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I'm writing a survey that involves Levy processes and wanted to mention the different forms of the Levy-Khintchine formula found in literature.

The most common version seems to give the Levy symbol as

$ \Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - i\langle u,y \rangle\mathbf{1}_{|y|\le1}{)}\, dK(y) $

while in other versions it seems to be given as

$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - \frac{ i\langle u,y \rangle}{1+|y|^2}{)} \, dK(y)$ $\Psi$

while here http://almostsure.wordpress.com/2010/09/15/processes-with-independent-increments/ it is given as

$\Psi(u) = i\langle b,u \rangle - \frac{1}{2} \langle u,\Sigma u\rangle + \int_{\mathbb{R}^d} {(} e^{i\langle u,y \rangle}-1 - \frac{ i\langle u,y \rangle}{1+|y|}{)} \, dK(y).$

Are all of these correct and equivalent? If the last one is, does anyone know a published source I could cite that mentions it?

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  • $\begingroup$ I see you linked my blog...I had considered changing the formula I use there to something more common. It really doesn't matter as long as the correction term is bounded and cancels out the zeroth and first order terms of the integrand at $y=0$. They all equate to the same thing as long as you adjust $b$ appropriately. $\endgroup$
    – George Lowther
    Commented Apr 8, 2012 at 13:26
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    $\begingroup$ Crossposted to math.SE where Byron Schmuland provides a reference To Stroock's text making the same point as George: math.stackexchange.com/questions/129293 $\endgroup$
    – cardinal
    Commented Apr 8, 2012 at 14:22
  • $\begingroup$ As it is already asked (and answered) elsewhere, I voted to close. It is probably a better fit at math.stackexchange than here anyway. $\endgroup$
    – George Lowther
    Commented Apr 8, 2012 at 14:25
  • $\begingroup$ I'm voting to close this question because it was asked and answered elsewhere $\endgroup$
    – Yemon Choi
    Commented Feb 12, 2019 at 1:05

1 Answer 1

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First formula and second formula are equivalent. Since $$\frac{{{{\left| y \right|}^2}}}{{1 + {{\left| y \right|}^2}}} \le {\left| y \right|^2} \wedge 1 \le \frac{{{{2\left| y \right|}^2}}}{{1 + {{\left| y \right|}^2}}}$$ See p29 of "Levy processes and stochastic calculus (2ed., CUP, 2009) Applebaum D." and p38 of "Lévy Processes and Infinitely Divisible Distributions[Ken-iti_Sato]".

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