\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}
\parindent=0pt

\textbf{Question}

In this question you may assume a standard Black-Scholes world in
which there are no dividend yields, and hence $q=0$.

Suppose that an option, $W$, is written which is worthless on expiry,
but which pays out a continuous cash-flow of $K(S,t)dt$ during the
time interval $(t, t+dt)$, prior to expiry. Using the standard
Black-Scholes analysis, including the cash-flow $Kdt$, show that $W$
satisfies the problem
$$\frac{\partial W}{\partial t} + \frac{1}{2} \sigma^2 S^2
\frac{\partial^2 W}{\partial S^2} + rS \frac{\partial W}{\partial S} -
rW = - K(S,t),$$
$$t<T, \ \ \ \ W(S,T)=0.$$

Use financial arguments to show that if $K>0$ for $t<T$ the $W>0$ for
$t<T$ and that if $K<0$ for $t<T$ then $W<0$ for $t<T$.

What is meant by ``implied volatility'' of an option $V$? What is the
vega, $\varphi$,of an option, and how does it relate to the ``implied
volatility''?

Assuming that the payoff for the option is independent of volatility,
show that the vega satisfies the problem
$$\frac{\partial \varphi}{\partial t} + \frac{1}{2} \sigma^2 S^2
\frac{\partial^2 \varphi}{\partial \S^2} + rS \frac{\partial
\varphi}{\partial S}- r\varphi = -\sigma S^2 \frac{\partial^2
V}{\partial S^2}$$
$$\varphi(S,T)=0$$
Hence deduce that ``implied volatility'' is well defined if the gamma of
the option does not change sign.

\newpage
\textbf{Answer}

Since W expires worthless, $W(S,T)=0$.

Set up standard portfolio $\Pi=W+ \Delta S$ where $\Delta$ is
previsible, i.e. constant over interval $dt$, and assume usual
geometric Brownian motion.
$$\frac{dS}{S}=\mu dt+ \sigma dX \ \ \ \rm{where} \ \ dX^2=dt$$

Then It$\hat{\rm{o}}$ implies 
\begin{eqnarray*} dW & = & (W-t+ \frac{1}{2} \sigma^2 S^2
W_{SS}) dt+ W_S dS\\
\rm{and} \ \ \ d\Pi & = & dW- \Delta dS+ Kdt,
\end{eqnarray*}
since holding the option gives $Kdt$ in cash.

Thus $d\Pi=(W_t+\frac{1}{2} \sigma^2 S^2 W_{SS})dt+ (W_S- \Delta)dS+
Kdt$.

Only risk comes in through $dS$ term, so taking $\Delta=W_S$
eliminates it.As this makes $\Pi$ riskless, its return must equal risk
free rate so
$$d\Pi=\left ( W_t+\frac{1}{2} \sigma^2 S^2 W_{SS}+ K \right ) dt =
r\Pi dt = r(W- SW_S)dt$$
$$\Rightarrow W_t+ \frac{1}{2} \sigma^2 S^2 W_{SS}+ rSW_S -rW = - K$$

If $K>0$ then holder of the option receives a guaranteed positive cash
flow for $t<T$ $\Rightarrow \ W>0$ for $t<T$ (i.e. you'd pay a
positive amount to get this positive cash flow)

If $K<0$ then holder of the option has to pay out $-Kdt>0$ every
interval $dt$ prior to expiry $\Rightarrow \ W<0$ for $t<T$ (i.e. you
would pay a positive amount to escape having to pay out $K$)

Implied vol is the volatility you would have to put into the
Black-Scholes model to obtain an observed option price; given the
option price and all other parameters ($S,\ r,\ K,\ t,\ T$ in this
case)

Vega $=\varphi=\frac{\partial V}{\partial \sigma}$, i.e. the
sensitivity of the option price to small changes in vol,
$\sigma$. ($dV=\frac{\partial V}{\partial \sigma}d\sigma$, all other
parameters being fixed)

Implied vol is only uniquely defined if $\frac{\partial V}{\partial
\sigma}$ does not change sign; i.e. can find $\sigma$ as a function of
$V$ only if graph of $V$ against $\sigma$ is either

\begin{center}
$\begin{array}{c}
\epsfig{file=448-1998-6.eps, width=40mm}\\
(\varphi=\frac{\partial V}{\partial \sigma}>0)
\end{array}
\ \ \ \rm{or} \ \ \ 
\begin{array}{c}
\epsfig{file=448-1998-7.eps, width=40mm}\\
(\varphi=\frac{\partial V}{\partial \sigma}<0)
\end{array} $
\end{center}

but if $\varphi=\frac{\partial V}{\partial \sigma}$ changes sign we
must have something like
\begin{center}
$ \begin{array}{c}
\epsfig{file=448-1998-8.eps, width=50mm}
\end{array}
\ \ \ 
\begin{array}{c}
\rm{and\ } \sigma_1,\ \sigma_2 \rm{\ both\ give}\\
\rm{same\ value\ of\ } V
\end{array} $
\end{center}

Take $\displaystyle \frac{\partial}{\partial \sigma}$ of
Black-Scholes;
\begin{eqnarray*} \frac{\partial}{\partial \sigma} \left ( V_t +
\frac{1}{2} \sigma^2 S^2 V_{SS} +rSV_S - rV \right )  & = & 0\\
\Rightarrow \varphi_t+ \frac{1}{2} \sigma^2 S^2 \varphi_{SS}
+rS\varphi_S -r\varphi & = & -\sigma S^2 V_{SS}
\end{eqnarray*}
and
$$\frac{\partial}{\partial \sigma}(V(S,T)=\rm{something\ indep\ of\
}\sigma)\Rightarrow \varphi(S,T)=0.$$

From argument about $W$, we see that if $-\sigma S^2V_{SS}>0$, we must
have $\varphi=\frac{\partial V}{\partial \sigma}>0$ for $t<T$, so
implied vol is well defined.

Likewise, if $-\sigma S^2V_{SS}<0$, we have $\varphi=\frac{\partial
V}{\partial \sigma}<0$ for $ t<T$, so implied vol is well defined.

Now note that $S^2>0$, $\sigma>0$ by definitions and the gamma,
$\Gamma$, of an option is defined by $\Gamma=V_{SS}$.

Hence the result.


\end{document}