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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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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 '15 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 at 9:09
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Denizens of StatsExchange, First, the good news, this author copies the Wikipedia formula so all is well there. The description this author wrote was incorrect. He should have written the probability of getting r failures after k+r trails.
Note that in the first k+r-1 trials there are exactly r-1 failures and k successes. Hence the formula correctly includes (k+r-1 C r-1) p^k (1-p)^(r-1).
Then, by definition, the final trial, namely the k+r th trial, must be the r th failure. This event is independent so we simply multiply it’s probability 1-p to find the stated probability.

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