Approximating an Up-and-Out Digital Call with Nearby Calls

Summary

The response gives a starting point for hedging a digital call that expires in the money only if the underlying has never crossed an upper barrier. It approximates the terminal digital payoff by taking the difference between two vanilla call payoffs with strikes just below and just above the digital strike, divided by the strike spacing. Applying the same barrier condition to those calls gives a pair of up-and-out calls whose payoff difference approximates the up-and-out digital.

This is a finite-difference approximation, so the hedge depends on choosing a small strike interval and is not an exact replication at finite spacing. The answer assumes the reader already knows how to construct static hedges for standard knock-out calls; it does not explain that component, discuss trading frictions, or quantify approximation error. It offers a hedge construction idea rather than a complete implementation or empirical assessment.

Key ideas

• A digital payoff can be approximated by the payoff difference of calls struck just below and above its strike.
• Applying the same barrier condition to both calls yields a starting hedge for an up-and-out digital.
• The approximation uses a finite strike interval and is not exact for a nonzero interval.
• The construction assumes a separate method for statically hedging ordinary knock-out calls.

Tags

# Static hedge for up-and-out Digital Call


# Static hedge for up-and-out Digital Call












I am trying to come up with a static hedge for a Digital Call with strike K that knocks out when price > barrier H. I know it will involve non-knockout digital calls with strike K and strike H but I am not sure in what proportion and what other digitals I will need to include in the hedge.

Appreciate any advice and suggestions. Thank you in advance

## Answer by Antoine Conze (score 1)

https://quant.stackexchange.com/a/45011

\begin{equation*} \begin{split} \mathbb{1}_{S_T > K, \max_{[0,T]} S_t < H} &\approx \frac{(S_T - (K-\varepsilon))^+ - (S_T - (K+\varepsilon))^+}{2 \varepsilon} \mathbb{1}_{\max_{[0,T]} S_t < H} \\ &= \frac{(S_T - (K-\varepsilon))^+\mathbb{1}_{\max_{[0,T]} S_t < H} - (S_T - (K+\varepsilon))^+\mathbb{1}_{\max_{[0,T]} S_t < H} }{2 \varepsilon} \end{split} \end{equation*} so you can start with the static hedge of standard knock out calls $(S_T - (K-\varepsilon))^+\mathbb{1}_{\max_{[0,T]} S_t < H}$ and $(S_T - (K+\varepsilon))^+\mathbb{1}_{\max_{[0,T]} S_t < H}$ if you're already familiar with that.

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