Deriving Binary Option Prices from Black–Scholes Vega and Volatility Skew
Summary
This note explains how to connect a Black–Scholes call’s volatility sensitivity to the strike derivative used to price a cash-or-nothing digital option. It starts from the call price and its vega, then shows that the apparent mismatch between a term involving spot and the desired term involving strike disappears through a standard normal-density identity.
Using the definitions of d1 and d2, the derivation establishes that spot times the d1 density equals discounted strike times the d2 density. It then differentiates the Black–Scholes call with respect to strike; the density terms cancel, leaving the familiar discounted digital probability. With volatility varying by strike, the chain rule adds a correction involving vega and the slope of implied volatility. The argument is an analytic Black–Scholes derivation; it assumes the model’s standard inputs and does not discuss practical smile interpolation, market conventions, or broader model risks.
Key ideas
- A cash-or-nothing digital call can be related to the negative strike derivative of a vanilla call price.
- The Black–Scholes d1 and d2 normal densities are connected by spot and discounted strike.
- That density identity makes the extra terms cancel when differentiating the vanilla call price with respect to strike.
- A strike-dependent implied volatility adds a chain-rule correction proportional to the volatility skew.
Tags
Full text
# Pricing binary options under volatility smile
# Pricing binary options under volatility smile
I was asked to show that the price of a digital/binary option $D$ while a volatility smile $\sigma(K)$ is present is given by
> $$D= \exp(-rT)( \Phi(d_2) - K \sqrt{T} \phi(d_2) \sigma ' (K))$$
Where $\Phi$ is the standard normal CDF and $\phi$ is the standard normal pdf. I started with showing that the price of $D$ under B-S is given by $-\frac{\partial C}{\partial K}$, where $C$ is the standard B-S pricing for a call option. Additionally, I have also shown that this can also be expressed as (with no vol smile) $$-\frac{\partial C}{\partial K} = \Phi(d_2)$$
Finally, I was able to show the following $$D = -\frac{\partial C}{\partial K} - \frac{\partial C}{\partial \sigma} \frac{\partial \sigma}{\partial K}$$
Now, the $\frac{\partial C}{\partial \sigma}$ term is Vega, which is explicitly given by $$\frac{\partial C}{\partial \sigma} = S\phi(d_1)\sqrt{T}$$
And now I have absolutely no idea what to do, because the question involves a $K\phi(d_2)$ term, but I have an $S\phi(d1)$ term. Are they related somehow?
## Answer by RRL (score 3, accepted)
https://quant.stackexchange.com/a/69794
Note that $d_1 = d_2 +\sigma \sqrt{T}$, and
$$\tag{1}S\sqrt{T}\phi(d_1) = \frac{S\sqrt{T}}{\sqrt{2\pi}}e^{-(d_2 + \sigma \sqrt{T})^2/2} = \frac{S\sqrt{T}}{\sqrt{2\pi}}e^{-d_2^2/2}e^{-d_2\sigma\sqrt{T}}e^{-\sigma^2T/2}\\ = S\sqrt{T}\phi(d_2)e^{-d_2\sigma\sqrt{T}}e^{-\sigma^2T/2}$$
Since
$$d_2 = \frac{\log \frac{S}{K}+rT -\frac{1}{2}\sigma^2T}{\sigma \sqrt{T}},$$
we have
$$\tag{2}e^{-d_2\sigma\sqrt{T}}= \frac{K}{S}e^{-rT}e^{\sigma^2T/2}$$
Substituting into (1) with (2) we get
$$S\sqrt{T}\phi(d_1) = e^{-rT}K\sqrt{T} \phi(d_2)$$
Note also that for the Black-Scholes price $C = S \Phi(d_1)-Ke^{-rT}\Phi(d_2)$ we have $$\tag{3}\frac{\partial C}{\partial K}= - e^{-rT}\Phi(d_2)$$
The derivation is
$$\tag{4}\frac{\partial C}{\partial K}= S\Phi'(d_1)\frac{\partial d_1}{\partial K} - e^{-rT}\Phi(d_2) - Ke^{-rT}\Phi'(d_2)\frac{\partial d_2}{\partial K}\\ = -e^{-rT}\Phi(d_2) + S\phi(d_1)\frac{\partial d_1}{\partial K} - Ke^{-rT}\phi(d_2)\frac{\partial d_2}{\partial K}$$
Since $\frac{\partial d_1}{\partial K} = \frac{\partial d_2}{\partial K}$ and using the previous result $S\phi(d_1) = e^{-rT}K \phi(d_2)$, the second and third terms on the RHS of (4) cancel, and we get the desired result (3).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.