In Jim Gatheral's book The Volatility Surface Section Dependence on Skew and Curvature on page 138, he asserts that

We know that the implied volatility of an at-the-money forward option in the Heston model is lower than the square root of the expected variance (just think of the shape of the implied distribution of the final stock price in Heston).

I suspect he is talking about a Jensen's inequality somewhere. But I do not see it. The expected variance should be $\mathbf E[v]$ where $v$ is the variance process described by the Heston model. I do not see a proof of this inequality, even though Chapter 3 offers some approximation of the implied volatility directly in terms of the model variance parameters. Does anyone have a proof?


1 Answer 1


Below are my 2 cents only, but this was too long for a comment.

As he shows in the next lines (see also Variance Swaps chapter of Bergomi's book) $$ \sigma_{VS}^2(T) = \int_{-\infty}^{+\infty} \tilde{\sigma}^2(z,T) \phi(z) dz \tag{0} $$ where $\sigma_{VS}(T)$ denotes the volatility of a fresh-start variance swap of maturity $T$; $\phi(\cdot)$ the standard Gaussian pdf; $\sigma(k,T)$ the implied volatility of smile in log-forward moneyness and time to expiry space, and $\tilde{\sigma}(\cdot,T)$ (modified smile) directly related the true smile $\sigma(\cdot,T)$ as follows $$ f: (k,t) \rightarrow -\frac{k}{\sigma(k,t)\sqrt{t}} + \frac{\sigma(k,t)\sqrt{t}}{2} $$ $$ \tilde{\sigma} : (z,t) \to (\sigma \circ f^{-1})(z,t) $$

Equation $(0)$ is equivalent to writing that $$ \sigma_{VS}^2(T) = \Bbb{E} \left[ \tilde{\sigma}^2(z,T) \right],\,z \sim N(0,1)$$ I think that he is then referring to the fact that $$ \sigma_{VS}(T) = \sqrt{ \Bbb{E} \left[ \tilde{\sigma}^2(z,T) \right] } \geq \Bbb{E} \left[ \tilde{\sigma}(z,T) \right] $$ by Jensen's inequality (square root is a concave function). Now if you parametrise $\tilde{\sigma}$ as $$\tilde{\sigma}(z) = \tilde{\sigma}_0 + \alpha z \tag{1}$$ You indeed have that $$ \sigma_{VS}(T) \geq \tilde{\sigma}_0 $$ which shows that $\sigma_{VS}(T) $ is greater than $\tilde{\sigma}_0$ if the "modified" smile $\tilde{\sigma}$ can be parametrised as given by $(1)$. He concludes that skew does not contribute to this result.

Now of course, the problem is that $(1)$ is certainly too rigid in practice (it could be argued that close to the forward moneyness it could be a decent approximation though) and $\tilde{\sigma}_0 \ne \sigma_0$ the genuine ATMF vol. So IMO his assertion cannot be made in general.

Note that Lorenzo Bergomi and Julien Guyon, The Smile in Stochastic Volatility Models managed to derive accurate approximations tying VS volatilities and ATMF volatilities in very general stochastic volatility models. If you look at equation (12) of their paper you'll see that, already at first order, skew is the only thing which contributes to the discrepancy between ATMF vol and VS vol, which goes against what Gatheral obtains.

At the end of the day, I think that his assertion holds in the modified smile space $\tilde{\sigma}(\cdot,T)$ but not in the genuine smile space $\sigma(\cdot,T)$.

  • $\begingroup$ Complete answer and more, with the paper. +1 and accepted. It was quite puzzling to me that the statement was the first sentence of the paragraph. Even as I read the paragraph on and obtain the same Jensen's inequality as yours, I still can not get his assertion regarding the vol at the money. Now it is clear, thanks to you, as usual. :-) $\endgroup$
    – Hans
    Apr 13, 2018 at 19:33
  • $\begingroup$ Glad I could help! $\endgroup$
    – Quantuple
    Apr 13, 2018 at 19:38
  • $\begingroup$ You invariably do! Now I need to go read that Bergom & Guyon paper. $\endgroup$
    – Hans
    Apr 13, 2018 at 19:46
  • $\begingroup$ @Quantuple In the definition of $\tilde{\sigma}$, do you mean $\sigma\circ f$? Because $f^{-1}(z,t)$ doesn't seem to work, since the range of $f$ is only 1-dimensional. And in the definition of $f$, are we missing a minus in front of the first term? (cf. page 139 in Gatheral's book) $\endgroup$
    – Phil-ZXX
    Apr 16, 2018 at 9:28
  • $\begingroup$ Hi @Phil-ZXX. Absolutely for the minus sign. Good catch. Yes you are right the notations are a bit sloppy it should be something like $\tilde{\sigma}(z,T) = \sigma(k = f^{-1}(z ; T), T)$ in the sense that for a given $z$ and knowing $T$ you take the reciprocal function to find the corresponding log-forward moneyness level $k$. Or in other words, $z = f(k,T)$. $\endgroup$
    – Quantuple
    Apr 16, 2018 at 9:40

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