\documentclass[11pt]{article}
\author{Jeremy Morris}
\title{Homework \# 11}

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\begin{document}
\maketitle

\begin{enumerate}
\item \textit{Let the current stock price be denoted \[S_{n} = S_{0}\exp{\left( (r - \frac{1}{2}\sigma^{2})n\Delta t + \sigma\sqrt{\Delta t}M_{n}\right)}\]where the random walk is given by \[M_{n} = \sum_{j=1}^{n}X_{j}\] with $M_{0} = 0$.  Each walk $X_{j}$ is independent from the others and\[X_{j} = \left\{\begin{array}{rl} 1 & \omega_{j} = H \\ -1 & \omega_{j} = T \end{array}\right.\]}
	\begin{enumerate}
	\item \textit{Fix time $T = N\Delta t$ and let $N \rightarrow \infty$, $\Delta t \rightarrow 0$.  Using the central limit theorem, show that the distribution of $\log{S_{N}/S_{0}}$ converges to a normal distribution with mean $(r - \frac{1}{2}\sigma^{2})T$ and variance $\sigma^{2}T$.}

	We note here that
	\begin{align*}
	\log{\frac{S_{N}}{S_{0}}} & = (r - \frac{1}{2}\sigma^{2})T + \sigma\sqrt{T/N}M_{N}
	\end{align*}
	and that all values are fixed except $M_{N}$ which is random.  We know that by the central limit theorem $M_{N}/N$ is normally distributed with some mean and variance.  The mean is given as
	\begin{align*}
	E\left[\frac{1}{N}M_{N}\right] & = \frac{1}{N}E\left[M_{N}\right]  = \frac{1}{N}\sum_{j=1}^{N}E\left[X_{j}\right] = \frac{1}{N}\sum_{j=1}^{N} \frac{1}{2}(1) + \frac{1}{2}(-1) = 0
	\end{align*}
	and the variance is given by
	\begin{align*}
	\textrm{var}\left(\frac{1}{N}M_{N}\right) & = \left(\frac{1}{N}\right)^{2}\left(E[M_{N}^{2}] - E[M_{N}]^{2}\right)
	\end{align*}
	we know that $E[M_{N}] = 0$ so that we are left with
	\begin{align*}
	\textrm{var}\left(\frac{1}{N}M_{N}\right) & = \left(\frac{1}{N}\right)^{2}E[M_{N}^{2}].
	\end{align*}
	So we have
	\begin{align*}
	E[M_{N}^{2}] & = E\left[ \left(\sum_{i=1}^{N}X_{i}\right)^{2} \right] = E\left[ \sum_{i=1}^{N}X_{i}^{2} + 2\sum_{i=1}^{N}\sum_{j \neq i}X_{i}X_{j} \right] \\
	& = E\left[ \sum_{i=1}^{N}X_{i}^{2} \right] + 2E\left[\sum_{i=1}^{N}\sum_{j \neq i}X_{i}X_{j} \right] \\
	& = N + 0.
	\end{align*}
	So that $\frac{1}{N}M_{N} \sim N(0,1/N)$.  We will make the remark that if $X \sim N(\mu,\sigma^{2})$, then $aX + b \sim N(\mu + b,a^{2}\sigma^{2})$.  This means that $M_{N} \sim N(0,N)$.  From this fact, we also get that $\log{S_{N}/S_{0}}$ is normally distributed with the correct mean and variance specified in the question.

	\item \textit{Show that if $X \sim N(\mu,\beta^{2})$, the $E[e^{X}] = e^{\mu + \beta^{2}/2}$.  Does $e^{X}$ have a normal distribution?}

	$E[e^{X}] = e^{\mu + \beta^{2}/2}$ is the definition of the moment generating function of the normal distribution.  $e^{X}$ has the log-normal distribution and only takes positive values.\

	\item \textit{Using part (b), assume $T$ is fixed, show that \[E\left[\lim_{\Delta t \rightarrow \infty} S(T) \right] = S_{0}e^{rT}\]}

	Then we have
	\begin{align*}
	E\left[\lim_{\Delta t \rightarrow \infty} S(T) \right] & = E\left[\lim_{\Delta t \rightarrow 0} S_{0}e^{(r - \frac{1}{2}\sigma^{2})T + \sigma\sqrt{\Delta t}M_{N}} \right] \\
	& = S_{0}E\left[\lim_{\Delta t \rightarrow 0} e^{(r - \frac{1}{2}\sigma^{2})T + \sigma\sqrt{\Delta t}M_{N}} \right]
	\end{align*}
	From part (a), we have that, in the limit as $N \rightarrow \infty$ (or $\Delta t \rightarrow 0$), we have that $(r - \frac{1}{2}\sigma^{2})T + \sigma\sqrt{T/N}M_{N} \sim N((r - \frac{1}{2}\sigma^{2})T,\sigma^{2}T)$.  And from part(b), we get that taking the expectation gives us
	\begin{align*}
	S_{0}E\left[\lim_{\Delta t \rightarrow \infty} e^{(r - \frac{1}{2}\sigma^{2})T + \sigma\sqrt{\Delta t}M_{N}} \right] = S_{0}e^{(r-\frac{1}{2}\sigma^{2})T - \frac{1}{2}\sigma^{2}T} = S_{0}e^{rT}
	\end{align*}

	\end{enumerate}

\item \textit{Let $S_{n}$ be the above process, show that the process\[\tilde{S}_{n} = \frac{S_{n}}{D^{n}}\]where\[D = \frac{1}{2}(e^{\sigma\sqrt{\Delta t}} + e^{-\sigma\sqrt{\Delta t}})e^{(r - \frac{1}{2}\sigma^{2})\Delta t}\]is a martingale.  When $\Delta t$ is small enough show that \[D \approx e^{r\Delta t} \approx (1 + r\Delta t).\]}

We show from the martingale property that
\begin{align*}
\tilde{E}_{n}\left[\tilde{S}_{n+1}\right] & = \tilde{S}_{n}.
\end{align*}
To begin we have
\begin{align*}
\tilde{E}_{n}\left[\tilde{S}_{n+1}\right] & = \tilde{E}_{n}\left[\frac{S_{n+1}}{D^{n+1}}\right] = \tilde{E}\left[ \frac{S_{0}e^{(r - \frac{1}{2}\sigma^{2})(n+1)\Delta t + \sigma\sqrt{\Delta t}M_{n+1}}}{(\frac{1}{2}(e^{\sigma\sqrt{\Delta t}} + e^{-\sigma\sqrt{\Delta t}})e^{(r - \frac{1}{2}\sigma^{2})\Delta t})^{n+1}} \right] \\
&= \frac{S_{0}e^{(r-\frac{1}{2}\sigma^{2})n\Delta t}e^{(r-\frac{1}{2}\sigma^{2})\Delta t}e^{\sigma\sqrt{\Delta t}M_{n}}}{\left(\frac{1}{2}(e^{\sigma\sqrt{\Delta t}} + e^{-\sigma\sqrt{\Delta t}})e^{(r - \frac{1}{2}\sigma^{2})\Delta t}\right)^{n+1}} \tilde{E}_{n}\left[ e^{\sigma\sqrt{\Delta t}X_{n+1}} \right]\\
&= \frac{\left(S_{0}e^{(r-\frac{1}{2}\sigma^{2})n\Delta t + \sigma\sqrt{\Delta t}M_{n}}\right)\frac{1}{2}(e^{\sigma\sqrt{\Delta t}} + e^{-\sigma\sqrt{\Delta t}})e^{(r-\frac{1}{2}\sigma^{2})\Delta t}}{\left(\frac{1}{2}(e^{\sigma\sqrt{\Delta t}} + e^{-\sigma\sqrt{\Delta t}})e^{(r - \frac{1}{2}\sigma^{2})\Delta t}\right)^{n+1}} \\
&= \frac{\left(S_{0}e^{(r-\frac{1}{2}\sigma^{2})n\Delta t + \sigma\sqrt{\Delta t}M_{n}}\right)}{\left(\frac{1}{2}(e^{\sigma\sqrt{\Delta t}} + e^{-\sigma\sqrt{\Delta t}})e^{(r - \frac{1}{2}\sigma^{2})\Delta t}\right)^{n}} = \tilde{S}_{n}
\end{align*}

\end{enumerate}

\end{document}
