# 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, 2016 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, 2016 at 8:04
• I tried do this: $$\sum_{k=m}^{\infty} \frac{(r+k-1)!}{(k-m)!} \cdot p^k = \sum_{k=0}^{\infty} \frac{(m+r-1+k)!}{k!} \cdot p^{k+m} = \sum_{k=0}^{\infty} (m+r-1)! \cdot \frac{(m+r-1+k)!}{k! \cdot (m+r-1)!} \cdot p^k \cdot p^m = p^m \cdot (m+r-1)! \cdot \sum_{k=0}^{\infty} \binom{m+r-1+k}{k} \cdot p^k.$$ Jan 11, 2016 at 8:10
• user304251, you should put your attempt in the question itself rather than in the comment. Jan 11, 2016 at 8:48

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 $$E[(X)_m] = \frac{p^m \prod_{i=0}^{m-1} r + i}{(1-p)^m}.$$

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.