{c}

In the <Black-Scholes model>, a <digital call option> and <digital put option> with remaining maturity $\tau$ have values
$$
D_{\rm call}(t,S)=e^{-r\tau}\Phi(d_-),
\qquad
D_{\rm put}(t,S)=e^{-r\tau}\Phi(-d_-),
$$
where $d_-$ is defined in the <Black-Scholes formula>. For $t<T$, the digital-call <delta hedge> is
$$
\partial_SD_{\rm call}(t,S)
=\frac{e^{-r\tau}\phi(d_-)}{S\sigma\sqrt\tau}.
$$

Black-Scholes digital option formula

Put
$$
d_\pm
=\frac{\log(S_0/K)+(r\pm\sigma^2/2)T}
{\sigma\sqrt T}.
$$
The call payoff is $(S_T-K)^+$, and it is positive when
$$
y\geq
\frac{\log(K/S_0)-(r-\sigma^2/2)T}{\sigma\sqrt T}
=-d_-.
$$
Writing $\phi$ for the <standard normal density>, the given <risk-neutral pricing> formula becomes
$$
\begin{aligned}
C_0
&=e^{-rT}\int_{-d_-}^{\infty}
\left[S_0e^{\sigma\sqrt T y+(r-\sigma^2/2)T}-K\right]
\phi(y)\,dy.
\end{aligned}
$$
Completing the square gives
$$
e^{\sigma\sqrt T y}\phi(y)
=e^{\sigma^2T/2}\phi(y-\sigma\sqrt T),
$$
so the two tail integrals are $\Phi(d_+)$ and $\Phi(d_-)$, where $\Phi$ is the <standard normal distribution function>. Therefore the <Black-Scholes formula> is
$$
\boxed{C_0=S_0\Phi(d_+)-Ke^{-rT}\Phi(d_-).}
$$

The payoff identity
$$
(S_T-K)^+-(K-S_T)^+=S_T-K
$$
replicates the call-minus-put position by one stock and borrowing the present value of $K$. Hence <put-call parity> gives
$$
C_0-P_0=S_0-Ke^{-rT},
$$
and therefore
$$
\boxed{
P_0=Ke^{-rT}\Phi(-d_-)-S_0\Phi(-d_+).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 29K / a / Solution

Under $\mathbb Q$,
$$
\log S_T\sim N\left(\log S_0+(r-\tfrac12\sigma^2)T,\sigma^2T\right).
$$
Define
$$
d_+=\frac{\log(S_0/K)+(r+\tfrac12\sigma^2)T}{\sigma\sqrt T},
\qquad
d_-=d_+-\sigma\sqrt T
=\frac{\log(S_0/K)+(r-\tfrac12\sigma^2)T}{\sigma\sqrt T}.
$$
The <standard normal cumulative distribution function> $\Phi$ then gives
$$
\mathbb Q(S_T>K)=\Phi(d_-),
\qquad
\mathbb E_{\mathbb Q}[S_T\mathbf1_{\{S_T>K\}}]=S_0e^{rT}\Phi(d_+).
$$
Substitution into the discounted expected payoff yields the <Black-Scholes formula>
$$
\boxed{V_0=S_0\Phi(d_+)-e^{-rT}K\Phi(d_-)}.
$$

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 29K / iv / Solution

{c}

For a non-dividend-paying stock, a European call with strike $K$ and remaining maturity $\tau$ has value
$$
C(t,S)=S\Phi(d_+)-Ke^{-r\tau}\Phi(d_-),
\qquad
d_\pm=\frac{\log(S/K)+(r\pm\sigma^2/2)\tau}{\sigma\sqrt\tau}.
$$


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