\documentclass[12pt,oneside]{amsart}
\usepackage{amsmath,amssymb,verbatim,setspace,geometry}
\geometry{top=1in,bottom=1in,includeheadfoot}
\doublespacing

\begin{document}
\noindent MTH 460/560 \hfill Problem 22 page 249 \hfill Spring 08

\noindent
Prof. Cowen
\vskip 18 pt

Using the notation of pages 238 and 239, we have $A \tilde q$ is in
both $G$ and $H$, where $G = \{A q \mid q \in Q\}$ and $H =\{x \in
\mathbf R^{5} \mid x \le v e\}$. Since in our problem the intersection
of $G$ with $H$ consists of only one point, namely $(3,3)^t$, then $A
\tilde q = (3,3)^t$. 

Let $\tilde q= (q_{1},\ldots,q_{5})^t$. Then we
have:
\begin{align}
1 q_{1} + 2 q_{2} + 3 q_{3} + 4 q_{4} + 5 q_{5} &= 3 \text{ and}\\
5 q_{1} + 4 q_{2} + 3 q_{3} + 2 q_{4} + 1 q_{5} &= 3
\end{align}
Subtracting the first equation from the second gives:
\[
4 q_{1} + 2 q_{2} - 2 q_{4} - 4 q_{5} = 0.
\]
Dividing by 2 gives that
\[
2 q_{1} + q_{2} = q_{4} + 2 q_{5}.
\]
\noindent (so what I said in class on Monday about $q_{1} = q_{5}$ and $q_{2} =
q_{4}$ is not the whole the story).

So the answer is: The set of all security strategies $\tilde q$ for 
Player 2 is the set of all $(q_{1},...,q_{5})^t$
such that $q_{1} + ... + q_{5} = 1$, $q_{i} \ge 0$ for $i = 1,\ldots,5$, and $2 q_{1} +
q_{2} = q_{4} + 2 q_{5}$.

For example, $q = (0.1, 0.3, 0.15, 0.4, 0.05)^t$ is a security strategy for
Player 2. (You can check that if Player 2 uses this strategy then
Player 1's payoff matrix is $(3,3)^t$, so no matter how he mixes he
cannot do better than 3, the value of the game.
\end{document}