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\begin{document}
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%\titlepage
%\vspace*{12pt}
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%\begin{center}
\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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%\bigskip
\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 $0v_1(0,0)$ 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}
Assumption (\ref{assump}1) is a set of technical assumptions ensuring payoffs are well defined and that certain approximation arguments hold. Assumption (\ref{assump}2) guarantees that simultaneous bids of 0 are not pure strategy equilibria, and Assumptions (\ref{assump}3)-(\ref{assump}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 \eqref{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, in an interval equilibrium every nonempty open subinterval of $[a,b]$ and $[c,d]$ has positive probability for their corresponding measures.
A key feature of any 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 the players actually choose the higher of their two bids (hawk) with the highest probability ($1-\mu(a)>1/2$) if $a<1/4$, 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 $a**1$ (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, $f(p) =\frac{a}{(1-p)^2}.$ Thus, each player chooses to bid $x\in (a,1-a]$ with density $f(x)=\frac{a}{1-x^2}$ and $x=a$ with probability $K=\frac{a}{1-a}$.
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 include 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 \ref{prop:1}) and find an explicit formula for their mixing distributions (Proposition \ref{prop:main}).
\noindent\fbox{%
\parbox{\textwidth}{%
\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: $00$ (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, $d =1-a$.
% 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 $00.
\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 \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 of Lemma \ref{lem:bounds}.
\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 we are using to prove Proposition \ref{prop:main}:
\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**