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\textbf{Question}

In this question you must assume zero dividend yields, $q=0$.

Show that $V=\alpha S$, where $\alpha$ does not depend on $S$ or $t$,
is a solution of the Black-Scholes equation.

Let $C(S,t;T_1,K)$ denote the Black-Scholes value of a European
vanilla call option with strike $K$ and expiry $T_1$ at time $t$ and
spot price $S$. Show that the value of an ``at the money'' call is
proportional to the spot (or strike).

A (European) forward-start call is a call option whose strike is not
known at the start of the contract but, rather, is agreed to be the
spot price $S_0$ a given time $T_0$ prior to expiry $T_1$. The option
can not be exercised prior to expiry $T_1$.

Show that the value $v$ of a European forward-start call option is
given by
$$V(S,t) = \left \{ \begin{array}{ll} \alpha S &amp; \textrm{if } t \le
T_0,\\
C(S,t;T_1,S_0) \ \ &amp; \textrm{if } T_0 \le t \le T_1
\end{array} \right.$$
where $S_0$ is the spot price at time $T_0$ and $\alpha$ is a function,
which you should determine, which depends only on $T_1-T_0$, $\sigma$
and $r$.

Briefly justify the $\Delta$-hedging strategy implied by this price
for $t \le T_0$.

\newpage
\textbf{Answer}

BS equation is $V_t+\frac{1}{2}\sigma^2S^2V_{SS}+ rSV_S- rV =0$

If $V=\alpha S$, $V_t=0$, $V_{SS}=0$, $V_S=1$, so we get
$$V_t+\frac{1}{2}\sigma^2S^2V_{SS}+ rSV_S -RV =rS-rS=0$$
$\Rightarrow \alpha S$ is a solution.

Value of call is $$C=SN(d_1)-Ke^{-r(T-t)}N(d_2)$$
where
\begin{eqnarray*} d_1 &amp; = &amp; \displaystyle \frac{\log(S/K)+ (r+
\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}\\
d_2 &amp; = &amp; d_1-\sigma\sqrt{T-t}
\end{eqnarray*}

At the money means, $K=S$ hence $\log(K/S)=0$ and 
$$d_1=d_1(T,t),\ \ \ d_2=d_2(T,t)$$

$\left. \begin{array}{rl} C = &amp; K(N(d_1(T,t))-e^{-r(T-t)} N(d_2
(T,t))\\
\propto &amp; K
\end{array} \right \}$ or replace $K$ by $S$

Suppose $S_0$ is the spot price at $T_0$. For $t&gt;T_0$ we know the
strike, it is $K+S_0$ and hence
$$V(S,t)=C(S,t;T_1,S_0)$$
At time $t=T_0$ we have
$$V(S_0,T_0)=S_0\alpha(T_1,T_0) =\alpha S_0$$
Now solve the BS equation back from $t=T_0$, i.e. solve
$$V_t+\frac{1}{2}\sigma^2S^2V_{SS}+ rSV_S -rV =0,$$
$$V(S,T_0)=\alpha S$$
$\Rightarrow V(S,t)=\alpha S$ for $t&lt;T_0$

$\alpha$ is given by substituting $K=S_0$, $t=T_0$ into usual formula
for a call, i.e.
$$\displaystyle \alpha = S_0 \left [ N \left (
\frac{(r+\frac{1}{2}\sigma^2)( T_1-T_0)}{\sigma\sqrt{T_1-T_0}} \right
) \right.$$ $$\left. -e^{-r(T_1-T_0)}N \left (
\frac {(r-\frac{1}{2}\sigma^2)(T_1-T_0)}{\sigma\sqrt{T_1-T_0} } \right )
\right ]$$ 
The $\Delta$ hedging strategy is to hold $\alpha$ assets up to $T_0$,
where $\alpha$ is the number of assets you need to hedge the
at-the-money call into which the option turns at $T_0$. i.e. You know
that at $T_0$ you will need $\alpha$ assets, so you hold them from
time $t=0$.

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