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\author{Jeremy Morris}
\title{Homework 8}

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\begin{description}
\item[Exercise 4.1] \textit{In the three-period model of Figure 1.2.2 of Chapter 1, let the interest rate be $r = \frac{1}{4}$ so the risk-neutral probabilities are $\tilde{p} = \tilde{q} = \frac{1}{2}$.}
	\begin{enumerate}
	\item \textit{Determine the price at time zero, denoted $V_{0}^{P}$, of the American put that expires at time three and has intrinsic value $g_{P}(s) = (4-s)^{+}$.}

	The American put has price $V_{0}^{P} = 2.56$.  The answer is achieved through modifying the code from project 2, code is attached.

	\item \textit{Determine the price at time zero, denoted $V_{0}^{C}$, of the American call that expires at time three and has intrinsic value $g_{C}(s) = (s-4)^{+}$.}

	The American call has price $V_{0}^{C} = 0.928$.

	\item \textit{Determine the price at time zero, denoted $V_{0}^{S}$, of the American straddle that expires at time three and has intrinsic value $g_{C}(s) = g_{P}(s) + g_{C}(s)$.}

	The American straddle has price $V_{0}^{S} = 3.296$.

	\item \textit{Explain why $V_{0}^{S} < V_{0}^{P} + V_{0}^{S}$.}

	When we look at the end of the tree, we get
	\begin{align*}
	V_{3}^{S}(\omega) = V_{3}^{P}(\omega) + V_{3}^{S}(\omega)
	\end{align*}
	and in the next step, we also get
	\begin{align*}
	V_{2}^{S}(\omega) = V_{2}^{P}(\omega) + V_{2}^{S}(\omega).
	\end{align*}
	Then in the next to last step we have
	\begin{align*}
	V_{1}^{S}(\omega) < V_{1}^{P}(\omega) + V_{1}^{S}(\omega).
	\end{align*}
	because
	\begin{align*}
	V_{1}^{P}(T) = 2
	\end{align*}
	and
	\begin{align*}
	V_{1}^{S}(T)=2.16 < V_{1}^{P}(T) + V_{1}^{S}(T) = 2 + 0.64.
	\end{align*}
	What we see is that there is an added advantage to having the American put option that is not transferred to the straddle.

	\end{enumerate}

\item[Exercise 4.5] \textit{In equation (4.4.5), the maximum is computed over all stopping times in $S_{0}$.  List all the stopping times in $S_{0}$ (there are 26), and from among them, list the stopping times that never exercise when the option is out of the money (there are 11).  For each stopping time $\tau$ in the latter set, compute $E[I_{\{\tau \leq2\}}\frac{4}{5}^{\tau}G_{\tau}]$.  Verify that the largest value for this quantity is given by the stopping time of (4.4.6), the one that makes this quantity equal to the 1.36 computed in (4.4.7).}

The list of all stopping times follows.
\begin{equation*}
\begin{array}{ll}
 (0,0,0,0) & (\infty,2,\infty,\infty) \\
 (1,1,1,1) & (\infty,\infty,2,\infty) \\
 (2,2,2,2) & (\infty,\infty,\infty,2) \\
 (\infty,\infty,\infty,\infty) & (2,2,\infty,\infty) \\
 (1,1,\infty,\infty) & (2,\infty,2,\infty) \\
 (1,1,2,\infty) & (2,\infty,\infty,2) \\
 (1,1,\infty,2) & (\infty,2,2,\infty) \\
 (1,1,2,2) & (\infty,2,\infty,2) \\
 (\infty,\infty,1,1) & (\infty,\infty,2,2) \\
 (2,\infty,1,1) & (2,2,2,\infty) \\
 (\infty,2,1,1) & (2,2,\infty,2) \\
 (2,2,1,1) & (2,\infty,2,2) \\
 (2,\infty,\infty,\infty) & (\infty,2,2,2)  
\end{array}
\end{equation*}
The stopping times that never exercise when the option is out of the money are
\begin{align*}
 & (0,0,0,0) \\
 & (\infty,\infty,2,\infty) \\
 & (\infty,\infty,\infty,2) \\
 & (\infty,\infty,\infty,\infty) \\
 & (\infty,\infty,1,1) \\
 & (\infty,\infty,2,2) \\
 & (\infty,2,\infty,\infty) \\
 & (\infty,2,2,\infty) \\
 & (\infty,2,\infty,2) \\
 & (\infty,2,2,2) \\
 & (\infty,2,1,1)
\end{align*}
With corresponding values for $V_{0}$ for each given by
\begin{align*}
V_{0} & = 1 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + 0\cdot1 + \left(\frac{4}{5}\right)^{2}1 + 0\cdot4\right] = \frac{1}{4}\left(\frac{4}{5}\right)^{2} = 0.16 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + 0\cdot1 + 0\cdot1 + \left(\frac{4}{5}\right)^{2}4\right] = \frac{1}{4}\left(\frac{4}{5}\right)^{2}4 = 0.64 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + 0\cdot1 + 0\cdot1 + 0\cdot4\right] = 0 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + 0\cdot1\right] + \frac{1}{2}\left(\frac{4}{5}\right)3 = 1.2 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + 0\cdot1 + \left(\frac{4}{5}\right)^{2}1 + \left(\frac{4}{5}\right)^{2}4\right] = 0.8 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + 0\cdot1 + \left(\frac{4}{5}\right)^{2}1 + 0\cdot4\right] = 0.16 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + \left(\frac{4}{5}\right)^{2}1 + \left(\frac{4}{5}\right)^{2}1 + 0\cdot4\right] = 0.32 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + \left(\frac{4}{5}\right)^{2}1 + 0\cdot1 + \left(\frac{4}{5}\right)^{2}4\right] = 0.8 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + \left(\frac{4}{5}\right)^{2}1 + \left(\frac{4}{5}\right)^{2}1 + \left(\frac{4}{5}\right)^{2}4\right] = 0.96 \\
V_{0} & = \frac{1}{4}\left[0\cdot0 + \left(\frac{4}{5}\right)^{2}1\right] + \frac{1}{2}\left[\left(\frac{4}{5}\right)3\right] = 1.36 \\
\end{align*}
Notice that we achieve the maximum when the text says we should.

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