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\author{Jeremy Morris}
\title{Homework 6}

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\begin{description}
\item[Exercise 3.2] \textit{Let $P$ be a probability measure on a finite probability space $\Omega$.  In this problem, we allow the possibility that $P(\omega) = 0$ for some values of $\omega \in \Omega$.  Let $Z$ be a random variable on $\Omega$ with the property that $P(Z \geq 0) = 1$ and $EZ = 1$.  For $\omega \in \Omega$, define $\tilde{P}(\omega) = Z(\omega)P(\omega)$, and for events $A \subset \Omega$, define $\tilde{P}(A) = \sum_{\omega \in \Omega}\tilde{P}(\omega)$.  Show the following.}

	\begin{enumerate}
	\item \textit{$\tilde{P}$ is a probability measure.}

	From the definitions
	\begin{align*}
	\sum_{\omega \in \Omega}\tilde{P} & = \sum_{\omega \in \Omega}Z(\omega)P(\omega) = EZ = 1.
	\end{align*}

	\item \textit{If $Y$ is a random variable, the $\tilde{E}[Y] = E[ZY]$.}

	Again, from the definitions
	\begin{align*}
	\tilde{E}[Y] & = \sum_{\omega \in \Omega}\tilde{P}(\omega)Y(\omega) = \sum_{\omega \in \Omega}\tilde{P}(\omega)Y(\omega) \frac{P(\omega)}{P(\omega)} \\
	 & = \sum_{\omega \in \Omega}P(\omega)Z(\omega)Y(\omega) = E[ZY].
	\end{align*}

	\item \textit{If $A$ is an event with $P(A) = 0$, then $\tilde{P}(A) = 0$.}

	From the definitions we have
	\begin{align*}
	\tilde{P}(A) & = \sum_{\omega \in A}\tilde{P}(\omega) = \sum_{\omega \in A}Z(\omega)P(\omega).
	\end{align*}
	Where we are given that for every $\omega \in A$, $P(A) = 0$, so that $\tilde{P}(A) = 0$.

	\item \textit{Assume that $P(Z > 0) = 1$.  Show that if $A$ is an event with $\tilde{P}(A) = 0$, then $P(A) = 0$.}

	As with the previous exercise
	\begin{align*}
	P(A) & = \sum_{\omega \in A}P(\omega) = \sum_{\omega \in A}\frac{\tilde{P}(\omega)}{Z(\omega)}
	\end{align*}
	Because $\tilde{P}(\omega) = 0$ for every $\omega \in A$ and $P(Z > 0) = 1$, $P(A) = 0$.

	\item \textit{Show that if $P$ and $\tilde{P}$ are equivalent, then they agree which events have probability one.}

	First, assume $P(A) = 1$, then
	\begin{align*}
	\tilde{P}(A) & = \sum_{\omega \in A}\tilde{P}(\omega) = \sum_{\omega \in A}Z(\omega)P(\omega) = E[Z] = 1.
	\end{align*}
	Next assume $\tilde{P}(A) = 1$, then
	\begin{align*}
	P(A) & = 1 - P(A^{C}) = 1 - \sum_{\omega \in A^{C}}P(\omega) = 1 - \sum_{\omega \in A^{C}}\frac{\tilde{P}(\omega)}{Z(\omega)} = 1.
	\end{align*}

	\item \textit{Construct an example in which we have only $P(Z \geq 0) = 1$ and $P$ and $\tilde{P}$ are not equivalent.}

	The following densities on three tosses demonstrates non-equivalent measures.
	\begin{align*}
	P & = \left\{ \begin{array}{lr}
		1/8 & \textrm{all heads} \\
		3/8 & \textrm{one tail} \\
		3/8 & \textrm{one head} \\
		1/8 & \textrm{all tails}
	\end{array} \right. \\
	\tilde{P} & = \left\{ \begin{array}{lr}
		1/3 & \textrm{all heads} \\
		1/3 & \textrm{one tail} \\
		1/3 & \textrm{one head} \\
		0 & \textrm{all tails}
	\end{array} \right. 
	\end{align*}

	\end{enumerate}

\item[Exercise 3.4] \textit{This problem refers to the model of Example 3.1.2, whos Radon-Nikodym process $Z_{n}$ appears in Figure 3.2.1.}
	\begin{enumerate}

	\item \textit{Compute the state price densities explicitly.}

	The state prices are given by the equation
	\begin{align*}
	\zeta_{n}(\omega) & = \frac{Z(\omega)}{(1+r)^{n}}.
	\end{align*}
	The values for $Z(\omega)$ are given in Example 3.1.2, so that the density of the state prices is given as
	\begin{align*}
	\zeta_{3} & = \left\{ \begin{array}{ll}
		27/125 & \omega = (HHH) \\
		54/125 & \omega = (THH,HTH,HHT) \\
		108/125 & \omega = (TTH,HTT,THT) \\
		216/125 & \omega = (TTT)
		\end{array} \right.
	\end{align*}

	\item \textit{Use the number computed above in formula (3.1.10) to find the time-zero price of the Asian option of Exercise 1.8 of Chapter 1.}

	Equation (3.1.10) is given as
	\begin{align*}
	V_{0} & = \sum_{\omega \in \Omega}V_{N}(\omega)\zeta(\omega)P(\omega),
	\end{align*}
	the values for $V_{N}$, taken from the previous homework assignment are
	\begin{align*}
	V_{3} & = \left\{ \begin{array}{ll}
		11 & \omega = (HHH) \\
		5 & \omega = (HHT) \\
		2 & \omega = (HTH) \\
		0.5 & \omega = (HTT) \\
		0.5 & \omega = (THH) \\
		0 & \omega = (THT) \\
		0 & \omega = (TTH) \\
		0 & \omega = (TTT)
		\end{array} \right.
	\end{align*}
	and the values for $P(\omega)$ are given in the text.  Then we have the following as the time-zero price :
	\begin{align*}
	V_{0} & = 11\left(\frac{27}{125}\right)\frac{8}{27} + 5\left(\frac{54}{125}\right)\frac{4}{27} + 2\left(\frac{54}{125}\right)\frac{4}{27} + \frac{1}{2}\left(\frac{108}{125}\right)\frac{2}{27} + \frac{1}{2}\left(\frac{108}{125}\right)\frac{2}{27} \\
	& = \frac{1}{125}(88 + 40 + 16 + 4 + 4) \\
	& = 1.216
	\end{align*}

	\item \textit{Compute also the state densities $\zeta_{2}(HT) = \zeta_{2}(TH)$.}

	From the text, we have
	\begin{align*}
	Z_{2}(HT) = Z_{2}(TH) & = \frac{9}{8}
	\end{align*}
	so that
	\begin{align*}
	\zeta_{2}(HT) = \zeta_{2}(TH) & = \frac{9}{8}\left(\frac{4}{5}\right)^{2} = \frac{18}{25}
	\end{align*}

	\item \textit{Use the risk-neutral pricing formula (3.2.6) in the form
	\begin{align*}
	V_{2}(HT) & = \frac{1}{\zeta_{2}(HT)}E_{2}[\zeta_{3}V_{3}](HT) \\
	V_{2}(TH) & = \frac{1}{\zeta_{2}(TH)}E_{2}[\zeta_{3}V_{3}](TH)
	\end{align*}
	to compute $V_{2}(HT)$ and $V_{2}(TH)$.}

	We begin with $V_{2}(HT)$, using values given in the previous part of this exercise, we have
	\begin{align*}
	V_{2}(HT) & = \frac{1}{\zeta_{2}(HT)}\left[ P(H)\zeta_{3}(HTH)V_{3}(HTH) + P(T)\zeta_{3}(HTT)V_{3}(HTT) \right] \\
	 & = \frac{25}{18}\left(2\frac{2}{3}\frac{54}{125} + \frac{1}{2}\frac{1}{3}\frac{108}{125}\right) = 1
	\end{align*}
	\begin{align*}
	V_{2}(TH) & = \frac{1}{\zeta_{2}(TH)}\left[ P(H)\zeta_{3}(THH)V_{3}(THH) + P(T)\zeta_{3}(THT)V_{3}(THT) \right] \\
	V_{2}(TH) & = \frac{25}{18}\left(\frac{1}{2}\frac{2}{3}\frac{54}{125} + 0\frac{1}{3}\frac{108}{125}\right) = 0.2
	\end{align*}

	\end{enumerate}

\end{description}

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