Digitals and barriers: why the hedging is the hard part

Exotic options are usually easy to price and hard to hedge, and the interview is almost always about the second half. A digital is a limit of a call spread, which prices it immediately and also shows exactly why nobody wants to be short one into expiry.

Definition 1.1

The digital

Cash-or-nothing digital, $\text{payoff} = \mathbf{1}\{S_T > K\}$ — Pays one unit if the underlying finishes above the strike and nothing otherwise. Under Black–Scholes it is worth $e^{-rT}\Phi(d_2)$ — the discounted risk-neutral probability of exercise, which is the cleanest possible illustration of what $\Phi(d_2)$ means.

Derivation 1.2

A digital is the limit of a call spread

Buy a call at $K - \varepsilon$ , sell one at $K + \varepsilon$ , and scale.

$\frac{C(K-\varepsilon) - C(K+\varepsilon)}{2\varepsilon}$
The spread pays 1 above $K + \varepsilon$ , 0 below $K - \varepsilon$ , and ramps between.
$\to -\frac{\partial C}{\partial K} \quad \text{as } \varepsilon \to 0$
The digital is minus the derivative of the call price with respect to strike.
$-\frac{\partial C}{\partial K} = e^{-rT}\Phi(d_2)$

D = e^{-rT}\Phi(d_2)

Figure 1.3 · A finite call spread approximates the digital step. Half of the 99–101 call spread pays zero below 99, ramps between 99 and 101, and pays one above 101. Shrinking the strike gap makes the ramp steeper and the digital hedge more sensitive near 100.

Why the hedge misbehaves. The payoff jumps, so near expiry the delta is enormous over a tiny price range and zero everywhere else — the replicating portfolio has to go from nothing to a full position as the underlying crosses the strike. In practice nobody tries. A desk sells the digital and hedges with a call spread of finite width, deliberately over-hedging so that the residual risk is a known cost rather than an unbounded one. The width of that spread *is* the price of the risk, and quoting it is the actual skill.

Example 1.4

You sell a digital paying $\$1$ above a strike of $100$ . You hedge with a call spread between $99$ and $101$ . How many spreads do you need, and what is the residual risk?

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Worked solution

Formula $\text{spread payoff at } S \ge 101 = 2 \Rightarrow \text{notional} = \frac{1}{2}$
Substitute $\text{buy } \tfrac12 \text{ of the } 99\text{--}101 \text{ call spread}$
Solve $S \le 99:\ \text{both pay } 0$
$S \ge 101:\ \text{hedge pays } 1, \text{ digital pays } 1$
$99 < S < 101:\ \text{hedge pays } \tfrac{S-99}{2}, \text{ digital pays } 0 \text{ or } 1$
Answer $\text{half a spread; risk confined to } (99, 101)$

Sanity check. Widening the spread over-hedges and costs more; narrowing it leaves a bigger jump. The choice is a price for a known, bounded risk, which is exactly how a desk wants risk to look.

Digitals and barriers: why the hedging is the hard part

After this lesson you should be able to

The digital

A digital is the limit of a call spread

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