# Realised variance under simple rough volatility model

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{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}

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:

$$$$\tilde{W}^{P}:=\sqrt{2 H} \int_{t}^{u} \frac{d W_{s}^{\mathbb{P}}}{(u-s)^{\gamma}}$$$$ On this step I don't know why we separated $$\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}$$ and so : $$$$\mathbb{E}^{\mathbb{P}}\left[v_{u} \mid \mathcal{F}_{t}\right]=v_{t} \exp \left\{2 \nu C_{H} Z_{t}(u)+\frac{1}{2} \eta^{2} \mathbb{E}\left|\tilde{W}_{t}^{\mathbb{P}}(u)\right|^{2}\right\}$$$$ However,I do not clearly understand this passage.is it because $$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{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}

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

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

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

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.