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\textbf{Question}

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%Question 3a
\item{(a)}
Draw payoff diagrams (from the holder's point of view) for a European
Vanilla Call and a European Vanilla Put, both with strike price $E$ and
expiry $T$. On financial grounds, would you expect American Vanilla
Calls and Puts to be more or less expensive than their European
equivalents?

%Question 3b
\item{(b)}
A POWER OPTION is an option whose payoff is given by
$$V(S,T) = AS^n$$
where $A$ and $n$ are constants. Show that the Black-Scholes equation
admits solutions of the form
$$V(S,t) = g(t)S^n$$
provided that $g(t)$ satisfies the ordinary differential equation
$$\frac{dg}{dt} + \left ( \frac{1}{2} \sigma^2 n + r \right ) (n-1) g
=0.$$

Hence or otherwise find the fair value for a power option. What
financial product does the option become in the special cases $n=0$
and $n=1$?
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\textbf{Answer}

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%Question 3a
\item{(a)}
Payoff for Euro Vanilla Call is max$(S-E,0)$

$\Rightarrow$ Payoff
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Payoff for Euro Vanilla Put is max$(E-S,0)$

$\Rightarrow$ Payoff
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The ability to exercise early that is passed by an American option is
evidently a great advantage to have. $\Rightarrow$ American Vanilla
should be more expensive than their European counterparts.


%Question 3b
\item{(b)}
We have $V_t+\frac{1}{2} \sigma^2 S^2 V_{SS} +rSV_S -rV =0$

Now try
$$V(S,t) = g(t)S^n.$$
\begin{eqnarray*}
\Rightarrow g'S^n+ \frac{1}{2} \sigma^2 S^2 n(n-1) S^{n-2} g+
rSgnS^{n-1} -rgS^n & = & 0\\
g'+\frac{1}{2} \sigma^2 n(n-1)g + rgn -rg & = & 0\\
g'+ \left [ \frac{1}{2} \sigma^2 n(n-1) +rn -r \right ] g & = & 0\\
\rm{i.e.}\ \ \ \ g' +n(n-1) \left [ \frac{1}{2} n\sigma^2 +r \right ]
g & = & 0.
\end{eqnarray*}

Thus
$$g(t) = \tilde{A}e^{(1-n)\left [ r+ \frac{1}{2} n\sigma^2 \right
]t}$$
where $\tilde{A}$ is an arbitrary constant, and so in all
$$V=S^n\tilde{A}e^{(1-n)\left [ r +\frac{1}{2} n \sigma^2 \right ]
t}.$$

Now we know that at expiry $V(S,T)=AS^n$ where $A$ and $n$ are given.

Thus
$$AS^n=S^n\tilde{A}e^{(1-n)\left [ r+ \frac{1}{2} n\sigma^2 \right ]
T}$$
$$\Rightarrow \tilde{A} =Ae^{-(1-n) \left [ R + \frac{1}{2} n\sigma^2
\right ] T}.$$
and so the value of the option is
$$V=Ae^{(1-n)\left [ r+ \frac{1}{2} n\sigma^2 \right ] (t-T)}S^n.$$

When $n=0$, $V=Ae^{r(t-T)}$ and the option is just the same as an
initial amount $A$ of cash growing at the risk-free rate.

When $n=1$, $V=AS$ and the option is just the same as holding $A$ of
the underlying asset.

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