Solution: Delta of a Digital Option vs. Vanilla

Exercise: Delta of a Digital Option vs. Vanilla

Part 1

\Delta_D = \frac{\partial}{\partial S}\left[e^{-r(T-t)}\Phi(d_2)\right] = e^{-r(T-t)}\phi(d_2)\cdot \frac{\partial d_2}{\partial S} = \frac{e^{-r(T-t)}\phi(d_2)}{S\sigma\sqrt{T-t}}.

Part 2

$S = 100, K = 100, T = 0.5, r = 0.05, \sigma = 0.2$ :

d_2 = \frac{0 + (0.05 - 0.02)\cdot 0.5}{0.2\sqrt{0.5}} = \frac{0.015}{0.1414} \approx 0.106.

$\phi(0.106) \approx 0.3969$ . Then:

\Delta_D = \frac{e^{-0.025}\cdot 0.3969}{100\cdot 0.2\cdot \sqrt{0.5}} \approx \frac{0.9753\cdot 0.3969}{14.14} \approx 0.0274.

Vanilla call delta (computed before): $\approx 0.598$ .

Digital delta is about 21× smaller — the digital pays out at most $1, while a vanilla call pays

(S_T - K)_+

with unbounded upside. So per dollar of underlying move, the digital's value changes far less than the vanilla's.

Part 3

At $t \to T$ , $T - t \to 0$ . The $1/\sqrt{T-t}$ factor blows up, and $\phi(d_2)$ at $S = K$ (ATM) stays around $1/\sqrt{2\pi} \approx 0.399$ :

\Delta_D(S = K, T - t = 0.001) = \frac{e^{-0.0001}\cdot 0.399}{100\cdot 0.2\cdot \sqrt{0.001}} \approx \frac{0.399}{0.632} \approx 0.63.

And for $T - t = 0.0001$ : $\Delta_D \approx 2.0$ . As $t \to T$ with $S = K$ : $\Delta_D \to \infty$ .

Why digital hedging is hard near expiry. At expiry, the digital's value is a step function:

\mathbf{1}_{S_T > K}

. Its derivative is a delta function at

K

— unbounded. Dynamic hedging requires buying/selling arbitrarily large amounts of the underlying as

S

crosses

K

near expiry. In practice, market makers cannot execute infinite-size trades, and the underlying's bid-ask spread makes this prohibitively expensive — so digital hedging breaks down near expiry.

Part 4

A tight call spread $(\text{Call}(K - \epsilon) - \text{Call}(K + \epsilon))/(2\epsilon)$ approximates a digital payoff: it's $\approx 1$ for $S \gg K + \epsilon$ , $\approx 0$ for $S \ll K - \epsilon$ , and linear between $K - \epsilon$ and $K + \epsilon$ . The key difference:

The spread's delta is bounded, even near expiry, because it's a difference of two vanilla options (each with bounded delta).
At the cost of an ambiguity of at most $\epsilon$ in the strike location, the MM gets a hedgeable product.

Trade-off:

\epsilon

smaller

\Rightarrow

closer to the true digital but harder to hedge near expiry;

\epsilon

larger

\Rightarrow

easier to hedge but the payoff profile differs from the digital. Typically

\epsilon

is set to around

0.5-2\%

K

This is a classic over-hedging: the market maker sells the client a product slightly "more generous" than the true digital, pricing in an extra premium to compensate for the hedging cost and model risk.

Takeaways

Digital delta is $e^{-r(T-t)}\phi(d_2)/(S\sigma\sqrt{T-t})$ — a bell-shaped function of $S$ centred around $K$ , unlike the monotone delta of a vanilla call.
Digital delta blows up near expiry. Unlike vanilla options (delta bounded in $[0, 1]$ ), digital options have unbounded delta as $t \to T$ near the strike. This is the mathematical manifestation of the pinning-risk problem.
Call-spread replication is the practitioner's standard trick to make an exotic product hedgeable. It trades payoff exactness for hedge robustness.
Exotic hedging often reduces to vanilla composition. Every first-generation exotic can be approximated by a combination of vanilla options, each of which has a bounded, well-behaved delta.