# 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?

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 ($\frac {\partial \sigma}{\partial K}=0$).

• Hi Quantuple ,where were you? :) Also, nice answer +1
– user16651
Commented Jul 26, 2016 at 16:43