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\textbf{Question}

This question concerns the frequently-traded derivative product known
as a ``Perpetual American Put''.

\begin{description}
%Question 6a
\item{(a)}
Define the terms PERPETUAL, AMERICAN and PUT.

%Question 6b
\item{(b)}
Explain why, for a Perpetual American Put, the Black-Scholes equation
reduces to
$$\frac{1}{2} \sigma^2 S^2 V_{SS} + rSV_S - rV =0$$
where $\sigma$ denotes volatility, $r$ denotes interest rate, $S$
denotes the asset value and $V$ denotes the value of the option.

%Question 6c
\item{(c)}
Given a simple arbitrage argument to show that the option value can
never be less than the early-exercise price and thus
$$V \ge \textrm{max}(E-S,0).$$
Explain why $V$ must also satisfy $V \to 0$ as $S \to \infty$.

%Question 6d
\item{(d)}
Denote the optimal exercise price (i.e. the price which should
automatically trigger exercise) by $S^*$. Explain why
\begin{description}
\item{(i)}
$V(S^*) = E-S^*$
\item{(ii)}
$\partial V/\partial S^* = 0$
\end{description}

Hence or otherwise, show that the value of the option is given by
$$V(S) = \left ( \frac{E}{1+\frac{\sigma^2}{2r}} \right
)^{1+2r/\sigma^2} \frac{\sigma^2}{2r} S^{-2r/\sigma^2}.$$
\end{description}

\newpage
\textbf{Answer}

\begin{description}
%Question 6a
\item{(a)}
\begin{description}
\item{PERPETUAL:-} The option has no expiry and `goes on forever'.
\item{AMERICAN:-} The option may be exercised at any time of the
holder's choosing.
\item{PUT:-} The payoff at exercise is determined by max$(E-S,0)$
\end{description}

\item{(b)}
The standard Black-Scholes equation is
$$V_t +\frac{1}{2}\sigma^2 S^2 V_{SS} +rSV_S -rV =0,$$
but if the option is perpetual then by definition it cannot have a
value that depends on time. Thus $V_t=0$ and Black-Scholes becomes
$$\frac{1}{2}\sigma^2 S^2 V_{SS} +rSV_S -rV =0$$

\item{(c)}
Suppose it were that case that $V<\rm{max}(E-S,0)$.

Then buy the option (cost $V$) and exercise straight away (receiving
$E-S$). Then the payoff will be
$$-V+(E-S)=(E-S)-V>0$$
which is a risk free profit.

Thus $V \ge \rm{max}(E-S,0)$.

Also, the option is a put, and so profits if the share price
reduces. If $S\to\infty$ therefore the option value must become zero.


\item{(d)}
If the situation $S=S^*$ ensures that we exercise the option, then
when this happens the value of the option must be the payoff.

Thus $V(S^*)=E_S^*$.

Also, The OPTIMAL EXERCISE PRICE must, by definition, be optimal and
thus maximize the option value. So regarded as a function of $S$ and
$S^*$, at optimal conditions
$$\frac{\partial V}{\partial S^*}(S,S^*) =0$$

Now we must solve
$$\frac{1}{2}\sigma^2 S^2 V_{SS} +rSV_S -rV=0$$
subject to 
\begin{eqnarray*}
V\to 0 & as & S \to\infty\\
V(S^*) & = & E-S^*\\
\frac{\partial V}{\partial S^*} & = & 0
\end{eqnarray*}
To solve the ODE (Euler's equation), try
$$V \propto S^n$$
(n to be determined)

\begin{eqnarray*}
\frac{1}{2}\sigma^2 n(n-1) +rn-r & = & 0\\
\Rightarrow n & = & 1, \ \ n=-2r/\sigma^2
\end{eqnarray*}

Thus $V=AS+BS^{-2r/\sigma^2}$, (A,B arbitrary constants).

Now we need $V\to0$ as $S\to\infty$ $\Rightarrow A=0$.

Also, since $V(S^*)=E_S^*$
\begin{eqnarray*}
B(S^*)^{-2r/\sigma^2} & = & E-S^*\\
\Rightarrow B & = & (S^*)^{2r/\sigma^2}(E-S^*)
\end{eqnarray*}
Thus
$$V=(E-S^*)(S^*)^{2r/\sigma^2}S^{-2r/\sigma^2}$$
Now
$$\frac{\partial V}{\partial S^*} = - \left ( \frac{S^*}{S} \right
)^{\frac{2r}{\sigma^2}} +(E-S^*)\frac{2r}{\sigma^2} \left (
\frac{S^*}{S} \right )^{\frac{2r}{\sigma^2}}\frac{1}{S^*}$$

\begin{eqnarray*}
\Rightarrow S^* & = & (E-S^*_\frac{2r}{\sigma^2}\\
& = & \frac{2Er}{\sigma^2 \left ( 1+ \frac{2r}{\sigma^2} \right ) }\\
& = & \frac{2Er}{\sigma^2 + 2r}
\end{eqnarray*}

\begin{eqnarray*}
\Rightarrow V & = & \left ( \frac{2Er}{\sigma^2 + 2r} \right
)^{\frac{2r}{\sigma^2}} S^{-\frac{2r}{\sigma^2}} \left (
E-\frac{2Er}{\sigma^2 + 2r} \right )\\
& = & \left ( \frac{E}{1+ \frac{\sigma^2}{2r}} \right
)^{\frac{2r}{\sigma^2}+ 1} S^{-\frac{2r}{\sigma^2}} E \left (
\frac{\sigma^2}{ \sigma^2 +2r} \right )\\
& = & \left ( \frac{E}{1+\frac{\sigma^2}{2r}} \right
)^{\frac{2r}{\sigma^2}} S^{-\frac{2r}{\sigma^2}} \frac{E\sigma^2}{2r}
\left ( \frac{2r}{\sigma^2 +2r} \right )\\
& = & \left ( \frac{E}{1+\frac{\sigma^2}{2r}} \right
)^{\frac{2r}{\sigma^2}+1} \frac{\sigma^2}{2r} S^{-\frac{2r}{\sigma^2}}
\end{eqnarray*}

\end{description}



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