Consider a Poisson process $N$ and an independent Gamma process $\Gamma$, with parameters $\gamma$ and $\lambda$.

Given $t>0$, the distribution of $N_{\Gamma_t}$ is a mixture of Poisson distributions whose parameter is randomly chosen according to a Gamma distribution.

The processes $\Gamma$ and $N$ are subordinators without drift, so $$\Gamma_t = \sum_{0 \le s \le t}\Delta\Gamma_s \text{ and } N_{\Gamma_t} = \sum_{0 \le s \le t}(N_{\Gamma_s}-N_{\Gamma_{s-}}).$$ Moreover, the process $(\Delta\Gamma_t)_{t \ge 0}$ is a Poisson point process with values in $[0,+\infty[$ and with intensity $\nu$ on $]0,+\infty[$ equal to the Levy measure of the subordinator $\Gamma$: $$\mathrm{d}\nu(x) = \gamma x^{-1}\exp(-\lambda x).$$

Conditionally on the process $\Gamma$, the random variables $(N_{\Gamma_s}-N_{\Gamma_{s-}})_{s>0}$ are independent and for each $s>0$, the distribution of $N_{\Gamma_s}-N_{\Gamma_{s-}}$ is Poisson with parameter $\Delta\Gamma_s$. Thus $$P[N_{\Gamma_s}-N_{\Gamma_{s-}}>0|\Gamma] = 1-e^{-\Delta\Gamma_s},$$ and for each $n \ge 1$, $$P[N_{\Gamma_s}-N_{\Gamma_{s-}}=n|\Gamma] = \frac{s^n}{n!}e^{-\Delta\Gamma_s}.$$

Therefore, the jump-times of the process $N_{\Gamma_\cdot}$ are given by a Poisson process $J$ with intensity $$\int_0^\infty(1-e^{-x})\mathrm{d}\nu(x) = \gamma\int_0^\infty(e^{-\lambda x}-e^{-(\lambda+1)x})\frac{\mathrm{d}x}{x} = \gamma \ln\frac{\lambda+1}{\lambda}.$$ For each $n \ge 1$, the times $s$ such that $N_{\Gamma_s}-N_{\Gamma_{s-}}=n$ are given by a Poisson process $J^{(n)}$ with intensity $$\int_0^\infty \frac{x^n}{n!}e^{-x}\mathrm{d}\nu(x) = \gamma\int_0^\infty\frac{x^{n-1}}{n!}e^{-(\lambda+1)x}\mathrm{d}x = \frac{\gamma}{n}(\lambda+1)^{-n}.$$ And since the Poisson process $J$ is the superposition of the independent Poisson processes $(J^{(n)})_{n \ge 1}$, the size of the successive jumps of the process $N_{\Gamma_\cdot}$ form an i.i.d. sequence, independent of $\Gamma$, whose distribution is given by $$p(n) := \frac{\frac{\gamma}{n}(\lambda+1)^{-n}}{\gamma \ln\frac{\lambda+1}{\lambda}} = \frac{q^n/n}{-\ln(1-q)} \text{ with } q = (\lambda+1)^{-1}.$$

Consider a Poisson process $N$ and an independent Gamma process $\Gamma$, with parameters $\gamma$ and $\lambda$.

Given $t>0$, the distribution of $N_{\Gamma_t}$ is a mixture of Poisson distributions whose parameter is randomly chosen according to a Gamma distribution.

The processes $\Gamma$ and $N$ are subordinators without drift, so $$\Gamma_t = \sum_{0 \le s \le t}\Delta\Gamma_s \text{ and } N_{\Gamma_t} = \sum_{0 \le s \le t}(N_{\Gamma_s}-N_{\Gamma_{s-}}).$$ Moreover, the process $(\Delta\Gamma_t)_{t \ge 0}$ is a Poisson point process with values in $[0,+\infty[$ and with intensity $\nu$ on $]0,+\infty[$ equal to the Levy measure of the subordinator $\Gamma$: $$\mathrm{d}\nu(x) = \gamma x^{-1}\exp(-\lambda x).$$

Conditionally on the process $\Gamma$, the random variables $(N_{\Gamma_s}-N_{\Gamma_{s-}})_{s>0}$ are independent and for each $s>0$, the distribution of $N_{\Gamma_s}-N_{\Gamma_{s-}}$ is Poisson with parameter $\Delta\Gamma_s$. Thus $$P[N_{\Gamma_s}-N_{\Gamma_{s-}}>0\mid\Gamma] = 1-e^{-\Delta\Gamma_s},$$ and for each $n \ge 1$, $$P[N_{\Gamma_s}-N_{\Gamma_{s-}}=n\mid\Gamma] = \frac{s^n}{n!}e^{-\Delta\Gamma_s}.$$

Therefore, the jump-times of the process $N_{\Gamma_\cdot}$ are given by a Poisson process $J$ with intensity $$\int_0^\infty(1-e^{-x}) \, \mathrm{d}\nu(x) = \gamma\int_0^\infty(e^{-\lambda x}-e^{-(\lambda+1)x}) \, \frac{\mathrm{d}x}{x} = \gamma \ln\frac{\lambda+1}{\lambda}.$$ For each $n \ge 1$, the times $s$ such that $N_{\Gamma_s}-N_{\Gamma_{s-}}=n$ are given by a Poisson process $J^{(n)}$ with intensity $$\int_0^\infty \frac{x^n}{n!}e^{-x} \, \mathrm{d}\nu(x) = \gamma\int_0^\infty\frac{x^{n-1}}{n!}e^{-(\lambda+1)x} \, \mathrm{d}x = \frac{\gamma}{n}(\lambda+1)^{-n}.$$ And since the Poisson process $J$ is the superposition of the independent Poisson processes $(J^{(n)})_{n \ge 1}$, the size of the successive jumps of the process $N_{\Gamma_\cdot}$ form an i.i.d. sequence, independent of $\Gamma$, whose distribution is given by $$p(n) := \frac{\frac{\gamma}{n}(\lambda+1)^{-n}}{\gamma \ln\frac{\lambda+1}{\lambda}} = \frac{q^n/n}{-\ln(1-q)} \text{ with } q = (\lambda+1)^{-1}.$$