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I'm new here. I was wondering what the well-known ATMF implied vol approximation mentioned on page 2 in Bergomi Smile Dynamics IV: $$S_T = \frac{s_T}{6\sqrt{T}}.$$

I cannot find any reference about this.

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Let $$\ln\left(S_T/S_t\right) $$

have mean $\mu_\tau$ and standard deviation $\sigma_\tau$, where $\tau=T-t$, and density of its standardized form $$ X= \frac{\ln(S_T/S_t)-\mu_\tau}{\sigma_\tau} $$

approximated by Gram-Charlier expansion

$$ f_X(x) = \phi(x) - \gamma_{1\tau} \frac{1}{3!} D^3 \phi(x) + \gamma_{2\tau} \frac{1}{4!} D^4 \phi(x), $$

with $\phi$ being standard normal density and $\gamma_{1\tau}$ and $\gamma_{2\tau}$ being third (skewness) and fourth (kurtosis) cumulants.

One can then price a call option with strike $K$ against density $f_X$ and then imply, via Black-Scholes formula, standard deviation:

$$ \hat{\sigma}_{K\tau} = \sigma_\tau\left[1- \gamma_{1\tau} \frac{1}{3!} d_{K\tau} - \gamma_{2\tau} \frac{1}{4!} (1- d_{K\tau}^2) \right] $$

with

$$ d_{K\tau} = \frac{\ln(S_t/K)-r\tau +0.5\sigma_\tau^2}{\sigma_\tau}. $$

Detailed proof is available here.

This in turn gives:

$$ \frac{\partial \hat{\sigma}_{K\tau}}{\partial \ln K}\bigg|_{K=S_t\mathrm{e}^{rt}} = \gamma_{1\tau} \frac{1}{3!}. $$

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  • $\begingroup$ Thank you so much! Basically, we approximate the standardized r.v. by the Gram-Charlier expansion, and then use that one to price the option and eliminate some high order terms to obtain the iv. $\endgroup$ Commented Jul 25, 2020 at 20:05
  • $\begingroup$ I also read the book, Stochastic Volatility Modelling, written by Bergomi. In Chapter 5 Appendix B, it provides another approach, but a lot of steps are skipped: It mentioned that use the order 1 vol., and express the derivatives of $P_0$, which is the price, in terms of derivative of $\rho_0$, which is the density of $log(S_T/F_T)$, and use the BS expression of gamma yield then obtain the $\mathcal{S}_T$. Any comments on this would be appreciated! $\endgroup$ Commented Jul 25, 2020 at 20:14
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    $\begingroup$ @JuniorQuant I think it’s best you make that another question. $\endgroup$
    – Bob Jansen
    Commented Jul 26, 2020 at 10:31
  • $\begingroup$ @BobJansen I see. I will make another one. Thanks for pointing out this! $\endgroup$ Commented Jul 26, 2020 at 14:17
  • $\begingroup$ I have two questions for the proof in here: 1. For equation (31), the authors mentioned the arbitrage condition. I was wondering what's that condition; 2. For the identity (32): $SN'(d_1) = Ke^{-rT}N'(d_2)$, is there any explanation for this equality? Thank you for your time and help! $\endgroup$ Commented Jul 26, 2020 at 23:26

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