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QUESTION

Consider the Black-Scholes equation for the continuous-time value
of an option $V(S,t)$,

$$\frac{\partial V}{\partial
t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial
S^2}+rS\frac{\partial V}{\partial S}-rV=0,$$

where $S$ is the price of the asset at time $t,\ \sigma$ is the
volatility and where the risk-free rate of return is $r$.

\begin{description}

\item[(a)]
Show that the equation, satisfied by the discounted option value
$U(S,t)$ relative to the maturity date of the option $T$, where,

$$V(S,t)=\exp(-r(T-t))U(S,t),$$

is given by

\begin{equation}
\frac{\partial U}{\partial
t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2U}{\partial
S^2}+rS\frac{\partial U}{\partial S}=0
\end{equation}

\item[(b)]
Hence show that under a change of variables to backwards time
$\tau=T-t$ and log-prices $\xi=\log S$,

\begin{equation}
\frac{\partial U}{\partial
\tau}=\frac{1}{2}\sigma^2\frac{\partial^2U}{\partial\xi^2}+
\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial U}{\partial\xi}.
\end{equation}

\item[(c)]
Hence show that a further change of variables, which you should
specify, can convert equation (2) into the diffusion equation,

\begin{equation}
\frac{\partial
W}{\partial\tau}=\frac{1}{2}\sigma^2\frac{\partial^2W}{\partial
x^2},\ W=W(x,\tau).
\end{equation}

\item[(d)]
Show by substitution that,

\begin{equation}
W(x,\tau)=\frac{1}{\sigma\sqrt{2\pi\tau}}\exp\left(-\frac{(x-x')^2}{2\sigma^2\tau}\right),
\end{equation}

(where $x'$ is an arbitrary constant) satisfies (3) with the
condition,

\begin{equation}
\int_{-\infty}^{+\infty}W(x,\tau)\,dx=1
\end{equation}

\item[(e)]
Hence write down the solution of the Black-Scholes equation (1)
which corresponds to (4).

\item[(f)]
Briefly indicate how this result can be used to derive a specific
solution, given boundary data at the maturity date.

\textbf{Hint:}$\int_{-\infty}^{+\infty}\exp(-\alpha
x^2)\,dz=\sqrt{\frac{\pi}{\alpha}}.$

\end{description}


ANSWER

\begin{description}

\item[(a)]
$$V(S,t)=e^{-t(T-t)}U(S,t)$$

$\frac{\partial V}{\partial
t}=re^{-r(T-t)}U(S,t)+e^{-r(T_t)}\frac{\partial U}{\partial t}\\
\frac{\partial V}{\partial S}=e^{-r(T-t)}\frac{\partial
U}{\partial S}\\ \frac{\partial^2V}{\partial
S^2}=e^{_r(T_t)}\frac{\partial^2U}{\partial S^2}$

Putting these into Black-Scholes

$$re^{-r(T-t)}U+e^{-r(T-t)}\frac{\partial U}{\partial
t}+\frac{1}{2}\sigma^2S^2e^{-r(T-t)}\frac{\partial^2U}{\partial
S^2}+rSe^{-r(T-t)}\frac{\partial U}{\partial S}-rS\frac{\partial
U}{\partial S}=0$$

\begin{equation}
\frac{\partial U}{\partial
t}+\frac{1}{2}\sigma^2S^2\frac{\partial^2U}{\partial
S^2}+rS\frac{\partial U}{\partial S}=0
\end{equation}

\item[(b)]

\begin{eqnarray*}
\tau=T-t\Rightarrow\frac{\partial U}{\partial
t}&\rightarrow&-\frac{\partial U}{\partial \tau}\\ \xi=\log
S\Rightarrow s=e^{\xi}\Rightarrow \frac{\partial U}{\partial
S}&=&\frac{\partial\xi}{\partial S}\frac{\partial
U}{\partial\xi}=\frac{1}{S}\frac{\partial U}{\partial
\xi}=e^{-\xi}\frac{\partial U}{\partial \xi}\\
\frac{\partial^2U}{\partial
S^2}&=&e^{-\xi}\frac{\partial}{\partial\xi}\left(e^{-\xi}\frac{\partial
U}{\partial \xi}\right)\\
&=&e^{-2\xi}\frac{\partial^2U}{\partial\xi^2}-e^{-2\xi}\frac{\partial
U}{\partial \xi}
\end{eqnarray*}

Substituting this in (1)

$$-\frac{\partial
U}{\partial\tau}+\frac{1}{2}\sigma^2e^{2\xi}\left(e^{-2\xi}\frac{\partial^2U}{\partial\xi^2}=e^{-2\xi}\frac{\partial
U}{\partial\xi}\right)+re^{\xi}e^{-\xi}\frac{\partial
U}{\partial\xi}=0$$

Therefore

\begin{equation}
\frac{\partial
U}{\partial\tau}=\frac{1}{2}\sigma^2\frac{\partial^2U}{\partial\xi^2}
+\left(r-\frac{1}{2}\sigma^2\right)\frac{\partial U}{\partial\xi}
\end{equation}

\item[(c)]
Set $x=\xi\left(r-\frac{\sigma^2}{2}\right)\tau,\
\overline{\tau}=\tau$

$$\frac{\partial}{\partial\xi}=\frac{\partial x}{\partial
\xi}\frac{\partial}{\partial
x}+\frac{\partial\overline{\tau}}{\partial\xi}{\partial}{\partial\overline{\tau}}
=\frac{\partial}{\partial x}+0=\frac{\partial}{\partial
x}\Rightarrow\frac{\partial^2}{\partial\xi^2}=\frac{\partial^2}{\partial
x^2}$$

$$\frac{\partial \tau}=\frac{\partial x}{\partial
\tau}\frac{\partial}{\partial
x}+\frac{\partial\overline{\tau}}{\partial\tau}\frac{\partial}{\partial\overline{\tau}}
=\left(r-\frac{\sigma^2}{2}\right)\frac{\partial}{\partial
x}+\frac{\partial}{\partial\overline{\tau}}$$

Set $U=W(x,\overline{\tau})$

and substitute into (2)

$$\left(\left(r-\frac{\sigma^2}{2}\right)\frac{\partial}{\partial
x}+\frac{\partial}{\partial\overline{\tau}}\right)W=
\frac{1}{2}\sigma^2\frac{\partial^2W}{\partial x^2}+
\left(r-\frac{\sigma^2}{2}\right)\frac{\partial W}{\partial x}$$

Therefore

$$\frac{\partial
W}{\partial\overline{\tau}}=\frac{1}{2}\sigma^2\frac{\partial^2w}{\partial
x^2}$$

or

$$\frac{\partial W}{\partial
\tau}=\frac{1}{2}\sigma^2\frac{\partial^2W}{\partial x^2}$$

removing the irrelevant ($\tau\rightarrow\overline{\tau}$)
transformation.

\item[(d)]

$$W(x,\tau)=\frac{1}{\sigma\sqrt{2\pi\tau}}e^{-\frac{(x-x')^2}{2\sigma^2\tau}}$$

\begin{eqnarray*}
\frac{\partial
W}{\partial\tau}&=&\frac{1}{\sigma\sqrt{2\pi\tau}}\times\left(
\frac{(x-x')^2}{2\sigma^2\tau^2}\right)
e^{-\frac{(x-x')^2}{2\sigma^2\tau}-\frac{1}{2\sigma}\frac{
e^{-\frac{(x-x')^2}{2\sigma^2\tau}}}{\sqrt{2\pi\tau}^3}}\\
\frac{\partial W}{\partial
x}&=&\frac{1}{\sigma\sqrt{2\pi\tau}}\times
\left(-\frac{2(x-x')}{2\sigma^2\tau}\right)
e^{-\frac{(x-x')^2}{2\sigma^2\tau}}\\ \frac{\partial^2W}{\partial
x^2}&=&\frac{1}{\sigma\sqrt{2\pi\tau}}\times\left[-\frac{1}{\sigma^2\tau}
+\frac{(x-x')^2}{\sigma^4\tau^2}\right]
e^{-\frac{(x-x')^2}{2\sigma^2\tau}}
\end{eqnarray*}

Therefore

\begin{eqnarray*}
\frac{1}{2}\sigma^2\frac{\partial^2W}{\partial
x^2}&=&\frac{\sigma}{2\sqrt{2\pi\tau}}\left[-\frac{1}{\sigma^2\tau}+
\frac{(x-x')^2}{\sigma^4\tau^2}\right]
e^{-\frac{(x-x')^2}{2\sigma^2\tau}}\\
&=&-\frac{1}{2}\frac{e^{-\frac{(x-x')^2}{2\sigma^2\tau}}}{\sqrt{2
\pi}\tau^{\frac{3}{2}}\sigma}+\frac{1}{\sigma\sqrt{2\pi\tau}}\frac{
(x-x')^2}{2\sigma^2\tau^2} e^{-\frac{(x-x')}{2\sigma^2\tau}}\\
&=&\frac{\partial W}{\partial t}
\end{eqnarray*}

(see above). Therefore

$$W(x,\tau)=\frac{1}{\sigma\sqrt{2\pi\tau}}e^{-\frac{(x-x')^2}{2\sigma^2\tau}}$$

is a solution of $\frac{1}{2}\sigma^2\frac{\partial^2W}{\partial k
x^2}=\frac{\partial W}{\partial\tau}$

But

\begin{eqnarray*}
\int_{-\infty}^{+\infty}W(x,\tau)\,dx&=&
\frac{1}{\sigma\sqrt{2\pi\tau}}\int_{-\infty}^{
+\infty}e^{-\frac{(x-x')^2}{2\sigma^2\tau}}\,dx\\
&=&\frac{1}{\sigma\sqrt{2\pi\tau}}\sqrt{\frac{\pi}{\frac{1}{2\sigma^2\tau}}}\\
&=&\frac{1}{\sigma\sqrt{2\pi\tau}}\sigma\sqrt{2\pi\tau}\\ &=&1
\end{eqnarray*}

Therefore this is the solution required.

\item[(e)]

Back substitute the variables:
$$x=\xi+\left(r-\frac{\sigma^2}{2}\right)\tau=\log
S+\left(r-\frac{\sigma^2}{2}\right)(T-t)$$ $$\tau=T-t,\
U\leftrightarrow W$$

therefore

$$V(s,t)=e^{-r(T-t)}\times\frac{1}{\sigma\sqrt{2\pi(T-t)}}e^{
-\frac{\left[\log\left(\frac{s}{s'}\right)
+\left(r-\frac{\sigma^2}{2}\right)(T-t)\right]}{2\sigma^2(T-t)}}$$

$\log s'=x'$ is required solution.

\item[(f)]

Equation (4) is the fundamental solution of the diffusion equation
(3), effectively a Green's-function (or derivative). To find
specific solution can write it as

$$W(x,\tau)=\frac{1}{\sigma\sqrt{2\pi\tau}}\int_{-\infty}^{+\infty}
e^{-\frac{(x-x')^2}{2\sigma^2\tau}}g(x')\,dx$$

where $g(x)$ is the boundary data for $\tau=0$ (i.e. at maturity).

This satisfies (3) by differentiation under integral sign and also
becomes a $delta$-function integration as $\tau\rightarrow0$
therefore satisfying boundary data).

Feeding in substitutes we arrive at

$$V(S,t)=\frac{e^{-r(T-t)}}{\sigma\sqrt{2\pi(T-t)}}
\int_0^\infty\exp\left\{-\frac{\left(\log\left(\frac{s}{s'}\right)
+\left(r-\frac{\sigma^2}{2}\right)(T-t)\right)^2}{2\sigma^2(T-t)}\right\}$$
$$\times\mathit{Payoff}(S')\frac{dS'}{S'}$$

where $\mathit{payoff}(S)$=payoff function at maturity $t=T$.

\end{description}




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