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\textbf{Question}

\begin{description}
%Question 2a
\item{(a)}
On the last day of 1999, two investors (Asif and Barbara) decided to
spend money in buying options. On this day, the share price of BOOTS
was 495p, the share price of BP AMOCO was 495p and the share price of
SHELL was 438p. The expiry date for all of the options that they
bought was April 13th. Options were available as follows

\begin{center}
\begin{tabular}{|l|c|c|}
\hline\\
Option & Strike Price & Available for\\
\hline \hline\\
BOOTS PUTS & 460p & 20p\\
BOOTS CALLS & 460p & 60p\\
BP PUTS & 500p & 37p\\
BP CALLS & 500p & 34p\\
SHELL PUTS & 460p & 45p\\
SHELL CALLS & 460p & 20p\\
\hline
\end{tabular}
\end{center}

Asif bought 1000 BOOTS calls, 1000 BP calls and 500 SHELL puts, and
Barbara bought 1000 BOOTS puts, 2000 BP puts and 2000 SHELL calls.

Assuming that there was no bid/ask spread and no dealing charges, and
that on April 13th 2000 the share prices were given by:

\begin{center}
\begin{tabular}{lc}
BOOTS & 555p\\
BP AMOCO & 410p\\
SHELL & 500p
\end{tabular}
\end{center}

determine how much Asif and Barbara paid for their options, and what
the total profit or loss was for each investor after expiry.

%Question 2b
\item{(b)}
Now YOU MAY ASSUME that small charges $df$ in the function $f(S,t)$
are related to small changes in $S$ and $t$ by Taylor's theorem and
that the asset prices $S$ of a share follows the lognormal random walk
$$dS = rS dt + \sigma S dX$$
where $X$ is a random variable, $r$ and $\sigma$ are constants, and
$dX^2 \to dt$ as $dt \to 0$.

By considering a portfolio $\Pi = V- \Delta S$ (where $\Delta$ is to
be determined), show that the fair value $V$ of an option satisfies
the Black-Scholes equation
$$V_t + \frac{1}{2}\sigma^2 S^2 V_{SS} + rSV_S - rV =0.$$

\end{description}
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\textbf{Answer}

\begin{description}
%Question 2a
\item{(a)}
As instructed we assume no dealing costs or bid/ask
spreads. Proceeding with the given amounts and prices, we find that
they SPENT
\begin{description}
\item{(A)}
$$(1000)(0.60)+(1000)(0.34)+(500)(0.45)=\pounds 1165$$
\item{(B)}
$$(1000)(0.20)+(2000)(0.37)+(2000)(0.20)=\pounds 1340$$
\end{description}

Now at expiry we have Boots$=555p$, BP$=410p$, Shell$=500p$.

The payoffs are thus:-

\begin{tabular}{lll}
Boots:- & Puts worthless&\\
& Calls$=555-460$&$=\pounds 0.95$\\
BP:- & Puts$=500-410$&$=\pounds 0.90$\\
& Calls worthless\\
Shell:- & Puts worthless\\
& Calls$=500-460$&$=\pounds 0.40$
\end{tabular}

At expiry therefore the positions are
\begin{description}
\item{(A)}
$$1000\times0.95+1000\times0+500\times0=\pounds 950$$
\item{(B)}
$$1000\times0+2000\times0.90+2000\times0.40=\pounds 2600$$
\end{description}

Thus $\underline{\rm{ASIF\ lost\ }\pounds 215}$,
$\underline{\rm{BARBARA\ gained\ }\pounds 1260}$.

%Question 2b
\item{(b)}

We have $dS=rSdt+\sigma dX$ and 
$$df=f_SdS+f_tdt+\frac{1}{2}F_{SS}dS^2+f_{St}dtdS+ \frac{1}{2}f_{tt}
dt^2 +\cdots $$

Now consider the given portfolio $\Pi=V-\Delta S$.

Then
$$d\Pi = dV -\Delta dS.$$
Now to find $dV$ we have
\begin{eqnarray*}
dV & = & V_SdS+ V_tdt+ \frac{1}{2}V_{SS}dS^2+ V_{St}dtdS+
\frac{1}{2}V_{tt}dt^2+ \cdots\\
& = & V_S (rSdt +\sigma SdX)+ V_tdt\\
& & + \frac{1}{2} V_{SS} (r^2S^2dt^2+
\sigma^2 S^2 dX^2+ 2r\sigma S^2dtdX)\\
& & + V_{St}dt(rSdt+ \sigma S dX) +\frac{1}{2}V_{tt}dt^2+ \cdots
\end{eqnarray*}

But since we have $dX^2\to dt$ as $dt\to 0$
$$\Rightarrow dV=V_SrSdt+ V_S\sigma S dX+ V_tdt+ \frac{1}{2} \sigma^2
S^2 V_{SS} dX^2$$
$$+ O(dXdt)+ O(dt^2)+ \cdots$$

So $dV=rSV_sdt+ \sigma SV_SdX+ V_tdt+ \frac{1}{2}\sigma^2 S^2
V_{SS}dt$ to leading order.

Thus
\begin{eqnarray*}
d\Pi & = & dV- \Delta dS\\
& = & rSV_Sdt+ \sigma SV_SdX+ V_tdt+ \frac{1}{2} \sigma^2 S^2 V_{SS}
dt- \Delta dS\\
& = & rSV_Sdt+ \sigma SV_SdX+ V_tdt+ \frac{1}{2} \sigma^2 S^2 V_{SS}
dt\\
& &  -\Delta rSdt- \Delta \sigma S dX
\end{eqnarray*}

The randomness may now be eliminated by setting $\Delta =V_s$, giving
$$d\Pi =V_tdt+ \frac{1}{2} \sigma^2 S^2 V_{SS} dt.$$

The usual arbitrage argument now says that this must be equivalent to
putting money in the bank, 

So
$$d\Pi =r\Pi dt = rdt(V-\Delta S) = rdt(V-V_SS)$$

$$\Rightarrow V_tdt+ \frac{1}{2} \sigma^2 S^2 V_{SS}dt = rVdt-
rSV_Sdt$$
$$\Rightarrow V_t+ \frac{1}{2} \sigma^2 S^2 V_{SS} +rSV_S -rV = 0$$

\end{description}


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