For a Levy process $(X_t)_{t\geq 0}$, we have $\mathbb{E}[X_t]=t\mathbb{E}[X_t^1]$ and $\text{Var}(X_t)=t\text{Var}(X_t^1)$. Does the same hold for the first absolute moment, i.e. does $\mathbb{E}[|X_t|]=O(t)$ hold? Assume there is no continuous part, and that the process has bounded jumps (the absolute moment of the compound Poisson part is O(t)), so: \begin{align*} X_t = \lim_{\epsilon\to 0}\int_{\epsilon<x<1,s\leq t}x\bar J(ds,dx) \end{align*} where $\bar J(ds,dx)=J(ds,dx)-\nu(dx)ds$ is the compensated jump measure.


For a symmetric $\alpha$-stable process $(X_t)_{t \geq 0}$, the scaling property $X_t \sim t^{1/\alpha} X_1$ implies

$$\mathbb{E}(|X_t|) = t^{1/\alpha} \mathbb{E}(|X_1|)$$

for any $\alpha>1$. This shows that we cannot expect $\mathbb{E}(|X_t|) = O(t)$ for any Lévy process $(X_t)_{t \geq 0}$. Even if we assume additionally that the process has no drift and diffusion part and that the process has bounded jumps, this is, in general, not true (see the counterexample below). There is the following statement on fractional moments:

Theorem: Let $(X_t)_{t \geq 0}$ be a Lévy process with Lévy triplet $(0,0,\nu)$. Suppose that there exist $\alpha,\beta>0$ such that $$\int_{|y| \leq 1} |y|^{\alpha} \, \nu(dy) <\infty \quad \text{and} \quad \int_{|y|>1} |y|^{\beta} \, \nu(dy)<\infty.$$ Then $$\mathbb{E}(|X_t|^{\kappa}) \leq C t^{\min\{\kappa/\alpha,1\}} \qquad \text{for all $t \leq 1$}$$ where $\kappa \in (0,\beta)$, $\kappa \neq \alpha$.

Roughly speaking, the behaviour of the Lévy measure $\nu$ at $\infty$ gives the existence of (fractional) moments whereas the behaviour of $\nu$ at $0$ determines the asymptotic of the fractional moments (for small $t$).

So, for a Lévy process $(X_t)_{t \geq 0}$ with bounded jumps (without drift and diffusion part), we have $\mathbb{E}(|X_t|) = O(t)$ as $t \to 0$ if $\int_{|y| \leq 1} |y|^{\alpha} \, \nu(dy)<\infty$ for some $\alpha \in (0,1)$.

The following lemma gives an example for a Lévy process with bounded jumps which does not satisfy $\mathbb{E}(|X_t|) = O(t)$.

Example: Let $(X_t)_{t \geq 0}$ be a truncated $\alpha$-stable process for some $\alpha \in (1,2)$, i.e. a Lévy process with Lévy measure $$\nu(dy) = \frac{1}{|y|^{1+\alpha}} 1_{(0,1)}(|y|) \, dy.$$ Then $\mathbb{E}(|X_t|) = O(t)$ does not hold.

Proof: Let $(Y_t)_{t \geq 0}$ be an independent Lévy process with Lévy measure $\frac{1}{|y|^{1+\alpha}} 1_{[1,\infty)}(|y|) \, dy$. Then $Z_t := X_t + Y_t$ defines a symmetric $\alpha$-stable process. By the scaling property, we have

$$\mathbb{E}(|Z_t|) = t^{1/\alpha} \mathbb{E}(|Z_1|).$$

Moreover, by the above theorem, we find

$$\mathbb{E}(|Y_t|) \leq C t$$

for all $t \leq 1$ for some constant $C>0$. Consequently, by the reverse triangle inequality,

$$\begin{align*} \mathbb{E}(|X_t|) &= \mathbb{E}(|Z_t-Y_t|) \geq \mathbb{E}(|Z_t|) - \mathbb{E}(|Y_t|) \\ &\geq t^{1/\alpha} \mathbb{E}(|Z_1|) - C t. \end{align*}$$

As $\alpha>1$ this clearly shows that $\mathbb{E}(|X_t|) = O(t)$ does not hold true for small $t$.

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