\documentclass{article}
\input{15401-preamble}
\psetnum{5}
\date{2011/11/9}

\usepackage{pst-qtree}

\begin{document}
\begin{pset}
  \begin{problem}

    \begin{subproblem}
      \begin{enumerate}
      \item Buy 1 call option at 15; sell 1 call option at 25
      \item Buy 1 call option at 15; sell 2 call options at 25; buy 1
        call option at 30
      \end{enumerate}
    \end{subproblem}

    \begin{subproblem}
      The trader expects the current price to remain steady over the
      next three months.
    \end{subproblem}
  \end{problem}

  \begin{problem}

    \begin{subproblem}

      \includegraphics{ps5-2-1.ps}

      The trader expects the stock price to change dramatically, but
      doesn't know the direction.
    \end{subproblem}

    \begin{subproblem}

      \includegraphics{ps5-2-2.ps}

      The trader expects the stock price to increase to a price higher
      than \$30, but wants to hedge the risk that it will not.
    \end{subproblem}
  \end{problem}

  \newpage
  
  \begin{problem}

    \begin{subproblem}
      Consider first the price of the option at the node ($t=1$,
      up). Here,
      \begin{align*}
        156.25a + 1.05b &= 46.25\\
        93.75a + 1.05b &= 0
      \end{align*}
      which is solved by $a= 0.740, b \approx -66.071$. The replicating
      portfolio at this node is long $0.740$ shares of stock and short
      $\$66.071$ in bonds, which has a cost $0.740 \cdot \$125 - \$66.071
      = \$26.43$. %This is higher than the exercise value \$15, so it
      %is the price of the option at this node.

      At the node ($t=1$, down), the value is zero.

      Now, at period 0,
      \begin{align*}
        125a + 1.05b &= 26.43 \\
        75a + 1.05b &= 0
      \end{align*}
      which is solved by $a=0.529, b=-37.76$. The replicating
      portfolio is long $0.529$ shares of stock and short $\$37.76$
      in bonds, which has a value $0.529 \cdot \$100 - \$37.76 =
      \$15.10$. %The option has no exercise value at this point, so its
      %value is $\$16.50$.
    \end{subproblem}

    \begin{subproblem}
      By put-call parity, $S + P = PV(K) + C$, so
      \begin{align*}
        P &= PV(K) + C - S \\
        &= \frac{\$110}{1.05^2} + \$15.10 - \$100 \\
        &= \$14.87
      \end{align*}
    \end{subproblem}

    \begin{subproblem}
      The exercise value of the put option at each node in the tree is
      
      \psset{treemode=R}
      \Tree [.0 [.0 0 16.75 ] [.35 16.25 53.75 ] ]

      Consider the ($t=1$, up) node in the tree. Here,
      \begin{align*}
        156.25a + 1.05b &= 0\\
        93.75a + 1.05b &= 16.25
      \end{align*}
      so the replicating portfolio is short $0.2$ shares of the
      stock and long $\$38.69$ in bonds, giving a value of $\$6.19$.

      At the ($t=1$, down) node,
      \begin{align*}
        93.75a + 1.05b &= 16.75 \\
        56.25a + 1.05b &= 53.75
      \end{align*}
      so the replicating portfolio is short $0.987$ shares and long
      $\$104.05$ in bonds, giving a value of $\$30.05$. However, this
      is less than the exercise value of \$35, so we would choose to exercise
      the option instead, and the value of the option at this node is
      \$35.

      At the $t=0$ node,
      \begin{align*}
        125a + 1.05b &= 6.19\\
        75a + 1.05b &= 35\\
      \end{align*}
      The replicating portfolio is short $0.576$ shares and long
      $\$74.49$ in bonds, giving a value of $\$16.87$.
    \end{subproblem}

    \begin{subproblem}
      The American put option is worth more than the European one. In
      the ($t=1$, down) node we chose to exercise the option early
      because it gave a higher payoff. The ability to do that makes
      the American option worth more.
      
    \end{subproblem}
  \end{problem}

  \begin{problem}

    \begin{subproblem}
      \begin{align*}
        x &= \frac{\ln \left( \frac{S}{KR^{-T}} \right)}{\sigma
          \sqrt{T}} + \frac{1}{2}\sigma\sqrt{T} \\
        &= \frac{\ln \left( \frac{80}{80\cdot1.02^{-1}} \right)}{\sigma
          \sqrt{1}} + \frac{1}{2}\sigma
        \sqrt{1}\\
        &= \frac{-0.0198}{\sigma} + \frac{1}{2}\sigma = 0.2104
        C&= SN(x) - KR^{-T}N\left(x - \sigma\sqrt{T}\right) \\
        &= 80 \cdot N(x) - 80 \cdot 1.02^{-1} N(x - 0.5) \\
        &\approx \$16.38
      \end{align*}
    \end{subproblem}

    \begin{subproblem}
      The option price is shown below:
      
      \includegraphics{ps5-4.ps}

      As the volatility increases, the value of the option
      increases. This makes sense, because the option is more valuable
      as the stock price increases, but has no effect if the stock
      price drops below its current value. A more volatile stock is
      more likely to have higher prices, so the option is worth more.
      
    \end{subproblem}
  \end{problem}
\end{pset}
\end{document}