Why is a continuous Lévy process twice integrable?

Question

In his textbook "Wahrscheinlichkeitstheorie" (de Gruyter, 2008), Jochen Wengenroth shows (p. 144) that if $(X_t)_{t\in[0,\infty)}$ is a continuous, real-valued Lévy process with $X_t\in \mathcal{L}_2$ for all $t \in [0, \infty)$, then $X$ is a Brownian motion with drift. He remarks, but does not prove, that the stipulation that $X_t \in \mathcal{L}_2$ is redundant. Why is it redundant? Are all Lévy processes twice integrable, or only the continuous ones?

Answer 1

The only continuous Lévy processes are Brownian motion (with or without drift). There are Lévy processes that are not square-integrable; viz. the Cauchy processes.

Answer 2

Let me try to indicate an answer to what I think is your real question ("Is there a direct way to show that a continuous Lévy process is square integrable?") A detailed discussion of this and much more, using stochastic calculus, can be found on Geo. Lowther's blog, specifically https://almostsure.wordpress.com/2010/09/15/processes-with-independent-increments/ .

Here is a sketch of the portion of Lowther's argument relevant to your question. First, by virtue of the stationary independent increments of $X$, there is a function $\psi:\Bbb R\to\Bbb C$ such that $\Bbb E[\exp(ibX_t)]=\exp(t\psi(b))$ for all real $b$ and all $t>0$. Moreover, $U_t:=\exp(ibX_t-t\psi(b)$ is a martingale. From this, by taking logarithms and applying Ito's formula, is follows that $X$ is a continuous semimartingale. Notice that $|U_t|=|\exp(-t\psi(b))|$ is uniformly bounded on finite time intervals, so $U$ is even a square-integrable martingale. Likewise, $|U_t^{-1}|$ is uniformly bounded on finite time intervals, so (using Ito's formuula for the second equality below) $$ M_t:=ibX_t-t\psi(b)-{1\over 2}\langle X\rangle_t =\int_0^t U^{-1}_s\,dU_s $$ is a square-integrable martingale. Finally, $$ X_t=-ib^{-1}\left[M_t+t\psi(b)+{1\over 2}\langle X\rangle_t\right] $$ is square integrable. (Observe that $\langle X\rangle_t$ is non-random because $X$ has stationary independent increments.)

I do not know of a more direct argument.