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I am using the definition of the negative binomial distribution from here. This is the same definition that Matlab uses. For convenience,

$$P(k) = {r + k -1 \choose k}p^r(1-p)^k ,$$

where $p$ is the probability of success. $P(k)$ is the probability of $k$ failures before $r$ successes. The probability generating function is supposed to be,

$$ g(x) = \left(\frac{p}{1-(1-p)x}\right)^r.$$

However, I am trying to prove this. Steps:

$$g(x) = \sum^{\infty}_{k=0} P(k)\, x^k$$

$$= \sum^{\infty}_{k=0} {r + k -1 \choose k}p^r(1-p)^k \, x^k $$

$$ = p^r \sum^{\infty}_{k=0} {r + k -1 \choose k}(x(1-p))^k. $$

I suppose the next step would be to show that,

$$ \sum^{\infty}_{k=0} {r + k -1 \choose k}(x(1-p))^k = \frac{1}{(x(1-p))^r}.$$

Is there a formula (or theorem) for infinite sums involving binomial coefficients that I can apply to get this?

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up vote 5 down vote accepted

Actually, what you want to show is

$$(1-y)^{-r} = \sum_{k=0}^{\infty} \binom{k+r-1}{k} y^k$$

You can see this using the generic rule:

$$(1-y)^{-r} = 1 + (-r) (-y) + \frac{1}{2!} (-r)(-r-1) (-y)^2 + \frac{1}{3!} (-r)(-r-1)(-r-2) (-y)^3+\ldots$$

This is really just a Maclurin expansion of $(1-y)^{-r}$. Take a look at the coefficient of $y^k$:

$$\begin{align}\frac{1}{k!} (-1)^k (-r)(-r-1)\ldots(-r-k+1) &= \frac{(r+k-1)(r+k-2)\ldots(r+1)(r)}{k!}\\ \end{align}$$

Now compare that to

$$\binom{k+r-1}{k} = \frac{(k+r-1)!}{k! (r-1)!} = \frac{(r+k-1)(r+k-2)\ldots(r+1)(r)}{k!} $$

Set $y=x (1-p)$ and you are done.

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Ah yes, I made a mistake typing that out. Will work it out later and accept your answer. Thanks! – Legendre Mar 28 '13 at 21:22

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