Let $X,Y$ be two independent negative binomial distributed random variables.

$X$ ~ $NB(r,p)$ and $Y$ ~ $NB(s,p)$

Show that:

$ X+Y $ ~ $ NB(r+s,p) $.

Remark: So where I'm stucked? I failed to show that $ \sum_{j=0}^{k} \binom{j+r-1}{j} \cdot \binom{k-j+s-1}{k-j} = \binom{k+r+s-1}{k}$. If I have this identity I can solve this exercise. First I thought that this is the vandermonde identity, but it isn't. So how can I show this identity? I know that I can solve this exercise by using the fact that a negative binomial distributed RV is a sum of geometric distributed RV, but i want to show it with my attempt.

Thank you for your help.


1 Answer 1


In general we have:$$\sum_{i+j=k}\binom{i}{r}\binom{j}{s}=\binom{k+1}{r+s+1}\tag1$$

This under the convention that $\binom{n}{k}=0$ if $k\notin\{0,1,\dots,n\}$.

For a combinatorial proof think of a row of $k+1$ balls.

Selecting $r+s+1$ of them can be done on $\binom{k+1}{r+s+1}$ ways (RHS).

Doing so we first pick out a ball that cuts the row in a left row of length $i\geq r$ balls and a right row of length $j\geq s$ balls. So this under the condition that $i+j=k$. Then from the left row we select $r$ balls and from the right row we select $s$ balls. This process reveals the LHS.

Your summation can be rewritten by:$$\sum_{i+j=k}\binom{j+r-1}{r-1}\binom{i+s-1}{s-1}=\sum_{i+j=k+r+s-2}\binom{j}{r-1}\binom{i}{s-1}$$

Then applying $(1)$ gives:$$=\binom{k+r+s-1}{r+s-1}=\binom{k+r+s-1}{k}$$


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