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When Implied Volatility Skew Has a Zero-Slope Strike

Article Quant Q&A · Author: user34971

Summary

The document derives a condition for a vanilla option’s implied volatility curve to have zero slope with respect to strike at a given maturity. Differentiating the Black–Scholes call price, expressed using strike-dependent implied volatility, and comparing it with the risk-neutral derivative of the call’s expected payoff yields an equality: the Black–Scholes probability of finishing in the money must match the probability under the underlying stochastic-volatility model.

The model lets the asset’s volatility evolve with drift and diffusion, with its diffusion correlated to the asset’s return shock. The resulting condition can be expressed through the distribution of the asset’s integrated stochastic-volatility return. The derivation identifies candidate stationary points, but a zero slope by itself does not establish whether a point is a minimum or maximum, nor whether it is global. A separate answer offers intuition about volatility-of-volatility and correlation, while explicitly leaving those claims unproved.

Key ideas

  • A zero strike slope in implied volatility requires matching risk-neutral in-the-money probabilities under Black–Scholes and the model.
  • The condition applies maturity by maturity and allows the candidate strike to vary with maturity.
  • Stochastic volatility and its correlation with the asset return shape the relevant terminal probability.
  • A stationary point condition alone does not classify a turning point as a minimum or maximum.

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Full text
# Conditions for implied volatility to attain a minimum (or maximum) value


# Conditions for implied volatility to attain a minimum (or maximum) value












Suppose

$$ dS = \sigma S \left(\rho dW + \sqrt{1-\rho^2} dZ \right) $$

and

$$ d\sigma = a(\sigma,t) dt + b (\sigma,t) dW $$

with $dW dZ = 0$.

What are the conditions necessary such that the implied volatility skew of vanilla options attains a (global/local) minimum or maximum value for each maturity date $T$? In other words that the slope of the implied volatility skew is zero at some strike $K$ for each $T$, where $K = K(T)$, i.e. the strike could be different depending on $T$.

## Answer by fes (score 2)

https://quant.stackexchange.com/a/57185

Ok here is my shot at this. Let me know if I am missing something. For each $T$ and other relevant parameters implied volatility is defined via

$$C^{BS}(\sigma^{IV}(K),K)=e^{-rT}\mathbb{E}^{Q}[(S_T-K)^{+}]$$

Taking derivative wrt. to $K$ gives

$$\frac{\partial}{\partial \sigma}C^{BS}(\sigma^{IV}(K),K)\frac{d\sigma^{IV}(K)}{dK}+\frac{\partial}{\partial K}C^{BS}(\sigma^{IV}(K),K)=e^{-rT}\mathbb{E}[-I(S>K)]=-e^{-rT}\mathbb{P}(S>K)$$

Using the fact that BS vega is always positive $\frac{d\sigma^{IV}(K)}{dK}=0$ iff

$$\frac{\partial}{\partial K}C^{BS}(\sigma^{IV}(K),K)=-e^{-rT}\mathbb{P}(S>K)$$

Using a standard result for the derivative of BS price wrt. to $K$ gives

$$-e^{-rT}N(d_2)=-e^{-rT}\mathbb{P}(S>K)$$

or

$$N(d_2)=\mathbb{P}(S>K)$$

This means that we need to find a point such that the risk neutral probability that the option ends in the money is the same under BS and your model. Using Ito's lemma for $\log(S_t)$ we have

$$S_t=S_0\exp(\int_{0}^{t}\sigma_sdW_s-\frac{1}{2}\int_{0}^{t}\sigma_s^2ds)$$

Hence our condition is

$$N(d_2)=\mathbb{P}(S_0\exp(\int_{0}^{T}\sigma_tdW_t-\frac{1}{2}\int_{0}^{T}\sigma_t^2dt)>K)$$

or

$$N(d_2)=\mathbb{P}(\log(S_0)+\int_{0}^{T}\sigma_tdW_t-\frac{1}{2}\int_{0}^{T}\sigma_t^2dt>\log(K))$$

Hence a necessary and sufficient condition is that for each $T$ there is some $K(T)$ such that

$$N(d_2(K(T))=\mathbb{P}(\int_{0}^{T}\sigma_tdW_t-\frac{1}{2}\int_{0}^{T}\sigma_t^2dt>\log(K(T)/S_0))$$

## Answer by dm63 (score 0)

https://quant.stackexchange.com/a/51428

Intuitively, $b>0$ should be sufficient for the existence of a global minimum implied vol. The minimum should be more pronounced if $b$ is large , and the strike price with the minimum implied volatility should be close to ATM when $\rho $ is close to zero. I’m pretty sure these assertions could be proven from the Hagen approximation to the SABR model.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.