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QUESTION

\begin{description}

\item[(a)]
An investor wishes to trade in options on an asset whose current
price one year from the maturity date of an option is \$25, the
exercise price of the option is \$20, the risk-free interest rate
is 5\% per annum and the asset volatility is 20\% per annum.
Calculate by what amount the asset price has to change for the
purchaser of a European call option to break even giving your
answer to 4 decimal places?

\item[(b)]
Write down the call-put parity formula for European options. Hence
repeat part (a) but for a European put.

\item[(c)]
Sketch the qualitative behaviour of the European call and put
values over the lifetime of the option as a function of the
underlying asset price.

\item[(d)]
Calculate the initial price of the call option in part (a) if the
asset pays a continuous dividend of $DS$ where $S$ is the asset
price and $D=0.01$.

\end{description}

You may assume that the solution of the Black-Scholes equation for
a European call option, paying no dividends, is given by,

$$c(S,t)=SN(d_1)-K\exp(-r(T-t))N(d_2),$$

$$d_1=\frac{\log\left(\frac{S}{K}\right)+
\left(r+\frac{\sigma^2}{2}\right)(T-t)}{\sigma\sqrt{T-t}},$$

$$d_2=\frac{\log\left(\frac{S}{K}\right)+
\left(r-\frac{\sigma^2}{2}\right)(T-t)}{\sigma\sqrt{T-t}}.$$


ANSWER

\begin{description}

\item[(a)]
$T=1,\ S_0=25,\ K=20,\ r=0.05,\ \sigma=0.2$

For the holder of a Eurocall, the asset price must rise by the
following to break even:

\setlength{\unitlength}{.5in}
\begin{picture}(6,4)
\put(1,0){\vector(0,1){4}}

\put(1,2){\vector(1,0){5}}

\put(1,1.5){\line(1,0){3}}

\put(4,1.5){\line(1,1){2}}

\put(.5,1.6){$C_0$}

\put(3.9,2.1){$K$}

\put(4.4,1.6){$K+C_0$}

\put(6,2){$S$}

\put(0,3.6){$C(S,t)$}

\end{picture}

Payoff at $t=T$.

Therefore the price must rise to $K+C$ for the holder to break
even. If the initial asset price is $S_0$, requires final asset
price is $K+C_0$ so the rise must be $K+C_0-S_0$ Therefore we need
to know the initial premium at $S_0$.

Use the formula given at $t=0$.

\begin{eqnarray*}
C(S_0,0)&amp;=&amp;S_0N(d_1(0))-Ke^{-rT}N(d_2(0))\\
d_1(0)&amp;=&amp;\frac{\left(\log\left(\frac{S_0}{K}\right)+
\left(r+\frac{\sigma^2}{2}T\right)\right)}{\sigma\sqrt{T}}\\
d_2(0)&amp;=&amp;\frac{\left(\log\left(\frac{S_0}{K}\right)+
\left(r-\frac{\sigma^2}{2}T\right)\right)}{\sigma\sqrt{T}}
\end{eqnarray*}

Feed in the above data to get

$\left.\begin{array}{c}d_1=1.47\\d_2=1.27\end{array}\right\}$ to 2
d.p.

We need to find $N(1.47)$ and $N(1.27)$. From the tables,
$N(1.47)=0.9297,\ N(1.27)=0.8980$

$$C(S,0)=25\times0.9292-20e^{-0.05}\times0.8980=6.1459$$

Therefore to break even they need a new price of
$K+C-0=20+6.1459=26.1459$

Therefore the current price needs to rise by $K+C_0-S=1.1459$.

\item[(b)]
The call-put parity formula is $$C(S,t)-P(S&lt;t)=S-Ke^{-r(T-t)}$$

Therefore

\begin{eqnarray*}
P(S_0,0)&amp;=&amp;C(s_0,0)-S_0+Ke^{-rT}\\ &amp;=&amp;1.1459-25+20e^{-0.05}\\
&amp;=&amp;0.1705
\end{eqnarray*}

\setlength{\unitlength}{.5in}
\begin{picture}(6,4)
\put(1,0){\line(0,1){4}} \put(1,2){\line(1,0){5}}

\put(1,3){\line(1,-1){1.5}} \put(2.5,1.5){\line(1,0){3.5}}

\put(.5,3.5){$K$} \put(0,3){$K-P_0$} \put(.3,1.5){$-P_0$}

\put(1.5,2.2){$K-P_0$} \put(2.4,1.7){$K$} \put(6,1.7){$P_0$}

\end{picture}

So need price to fall to $K-P_0$ to break even
=$20-0.1705=19.8295$

i.e. needs to fall by $25-19.8295=5.1705$.

\item[(c)]

\begin{center}
\epsfig{file=3088-1999-1.eps, width=100mm}
\newline

\epsfig{file=3088-1999-2.eps, width=100mm}
\end{center}

\item[(d)]
Result of part (a) changes by converting $r\rightarrow r-D$ (as
per example in lecture notes) in all equations. Thus the effective
interest rate is $0.05-0.01=0.04$

$$d_1(0)=\frac{\left(\log\left(\frac{S_0}{K}\right)+\left(r-D+\frac{\sigma^2}{2}\right)T\right)}{\sigma\sqrt{T}}$$

etc.

With

\begin{eqnarray*}
C(S,t)&amp;=&amp;Se^{-D(T-t)}N(d_1)-Ke^{-r(T-t)}N(d_2)\\ d_1(0)&amp;=&amp;1.42\\
d_2(0)&amp;=&amp;1.22\\ C(s_0,0)&amp;=&amp;25e^{-0.01}N(1.42)-2oe^{-0.05}N(1.22)\\
N(1.42)&amp;=&amp;0.9222\\ N(1.22)&amp;=&amp;0.8888\\
C(S_0,0)&amp;=&amp;22.8256-16.9091=5.9165
\end{eqnarray*}

\end{description}



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