A measure $\nu$ is a Levy measure if:
\begin{equation}
\nu(\{0\}),\quad \int_{|x|<1}|x|^{2} \nu (dx) <\infty, \quad \int_{|x| \geq 1} \nu (dx)<\infty.
\end{equation}
At first, I thought that necessarily for $\epsilon >0$:
\begin{equation}\label{prefinite_condition_measure0}\tag{I}
\nu( (-\epsilon, \epsilon) ) = \infty
\end{equation}
But this is not true.
For example, by the Levy-Khintchine representation of infinitely divisible distributions $X$ , we have that $X\sim (b, \sigma, \nu)$, where $\nu$ is a Levy measure. If $X$ is a Compound Poisson random variable, $X \sim CP(\lambda, dF)$
$$X = \sum_{j=1}^N Z_j, \quad N \sim \hbox{Poisson}(\lambda), \quad Z \sim dF$$
we have that $X \sim (0,0, \lambda dF)$. So, certainly we have $d\nu = \lambda dF$ with:
$$\nu( (-\epsilon, \epsilon) ) < \infty, \quad \epsilon < 1 .$$
So I am looking for some example of Levy measure that satisfy (\ref{prefinite_condition_measure0}).
Could you give me some example?
(We certainly cannot build on an example coming from a Compound Poisson distribution)
Best Answer
$d\nu (x)=\frac 1 {x^{\alpha}} \chi_{(0,\infty)}dx$ with $ 1<\alpha <3$ is such an example.