Let N be a random variable following Poisson with parameter a. A new random variable X is defined as
X ~ Binom(N,p) where p is the probability. I am supposed to find the probability of X
I know that here X|N is given. So, to find the the marginal density of X, i have todo the following:
f(x) = $\sum$ f(X|N=n) $\cdot$ f(N=n) with respect to N
However, I am getting confused about the limits of summation as I have two conditions $0$<x<n and $0$<n<$\infty$ and I am not sure how to use these.
The joint support may be written as such:
$${\quad\{(x,n)\in\Bbb N^2:0\leq x\leq n\lt\infty\}\\=\{(x,n)\in\Bbb N^2:0\leq n<\infty, 0\leq x\leq n\}\\=\{(x,n)\in\Bbb N^2:0\leq x\lt\infty, x\leq n\lt\infty\}}$$
The last is most relevant.
In other words: Because it is impossible for $N<X$ to occur, (ie: $f_{X\mid N}(x\mid n)=0$ when $n<x$), therefore the summation starts at $x$ and proceeds towards the infinite. Thus the formula you need is:
$$f_X(x) = \sum_{n=x}^\infty f_{X\mid N}(x\mid n)~f_N(n)\quad\big[x\in\Bbb N:0\leq x\lt\infty\big]$$
And you have been provided the distributions, so these terms are: $$\begin{align}f_{X\mid N}(x\mid n) &=\frac{n!~p^x(1-p)^{n-x}}{x!~(n-x)!}~&&\big[x,n\in\Bbb N:0\leq x\leq n\lt \infty\big]\\[2ex] f_N(n) &=\frac{a^n\mathrm e^{-a}}{n!}&&\big[n\in\Bbb N:0\leq n\lt\infty\big]\end{align}$$