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\noindent{\bf Math 496/889 \hfill Exam 1 \hfill Name:\rule{5.cm}{.25mm}} \\
\noindent{\bf Friday, November 7, 2003 \hfill SSN:\rule{5.cm}{.25mm}}\\
\bigskip
\begin{tabular}{||l|c|c|c|c|c|c|c||}
\hline
Problem & 1 & 2 & 3 & 4 & 5 & 6 &Total\\
\hline
Possible& 20 & 10 & 15 & 10 & 15 & 20 & 90 \\
\hline
Points & & & & & & & \\
\hline
\end{tabular}
\begin{enumerate}
\item (20 points)
Consider a two-time-stage binomial-outcome model for pricing an
financial derivative. Each time stage is a
year. A stock starts at 100.
In each year, the stock can go up by 8\% or down by
4\%. The continously compounded interest rate on a \$1 bond
is constant at 4\% each year. Find the
price of a \emph{put option} with exercise price 105, with
exercise date at the end of the second year. Also, find the
replicating portfolio at each node.
\begin{verbatim}
At node (1,1) At node (2,2)
phi = -0.11085 V = 0
psi = 11.414 S = 116.64
V = 0.414
S = 108
At node (0,0) At node (2,1)
phi = -0.372 V = 1.32
psi = 39.042 S = 103.68
V = 1.80
S = 100
At node (1,0) At node (2,0)
phi = -1 V = 12.84
\psi = 100.883 S = 92.16
V = 4.883
S = 96
Pi = 0.67342, 1 - pi = 0.32657
\end{verbatim}
\item (10 points) James Bond is determined to ruin the casino at Monte Carlo
by betting \$1 on Red at the roulette wheel. The
probability of Bond winning at one turn in this game
is $18/38 \approx 0.474$. James Bond, being Agent 007, is backed by
the full financial might of the British Empire, and so can be
considered to have unlimited funds. Approximately how
much money should the casino have to start with so that Bond has
only a ``one-in-a-million'' chance of ruining the casino?
\textbf{Solution:}
Here the casino is the ``gambler'', with probability of success at
any trial being $p = 0.526$ and the probability of failure at any
trial is $q = 1- p = 0.474$. The probability of ultimate ruin of a
gambler playing against an infinitely rich adversary is is (if $p >
q$), $ (q/p)^S_0$ and we wish to make this $10^(-6)$.
Solve $(0.474/0.526)^S_0 = 10^(-6)$ for $S_0 = 132.72$. So if the
casino has \$133 it has less than a 1-in-a-million chance of being
ruined.
\item (15 points)
Make a convincing argument that the net fortune $S_n$
coin-flipping game must be $0$ infinitely many times.
( That is, we consider the sum $S_n$
where the independent, identically
distributed random variables in the sum
$S_n = X_1 + ... + X_n$ are the Bernoulli random variables $X_i =
+1$ with probability $p = 1/2$ and $X_i = -1$ with probability $q =
1-p = 1/2$.) \emph{Suggestion:} Consider using the results of
the Law of the Iterated Logarithm.
\textbf{Solution:} Take $\lambda_{-1} = 1 - 1/2 < 1$.
The Law of the iterated logarithm says that that there is a sequence
times going to infinity when the fortune is greater than $(1 - 1/2)*
\sqrt{2 t \log(\log(t))}$ hence positive, and a sequence of times
going
to infinity when the fortune is less than $-(1 - 1/2)*
\sqrt{2 t \log(\log(t))}$ hence negative. By inter-leaving the
selection of times (possible because both sequences of times are going to
infinity) we can find a sequence of inerediate times when the value
must be zero.
This problem can also be done more rigorously and simply
with the Borel-Cantelli lemma.
\item (10 points) For two random variables $X$ and $Y$, statisticians call
$$
\operatorname{Cov}(X, Y) = E[ (X - E[X])(Y-E[Y]) ]
$$
the \emph{covariance} of $X$ and $Y$. If $X$ and $Y$ are
independent, then $\operatorname{Cov}(X,Y) = 0$. A positive
value of $\operatorname{Cov}(X,Y)$ indicates that $Y$ tends to
increase as $X$ increases, while a negative value indicates that $Y$
tends to decrease when $X$ increases. Thus,
$\operatorname{Cov}(X,Y)$ is an indication of the mutual
dependence of $X$ and $Y$. Show that
$$
\operatorname{Cov}(W(s),W(t)) = E[W(s) W(t)] = s
$$
for $0 < s < t$.
% Adapted from Karlin and Taylor, page 383, problem 3
\textbf{Solution:} Write
\begin{eqnarray*}
\operatorname{Cov}(W(s),W(t)) &=& E[W(s) W(t)] \\
&=& E[ W(s) \cdot(( W(t) - W(s)) + (W(s))] \\
&=& E[ W(s) \cdot(( W(t) - W(s))] + E[W(s) \cdot W(s))] \\
&=& 0 + s = s.
\end{eqnarray*}
\item (15 points)
Let $Z$ be a normally distributed random variable, with mean
$0$ and variance $1$, that is, $Z \sim N(0,1)$. Then consider the continuous
time stochastic process $X(t) = \sqrt{t} Z$. Show that the distribution of
$X(t)$ is normal with mean $0$ with variance $t$. Is $X(t)$ a
Brownian motion? Explain why or why not.
\textbf{Solution:}
No because $X(s) = \sqrt{s} Z$ is not independent of $X(t) - X(s) =
(\sqrt{t} - \sqrt{s}) Z$ Furthermore, the increments have
variance $( \sqrt{t} - \sqrt{s})^2 = t - 2 \sqrt{ts} + s \ne t- s$.
\item (20 points)
Simulate the solution of the stochastic differential equation
$$
dY(t) = Y(t) dt + Y(t) dW, \qquad Y(0) = 1
$$
on the interval $[0,1]$ with a step size $dt = 1/5$. Use
$W_{25}(t)$ with increments of $1/5$ to approximate Brownian Motion.
%% \begin{tabular}{||c|c|c|c|c|c|c|c||}
%% \hline
%% j & t_j & Y_j & Y_j \,dt & dW & Y_j dW & Y_j + Y_j dW & Y_j + Y_j\,dt + Y_j dW \\
%% \hline
%% 0 & 0 & 1 & 0.2 & 0 & & & \\
%% 1 & 0.2 & & & 0. & & & \\
%% 2 & 0.4 & & & 0. & & & \\
%% 3 & 0.6 & & & 0. & & & \\
%% 4 & 0.8 & & & 0. & & & \\
%% 5 & 1.0 & & & 0 & & & \\
%% \end{tabular}
\end{enumerate}
\end{document}