Exercise: Delta of a Digital Option vs. Vanilla

Prerequisites: Delta

Problem

A digital (binary) cash-or-nothing call pays $1 if

S_T > K

and $0 otherwise. Under Black-Scholes with

q = 0

, its price is

D(S, t) = e^{-r(T-t)}\Phi(d_2), \qquad d_2 = \frac{\ln(S/K) + (r - \tfrac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}.

Compute $\Delta_D = \partial D/\partial S$ in closed form.
Evaluate $\Delta_D$ at $S = 100, K = 100, T = 0.5, r = 0.05, \sigma = 0.2$ . Compare to $\Delta_{\text{call}}$ for a vanilla call with same parameters.
Limit as expiry approaches. Compute $\Delta_D$ at $t$ close to $T$ (e.g. $T - t = 0.001$ ) for $S = 100, K = 100$ . What happens? Why is digital hedging harder than vanilla hedging near expiry?
Practical mitigation. A market maker selling a digital typically over-hedges by replacing the digital with a tight call spread $(\text{Call}(K - \epsilon) - \text{Call}(K + \epsilon))/(2\epsilon)$ . Explain why this makes the position easier to hedge, even though it's no longer exactly a digital.

Hint

For part 1: $\partial\Phi(d_2)/\partial S = \phi(d_2)\cdot \partial d_2/\partial S = \phi(d_2)/(S\sigma\sqrt{T-t})$ . Be careful with the $e^{-r(T-t)}$ factor.

Jump to the solution when you're ready.