Probability generating function of negative binomial distributions

Question

I have recently took a course on probability theory and learned negative binomial distribution. The probability generating function (pgf) for negative binomial distribution under the interpretation that the the coefficient of $z^k$ is the number of trials needed to obtain exactly $n$ successes is $F(z)=(\frac{pz}{1-qz })^n=\sum_k\ \binom {k-1} {k-n} p^nq^{k-n}z^k$. But Donald Knuth gave another definition of the pgf of negative binomial distribution on Concrete Mathematics, which is $G(z)=(\frac{p}{1-qz})^n=\sum_k\binom {n+k-1} k p^nq^kz^k$, which's combinatorial interpretation is the distribution of $k$ failures when the number of successes $n$ is specified. So my question is which of these formulae is actually the pgf of negative binomial distribution, if not, what's the exact name for that pgf? And why there's inconsistency on different definitions? I’m very confused about this question. Actually the definition of negative binomial distribution on Wikipedia is different from that of on Introduction to Probability too.

Answer 1

There are indeed two different versions of the negative binomial distribution: (i) the distribution of the number (say $X$) of independent Bernoulli trials (with a given success probability $p$ in each trial) needed to have $k$ successes and (ii) the distribution of the number (say $Y$) of failures in an infinite series of such trials before the $k$th success. Obviously, $X=Y+k$. Each of these two "real-world" interpretations may have a certain appeal.

The existence of these two different (but closely and simply related) versions of the negative binomial distribution is perhaps unfortunate and confusing, but it is a fact. There is apparently nothing that can be done about it.

Answer 2

Others have pointed out that there are conflicting conventional definitions. Here I will argue that one of them is better.

The distribution of the number $X$ of trials needed to get $k$ successes is supported on the set $\{k,k+1,k+2,\ldots\},$ and the number of failures before getting $k$ successes is supported on the set $\{0,1,2,3,\ldots\}.$

$$\binom{n+k-1} k p^nq^k$$

That can be written as

$$ \binom{-n}{\phantom{+}k} p^n(-q)^k \tag1 $$

where the binomial coefficient is defined in such a way that, for example, $$\binom{2.3}4 = \frac{2.3\times1.3\times0.3\times(-0.7)}{4\times3\times2\times1}, \text{etc.}$$

That explains the name “negative binomial distribution.” From this we get a corollary: $$ 1 = \sum_{k\,=\,0}^\infty \Pr(X=k) = p^n \sum_{k\,\ge\,0} \binom{-n}{\phantom{+}k} (-q)^k $$ so that $$ \sum_{k\,\ge\,0} \binom{-n}{\phantom{+}k} (-q)^k = (1-q)^{-n}, \tag 2 $$ a binomial theorem with a negative exponent.

In $(1)$ and $(2)$ the exponent $n$ or $-n$ need not be an integer. In other words, there is a negative binomial distribution for every positive real number $n,$ not necessarily an integer. And if $X,Y$ are independent random variables with negative binomial distributions with the same values of $p$ but with $n,m$ respectively in the roles in which $n$ appears above, then $X+Y$ has a negative binomial distribution with the same parameter $p$ and $n+m$ in the role in which $n$ appears above. In other words, the family of negative binomial distributions is then infinitely divisible. You can't do that with the other definition, in which $k,k+1,k+2,\ldots,$ rather than $0,1,2,\ldots$ are the possible value.

This definition also makes the family of negative binomial distributions into a compound Poisson distribution, i.e. there is a discrete probability distribution supported on $\{1,2,3,\ldots\}$ for which, if $W_i$ are independent random variables with that distribution and $N\sim\operatorname{Poisson}(\lambda t),$ then $$ \sum_{\ell\,=\,1}^N W_i $$ has a negative binomial distribution. The distribution of $W$ is the logarithmic series distribution: $$ \Pr(W=w) = \text{constant} \times \frac{a^w} w, \text{ for } w=1,2,3,\ldots $$

Answer 3

As Iosif Pinelis noted there are two versions if the geometric distribution (the special negative binomial distribution with parameter $n = 1$ successes in your notation), which are related by the formular $X = Y + 1$. These have the pdf $p_X = \frac{pz}{1-qz}$ and $p_Y = \frac{p}{1-qz}$. Since the times between successes are independent and identically distributed, you get $F$ resp. $G$ simply by taking of $p_X$ and $p_Y$ the $n$-th power, as in your formulas.