<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION

 A European put option with a strike price \$50 matures
in one year. Divide the one year period into two six-month
intervals. The continuously compounded risk-free rate of return is
5\% and the volatility is 30\% per annum.

\begin{description}

\item[(a)]
Calculate the up and down factors of the binomial tree model for
the asset price.

\item[(b)]
Find the martingale probability, $q$, of an upstate occurring.

\item[(c)]
If the current stock price is \$40, determine the binomial value
of the option.

\item[(d)]
Hence provide the trading strategy.

\end{description}
Hence verify that the values of
the call and put satisfy the put-call parity formula.


ANSWER

\underline{Europut}: same prices U, D, $r,\ \sigma,\ e^{r\delta
t}$

\setlength{\unitlength}{.5in}

\begin{picture}(10,10)

\put(.5,5){40}

\put(1,5){\line(3,2){3}}

\put(1,5){\line(3,-2){3}}

\put(4,3){32.4352}

\put(5,3){\line(1,1){2}}

\put(5,3){\line(1,-1){2}}

\put(4,7){49.5764}

\put(5,7){\line(1,1){2}}

\put(5,7){\line(1,-1){2}}

\put(7,1){26.30105}

\put(7,5){40.20052}

\put(7,9){61.44549}

\end{picture}

Replicating portfolio at $t=\frac{1}{2}$

\underline{Up state}

\begin{picture}(10,4)

\put(0,2){$v_1^+=49.5764\phi_1^++\psi_1^+$}

\put(1.5,1.2){shares}

\put(2,1.5){\vector(0,1){.4}}

\put(2.5,1.2){cash}

\put(2.8,1.5){\vector(0,1){.4}}

\put(3.5,2){\vector(1,1){1}}

\put(4,2.3){up at $t=1$}

\put(4.6,3){$v_2^+=61.44549\phi_1^++1.025324\psi_1^+$}

\put(3.5,2){\vector(1,-1){1}}

\put(4,1.5){down at $t=1$}

\put(4.6,1){$v_2^0=40.20051\phi_1^++1.02532\psi_1^+$}

\end{picture}

$\left.\begin{array}{c}v_2^+=P_2^+=0\\v_2^0+P_2^0=9.79949\end{array}\right\}$
Therefore solve

\begin{eqnarray*}
0&amp;=&amp;61.44549\phi_1^++1.02532\psi_1^+\\
9.79949&amp;=&amp;40.20051\phi_1^++1.02532\psi_1^+
\end{eqnarray*}

$\Rightarrow\left\{\begin{array}{c}\phi_1^+=-0.46126\\\psi_1^+=27.64253\end{array}\right.$

\underline{Down state}

\begin{picture}(10,4)

\put(0,2){$v_1^-=32.4352\phi_1^-+\psi_1^-$}

\put(3.5,2){\vector(1,1){1}}

\put(4,2.3){up at $t=1$}

\put(4.6,3){$v_2^0=40.20051\phi_1^-+1.02532\psi_1^-$}

\put(3.5,2){\vector(1,-1){1}}

\put(4,1.5){down at $t=1$}

\put(4.6,1){$v_2^-=26.30105\phi_1^-+1.02532\psi_1^-$}

\end{picture}

$\left\{\begin{array}{c}v_2^0=P_2^0=9.79949\\v_2=P_2^-=23.69895\end{array}\right.$
Therefore solve

\begin{eqnarray*}
9.79949&amp;=&amp;40.20051\phi_1^-+1.02532\psi_1^-\\
23.69895&amp;=&amp;26.30105\phi_1^-+1.02532\psi_1^-
\end{eqnarray*}

$\Rightarrow\left\{\begin{array}{c}\phi_1^-=-1.0000\\\psi_1^-=48.76526\end{array}\right.$

So at $t=\frac{1}{2}$

$P_1^+=v_1^+=49.5764\times(-0.46126)+27.64253=4.77492$


$P_1^-=v_1^-=32.4352\times9-1.0000)+48.76526=16.33006$

At $t=0$

\begin{picture}(10,4)

\put(0,2){$v_0=40\phi_0+\psi_0$}

\put(2.5,2){\vector(1,1){1}}

\put(3,2.3){up at $t=\frac{1}{2}$}

\put(3.6,3){$v_1^+=49.5764\phi_0+1.02532\psi_0$}

\put(2.5,2){\vector(1,-1){1}}

\put(3,1.5){down at $t=\frac{1}{2}$}

\put(3.6,1){$v_2^-=32.4352\phi_0+1.02532\psi_0$}

\end{picture}

$v_1^+=P_1^+=4.77492$

$v_1^-=P_1^-=16.33006$

Therefore solve

\begin{eqnarray*}
4.77492&amp;=&amp;49.5764\phi_0+1.02532\psi_0\\
16.33006&amp;=&amp;32.4352\phi_0+1.02532\psi_0
\end{eqnarray*}

$\Rightarrow\left\{\begin{array}{c}\phi_0=-0.67411\\\psi_0=37.25190\end{array}\right.$

Therefore $v_0=40\times-0.67411+37.25190=10.28750=P_0$

\underline{Summary}

\setlength{\unitlength}{.5in}

\begin{picture}(10,10)

\put(0,5.5){$P_0=10.28750$} \put(0,5){$\phi_0=-0.67411$}
\put(0,4.5){$\psi_0=37.25190$}

\put(2,5){\line(3,2){3}}

\put(2,5){\line(3,-2){3}}

\put(5,3.5){$P_1^-=16.33006$} \put(5,3){$\phi_1^-=-1.0000$}
\put(5,2.5){$\psi_1^-=48.76526$}

\put(7,4){\line(1,1){1}}

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

\put(5,7.5){$P_1^+=4.77492$} \put(5,7){$\phi_1^+=-0.46126$}
\put(5,6.5){$\psi_1^+=27.64253$}

\put(7,8){\line(1,1){1}}

\put(7,6){\line(1,-1){1}}

\put(8,1){$P_2^-=23.69895$}

\put(8,5){$P_2^0=9.79949$}

\put(8,9.5){$P_2^+=0$}

\end{picture}

\begin{tabular}{crcr}
$t=0$&amp;sell short (don't own but trade on account) 0.67411\\&amp;
shares and receive $0.67411\times 40$&amp;=&amp;\$26.96440\\ &amp;invest an
amount&amp;=&amp;$-\$37.25190$\\&amp;Need an extra &amp;&amp;$-\$10.28750$\\ &amp;get from
owner of put option$\Rightarrow$\\&amp; self financing portfolio.\\
$t=\frac{1}{2}$&amp; If up then short 0.46126 shares, i.e. buy\\&amp;
(0.67411-0.46126) shares at 49.5764 at a cost of\\&amp;
$0.21285\times49.5764$&amp;=&amp;$\$10.55234$\\ &amp;Financed by investment
which has risen to\\&amp; $37.25190\times1.02552=\$38.19512$. Pay
$\$10.55234$ for\\&amp; shares to leave
$=38.19512-10.55234$&amp;=&amp;$\$27.64278$ \\&amp;If down then short 1.000
shares i.e. sell on\\&amp;account $(1-0.67411)=0.32589$ and receive
\\&amp;$0.32589\times32.4352$&amp;=&amp;\$10.57031\\&amp;Thus total investment is
now \\&amp; $37.25190\times1.02532+10.57031$&amp;=&amp;\$48.76543
\\ $t=1$&amp;3 possible outcomes, but to save the trees, just\\&amp; consider
the down state, $P_2^-=23.69895$\\&amp; Issuer has a liability of 1
short share,\\&amp; but has saved
$1.02532\times48.76526$&amp;=&amp;\$50.00\\&amp;Must accept 1 share from
owner\\&amp; of put (who exercises) at strike price of\\&amp; \$50. Cost
to issuer $=1\times\$50$&amp;=&amp;$-\$50.00$\\&amp;break even
&amp;$\rightarrow$&amp;0.00
\end{tabular}

What happens to the 1 share the issuer receives? He has to
complete his shorting contract at the end of the account period,
so uses it for that: liability 1 share, owns 1 share, net
result$\rightarrow$ break even. (NB receives no cash for
completing this contract as he's already received money for it at
$t=0,\ t=\frac{1}{2}$).

Call/put parity formula:

$$C(t)-P(t)=S_t-ke^{-r{T-t}}$$

Use $C,\ S,\ k,\ r$ to find $P=C-S+ke^{-r(T-t)}$

\underline{T=1} (Just do up states)

\begin{tabular}{cccc}
&amp;$t=0$&amp;$t=\frac{1}{2}$&amp;$t=1$\\ $C$&amp;2.72610&amp;5.58605&amp;11.44549\\
$S$&amp;40&amp;49.5764&amp;61.44549\\ $k$&amp;50&amp;50&amp;50\\ $T-t$&amp;1&amp;$\frac{1}{2}$&amp;0\\
$C-S+ke^{-r(T-t)}$&amp;$10.28757=P_0$&amp;$4.77515=P_1^+$&amp;$0=P_2^+$
\end{tabular}

The last row is the put prices (up to rounding errors.)



\end{document}
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