TITLE: Negative binomial distribution - sum of two random variables
QUESTION [10 upvotes]: Suppose $X, Y$ are independent random variables with $X\sim NB(r,p)$ and $Y\sim NB(s,p)$. Then $$X + Y \sim NB(r+s,p)$$
How do I go about proving this? I'm not sure where to begin, I'd be glad for any hint.

REPLY [4 votes]: Building upon the idea that NB(r,p) is the time to the r-th success in Bernoulli trials, and that the trials are independent, it is clear that NB(r+k,p) can be seen as the time to the r-th success and then to the next k-th success, giving the result directly with no algebra.