deletions | additions       diff --git a/sectionEquilibrium_o.tex b/sectionEquilibrium_o.tex  index 6e9c76a..f84ccf0 100644  --- a/sectionEquilibrium_o.tex  +++ b/sectionEquilibrium_o.tex 
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Suppose that the regulator relies on the lobbying efforts to decide whether the product can be marketed. We look for an equilibrium of the contest game, where lobby $I$ with type $\delta$ plays $x^{*}\left(\delta\right)$ and lobby $E$ plays $y^{*}$. We show that a unique equilibrium exists, where all players play interior strategies.  \bigskip{}  Consider first the equilibrium behavior of lobby $I$. Anticipating lobby $E$'s effort $y$, lobby $I$ of type $\delta$ chooses $x$ to maximize his expected utility  $$  \pi\left(x,y\right)\left(b-\min\left(\delta,k\right)\right)-x\mbox{.} 
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0, & \mbox{otherwise}.  \end{cases}  \label{eq:bestreply}  \end{equation}  \bigskip{}  Consider now the equilibrium behavior of lobby $E$. Anticipating lobby $I$'s effort $x\left(\delta\right)$, for all $\delta$, lobby $E$ chooses $y$ to maximize his expected utility  $$  \int\nolimits _{0}^{1}\pi\left(x\left(\delta\right),y\right)\left(\min\left(\delta,k\right)-\delta\right)d\delta-y\mbox{.}  $$  Given that $\min\left(\delta,k\right)-\delta=0$, for all $\delta\leq k$, and $\min\left(\delta,k\right)-\delta=k-\delta$, for all $\delta>k$, it can be rewritten as  $$  \int\nolimits _{k}^{1}\pi\left(x\left(\delta\right),y\right)\left(k-\delta\right)d\delta-y\text{.}  $$  Moreover, in equilibrium, every player $I$ with type $\delta\geq k$ plays the same strategy, that is, $x\left(\delta\right)=x\left(k\right)$, for all $\delta\geq k$, implying that lobby $E$'s expected utility  can be written as  $$  \pi\left(x\left(k\right),y\right)\int\nolimits _{k}^{1}\left(k-\delta\right)d\delta-y\text{.}  $$  If $x\left(k\right)=0$, it is clear that $y=0$ cannot be optimal. Indeed, since $\pi\left(0,0\right)=1$ and $\pi\left(0,y\right)=0$ for all $y>0$, an infinitesimal bid allows lobby $E$ to avoid the expected damage $\int\nolimits _{k}^{1}\left(\delta-k\right)d\delta$. If $x\left(k\right)>0$, we can differentiate lobby $E$'s expected utility with respect to $y$, to obtain the first-order condition describing lobby $E$'s optimal effort  $$  \left(\frac{x\left(k\right)}{\left(x\left(k\right)+y\right)^{2}}\right)  \int\nolimits _{k}^{1}\left(\delta-k\right)d\delta-1\leq0\text{,}  $$  which holds with equality for an interior solution. Then, substitute $x\left(k\right)=\sqrt{\left(b-k\right)y}-y$ to get  \begin{equation}  \left(\frac{1}{\sqrt{\left(b-k\right)y}}-\frac{1}{b-k}\right)  \int\nolimits _{k}^{1}\left(\delta-k\right)d\delta-1\leq0.  \label{eq:focE}  \end{equation}  Clearly, it will be positive if $y$ is small enough and negative if $y\geq b-k$, meaning that lobby $E$'s equilibrium strategy satisfies $00$.  \bigskip{}  Altogether, this shows that the Nash equilibrium of the contest game is unique and interior.\footnote{It is worth noting that this result is not specific to the uniform distribution of damage. In fact, it remains true for any cumulative distribution of damage $F\left(\delta\right)$ such that $\int\nolimits _{k}^{1}\left(\delta-k\right)dF\left(\delta\right)>0$. We thank David Malueg for pointing us this point.} Solving (\ref{eq:focE}), we obtain that the equilibrium efforts of lobbies $I$ and $E$ equal  \begin{equation}  x^{\ast}\left(\delta\right)=  \sqrt{\left(b-\min\left(\delta,k\right)\right)y^{*}}-y^{*}\mbox{, for all }\delta,  \label{eq:x*}  \end{equation}  \begin{equation}  y^{\ast}=  \left(\frac{\left(1-k\right)^{2}}{2\left(b-k\right)+\left(1-k\right)^{2}}\right)^{2}\left(b-k\right).  \label{eq:y*}  \end{equation}