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\author{Jeremy Morris}
\title{Homework 3}

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\begin{description}
\item[Exercise 2.2 ] \textit{Consider the stock price $S_{3}$ in Figure 2.3.1}
	\begin{enumerate}
	\item \label{2.2.1}\textit{What is the distribution of $S_{3}$ under the risk-neutral probabilities $\tilde{p}=1/2, \tilde{q} = 1/2$.}
	\begin{eqnarray*}
	P\{S_{3} = 32\} & = & \frac{1}{8}, \omega = (HHH) \\
	P\{S_{3} = 8\} & = & \frac{3}{8}, \omega = (HHT,THH,HTH) \\
	P\{S_{3} = 2\} & = & \frac{3}{8}, \omega = (TTH,HTT,THT) \\
	P\{S_{3} = 0.50\} & = & \frac{1}{8} , \omega = (TTT)
	\end{eqnarray*} 
	

	\item \label{2.2.2}\textit{Compute $\tilde{\mathbf{E}}S_{1}$, $\tilde{\mathbf{E}}S_{2}$ and $\tilde{\mathbf{E}}S_{3}$.  What is the average rate of growth of the stock price under $\tilde{\mathbf{P}}$?}

	\begin{eqnarray*}
	\tilde{\mathbf{E}}S_{1} & = & 2P\{S_{1} = 2\} + 8P\{S_{1} = 8\} = 2(1/2) + 8(1/2) = 5 \\
	\tilde{\mathbf{E}}S_{2} & = & 1P\{S_{2} = 1\} + 4P\{S_{2} = 4\} + 16\{S_{2} = 16\}\\
	& = & 1(1/4) + 4(2/4) + 16(1/4) = 6.25 \\
	\tilde{\mathbf{E}}S_{3} & = & 0.5P\{S_{3} = 0.5\} + 2P\{S_{3} = 2\} + 8\{S_{3} = 8\} + 32\{S_{3} = 32\}\\
	& = & 0.5(1/8) + 2(3/8) + 8(3/8) + 32(1/8) = 7.8125
	\end{eqnarray*}
	The average rate of growth $\delta$ is defined as
	\begin{align*}
	\delta & = \frac{1}{n-1}\sum_{i=2}^{n} \frac{ES_{i}}{ES_{i-1}}
	\end{align*}
	In this case we have
	\begin{align*}
	\delta & = \frac{1}{2}\left( \frac{6.25}{5} + \frac{7.8125}{6.25} \right) = 1.25
	\end{align*}

	\item \textit{Answer (\ref{2.2.1}) and (\ref{2.2.2}) again under the actual probabilities $p = 2/3$, $q = 1/3$.}
	
	The distribution from (\ref{2.2.1}) would be
	\begin{eqnarray*}
	P\{S_{3} = 32\} & = & \frac{8}{27}, \omega = (HHH) \\
	P\{S_{3} = 8\} & = & \frac{12}{27}, \omega = (HHT,THH,HTH) \\
	P\{S_{3} = 2\} & = & \frac{6}{27}, \omega = (TTH,HTT,THT) \\
	P\{S_{3} = 0.50\} & = & \frac{1}{27} , \omega = (TTT)
	\end{eqnarray*} 

	And the expected values from (\ref{2.2.2}) would be
	\begin{eqnarray*}
	\mathbf{E}S_{1} & = & 2P\{S_{1} = 2\} + 8P\{S_{1} = 8\} = 2(1/3) + 8(2/3) = 6 \\
	\mathbf{E}S_{2} & = & 1P\{S_{2} = 1\} + 4P\{S_{2} = 4\} + 16\{S_{2} = 16\}\\
	& = & 1(1/9) + 4(5/9) + 16(4/9) = 9.444 \\
	\mathbf{E}S_{3} & = & 0.5P\{S_{3} = 0.5\} + 2P\{S_{3} = 2\} + 8\{S_{3} = 8\} + 32\{S_{3} = 32\}\\
	& = & 0.5(1/27) + 2(6/27) + 8(12/27) + 32(8/27) = 13.5
	\end{eqnarray*}
	The average rate of growth is
	\begin{align*}
	\delta & = \frac{1}{2}\left( \frac{9.444}{6} + \frac{13.5}{9.444} \right) = 1.502
	\end{align*}
	\end{enumerate}

\item[Exercise 2.4] \textit{Toss a coin repeatedly.  Assume the probability of head on each toss is 1/2, as is the probability of a tail.  Let $X_{j} = 1$ if the $j^{th}$ tail results in a head and $X_{j} = -1$ if it results in a tail.  Consider the stochastic process $M_{0},M_{1},M_{2},\ldots$ defined by $M_{0} = 0$ and
\[M_{n} = \sum_{j=1}^{n}X_{j}, n \geq 1\]
This is called a symmetric random walk.}
	\begin{enumerate}
	\item \textit{Using the properties of Thm 2.3.2, show that $M_{0},M_{1},M_{2},\ldots$ is a martingale.}

	We can show this directly, the distribution of $M_{n}$ is
	\begin{align*}
	M_{n} & = \left\{ \begin{array}{rl}
		 M_{n-1} + 1& \textrm{if }\omega_{n} = H \\
		 M_{n-1} - 1& \textrm{if }\omega_{n} = T
	\end{array}
	\right.
	\end{align*}
	%where
	%\begin{align*}
	%M_{n-1} & = \sum_{i=0}^{n-1}X_{i}
	%\end{align*}
	%in this case $M_{n-1}$ is not random because the path is specified up to that point.  Then we look at
	Then we show that $E_{n}[M_{n+1}] = M_{n}$.
	\begin{align*}
	E_{n}[M_{n+1}] & = E\left[\sum_{i=0}^{n+1}X_{i} \mid \omega_{1}\omega_{2}\ldots\omega_{n} \right] = E\left[ M_{n-1} + \sum_{i=n}^{n+1}X_{i} \mid \omega_{n} \right] \\
	& = M_{n-1} + E\left[ \sum_{i=n}^{n+1}X_{i} \mid \omega_{n} \right] \\
	\end{align*}
	this proves that $M_{n}$ is a Martingale since
	\begin{align*}
	E\left[ \sum_{i=n}^{n+1}X_{i} \mid \omega_{n} \right] &=\left\{ \begin{array}{rl}
		 1 & \textrm{if }\omega_{n} = H \\
		-1 & \textrm{if }\omega_{n} = T 
	\end{array}
	\right.
	\end{align*}

	\item \textit{Let $\sigma$ be a positive constant and, for $n \geq 0$, define
	\[S_{n} = e^{\sigma M_{n}}\left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)^{n}\]
	Show that $S_{0},S_{1},S_{2},\ldots$ is a martingale.}

	Here we use the same method as in the previous question.  The distribution for $S_{n}$ is
	\begin{align*}
	S_{n} & = \left\{ \begin{array}{rl}
		 S_{n-1}\left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)e^{\sigma} & \textrm{if }\omega_{n} = H \\
		 S_{n-1}\left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)e^{-\sigma} & \textrm{if }\omega_{n} = T
	\end{array}
	\right.
	\end{align*}
	Then for the conditional expectation side of the equation, we get
	\begin{align*}
	E_{n}[S_{n+1}] & = E_{n}\left[ \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)^{n+1} \exp{\left(\sigma\sum_{i=0}^{n+1} X_{i}\right)} \right] \\
	& = E_{n}\left[ \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)^{n-1} \exp{\left(\sigma\sum_{i=0}^{n-1} X_{i}\right)} \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right) \exp{\left(\sigma X_{n+1}\right)} \right] \\
	& = \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)^{n-1} \exp{\left(\sigma\sum_{i=0}^{n-1} X_{i}\right)}E_{n}\left[  \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right) \exp{\left(\sigma X_{n+1}\right)} \right] \\
	& = S_{n}
	\end{align*}
	Where
	\begin{align*}
	E_{n}\left[  \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right) \exp{\left(\sigma X_{n+1}\right)} \right] & = 
	\left\{ \begin{array}{ll}
		 \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)e^{\sigma} & \textrm{if }\omega_{n} = H \\
		 \left(\frac{2}{e^{\sigma} + e^{-\sigma}}\right)e^{-\sigma} & \textrm{if }\omega_{n} = T
	\end{array}
	\right.
	\end{align*}

	\end{enumerate}

\item[Exercise 2.5]\textit{Let $M_{0},M_{1},\ldots$ be the symmetric random walk of Exercise 2.4, and define $I_{0}=0$ and
\[I_{n} = \sum_{j=0}^{n-1}M_{j}(M_{j+1} - M_{j}), \textrm{ } n = 1,2,\ldots\] }
	\begin{enumerate}
	\item \textit{Show that \[I_{n} = \frac{1}{2}M_{n}^{2} - \frac{n}{2}\]}

	To start we have
	\begin{align*}
	I_{n} & = \sum_{j=0}^{n-1} M_{j}(M_{j+1} - M_{j}) = \sum_{j=0}^{n-1} M_{j}M_{j+1} - M_{j}^{2}
	\end{align*}
	Then we add zero
	\begin{align*}
	I_{n} & = \sum_{j=0}^{n-1} M_{j}M_{j+1} - M_{j}^{2} + \frac{1}{2}M_{j+1}^{2} - \frac{1}{2}M_{j+1}^{2} + \frac{1}{2}M_{j}^{2} - \frac{1}{2}M_{j}^{2} \\
	& = \sum_{j=0}^{n-1} -\frac{1}{2}(M_{j+1} - M_{j})^{2} + \frac{1}{2}M_{j+1}^{2} - \frac{1}{2}M_{j}^{2} \\
	& = \frac{1}{2}\left(\sum_{j=0}^{n-1} M_{j+1}^{2} - M_{j}^{2} \right) - \frac{n}{2} \\
	& = \frac{1}{2}M_{n} - \frac{n}{2}
	\end{align*}
	where the sum in the last step is a telescoping series.

	\item \textit{Let $n$ be an arbitrary nonnegative integer, and let $f(i)$ be an arbitrary function of a variable $i$.  In terms of $n$ and $f$, define another function $g(i)$ satisfying \[E_{n}[f(I_{n+1})] = g(I_{n})\]}

	Here we just take the expectation
	\begin{align*}
	E_{n}\left[ f(I_{n+1}) \right] & = E_{n}\left[ f\left(\sum_{j=0}^{n} M_{j}(M_{j+1} - M_{j}) \right)\right] \\
	 & = E_{n}\left[ f(I_{n}) + f\left(M_{n}(M_{n+1} - M_{n}) \right)\right]
	\end{align*}
	Here we drop the function $f$, since it can be any arbitrary function, we take it to be $f(i) = i$.
	\begin{align*}
	E_{n}[I_{n+1}] & = E_{n}\left[ I_{n} + (M_{n}(M_{n+1} - M_{n}) \right] \\
	& = E_{n}\left[ \frac{1}{2}M_{n}^{2} - \frac{n}{2} + M_{n}M_{n+1} - M_{n}^{2} \right] \\
	& = -\frac{1}{2}M_{n}^{2} - \frac{n}{2} + E_{n}\left[ M_{n}M_{n+1} \right] \\
	& = -\frac{1}{2}M_{n}^{2} - \frac{n}{2} + M_{n}E_{n}\left[ M_{n+1} \right] \\
	& (M_{n+1} \textrm{ is a Martingale}) \\
	& = -\frac{1}{2}M_{n}^{2} - \frac{n}{2} + M_{n}^{2} \\
	& = \frac{1}{2}M_{n}^{2} - \frac{n}{2} \\
	& = I_{n}
	\end{align*}
	So that $g(I_{n}) = I_{n}$.

	\end{enumerate}

\end{description}

\end{document}
