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A Lévy measure $\nu$ on $\mathbb R^{d}$ is a measure satisfying $$\nu\{0\} = 0, \ \int_{\mathbb R^{d}} (|y|^{2}\wedge 1) \nu(dy) <\infty.$$

A Lévy process can be characterized by triples $(b, A, \nu)$ by Lévy-Itô decomposition, then $$X_{t} = bt + W_{A}(t) + \int_{B_{1}} x \tilde N(t, dx) + \int_{B_{1}^{c}} x N(t, dx)$$ where $N(t, B)$ is a Poisson measure with $\mathbb E N(1, B) = \nu(B)$ for a set $B$ bounded below, and $\tilde N(t, dx) = N(t, dx) - t \nu(dx)$ is its compensated one.

[Q.] If $(0, 0, \nu)$ is a triplet of a Lévy process $X$ whose first moment is finite, is the following always true? $$ \lim_{r\to 0^{+}}\int_{B_{1}\setminus B_{r}^{c}} x \nu(dx) < \infty.$$ Moreover, if $\nu(B_1^c) = 0$, then $$\mathbb E X_1 = \lim_{r\to 0^{+}}\int_{B_{1}\setminus B_{r}^{c}} x \nu(dx).$$ END.

Remark: If $\nu(dx) = x^{-2} dx$, then it corresponds to 1-stable process, and $ \lim_{r\to 0^{+}}\int_{B_{1}\setminus B_{r}^{c}} x \nu(dx) = 0$, while $ \int_{B_{1}} x \nu(dx) $ is not well-defined. [Q1.] Is there always a Lévy process corresponding to $(0, 0, \nu)$ for an arbitrary Lévy measure $\nu$?

Remark: Consider $\nu(dx) = x^{-2} I(x>0) dx$, it is a Lévy measure. But if there was an associated process $X_{t}$, then $\mathbb E[X_{1}] = \int_{0}^{\infty} x \nu(dx) = \infty$.

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  • $\begingroup$ For every Levy triple there exists a corresponding Levy process, see Chapter 2 of Kyprianou's book "Fluctuations of Lévy Processes with Applications: Introductory Lectures" or page 13 of Bertoin's book "Levy processes". Levy process $X_t$ might have infinite mean $\mathbf E[X_t]=\infty$ $\endgroup$ Commented Jun 10, 2016 at 19:21
  • $\begingroup$ @DenisDenisov Thanks, I adjusted the question accordingly. $\endgroup$
    – kenneth
    Commented Jun 11, 2016 at 0:29

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No, for a counterexample, just take $$ \nu(dx) = \left(C_1 x^{-2} I(0<x<1) +C_2 |x|^{-2} I(-1<x<0)\right)dx, $$ where $C_1\neq C_2$. Then, $$ \int_{r<|x|\le 1} x\nu(dx) = (C_2-C_1) \int_r^1 x^{-1} dx \to \infty,\quad r\to 0. $$ As Levy measure is $0$ for $x:|x|> 1$, all moments of the Lévy process $X_t$ are finite. The existence of the Lévy process with this Lévy measure is ensured by the references I gave earlier.

Your question seems to be connected to Lévy processes of bounded variation. The following is Lemma 2.21 in the second edition of Kyprianou's book cited above: A Lévy process with Lévy–Khintchine exponent corresponding to the triple $(a,\sigma,\nu)$ has paths of bounded variation if and only if $$ \sigma=0 \mbox{ and } \int_R (1\wedge |x|) \nu(dx)<\infty. $$

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  • $\begingroup$ Thanks, it corrects my original misunderstanding on Levy process. $\endgroup$
    – kenneth
    Commented Jun 11, 2016 at 15:34
  • $\begingroup$ Could you please give a reference to the fact that if the Levy measure is 0 for $|x|<1$ then moments of all order exists $\endgroup$
    – L--
    Commented Dec 12, 2023 at 17:03
  • $\begingroup$ @L-- For example, Sato, 'Lévy Processes and Infinitely Divisible Distributions', Chapter 26, Theorem 26.1 $\endgroup$ Commented Dec 13, 2023 at 9:07
  • $\begingroup$ I am quite new to this topic. But I think there is a typo. Shouldn't it be $|x|>1$ instead? $\endgroup$
    – L--
    Commented Dec 14, 2023 at 11:26
  • $\begingroup$ Yes, this was the typo, the answer should have had that $x: |x|>1$. I have corrected it now, thanks for pointing it out. $\endgroup$ Commented Dec 14, 2023 at 14:27
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Let us consider the one-dimensional case. By the Lévy-Khintchine formula, we have$$E(e^{i\theta X_1})= e^{-\phi(\theta)},$$ where $$\phi(\theta)=\int_{R}(1-e^{i\theta x}+i\theta x 1_{|x|<1})\nu(dx).$$ Then $$E(iX_1e^{i\theta X_1})= e^{-\phi(\theta)}\int_{R}(ixe^{i\theta x}-i x 1_{|x|<1})\nu(dx).$$ Letting $\theta\to 0$ we see $$E(X_1)= \int_{R}(x-x 1_{|x|<1})\nu(dx)= \int_{|x|\geq 1}x \nu(dx).$$ Then we see that the expectation is only determined by the “big jumps term” (provided there is neither a drift term nor a Brownian motion term).

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