Suppose that there exists two kinds of stars. The probability that a given volume contains $j$ stars of the first kind is $p(j; a)$, and the probability that it contains $k$ stars of the second kind is $p(k; b)$; the two events are assumed to be independent. Prove that the probability that the volume contains a total of $n$ stars is $p(n; a+b)$.
$p(j; a)$ is the Poisson distribution with parameter $a$. First, we have ${n \choose j}$ compositions of two stars. Therefore, this problem is reduced to finding $\sum_{j=0}^n {n \choose j} p(j; a) p(n-j;b)$. This is equal to $e^{-a-b} \sum_{j=0}^n {n \choose j} a^{j}b^{n-j} / (j! (n-j)!)$. I know that $\sum_{j=0}^n {n \choose j} a^j b^{n-j} = (a+b)^n$, and I found that $\sum_{j=0}^n {n \choose j} a^{j}b^{n-j} / (j! (n-j)!) = \frac1{n!}\sum_{j=0}^n {n \choose j}^2a^{j} b^{n-j}$. I wish that $\sum_{j=0}^n {n \choose j}^2a^{j} b^{n-j} = (a+b)^n$, but I think that this is not true.
I would appreciate if you give some hint.
