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

  • 2
    $\begingroup$ No, the delta of a vanilla is always finite $-1<=\Delta<=1$, the delta of a binary can go to infinity $\endgroup$
    – nbbo2
    Jan 12, 2018 at 16:55
  • $\begingroup$ thanks i just i figured it out. delta of binary option looks like gamma of normal euro option. p/l graph of binary option looks like delta of normal euro option. $\endgroup$
    – confused
    Jan 12, 2018 at 17:22

2 Answers 2


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.


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 $\epsilon \to 0$.


$$C_{\text{binary}}(S,K) = \lim_{\epsilon \to 0}\frac{C(S, K - \epsilon) - C(S,K)}{\epsilon} = \frac{\partial C}{\partial K}(S,K),$$

and we see the distinction between the delta of the binary and vanilla call options:

$$\Delta_{\text{binary}} = \frac{\partial C_{\text{binary}}}{\partial S} = \frac{\partial^2 C}{\partial S \partial K} \neq \frac{\partial C}{ \partial S}= \Delta$$


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