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

For $0\leq t<T$, set $\tau=T-t$ and
$$
d_-(t,S_t)
=\frac{\log(S_t/K)+(r-\sigma^2/2)\tau}
{\sigma\sqrt\tau}.
$$
Under the <equivalent martingale measure>, conditional log-normality gives
$$
\mathbb Q(S_T\geq K\mid\mathcal F_t)
=\Phi(d_-(t,S_t)).
$$
The <risk-neutral pricing> value of the <digital call option> is consequently
$$
\boxed{
D_{\rm call}(t,S_t)
=e^{-r(T-t)}\Phi(d_-(t,S_t)).}
$$
At $t=T$ this converges to the stated payoff away from $S_T=K$, with the payoff convention specifying the boundary value.

The <delta hedge> holds the derivative of the claim value with respect to the current stock price. Since
$$
\frac{\partial d_-}{\partial S}=\frac1{S\sigma\sqrt\tau},
$$
the number of risky-asset units for $t<T$ is
$$
\boxed{
\Delta_t
=\frac{e^{-r(T-t)}\phi(d_-(t,S_t))}
{S_t\sigma\sqrt{T-t}}.}
$$
This is the <Black-Scholes digital option formula>. The hedge becomes singular close to maturity near the strike, reflecting the discontinuity of the payoff.

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

For a <European call option>, the pointwise inequality $(S_T-K)^+\geq S_T-K$ and <risk-neutral pricing> give
$$
V_0\geq e^{-rT}(\mathbb E_{\mathbb Q}S_T-K)
=e^{-rT}(S_0e^{rT}-K)
=\boxed{S_0-e^{-rT}K}.
$$

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

= Risk-neutral valuation
{synonym}

Under an <equivalent martingale measure> $Q$, the no-arbitrage value of a replicable payoff is its discounted conditional expectation under $Q$.

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 Incoming links: Risk-neutral pricing