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\textbf{Question}

A cash-or-nothing call is an option which pays at expiry \$1 at expiry
$T$ if the spot price is above the strike $K$ and nothing if $S \le
K$. A cash-or-nothing put is an option which pays out nothing if $S>K$
and \$1 if $S \le K$. Let $C_b$ and $P_b$ denote the values of
cash-or-nothing calls and puts respectively.

Assuming both options have the same expiry date $T$, derive the
put-call parity relation 
$$C_b + P_b=e^{-r(T-t)}.$$
By considering relevant payoff diagrams, show that the payoff for
$C_b$ is equivalent to the delta of a vanilla European call option,
with the same strike, at expiry. By differentiating the Black-Scholes
equation with respect to $S$ show that the value of a cash-or-nothing
call option on an underlying which pays no dividend yield is equal to
the delta of a vanilla European call option written on an underlying
which pays a continuous dividend yield $q^*$ and different interest
rate $r^*$ and determine $q^*$ and $r^*$.

Hence show that $C_b$ is given by
$$C_b(S,t) = N(d_2).$$

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\textbf{Answer}

If we hold both $C_b$ and $P_b$ then we are guaranteed $ \$ 1 $ at
expiry. Hence $C_b+P_b=$present value of $\$1$,
i.e. $$C_b+P_b=e^{-r(T-t)}$$ 


Payoff for $C_b$ is
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Payoff for normal call is
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and $\Delta =\frac{\partial C}{\partial S}$ at expiry is
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i.e. at expiry, $C_b=\Delta$.

Now $C_b$ satisfies 
$$\displaystyle \frac{\partial C_b}{\partial t}
=\frac{1}{2} \sigma^2S^2\frac{\partial^2C_b}{\partial S^2}
+rS\frac{\partial C_b}{\partial S} -rC_b=0\longleftarrow(1)$$
and call satisfies, say, 
$$\displaystyle \frac{\partial V}{\partial t} +\frac{1}{2}\sigma^2
S^2\frac{\partial^2 V}{\partial S^2}+(r^*-q^*)S\frac{\partial
V}{\partial S} -r^*V =0\longleftarrow(2) $$
and at $t=T$, $C_b=\frac{\partial V}{\partial S}=\Delta$

$\begin{array}{rl}
\displaystyle \frac{\partial}{\partial S}(2) \Rightarrow &
\displaystyle \frac{\partial \Delta}{\partial t} +\frac{1}{2}\sigma^2
S^2 \frac{\partial^2 \Delta}{\partial S^2} +\sigma^2S\frac{\partial
\Delta}{\partial S}\\ 
& \displaystyle +(r^*-q^*)S\frac{\partial \Delta}{\partial S}
+(r^*-q^*)\Delta -r^*\Delta =0
\end{array}$

where $\Delta=\frac{\partial V}{\partial S}$, i.e.
$$ \displaystyle \frac{\partial \Delta}{\partial t}+
\frac{1}{2}\sigma^2S^2 \frac{\partial^2\Delta}{\partial S^2}
+(\sigma^2 +r^*-q^*)S\frac{\partial \Delta}{\partial S}- q^*\Delta
\longleftarrow (3)$$
Can make $(1) \equiv (3)$ by taking $ q^* = r $, $ \sigma^2+r^*-q^* = r$

So $$ q^*-r,\ \ \ r^*=r-\sigma^2+q^*=2r-\sigma^2.$$

Then then ensures that $C_b=\Delta$ since they are equal at expiry and
satisfy the same PDE.

$\Delta$ for the regular call is (formula sheet)
\begin{eqnarray*} \Delta = N(d_1^*) \ \ \ d_1^* & = & \displaystyle
\frac{\log(S/E)
+(r^*-q^*+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}\\
& = & \displaystyle \frac{\log(S/E) + (2r-\sigma^2 -r
+\frac{1}{2}\sigma^2)(T- t)}{\sigma \sqrt{T-t}}\\
& = & \displaystyle \frac{\log(S/E) +(r-\frac{1}{2}\sigma^2)(T
-t)}{\sigma \sqrt{T-t}}\\
& = & d_2 \ \ \ \rm{for\ }q=0 \ \rm{case}
\end{eqnarray*}

$\Rightarrow C_b=N(d_2)e^{-r(T-t)}$.


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