The Lévy-Khintchine formula for the log of the characteristic function of an infinitely divisible random variable is

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

for some $(a,\sigma,\nu)$.

The Cauchy distribution has characteristic function $\exp(-|s|)$.

How can this be rewritten in the form of the Lévy-Khintchine formula?

  • $\begingroup$ You have forgotten to exponentiate the entire right hand side. $\endgroup$ – Alex R. Jan 20 '13 at 1:52
  • $\begingroup$ Thanks Alex. I've made a correction. $\endgroup$ – Digital Gal Jan 20 '13 at 1:58
  • $\begingroup$ Lévy with an é. $\endgroup$ – Did Jan 20 '13 at 11:04

Depending on how you define $\Psi$ (something that you should make clear), your general formula is or is not flawed. For the choice which corresponds to $$ \Psi(s)=\mathrm ias - \frac{1}{2}\sigma^2s^2 + \int_{\mathbb{R}}(\mathrm e^{\mathrm isx}-1-\mathrm isx\mathbb{1}_{|x|<1})\mathrm d\nu(x), $$ and for the standard Cauchy distribution, $$ a=\sigma^2=0,\qquad\mathrm d\nu(x)=|x|^{-2}\mathbf 1_{x\ne0}\mathrm dx. $$

  • $\begingroup$ thanks im able to redervive that,although i had to use the fact that there is a density for a cauchy rv. $\endgroup$ – Digital Gal Jan 21 '13 at 3:18

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