I have a Bayesian stat question where I am a little confused.
Let $N_{1}$ and $N_{2}$ independent having $\mbox{Poisson}(\Lambda)$ distribution and $\Lambda \sim \mbox{Gamma}(\alpha,\theta)$. Then how we prove that $N_{1} + N_{2} \sim \mbox{Poisson}(2 \Lambda)$ because it could be the case that $N_{1} \sim \mbox{Poisson}(2)$ and $N_{2} \sim \mbox{Poisson}(17)$?
Thank you!
To avoid ambiguity, it would be preferable to write the given as
$$N_i|\Lambda=\lambda \quad \overset{iid}{\sim } Poiss(\lambda),$$
and the claim you are trying to show as
$$N_1+N_2|\Lambda=\lambda \quad \overset{iid}{\sim } Poiss(2\lambda).$$
Then the claim is easily shown using characteristic functions:
$$\varphi_{N_1+N_2|\Lambda =\lambda}(t)=\varphi_{N_1|\Lambda =\lambda}(t)\varphi_{N_2|\Lambda =\lambda}(t)=(\exp(\lambda (e^{it}-1)))^2=\exp(2\lambda (e^{it}-1)).$$
The distribution of $\Lambda$ is irrelevant.