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We know that a random vector $X$ is infinitely divisible (ID) if for all $n \in \mathbb N$, there exist $X_1^n,..., X_{n}^n$ i.i.d. random vectors such that: \begin{equation} X = X_1^n + ...+ X_n^n. \end{equation} On the other hand, every ID random vector is such that the characteristic function has a Lévy–Khintchine representation: \begin{equation}\label{aa}\tag{I} \ln\varphi_X(s)= i b_c^\intercal s - \frac{s ^\intercal \sigma s}{2} + \int_{\mathbb R^p} \left[e^{i s^\intercal x } - 1 - i s^\intercal x c(x) \right] dm(x) . \end{equation} where $c(x)$ is an integrable truncation function, $b_c$ is a drift term, $\sigma$ is a non-negative definite matrix and $m$ is a Lévy measure, i.e.: \begin{equation}\label{bbb} m(\{0\})=0,\quad \int_{|x|<1}|x|^2 m (dx) <\infty, \quad \int_{|x| \geq 1} m (dx)<\infty. \end{equation} Lévy measures have some properties:

  1. $m((-\epsilon,\epsilon)^c)< \infty$ for all $\epsilon >0$;
  2. $m$ may have infinite mass around zero: $m((-\epsilon,\epsilon))=\infty$

There is an interpretation of a Lévy measure in the context of Lévy process: let $(X_t)_{t \geq 0}$ be a Lévy process. Each Lévy process is uniquely determined by the Lévy–Khintchine triplet $(b_c, \sigma, m)$ such that \begin{equation}\label{c}\tag{II} \ln\varphi_{X_t}(s)= t \left[ i b_c s - \frac{s ^\intercal \sigma s}{2} + \int_{\mathbb R^p} \left[e^{i s^\intercal x } - 1 - i s^\intercal x c(x) \right] dm(x) \right]. \end{equation} The converse is also true, for any triplet $(b_c, \sigma, m)$, there is a Levy process $(X_t)_{t \geq 0}$ such that the characteristic function of $X_t$ is given by (\ref{c})

In this context, the Lévy measure $m$ can be interpreted as the distribution of the jumps of the Lévy process $(X_t)_{t \geq 0}$. See this question for more details.

Question

Given $X$ a ID random vector (not stochastic process) with characteristic function determined by the triplet $(b_c, \sigma, m)$ as in (\ref{aa}). So, what is the interpretation of the Lévy measure $m$?

For example, what properties does $X$ have if $m((-\epsilon,\epsilon))=\infty$?

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    $\begingroup$ For any infinitely divisible random vector $X$, you can always have a Lévy process $(Y_t)_{t\ge0}$ such that $Y_1$ equals $X$ in distribution. So, the Lévy measure for the distribution of $X$ is the same as that for the corresponding Lévy process $(Y_t)_{t\ge0}$. Hence, the desired interpretation. $\endgroup$ Commented Feb 12, 2023 at 12:56
  • $\begingroup$ Yes, it is an interpretation, but this interpretation tells me more about the constructed Lévy process and nothing about the random variable $X$. $\endgroup$
    – PSE
    Commented Feb 12, 2023 at 16:15

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