\documentclass[11pt]{article}
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\usepackage{amsmath}
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\newcommand{\ds}{\displaystyle}
\newcommand{\pd}[2]{\frac{\partial{#1}}{\partial{#2}}}
\newcommand{\pfrac}[2]{\left(\frac{{#1}}{{#2}}\right)}
\begin{document}
\begin{enumerate}
\item[1.] If each robber has a $F(y)$ chance of being caught, then $NxF(y)$ of the $Nx$ who try to steal are caught; the rest steal successfully. We assume this means that $NxF(y)$ cops catch one robber each, while the rest of the $Ny$ sent to patrol return emptyhanded. \\
Denote the collective utility functions $U_r(x,y)$ and $U_c(x,y)$. Then
\begin{align*}
U_r(x,y) &= N(1-x)(0) + NxF(y)(-5) + [Nx-NxF(y)](5) \\
&= 5Nx[1-2F(y)] \\
U_c(x,y) &= N(1-y)(1) + NxF(y)(5) + [Ny-NxF(y)](-1) \\
&= N[1-2y+6xF(y)
\end{align*}
\begin{em}
Note that it is possible that $Ny-NxF(y)<0$, in which case the last term of $U_c$ doesn't really make sense. We feel, however, that our model is the most faithful interpretation of the problem statement. One interpretation might be that when$Ny-NxF(y)<0$, each cop has caught more than one robber each, so there is an additional utility bonus.\\
\end{em}
It is easily verified that sending all or none of the employees cannot be an equilibrium strategy for either Fagin or Gordon. We are therefore looking for interior equilibria ($x,y\in(0,1)$ so we require
\begin{align*}
\ds\pd{}{y}U_c(x,y) = \ds\pd{}{x}U_r(x,y) = 0
\end{align*}
Solving these FONCs yields
\begin{eqnarray*}
F(y) = \frac{1}{2} & \text{and} & F'(y) = \frac{1}{3x},
\end{eqnarray*}
which we can use to solve for $x$ and $y$ given any $F(y)$.\\
\begin{enumerate}
\item For $F(y)=y$, the utility functions are the same as the game discussed in class. As expected, this yields the same pure NE, $(x,y)=\left(\dfrac{1}{3},\dfrac{1}{2}\right)$.
\item For $F(y)=y^2$, the FONC's are satisfied at $\left(\dfrac{1}{3\sqrt{2}},\dfrac{1}{\sqrt{2}}\right)$. However, $\ds\pd{^2U_c}{y^2} > 0$, so the cops' utility function is convex, and $y=1/\sqrt{2}$ is a minimum. Thus, there is no NE in pure strategies.
\item For $F(y)=\sqrt{y}$, the FONC are satisfied at $\left( \dfrac{1}{3}, \dfrac{1}{4} \right)$.
\end{enumerate}
\end{enumerate}
\end{document}