Basic Concept of Probability Distributions 4: Negative Binomial Distribution

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PMF
Suppose there is a sequence of independent Bernoulli trials, each trial having two potential outcomes called "success" and "failure". In each trial the probability of success is $p$ and of failure is $(1-p)$. We are observing this sequence until a predefined number $r$ of failures has occurred. Then the random number of successes we have seen, $X$, will have the negative binomial (or Pascal) distribution: $$f(x; r, p) = \Pr(X=x) = {x + r-1\choose x}p^{x}(1-p)^{r}$$ for $x = 0, 1, 2, \cdots$.

Proof:
$$
\begin{align*}
\sum_{x =0}^{\infty}P(X = x) &= \sum_{x= 0}^{\infty} {x + r-1\choose x}p^{x}(1-p)^{r}\\
&= (1-p)^{r}\sum_{x=0}^{\infty} (-1)^{x}{-r\choose x}p^{x}\;\;\quad\quad (\mbox{identity}\ (-1)^{x}{-r\choose x}= {x+r-1\choose x})\\
&= (1-p)^r(1-p)^{-r}\;\;\quad\quad\quad\quad\quad\quad (\mbox{binomial theorem})\\
&= 1
\end{align*}
$$
Using the identity $(-1)^{x}{-r\choose x}= {x+r-1\choose x}$:
$$
\begin{align*}
{x+r-1\choose x} &= {(x+r-1)!\over x!(r-1)!}\\
&= {(x+r-1)(x+r-2) \cdots r\over x!}\\
&= (-1)^{x}{(-r-(x-1))(-r-(x-2))\cdots(-r)\over x!}\\
&= (-1)^{x}{(-r)(-r-1)\cdots(-r-(x-1))\over x!}\\
&= (-1)^{x}{(-r)(-r-1)\cdots(-r-(x-1))(-r-x)!\over x!(-r-x)!}\\
&=(-1)^{x}{-r\choose x}
\end{align*}
$$
Mean
The expected value is $$\mu = E[X] = {rp\over 1-p}$$
Proof:
$$
\begin{align*}
E[X] &= \sum_{x=0}^{\infty}xf(x; r, p)\\
&= \sum_{x=0}^{\infty}x{x + r-1\choose x}p^{x}(1-p)^{r}\\
&=\sum_{x=1}^{\infty}{(x+r-1)!\over(r-1)!(x-1)!}p^{x}(1-p)^{r}\\
&=\sum_{x=1}^{\infty}r{(x+r-1)!\over r(r-1)!(x-1)!}p^{x}(1-p)^{r}\\
&= {rp\over 1-p}\sum_{x=1}^{\infty}{x + r-1\choose x-1}p^{x-1}(1-p)^{r+1}\\
&={rp\over 1-p}\sum_{y=0}^{\infty}{y+(r+1)-1\choose y}p^{y}(1-p)^{r+1}\quad\quad\quad \mbox{setting}\ y= x-1\\
&= {rp\over 1-p}
\end{align*}
$$
where the last summation follows $Y\sim\mbox{NB}(r+1; p)$.

Variance
The variance is $$\sigma^2 = \mbox{Var}(X) = {rp\over(1-p)^2}$$
Proof:
$$
\begin{align*}
E\left[X^2\right] &= \sum_{x=0}^{\infty}x^2f(x; r, p)\\
&= \sum_{x=0}^{\infty}x^2{x + r-1\choose x}p^{x}(1-p)^{r}\\
&=\sum_{x=1}^{\infty}x{(x+r-1)!\over(r-1)!(x-1)!}p^{x}(1-p)^{r}\\
&=\sum_{x=1}^{\infty}rx{(x+r-1)!\over r(r-1)!(x-1)!}p^{x}(1-p)^{r}\\
&= {rp\over 1-p}\sum_{x=1}^{\infty}x{x + r-1\choose x-1}p^{x-1}(1-p)^{r+1}\\
&={rp\over 1-p}\sum_{y=0}^{\infty}(y+1){y+(r+1)-1\choose y}p^{y}(1-p)^{r+1}\quad\quad\quad (\mbox{setting}\ y= x-1)\\
&= {rp\over 1-p}\left(\sum_{y=0}^{\infty}y{y+(r+1)-1\choose y}p^{y}(1-p)^{r+1}+\sum_{y=0}^{\infty}{y+(r+1)-1\choose y}p^{y}(1-p)^{r+1} \right)\\
&= {rp\over 1-p}\left({(r+1)p\over 1-p} + 1\right)\quad\quad\quad\quad\quad\quad(Y\sim\mbox{NB}(r+1; p),\ E[Y] = {(r+1)p\over1-p})\\
&= {rp\over 1-p}\cdot{rp+1\over 1-p}
\end{align*}
$$
Thus the variance is
$$
\begin{align*}
\mbox{Var}(X) &= E\left[X^2\right] - E[X]^2\\
&= {rp\over 1-p}\cdot{rp+1\over 1-p}- \left({rp\over 1-p}\right)^2\\
&= {rp\over 1-p}\left({rp+1\over 1-p} - {rp\over 1-p}\right)\\
&= {rp\over(1-p)^2}
\end{align*}
$$
Examples
1. Find the expected value and the variance of the number of times one must throw a die until the outcome 1 has occurred 4 times.

Solution:
Let $X$ be the number of times and $Y$ be the number of success in the trials. Obviously, we have $X = Y+4$. Then the problem can be rewritten as ``the expected value and the variance of the number of times one must throw a die until the outcome 1 has NOT occurred 4 times''. That is, $r = 4$, $p = {5\over 6}$ and $Y\sim\mbox{NB}(r; p)$. Thus $$E[X] = E[Y+4]= E[Y] + 4 = {rp\over 1-p}+4 = 24$$ $$\mbox{Var}(X) = \mbox{Var}(Y+4) = \mbox{Var}(Y) = {rp\over(1-p)^2}= 120$$




Reference

1. Ross, S. (2010). A First Course in Probability (8th Edition). Chapter 4. Pearson. ISBN: 978-0-13-603313-4.
2. Chen, H. Advanced Statistical Inference. Class Notes. PDF