Lévy-Khintchine formula for Cauchy distribution

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?

• You have forgotten to exponentiate the entire right hand side. – Alex R. Jan 20 '13 at 1:52
• Thanks Alex. I've made a correction. – Digital Gal Jan 20 '13 at 1:58
• Lévy with an é. – 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, $$a=\sigma^2=0,\qquad\mathrm d\nu(x)=|x|^{-2}\mathbf 1_{x\ne0}\mathrm dx.$$