Negative Binomial distribution

A generalization of the Geometric distribution. Consider a sequence of independent Bernoulli trials, each with success probability $0<p<1$. Assume the trials continue until at least the $r$th success, where $r\in {\mathbb{Z}}^{+}$.

Let $X$ be the r.v. that equals the number of failures before the $r$th success. Then $X$ has the Negative Binomial distribution and the textbook writes this as $X\sim \text{NBin}(r,p)$.

Notice that when $r=1$, then this is just the Geometric distribution. Hence it’s a generalization.

PMF

Similar to the intuition for the Geometric distribution’s PMF, we now have multiple successes $r$ along with our $k$ failures. However, the order in which the $r$ successes matters, because the sample space is all the sequences of trials that lead to the $r$th success.

Given $X=k$, this means we have $k+r$ total trials. The $k+r$th trial is the one where we get our $r$th success, so we need to place the $r-1$ successes among the other $k+r-1$ trials that occur before it. We fill in the remaining $k$ spots with our failures.

Let $q=1-p$. the PMF is given by

$$P(X=k)=\left(\genfrac{}{}{0ex}{}{k+r-1}{r-1}\right)p^r{q}^k$$

Expectation

Consider $X\sim \text{NBin}(3,p)$. We can define it in terms of 3 new r.v.s:

• $X_1$ equals the number of failures until the 1st success.
• $X_2$ equals the number of failures between the 1st and 2nd success.
• $X_3$ equals the number of failures between the 2nd and 3rd success.

Then it must be true that $X=X_1+X_2+X_3$. In addition, they are all independent of one another, and all have the Geometric distribution.

By LOE, we have $\mathbb{E}(X)=\mathbb{E}(X_1)+\mathbb{E}(X_2)+\mathbb{E}(X_3)$, and so the expectation of $X$ is given by

$$\mathbb{E}(X)=r\cdot \frac{1-p}{p}$$

for $r$ successes and success probability of $p$.