Write the martingale part of this semimartingale as an Itō integral process with respect to a one-dimensional Brownian motion

109 Views Asked by At

Let $H$ be a separable $\mathbb R$-Hilbert space and $(X_t)_{t\ge0}$ be an $H$-valued semimartingale with $$\int_0^t\left\|X_s\right\|_H^2\:{\rm d}s<\infty\;\;\;\text{almost surely for all }t\ge0\tag1$$ and assume the local martingale part of $(X_t)_{t\ge0}$ is of the form $$M_t:=QW_t\;\;\;\text{for }t\ge0,$$ where $Q$ is a linear operator from $\mathbb R^d$, $d\in\mathbb N$, to $H$ and $W$ is a cylindrical $\mathbb R^d$-valued Wiener process (i.e. an ordinary $d$-dimensional Brownian motion).

By the Itō formula, the local martingale part of $\left\|X_t\right\|_H^2$ is given by $$N_t:=2\int_0^t\langle X_s,Q\rangle_H\:{\rm d}W_t\;\;\;\text{for }t\ge0.$$ Are we able to rewrite $(N_t)_{t\ge0}$ as an Itō integral process with respect to a one-dimensional Brownian motion? If not, are we able to give a suitable form of $Q$ which allows for the desired result?

EDIT: Note that $\tilde W:=QW$ is again a Wiener process, but with covariance operator $QQ^\ast$.

Remark: Let $(e_1,\ldots,e_d)$ denote the standard basis of $\mathbb R^d$. We know that $B^{(i)}:=\langle W,e_i\rangle$ is a Brownian motion for all $i\in\{1,\ldots,d\}$ and $B^{(1)},\ldots,B^{(d)}$ are independent. Moreover, $$\int_0^t\Phi_s\:{\rm d}W_s=\sum_{i=1}^d\int_0^t\Phi_se_i\:{\rm d}B^{(i)}\tag2$$ for all $t\ge0$ almost surely for all predictable $\mathfrak L(\mathbb R^d,H)$-valued processes $(\Phi_t)_{t\ge0}$ with $$\int_0^t\left\|\Phi_s\right\|_{\mathfrak L(\mathbb R^d,\:H)}^2\:{\rm d}s<\infty\;\;\;\text{almost surely for all }t>0\tag3.$$