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\begin{align} H &= \frac{1}{2} \ln (2\pi\sigma^2) + \frac{1}{2}\\ &= \frac{1}{2} \ln (2\pi e \sigma^2) \end{align} is the analytical solution for the entropy of a Gaussian random variable, such as a returns series, in the textbook Financial Machine Learning.

The author then re-works one of the above into a formula for entropy-implied volatility by isolating $\sigma$ by itself, and renaming it $\sigma_H$:

$$\sigma_{H} = \frac{e^{H} - \frac{1}{2}}{\sqrt{2\pi}}$$

but isn't this useless because, in order to calculate $H$, you need sample volatility $\sigma$ anyway? In other words, to calculate $\sigma_H$, you need $H$, which itself in turn requires $\sigma$, meaning $\sigma = \sigma_H$. It's fair enough that $H$ is a function of $\sigma$, but $\sigma_H$ is a function of $H$ as well as its own self, $\sigma$

Is the $\sigma_H$ formula just a small exercise in algebra that is circularly redundant given that it itself ($\sigma$) is required as an input?

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  • $\begingroup$ I think you estimate entropy from the data ($H$) and plug it into the estimator. $\endgroup$ Commented Nov 11, 2020 at 7:09
  • $\begingroup$ of course, but $H$ itself requires $\sigma$ as an input. That's the whole point of the question $\endgroup$
    – develarist
    Commented Nov 11, 2020 at 11:20
  • $\begingroup$ I think that $H$ is assessed via an empirical entropy estimator from raw data, no? $\endgroup$ Commented Nov 11, 2020 at 11:52
  • $\begingroup$ but $H$ is the analytical solution for the Gaussian (not empirical) case, which in turn means $\sigma_H$ only admits the Gaussian (not empirical) $H$. $\endgroup$
    – develarist
    Commented Nov 11, 2020 at 11:53
  • $\begingroup$ I think you would use some empirical estimator for $\hat{H}\equiv -\int f(x)log(f(x))dx$ and plug it in, no? Everything else would make no sense. $\endgroup$ Commented Nov 11, 2020 at 12:36

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