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\textbf{Question}

Explain the major difference between an ASIAN option and a LOOKBACK
option. Based on general financial arguments, would you expect a
Lookback option to more or less valuable than its European vanilla equivalent?

A STOP-LOSS option may be regarded as a perpetual barrier lookback
Put which pays a fixed proportion $\lambda$ of the maximum realized
asset price $J$ if ever the asset price falls below some barrier which
will be denoted by $\lambda J$, $(\lambda<1)$. Assume that a stop-loss
option is written on an underlying that pays a continuous dividend
yield $D_0$. By seeking a time-independent solution to the
Black-Scholes equations (with suitable boundary conditions) of the
form
$$V = JW(S/J)$$
value this option. What is the major financial reason for buying such
an option?

What value does the option have when $D_0=0$? Give a financial
interpretation of this result.

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\textbf{Answer}

For a lookback option the payoff depends not only on the asset price
at Expiry, but also on the min or max of the asset price over some
period prior to expiry. They are similar to Asians save for the fact
that an Asian depends on an AVERAGE not a max or a min.

Essentially a Lookback allows the holder to sell at the best price of
the asset. This is a huge advantage, so the lookback should be more
expensive than its Euro Vanilla equivalent.

Now consider a stop-loss option. Since it is a perpetual,
$\frac{\partial}{\partial t}=0$ and $V$ satisfies
$$\frac{1}{2} \sigma^2 S^2 V_{SS} +(r-D_0)SV_S -rV=0.$$
Now if ever $S$ reaches $\lambda J$, the payoff is $\lambda J$.
$$\Rightarrow V(\lambda J,J)=\lambda J$$
So now try $v=JW(\eta)$, $\ \ (\eta=S/J)$ and use dash for
$\frac{\partial}{\partial \eta}$.

\begin{eqnarray*}
\Rightarrow \frac{1}{2} \sigma^2 S^2 W''/J +(r-D_0)SW'-rJW & = & 0\\
\Rightarrow \frac{1}{2} \sigma^2 \eta^2 W'' +(r-D_0)\eta W'-rW & = & 0
\end{eqnarray*}
(also $V(\lambda J,J)=\lambda J)\to W(\lambda)=\lambda.)$

Euler's equation:- try $W\propto \eta^{\alpha}$
$$\Rightarrow \frac{1}{2} \sigma^2 \alpha(\alpha -1) +(r-D_0)\alpha -r
=0$$
$$\alpha^{\pm} = \frac{-(r-D_0)+\frac{\sigma^2}{2} \pm
\sqrt{\left ( r-D_0-\frac{\sigma^2}{2} \right )^2
+2r\sigma^2}}{\sigma^2}$$

We need a final B/C. Once a new max had been reached, we have $V_J=0$
when $S=J$ since `old' maxima are forgotten instantly.
\begin{eqnarray*}
\Rightarrow W=\frac{JS}{J^2}W' & = & 0\\
\Rightarrow W(1) & = & W'(1)
\end{eqnarray*}

Now $W=A\eta^{\alpha^+}+B\eta^{\alpha^-}$. 

Now impose the 2 B/C's:-
\begin{eqnarray*}
\lambda & = & A\lambda^{\alpha^+}+B\lambda^{\alpha^-}\\
A+B & = & \alpha^+A+\alpha^-B\\
\Rightarrow A & = &
\frac{\lambda(\alpha^--1)}{\lambda^{\alpha^+}(\alpha^--1)+
(1-\alpha^+) \lambda^{\alpha^-}}\\
B & = & \frac{\lambda (1-\alpha^+)}{\lambda^{\alpha^+}(\alpha^--1)+
\lambda^{\alpha^-(1 - \alpha^+)}}
\end{eqnarray*} 

$$\Rightarrow W = \lambda \left [ \frac{\eta^{\alpha^+}(1-\alpha^-)-
\eta^{\alpha^-} (1-\alpha^+)}{\lambda^{\alpha^+}(1-\alpha^-) -
\lambda^{\alpha^-} (1-\alpha^+)} \right ]$$

The major financial reason for buying this option is to safeguard
one's earlier success against a sudden sharp fall in the asset price.

When $D_0=0$
$$\alpha^{\pm}=\frac{\frac{\sigma^2}{2} -r \pm \sqrt{ \left ( r+
\frac{\sigma^2}{2} \right )^2}}{\sigma^2}$$
$\Rightarrow \alpha^+=1$, $\alpha^-=-2r/\sigma^2$
$$W = \lambda \left [ \frac{\eta (1+2r) -
\eta^{-\frac{2r}{\sigma^2}}(1-1)}{\lambda (1+2r) -
\eta^{-\frac{2r}{\sigma^2}}(1-1) } \right ] = \eta$$

Just the asset. Obvious since with no dividends you might just as well
buy the asset.

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