. Hence it is negative for all integers $n\geq 3$. 
The decomposition of the sum we used  also holds only when $n\geq 4$.   We have checked the $n=0$ base case already, so it remains to check the inductive step for $n=1,2,$ and $3$. %\marg{Check that the number is right. Get rid of the excess numbers in the laid out equations.}

Recall that $a\in(0,1)$. For $n=1$ we have 
\begin{align*}
m^{(1)}(a)&<\frac{(1+3)!(1+1)}{(1-a)^{1+1}} \left( \frac{(1-a)^1}{(1+3)(1+2)(1+1)} + \sum_{i=0}^{1-1} (i+1)\binom{1+3}{i+3}^{-1}\right)\\
&= \frac{(1+3)!(1+1)}{(1-a)^{1+1}} \left( \frac{(1-a)}{24}+\frac{1}{4}\right)\\
&< \frac{(1+3)!(1+1)}{(1-a)^{1+1}}.
\end{align*}

For $n=2$ we have from \eqref{e34},
\begin{align*}
m^{(2)}(a)&<\frac{(2+3)!(2+1)}{(1-a)^{2+1}} \left( \frac{(1-a)^2}{(2+3)(2+2)(2+1)} + \sum_{i=0}^{2-1} (i+1)\binom{2+3}{i+3}^{-1}\right)\\
&= \frac{(2+3)!(2+1)}{(1-a)^{2+1}} \left( \frac{(1-a)^2}{60}+\frac{1}{2}\right)\\
&< \frac{(2+3)!(2+1)}{(1-a)^{2+1}}.
\end{align*}

Finally, for $n=3$ we have 
\begin{align*}
m^{(3)}(a)&<\frac{(3+3)!(3+1)}{(1-a)^{3+1}} \left( \frac{(1-a)^3}{(3+3)(3+2)(3+1)} + \sum_{i=0}^{3-1} (i+1)\binom{3+3}{i+3}^{-1}\right)\\
&= \frac{(3+3)!(3+1)}{(1-a)^{3+1}} \left( \frac{(1-a)^3}{120}+\frac{41}{60}\right)\\
&< \frac{(3+3)!(3+1)}{(1-a)^{3+1}}.
\end{align*}

Thus we obtain the upper bound of Lemma 1, equation (\ref{lemma1equation}): 
\begin{equation}
m^{(n)}(a)< \frac{(n+3)!(n+1)}{(1-a)^{n+1}} \quad\text{for all}\quad n\geq 0.
\end{equation}

For the lower bound we proceed similarly. First,  we consider the general case assuming the inductive hypothesis.
\begin{align*}
m^{(n)}(a)&=\frac{(n+1)!}{ 1-a }+\sum_{i=0}^{n-1} \frac{ (i+1)(n-1-i)! +(n-i)! a }{1-a} m^{(i)}(a)\\
&\geq\frac{(n+1)!}{ 1-a }+\sum_{i=0}^{n-1} \frac{ (i+1)(n-1-i)! +(n-i)! a }{1-a}\cdot \frac{i!}{(1-a)^i}\\
&=\frac{ (n+1)!}{ 1-a }+ \frac{n!+a(n-1)!}{(1-a)^n} +\sum_{i=0}^{n-2} \frac{((i+1)(n-1-i)! +(n-i)! a)i!}{(1-a)^{i+1}} \\
& >  \frac{n!}{(1-a)^n} 
\end{align*}
where we have simply pulled the $n-1$ term out of the sum in the third line.
Note that we will need to check just the base case, $n=0$, as the above calculation holds for all $n\geq 1$.

For the $n=0$ case we check,
\begin{equation}
m^{(0)}(a)=m(a)=\frac{1}{(1-a)}> \frac{(0)!}{(1-a)^{0}}=1,
\end{equation}
as desired. This completes the lower bound and the proof.  
\end{proof}

\bigskip

%Note  also that Lemma \ref{lem:bounds}'s inequality  (\ref{lemma1equation}) implies that\marg{Why note this? How is it relevant?} 
%\begin{align}
%m^{(n)}(a)\frac{({p}-a)^n}{n!}&\leq ka \frac{(n+3)!}{n!} \frac{(1+a)^{n+2}}{(1-a)^{n+2}} ({p}-a)^n \\
% &\leq \pi_0  \frac{1+a}{1-a}(n+1) \left(\frac{(1+a)({p}-a)}{1-a}\right)^{n} \nonumber
%\end{align}


Lemma \ref{lem:bounds} has two important corollaries: 

\begin{Corollary}\label{cor:power-series}
The power series,
\begin{equation}
\sum_{i\geq 0} \frac{m^{(i)}(a)}{i!}({p}-a)^i
\end{equation}
converges uniformly for all $p\in (-1+2a,1)$. Moreover, it has a pole at $p=1$.
\end{Corollary}

\begin{proof}
By Lemma \ref{lem:bounds}, whenever $-(1-a)<{p}-a< 1-a$ or equivalently, $-1+2a<{p}<1$, we have 
\[
\abs{\frac{m^{(i)}(a)}{i!}({p}-a)^i}< \frac{(i+3)(i+2)(i+1)^2}{ 1-a }\eps^i
\] 
where $\eps=\abs{\frac{{p}-a}{1-a}}<1$. In particular, this is the summand of an absolutely (and exponentially) convergent power series. 

On the other hand, the lower bound of Lemma \ref{lem:bounds} shows that
\[
\frac{m^{(i)}(a)}{i!}({1}-a)^i > 1
\]
and hence the series diverges at $p=1$.
\end{proof}

\bigskip

\begin{Corollary}\label{cor:1.2}
The function $m$ defined on the interval $[a,b]$ for any $0<a<b<1$ given by
\[
m({p})=\sum_{i\geq 0} \frac{m^{(i)}(a)}{i!}({p}-a)^i
\]
where the $m^{(i)}(a)$ are computed from \eqref{eq:lin2} is real-analytic on $[a,b]$ and yields the unique measurable function solving \eqref{eq:main_int}.
\end{Corollary}

\begin{proof}
First observe that $[a,b]\subset  (-1+2a,1)$.
Since the series converges on the domain of definition by Lemma \ref{lem:bounds}, it is real-analytic there. (Indeed it converges and is real-analytic on a larger open interval.)

We have established earlier in the proof of Proposition \ref{prop:main} that any solution to \eqref{eq:main_int} is a real-analytic function $m(p)$ satisfying \eqref{eq:coeffrecur2}. Any such function $m(p)$ is uniquely determined by the values of $m^{(i)}(x)$ at $x=a$ which may be computed from the recursion \eqref{eq:lin2}. 

%
% Let $m_1=m$. Any other solution $m_2$ that is infinitely differentiable at $v=a$ must have the same derivatives at $v=a$ by the construction above. However, an alternate solution $m_2$ may not have this regularity. Nevertheless, taking the difference of equation \eqref{eq:main_int} for $m=m_1$ and $m=m_2$ shows that we must have for $f=m_1-m_2$,
%\[
%\int_a^{1-{q}} \frac{{q}}{{q}+{p}} f({p})d{p}=0
%\]
%However the functions $\frac{{q}}{{q}+{p}}$ in $p$ are linearly independent in $L^1([a,1-q])$ for all $q<1-a$ and hence $f=0$ almost everywhere, yielding equivalent measures in the Lebesgue class. 
\end{proof}

Corollary \ref{cor:power-series} having established convergence and Corollary \ref{cor:1.2} uniqueness, the proof of Proposition \ref{prop:main} is complete. %\marg{Check numbers here. The big-picture logic in the proof relating all the lemmas, etc.  needs work. }
% 
%We alert the reader that Lemma \ref{lem:bounds}'s proof runs a few pages. However, together with its corollaries, it yields an immediate proof of Proposition \ref{prop:main} which gives the explicit formula for the equilibrium strategy.


\subsection{Calculating the   $ \mathbf{ m^{(i)}(a)}$ }
The  terms $m^{(i)}(a)$ in Proposition \ref{prop:main} are calculated recursively. To obtain formula \eqref{eq:exampExpand}, we start with $m^{(0)}(a)  = 1/(1-a)$, and then using \eqref{eq:coeffrecur2} we continue,
\begin{align*}
m^{(1)}(a)&  =  \frac{ (1+1)!}{ 1-a }+  \frac{ (0+1)(1-1-0)! +(1-0)! a }{1-a}  \frac{1}{1-a}\\ 
&= \frac{  3-a}{(1-a)^2} \\
 \\
m^{(2)}(a) &= \frac{ (2+1)!}{ 1-a }+    \frac{ (0+1)(2-1-0)! +(2-0)! a }{1-a} \frac{1}{1-a} +  \frac{ (1+1)(2-1-1)! +(2-1)! a }{1-a}  \frac{  3-a}{(1-a)^2} \\
 &= \frac{13-10a +3a^2 }{(1-a)^3} \\ 
 \\
m^{(3)}(a) = &\frac{ (3+1)!}{ 1-a }+    \frac{ (0+1)(3-1-0)! +(3-0)! a }{1-a}  \frac{1}{1-a}\\
 &+    \frac{ (1+1)(3-1-1)! +(3-1)! a }{1-a} \frac{  3-a}{(1-a)^2} +    \frac{ (2+1)(3-1-2)! +(3-2)! a }{1-a} \frac{13-10a +3a^2 }{(1-a)^3}\\
& = \frac{71-89 a + 55 a^2-13 a^3}{(1-a)^4}.
\end{align*}
 

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\section*{References}
%\begin{thebibliography}{asdas}


%\noindent
% {\sc References}
 
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 Andersson, Ola, Cedric Argenton  \& Jorgen W. Weibull  (2018)\href{https://doi.org/10.1016/j.mathsocsci.2017.10.007 }{``Robustness to Strategic Uncertainty in the Nash Demand Game?''} {\it Mathematical Social Sciences}, 91: 1--5.


 
Ausubel,  Lawrence M., Peter Cramton \&  Raymond J. Deneckere  (2002)\href{https://d1wqtxts1xzle7.cloudfront.net/44344673/ausubel-cramton-deneckere-bargaining-with-incomplete-information.pdf?1459660015=&response-content-disposition=inline%3B+filename%3DBargaining_with_incomplete_information.pdf&Expires=1601864066&Signature=WCv3wY3xqmVUJeufv2CBS6-vmG3WccRgtxI0U0L22lQ~HnLHwbdHWIFyPeDBAZYJ9iXBhqBdfoTAkf5kEzcAjFycV5SZBRWYAA03-dlkmRC3mwGK4oQCO9J~mdezblpyLp4Mc49l3yyg66q1CLy8CUWWmpIK2pu5L1FgUJ1UAReGWVT4An37FhOxLgY5T4dtQH5eSMnOp1a4Mfaw~w7DsgaHjUK-BysVQkHb9E8S5BeOP-Urt55dUQYfr1ivxJlufhTPD3tG5QTDv220ayvXgPzcEQzzUntiCHnWim3LwsATlR3~rixY5LuyzuTanTm8GRLDFnFriY06TYh-rlETcQ__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA }{
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Binmore, Kenneth,  Ariel Rubinstein  \& Asher Wolinsky  (1986)\href{ https://www.researchgate.net/profile/Ariel_Rubinstein2/publication/24048549_The_Nash_Bargaining_Solution_in_Economic_Modeling/links/00b7d530b3f0a2fe94000000/The-Nash-Bargaining-Solution-in-Economic-Modeling.pdf}{``The Nash Bargaining Solution in
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%  If the Tullock function is convex enough, there is no pure strategy equilibrium. Taht's because there is no breakdown when effort is too big, just a benefit. but if the MB is high, then one player will prefer to give up. 
 

Friedenberg, Amanda (2019) \href{http://u.arizona.edu/~afriedenberg/bargaining.pdf }{``Bargaining under Strategic Uncertainty: The Role of Second-Order Optimism,''}
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Kurtz, Thomas G. \&  Stewart N. Ethier (2009) {\it Markov Processes: Characterization and Convergence.} Germany: Wiley.
  
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Muthoo, Abhinay (1999) {\it Bargaining Theory with Applications}, Cambridge: Cambridge University Press.

 Nash, John F.    (1950)  \href{ https://www.jstor.org/stable/1907266 }{``The Bargaining Problem,''}
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 Nash, John F.     (1953) \href{ https://www.jstor.org/stable/1906951}{``Two-Person Cooperative Games,''}
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Econometrica},  
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 Roth, Alvin E.  (1985) ``Toward a Focal-Point Theory of Bargaining,"  in  Roth, A. E. (ed.) {\it Game-Theoretic Models of Bargaining}, chap. 12, Melbourne: Cambridge University Press (1985).



 Rubinstein,  Ariel  (1982) \href{ https://www.jstor.org/stable/1912531 }{``Perfect Equilibrium in a Bargaining Model,''}
 {\it Econometrica},   50: 97--109.
 

 

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