Given $r=0$, $\sigma(K)=\text{const}$ and:

$$ \text{Binary} = \lim_{ε → 0} \frac{(C(K,\sigma (K))-C(K+ε,\sigma(K+ε)))}{ε} $$

I have to find the analytical expression for the above.

Since $σ(K)=\text{const}$, I know that we can write the above as:

$$ \text{Binary} = \lim_{ε → 0}\frac{(C(K)-C(K+ε))}{ε} $$

Do I take the derivative next or use the Taylor's theorem?


1 Answer 1


As you say, you simply differentiate with respect to $K$. Assuming your binary's maturity is $T$, note that in a Black-Scholes framework with constant risk-free rate $r$, by the Breeden-Litzenberger equations:

$$ \begin{align} \text{Binary}&=\lim_{\epsilon \rightarrow 0}\frac{-C(K+\epsilon)+C(K)}{\epsilon} \\[6pt] &=-\frac{\partial C}{\partial K}(K) \\[9pt] &=e^{-rT}(1-Q(K)) \end{align}$$

where $Q(\cdot)$ is the cumulative, risk-neutral distribution and $(1-Q(K))$ gives the probability that the underlying asset's price is below $K$ at time $T$, which is consistent with the payoff of a binary option.

  • $\begingroup$ thanks for the explanation. This is the same as saying that Binary = e^-rT * N(d2) right? $\endgroup$
    – miababy
    Commented Nov 21, 2017 at 21:21
  • $\begingroup$ Indeed. Note that my earlier answer was slightly wrong, I have updated it. $\endgroup$ Commented Nov 21, 2017 at 21:26
  • $\begingroup$ Hi! thanks. Just another question, if now σ(K) is not constant, how do I find the analytical solution using chain rule? $\endgroup$
    – miababy
    Commented Nov 21, 2017 at 22:09
  • $\begingroup$ In that case you would get: $\text{Binary}=-\frac{\partial \sigma}{\partial K}(K)\frac{\partial C}{\partial K}(\sigma(K),K)$. $\endgroup$ Commented Nov 21, 2017 at 22:32

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