Using the Mandelbrot-Vann Ness representation of fractional Brownian motion in terms of Wiener integrals, increments of the logarithm of realized variance $v = \sigma^{2}$, under the physical measure $\mathcal{P}$, are expressed as

\begin{equation} \begin{aligned} \log v_{u}-\log v_{t} &=2 \nu C_{H}\left(W_{u}^{H}-W_{t}^{H}\right) \\ &=2 \nu C_{H}\left(\int_{-\infty}^{u}|u-s|^{H-\frac{1}{2}} d W_{s}^{\mathbb{P}}-\int_{-\infty}^{t}|t-s|^{H-\frac{1}{2}} d W_{s}^{\mathbb{P}}\right) \\ &=2 \nu C_{H}\left(\int_{t}^{u}|u-s|^{H-\frac{1}{2}} d W_{s}^{\mathbb{P}}+\int_{-\infty}^{t}\left[|u-s|^{H-\frac{1}{2}}-|t-s|^{H-\frac{1}{2}}\right] d W_{s}^{\mathbb{P}}\right) \\ &=: 2 \nu C_{H}\left[M_{t}(u)+Z_{t}(u)\right] \end{aligned} \end{equation}

With $H$,our hurst parameter that determines the roughness of the fractional Brownian motion .In this expression, the left integral $M_{t}(u)$ is independent of $\mathcal{F}_{t}$ and the right integral $Z_{t}(u)$ is $\mathcal{F}_{t}$-measurable.Note that $\tilde{W}^{P}$ is defined as:

\begin{equation} \tilde{W}^{P}:=\sqrt{2 H} \int_{t}^{u} \frac{d W_{s}^{\mathbb{P}}}{(u-s)^{\gamma}} \end{equation} On this step I don't know why we separated $\tilde{W}^{P}$ from $C_{H}$,shouldn't the term $C_{H}$ be mandatory to have a proper fractional Brownian motion of parameter $H$ .Moreover I don't get why we added a $\sqrt{2H}$ in the expression.To continue it is said that $\tilde{W}^{P}$ has the same properties as $M_{t}(u)$, only with variance $(u − t)^{2H}$ . With $\eta:=\frac{2\nu C_{H}}{\sqrt{2H}}$ we have $2\nu M_{t}(u) C_{H}= \eta \tilde{W}^{P}$ and so : \begin{equation} \mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right]=v_{t} \exp \left\{2 \nu C_{H} Z_{t}(u)+\frac{1}{2} \eta^{2} \mathbb{E}\left|\tilde{W}_{t}^{\mathbb{P}}(u)\right|^{2}\right\} \end{equation} However,I do not clearly understand this passage.is it because $Z_{t}(u)$ depends only on historical values,which makes it non Markovian that we do not treat it as a random variable here ? After that ,the last step is straightforward to derive as : \begin{equation} \begin{aligned} v_{u} &=v_{t} \exp \left\{\eta \tilde{W}_{t}^{\mathbb{P}}(u)+2 \nu C_{H} Z_{t}(u)\right\} \\ &=\mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right] \mathcal{E}\left(\eta \tilde{W}_{t}^{\mathbb{P}}(u)\right) \end{aligned} \end{equation}

With $\mathcal{E}$ being the Wick stochastic integral such :

\begin{equation} \mathcal{E}(\Psi)=\exp \left(\Psi-\frac{1}{2} \mathbb{E}\left[|\Psi|^{2}\right]\right) \end{equation}

Lastly I also don't understand why in the rough volatility models we have $\mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right] \neq \mathbb{E}^{\mathbb{P}}\left[v_{u} \mid v_{t}\right]$.

Thank you for your help.

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    $\begingroup$ This one calls for Kevin, Gordon, MagicIsInTheChain, IR7, or possibly DaneelOlivaw, Quantuple, Dimitri, Pleb or AKDemy. Let's see who answers first :) $\endgroup$ Aug 4, 2021 at 12:18
  • 4
    $\begingroup$ Cross-posted on MSE: math.stackexchange.com/questions/4224954/… $\endgroup$ Aug 16, 2021 at 4:19

1 Answer 1


My answer on MSE has details on the computations of $\mathbb{E}(v_s \, | \, \mathcal{F}_t)$ and $\mathbb{E}(v_s \, | \, \mathcal{v}_t)$, which answers why the rough Bergomi model is not Markovian. See here.

However, on this post you have an extra question: why do we rip out $C_H$ in our definition of $\tilde{W}$?

The answer is simple: it's just a normalisation constant. This affects the variance of the process, but it does not affect the roughness of the process. In particular, the Hölder exponent of $$\tilde{W}_t(u) = \sqrt{2H}\int_t^u \frac{dW_s}{(u-s)^\gamma}$$ is solely dependent on the choice of $\gamma$ in the power law kernel $(u-s)^{-\gamma}$. To see this, you may apply Kolmogorov's continuity criterion to get that $\tilde{W}_t(u)$ admits a.s. $(\frac{1}{2} - \gamma - \epsilon)$-Hölder continuous paths. Since we set $H = \frac{1}{2} - \gamma$, this is equivalent to a.s. $(H-\epsilon)$-continuous paths, which is the expected roughness for fBM.


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