# Greeks for binary option?

How to derive an analytic formula of greeks for binary option?

We know a vanilla option can be constructed by an asset-or-nothing call and a cash-or-nothing call, does that help us?

Wikipedia states

Since a binary call is a mathematical derivative of a vanilla call with respect to strike, the price of a binary call has the same shape as the delta of a vanilla call, and the delta of a binary call has the same shape as the gamma of a vanilla call.

Does that mean the delta of a binary call is also the gamma of a vanilla call? Can we use the analytical formula for gamma of vanilla call for binary option?

• Your question reminded me of this MIT OCW lecture partly regarding binaries that I found very insightful: ocw.mit.edu/courses/mathematics/… – bcf Dec 4 '15 at 15:06

For a digital option with payoff $1_{S_T > K}$, note that, for $\varepsilon > 0$ sufficiently small, \begin{align} 1_{S_T > K} &\approx \frac{(S_T-(K-\varepsilon))^+ - (S_T-K)^+}{-\varepsilon}.\tag{1} \end{align} That is, The value of the digital option \begin{align*} D(S_0, T, K, \sigma) &= -\frac{d C(S_0, T, K, \sigma)}{d K}, \end{align*} where $C(S_0, T, K, \sigma)$ is the call option price with payoff $(S_T-K)^+$. Here, we use $d$ rather than $\partial$ to emphasize the full derivative.

If we ignore the skew or smile, that is, the volatility $\sigma$ does not depend on the strike $K$, then \begin{align*} D(S_0, T, K, \sigma) &= -\frac{d C(S_0, T, K, \sigma)}{d K}\\ &= N(d_2)\\ &= N\big(d_1-\sigma \sqrt{T}\big). \tag{2} \end{align*} That is, the digital option price has the same shape as the corresponding call option delta $N(d_1)$. Similarly, the digital option delta $\frac{\partial N(d_1-\sigma \sqrt{T})}{\partial S_0}$ has the same shape as the call option gamma $\frac{\partial N(d_1)}{\partial S_0}$. Here, we note that they have the same shape, but they are not the same.

However, if we take the volatility skew into consideration, the above conclusion does not hold. Specifically, \begin{align*} D(S_0, T, K, \sigma) &= -\frac{d C(S_0, T, K, \sigma)}{d K}\\ &= -\frac{\partial C(S_0, T, K, \sigma)}{\partial K} - \frac{\partial C(S_0, T, K, \sigma)}{\partial \sigma} \frac{\partial \sigma}{\partial K}\\ &= N(d_2) - \frac{\partial C(S_0, T, K, \sigma)}{\partial \sigma} \frac{\partial \sigma}{\partial K},\tag{3} \end{align*} which may not have the same shape as $N(d_2)=N(d_1-\sigma \sqrt{T})$. In this case, we prefer to value the digital option using the call-spread approximation given by (1) above instead of the analytical formula (2) or (3).

Detla (European binary call) = Gamma (European vanilla Call), Indeed.

For the derivation, have a look at:

****Ps: Thus, the answer speaks from itself (cateris paribus)****

You can look at the relationship between vanilla and digital options in a model free setting: Let $C$ be the vanilla call, $D$ be the digital call, $E[]$ the expectation under the $T$-forward measure, and assume that the maturity $T$ discount factor is 1 to keep things simple. Then

\begin{eqnarray*} C&=& E[(S_T - K)^+] \\ D &=& E[1_{S_T > K}] \\ \frac{\partial C}{\partial S_0} &=& E[\frac{\partial S_T}{\partial S_0} 1_{S_T > K}] \\ \frac{\partial^2 C}{\partial S_0^2} &=& E[\frac{\partial^2 S_T}{\partial S_0^2} 1_{S_T > K}] + E[(\frac{\partial S_T}{\partial S_0})^2 \delta_{S_T = K}] \\ \frac{\partial D}{\partial S_0} &=& E[\frac{\partial S_T}{\partial S_0} \delta_{S_T = K}] \\ \end{eqnarray*}

Further, under the $T$-forward measure you have $S_T = S_0 M_T$ where $M_t$ is a martingale with value 1 at origin.

Now if you assume that $M_T$ does not depend on $S_0$, such as in the Black & Scholes model, or any homogeneous model, you get

\begin{eqnarray*} \frac{\partial C}{\partial S_0} &=& E[M_T 1_{S_T > K}] \\ \frac{\partial^2 C}{\partial S_0^2} &=& E[M_T^2 \delta_{S_T = K}] \\ \frac{\partial D}{\partial S_0} &=& E[M_T \delta_{S_T = K}] \\ \end{eqnarray*}

Since $M_T$ can be used as a Radon-Nikodym derivative, you see that $D$ and $\frac{\partial C}{\partial S_0}$ are expectations of the same quantity under different measures, hence they have the same shape but are not equal, and likewise for $\frac{\partial D}{\partial S_0}$ and $\frac{\partial^2 C}{\partial S_0^2}$.