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I don't understand why the "negative binomial" random variable has that name. What is negative about it? What is binomial about it? What is negative-binomial about it?

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    $\begingroup$ Also see the comments under this more general question -- which really deserves a proper answer, mea culpa. $\endgroup$
    – Glen_b
    Aug 28, 2015 at 1:14

3 Answers 3

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It's a reference to the fact that a certain binomial coefficient that appears in the formula for that distribution can be written more simply with negative numbers.

When you conduct a series of experiment with success probability $p$, the likelihood that you will see $r$ failures after exactly $k$ trials is

${k+r−1}\choose {k}$ $p^k(1−p)^r$.

This can also be written as

$(−1)^k$${−r}\choose {k}$$p^k(1−p)^r$

and the word "negative" refers to that $−r$ in that binomial coefficient. Observe how this formula looks just like the formula for the ordinary binomial distribution except for that sign coefficient.

Another name for the negative binomial distribution is Pascal's distribution so there is that too.

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More detailed answer according to Wikipedia:

The probability mass function of the negative binomial distribution is

$f(k; r, p) \equiv \Pr(X = k) = \binom{k+r-1}{k} p^k(1-p)^r \quad\text{for }k = 0, 1, 2, \dotsc $

Here the quantity in parentheses is the binomial coefficient, and is equal to

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

This quantity can alternatively be written in the following manner, explaining the name “negative binomial”:

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

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    $\begingroup$ I don't understand your statement "When you conduct a series of experiment with success probability p, the likelihood that you will see r failures after exactly k trials is ...". It seems to me the formula should be ${k \choose r} p^{k-r} (1-p)^r$. Where did you get the formula you listed? I suspect maybe you are not describing the random process quite right. Do you mean the probability of getting exactly $r$ failures after conducting $k+r-1$ trials? If so, shouldn't the $p^k$ be $p^{k-1}$? What's going on here? Can you define the event that you're referring to, more carefully? $\endgroup$
    – D.W.
    Aug 28, 2015 at 5:42
  • $\begingroup$ @D.W. It was an unfortunate formulation. What is meant, is not a likelihood to see $r$ failures given that $k$ trials have been conducted, but a likelihood to need $k$ trials in order to observe $r$ failures. $\endgroup$
    – amoeba
    Jun 28, 2019 at 9:09
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I think the most likely origin is that if you take any moment formula from $\mathrm{Bin}(n,q)$ and you replace the parameters $n$ and $q$ in that formula with negative values $-\alpha$ and $-\theta$, then the result will be the equivalent moment formula for a Negative Binomial distribution with parameters $\alpha$ and $\theta$. For short, it looks like as if the Negative Binomial distribution is a Binomial distribution with negative coefficients.

Some background:

In actuarial science (where the distribution was discovered), they use a different parameterization for the Negative Binomial distribution. Actuaries use $\,\alpha=r\,$ and $\,\theta=\frac{p}{1-p}\,$. Actuaries define the Negative Binomial distribution as the mixture distribution of the Poisson family over a Gamma distribution with parameters $\,\alpha\,$ and $\,\theta$, i.e. the distribution with density

$f(x)=\frac{1}{\theta^\alpha\Gamma(\alpha)}x^{\alpha-1}\mathrm{e}^{x/\theta}.$

That definition makes it extremely easy to compute the probability generating function, $\mathrm{Pgf}$, of the Negative Binomial distribution in terms of the moment generating function, $\mathrm{Mgf}_\Gamma$, of the Gamma distribution. Let $N$ be Negative Binomial and $\Gamma$ the Gamma distributed parameter, then:

$\mathrm{Pgf}(z)=\mathbb{E}[z^N]=\mathbb{E}[\mathbb{E}[z^N|\Gamma]]=\mathbb{E}[\mathrm{Pgf}_{\mathrm{Pois}(\Gamma)}(z)]=\mathbb{E}[\mathrm{e}^{\Gamma(z-1)}]=\mathrm{Mgf}_{\Gamma}(z-1)=(1-\theta(z-1))^{-\alpha}.$

And, if you compare the above probability generating function of the Negative Binomial distribution with the probability generating function of $\mathrm{Bin}(n,q)$, you will see that they are the same, only that $n$ and $q$ have been replaced with $-\alpha$ and $-\theta$. As the probability generating function determines the moments uniquely through its derivatives (evaluated at 1) and $n,q,-\alpha,-\theta$ are just constants, you will get the same moment formulas for Binomial and Negative Binomial just with negative parameters.

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Below are some alternative explanations.

They relate to the currently accepted answer. It boils down to a negative index somewhere that contrasts to the positive index for the 'regular' binomial distribution. But, in these historic cases, there is a different focus in relation to the place where the 'negative' index occurs.

Alternative explanation 1

  • The typical binomial distribution relates to the term with $p^r$ in the expansion $(q+p)^k$

  • The negative binomial distribution relates to the term with $p^r$ in the expansion $(q - p)^{-k}$ and rewriting $p' = -p$ and $k' = -k$

    $$\begin{array}{}(q-p)^{-k} &=& \sum_{r=0}^\infty {-k\choose r} q^r (-p)^{-k-r}\\ &= &\sum_{r=0}^\infty \frac{k'(k'-1)\dots(k'-r+1)}{r!} q^r (p')^{k'-r} \end{array}$$

This stems from Haldane 1941 "The fitting of binomial distributions"

In other cases a good fit may be obtained to a binomial distribution where $P_r$, is the coefficient of $t^r$ in the expansion $(1 -p +pt)^k$. Here $p$ and $k$ are both positive or both negative. Where they are negative it is convenient to write $p' = -p$, $k' = - k$.

In Fisher's 1941 The negative binomial distribution the expressions $(q+p)^n$ and $(q-p)^{-k}$ are used.

Alternative explanation 2

In Yule 1910 "On the Distribution of Deaths with Age when the Causes of Death Act Cumulatively, and Similar Frequency Distributions"

The expression is related with the binomial expansion of $p^r(1-q)^{-r}$ on page 28

But this is the binomial expansion of $p^r (1 - q)^{-r}$, and we have therefore reached the law that the proportions of the original population dying during the successive exposures, if $r$ unfavourable exposures are fatal, are given by the successive terms of the binomial expansion of $p^r(1-q)^{-r}$

also later at the end of the page

As our discontinuous series is a binomial series, the form of curve to which we will pass will evidently be the binomial curve of Professor Pearson (type iii of his memoir*), though he derived it from a binomial series with positive and not with negative index.

[link added by me]

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