$\eta$ is non-deterministic.

1.  If $\sigma_\xi=0$, $\gamma_0>0$ and $\nu_\xi((0,\infty))=0$, then a density $f(t)$, $t\geq 0$, of $\mu=\Phi_\xi({\mathcal{L}}(\eta_1))$ exists, which is continuous on ${\mathbb{R}}_+\setminus \{\frac{a_\eta}{\gamma_0} \}$, and fulfills $$\begin{aligned}
    \label{eq:densityboundedbelow}
    f(t)&=&0, \quad t<\frac{a_\eta}{\gamma_0}, \\ 
    (a_\eta - \gamma_0  t) f(t) 
    & = & - \int_{\frac{a_\eta}{\gamma_0}}^t \left(\nu_\xi((-\infty, \log\frac{s}{t}))  + \nu_\eta((t-s, \infty)) \right) f(s) ds, \quad t\geq \frac{a_\eta}{\gamma_0}. \nonumber\end{aligned}$$

2.  If $\sigma_\xi=0$, $\gamma_0>0$, $\nu_\xi((0,\infty))>0$, $\nu_\xi((-\infty,0))=0$ and $\nu_\eta\equiv 0$, then a density $f(t)$, $t\geq 0$, of $\mu=\Phi_\xi({\mathcal{L}}(\eta_1))$ exists, which is continuous on ${\mathbb{R}}_+\setminus \{\frac{a_\eta}{\gamma_0} \}$, and fulfills $$\begin{aligned}
    \label{eq:densityboundedabove}
    f(t)&=&0, \quad t>\frac{a_\eta}{\gamma_0}, \\ 
    (a_\eta - \gamma_0  t) f(t) 
    & = & \int_t^{\frac{a_\eta}{\gamma_0}} \nu_\xi((\log\frac{s}{t}, \infty)) f(s) ds, \quad t\leq \frac{a_\eta}{\gamma_0}. \nonumber\end{aligned}$$

3.  Otherwise, assume that $\mu=\Phi_\xi({\mathcal{L}}(\eta_1))$ is absolutely continuous (with differentiable density $f(t)$, $t\geq 0$, such that $\lim_{t\to 0} t^2 f(t)=0$ if $\sigma_\xi>0$), then $f$ fulfills $\lambda$-a.e. (with $\lambda$ the Lebesgue measure) $$\begin{aligned}
    \label{eq:density}
     \lefteqn{a_\eta f(t)- \left(\gamma_0+\frac{\sigma_\xi^2}{2} \right) t f(t) - \frac{\sigma_\xi^2}{2} t^2 f'(t) }\\
    & = &\int_t^\infty \nu_\xi((\log\frac{s}{t}, \infty)) f(s) ds - \int_0^t \left(\nu_\xi((-\infty, \log\frac{s}{t})) + \nu_\eta((t-s, \infty)) \right) f(s) ds, \quad t\geq 0. \nonumber\end{aligned}$$

Conversely, if $f(t)$, $t\geq 0$, is a probability density which fulfills , or $\lambda$-a.e. for some Lévy characteristics $\gamma_0, \sigma^2_\xi, \nu_\xi, a_\eta$ and $\nu_\eta$, then it is a density of 