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\textbf{Question}

\begin{description}
%Question 5a
\item{(a)}
Suppose that $U(S,t)$ satisfies the Black-Scholes equation. Show that
if $V$ is defined by
$$U(S,t) = S^nV(\eta,t)$$
where $\eta=K/S$ and $K$ and $n$ are constant then
$$V_t + \frac{1}{2} \sigma^2 V_{\eta\eta} + r \eta V_{\eta} - rV =0$$
provided $n$ takes a particular value (which you should determine).

%Question 5b
\item{(b)}
A European DOWN-AND-OUT Call with strike $E$, expiry $T$ and barrier
$X$ is identical to a European Call option \textit{except} for the
fact that the option cannot be exercised if the price of the
underlying ever drops below $X$. Explain briefly why the value
$D(S,t)$ of such an option must satisfy $D(X,t)=0$ and $D(X,T)=
\textrm{max}[S-E,0]$. Using the result of part (a) or otherwise show
that, if $C(S,t)$ denotes the value of a European Call option with
strike $E$ and expiry $T$, then
$$D(S,t) = C(S,t) - AS^{1-2r/\sigma^2} C(K/S,t)$$
where $A$ and $K$ are constants (which you should determine).

%Question 5c
\item{(c)}
By considering the payoff of a portfolio which is long one
down-and-out Call and long one down-and-in Call, determine the value
of a down-and-in Call.

\end{description}
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\textbf{Answer}

\begin{description}
%Question 5a
\item{(a)}
Since $U$ satisfies Black-Scholes we have

$U_t+\frac{1}{2} \sigma^2 S^2 U_{SS} +rSU_S -rU =0$.

Now put $U=S^nV(\eta, t)$ where $\eta = K/S$.

Then
\begin{eqnarray*}
U_t & = & (S^nV)_{\eta}\eta_t +(S^nV)_t t_t = 0 +S^nV_t\\
& = & S^nV_t\\
U_S & = & nS^{n-1}V +S^nV_S = nS^{n-1}V +S^nV_{\eta}\eta_S\\
& = & nS^{n-1}V -\eta S^{n-1} V_{\eta}\\
\rm{also} & &\\
U_{SS} & = & n(n-1)S^{n-2}V +nS^{n-1}V_{\eta} \left ( -\frac{K}{S^2}
\right )\\ 
& & -(n-2)KS^{n-3}V_{\eta} -KS^{n-2}V_{\eta \eta} \left (
-\frac{K}{S^2} \right )\\
& = & n(n-1)S^{n-2}V -\eta S^{n-2} nB_{\eta} -(n-2)S^{n-2} \eta
V_{\eta}\\
& & +S^{n-2} \eta^2 V_{\eta\eta}
\end{eqnarray*} 

$\Rightarrow$
\begin{eqnarray*} S^nV_t & + & \frac{1}{2} \sigma^2 S^2 [
n(n-1)S^{n-2}V -S^{n-2}n\eta V_{\eta} -(n-2)S^{n-2} \eta V\\
& + & {\eta} +
S^{n-2}\eta^2V_{\eta\eta} ]\\
& + & rS[ nS^{n-1}V -S^{n-1}\eta V_{\eta}] -rS^nV =0
\end{eqnarray*}

Canceling $S^n$ and re-arranging gives
$$V_t + V_{\eta\eta} \left [ \eta^2 \frac{1}{2} \sigma^2 \right ] +
\left [ -\frac{n\sigma^2}{2} -\frac{\sigma^2}{2}(n-2) -r \right ]\eta
V_{\eta}$$
$$+V \left [ \frac{1}{2} \sigma^2 (n-1)N +rn -r \right ] =0$$

So to get back to Black-Scholes again we need $n$ such that
\begin{eqnarray*}
-\frac{n\sigma^2}{2}- \frac{\sigma^2}{2}n + \sigma^2 & = & 2r\\
\Rightarrow n & = & 1-\frac{2r}{\sigma^2}.
\end{eqnarray*}
With this value, the coefficient of V becomes
$$\frac{\sigma^2}{2} \left ( 1 -\frac{2r}{\sigma^2} \right ) \left (
-\frac{2r}{\sigma^2} \right ) - \frac{2r^2}{\sigma^2} = -r $$

Thus with $n=1-2r/\sigma^2$ we DO get back to B/Scholes.

%Question 5b
\item{(b)}
For a European Down-and-out call the payoff at expiry is the same as a
vanilla call IF the option is exercised. Thus at expiry T
$$D(S,T) = \rm{max}(S-E,0)$$
Also, the option becomes worthless $\forall t$ the instant that the
share price hits $S=X$ and thus
$$D(X,t) = 0.$$

Now consider $D(S,t)=C(S,t) -AS^{1-\frac{2r}{\sigma^2}} C\left (
\frac{K}{S}, t \right )$.

We have to show that this satisfies 3 things:-
\
\begin{description}
\item{(i)}
Must satisfy Black-Scholes. Well $C(S,t)$ does by definition and by
part (a) of this question so does $S^{1-\frac{2r}{\sigma^2}}C(K/S,
t)$.

The linearity of Black-Scholes now ensures that we may add
solutions $\Rightarrow$ Black-Scholes is satisfied.
\item{(ii)}
We must ensure that $D(X,t) =0 \ \forall t$.
Now
$$D(X,t) = C(X,t) -AX^{1-\frac{2r}{\sigma^2}} C(K/X, t)$$
and clearly we can fix this up to be zero if we choose
$$A=X^{- \left ( 1- \frac{2r}{\sigma^2} \right )}, \ \ K=X^2$$
\item{(iii)}
We must ensure finally that $B(S,T)=\rm{max}(S-E,0)$.

But
\begin{eqnarray*}
B(S,T) & = & C(S,T) - (S/X)^{1 -\frac{2r}{\sigma^2}} C(X^2/S, T)\\
& = & \rm{max}(S-E,0) - (S/X)^{ 1- \frac{2r}{\sigma^2}}
\rm{max}(X^2/S-E, 0)
\end{eqnarray*}
But for sure $S>X$ and $E>X$ so $X^2/S-E<)$
$$\Rightarrow B(S,T)=\rm{max}(S-E,0)$$
as it should.
\end{description}


%Question 5c
\item{(c)}
Let $\Pi - D_i(S,t) +D_0(S,t)$

Then obviously
$$\Pi =C(S,t)$$
since whether the barrier is triggered or not the option will be the
same as a European call.

\begin{eqnarray*}
\Rightarrow D_i(S,t) + D_0(S,t) & = & C(S,t)\\
\Rightarrow D_i(S,t) & = & C(S,t)\\
& &  - \left [ C(S,t) -(S/X)^{1
-\frac{2r}{\sigma^2}} C(X^2/S, t) \right ]\\
D_i(S,t) & = & {\frac{S}{X}}^{1- \frac{2r}{\sigma^2}} C(X^2/S,t)
\end{eqnarray*} 

\end{description}


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