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{\large {\bf    The Rubinstein Bargaining Model with Both Discounting and Fixed Per-Period Costs}
}
February 12,  2020

\bigskip



  Eric B.  Rasmusen

{\it Abstract}

\end{center}

\begin{small}

 Rubinstein (1982) describes two bargaining models which  reach opposite conclusions.   Model I, now the canonical model of bargaining, has positive discount rates  and the   split is close to 50-50  when  discount rates  are small and almost equal.  Model II   has a fixed cost for each period of bargaining, and  the split is 100-0 when bargaining costs are small and almost equal.   Rubinstein does not say what happens in a model with both discounting and bargaining costs. If the mixed  model were to behave more like Model II, the Rubinstein model would be a   poor fit to reality. In fact, the mixed model behaves more like  Model I, so the canonical model is safe from this criticism. 


\noindent
   Professor of 
Business Economics and Public Policy, Kelley School of Business, Indiana
University,   1309 East Tenth Street, Bloomington, Indiana,  USA 47405, (812) 855-9219 (secretary),   Erasmuse@indiana.edu,
  \url{http://www.rasmusen.org/papers/fixedcost.pdf}.

 

%Send this to econmoics letters. 

 
 
 


\end{small}
 


\newpage

\noindent
 {\bf 1. Introduction}

 Rubinstein's 1982 {\it Econometrica} article on bargaining has been exceptionally influential, receiving 2,003 cites in Google Scholar  as of June 2008.  It  is a simple and elegant model of bargaining   in which agreement is reached immediately, the split is close to 50-50 between identical players, and the split diverges from 50-50 depending on a simple measure of bargaining power--- the discount rate. The model is useful not only in direct application, but as a component of more complex models of interactions in which players split  a surplus at some point in the game they are playing.  Rather than simply assume that the bargaining split is 50-50, or is $\lambda, (1-\lambda)$ for some reduced-form  bargaining parameter, the modeller can build a model from primitive assumptions about payoffs and actions. 
 

 A complete reading of Rubinstein (1982), however,  raises a nagging doubt. Rubinstein (1982) actually contains two simple models of bargaining, and they reach opposite conclusions. 
  In Model I, the  much-cited model  based on time discounting,  the split is about 50-50 when    discount rates are small and  almost equal. In  Model II,  which   has a fixed cost for each period of bargaining, the split is 100-0  when bargaining costs are  small and  almost-- but not quite--  equal. Rubinstein does not say what happens in a model with both discounting and per-offer bargaining costs. If that model were to behave more like Model II, the Rubinstein model would be a   poor fit to reality.  We would wonder whether Model I might really be a special case that we have adopted  because we like to think we are using structural models but which really is no advance over simply assuming that bargainers split the surplus 50-50.  So let's take a look at  the model that combines both discounting and per-offer bargaining costs. 


\bigskip

\noindent
 {\bf 2. The Model}

 Players 1 and 2  are splitting a pie of size 1. We will use $x_t$ to denote an offer in period $t$ that gives share $x_t$ to player 1.   Player 1 starts with an offer of a split of $x_1$ for himself and $(1-x_1)$ for player 2. If player 2 accepts, the game is over and the payoffs are  
\begin{equation} \label{a1}
 \pi_1= x_1 -c_1 \hspace{1in}  \pi_2= 1- x_1 - c_2.
 \end{equation} 

  If  player 2 rejects, a period of time passes, at the end of which he pays $c_2$ and  player 1 pays $c_1$. 
 He then makes an offer of a split of   $x_2 $ for player 1 and $1-x_2$ for himself. If player 1 accepts, the game is over and the payoffs are (viewed from the start of the game)  
\begin{equation} \label{a2}
 \pi_1= -c_1 + \delta_1[- c_1 + x_2 ]   \hspace{1in}  \pi_2= -c_2+  \delta_2[- c_2 +  (1-x_2)] 
 \end{equation} 
 because   payoffs are discounted by discount factors of $\delta_1 = \frac{1}{1+\rho_1}<1$ and $\delta_2 = \frac{1}{1+\rho_2}<1$ each period.  Our main interest is in what happens when the periods are short, so the discount rates approach $\rho_1= \rho_2=0$ and the discount factors approach   $\delta_1 =  \delta_2 = 1$. 

 If player 1 rejects rejects  player 2's offer, another period of time elapses, the players incur costs $c_1$ and $c_2$ again,  and  player 1 makes the next offer, $x_3$. The two players make alternating offers until agreement is reached, or forever if agreement is not reached. 

Rubinstein's Theorem established existence, though not uniqueness, for a more general version of payoffs that includes this particular case. He does show uniqueness and the particular equilibrium share for the two cases  of  just  discounting and just fixed costs. 

\bigskip

\noindent
 {\bf 3. The Equilibrium}


In  any subgame perfect equilibrium, player 1's first offer will be accepted immediately. The only reason to wait is to  get a bigger share  $x_1= x_1'$ later. Player 1 would prefer to offer  $x_1'-c_1 + \epsilon$ now.  

It follows that the equilibrium is stationary.  The players' payoffs cannot get bigger or smaller, or they would make more generous offers now that would equal  any bigger payoffs without having to pay the bargaining costs.  So let us proceed assuming that player 1 always chooses the same $x_1$ and player 2 always chooses the same $x_2$.    

 Let Player 1 take $x_2$, the share player 2 will offer player 1, as given. 
 Player 1 knows that whatever value $x_2$ may take, there is always some value $x_1$ he can choose that will induce player 2 to accept  a share of $1-x_1$ now rather than wait and get $1-x_2$.    Player 1 thus wishes to 
choose $x_1$ low enough for player 2 to   accept it now rather than wait a period, incur the additional cost $c_2$, and have $x_2$ accepted.  He can do this by   choosing $x_1$ so that  $(1-x_1)$  is at least as big as  $\delta_2[ (1-x_2 ) - c_2]$: 
\begin{equation} \label{a3}
  (1-x_1 )  \geq    \delta_2[ (1-x_2 ) - c_2]
 \end{equation} 
  
    Unless the resulting value of $x_1$ equals 1, giving him the entire pie, player 1 will want to choose $x_1$ to  make inequality (\ref{a3}) an equation.   This gives us player 1's reaction equation, 
\begin{equation} \label{a4}
    x_1 (x_2 )  =  Max\{ 1- \delta_2 [(1-x_2 )- c_2], 1\}    
 \end{equation}
     If $x_2=0$, player 1 will choose $x_1=  1- \delta_2   + \delta_2 c_2$, because he believes that if player 2's equilibrium offer is always to keep the entire pie, player 1 can still hold out for a small $x_1$ that player 2 will accept rather than waiting and making his own offer.   As $x_2 $ rises, $x_1(x_2) $ rises linearly  at a rate  of $\delta_2$ as player 1 can get away with asking for a larger and larger share. When $x_2= 1-  c_2 $, player 1 responds with $x_1=1$,  asking for the entire pie because player 2 will not  benefit from waiting a period only to offer  all but $c_2$ of the pie to player 1 anyway.  Any higher values of $x_2$ also would cause player 1 to offer $x_1=1$ because player 2 would obtain even less benefit by waiting.   

  
 Now let us think about player 2's reaction curve. 
Taking player 1's next-period offer of $x_1$ as given, player  1 would like to offer $x_2=0$ if player 1 would accept it, but otherwise as little as possible.    
Player 1 will compare $x_2$ from accepting to $\delta_1 (x_2-c_1)$ from rejecting and making his own offer later. Equating these yields, 
\begin{equation} \label{a5}
 x_2  \leq \delta_1 (x_2-c_1)
 \end{equation} 
 Solving for $x_1$ yields player 2's reaction function: 
\begin{equation} \label{a6}
    x_2 (x_1 )  =   Min \{0,   \delta_1(x_1 -c_1)\}  
 \end{equation} 
     If $x_1=0$, player 2 can offer $x_2=0$ and player 1 will clearly do better to accept rather than wait and get the same zero share. If $x_1$ rises to $x_1=c_1$, player 1 becomes indifferent about accepting or rejecting, and for higher $x_1$, player 2 must increase the value of his offer linearly  at a  rate of $\delta_1$. When $x_1=1$, player 2 must offer $x_2= \delta_1(1 -c_1)$ to get player 1 to accept the offer. 



 \hspace*{-48pt} \begin{minipage}[c]{ \linewidth}
   \begin{center} 
     
       {\sc    Figure 1:\\
         Reaction Curves  $x_1$ and $x_2$   } \label{figure1.png}

  \includegraphics[width=3in]{figure1.png}
            \end{center}
   \end{minipage}
 
Figure 1 illustrates the reaction  functions when $\delta_1=\delta_2 = .7$ and $c_1=c_2 = .08$.   Note the  intercepts on each axis that result in an interior solution, which requires:
\begin{equation} \label{a6a}
 c_1< 1 - \delta_2(1-c_2) 
 \end{equation} 
 and 
\begin{equation} \label{a6b}
 1-c_2 > \delta_1 (1-c_1) 
 \end{equation} 

 


If these  two  reaction equations  do have an interior crossing  as  in Figure 1,  they solve to
\begin{equation} \label{a7}
  x_1^*=   \frac{ 1   -   \delta_2  +  c_2  \delta_2  - c_1  \delta_1 \delta_2 }{1-
        \delta_1 \delta_2 } 
 \end{equation} 
and
  \begin{equation} \label{a8}
   x_2^*=   \frac{  \delta_1 -\delta_1 \delta_2  + c_2 \delta_1 \delta_2  -  c_1\delta_1  }{1-
       \delta_1 \delta_2 } 
 \end{equation} 


Rubinstein looks at  the  two cases  of Model I, where $c_1=c_2=0$  (but $\delta_1, \delta_2<1$,  and they  might be unequal) and Model II, where $\delta_1=\delta_2=1$ (but   $c_1  \geq 0,  c_2 \geq  0$,  and they  might be unequal). In  Model I,  the reaction functions have a unique interior solution.  In Model II, the  reaction functions do not have an interior solution unless $c_1=c_2$, in which case they have a continuum of solutions. 


When  both forces are at work  in the present model, the reaction functions do   intersect at a unique point, but  they  might not cross at an interior solution. This leaves open the possibility that with 
  equal transaction costs, player 1's share is 100\% because of a first-mover advantage.  (It is easy to see that the first mover can earn at least some positive payoff because the second player does not want to incur the fixed offer cost). 
Thus, let us explore when the reaction curves do not cross, and instead intersect an an extremum. 

\hspace*{-48pt} \begin{minipage}[c]{ \linewidth}
   \begin{center} 
     \begin{tabular}{cc}
       {\sc    Figure 2: } &        {\sc    Figure 3:} \\
            {\sc   Player 1 Stronger: $x_1=1$  } \label{figure2.png}&
     {\sc Player 2 Stronger: $x_1= 1 -  \delta_2(1-c_2)    $  } \label{figure2.png}\\
   \includegraphics[width=3in]{figure2.png} & \includegraphics[width=3in]{figure3.png}
\end{tabular}
            \end{center}
   \end{minipage}


  The  reaction curves will not cross at an interior solution  if $c_1> 1- \delta_2 + \delta_2c_2$ (in which case player 2 gets almost the entire pie; $x_1 =1 - \delta_2(1-c_2)$ ) or if $\delta_1 (1-c_1)>  1-c_2  $  and    $c_1 < 1-1/\delta_1+ \frac{c_2}{\delta_1  }$ (in which case player 1 gets the entire pie; $x_1=1$).  We can rule out $1- \delta_2 + \delta_2c_2 <1-1/\delta_1+ \frac{c_2}{\delta_1  }$, so it is not possible to have both corner solutions at once:  multiple equilibria. (The previous sentence's inequality implies $\delta_2 + \delta_2 c_2 <1/\delta_1 - c_2/\delta_1$, so $\delta_1(\delta_2 - \delta_2 c_2) < 1 -  c_2$, which is false since $\delta_1$ and $\delta_2$ are less than 1.)  Let us consider the salient cases. 
 
\noindent
(1)   Suppose both players have the same positive fixed cost and the same  positive discount factor, so  If $\delta_1=\delta_2 =\delta>0$ and $c_1=c_2 =c>0$. In this case, illustrated in Figure 1, there is always a unique corner solution, not a continuum as happens  in Model II, where  just the offer costs are positive and equal.  Here,  $x_1^*= \frac{1 - \delta_1  + c_1 \delta_1 - c_1 \delta_1^2}{1 - \delta_1^2}$, which is less than 1.  To understand this, first think of the intuition for why  in Model II, with $\delta_1=\delta_2=0$, it would be an equilibrium (one of many) for player 1 to offer $x_1=1$ and have it accepted.  This is so because in that equilibrium it expected that player 1 will offer $x_1=1$ whenever it is his turn to offer, so the least player 2 can acceptably offer  for player 1's share is $1-c$, but in that case player 2, having incurred the cost $c$ by refusing player 1's initial offer, has gained nothing by delay.  If $\delta >0$, on the other hand, then player 2 can offer something a little less generous than $x_1=1 -c$ and player 1 will still accept--- so player 1 must forestall this by being a little more generous and set $x_1 <1$.   
 
 


\noindent
 (2) Suppose  both 
 players have the same fixed cost $c_1=c_2 = c$, but   different discount factors.    In that case,  there is an interior solution no matter how much their discount factors differ.   A corner solution giving player 1 the entire pie  would require, from (\ref{a6b}),  that $1-c <\delta_1(1-c )$, which is impossible.  A corner solution giving player 2 almost the entire pie requires, from (\ref{a6a}),   that $c  > 1 -\delta_2 + \delta_2 c  $, which is equivalent to $c  > (1 -\delta_2)(1) + \delta_2 c $, which is also  impossible if  $c <1$. 

\noindent
(3) Suppose both players have the  same discount rate, $\delta_1=\delta_2 =\delta$, but  different  offer costs. This  generates  Figure 2's corner solution of $x_1 =1$   if $c_2$ is big enough, and  Figure 3's $x_1 =  1 - \delta_2(1-c_2)$ if $c_2$ is  small enough. 
 Let's focus on  the $x_1=1$ outcome, since it is possible for smaller differences in offer costs. Inequality (\ref{a6b}) tells us that $x_1=1$ if $\delta >\frac{ 1-c_2}{1-c_1}$, i.e.,  if the  offer costs and their difference are big enough relative to the amount of discounting. 
How big is big enough? 
 An instructive special case  is where the discounting cost per player  is   the same magnitude as the offer cost, so we can see which kind of cost as more impact on the qualitative outcome.    Therefore, let's consider the case of $c_1 = .5(1-\delta)$, with $c_2 = \gamma c_1$ for some $\gamma >1$.    In that case, our  inequality is, after simplification, $ \gamma >  \frac{2- \delta - \delta^2}{1-\delta}$.  With even as extreme strong as $\delta =.5$, this says that player 2's offer cost would have to 2.5 times as big as  player 1's  to have an equilibrium in which player 1 asks for a 100\% share.   
 
   Thus,  the canonical Rubinstein's Model 1  is reasonably robust to the addition of fixed offer costs of bargaining, whereas  Model 2 changes drastically with the addition of discounting. The key difference between discounting and fixed offer costs is that discounting has a higher absolute cost to a player who expects a greater bargaining share, whereas the fixed offer costs are independent of one's share. As  a result, there is a strong tendency for discounting to drive the equilibrium towards an even split, because if one player's equilibrium share is larger, so is his cost of delay, which tends to reduce his equilibrium share. This tendency can be reversed only if the players' fixed offer costs are considerably different from each other. 
  


   

 



\bigskip


\noindent
{\bf References} 
 
 
Cramton,    Peter C.  (1991) 
``Dynamic Bargaining with Transaction Costs,''
 {\it  Management Science} ,   37 (10): 1221-1233 (Oct.  1991).
 


Rubinstein, Ariel (1982)   
``Perfect Equilibrium in a Bargaining Model,''
 {\it Econometrica},  50 (1): 97-109  (Jan. 1982). 

\bigskip

\newpage
 Calculations. 
 


Solving, 
\begin{equation} 
  1  = -\delta_2 c_2 + \delta_2 x_2^*  + x_1^* 
 \end{equation} 

\begin{equation} \ 
  1 + \delta_2 c_2 - x_1^* =   \delta_2 x_2^*    
 \end{equation} 


\begin{equation} 
    x_2^* = \frac{ 1 + \delta_2 c_2 - x_1^*  }{ \delta_2  }
 \end{equation} 

So

\begin{equation} 
  1- \frac{ 1 + \delta_2 c_2 - x_1^*  }{ \delta_2  } = \delta_1[-c_1 + x_1^*].
 \end{equation}

\begin{equation} 
 \delta_2-   1 - \delta_2 c_2 + x_1^*    =\delta_2  \delta_1[-c_1 + x_1^*].
 \end{equation}


\begin{equation}  
  x_1^*    = - \delta_2  \delta_1 c_1 + \delta_2  \delta_1 x_1^*  -  \delta_2+  1 + \delta_2 c_2
 \end{equation}



\begin{equation} 
  x_1^*  (1-  \delta_2  \delta_1)  = - \delta_2  \delta_1 c_1  -  \delta_2+  1 + \delta_2 c_2
 \end{equation}


\begin{equation} 
  x_1^*    = \frac{ - \delta_2  \delta_1 c_1  -  \delta_2+  1 + \delta_2 c_2 }{(1-  \delta_2  \delta_1)}
 \end{equation}

 
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