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\author{Jeremy Morris}
\title{Homework 1}

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\item[Exercise 1.1] \textit{Assume in the one-period binomial market of Section 1.1 that both $H$ and $T$ have positive probability of occurring.  Show that condition (1.1.2) precludes arbitrage.  In other words, show that if $X_{0} = 0$ and
\begin{align*}
X_{1} &= \Delta_{0}S_{1} + (1+r)(X_{0} - \Delta_{0}S_{0}) 
\end{align*}
then we cannot have $X_{1}$ strictly positive with positive probability unless $X_{1}$ is strictly negative with positive probability as well, and this is the case regardless of the choice of the number $\Delta_{0}$.}

If $X_{0} = 0$, then we get
\begin{align}
X_{1} & = \Delta_{0}\left[S_{1} - (1+r)S_{0}\right] \label{eq1}
\end{align}
Then, if the stock price goes up (which happens with positive probability), we have
\begin{align*}
X_{1} & = \Delta_{0}\left[uS_{0} - (1+r)S_{0}\right] \\
      & = S_{0}\Delta_{0}\left[u - (1+r)\right]
\end{align*}
where, from condition (1.1.2), we have $u > 1+r$ and (\ref{eq1}) is positive.  Likewise, if the stock price goes down, we have
\begin{align*}
X_{1} & = \Delta_{0}\left[dS_{0} - (1+r)S_{0}\right] \\
      & = S_{0}\Delta_{0}\left[d - (1+r)\right]
\end{align*}
and condition (1.1.2) tells us that $d < 1+r$ and (\ref{eq1}) is negative.  Since $S_{0}$ is always positive, the number $\Delta_{0}$ only serves to switch the signs in both cases.  Therefore no arbitrage holds regardless of the choice of $\Delta_{0}$.

\item[Exercise 1.4] \textit{In the proof of Theorem 1.2.2, show under the induction hypothesis that
\[X_{n+1}(\omega_{1}\omega_{2}\ldots\omega_{n}T) = V_{n+1}(\omega_{1}\omega_{2}\ldots\omega_{n}T)\]
}

From the text, we use the equations
\begin{align}
X_{n+1}(T) & = \Delta_{n}dS_{n} + (1+r)(X_{n} - \Delta_{n}S_{n}) \label{xn1} \\
\Delta_{n} & = \frac{V_{n+1}(H) - V_{n+1}(T)}{S_{n+1}(H) - S_{n+1}(T)} \label{delta}
= \frac{V_{n+1}(H) - V_{n+1}(T)}{(u-d)S_{n}}
\end{align}
Then substituting (\ref{delta}) into (\ref{xn1}) and taking into account the event $T$, we get
\begin{align*}
X_{n+1}(T) & = (1+r)X_{n} + (1+r)(\Delta_{n}dS_{n} - \Delta_{n}S_{n}) \\
 & = (1+r)X_{n} + \Delta_{n}S_{n}(d - (1+r)) \\
 & = (1+r)V_{n} + \left(\frac{V_{n+1}(H) - V_{n+1}(T)}{(u-d)S_{n}}\right)S_{n}(d - (1+r)) \\
% & = (1+r)V_{n} + \frac{\left(V_{n+1}(H) - V_{n+1}(T)\right)(d - (1+r))}{(u-d)} \\
 & = (1+r)V_{n} + \left(\frac{d - (1+r)}{u-d}\right)V_{n+1}(H) - \left(\frac{d - (1+r)}{u-d}\right)V_{n+1}(T) \\
 & = (1+r)V_{n} - \left(\frac{1+r-d}{u-d}\right)V_{n+1}(H) + \left(\frac{1+r-d}{u-d}\right)V_{n+1}(T) \\
 & = \tilde{p}V_{n+1}(H) + \tilde{q}V_{n+1}(T) - \tilde{p}V_{n+1}(H) + \tilde{p}V_{n+1}(T) \\
 & = \tilde{q}V_{n+1}(T) + \tilde{p}V_{n+1}(T) \\
 & = V_{n+1}(T)
\end{align*}

\item[Exercise 1.6 (Hedging a long position-one period)] \textit{Consider a bank that has a long position in the European call written on the stock price in Figure 1.1.2.  The Call expires at time one and has strike price $K=5$.  In Section 1.1, we determined the time-zero price of this call to be $V_{0} = 1.20$.  At time zero, the bank owns this option, which ties up capital $V_{0} = 1.20$.  The bank wants to earn the interest rate $25\%$ on this capital until time one (i.e., without investing any more money, and regardless of how the coin tossing turns out, the bank wants to have
\[\frac{5}{4} \cdot 1.20 = 1.50\]
at time one, after collecting the payoff from the option (if any) at time one).  Specify how the bank's trader should invest in the stock and money markets to accomplish this.}

Where $S_{0} = 4, S_{1}(H) = 8, S_{1}(T) = 2, r = 0.25, k = 5, X_{1} = 1.50$ and $V_{1} = (S_{1} - k)^{+}$, solve
\begin{align*}
X_{1}(H) & = \Delta_{0}S_{1}(H) + (1+r)(X_{0} - \Delta_{0}S_{0}) + V_{1}(H) \\
X_{1}(T) & = \Delta_{0}S_{1}(T) + (1+r)(X_{0} - \Delta_{0}S_{0}) + V_{1}(T)
\end{align*}
for $\Delta_{0}$ and $X_{0}$.  This will tell us how much stock will need to be purchased and how much needs to be invested in the money market.  Plugging in all the numbers, we get
\begin{align*}
1.5 & = 8\Delta_{0} + 1.25(X_{0} - 4\Delta_{0}) + 3 \\
1.5 & = 4\Delta_{0} + 1.25(X_{0} - 4\Delta_{0})
\end{align*}
Which reduces to the system of equations
\begin{align*}
\left(\begin{array}{r} -1.5\\1.5 \end{array}\right) & = 
\left(\begin{array}{rr} 1.25 & 3 \\ 1.25 & -1 \end{array}\right)
\left(\begin{array}{c} X_{0} \\ \Delta_{0} \end{array}\right)
\end{align*}
This gives $X_{0} = 0.60$ and $\Delta_{0} = -0.75$.  This means that we put $0.60$ in a bank account and take a short position in $0.75$ shares of the stock assuming that we own a European option on the stock.

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