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.

Solution

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

{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
$$
\partial_SD_{\rm call}(t,S)
=\frac{e^{-r\tau}\phi(d_-)}{S\sigma\sqrt\tau}.
$$

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