Why is the negative binomial distribution defined as $$P(X=x|r,p)= \binom{x-1}{r-1}p^{r}(1-p)^{x-r}$$

Basically this is the probability that $x$ Bernoulli trials are needed for $r$ successes. So we need $r-1$ successes in the first $x-1$ trials. Then success on the $r^{th}$ trial happens with probability $p$. Why can't we write it as the following:

$$P(X = x|r,p) = \binom{x}{r}p^{r} (1-p)^{x-r}$$

This means that you have $r$ successes in the first $x$ trials.


You wrote that $x$ trials are needed for $r$ successes. That means in particular that $x-1$ trials were not enough. So we hit our goal of $r$ successes at the $x$-th trial.

Suppose for example that we are tossing a fair coin until we get the first head. What is the probability that $2$ tosses are needed for $1$ success? Here $x=2$ and $r=1$. Two tosses are needed precisely if we get TH. This has probability $1/4$. By way of contrast, the probability of exactly one head in two tosses is $1/2$.

  • $\begingroup$ @David Mitra: Thanks for spotting the missing not! Fixed. $\endgroup$ – André Nicolas Jan 27 '12 at 20:34

One reason the negative binomial distribution is written that way is that $$\sum_{x=r}^\infty \binom{x-1}{r-1}p^{r}(1-p)^{x-r} =1$$ while $\sum_{x=r}^\infty \binom{x}{r}p^{r}(1-p)^{x-r} \gt 1$ and so is not a probability distribution.

The cause of this is non-mutually exclusive events: if you take a sequence of trials, then that sequence can include both 4 trials and 2 successes, and (if the fifth trial is then a failure) 5 trials and 2 successes; you only get mutually exclusive events if you stop when you first have 2 successes.

An alternative approach would be to look at $Y$ the number of failures before the $r$th success with $\Pr(Y=y) = \binom{y+r-1}{r-1}p^{r}(1-p)^{y}$ noting $$\sum_{y=0}^\infty \binom{y+r-1}{r-1}p^{r}(1-p)^{y} =1.$$

You can of course look at the binomial distribution, taking the sum over $r$ (with $x$ fixed) using your formulation and getting $$\sum_{r=0}^x \binom{x}{r}p^{r}(1-p)^{x-r} = 1.$$


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