Independent increment of Lévy process

54 Views Asked by At

I'm a beginner on Lévy process and I got a question as follows. Let $X=(X_t)_{t\geqslant 0}$ be a zero mean Lévy process having form $X_t=\sigma B_t + J_t$, where $B$ and $J$ are the Brownian part and jump part of $X$. For $0\leqslant t \leqslant 1$, denote $\mathcal F_t^B=\sigma(B_s, s\leqslant t)$, $\mathcal F_t^J=\sigma(J_s, s\leqslant t)$ and $\mathcal G_t=\mathcal F_t^B \vee \mathcal F_1^J$. In my intuition, it seems $B_1-B_t$ is independent of $\mathcal G_t$, however I'm stuck for proving or disproving it. Any hint or reference will be appreciate.