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I've read a lot of derivations about VIX formula. I can say it is -adjusted- fair strike of variance swap. But I can't see how it goes from variance swap rate to VIX formula. In particular I can't see the last part of VIX formula hosted here on page 4.

Could you please lead me from Hull Technical Note 22:

\begin{equation} \ E(V)= \frac{2}{T}ln\frac{F_{0}}{S^{*}} - \frac{2}{T}\left[ \frac{F_{0}}{S^{*}}-1\right] +\frac{2}{T}\left[\int_{K=0}^{S^{*}} \frac{1}{K^{2}}e^{RT}p(K)dK + \int_{K=S^{*}}^{\infty} \frac{1}{K^{2}}e^{RT}c(K)dK\right] \end{equation}

to VIX Formula \begin{equation} \sigma^{2}= \frac{2}{T}\sum_i^{}\frac{\triangle K_{i}}{K_{i}^{2}}e^{RT}Q(K_{i}) - \frac{1}{T}\left[ \frac{F}{K_{0}}-1\right]^{2} \end{equation}

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  • $\begingroup$ The determination of Vix is quite involved, using the implied volatility of listed options and weighs them by how far they are OTM. The CBOE website has the calculation methodology. $\endgroup$
    – AlRacoon
    Mar 3, 2019 at 17:25
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    $\begingroup$ @AlRacoon yes it has methodology as I attached to this post. But it doesn’t involve derivation. $\endgroup$
    – TryingtobeQuant
    Mar 3, 2019 at 17:45
  • $\begingroup$ It is clear that the $\Delta K_i$ come discretizing the integral - is it? $\endgroup$
    – Richi W
    Mar 3, 2019 at 19:29
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    $\begingroup$ @Richard Yes it is. What about first part of Hull's equation and last term in VIX white? Can we say it is just an approximation ? $\endgroup$
    – TryingtobeQuant
    Mar 3, 2019 at 19:32
  • $\begingroup$ @TryingtobeQuant yes, it is the second order approximation of the logarithm of 1+x as discussed below in the answer $\endgroup$
    – Richi W
    Mar 4, 2019 at 10:43

1 Answer 1

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The piece you are missing is an approximation via the Taylor formula of the logarithm:

$$\ln(1+x) \approx x-\frac{x^2}{2} \; .$$

Apply this to the first term in the final formula of the technical paper:

$$\frac{2}{T}\ln\frac{F_{0}}{S^{*}} = \frac{2}{T}\ln\left(1+\left(\frac{F_{0}}{S^{*}}-1\right)\right) \approx \frac{2}{T}\left(\left(\frac{F_{0}}{S^{*}}-1\right) - \frac{1}{2}\left(\frac{F_{0}}{S^{*}}-1\right)^2\right) \;.$$

Now, the first term of this approximation cancels with the second term of the technical paper formula. You're left with the quadratic term.

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