\nu_\eta((t-s, \infty)) f(s) ds$$ which is due to the fact that convolutions become multiplications under the Laplace transform. Now, putting all terms together we easily derive .\
Observe that in the setting of case (i), it follows from [@BLMranges Lemma 1 and Thm. 1] that the measure $\mu$ has support $[\frac{a_\eta}{\gamma_0}, \infty)$. Further recall that in this case $\xi$ is a spectrally negative process $\xi$ and hence $\mu$ is selfdecomposable and has a continuous density on $(\frac{a_\eta}{\gamma_0}, \infty)$ (cf. [@steutelvanharn Thm. V.2.16]).\
By [@BLMranges Lemma 1 and Thm. 1] in the setting of case (ii) $\mu$ has support $[0,\frac{a_\eta}{\gamma_0}]$ and otherwise $\mu$ has full support on $[0,\infty)$. Hence we derive the corresponding formulas from . Existence of a density in case (ii) follows from [@bertoinlindnermaller08 Thm. 3.9], continuity has been proven in [@CarmonaPetitYor].\
For the converse assume that $f$ is a density which fulfills , then reverting the above we see that its Laplace transform fulfills which yields the claim by [@BLMranges Thm. 3].

\[rem:lindnersato\] In [@lindnersato09] the exponential functional $V$ as in has been studied in the case where $(\eta_t)_{t\geq 0}$ is a Poisson process with jump intensity $v>0$, and $\xi_t= (\log c) N_t$ for $c>1$ and another (independent) 