# 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?

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