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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

$\Psi(s)=ias + \frac{1}{2}\sigma^2s^2 + \int_{\mathbb{R}}(1-e^{isx} + is\mathbb{1}_{|x|&lt1})d\nu(x)$

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

SBF
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Fantastic Mr. Fox
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1 Answers1

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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).

Stefan Hansen
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  • 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. – Fantastic Mr. Fox Apr 24 '12 at 13:33
  • I don't understand what you're asking, since $\nu$ is a measure. What do you mean by decreasing? – Stefan Hansen Apr 24 '12 at 14:07
  • Sorry if $\nu(x)$ is negative for certain x? – Fantastic Mr. Fox Apr 24 '12 at 15:22
  • No, it's a measure, i.e. $\nu(A)\geq 0$ for every Borel set $A$. – Stefan Hansen Apr 25 '12 at 06:30
  • So it is possible that there could exist a function $\nu_2(x)$ which satisfies the Levy-Khintchine characteristic exponent as given in the original post but is not a measure i.e. $\nu_2(x)<0$ for some $x$? While a unique measure $\nu_1(x)\geq0$ $\forall x\geq0$ may also exist? – Fantastic Mr. Fox Apr 25 '12 at 14:02
  • Hello stefan, I am just wondering if you understand my question or if you are still there please? Thanks – Fantastic Mr. Fox Apr 30 '12 at 17:40
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    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). – Stefan Hansen Apr 30 '12 at 19:05
  • Thank you for your time Stefan. Yes I was curious whether $\nu$ could be a signed measure. – Fantastic Mr. Fox Apr 30 '12 at 20:30