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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?

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    $\begingroup$ Please include your thoughts and efforts in this and future posts. Formatting tips here. $\endgroup$
    – Em.
    Jan 11, 2016 at 8:02
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    $\begingroup$ 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, 2016 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, 2016 at 8:10
  • $\begingroup$ user304251, you should put your attempt in the question itself rather than in the comment. $\endgroup$
    – hermes
    Jan 11, 2016 at 8:48

2 Answers 2

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Just to give an actual answer for posterity, let us use the hint by @A.S. The probability generating function (PGF) is: $$ f(z) = \left(\frac{1-p}{1-pz}\right)^r, $$ and thus $$ f^{(k)}(z) = \frac{p^k \left(\frac{1-p}{1-pz}\right)^r \prod_{i=0}^{m-1} r + i}{(1-pz)^k}, $$ so finally $$ E[(X)_m] = \frac{p^m \prod_{i=0}^{m-1} r + i}{(1-p)^m}. $$

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While you can get the $k$th factorial moment from the generating function, you can also calculate the value directly-here's how.

First rewrite: $$\mathbb{E}[(X)_k] = \sum_{x\geq 0} {x+r-1 \choose x} (x)_k q^xp^r,$$ by taking out $q^k$ and $p^r$ to get $$p^rq^k\sum_{x\geq 0} {x+r-1 \choose x} (x)_k q^{x-k}.$$ Note here for brevity, I write $q=1-p$.

Next use $D^kx^a=(a)_kx^{a-k}$ where $D$ is the derivative operator.

$$p^rq^k\sum_{x\geq 0} {x+r-1 \choose x} (x)_k q^{x-k} = p^rq^k\sum_{x\geq 0} {x+r-1 \choose x} (-D)^kq^{x}.$$

Now take interchange the sum and derivative operator-you can show that this is a valid operation in this context via Fubini's theorem. Now consider,

$$p^rq^k(-D)^k \underbrace{\sum_{x\geq 0} {x+r-1 \choose x} q^{x}}_{=p^{-r}}=p^rq^k(-D)^kp^{-r}.$$

Now you can also show that $(-D)^kp^{-r} = r^{(k)}p^{-(r+k)}$, where the $r^{(k)}$ denotes a rising factorial. Using this fact gives us

$$p^rq^kr^{(k)}p^{-(r+k)},$$

which simplifies to

$$\left(\frac{q}{p}\right)^kr^{(k)}.$$

For $r=1$ this gives the formula for the $k$th factorial moment of a geometric random variable, $k!\left(\frac{q}{p}\right)^k.$

This is a-slightly-different approach than the generating function but gives you the same formula. Notice that I wrote the negative binomial as failures before the $r$th success and the OP and the other answer written give the formula where the definition is number of successes before the $r$th failure so to compare you need to change my $p$s and $q$ to match.

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