1
$\begingroup$

On his book https://www.amazon.fr/dp/B019FNKQS8/ref=dp_kinw_strp_1 Bergomi derives a multifactor mean reversible volatility of the volatility such that :

\begin{equation*} d \xi_{t}^{T}=\omega(\tau) \xi_{t}^{T} d X_{t} \end{equation*}

$\xi_{t}^{T}$ is the instantaneous forward variance. $\omega(\tau)$ is a weighted average of forward volatility at time $t$ for an horizon of $T$ .

$X_{t}$ is a mean revertible OU process with length $t$ such : \begin{align} d X_t = -k \omega X_t dt + \omega d W_t \, , \, X_0 = 0 \end{align} We also have $\log(\xi_{t}^{T}) \sim N\Big(\log(\xi_{0}^{T})- \frac{1}{2}\int^{t}_{0}\omega^{2}(T -s ) d s , \int^{t}_{0} \omega^{2}(T -s ) d s\Big)$.The proof is trivial and similar to the one of the Black and Scholes.

The equation becomes equal to :

\begin{align} d \xi_{t}^{T} = w e^{-k(T-t)}\xi_{t}^{T} dX_t \label{eq:eq46} \end{align}

when setting $w$ = $2v$ with $v$ being the volatility of the volatility(which is more natural than the volatility of the variance )

What I don't understand is when translating to multifactor model with $N$ Brownian motions such :

$$ d \xi_{t}^{T}=\omega \alpha_{w} \xi_{t}^{T} \sum_{i} w_{i} e^{-k_{i}(T-t)} d X_{t}^{i} $$ where $\alpha_{w}$ is a normalizing factor and I guess $w_{i}$ is the covariance of the OU processes.However I am not sure of that claim. $$ \begin{aligned} \omega &=2 \nu \\ \alpha_{w} &=\frac{1}{\sqrt{\sum_{i j} w_{i} w_{j} \rho_{i j}}} \end{aligned} $$ I don't understand why and how we used such normalisation because at the end we will have :

\begin{equation} \xi_{t}^{T}=\xi_{0}^{T} \exp ( \sum_i \omega e^{-k_{i}(T-t)} X_{it}- \omega^{2}\sum_{ij}\frac{e^{-(k_{i}+k_{j})(T-t)}}{2} E[X_{it}X_{jt}]) \end{equation}

and will afterward retrieve the initial mean reversible SDE of the lognormal volatility of the variance such :

\begin{equation} \omega(T-t) = \sum_i e^{-k_{i} (T-t) } \omega d W_t \end{equation}

I also find other illustrations on internet that write the multifactor model SDE such :

\begin{equation} d \xi_t^T = w \xi_t^T \sum_k \lambda_{kt}^T \xi_t^T dW_t^k \end{equation}

However I do not understand what is $ \lambda_{kt}^T$ and how the extrapolation has been made from the one factor model .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.