What Is the MGF of the Negative Binomial Distribution?

The negative binomial distribution has more than one common convention, so its moment-generating function (MGF) depends on what the random variable counts. The formula below uses the convention that $X$ is the number of failures before the $r$ th success, with success probability $p$ on each independent trial.

With that convention, the MGF is finite for $t<-\ln(1-p)$ . If your variable counts total trials instead, the adjustment is a factor of $e^{rt}$ .

MGF for failures before the rth success

Let $r$ be a positive integer and let $0<p<1$ . The probability mass function is

P(X=x)=\binom{x+r-1}{x}p^r(1-p)^x,\qquad x=0,1,2,\ldots.

The MGF is $M_X(t)=\mathbb{E}[e^{tX}]$ . Substituting the probability mass function gives

\begin{aligned} M_X(t) &=\sum_{x=0}^{\infty}e^{tx}\binom{x+r-1}{x}p^r(1-p)^x\\ &=p^r\sum_{x=0}^{\infty}\binom{x+r-1}{x}\bigl((1-p)e^t\bigr)^x. \end{aligned}

The binomial-series identity

\sum_{x=0}^{\infty}\binom{x+r-1}{x}z^x=(1-z)^{-r},\qquad |z|<1,

then yields

\boxed{M_X(t)=\left(\frac{p}{1-(1-p)e^t}\right)^r}.

For a real value of $t$ , the series converges when $(1-p)e^t<1$ , which is equivalent to

t< -\ln(1-p).

This domain includes $t=0$ , as an MGF must, and $M_X(0)=1$ . When $p=1$ , there are no failures and $X=0$ with probability one, so its MGF is $1$ for every real $t$ . The usual negative-binomial model assumes $0<p<1$ .

Worked example

Suppose you count failures before the third success, and each trial succeeds with probability $p=0.4$ . Then $r=3$ and $1-p=0.6$ , so

M_X(t)=\left(\frac{0.4}{1-0.6e^t}\right)^3, \qquad t< -\ln(0.6)\approx 0.511.

At $t=0$ , the expression equals $(0.4/0.4)^3=1$ . The mean and variance can also be checked by differentiating the logarithm of the MGF:

\begin{aligned} \frac{d}{dt}\ln M_X(t)&=\frac{r(1-p)e^t}{1-(1-p)e^t},\\ \mathbb{E}[X]&=M_X'(0)=\frac{r(1-p)}{p}=4.5,\\ \operatorname{Var}(X)&=\left.\frac{d^2}{dt^2}\ln M_X(t)\right|_{t=0} =\frac{r(1-p)}{p^2}=11.25. \end{aligned}

The mean of 4.5 is the expected number of failures before the third success; it need not be an integer.

If the variable counts total trials

Let $N$ be the number of trials up to and including the $r$ th success. Since $N=X+r$ , its MGF is

M_N(t)=\mathbb{E}[e^{t(X+r)}]=e^{rt}M_X(t) =\left(\frac{pe^t}{1-(1-p)e^t}\right)^r.

The domain is unchanged. For the example above, $N$ has mean $r/p=7.5$ and variance $r(1-p)/p^2=11.25$ . The fixed shift by $r$ increases the mean by $r$ but does not change the variance.

A different convention counts successes before a specified number of failures. If the target is $r$ failures and each trial succeeds with probability $p$ , then the failure probability is $1-p$ , and the MGF for the success count $Y$ is

M_Y(t)=\left(\frac{1-p}{1-pe^t}\right)^r, \qquad t< -\ln(p).

The formula changes because the counted outcome and the event being repeated until are different.

Common mistakes

Mixing up $p$ and $1-p$ . State which outcome has probability $p$ and which outcome your variable counts before using a formula.
Forgetting the fixed shift. Counting total trials rather than failures adds $r$ to the random variable, so its MGF needs the factor $e^{rt}$ .
Ignoring the convergence condition. The expression is not finite for every real $t$ ; for failures before the $r$ th success, require $t< -\ln(1-p)$ .
Treating conventions as interchangeable. A formula from a source may use the number of successes before a fixed number of failures. Compare the random variable’s definition and support before applying it.

References

A First Course in Probability, Sheldon Ross

At a glance

MGF for failures before the rth success

Worked example

If the variable counts total trials

Common mistakes

References