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\textbf{Question}

A PATH-DEPENDENT European option with expiry $T$ is a European option
whose payoff is dependent on $S(T)$ and the quantity
$$\int_0^T f(S(\tau), \tau) \,d\tau$$
where $f$ is a given function of $S$ and $t$. By defined a new
independent variable
$$I=\int_0^t f(S(\tau), \tau) \,d\tau$$
show that the stochastic differential equation satisfied by $I$ is
$$dI = f(S,t)dt.$$

Use this result and an appropriate form of Ito's lemma to show that
the partial differential equation satisfied by such options is
$$V-t + f(S,t)V_I + \frac{1}{2} \sigma^2 S^2 V_{SS} + rSV_S - rV =
0.$$

Now consider the EUROPEAN AVERAGE STRIKE option where
$$f(S,t) = S(t).$$

Show that the partial differential equation is satisfied by solutions
of the form
$$V=SU(\eta, t) \ \ \ \ (\eta = I/S)$$
provided that $U$ satisfies a given partial differential equation
(which you should derive).

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\textbf{Answer}

For the PATH-DEPENDENT option we have a payoff dependent on 

$S$ AND $\int_0^T f(S(\tau),\tau) \,d\tau$.

Define the new indpt variable
$$I=\int_0^t f(S(\tau),\tau ) \,d\tau$$
Then
\begin{eqnarray*}
I(t+dt) & = & I + dI = \int_0^{t+dt} f(S(\tau), \tau) \,d\tau\\
& = & \int_0^t f(S(\tau), \tau) \,d\tau +\int_t^{t+dt} f(S(\tau),
\tau) \,d\tau\\ & = & I + f(S(t),t)dt\\
\Rightarrow dI & = & f(S,t)dt
\end{eqnarray*}

Now Ito's lemma will be exactly the same as normal, save for the
addition of an $f(S,t)V_I$ term (since $dI$ has no random component).

$$dV=\sigma SV_SdX+ \left ( \frac{1}{2}\sigma^2 S^ V_{SS} + rSV_S+ V_t
+f(S,t)V_I \right ) dt.$$
Now consider a portfolio $\Pi = V-\Delta S$ as usual.

We have
\begin{eqnarray*}
d\Pi & = & dV- \Delta dS\\
& = & -\Delta (\sigma SdX +rSdt) +dV\\
& = & (\sigma SV_S -\Delta \sigma S)dX\\
& &  + \left ( \frac{1}{2} \sigma^2
S^2 V_{SS} + rSV_S +V_t + f(S,t)V_I - rS\Delta \right ) dt
\end{eqnarray*}

As usual, eliminate randomness by choosing $\Delta =V_S$
$$\Rightarrow d\Pi = \left ( \frac{1}{2}\sigma^2 S^2 V_{S} + V_t+
f(S,t)V_I \right ) dt = r\Pi dt$$

by using the usual arbitrage argument that since $d\Pi$ is riskless,
it must grow at the risk free rate.
$$\Rightarrow \frac{1}{2}\sigma^2 S^2 V_{SS} +V_t +f(S,t)V_I =
r(V-SV_S)$$
$$\Rightarrow V_t +\frac{1}{2}\sigma^2 S^2 V_{SS} +f(S,t)V_I +rSV_S
-rV =0$$
- Black-Scholes for a path dependent option.

Now consider $V=SU(\eta,t) \ \ \ \eta=I/S$

We have
\begin{eqnarray*}
V_t & = & SU_t\\
V_I & = & SU_{\eta}/S = U_{\eta}\\
V_S & = & U + SU_{\eta} \left ( -\frac{I}{S^2} \right ) = U - \left (
\frac{I}{S} \right )U_{\eta}\\
V_{SS} & = & U_S + \left ( \frac{I}{S^2} \right ) U_{\eta} - \left (
\frac{I}{S} \right ) U_{\eta\eta} \left ( -\frac{I}{S^2} \right )\\
& = & U_{\eta} \left ( -\frac{I}{S^2} \right ) + \left ( \frac{I}{S^2}
\right ) U_{\eta} + \left ( \frac{I^2}{S^2} \right ) U_{\eta\eta}\\
& = & \left ( \frac{I^2}{S^3} \right ) U_{\eta\eta}
\end{eqnarray*}

So in the previous equation we get
\begin{eqnarray*}
SU_t +\frac{1}{2}\sigma^2 S^2 \frac{S^2I^2}{S^3}U_{\eta\eta}
+SU_{\eta} +rS \left ( U- \frac{I}{S}U_{\eta} \right ) -rSU & = & 0\\
U_t+\frac{1}{2}\sigma^2 \frac{I^2}{S^2}U_{\eta\eta} +U_{\eta} \left [
1-\frac{rI}{S} \right ] +rU-rU & = & 0\\
\Rightarrow U_t +\frac{1}{2} \sigma^2 \eta^2 U_{\eta\eta} +
[1-r\eta]U_{\eta} & = & 0
\end{eqnarray*}


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