Normalizing a Riemann Approximation for Rough Volatility Prediction

Summary

This exchange addresses a question about a Riemann-sum approximation to a rough-volatility prediction formula. The question concerns why a term involving the cosine of the Hurst parameter, divided by pi and multiplied by a power of the time increment, is omitted in a simulation. The answer explains that this factor is dropped along with the discretized increment term, while the approximation is divided by a normalization factor formed by summing the integrand without the log-variance component.

The response supports this explanation with an integral identity: integrating the kernel with the prefactor over its full domain gives one. This makes the prefactor function as a normalization of the kernel, so an explicit normalization in the discrete approximation can absorb its effect. The exchange is brief and focuses on this implementation detail; it does not lay out the full prediction formula, simulation procedure, or broader empirical evidence. The explanation is tied to the approximation described in the question.

Key ideas

• The simulation omits the prefactor and time-increment term from the discretized expression.
• The answer says the discrete sum is normalized by the sum of the kernel without the log-variance term.
• The continuous kernel, including the prefactor, integrates to one over its domain.
• The exchange explains an approximation detail but does not present the full rough-volatility prediction method.

Tags

# Rough Volatility Prediction - Gatheral, Jaisson, Rosenbaum Paper


# Rough Volatility Prediction - Gatheral, Jaisson, Rosenbaum Paper












I just read through the paper "Volatility Is Rough" by Gatheral, Jaisson and Rosenbaum. There is a website (link: http://tpq.io/p/rough_volatility_with_python.html) that details the simulations they have done in their paper which I found quite useful. There is one point, however, where I seem to miss something. They are approximating the prediction formula (5.1 in the paper) via its Riemann sum but they drop the term $\frac{cos(H\pi)}{\pi}$. Any reason why? Apologies if this is obvious.

Thanks.

## Answer by DangerousMouse (score 1)

https://quant.stackexchange.com/a/47022

1) They drop the $$ \frac{\cos (H\pi)}{\pi} \cdot \Delta^{H + 1/2}$$

2) They divide by a normalization factor, which is the sum of the integrand (without the $\log v_s$).

If you integrate: $$ \frac{\cos (H\pi)}{\pi} \cdot \Delta^{H + 1/2} \cdot \frac{1}{(x + \Delta) \cdot x^{H + 1/2}}$$ from zero to infinity, you will get 1.

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