\documentclass[11pt]{article}

\author{Jeremy Morris}
\title{Homework 4}

\usepackage{longtable,amssymb,amsmath,amsfonts}
\usepackage[dvips]{graphicx}
\addtolength{\hoffset}{-1.05cm}
\addtolength{\textwidth}{2.5cm}
\addtolength{\textheight}{2.5cm}

\begin{document}
\maketitle

\begin{description}
\item[Exercise 2.9 (Stochastic volatilty, random interest rate)] \textit{Consdier a two-period stochastic volatility, random interest rate model of the type described  in Exercise 1.9 of Chapter 1.  The stock prices and interest rates are shown in Figure 2.8.1}
	\begin{enumerate}
	\item \textit{Determine risk-neutral probabilities\[\tilde{P}(HH), \tilde{P}(HT), \tilde{P}(TH), \tilde{P}(TT)\]such that the time-zero value of an option that pays off $V_{2}$ at time two is given by the risk-neutral pricing formula \[V_{0} = \tilde{E}\left[ \frac{V_{2}}{(1+r_{0})(1+r_{1})}\right]\]}

	We can use conditional probabilities to calculate the risk-neutral probabilities.
	\begin{align*}
	\tilde{P}\{H_{1}H_{2}\} & = \tilde{P}\{H_{1}\}\tilde{P}\{H_{2}\mid H_{1}\} \\
	\tilde{P}\{H_{1}\} & = \frac{1 + r_{0} - d_{0}}{u_{0} - d_{0}} = \frac{1 + 1/4 - 1/2}{2 - 1/2} = \frac{1}{2} \\
	\tilde{P}\{H_{2}\mid H_{1}\} & = \frac{1 + r_{1}(H) - d_{1}}{u_{1} - d_{1}} = \frac{1 + 1/4 - 1}{3/2 - 1} = \frac{1}{2} \\
	\tilde{P}\{H_{1}H_{2}\} & = \frac{1}{2}\frac{1}{2} = \frac{1}{4} \\
	\end{align*}
	\begin{align*}
	\tilde{P}\{T_{2}\mid H_{1}\} & = \frac{u_{1} - 1 - r_{1}(H)}{u_{1} - d_{1}} = \frac{3/2 - 1 - 1/4}{3/2 - 1} = \frac{1}{4} \\
	\tilde{P}\{H_{1}T_{2}\} & = \frac{1}{2}\frac{1}{2} = \frac{1}{4} \\
	\end{align*}
	\begin{align*}
	\tilde{P}\{T_{1}\} & = \frac{u_{0} - 1 - r_{0}}{u_{0} - d_{0}} = \frac{2 - 1 - 1/4}{2 - 1/2} = \frac{1}{2} \\
	\tilde{P}\{H_{2} \mid T_{1}\} & = \frac{1 + r_{1}(T) - d_{1}}{u_{1} - d_{1}} = \frac{1 + 1/2 - 1}{4 - 1} = \frac{1}{6} \\
	\tilde{P}\{T_{1}H_{2}\} & = \frac{1}{2}\frac{1}{6} = \frac{1}{12} \\
	\end{align*}
	\begin{align*}
	\tilde{P}\{T_{2} \mid T_{1}\} & = \frac{u_{1} - 1 - r_{1}(T)}{u_{1} - d_{1}} = \frac{4 - 1 - 1/2}{4 - 1} = \frac{5}{6} \\
	\tilde{P}\{T_{1}T_{2}\} & = \frac{1}{2}\frac{5}{6} = \frac{5}{12} 
	\end{align*}

	\item \textit{Let $V_{2} = (S_{2} - 7)^{+}$.  Compute $V_{0}, V_{1}(H)$ and $V_{2}(T)$.}

	\begin{align*}
	V_{1}(H) & = \frac{1}{1+r_{1}(H)}\left( \tilde{P}\{HH\}V_{2}(HH) + \tilde{P}\{HT\}V_{2}(HT)\right) \\
	 & = \frac{1}{1+1/4} \left( \frac{1}{4}5 + \frac{1}{4}1 \right) = \frac{6}{5}
	\end{align*}
	\begin{align*}
	V_{1}(T) & = \frac{1}{1+r_{1}(T)}\left( \tilde{P}\{TH\}V_{2}(TH) + \tilde{P}\{TT\}V_{2}(TT)\right) \\
	 & = \frac{1}{1+1/2} \left( \frac{1}{12}1 + \frac{5}{12}0 \right) = \frac{1}{18}
	\end{align*}
	\begin{align*}
	V_{0} & = \frac{1}{1+r_{0}} \left( \tilde{P}\{H\}V_{1}(H) + \tilde{P}\{T\}V_{1}(T) \right) \\
	 & = \frac{1}{1+1/4} \left(\frac{1}{2}\frac{6}{5} + \frac{1}{2}\frac{1}{18} \right) = 0.50222
	\end{align*}

	\item \textit{Suppose an agent sells the option described above for $V_{0}$ at time zero.  Compute the position $\Delta_{0}$ she should take in the stock at time zero so that at time one, regardless of whether the first coin toss results in head or tail, the value of her portfolio is $V_{1}$.}

	\begin{align*}
	\Delta_{0} & = \frac{V_{1}(H) - V_{1}(T)}{S_{1}(H) - S_{1}(T)} \\
	 & = \frac{6/5 - 1/18}{8 - 2} = 0.1907407
	\end{align*}

	\item \textit{Suppose that the first coin toss results in head.  What position $\Delta_{1}(H)$ should the agent now take in the stock to be sure that, regardless of the second toss, the value of her portfolio at time two will be $(S_{2} - 7)^{+}$.}

	\begin{align*}
	\Delta_{1}(H) & = \frac{V_{2}(HH) - V_{2}(HT)}{S_{2}(HH) - S_{2}(HT)} \\
	 & = \frac{5 - 1}{12 - 8} = 1
	\end{align*}

	\end{enumerate}

\item[Exercise 2.10 (Dividend-paying stock)]\textit{An agent who begins with initial capital $X_{0}$ and at each time $n$ takes a position of $\Delta_{n}$ shares of stock, where $\Delta_{n}$ depends only on the first $n$ coin tosses, has a portfolio value governed by the wealth equation \[X_{n+1} = \Delta_{n}Y_{n+1}S_{n} + (1+r)(X_{n} - \Delta_{n}S_{n})\]}
	\begin{enumerate}
	\item \textit{Show that the discounted wealth process is a martingale under the risk-neutral measure.}

	The discounted wealth process is given as
	\begin{align*}
	\frac{X_{n}}{(1+r)^{n}} & = \tilde{E}_{n}\left[\frac{X_{n+1}}{(1+r)^{n+1}}\right]
	\end{align*}
	Then we show that it holds for the particular form of the wealth equation given in this problem.
	\begin{align*}
	\tilde{E}_{n}\left[\frac{X_{n+1}}{(1+r)^{n+1}}\right] & = \tilde{E}_{n}\left[ \frac{\Delta_{n}Y_{n+1}S_{n} + (1+r)(X_{n} - \Delta_{n}S_{n})}{(1+r)^{n+1}} \right] \\
	 & = \tilde{E}_{n}\left[ \frac{\Delta_{n}Y_{n+1}S_{n}}{(1+r)^{n+1}} + \frac{X_{n} - \Delta_{n}S_{n}}{(1+r)^{n}} \right] \\
	 & = \frac{\Delta_{n}S_{n}}{(1+r)^{n+1}} \tilde{E}_{n}\left[ Y_{n+1} \right] + \frac{X_{n} - \Delta_{n}S_{n}}{(1+r)^{n}}
	\end{align*}
	Then all we have to do is show that $\tilde{E}_{n}\left[ Y_{n+1} \right] = (1+r)$.
	\begin{align*}
	\tilde{E}_{n}\left[ Y_{n+1} \right] & = \tilde{p}u + \tilde{q}d \\
	& = \frac{1+r-d}{u-d}u + \frac{u-1-r}{u-d}d \\
	& = \frac{(1+r)-d}{u-d}u + \frac{u-(1+r)}{u-d}d \\
	& = \frac{u(1+r)-ud}{u-d} + \frac{ud-d(1+r)}{u-d} \\
	& = \frac{(u-d)(1+r)}{u-d} = (1+r)
	\end{align*}
	Then
	\begin{align*}
	\tilde{E}_{n}\left[\frac{X_{n+1}}{(1+r)^{n+1}}\right] & = \frac{\Delta_{n}S_{n}}{(1+r)^{n+1}} (1+r) + \frac{X_{n} - \Delta_{n}S_{n}}{(1+r)^{n}} \\
	& = \frac{\Delta_{n}S_{n}}{(1+r)^{n}} + \frac{X_{n} - \Delta_{n}S_{n}}{(1+r)^{n}} \\
	& = \frac{X_{n}}{(1+r)^{n}}
	\end{align*}

	\item \textit{Show that the risk-neutral pricing formula still applies.}

	This fact follows from the previous exercise and Theorem 1.2.2.

	\item \textit{Show that the discounted stock price is not a martingale under the risk-neutral measure.  However, if $A_{n+1}$ is a constant $a \in (0,1)$, regardless of the value of $n$ and the outcome of the coin tossing $\omega_{1}\ldots\omega_{n+1}$, then $\frac{S_{n}}{(1-a)^{n}(1+r)^{n}}$ is a martingale under the risk-neutral measure.}

	The stock price is given as follows
	\begin{align*}
	S_{n+1} & = (1 - A_{n+1})Y_{n+1}S_{n}
	\end{align*}
	Then we have to show
	\begin{align*}
	\frac{S_{n}}{(1+r)^{n}} & \neq \tilde{E}_{n}\left[\frac{S_{n+1}}{(1+r)^{n+1}}\right]
	\end{align*}
	\begin{align*}
	\tilde{E}_{n}\left[\frac{S_{n+1}}{(1+r)^{n+1}}\right] & = \frac{1}{(1+r)^{n+1}}\tilde{E}_{n}\left[(1-A_{n+1})Y_{n+1}S_{n}\right] \\
	& = \frac{S_{n}}{(1+r)^{n+1}}\tilde{E}_{n}\left[(1-A_{n+1})Y_{n+1}\right] \\
	& = \frac{S_{n}}{(1+r)^{n}}\tilde{E}_{n}\left[(1-A_{n+1})\right] \\
	& = \frac{S_{n}}{(1+r)^{n}} - \tilde{E}_{n}\left[A_{n+1}\right]
	\end{align*}
	If $a \in (0,1)$ is some constant and we take a different discounted stock price, this will become a martingale.
	\begin{align*}
	\frac{S_{n}}{(1-a)^{n}(1+r)^{n}} & = \tilde{E}_{n}\left[\frac{S_{n+1}}{(1-a)^{n+1}(1+r)^{n+1}}\right]
	\end{align*}
	\begin{align*}
	& = \frac{1}{(1-a)^{n+1}(1+r)^{n+1}}\tilde{E}_{n}\left[(1-a)Y_{n+1}S_{n}\right] \\
	& = \frac{1}{(1-a)^{n+1}(1+r)^{n+1}} (1+r)(1-a)S_{n} \\
	& = \frac{S_{n}}{(1-a)^{n}(1+r)^{n}}
	\end{align*}
	\end{enumerate}

\end{description}

\end{document}
