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A Borel measure $\nu$ on $\mathbb{R}$ is called a Lévy measure if $\nu({0})=0$ and $\int_\mathbb{R}(1\wedge|x^2|) \, \nu(dx) < \infty .$ (https://en.wikipedia.org/wiki/Financial_models_with_long-tailed_distributions_and_volatility_clustering#Infinitely_divisible_distributions)

So, what exactly is $(1\wedge|x^2|)$? (Or rather correctly, what is the definition of levy measure saying?)

Edit: OK, but then why is levy measure needed?

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    What do you mean by "why is it needed"? The Lévy-Khintchine decomposition provides a way of characterizing any Lévy process in terms of three components (the Lévy triplet) - one of which is a measure called the Lévy measure. – Stefan Hansen Mar 07 '13 at 08:20
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    Levy measure describes the distribution of the jumps of the process. For example, a Poisson process of parameter $c>0$ has the Levy measure $c\delta(x-1)$, implying that the jump of size 1 occurs with intensity $c$. (In general, the jump part of a Levy process is a compensated sum of the Poisson point process with characteristic measure as the Levy measure.) – Sangchul Lee Mar 07 '13 at 08:30
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    @SangchulLee You mentioned that the levy measure of poisson process with rate c has levy measure $c\delta(x-1)$, could you please show the calculation explicitly? – math101 Feb 25 '16 at 11:04
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    @math101, Following the definition of Lévy measure as in the Lévy-Kintchine formula, you can easily check that $\nu(x) = \lambda \delta(x-1)$ gives $$ \phi_{X_1}(\xi) = \mathrm{e}^{\lambda (e^{i \xi} - 1)}, $$ which is the characteristic function of the Poisson process. – Sangchul Lee Feb 26 '16 at 00:15
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    yes. To match $c(e^{i\xi}-1)$ with $\int(e^{i\xi x}-1)\nu(dx)$. The first part, $e^{i\xi}=\int e^{i\xi x}\delta_1(dx)=\int e^{i\xi x} d\delta_1=e^{i\xi 1}=e^{i\xi}$. The 2nd part, we have $1=\int\delta_1(dx).$ – math101 Feb 26 '16 at 08:26

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$$1\wedge|x^2|=\begin{cases}1 & \text{if} & |x|\gt1, \\ x^2 & \text{if} & |x|\leqslant1. \end{cases}$$

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