Can somebody show me how to use the convolution formula to see that $X+Y$, where $X \sim NB(r,p), (P(X=k) = \binom{k + r -1}{k}p^r(1-p)^k), Y \sim NB(s,p), r, s \in \mathbb N, p \in (0,1)$ are independent random variables, has also negative binomal distribution?
According to my textbook binomial theorem and Cauchy product for power series could be helpful.
Best Answer
\begin{align} & \sum_{x=0}^y\binom{x+r-1}{x}p^r(1-p)^x \cdot \binom{(y-x)+s-1}{y-x}p^s(1-p)^{y-x} \\[10pt] = {} & p^{r+s} (1-p)^y \sum_{x=0}^y \binom{x+r-1}{x} \binom{(y-x)+s-1}{y-x} \\[10pt] = {} & p^{r+s} (1-p)^y \binom{y+r+s-1}{y} \end{align} So the question now is how to prove that $$ \sum_{x=0}^y \binom{x+r-1}{x} \binom{(y-x)+s-1}{y-x} = \binom{y+r+s-1}y. $$ I'd rather write a combinatorial argument for this than an algebraic one, although either should work. I'll be back in a hour or so to finish this off. Maybe someone else will have posted the rest by then, or maybe not$\ldots\ldots\ldots$