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?


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
    Nov 21 '17 at 21:21
  • $\begingroup$ Indeed. Note that my earlier answer was slightly wrong, I have updated it. $\endgroup$ Nov 21 '17 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
    Nov 21 '17 at 22:09
  • $\begingroup$ In that case you would get: $\text{Binary}=-\frac{\partial \sigma}{\partial K}(K)\frac{\partial C}{\partial K}(\sigma(K),K)$. $\endgroup$ Nov 21 '17 at 22:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.