I was asked in an interview about how the vol smile affect the price of a binary option, which is essentially the Prob(ITM) under risk neutral measure. My thought is that the implied vol at spot which makes the option OTM is high, that means the prob(ITM) at that region is higher and so the price of binary option. Please correct me or give a more rigorous extension to my thoughts.
1 Answer
First note that the price of binary call is related to the price of an ordinary call in any model by
$$
BinC(T,K) = e^{-rT}\mathbb{E}^{\mathbb{Q}}[1_{S_T>K}] = - \frac{\partial}{\partial K}e^{-rT}\mathbb{E}^{\mathbb{Q}}[(S_T-K)_+] = - \frac{\partial}{\partial K}C(T,K)
$$
Now the volatility smile is implicitly defined by
$$
C(T,K) = C_{BS}(T,K,\Sigma(T,K))
$$
So, by taking minus the derivative wrt $K$, we get
$$
BinC(T,K) = BinC_{BS}(T,K,\Sigma) - \partial_\sigma C_{BS}(T,K,\Sigma )\partial_K\Sigma(T,K)
$$
In other words, the volatility smile leads to a corrective term to the price of a binary call which is the BS vega times the skew.
Conversely this formula can be used to calculate the skew based on the price of digitals.
To get a less formal intuition, replace your digital call by a small call spread $\frac{1}{\Delta K}(C(T,K)-C(T,K+\Delta K))$. In BS the vol at $K+\Delta K$ is the same than at $K$. If the volatility decreases with the strike (negative skew) then the volatility at $K+\Delta K$ is lower than the volatility at $K$. So the second call loses value and the call spread's value increases. Since a digital is nothing but the limit of the call spread when $\Delta K \to 0$, you see that its price increases when the skew becomes more negative.
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$\begingroup$ Hi, this is very interesting to me. Do you perhaps know of a resource (textbook perhaps) that goes into this methodology of pricing binary options using the corresponding BS implied volatility and the corrective term? $\endgroup$– OscarCommented Oct 26, 2021 at 19:06