Negative binomial distribution

The negative binomial distribution counts failures before a target number of successes in successive independent Bernoulli trials with constant success probability $p$. In this stopping model, $r$ is a positive integer and $0<p\le 1$. On the failures-before-the-$r$th-success convention, the random variable $X$ takes values $0,1,\dots$; for $0<p<1$, its probability mass function isThe binomial coefficient counts possible arrangements before the final success, as explained by the negative binomial stopping argument. At $p=1$, every trial succeeds: $X=0$ with probability one, so its expected value and variance are both zero. This handles the endpoint without needing to interpret $0^0$. At $p=0$, the target is never reached and there is no proper finite waiting-count law.

For a worked example, toss a fair coin until the second head, calling heads $S$ and tails $F$. Thus $r=2$ and $p=1/2$. Exactly two failures means that the second success is on trial four. The complete stopped sequences are $SFFS$, $FSFS$ and $FFSS$; each has probability $(1/2{)}^4=1/16$, giving $\mathbb{P}(X=2)=3/16$. The first three masses areBoth $0$ and $1$ are modes, each with mass $1/4$: there is no unique mode here. Later masses decrease, since $\mathbb{P}(X=x+1)/\mathbb{P}(X=x)=(x+2)/(2x+2)<1$ for $x\ge 1$.

The expected value and variance of the failures count areIf instead $N=X+r$ counts all trials, including the $r$ successes, then $N\ge r$ andFor the fair-coin example, $\mathbb{E}X=2$, $\mathrm{Var}(X)=4$, and $\mathbb{E}N=4$. Shifting the count changes its expected value but preserves its variance. When $r=1$, $X$ has the failures-count geometric distribution, while $N=X+1$ has the all-trials geometric distribution. The two conventions and the moments are treated in Kyle Siegrist, The Negative Binomial Distribution, Random Services.

The negative binomial distribution is useful when a procedure runs until it collects a target number of successes: for example, obtaining $r$ usable samples from independent attempts with a stable acceptance probability. A binomial distribution instead counts successes in a fixed number of trials. The attractive connection is that these two experimental questions can be answered from the same trial sequence, using the negative binomial stopping argument. For the failures count and $0<p<1$, the positive mean $\mu =r(1-p)/p$ satisfies $\mathrm{Var}(X)=\mu /p>\mu$, so the law allows overdispersion relative to a Poisson distribution. At $p=1$, both quantities vanish and the inequality is not strict. The separate Poisson-gamma mixture explains another way this count law arises; its noninteger shape parameters do not represent a literal target number of successes.

When each observation counts untagged captures before the $r$th tagged capture with immediate replacement, its negative binomial distribution has success probability $p=k/N$. For $n$ independent runs, the log-likelihood is $nr\log p+({\sum }_i{x}_i)\log (1-p)$ plus a constant. Differentiation gives the displayed estimator and expected Fisher information. For an interior true parameter, its asymptotic variance is $p^2(1-p)/(nr)$. A confidence interval $(L,U)$ for $p$ transforms to $(k/U,k/L)$ for the population size, with endpoint order reversed.

For the negative binomial distribution with positive integer target $r$ and success probability $0<p<1$, the $r$th success occurs after exactly $x$ failures only if the final trial is a success, while the preceding $x+r-1$ trials contain $x$ failures and $r-1$ successes. Choosing the failure locations gives the binomial coefficient $\left(\genfrac{}{}{0ex}{}{x+r-1}{x}\right)$. Each such stopped sequence has probability $(1-p{)}^x{p}^r$ by independence, and the sequences describe mutually exclusive events, so their probabilities add. This proves the probability mass function. Counting arrangements among all $x+r$ positions would wrongly include sequences that already reached the target earlier. The same argument is used for three heads in MIT 18.05, spring 2022, problem set 9 solutions, problem 2(b).

A short route to normalization and moments is to split the failures count into successive waiting blocks:where $F_i$ counts failures after success $i-1$ and before success $i$, with $F_1$ starting at the first trial. Fresh independent Bernoulli trials make the blocks independent, each with the failures-count geometric distribution. Each block has total mass $p{\sum }_{j=0}^{\infty }(1-p{)}^j=1$ by the geometric series, so their sum also defines a normalized probability distribution. The linearity of expectation and variance additivity for independent random variables give $\mathbb{E}X=r(1-p)/p$ and $\mathrm{Var}(X)=r(1-p)/p^2$. This makes the moments transparent without a long calculation of sums. For $p=1$, every block is zero.

The same stopping rule connects a waiting count to a fixed-length count. Let $N=X+r$ be the trial number of the $r$th success, and let $B_n$ count successes in the first $n$ trials, so $B_n$ has a binomial distribution. For any nonnegative integer $n$,The target is still unmet after $n$ trials exactly when fewer than $r$ successes have occurred. With $r=2$, $p=1/2$ and $n=4$, this gives $\mathbb{P}(N>4)=(1+4)/16=5/16$; equivalently, $\mathbb{P}(X\le 2)=1-5/16=11/16$, matching $1/4+1/4+3/16$. At $p=1$, the event identity still holds, with $N=r$ and $B_n=n$ deterministically.

In independent Bernoulli trials with success probability $p$ and failure probability $q=1-p$, let $T_n$ be the trial of success $n$ and $U_m$ the trial of failure $m$, for positive integers $n,m$. If success $n$ occurs after $k<m$ failures, its final trial is a success and the preceding trials have $n-1$ successes and $k$ failures. The negative binomial stopping argument therefore gives the title sum.

Put $F=n+m$. By trial $F-1$, at least one target must have been reached, so the same event is that a binomial distribution with $F-1$ trials has at least $n$ successes. Conditioning instead on the final trial of a block of $F$ trials givesIndeed, more than $n$ successes in the full block guarantee the success target won; with exactly $n$ successes it won only when the final trial was a failure. These disjoint cases explain the extra boundary term. A zero success target wins immediately; a zero failure target has already been reached. To include $n=0$ in the displayed finite sum requires the explicit generalized convention $\left(\genfrac{}{}{0ex}{}{-1}{0}\right)=1$; an upper summation limit of $-1$ denotes an empty sum.

Suppose independent Bernoulli trials give a player a win with probability $p$, and the match ends when either player reaches $r$ wins. The player's match-win probability isFor a win with $n$ losses, the final trial must be a win. The preceding $r+n-1$ trials contain exactly $r-1$ wins and $n$ losses; the binomial coefficient counts their arrangements. Because $n<r$, the opponent cannot already have won, and because the player has only $r-1$ previous wins, the match has not ended earlier. Summing the disjoint terminal events gives the formula, a truncated negative binomial distribution sum. The constant game-win probability must describe independent games, or equivalently remain the same conditional on every previous history.

If $Y\mid \Lambda \sim \mathrm{Poisson}(\Lambda )$ and $\Lambda$ has a gamma distribution, then $Y$ has a negative binomial distribution. For gamma shape $\nu$ and scale $\alpha$,