Poisson process $(N_t)_{t\geq 0}$ with jump intensity $u>0$.\
From Theorem \[thm:density\] above, we observe that in this setting, if a density of $V$ exists, then it fulfills $\lambda$-a.e. $$v\int_{(t-1)\vee 0}^t f(s) ds = u \int_t^{ct} f(s) ds, \quad t\geq 0$$ or in terms of the cumulative distribution function $F(t)=\int_0^t f(s) ds$ and the parameter $q=\frac{v}{u+v}\in (0,1)$ $$\label{eq:selfsimcumulant}
 F(t)= (1-q) F(ct) + q F(t-1), \quad t>0, \quad \text{ where } F(t)=0, \quad  t\leq 0.$$ Actually, can be shown to hold even if $\mu={\mathcal{L}}(V)$ is not absolutely continuous, by a similar proof as for Theorem \[thm:density\]. Further, from we deduce the self-similarity relation $$\mu=(1-q)\, \mu\circ T_0^{-1} + q\, \mu \circ T_1^{-1}$$ for $\mu$ with weights $\{1-q,q\}$ and $$T_0:x\mapsto \frac{x}{c}, \quad T_1:x\mapsto x+1.$$ Remark that $T_1$ is not a contraction and hence $\mu$ is not a self-similar measure in the classical and well-studied sense of [@Hutchinson].\
Nevertheless, in [@lindnersato09], the authors proved that $\mu$ shares some properties with self-similar measures. In particular, $\mu$ is continuous singular if $c$ is a Pisot-Vijayaraghavan number, but for Lebesgue a.a. $c>1$ there exists $\bar{q}<1$ such that $\mu$ is absolutely continuous for all $q\in (\bar{q},1)$.

From the theorem above, we can derive characterizations of densities of selfdecomposable distributions 