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Christopher Connell and Eric   Rasmusen\\
March 3, 2021
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\vspace{32pt}
 
\noindent
{  {\bf  Splitting a Pie: Mixed Strategies in Bargaining under Complete Information
}}\\
(\href{https://www.rasmusen.org/rasmapedia/index.php?title=Talks:_Polarization_and_Splitting_a_Pie_(January_19,_2021)}{link to seminar web page with slides and handouts and the paper}  or go via \href{https://www.rasmusen.org/rasmapedia/}{https://www.rasmusen.org/rasmapedia}   )
 
 

\noindent
 {\sc   The Model  }
Players 1 and 2 simultaneously choose bids $p_1$ and $p_2$ in the interval [0,1].  If their bids add up to more than 100\%, both get zero and we say  the pie ``explodes''.   
Otherwise, when the bids are $p_1+p_2\leq 1$,   player 1 gets  share  $p_1/  (p_1+p_2)$ as his payoff  and player 2 gets share $p_2/  (p_1+p_2)$.  

  This game has a continuum of pure strategy Nash  equilibria, every permutation such that $p_1+p_2 = 1$,  and the pie never explodes.  
  
  \noindent
  {\bf  One of the Hawk-Dove Equilibria}:
    Each player chooses $.30$ with probability $.6$ and $.70$ with probability $.4$ The expected payoff is .42. With probability .16, they both choose .70 and the pie explodes. 
    
  
  
  We will focus on a particularly interesting  kind of equilibrium,  that in   which the players bid by  mixing over the intervals $[ a, b]$ and  $[c,d]$ using strictly positive probability densities $f_1(p_1)$ and $f_2(p_2)$, possibly with atoms of probability on particular bids.
  
 Most simple is the symmetric equilibrium, in which both mix over the same interval $[a,1-a]$ using the same density $f(p)$. For example, both might mix over [.4, .6], with an atom at .4 and a density rising from .4 to .6, as in the figure. 
 
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 {\sc    The  Symmetric-Equilibrium Mixing Density    $f_1({p})$  and Atom $K_1$  for ${a}=.10,.20,.30,.40 $ } 
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\noindent
{\bf Proposition 1:} Any equilibrium in which  each player mixes over an interval\\
 (a)    consists  of  probability atom $K_1$ at $a$ and density $f_1({p})$ on $[ a, b]$ for Player 1 \\
 (b) 
     and  of  probability atom $K_2$ at $1-b$ and density $f_2({q})$  on $[1-b, 1-a]$  for Player 2. \\
(c)     The densities are positive throughout, and if their derivatives exist,  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 $\\
(d)      and  increase convexly \\
 (e)         with positive derivatives of every order.  
 
  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)  \;\;\; ({\it the\; crucial \;equation})
\end{equation}
Then: 
 \begin{equation}
\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})   =0.
\end{equation}
  
 

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