$P(q|N)=\frac{e^{-\eta N}\cdot (\eta N)^q}{q!}$ complies with conditional Poisson distribution where $\eta$ is constant parameter and $P(N)\sim \mathrm{Poisson}(\lambda)$. I'm trying to prove $P(q)=\sum_{N=0}^{\infty}P(q|N)P(N)\sim \mathrm{Poisson}(\eta \lambda)$.
After simple calculation this problem reduces to the proof of equation $\sum_{N=0}^{\infty}\frac{e^{-\eta N}\cdot (\eta N)^q \lambda^N}{N!}=e^{-\eta \lambda}\cdot (\eta \lambda)^q\cdot e^\lambda$.
Can anyone help me to prove or disprove it?
Thank you very much!
We have \begin{align*} \sum_{N = 0}^{\infty} \frac{\left( e^{-\eta} \lambda \right)^{N}}{N!} = \exp\!\left( \exp\!\left( -\eta\right) \lambda \right) \end{align*} which we can differentiate $q$ times with respect to $\eta$ since it's a power series (uniform convergence) to obtain \begin{align*} \sum_{N = 0}^{\infty } \frac{1}{N!}{\frac{\partial^{q} }{\partial \eta^{q}} \left( e^{-\eta N}\right)\lambda ^{N}} = \sum_{N = 0}^{\infty}\frac{1}{N!} {e^{-\eta N}N^{q}( -\eta)^{q}\lambda ^{N}} = \frac{\partial ^{q}}{\partial \eta^{q}} \exp\!\left( \exp\!\left( -\eta\right) \lambda \right) .\end{align*} Does this help answering your question?