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\textbf{Question}

Consider the EUROPEAN AVERAGE RATE option with expiry $T$ whose value
is governed by the partial differential equation
$$V_t + (\log S) V_I + \frac{1}{2} \sigma^2 S^2 V_{SS} + rSV_S -rV=0$$
where the independent variable $I$ is defined by
$$I = \int_0^t \log S(\tau) \,d\tau .$$

\begin{description}
%Question 8a
\item{(a)}
By considering definition of $I$, and the quantity
$$Q = \left ( \sum_{i=1}^n S(t_i) \right )^{\frac{1}{n}}$$
in the limit $n\to\infty$ or otherwise, explain what kind of average
is being used in the average rate option.

%Question 8b
\item{(b)}
If the payoff of the option is a function of $I$ only, show that
solutions of the form
\begin{eqnarray*}
V &amp; = &amp; F(\theta, t)\\
\theta &amp; = &amp; \frac{I + (T-t) \log S}{T}
\end{eqnarray*}
exist provided $F$ satisfies
$$F_t + a(t) F_{\theta\theta} + b(t) F_{\theta} -rF = 0 \
\longrightarrow \ (1)$$
where $a(t)$ and $b(t)$ are functions that should be determined.

%Question 8c
\item{(c)}
Explain briefly why $(1)$ is easier to solve than the original
problem. If the payoff of the option depends on $S$ as well as $I$,
does this method still work?

\end{description}
\newpage
\textbf{Answer}

We have 
$$V_t +(\log S)V_I+ \frac{1}{2} \sigma^2 S62 V_{SS} +rSV_S -rV = 0$$
$$I=\int_0^t \log S(\tau) \,d\tau$$

\begin{description}
%Question 8a
\item{(a)}
Consider $\displaystyle  Q = \left ( \prod_{i=1}^n S(_i) \right
)^{\frac{1}{n}}$.

Now
$$Q = \frac{1}{n}\log \left ( \prod_{i=1}^n S(_i) \right )$$
But since $\log (a_1a_2 \cdots a_n) = \sum \log a_i$

$$\Rightarrow \log Q = \frac{1}{n} \sum_{i=1}^n \log (S(t_i)).$$
Now the $\underline{\rm{total}}$ time of the option is finite so say
\begin{eqnarray*}
ndt &amp; = &amp; T=t_n\\
\Rightarrow \log Q &amp; = &amp; \frac{1}{T} \sum_{i=1}^{n} \log S(ndt)dt
\end{eqnarray*}

In the limit as $n\to\infty$ we therefore find that
$$\log Q \to \frac{1}{T} \int_0^T \log S(\tau) \,d\tau$$
i.e. in the limit as $n\to\infty$
$$Q\to\exp \left ( \frac{1}{T} \int_0^T \log S(\tau) \,d\tau \right
).$$

Now Q is clearly the GEOMETRIC average of $S$. So the option is a
average strike with a GEOMETRIC AVERAGE.

%Question 8b
\item{(b)}
Now we have
$$V=F(\theta,t), \ \ \ \theta = \frac{I+(T-t)\log S}{T}$$

\begin{eqnarray*}
V_t &amp; = &amp; F_{\theta}\theta_t +F_tt_t = F_{\theta} \left ( -\frac{1}{T}
\log S \right ) +F_t\\
V_I &amp; = &amp; F_{\theta}\theta_I +F_tt_I = \frac{1}{T}F_{\theta} + 0\\
&amp; = &amp; \frac{1}{T} F_{\theta}\\
V_S &amp; = &amp; F_{\theta}\theta_S +F_tt_S = F_{\theta} \left (
\frac{T-t}{ST} \right ) + 0\\
&amp; = &amp; F_{\theta} \left ( \frac{T-t}{ST} \right )\\
V_SS &amp; = &amp; \left [ F_{\theta} \left ( \frac{T-t}{ST} \right ) \right ]_S =
F_{\theta\theta}\theta_S \left ( \frac{T-t}{ST} \right ) - F_{\theta}
\left ( \frac{T-t}{TS^2} \right )\\
&amp; = &amp; F_{\theta\theta} \left [ \frac{T-t}{ST} \right ]^2 - F_{\theta}
\left [ \frac{T-t}{TS^2} \right ] .
\end{eqnarray*}

Substitute in the equation:-
$$F_t -\frac{1}{2}(\log S)F_{\theta} +(\log S)\left ( \frac{1}{T}
F_{\theta} \right ) -rF +rSF_{\theta} \left ( \frac{T-t}{ST} \right
)$$
$$+ \frac{1}{2} \sigma^2 S^2 \left [ F_{\theta\theta} \left (
\frac{T-t}{ST} \right )^2 -F_{\theta} \left ( \frac{T-t}{TS^2} \right
) \right ] =0$$

Now rearranging gives
\begin{eqnarray*}
F_t+\frac{\sigma^2}{2}\frac{(T-t)^2}{T^2}F_{\theta\theta} +F_{\theta}
\left ( \frac{r(T-t)}{T} -\frac{1}{2}\sigma^2 \left ( \frac{T-t}{T}
\right ) \right )&amp; &amp;\\
 - rF &amp; = &amp; 0\\
F_t + \frac{\sigma^2}{2}\frac{(T-t)^2}{T^2} F_{\theta\theta} + \left (
\frac{T-t}{T} \right ) \left ( r - \frac{\sigma^2}{2} \right )
F_{\theta} -rF &amp; = &amp; 0\\
\rm{i.e.}\ \ F_t+a(t)F_{\theta\theta} +b(t)F_{\theta} -rF &amp; = &amp; 0
\end{eqnarray*} 
Where
$$a(t)=\frac{\sigma^2}{2}\frac{(T-t)^2}{T^2}, \ \ \ b(t)=\left (
\frac{T-t}{T} \right ) \left ( r- \frac{\sigma^2}{2} \right ).$$


%Question 8c
\item{(c)}
The equation just derived is a PDE, but it only has 2 independent
variable, as opposed to original equation which has 3 ($t$, $I$ and
$S$). The new problem is thus `one dimension easier' to solve.

If the payoff depends on $S$ as well as I, then this transformation
cannot work, for at $t=T$ (i.e. at expiry when the payoff takes place)
$$\theta = \frac{I+(T-t)\log S}{T}=\frac{I}{T}$$
Thus $V=f(I/T,t)$ and cannot therefore be made to depends upon $S$. 





\end{description}


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