# Ordinary-Stochastic Levy Triplet Relation for Tempered Stable Process

I'm trying to verify a result in the paper of https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1700305 with regards to the tempered stable or CGMY process in option pricing literature, which has Levy measure in ordinary exponential form as

$\nu_L=\frac{Ce^{-Mx}}{x^{Y+1}}\chi_{\{x>0\}}dx+\frac{Ce^{-G|x|}}{|x|^{Y+1}}\chi_{\{x<0\}}dx$

Considering the theorem for the relation between the triplet of ordinary and stochastic exponentials,

If $L$ is a Levy process with characteristic triple $(\mu,\sigma^2,\nu$, then $\mathcal{E}_L(t)=e^{L_1(t)}$ where $L_1$ is a L\'evy process with characteristic triplet $(\mu_1,\sigma_1^2,\nu_1)$, given by

$\nu_1 = \nu\circ f^{-1},\quad f(x)=\log(1+x),$

$\mu_1=\mu-\frac{1}{2}\sigma^2+\int_{\mathbb{R}-\{0\}}[\log(1+x)\chi_{(-1,1)}(\log(1+x))-x\chi_{(-1,1)}(x)]\nu(dx),$

$\sigma_1 = \sigma$

Conversely, there exists a Levy process $L_2$ with characteristic triplet $(\mu_2,\sigma_2^2,\nu_2)$ such that there exists $e^{L_t}=\mathcal{E}_{L_2}(t)$, where

$\nu_2 = \nu\circ g^{-1},\quad g(x)=e^x-1,$

$\mu_2 = \mu+\frac{1}{2}\sigma^2+\int_{\mathbb{R}-\{0\}}[(e^x-1)\chi_{(-1,1)}(e^x-1)-x\chi_{(-1,1)}(x)]\nu(dx),$

$\sigma_2 = \sigma$

The paper gives the Levy measure under the second scenario (that is, the composite function of the ordinary exponential Levy measure with $log(1+x)$), as

$\nu = \frac{C(1+x)^{-M-1}}{log(1+x)^{Y+1}}\chi_{\{x>0\}}dx+\frac{C(1+x)^{G-1}}{-log(1+x)^{Y+1}}\chi_{\{-1<x<0\}}dx$

I've been looking at this for days and can't for the life of me figure out why the exponential tempering factors are now subtracting 1 under the new representation. Can someone clarify this for me? Thank you!