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\textbf{Question}

Assume that an asset $S$ has growth rate $\mu$, volatility $\sigma$
and pays a continuous dividend yield $q$ and that it evolves according
to the stochastic differential equation
$$\frac{dS}{S} = (\mu - q)dt + \sigma dX$$
where $dX$ is a Wiener process with the properties that
\begin{eqnarray*}
\textrm{\Large{$\varepsilon$}} (dX) & = & 0\\
\textrm{\Large{$\varepsilon$}} (dX^2) & = & dt\\
\lim_{dt\to 0} dX^2 = dt
\end{eqnarray*}

Give a heuristic derivation of It$\underline{\textrm{o}}$'s lemma for
a sufficiently differentiable function $V(S,t)$ which depends on both
$S$ and $t$.

Suppose that an option is written on this asset with the properties
that at expiry it is equal to the asset, and prior to its expiry it
pays out a known sum $K(S,t)dt$ during each time interval $(t,
t+dt)$. By constructing an instantaneously risk-free portfolio and
considering cash flows, show that it value $V$ must satisfy the
problem
$$\frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2
\frac{\partial^2 V}{\partial S^2} + (r-q)S \frac{\partial V}{\partial
S} -rV = -K(S,t)$$
$$t<T, \ \ \ \ V(S,T)=S.$$

Show that if $K(S,t)$ has the form $g(t)S$ where $g(t)$ is a known
function of time, then there are solutions of the form
$V=f(t)S$. Assuming that $V$ does have this form find $V(S,t)$. Hence
show that the delta for such an option is 

$$\Delta (S,t) = e^{-q(T-t)} + \int_t^T e^{-q(s-t)}g(s) \,ds.$$

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\textbf{Answer}

It$\hat{\rm{o}}$ asserts that if $f=f(S,t)$ then
\begin{eqnarray*}
df & = & \displaystyle \frac{\partial V}{\partial t}dt+ \frac{\partial
V}{\partial S}dS+ \frac{1}{2}\frac{\partial^2V}{\partial S^2}dS^2+
O(dt) \ \ (\rm{Taylor\ series!})\\ & = & \displaystyle \frac{\partial
v}{\partial t}dt+ \frac{\partial V}{\partial S}dS+
\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}dt + O(dt)
\end{eqnarray*}

Since $dS^2 =S^2((\mu-q)dt+ \sigma dX)^2= S^2dX^2+\cdots =S^2dt$

Set up portfolio $\Pi=V-\Delta S$ where $\Delta$ is previsible,
(i.e. $d(\Delta S)=\Delta dS$), i.e. $\Delta$ is fixed during time
step $dt$. Then
\begin{eqnarray*}
d\Pi & = & dV- \Delta dS+K(S,t)dt(\leftarrow\rm{cash\ flow\ from\
option})\\ & & -\Delta qSdt(\leftarrow\rm{cash\ flow\ from\
dividend})\\ & = & \displaystyle \frac{\partial V}{\partial t}dt+
\frac{\partial V}{\partial S}dS+
\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}- \Delta dS+
Kdt- \Delta qSdt
\end{eqnarray*} 

Make $\Pi$ risk free by putting $\Delta =\displaystyle \frac{\partial
V}{\partial D}$, so all $dX$ terms are eliminated;
\begin{eqnarray*} d\Pi & = & \displaystyle \left ( \frac{\partial
V}{\partial t}+ \frac{1}{2}\sigma^2S^2\frac{\partial^2v}{\partial S^2}
+K- \Delta qS \right ) dt\\ & = & r\Pi dt
\end{eqnarray*}
(interest earned on $\Pi$, since $\Pi$ is riskfree and must grow at
risk free rate.)

Thus
\begin{eqnarray*}
\displaystyle \frac{\partial V}{\partial t}+ \frac{1}{2}\sigma^2S^2
\frac{\partial^2 V}{\partial S^2}+ K- \Delta qS & = & r(V-\Delta S)\\
\Rightarrow \displaystyle \frac{\partial V}{\partial t}
+\frac{1}{2}\sigma^2 S^2\frac{\partial^2}{dS^2} +(r-q)S\frac{\partial
V}{\partial S} -rV & = & -K \ \ \ \rm{as\ }\Delta =\frac{\partial
V}{\partial S}
\end{eqnarray*}

If $K=g(t)S$ then we have
$$V_t+\frac{1}{2}\sigma^2S^2V_{SS}+ (r-q)SV_S-rV = -g(t)S$$
so if we try $V=f(t)S$ we get
$$\dot{f}(t)S +(r-q)f(t)S- rf(t)S =-g(t)S$$
which reduces to the ODE
$$\dot{f}(t)-qf(t)=-g(t)$$

So the form $V=f(t)S$ is consistent. From $V(S,T)=S$, we see that
$f(T)=1$. Thus we have to solve $$\dot{f}-qf=-g \ \ \ f(T)=1$$
i.e
\begin{eqnarray*} \frac{d}{dt}(e^{-qt}f) & = & -ge^{-qt}, \ \ \
f(T)=1\\ \Rightarrow \int_t^T \frac{d}{dS}(e^{-qs}f(s) \,ds & = &
e^{-qT}f(T) -e^{-qt}f(t) = \int_t^T e^{-qs}g(s) \,ds\\ \Rightarrow
f(t) & = & e^{-q(T-t)}+ \int_t^T e^{-q(s-t)}g(s) \,ds
\end{eqnarray*}

Obviously if $V=f(t)S$, $\Delta =\frac{\partial V}{\partial S} =f(t)$,
and
$$V= \left ( e^{-q(T-t)}+ \int_t^T e^{-q(s-t)}g(s) \,ds \right ) S.$$ 


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