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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.

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    $\begingroup$ The votes to close as off-topic are absurd, and yet there are three of them. Why? ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 23 '14 at 17:44
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\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$\ldots\ldots\ldots$

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  • $\begingroup$ Thanks so far! Could you additionally show why the equation holds? I would really appreciate that! $\endgroup$ – xxx Nov 23 '14 at 21:07
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Hint: Power series, indeed. Say, what is the probability generating function of a negative binomial distribution, already?

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