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?

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

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