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\textbf{Question}

Briefly describe the main features of a EUROPEAN AVERAGE STRIKE
option, explaining in particular how average strike options differ
from European average rate and European vanilla options.

YOU MAY ASSUME that it the payoff of a path-dependent option is
dependent on the quantities $S(T)$ and
$$\int_0^T f(S(\tau), \tau) \,d\tau$$
then the independent variable
$$I = \int_0^t f(S(\tau), \tau) \,d\tau$$
satisfies the stochastic differential equation
$$dI = f(S,t)dt.$$

Use this fact and Ito's lemma (which you may use without proof) to
show that the value $V=V(S,I,t)$ of such an option satisfies the
partial differential equation
$$V_t + f(S,t)V_I + \frac{1}{2} \sigma^2 S^2 V_{SS} + rSV_S -rV = 0$$
where, as usual, $V$, $S$, $t$, $\sigma$ and $r$ denote respectively
the option value, the asset price, time, the volatility and the
interest rate.

Show that in the case of the continuously sampled arithmetic average
strike the governing differential equation may be reduced to one
involving only two independent variables by setting
$$V(S,R,t) = IW(R,t)$$
where
$$R = \frac{S}{\int_0^t S(\tau) \,d\tau}$$
and
$$R=\frac{S}{I}.$$

\newpage
\textbf{Answer}

For a European Vanilla with strike $E$, the payoff is max$(S-E,0)$
(CALL) or max$(E-S,0)$ (PUT), which depends only on $S$ AT EXPIRY. The
payoff for a Euro Av Strike is the same as this, except that $E$ is
replaced by the average of $S$ over the contract period. So for an
average strike thee os $\underline{\rm{no}}$ ``Exercise'' price. For
an AVERAGE RATE option the same is true, but now $S$ is replaced by
the average.

Now assume
\begin{eqnarray*}
V &amp; = &amp; V(I,S,t)\\
I &amp; = &amp; \int_0^t f(S(\tau),\tau) \,d\tau \\
dI &amp; = &amp; f(S,t)dt
\end{eqnarray*}

Consider the portfolio $\Pi= V-\Delta S$.

Then as usual $d\Pi = dV - \Delta dS$.
\begin{eqnarray*}
\rm{Also\ \ } dS &amp; = &amp; \sigma Sdt + \sigma DdX \ \ \ (\rm{as\
usual})\\
\rm{and\ \ } dI &amp; = &amp; f(S,t)dt
\end{eqnarray*}

Also in the usual way $dX\sim \sqrt{dt}$ as $dt\to 0$

$\Rightarrow$ in the limit $dS^2=\sigma^2 S^2 dt$

Taylor $\Rightarrow$
$$dV=V_SdS+V_tdt +V_IdI +\frac{1}{2}V_{SS}dS^2$$
\begin{eqnarray*}
d\Pi &amp; = &amp; V_SdS + V_tdt + V_IdI +\frac{1}{2}V_{SS} \sigma^2 S^2 dt
-\Delta (\sigma Sdt +\sigma SdX)\\
&amp; = &amp; V_S(\sigma Sdt +\sigma SdX) +V_tdt +V_If(S,t)dt\\
&amp; &amp; +\frac{1}{2} V_{SS} \sigma^2 S^2 st - \Delta \sigma S dt - \Delta
\sigma SdX
\end{eqnarray*}

As usual, the randomness vanishes if we choose $\Delta=V_S$
\begin{eqnarray*}
\Rightarrow d\Pi &amp; = &amp; V_S\sigma Sdt +V_tdt +V_I fdt
+\frac{1}{2}V_{SS} \sigma^2 S^2 dt -V_S \sigma S dt\\
&amp; = &amp; \left ( V_t +f(S,t)V_I +\frac{1}{2}V_{SS}\sigma^2 S^2 \right )
dt
\end{eqnarray*}

Now by the usual arbitrage argument
$$d\Pi = r\Pi dt = r(V-\Delta S)dt = r(V-SV_S)dt$$
$$\Rightarrow V_t +f(S,t)V_I +\frac{1}{2}\sigma^2 S^2 V_{SS} +rSV_S
-rV -0.$$

Now the average is $\displaystyle A(t)=\frac{1}{t} \int+0^t S(\tau)
\,d\tau$

and we are to try $V=IW$, $\ \ (R+S/I)$.
\begin{eqnarray*}
\Rightarrow V_t &amp; = &amp; IW_t\\
V_I &amp; = &amp; W -\frac{S}{I}W_R\\
V_S &amp; = &amp; \frac{I}{I}W_R\\
V_{SS} &amp; = &amp; \frac{1}{I}W_{RR}
\end{eqnarray*}

$$\Rightarrow IW_t+f(S,t)(W-SW_R/I) +\frac{1}{2} \frac{\sigma^2
S^2}{I} W_{RR} +rSW_R -rIW =0$$
but $R=S/I$
$$\Rightarrow W-t +\frac{f}{I}(W-RW_R) \frac{1}{2}\sigma^2 R^2 W_{RR}
+rRW_R -rW =0$$

Now $f(S,t)=IR$ since $I=\int_0^t S(\tau) \,d\tau$.

\begin{eqnarray*}
\Rightarrow W_t+\frac{1}{2}\sigma^2 R^2 W_{RR} -R^2 W_R +RW +rRW_R-rW
&amp; = &amp; 0\\
\Rightarrow W_t +\frac{1}{2}\sigma^2 R^2 W_{RR} +R(r-R)W_R -(r-R)W &amp; =
&amp; 0
\end{eqnarray*}  


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