How Binary Option Delta Differs from Vanilla Call Delta
Summary
The document explains why a binary call and a standard European call with matching contract inputs do not have the same delta. A vanilla call has a payoff that rises continuously with the underlying above its strike, while a cash-or-nothing binary pays a fixed amount only if the terminal condition is met. As the binary approaches its strike or expiry, its value can become especially sensitive to changes in the underlying because its payoff changes sharply around the threshold.
It also derives a replication relationship: a binary call can be approximated by a narrow call spread formed from vanilla calls with nearby strikes. In the limit, the binary price corresponds to the vanilla call's strike derivative, so the binary delta is a mixed derivative with respect to spot and strike, rather than the vanilla call's spot derivative. This is a theoretical limiting argument; the document does not quantify deltas for specific market inputs or discuss practical spread costs.
Key ideas
- A cash-or-nothing binary call and a vanilla call have different payoff shapes and therefore different deltas.
- Binary option sensitivity can rise sharply near the strike and expiry.
- A narrow call spread can approximate a binary call payoff as the strike gap shrinks.
- The binary delta is related to a mixed spot-and-strike derivative of the vanilla call price.
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Full text
# Is the delta of a binary option the same as the delta for a regular European option?
# Is the delta of a binary option the same as the delta for a regular European option?
Assume both options have strike of 100, same time to expo, no dividend, same interest rate, same vol and lets say underlying is trading 95. Do both have the same deltas?
I read this and still don't get it: Delta of binary option
## Answer by AlRacoon (score 5, accepted)
https://quant.stackexchange.com/a/37704
No. The deltas are very different particularly when they are approaching the strike and expiry. You have one instrument that pays off linearly with the underlying and another that pays off either 0 or some fixed amount. The binary would therefore have much more sensitivity to the underlying price as it moves in the money as the payoff steps up instantaneously from 0 to the fixed amount, where as the other goes up linearly from 0 to the ending payoff.
## Answer by RRL (score 2)
https://quant.stackexchange.com/a/37710
A binary call option with strike $K$ that pays either $0$ or $1$ at expiry can be replicated approximately by a call spread. For some small $\epsilon > 0$ go long $\epsilon^{-1}$ calls with strike $K - \epsilon$ and go short $\epsilon^{-1}$ calls with strike $K$.
The payoff of this call spread will dominate and approach exactly the payoff of the binary option (theoretically) in the limit as $\epsilon \to 0$.
Hence,
$$C_{\text{binary}}(S,K) = \lim_{\epsilon \to 0}\frac{C(S, K - \epsilon) - C(S,K)}{\epsilon} = \frac{\partial C}{\partial K}(S,K),$$
and we see the distinction between the delta of the binary and vanilla call options:
$$\Delta_{\text{binary}} = \frac{\partial C_{\text{binary}}}{\partial S} = \frac{\partial^2 C}{\partial S \partial K} \neq \frac{\partial C}{ \partial S}= \Delta$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.