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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 $T$.

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  • $\begingroup$ Could your problem be answered with the insights of the paper 'The Moment Formula for Implied Volatility at Extreme Strikes' by Roger W. Lee.? $\endgroup$
    – Quantuple
    Aug 6, 2020 at 12:42
  • $\begingroup$ Not sure. Lee gives bounds on the skew, but I don't think he proves or disproves that for each time slice (for certain volatility models) the put skew is negative and the call skew positive (or the other way around). For homogeneous SV models of degree 1 what needs to be shown (or shown to be false) is that for each $T$ there is a strike such that the price of a digital under SV is equal to the price of a digital under BS (using the IV at that strike), or equivalently that the SV delta is equal to BS delta at some strike. $\endgroup$
    – user34971
    Aug 6, 2020 at 17:07
  • $\begingroup$ Thanks for the clarifications. Then shouldn't the strike you are looking just be the one where the Black-Scholes Vega is zero? The standard chain rule indeed yields $$\Delta_{SV} = \Delta_{BS} + \nu_{BS} \frac{\partial \Sigma}{\partial S}$$ The fact that the SV model is homogeneous allows you to write that the partial derivative on the RHS is indeed proportional to the skew $\frac{\partial \Sigma}{\partial K}$ but is that really relevant here if you are just trying to match the deltas? $\endgroup$
    – Quantuple
    Aug 6, 2020 at 19:58
  • $\begingroup$ @Quantuple Thanks, indeed you can relate the two deltas as above. But for vanillas there is no strike $K \in (0,\infty)$ where $\nu^{BS}$ is zero. $\endgroup$
    – user34971
    Aug 7, 2020 at 5:56
  • $\begingroup$ Right, not strictly zero, but maybe with the asymptotic rate of convergence of the Vega plus that of the skew given by Lee you can prove the convergence of the Deltas? But doesn't exactly answer your question, I'll think about it a little more :) $\endgroup$
    – Quantuple
    Aug 7, 2020 at 6:24

2 Answers 2

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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

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

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  • $\begingroup$ Agree with the expression, but isn't that the same as saying $N(d_2(K)) = P( S_T>K)$ which I'm not sure opens a way to an answer. Or am I missing something? Alternatively, directly working with the volatility process as in your final expression, doesn't I believe lead to a general condition whether/when such strikes exist as you'd have to check it volatility model by volatility model? $\endgroup$
    – user34971
    Aug 7, 2020 at 19:25
  • $\begingroup$ @ilovevolatility Yes for each $T$ there needs to be a strike such that the RN probability the option ends in the money is equal under BS and your model. Agree this is not a full solution but simplifies things at least to me. $\endgroup$
    – fes
    Aug 7, 2020 at 20:44
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    $\begingroup$ @StackG Thanks for editing! $\endgroup$
    – fes
    Aug 8, 2020 at 5:59
  • $\begingroup$ @ilovevolatility So e.g. for the Heston model characterised by (15) and (16) here (web.math.ku.dk/~rolf/teaching/ctff03/Gatheral.1.pdf) you need the additional condition that for some strike the $P_0$ term is equivalent under BS. $\endgroup$
    – fes
    Aug 9, 2020 at 8:33
  • $\begingroup$ Too specific to only consider Heston. But let's take $\rho=0$ then we know that such a $K$ always exists for the SV models under consideration. Can we show that using your approach? I can show that using put-call symmetry very easily, but perhaps your approach is more general. $\endgroup$
    – user34971
    Aug 10, 2020 at 8:23
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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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  • $\begingroup$ Thanks - would like to avoid perturbative arguments if possible though. $\endgroup$
    – user34971
    Mar 1, 2020 at 13:37

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