Show that every continuous Lévy process is a Wiener process Let $H$ be a $\mathbb R$-Hilbert space (assume $H=\mathbb R^d$ for some $d\in\mathbb N$, if this is helpful for you to understand the following) and $(X_t)_{t\ge0}$ be an $H$-valued continuous Lévy process on a probability space $(\Omega,\mathcal A,\operatorname P)$.

How can we show that $X$ is a Wiener process?

Let $$M^y_t:=\langle\underbrace{X_t-\operatorname E[X_t]}_{=:\:\tilde X_t},y\rangle_H\;\;\;\text{for }t\ge0$$ for $y\in H$. We can easily show that $M^y$ and $\left(\left|M^y_t\right|^2-t\operatorname{Var}\left[M^y_1\right]\right)$ are martingales with respect to the filtration $\mathcal F^X$ generated by $X$.
So, by Lévy characterization of Brownian motion, $M^y$ is an $\mathcal F^X$-Brownian motion for all $y\in H$.

How do we conclude that $X$ is a Wiener process? Is it a $Q$-Wiener process (if so, how do we construct the covariance operator $Q$) or a cylindrical Wiener process? If necessary, assume $H$ is finite dimensional.


$M^y$ is not a standard Brownian motion, but a Brownian motion with quadratic variation $$\langle M^y\rangle_t=t\operatorname E[M^y_1]=t\langle Qy,y\rangle_H\;\;\;\text{for all }t\ge0\tag1$$ for all $y\in H$, where $$Q:=\operatorname E\left[\tilde X_1\otimes\tilde X_1\right].$$ From this the desired result is obvious.
