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How the Log Approximation Produces the VIX Variance Correction

Article Quant Q&A · Author: TryingtobeQuant

Summary

The document shows how a Taylor approximation connects a variance swap expression to the correction term in the VIX variance formula. It begins with the logarithmic forward-to-reference-level ratio in the continuous-strike expression and rewrites the ratio as one plus its deviation from unity.

Approximating the logarithm to second order yields a linear component and a quadratic component. The linear component cancels the separate linear term in the original expression, leaving the negative squared deviation that appears as the final correction. This explains the algebraic source of that part of the VIX formula. The derivation is approximate: the logarithm is truncated after the quadratic term, so the explanation does not establish the accuracy of the approximation in every market condition or reproduce the full conversion from option integrals to a discrete strike sum.

Key ideas

  • The VIX correction can be traced to a second-order Taylor expansion of a logarithm.
  • The forward-to-reference-level ratio is expressed as one plus its deviation from unity.
  • The expansion’s linear term cancels the separate linear component in the variance expression.
  • The remaining quadratic term gives the negative squared correction in the VIX formula.
  • The argument is an approximation and does not derive the full discrete option-strip formula.

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Full text
# Derivation of VIX Formula


# Derivation of VIX Formula












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}

## Answer by Raskolnikov (score 14, accepted)

https://quant.stackexchange.com/a/44404

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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