Bergomi Volatility Model

Question

I was studying on the Bergomi volatility model(using forward variance represented as $\xi_{t}^{T}$).However I don't understand how the author passes from the sde to the first step by only integrating respectively over $\xi_{t}^{T}$ and $W_{t}^{k}$.

$\begin{array}{l} \text {The dynamics are represented as : }\left\{\begin{array}{l} d S_{t}^{\omega}=(r-q) S_{t}^{\omega} d t+\sqrt{\xi_{t}^{t}} S_{t}^{\omega} d Z_{t} \\ d \xi_{t}^{T}=\omega \xi_{t}^{T} \sum_{k} \lambda_{k t}^{T}\left(\xi_{t}\right) d W_{t}^{k} \end{array}\right.\\ \text { At order } 1(\text { Using one factor}) \text { in } \omega: \quad \xi_{t}^{T}=\xi_{0}^{T}\left(1+\omega \int_{0}^{t} \sum_{k}\left(\lambda_{k \tau}^{T}\right)_{0} d W_{\tau}^{k}\right) \end{array}$

With The instantaneous variance of the spot process such $\xi_{t}^{t}$, $S_t$ the stock price, $w$ a scaling factor,$d W_{\tau}^{k}$ correlated standard brownian motions.

Could someone help me to understand this step thank you.

Answer 1

SDE for forward variance: $$ d \xi_t^T = w \xi_t^T \sum_k \lambda_{kt}^T(\xi_t^T) dW_t^k $$

Integrate: $$ \xi_t^T = \xi_0^T + w \int_0^t \xi_{\tau}^T \sum_k \lambda_{k{\tau}}^T(\xi_{\tau}^T) dW_{\tau}^k $$

Plug into RHS of SDE: $$ d \xi_t^T = w \left(\xi_0^T + w \int_0^t \xi_{\tau}^T \sum_k \lambda_{k{\tau}}^T(\xi_{\tau}^T) dW_{\tau}^k\right) \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \sum_k \lambda_{kt}^T\left(\xi_0^T + w \int_0^t \xi_{\tau}^T \sum_k \lambda_{k{\tau}}^T(\xi_{\tau}^T) dW_{\tau}^k\right) dW_t^k $$

Expand at first order in $w$: $$ d \xi_t^T \approx w \xi_0^T \sum_k \lambda_{kt}^T\left(\xi_0^T\right) dW_t^k $$

Integrate: $$ \xi_t^T \approx \xi_0^T + \int_0^t w \xi_0^T \sum_k \lambda_{k{\tau}}^T\left(\xi_0^T\right) dW_{\tau}^k = \xi_0^T \left(1 + w \int_0^t \sum_k \lambda_{k{\tau}}^T\left(\xi_0^T\right) dW_{\tau}^k \right) $$