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\textbf{Question}

\begin{description}
%Question 6a
\item{(a)}
If $A$ is a constant, show that $V=AS$ is a solution of the
Black-Scholes equation. What ``option'' has this value?

Call options with strike $E$ and expiry $T$ are to be written on a
share that pays a dividend. The structure of the dividend payment is
as follows; a single payment with yield $y$ (so that the amount
received by the holder is $yS$) will be made at a time $t_d,T$. The
fair value of such options is denoted by $C(S,T;E,T)$.

%Question 6b
\item{(b)}
Explain why the option price remains continuous as the dividend date
is crossed, but the share price drops from $S$ to $(1-y)S$. Give
details of the arbitrage possibilities that would exist if $S$ did
\textrm{not} jump to $(1-y)S$ across $t=t_d$.

%Question 6c
\item{(c)}
Let $V(S,t;E,T)$ denote the fair (Black-Scholes) price for a Call
option on a share that pays no dividends with strike $E$ and expiry
$T$. Show that
$$C(S,t;E,T) = \left \{ \begin{array}{ll} V(S,t;E,T) & (t_d \le t \le
T)\\
(1-y)V(S,t;E/(1-y),T) \ \ \ & () \le t \le t_d)
\end{array} \right.$$

%Question 6d
\item{(d)}
Using a financial argument or otherwise, determine whether 

$C(S,t;E,T)$ is larger or smaller than $V(S,t;E,T)$.

\end{description}
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\textbf{Answer}

\begin{description}
%Question 6a
\item{(a)}
Consider $V=AS$ in Black-Scholes.
\begin{eqnarray*}
\Rightarrow 0 + \frac{1}{2} \sigma^2 S^2(0) +rSA -rAS & = & 0\\
\Rightarrow rSA-rSA & = & 0
\end{eqnarray*}
So that $V=AS$ clearly satisfies Black-Scholes. The ``option'' with
this value consists simply of $A$ shares in the underlying - which
eventually had value $AS$.

%Question 6b
\item{(b)}
Underlying pays $yS$ at $t_d<T$: strike$=E$, expiry$=T$.

The option itself pays no dividends, and so there are no abrupt
changes in its value as $t=t_d$ is crossed and it is therefore
continuous.

Now suppose that $S$ DID NOT jump to $(1-y)S$ across $t=t_d$. If it
jumped to a value $S^+>(1-y)S$ then we arbitrage by short selling $S$
before $t_d$, pay the dividend $qS$ and buying back straight after
$t_d$.

If $S^+<(1-y)S$ then buy the asset before $t_d$, collect the dividend
and then sell $\Rightarrow$ risk free profit.


%Question 6c
\item{(c)}
Now $V(S,t;E,T)$ is the value of a call option on a share paying no
dividends. Since $C$ is continuous we have
$$C(S^-,t_d^-) = C(S^+,t_d^+)$$
but
\begin{eqnarray*}
S^+ & = & (1-y)S^-\\
\Rightarrow C(S^-, t_d^-) & = & c((1-y)S^-, t_d^+)
\end{eqnarray*}
i.e. the required jump condition is
$$C(S,t_d^-) = C( (1-y)S, t_d^+).$$

Now for $t>t_d$ $C$ satisfies
$$C_t +\frac{1}{2} \sigma^2 S^2 C_{SS} +rSC_S -rC =0$$
with $C(S,T) =\rm{max}(S-E,0)$. By $\underline{\rm{definition}}$ the
solution to this is
$$V(S,t;E,T).$$
Thus
$$C(S,t;E,T)=V(S,t;E,T) \ \ \ \ (t_d \le t \le T).$$

Now we can use the jump condition:- at $t_d$
$$C(S,t_d^-;E,T) = C((1-y)S,t_d^+)=V((1-y)S,t_d^+;E,T)$$

Now as we saw in the first part of the question that $AS$ is a
solution of Black-Scholes for any $A$, so certainly $V((1-y)S,t;E,T)$
is.

What does $V((1-y)S,t)$ do at expiry? 

Well
\begin{eqnarray*}
V((1-y)S,T) & = & \rm{max}((1-y)S-E,0)\\
& = & (1-y)\rm{max}(S-E/(1-y), 0)
\end{eqnarray*}
i.e. the payoff is the same as $1-y$ calls with a strike $E/(1-y)$.

Thus for $t<t_d$
$$C(S,t;E,T)=(1-y)C(S,t;E/(1-y),T)$$

and so finally

$C(S,t;E,T)= \left \{ \begin{array}{ll}
V(S,t;E,T) & (T \ge t \ge t_d)\\
(1-y)\rm{max}(S,t;E/(1-y),T) \ \ & (t_s \ge t)
\end{array}
\right. $


%Question 6d
\item{(d)}
C $\underline{\rm{must}}$ be less than $V$; the option pays no
dividend, but the underlying suffers a fall in prices because of the
dividend. Its upside potential must therefore be less and so $C<V$ 

\end{description}


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