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QUESTION

A European call option is to be priced using the binomial model
assuming the following data

\begin{center}
\begin{tabular}{l}
Strike \$50;\\ Maturity 1 year, two intervals;\\ Continuously
compounded annual risk free interest at 3\%;\\ Volatility of
underlying stick 30\%;\\ Current price \$50
\end{tabular}
\end{center}

\begin{description}

\item[(a)]
Show that the up and down factors for the share price $U,\ D$
respectively, over a six month period are $U=1.227073,\
D=0.802814$. What is the continuously compounded interest rate for
each six month period? Calculate the asset prices, representing
this information on a binomial tree.

\item[(b)]
By constructing a replicating portfolio of shares and cash and
working to 5 decimal places, calculate the initial premium for the
option.

\item[(c)]
Discuss briefly the trading strategy for the write of the option
if the underlying share always rises in value.

\end{description}


ANSWER

$K=50\,T=1,\ \delta t=\frac{1}{2},\ r=3\%=0.03,\ \sigma=30\%=0.3,\
S_0=50$

\begin{description}

\item[(a)]
$U=e^{\left(1-\frac{\sigma^2}{2}\right)\delta t+\sigma\sqrt{\delta
t}}=1.227073\\ D=e^{\left(1-\frac{\sigma^2}{2}\right)\delta
t-\sigma\sqrt{\delta t}}=0.802814$

\setlength{\unitlength}{.5in}

\begin{picture}(10,10)

\put(0,4.9){$s_0=50$}

\put(7.5,9){$U^2S_0=75.28546$}

\put(7.5,5){$UDS_0=49.2556$}

\put(7.5,1){$D^2S_0=32.22553$}

\put(3,7){$US_0=62.35367$}

\put(3,3){$DS_0=40.14071$}

\put(1,5){\line(1,1){2}}

\put(1,5){\line(1,-1){2}}

\put(5.5,7){\line(1,1){2}}

\put(5.5,7){\line(1,-1){2}}

\put(5.5,3){\line(1,1){2}}

\put(5.5,3){\line(1,-1){2}}

\end{picture}

Continuously compounded interest rate$=e^{r\delta
t}=e^{0.03\times0.5}=1.015113$

\item[(b)]
Calculate final value of option $=\max\{s-k,o\}$

\setlength{\unitlength}{.5in}

\begin{picture}(10,10)

\put(0,4.9){$C_0$}

\put(5,9){$C_2^+=\max(75.28546-50,0)=25.28546$}

\put(5,5){$C_2^0=\max(49.2556-50,0)=0$}

\put(5,1){$C_2^-=\max(32.22553-50,0)=0$}

\put(2.5,7){$C_1^+$}

\put(2.5,3){$C_1^-$}

\put(.5,5){\line(1,1){2}}

\put(.5,5){\line(1,-1){2}}

\put(3,7){\line(1,1){2}}

\put(3,7){\line(1,-1){2}}

\put(3,3){\line(1,1){2}}

\put(3,3){\line(1,-1){2}}

\end{picture}

Calculate $C_1^+,\ C_1^-,\ C_0$ by stepping back from $C_2^+,\
C_2^0,\ C_2^-$

Replicating portfolio matches values of $C_m^n$

\underline{$t=1$ UP}

Construct a portfolio at $t=1$ in up state, $\psi_1^+$ shares,
$\psi_1^+$ cash.

$$C_1^+=61.35367\phi_1^++\psi_1^+$$

so at $t=2$, if up-state ($U^2S_0$)

$C_2^+=75.28546\phi_1^++1.015113\psi_1^+=25.28546$

and at $t=2$, if in down-state ($UDS_0$)

$C_2^-=49.25556\phi_1^++1.0151134\psi_1^+=0$

Therefore

$\phi_1^+=\frac{25.28546}{(75.28546-49.2556)}=0.971402\\
\psi_1^+=-\frac{49.2556}{1.015113}\times0.971402=-47.134642$

Therefore

$C_1^+=61.35367\times0.971402-47.134642=12.464436$

\underline{$t=1$ DOWN}

Portfolio is $\phi_1^-$ shares and $\psi_1^-$ cash.

$$C_1^-=40.14071\phi_1^-+\psi_1^-$$

So at $t=2$ if in up-state ($UDS_0$)

$C_2^0=49.2556\phi_1^-+1.015113\psi_1^-=0$

and at $t=2$ if in down-state ($D^2S_0$)

$C_2^-=32.22553\phi_1^-+1.015113\psi_1^-=0\\ \Rightarrow
\phi_1^-=\psi_1^-=0$ (No portfolio needed)

Therefore $C_1^-=0$.

\underline{$t=0$}

Portfolio is $\phi_0$ shares, $\psi_0$ cash.

$$C_0=50\phi_0+\psi_0$$

So at $t=1$, if up-state ($US_0$)

$C_1^-=40.14071\phi_0+1.015113\psi_0=0$

Therefore

$\phi_0=\frac{12.464436}{(61.35367-40.1407)}=0.587586\\
\psi_0=-\frac{40.14071\times0.587586}{1.015113}=-23.234969$

Therefore

$C_0=50\times0.587586-23.234969=6.144331$ (initial price)


\item[(c)]\

\begin{tabular}{crc}
$t=0$& buy 0.587586 shares @ 50, financed by \\&23.234969
borrowing\\ & and 6.144331 premium\\ &Net
cost=$(50\times0.587586)-23.234969-6.14431=0$\\ $t=1$& Buy
$(0.971402-0.587586)$ shares @ 61.35367\\ & financed by borrowing
$0.383816\times61.35367$\\& which takes borrowing to
$23.548520+1.015113\times$\\&$23.234969=47.134642$\\ $t=0$& Buy
$(1-0.971402)$ shares @75.28546cost=&2.153014\\ &sell 1 share @ 50
to owner of all recoup=&-50\\ &Payback borrowing
$1.015113\times47.134642$cost=&47.846988\\&net=&0
\end{tabular}

\end{description}



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