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

\textbf{Question}

In this question YOU MAY ASSUME that the value $V$ of a European
Vanilla Call is given by
$$V-SN(d_1)-Ee^{-r(T-t)}N(d_2)$$
where
\begin{eqnarray*}
d_1 & = & \frac{\log (S/E) + \left ( r + \frac{1}{2}\sigma^2 \right )
(T-t)}{\sigma \sqrt{T-t}},\\
d_2 & = & \frac{\log (S/E) + \left ( r - \frac{1}{2}\sigma^2 \right )
(T-t)}{\sigma \sqrt{T-t}},\\
N(p) & = & \frac{1}{\sqrt{2\pi}}\int_{-\infty}^p e^{-q^2/2} \,dq
\end{eqnarray*}
and $S$, $E$, $t$, $T$, $\sigma$ and $r$ denote respectively the asset
price, the strike price, the time, the expiry, the volatility and the
interest rate.

\begin{description}
%Question 5a
\item{(a)}
Show that
$$\frac{\partial}{\partial S}(N(d_1)) =
\frac{1}{\sqrt{2\pi(T-t)}}\frac{e^{-d_1^2/2}}{\sigma S}$$
and calculate
$$\frac{\partial}{\partial S} (N(d_2)).$$
Hence show that the DELTA, defined by $\Delta_c=\partial V/\partial S$
for a European Vanilla Call is given by
$$\Delta_c = N(d_1).$$
What does the delta of an option measure?

%Question 5b
Denote the value of a call by $C$ and the value of a put by $P$. By
considering a portfolio which is long one asset, long one put and
short on call, show that
$$C-P=S-Ee^{-r(T-t)}$$
and hence show that the delta $\Delta_p$ for a European Vanilla Put is
$$\Delta_p = N(d_1)-1.$$
Give a financial reason why $\Delta_p<\Delta_c$.
\end{description}

\newpage
\textbf{Answer}

\begin{description}
%Question 5a
\item{(a)}
As advised, we assume that for a Euro Vanilla Call
$$V=SN(d_1)- Ee^{-r(T-t)}N(d_2).$$
Now
$$\frac{\partial}{\partial S} (N(d_1)) = N_{d_1}d_{1S}.$$
But
$$N(d_1)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_1} e^{-q^2/2} \,dq $$

\begin{eqnarray*}
\Rightarrow N_{d_1} & = & \frac{1}{\sqrt{2\pi}}e^{-d_1^2/2}\\
\rm{also\ } d_1S & = & \frac{1}{\sigma S \sqrt{T-t}}.
\end{eqnarray*}

Thus $(N(d_1))_S = \frac{e^{-d_1^2/2}}{(\sigma S \sqrt{2\pi (T-t)})}$

Similarly
\begin{eqnarray*}
(N(d_2))_S & = & N_{d_2}d_2S\\
& = & \frac{1}{\sqrt{2\pi}}e^{-d_2^2/2}\times \frac{1}{\sigma S
\sqrt{t-t}}
\end{eqnarray*}

So now
\begin{eqnarray*}
\Delta_c & = & \frac{\partial V}{\partial S} =
N(d_1)+\frac{e^{-d_1^2/2}}{\sigma\sqrt{ 2\pi (T-t)}}- \frac{E
e^{-r(T-t)} e^{-d_2^2/2}}{S \sigma \sqrt{2\pi(T-t)}}\\
& = & N(d_1)+ \frac{1}{\sigma\sqrt{2\pi (T-t)}} \left [ e^{-d_1^2/2}
-e^{-\log(S/E) -r(T-t) -d_2^2/2} \right ]
\end{eqnarray*}

Now $\displaystyle d_1-d_2 = \frac{\sigma^2(T-t)}{\sigma\sqrt{T-t}} =
\sigma \sqrt{T-t}$

\begin{eqnarray*}
\Rightarrow d_1^2 & = & d_2^2 +\sigma^2 (T-t) +2d_2\sigma \sqrt{T-t}\\
\Rightarrow \Delta & = & N(d_1)+\frac{e^{-d_2^2/2}}{\sigma\sqrt{2\pi
(T-t)}} \left [ e^{-\frac{\sigma^2}{2}(T-t)-d_2\sigma\sqrt{T-t}}
\right.\\
& & \left. -e^{-\log(S/E) -r(T-t)} \right ]\\
& = & N(d_1)\\
& & + \frac{e^{-d_2^2/2}}{\sigma\sqrt{2\pi(T-t)}} \left [
e^{-\frac{\sigma^2}{2} (T-t) - \log(S/E) - r(T-t) +\frac{\sigma^2}{2}
(T-t)}\right.\\
& & \left. -e^{-\log(S/E) -r(T-t)} \right ]\\
& = & N(d_1)
\end{eqnarray*} 

The delta of an option measures the sensitivity of the option value to
changes in the underlying and of course is the key hedging quantity.


%Question 5b
\item{(b)}
Now consider a portfolio $\Pi=S+P-C$. The payoff at expiry is
$$\rho = S+\rm{max}(E-S,0) -\rm{max}(S-E,0)$$

There are two cases:-
\begin{description}
\item{(i)}
$S \ge E \Rightarrow \rho = S + 0 - (S-E) \Rightarrow \rho = E$
\item{(ii)}
$S /le E \Rightarrow \rho = S + E - S - 0 \Rightarrow \rho = E$
\end{description}

So arbitrage
$$\Rightarrow \Pi = Ee^{-r(T-t)}$$

$\Rightarrow$ Put/Call Parity $S+P-C=Ee^{-r(T-t)}$

i.e.
$$C-P=S-Ee^{-r(T-t)}$$

Thus $\Delta_c=\frac{\partial C}{\partial S}$, but $\frac{\partial
C}{\partial S} - \frac{\partial P}{\partial S} =1 -0$, (from above).

\begin{eqnarray*}
\Rightarrow \Delta_P & = & \Delta_C -1\\
\rm{i.e.\ } \Delta_P & = & N)d_1)-1
\end{eqnarray*}

The delta of a Put is less as a Call has an unlimited upside.
\end{description}


\end{document}