Skip to content
All library documents

Why Digital Option Prices Reflect Implied Volatility Skew

Article Quant Q&A · Author: bcf

Summary

The document compares a digital call priced by Monte Carlo under a general stock model with a Black–Scholes digital priced using the implied volatility backed out from a vanilla call at the same strike and expiry. The central explanation is that matching one vanilla call price does not generally make the digital prices match: the vanilla-implied volatility can vary with strike, and the digital payoff is sensitive to that skew.

A digital call can be approximated by the strike derivative of call prices, using a narrow call spread. Applying the chain rule separates the Black–Scholes digital component from an adjustment involving Black–Scholes vega and the strike slope of implied volatility. In a flat-volatility Black–Scholes setting, that slope vanishes, so the prices agree in principle; Monte Carlo error and bias can still cause small differences. The relationship is a local approximation argument and assumes comparable payoff conventions and pricing inputs.

Key ideas

  • A vanilla call's implied volatility matches its price but does not generally determine the price of a different payoff.
  • A digital call can be represented as the limiting price sensitivity of a call spread around its strike.
  • The strike derivative of the call price includes an adjustment related to implied volatility skew and Black–Scholes vega.
  • With flat volatility under Black–Scholes dynamics, the skew adjustment vanishes and the two digital prices agree in theory.
  • Monte Carlo estimation can introduce sampling error and bias even when the models otherwise coincide.

Tags

Full text
# Implied Vol in Different Payoffs


# Implied Vol in Different Payoffs












Let's say I have a black box stock price model I run Monte Carlo on to estimate European call prices. For a given strike $K$ and expiration $T$, I then back out the Black-Scholes implied volatility $\sigma(K, T)$ from the Monte Carlo price $C_{MC}$, and this assumes the model is lognormal.

I now want to price a digital option using this black box model at the same $K$ and $T$ for which I computed the implied vol. I use Monte Carlo for this and obtain a price $D_{MC}$

Let's say I also compute the price of this digital using the closed-form Black-Scholes price and use $\sigma(K,T)$ as my vol. I obtain a price of $D_{BS}$.

Now, as I understand, $\sigma(K, T)$ is exactly that volatility I had to plug into my Black-Scholes price to obtain $C_{MC}$. I am now pricing another type of option (digital) for this $K$ and $T$ and use $\sigma(K, T)$ in my closed-form. My question is, what is the relation between $D_{MC}$ and $D_{BS}$? Would one expect these to be equal? Would they be equal only if the black box model was actually just the lognormal model?

## Answer by Quantuple (score 4, accepted)

https://quant.stackexchange.com/a/28304

They would only have been equal (up to the usual MC accuracy and bias) should the black-box model had assumed a GBM dynamics as in the classic Black-Scholes framework.

$D_{MC}$ and $D_{BS}$ will indeed differ in general because digital options are sensitive to the implied volatility skew, which is inexistent in a Black-Scholes world where $\sigma (K,T)=\sigma $ (flat vol surface).

To see this, remember the model-free approximation of a digital call $D(K,T)$ as a (normalised) bull call spread with strikes $K-\epsilon$ and $K+\epsilon$ in the limit as the wedge $\epsilon $ tends towards zero:

$$ D (K,T) = \lim_{\epsilon \rightarrow 0} \frac {C (K-\epsilon,T)-C (K+\epsilon,T)}{2\epsilon} $$ hence the following, \begin{align} D (K,T) &= -\frac {dC}{dK} \\ &= \underbrace {-\frac {\partial C}{\partial K}}_ {\text {BS digital price}} - \underbrace {\frac {\partial C}{\partial \sigma}}_{\text {BS Vega}} \underbrace {\frac {\partial \sigma}{\partial K}}_{\text {Vol skew}} \end{align}

In a nutshell, as soon as your black-box model allows for vol skew ($\frac {\partial \sigma}{\partial K} \ne 0$), its digital prices will differ from the theoretical BS price which does not ($\frac {\partial \sigma}{\partial K}=0$).

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.