Bayesian Statistics Question: Given $N$, $X$ is distributed as $\mathrm{B}(N,\theta)$. Derive the unconditional
distribution of X assuming N is distributed as $P(\lambda)$.
This is what I have tried so far:
$$x|N \sim \mathrm{B}(N,θ) \text{ so }\Pr(X=x|N) = \begin{pmatrix} N \\ x \end{pmatrix} \theta^x \,(1-\theta)^{N-x}$$ 
and also $$\Pr(N) = \exp(-\lambda)\dfrac{\lambda^N}{N!},$$  
so the joint distribution $\Pr(X=x,N=n)= \begin{pmatrix} N \\ x \end{pmatrix} \theta^x \,(1-\theta)^{N-x} \cdot  \exp(-\lambda)\dfrac{\lambda^N}{N!}$
then to find $Pr(X) = \sum_{n=0}^{+\infty}Pr(X=x,N=n).$
But how on earth do you this series if there's so many unknowns? And is there a trick (to recognize the distribution ?)
My professor has provided the answer as $X ~ \mathrm{Poisson} (\lambda\theta)$ so I think we have to recognize the joint as a the distribution of Poisson with parameter $\lambda\theta$. But I am not sure how to go about doing it. Help anyone?
 A: Please let's not hope for `easy tricks'. They play their part in math but we must not be afraid of doing computations. 


*

*First, noticed that you only need to care about the terms that contain $n$ in this sum. Everything that does not contain $n$ is "junk" you should cast outside the summation symbol (they multiply it).

*Also simplify everything that has $N$ on it. For example: write $\begin{pmatrix} N \\x\end{pmatrix}=\dfrac{N!}{(N-x)!x!}$ and simplify the expression. There are several othersimplifications you can do.

*You know what are the value of the expectation of a Poisson: $\sum_{k=0}^{+\infty} k\cdot \frac{\exp(-\lambda)\,\lambda^k}{k!}=\frac{1}{\lambda}$ and also that probabilities must add-up to one: $\sum_{k=0}^{+\infty}  \frac{\exp(-\lambda)\,\lambda^k}{k!}=1$.

A: $$\Pr(X=x) = \sum_{n=0}^{+\infty}\Pr(X=x,N=n)=\sum_{n\geqslant x} {n\choose x} \theta^x \,(1-\theta)^{n-x} \cdot  \mathrm e^{-\lambda}\dfrac{\lambda^n}{n!}$$
$$\Pr(X=x)= \frac{\theta^x}{x!} \lambda^x\mathrm e^{-\lambda}\sum_{n\geqslant x}\frac{(1-\theta)^{n-x}\lambda^{n-x}}{(n-x)!}= \frac{(\lambda\theta)^x}{x!}\mathrm e^{-\lambda}\mathrm e^{\lambda(1-\theta)}=\text{____}$$
A: This is a standard property of the Poisson process. You may want to look it up in these notes as example 3.5.3.
