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\textbf{Question}

\begin{description}
%Question 4a
\item{(a)}
An up-and-out barrier Put option is identical to a European Vanilla
Put, save for the fact that if any time during the life of the option
the asset price exceeds the barrier $B$, the option instantly becomes
(and remains) worthless. Assuming that the underlying asset pays no
dividends, explain briefly why the fair price $V$ of the option must
satisfy the boundary value problem
$$V_t + \frac{1}{2} \sigma^2 S^2 V_{SS} + rSV_S -rV = 0, \ \ \
(S&lt;B),$$
$$V(B,t)=0, \ \ \ V(S,T)=\textrm{max}(E-S,0), \ \ \ (S&lt;B).$$
(Here as usual the asset value, strike price, volatility and interest
rate are denoted by $S$, $E$, $\sigma$ and $r$ respectively.)

%Question 4b
\item{(b)}
Assume now that for a particular up-and-out Put B&gt;E. Show that if
U(S,t) satisfies the Black-Scholes equation and $V$ is defined by
$$U(S,t) = S^nV(\eta, t), \ \ \ \ \left ( \eta=\frac{K}{S} \right )$$
where $K$ is an arbitrary constant, then $V$ also satisfies the
Black-Scholes equation provided $n$ takes a specific value (which you
should determine).

Hence or otherwise show that the fair value of and up-and-out barrier
Put is given by
$$V=P_{BS}(S,t)- \left ( \frac{S}{B} \right
)^{1-2r/\sigma^2}P_{BS}(B^2/S, t)$$
where $P_{BS}$ denotes the value of a European Vanilla Put. 
\end{description}

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\textbf{Answer}

\begin{description}
\item{(a)}
%Question 4a
An up-and-out barrier option is just like a normal Put until the
barrier is reached, and so satisfies Black-Scholes.
$$\Rightarrow V_t+\frac{1}{2}\sigma^2 S^2 V_{SS} +rSV_S -rV = 0 \ \ \
(S&lt;B)$$

As soon as $S=B$ it is worthless

$$\Rightarrow V(B,t)=0$$
(as this must apply instantaneously and for all time.)

Provided $S&lt;B$ the payoff is just that for a Euro Vanilla Put and thus
$$V(S,T)=\rm{max}(E-S,0).$$

%Question 4b
\item{(b)}
Now $U(S,t)=S^n V(\eta ,t)$, $\ (\eta=K/S)$
\begin{eqnarray*}
\Rightarrow U_t &amp; = &amp; S^nV_t\\
U_S &amp; = &amp; nS^{n-1}V-KS^{n-2}V_{\eta} = nS^{n-1}V-\eta
S^{n-1}V_{\eta}\\
U_{SS} &amp; = &amp; n(n-1)S^{n-2}V- nS^{n-2}\eta V_{\eta}- (n-2)S^{n-2} \eta
V_{\eta}\\ &amp; &amp; + \eta^2S^{n-2}V_{\eta\eta}
\end{eqnarray*}
Now $U$ satisfies Black-Scholes so
$$U_t +\frac{1}{2} \sigma^2 S^2 U_{SS}+ rSU_S- rU = 0$$

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

Thence
$$V_t+\frac{1}{2}\sigma^2 \left [ (n^2-n)V- n\eta V_{\eta} -\eta (n-2) V_{\eta}
+\eta^2V_{\eta\eta} \right ]$$
$$+rn[V]- r\eta V_{\eta} -rV =0$$

so that
$$V_t+\frac{1}{2}\sigma^2\eta^2V_{\eta\eta}+ V_{eta} \left [
-\frac{1}{2} \sigma^2 n \eta -r\eta- \eta(n-2)\frac{1}{2}\sigma^2
\right ]$$
$$+V\left [ \frac{\sigma^2}{2}(N^2-n)+ r(n-1) \right ] =0$$

$\Rightarrow$
$$V_t +\frac{1}{2} \sigma^2 \eta^2 V_{\eta\eta} +\eta V_{\eta} \left
[-r- \sigma^2[n-1] \right ]$$
$$+ (n-1)\left [ r +\frac{1}{2}n\sigma^2 \right ] V =0 $$

To get Black-Scholes out of this we need
$$-r-\sigma^2(n-1) =r$$
$$\Rightarrow n=1-\frac{2r}{\sigma^2}$$
But then
$$(n-1) \left [ r+\frac{1}{2} n\sigma^2 \right ] =
-\frac{2r}{\sigma^2} \left [ r+ \frac{1}{2} \sigma^2 \left ( 1-
\frac{2r}{\sigma^2} \right ) \right ] = -r$$

Thus with $n=1-2r/\sigma^2$, V satisfies the Black-Scholes equation
$$V_t +\frac{1}{2}\sigma^2 \eta^2 V_{\eta\eta} + r\eta V_{\eta} -rV
=0.$$

Since Black-Scholes is linear we may add solution. So consider a
solution of the form
$$V=P_{BS}(S,t)+ AS^nP_{BS}(K/S,t)$$
where $A$ is a constant and $n$ is chosen as above.

We have:-
\begin{description}
\item{(i)} This satisfies Black-Scholes $\forall A$, $\forall K$

\item{(ii)} We need $V(B,t)=0$. Thus
$$0 = P_{BS}(B,t)+ AB^nP_{BS}(K/B,t).$$
Clearly this condition holds if we set $K=B^2$ and $A=-B^{-n}$
\end{description}

Thus
$$V=P_{BS}(S,t) -\left ( \frac{S}{B} \right )^n P_{BS}(B^2/S,t), \ \ \
(n=1-2r/\sigma^2)$$

Finally we must check the payoff.

At expiry
\begin{eqnarray*}
V(S,T) &amp; = &amp; P_{BS}(S,T) - \left ( \frac{S}{B} \right )^n
P_{BS}(B^2/S,T)\\
&amp; = &amp; \rm{max}(E-S,0)-0
\end{eqnarray*}
since if $E&lt;B$ the second term is max$(E-B^2/S,0)=0$.

\end{description}



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