# Factorial moment of negative binomial

What's the factorial moment of negative binomial distribution, if $$\Pr(X = k) = \binom{k+r-1}{k} p^k(1-p)^r$$

I tried it: $$E\left[ \frac{X!}{(X-m)!}\right] = \sum_{k=m}^{\infty} \frac{(r+k-1)!}{(k-m)!} \cdot p^k = \sum_{k=0}^{\infty} \frac{(m+r-1+k)!}{k!} \cdot p^{k+m} = \sum_{k=0}^{\infty} (m+r-1)! \cdot \frac{(m+r-1+k)!}{k! \cdot (m+r-1)!} \cdot p^k \cdot p^m = p^m \cdot (m+r-1)! \cdot \sum_{k=0}^{\infty} \binom{m+r-1+k}{k} \cdot p^k.$$

What's next?

• Please include your thoughts and efforts in this and future posts. Formatting tips here. – Em. Jan 11 '16 at 8:02
• PGF of $NB(r;p)$ is $f(z)=(\frac{1-p}{1-pz})^r$ and $E(X)_k=f^{(k)}(1)$. – A.S. Jan 11 '16 at 8:04
• I tried do this: $$\sum_{k=m}^{\infty} \frac{(r+k-1)!}{(k-m)!} \cdot p^k = \sum_{k=0}^{\infty} \frac{(m+r-1+k)!}{k!} \cdot p^{k+m} = \sum_{k=0}^{\infty} (m+r-1)! \cdot \frac{(m+r-1+k)!}{k! \cdot (m+r-1)!} \cdot p^k \cdot p^m = p^m \cdot (m+r-1)! \cdot \sum_{k=0}^{\infty} \binom{m+r-1+k}{k} \cdot p^k.$$ – user304251 Jan 11 '16 at 8:10
• user304251, you should put your attempt in the question itself rather than in the comment. – Math Wizard Jan 11 '16 at 8:48