The Brownian component has finite variance $bt$, and the compensated jump integral has variance
$$
t\int x^2K(dx)
$$
when this integral is finite. Conversely, a finite second moment forces the jump measure to have a finite second moment. Therefore $(X_t^2)$ is integrable exactly when
$$
\boxed{\int_{\mathbb R}x^2K(dx)<\infty}.
$$
These four equivalences are the <Path and moment criteria from a Lévy triplet>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 6 / d / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-201/6/d/iv/solution