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The Levy-Khintchine formula gives a triple $(a,\sigma,\nu)$ for the characteristic exponent $\Psi(s)$ of an infinitely divisible random variable where

$\Psi(s)=ias + \frac{1}{2}\sigma^2s^2 + \int_{\mathbb{R}}(1-e^{isx} + is\mathbb{1}_{|x|<1})d\nu(x)$

My question is whether $\nu$ is unique? Or whether it is only the unique Levy measure?

Are there references in the literature to this fact.

Thanks

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Yes, there is a one-to-one correspondance between infinitely divisble distributions and the triple $(a,\sigma,\nu)$ given by the Lévy-Khintchine formula. You can have a look at Lévy Processes and Infinitely Divisible Distributions by Ken Iti Sato (Theorem 8.1).

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  • $\begingroup$ Thanks you. Is it possible that $\nu(x)$ is decreasing for certain values of x? Obviously this would imply that the random variable is not infinitely divisible. $\endgroup$ – Fantastic Mr. Fox Apr 24 '12 at 13:33
  • $\begingroup$ I don't understand what you're asking, since $\nu$ is a measure. What do you mean by decreasing? $\endgroup$ – Stefan Hansen Apr 24 '12 at 14:07
  • $\begingroup$ Sorry if $\nu(x)$ is negative for certain x? $\endgroup$ – Fantastic Mr. Fox Apr 24 '12 at 15:22
  • $\begingroup$ No, it's a measure, i.e. $\nu(A)\geq 0$ for every Borel set $A$. $\endgroup$ – Stefan Hansen Apr 25 '12 at 6:30
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    $\begingroup$ I don't follow what you are asking. In the Levy-Khintchine setup $\nu$ is a measure and $d\nu(x)$ denotes integration with respect to the measure $\nu$. Writing $\nu(x)$ does not make any sense in the setup, because $\nu$ is function on the Borel sets of $\mathbb{R}$. Nevertheless, I have no idea whether this generalizes to $\nu$ being a signed measure (i.e. $\nu$ is allowed to be negative). $\endgroup$ – Stefan Hansen Apr 30 '12 at 19:05

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