The rough Bergomi model is defined as

\begin{cases} \frac{dS_t}{S_t} = \sqrt{v_t}dW_t^1 \\ v_t=\xi_0(t)\exp(\eta \tilde{W}_t^H-\frac{1}{2}\eta^2t^{2H}) \\ \tilde{W}_t^H = \int_0^t \sqrt{2H}(t-s)^{H-\frac{1}{2}}dW_t^2 \\ \langle dW_t^1,dW_t^2 \rangle = \rho dt \end{cases}

Ref: Bayer, 2016, Pricing under rough volatility, page 5, 10, 14, or Zhu, 2020, Markovian approximation of the rough Bergomi model for Monte Carlo option pricing, page 3

According to the assumption 1 of (Zhu, 2020), p.3

  • we assume throughout the paper that the initial forward variance curve $\xi_0(t)$ is flat. This simplification is common in the rBergomi literature; see, for example, Bayer et al. 1, ... . We henceforth use the notation $\xi_0(t) = \color{red}{\xi_0}$ for the constant initial forward variance curve.

All these papers I have read up to now say that this model has $3$ parameters : $H$, $\rho$ and $\eta$ and the initial forward variance $\xi_0$ is rarely mentionned. For example, in (McCrickerd, 2017, Turbocharging Monte Carlo pricing for the rough Bergomi model, page 5), the author used $\xi_0 = 0.235^2$ but didn't justify this value.

I guess the initial forward variance $\xi_0$ can be observed in the market, but if this is true, then why we need to suppose the initial forward variance curve $t \to \xi_0(t)$ is flat ($\xi_0(t) = \xi_0$) ? (in other words, people can check the form of this forward variance curve with observed data and don't need to make the assumption that the curve is flat)

Hence, my question is: how do we determine the initial forward variance $\xi_0$ in practice?

Thank you for your time reading this question.

Thanks to Quantuple's comment, I think now the forward variance curve $\xi_t(u)$ in the rough Bergomi model can be constructed from the variance swap strike $\sigma_{\text{strike}}^2(u)$. Indeed, by definition, $$\xi_t(u) := E(v_u|\mathcal{F}_t) \tag{1}$$ and from the payoff of a variance swap $N(\sigma_{\text{realised},s}^2(t) -\sigma_{\text{strike},s}^2(t))$, we have $$\sigma_{\text{strike},s}^2(t) =E\left(\frac{1}{t}\int_s^t v_udu |\mathcal{F}_s\right)\tag{2}$$

From $(1), (2)$ we can deduce

$$\sigma_{\text{strike},s}^2(t) =\frac{1}{t}\int_s^t E\left(v_u |\mathcal{F}_s\right)du=\frac{1}{t}\int_s^t \xi_s(u)du\implies \xi_s(t) = \frac{d}{dt} \left( t\sigma_{\text{strike},s}^2(t) \right)$$

In particular, with $s = 0$, we have $$\xi_0(t) = \frac{d}{dt} \left( t\sigma_{\text{strike},0}^2(t) \right) \tag{3}$$

To a certain extent, we can assume that $\sigma_{\text{strike},0}^2(t) \approx \text{const} = \sigma_{\text{strike}}^2$, then from $(3)$ we deduce

$$\color{red}{\xi_0} :=\xi_0(t) \approx \sigma_{\text{strike}}^2$$

  • 3
    $\begingroup$ In a pure diffusion setting (no jumps) the forward variance curve is the term structure of fair variance swaps strikes, which can be estimated in a model-free way from listed European vanilla options (see VIX construction methodology and related literature pertaining to log-contracts and Carr-Madan formula). $\endgroup$
    – Quantuple
    Apr 23, 2021 at 7:57
  • $\begingroup$ @Quantuple Thank you very much. I think now the initial forward variance curve can be approximated by the strike of variance swap. I added the proof in the end of the question. Could you see whether this proof is correct? $\endgroup$
    – NN2
    Apr 23, 2021 at 14:46
  • 1
    $\begingroup$ This is indeed what I had in mind! Note that the underlying assumptions are: (i) a pure diffusion process (no jumps) (ii) approximation the true (discrete) variance swap payoff in continuous time (basically meaning you are replacing the variance of the returns by the quadratic variation of the log-price process). $\endgroup$
    – Quantuple
    Apr 26, 2021 at 6:14


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